Orion, Mission to Alpha Centauri

Posted by: Momos

Orion, Mission to Alpha Centauri - 01/15/10 03:29 PM

http://www.scienceagogo.com/message_board7/messages/747.shtml

Ok, this thread is ancient smile

Nevertheless I'd like your opinion on using Nuclear Pulse Propulsion for manned interstellar travel (not for reaching orbit).

Let's assume a 100 year one-way journey to Alpha-Centauri A/B at 4.4 ly.
So the average speed should be 4,4 % c.

Deceleration could possibly be done by using km-long theters to produce an electric/magnetic field with many kilometers in diameter, once inside the heliopause of the target star.

Which figures could we assume?
(For start lets just say smooth deceleration, over 0.5ly during 5 years)

Energy could be provided by current day fission or "possible near future" fusion generator.

The ship would be assembled in space, maybe in one of the lagrange points. Let's assume a ship mass of 10.000 - 50.000 metric tons, which is half the mass of a Nimitz-class aircraft carrier.

So: if the plasma debris of a thermo-nuclear-explosion has a velocity of 2*10^7 m/s (?) and 40% of the blast could be used for momentum change: would this be possible?

And is there any other technology which might be better suited for such missions? With higher exhaust velocities? Maybe LAPPS?
Posted by: redewenur

Re: Orion, Mission to Alpha Centauri - 01/16/10 04:29 AM

Always an interesting topic. Wisely, you use the Sci-Fi section, even though respected eggheads began working on such proposals many decades ago.

Try to get back. Need to check out the available info on the net when/if I get time. There's probably plenty of it.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 01/16/10 04:48 AM

I think we should wait until we have the technology to travel to our closest planet before we take such a huge step.

dont worry if we survive the current oil age there are plenty of ways to get around to other planets and stars.

after the oil age is over those of us that have survived will be surviving using free energy or solar energy and travel to other planets will be more desireable if not necessary.

but for now you should concentrate on a way to travel to our closest planet using the only available fuel source we are allowed to use.

maybe try coal even or natural gas.

some feasible fuel supply that can be taxed.

that way no one will scoff at your ideas and there would be billions of funding dollars just waiting for you to waste them on your next idea.
Posted by: Momos

Re: Orion, Mission to Alpha Centauri - 01/18/10 11:18 AM

Originally Posted By: paul
I think we should wait until we have the technology to travel to our closest planet before we take such a huge step.


Well, by saying "the ship will be assembled in space" I meant "after we might have established an industrial capacity in space using/mining/refining resources in space".
Of course a One-Way-Mission to the nearest star (which hasn't got small planet in the habitable zone) implies we have to have the technology and experience to build and maintain a society in space.

My question is, if it is possible at all to reach the nearest Stars without referring to unproven hypothesis and technologies which probably will never be developed.
(Like FTL, "free Energy").

Originally Posted By: paul

some feasible fuel supply that can be taxed.


There is no such thing as nontaxable-stuff.
If need arises a parliament could invent taxes for IR-Emissions, regardless if the energy for this emissions comes from coal, nuclear fusion or "free energy"...
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 02/05/10 08:59 PM

Quote:
Quote:
My question is, if it is possible at all to reach the nearest Stars without referring to unproven hypothesis and technologies which probably will never be developed.
(Like FTL, "free Energy").



no , its not possible.

without using free energy theres no space travel to other stars.

using what we have now all at one shot might accelerate a ship carrying everything we now have on the earth such as oil , coal , natural gas , atomic fussion materials etc...

towards another star , but how big would that ship be?

remember the bigger it is the more energy will be needed for acceleration.

186 million miles per second.
thats travelling from the earth to our sun and then back to the earth in 1 second.
do that for 1 year and you have traveled 1 light year.


our closest neiboring star Proxima Centauri is 4.3 light years away.


the ship would eventually reach the star , if anybody survived the initial acceleration and nothing got in the way.

then you will need at least 1/4 the amount to slow the ship down.

unless you eject from the main ship and slow down a smaller landing ship.

not trying to burst the bubble just being realistic here.

thats why we should start going to our planets using our energy sources first.

then later maybe a better form of propulsion could be found made available.



Posted by: paul

Re: Orion, Mission to Alpha Centauri - 02/14/10 01:12 AM

Quote:
186 million miles per second.
thats travelling from the earth to our sun and then back to the earth in 1 second.
do that for 1 year and you have traveled 1 light year.


should have been 186 thousand miles per second.
sorry must have been either a brain fart ot alzheimers.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 05/19/10 07:51 PM

Originally Posted By: paul
Quote:
Quote:
My question is, if it is possible at all to reach the nearest Stars without referring to unproven hypothesis and technologies which probably will never be developed.
(Like FTL, "free Energy").



no , its not possible.

without using free energy theres no space travel to other stars.


Sorry, the above is just silly. Unmanned travel to other star systems is well within current technology, so long as you're not in a rush to get there. In fact, several human probes are already traveling faster than escape velocity and will (in thousands of years) pass other stars. No need for exotic energy sources, just 1960's rocket propulsion and the odd gravitation sling was all that was needed.

As for human travel, the generation ship idea's been around for a long time - out of our current technological capabilities, but is capable of moving a human colony using a Hohmann (minimal energy) transfer orbits that are well within the energy levels that could be provided by an orion drive, fusion drive, VASIMR, etc. Several other projects have been designed, but never implemented, that could potentially send humans to nearby stars within a human lifetime (Project Longshot and Daedalus, for example). These were based on existent tech, or tech expected to be developed in the near-term.

And people seem to forget, you can leave a significant portion of your "propulsion" at home - laser-powered solar sails are near tech that could "easily" get a space craft over the 1%C mark. Doesn't reduce the amount of energy needed to slow down on the other end (nor return, if that is your goal), but it does cut down on fuel use dramatically. It even makes a "fly buy" mission "fuel free" (ignoring reactors and steering).

BTW, what do you mean by "free energy"?

Bryan

EDIT: by "laser powered solar sails" I'm referring to where you have a solar-sail powered craft, and use orbital lasers to provide propulsion above what the sun provides. Because they lasers stay in earth orbit, the fuel needed to run them stays "at home" and need not be carried.
Posted by: Momos

Re: Orion, Mission to Alpha Centauri - 05/26/10 08:45 AM

"Unmanned travel to other star systems is well within current technology, so long as you're not in a rush to get there."

Well, I'm not very patient. So is the public, which has to do the funding.

I guess an unmanned probe (possibly a fly-by) would have a chance if the target is interesting enough and the necessary amount not to big.

My initial interest in this topic is the question if it is entirely impossible to get to the nearest star system broadly within the lifespan of a human.

Assuming a lifespan of 100 years for the carefully choosen crew.
A take-of-age of 25 years, "retirement age" of 85 years (!).
I would even consider a 1 to 2 Generation Ship, with new born childs replacing the aging crew.
Time for the journey could be 60 - 100 years.

Thats why I figured an average speed of 4.4% c should be sufficient to reach alpha centaury during a manageable time frame.

The crew would have to transport every machinery to build and sustain a space faring society around the target star, enough genetic material to allow for a stable population.... lot's of transportation smile

"laser powered solar sails" have several advantages, but the main disadvantage would be the political ramifications of Multi-Terrawat-Lasers in space.

Additionally I would kind of prefer to have the engine on board rather then depending on people billions of km away to maintain the equipment.

So far the only remotely feasible concept, given my wishes, seems to be nuclear pulse propulsion.
However I calculate the needed masses/energies are always to big.

Even the nearest stars are awfully far away :-/
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 05/26/10 02:22 PM

Originally Posted By: Momos
"Unmanned travel to other star systems is well within current technology, so long as you're not in a rush to get there."

Well, I'm not very patient. So is the public, which has to do the funding.


Sad, but true...

Originally Posted By: Momos
I guess an unmanned probe (possibly a fly-by) would have a chance if the target is interesting enough and the necessary amount not to big.


Not to mention the most practical. People need a lot of support equipment to survive, machines generally do not. So a lot of the mass issues become less when talking about unmanned missions.

Originally Posted By: Momos
My initial interest in this topic is the question if it is entirely impossible to get to the nearest star system broadly within the lifespan of a human.


With current tech I doubt it, but I'd bet some of the proposals out there could do it in the near-ish future (assuming enough money is spent on the R&D side of things).

Originally Posted By: Momos
"laser powered solar sails" have several advantages, but the main disadvantage would be the political ramifications of Multi-Terrawat-Lasers in space.


Stupid politics, getting in the way of our fun smile

Originally Posted By: Momos
So far the only remotely feasible concept, given my wishes, seems to be nuclear pulse propulsion.
However I calculate the needed masses/energies are always to big.


Did you look into some of the other techs like VASIMR? I believe they get ISP's similar to that (maybe even better) than that of orion.

Bryan
Posted by: Momos

Re: Orion, Mission to Alpha Centauri - 05/28/10 08:40 AM


well, somewhere I found the following figures, but I don't know how representative they are:

* Nuclear: 20.000 m/s – 50.000 m/s
* Ions (HiPEP, VASIMIR): 30.000 m/s – 50.000 m/s
* Ablative Laser Propulsion (ALP: LAPPS): 200.000 m/s - 32.000.000 m/s
* Anti-Matter (AIMStar): 600.000 m/s
* Nuclear Puls Propulsion (Orion, Deadalus): 100.000 m/s – 10.000.000 m/s

My first try was use of an engine with exhaust speed of 200.000 m/s:

For 100t of ship mass and 60 year flight time I got a 3,5*10^100 kg of reaction mass. ouch. eek

Second try with 32.000.000 m/s (which is itself 10% c, an extraordinary and probably unreliable value):
405 t of propellant (which, at present, is gold laugh )

for 60 years the needed acceleration/deceleration is: 2,322252 * 10^-2 m/s², which seems pretty small.
But to accelerate the overall mass of ~500 t we would need ~ 11.6 kN ? (at the beginning).

Somewhere it was stated LAPPS could achieve a thrust of 0,031 N per MW, I'm hard wishing for an advancement of the factor 100 (! blush )
So we get 3 N per MW, which means we would need ~366GW.

A nuclear reactor for 1.6 GW already has a mass of more then 10.000 t, 100times the load capacity, let alone the fission material for 60 years. cry


I don't know the figures for VASIMIR and co, but it all sounds very unlikely. Either we need lots of propellant or a really big nuclear power plant.

Thats why I thought an orion style propulsion, without the need for external energy might be more feasible.

Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 05/28/10 02:20 PM

Originally Posted By: Momos

I don't know the figures for VASIMIR and co, but it all sounds very unlikely. Either we need lots of propellant or a really big nuclear power plant.


But that's the point of vasimr - you use a nuclear reactor to provide the kinetic energy you impart on your fuel. You can get extremely high ISP's that way, thus cutting your fuel bill. At the highest efficiency settings (low-thrust, but high exhaust velocity), I've read predictions that VASIMR may have higher ISP's than any other potential tech aside from anti-matter.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 05/28/10 11:33 PM

Originally Posted By: momos
I'd like your opinion on using Nuclear Pulse Propulsion for manned interstellar travel


Originally Posted By: paul
no , its not possible.


Originally Posted By: imagegeek
Sorry, the above is just silly. Unmanned travel to other star systems is well within current technology,


manned interstellar travel

and how fast can you accelerate the ship , that is manned?
how many generations will pass before the ship approaches
the deceleration point?

the probes we launched in the 60's and 70's
pioneer 10 was passed by voyager 2 in 1998 becomming the
furtherest man made object in space.



it has just passed our planetary system.

what new magical propulsion do we now use that was not
used then that would get us there faster?

solar wind? not hardly.

solar power? !!! but the suns sooooo tiny there.


dont forget if your going to use nuclear power to do the
job... how much fuel will you need?

do we have that much fuel?

so one false move and its lights out.
do we already have a nuclear power station in space?

just what would you use to power such a short lived journey if you dont use free energy?

just suppose your distant future family members are still
living and actually have succeded in keeping there muscle and bone structure in tact , what would the effects of slowing the spacecraft down have on them , as for the last 30 or 40 generations in your family have never experienced any deceleration on their bodies.

and for the next 30 or 40 generations they will all
experience deceleration.

then they arrive and no deceleration.

will they never get a cold or a virus in the thousands of years that would wipe them all out?

what would be the odds that any of them would survive the first year?

now or course there are no supply shipps trailing them or manufacturing ships or farm ships.

so your going to have to make everything that might break down as you travel.

this includes every manufacturing process on earth.

again

no its not possible.

im jst being realistic , thats all.

Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 05/30/10 02:06 AM

Originally Posted By: paul
manned interstellar travel


Paul, I was referring to this comment:

"without using free energy theres no space travel to other stars."

That is, as I pointed out, silly. Even with 1960's tech we could send stuff to other stars - it just takes an eternity to get there. No need for any magical free energy, just chemical rockets, gravity and time.

I'd also point out that you've never defined what you mean by "free energy". I have a hunch, but I don't want to put words into your mouth.

Originally Posted By: paul
and how fast can you accelerate the ship , that is manned?


3-5G's is sustainable for modest periods of time, 1G for eternity.

At 1G acceleration you get to the authors asked for 4.4%C in just shy of 16 days.

Originally Posted By: paul
how many generations will pass before the ship approaches the deceleration point?


Depends on your speed of travel. At a fast enough speed, anywhere is reachable in a human's life-time, as measured by ship-time. Acceleration to near-C, at a constant 1G, takes about 1 year.

Originally Posted By: paul
what new magical propulsion do we now use that was not used then that would get us there faster?


We've had working ion engines for over a decade, and vasimr has been demonstrated. Both of those are far more fuel efficient, and enjoy ISPs thousands of times better, than chemical rockets. Heck, Orion was considered possible with 60's tech - and was developed to the point of experimental models flown on conventional explosives.

And I'd point out that this thread was never about today's tech, but rather the orion system and potential alternatives. There is no reason an orion drive couldn't reach 4.4% c, all using conventional fissile materials. You'd need a lot of it, but you could get there.

Originally Posted By: paul
dont forget if your going to use nuclear power to do the job... how much fuel will you need?


Fusion, fission or RTG's?. The answer varies wildly, depending on which one, and what fuel. And what drive you're using, how fast you're flying, how big your ship is, how far you're going, etc.

Originally Posted By: paul
do we have that much fuel?


Yep.

Originally Posted By: paul
so one false move and its lights out.
do we already have a nuclear power station in space?


Several RTG's.

Originally Posted By: paul
just what would you use to power such a short lived journey if you dont use free energy?


What free energy?

Originally Posted By: paul
just suppose your distant future family members are still living and actually have succeded in keeping there muscle and bone structure in tact , what would the effects of slowing the spacecraft down have on them , as for the last 30 or 40 generations in your family have never experienced any deceleration on their bodies.


What keeps you from generating centripetal force? Centripetal force = "artificial gravity" for all eternity when you're spinning in a vacuum.

Originally Posted By: paul
will they never get a cold or a virus in the thousands of years that would wipe them all out?


Almost guaranteed to be zero. Pathogens evolve in very well understood fashions, usually requiring passage between several separate populations or even multiple species. A small isolated population is not one which'll breed new pathogens. Small isolated populations make good victims of existing diseases, but poor incubators for making new ones.

Originally Posted By: paul
what would be the odds that any of them would survive the first year?


Depends on how well things are built, etc.

Originally Posted By: paul

now or course there are no supply shipps trailing them or manufacturing ships or farm ships.


Why not? If you can launch 1 frigging huge ship into space, there is no reason you cannot launch more.

Originally Posted By: paul
so your going to have to make everything that might break down as you travel...this includes every manufacturing process on earth.


Not really. You only need those manufacturing processes required to replace ship-board items. No need for 747 manufacturing on the ship. By standardizing parts as much as possible, manufacturing capacity can be minimized.

And you forget the possibility of self-replicating prototypers (simple version of which have been demonstrated several times in the 2000's). These devices produce any parts needed, and assemble them - including what is required to replicate themselves. All you need is energy and raw material.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 05/31/10 05:39 PM

Quote:
Acceleration to near-C, at a constant 1G, takes about 1 year


how much fuel will that acceleration cost?

what happens when you encounter a grain of sand traveling at
near c.

how much energy would it cost to detect that grain of sand
at a distance far enought to allow for a course change.

and we dont yet know if there is anything in the void between the solar systems that would allow for a course change.

how do you know that there are even enought ions in the void?
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 05/31/10 06:13 PM

Originally Posted By: paul
Quote:
Acceleration to near-C, at a constant 1G, takes about 1 year


how much fuel will that acceleration cost?


Depends on the mass of the vessel, and how close to C you try to get.

Originally Posted By: paul
what happens when you encounter a grain of sand traveling at
near c.


Depends. If you have a waffle shield, it gets a hole in it and the particle is deflected. If not, your ship experiences a collision with the force of a small nuclear explosion.

Originally Posted By: paul
how much energy would it cost to detect that grain of sand
at a distance far enought to allow for a course change.


Why bother? Deflecting a particle of sand is a much lower-energy option than moving the whole ship, and if a passive shield is used, requires no pre-detection of anything other than large objects.

Originally Posted By: paul
and we dont yet know if there is anything in the void between the solar systems that would allow for a course change.


You don't need anything to "allow for" a course change - its all Newton's 3rd law: every reaction has an opposite and equal reaction. So you throw some propellant out with a force of 1N, your ship'll experience an equal force in the opposite direction and thus be accelerated in a direction opposite that of the propellant.

That propellant doesn't need anything to "allow" for it to provide that opposite reaction.

Originally Posted By: paul
how do you know that there are even enought ions in the void?


Why do I need ions?

BTW, the density of hydrogen in interstellar space is already known, to a point where we've mapped its density both locally, and across much of the milky way:
http://en.wikipedia.org/wiki/Local_Bubble


Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 05/31/10 06:33 PM

Originally Posted By: ImagingGeek
If you have a waffle shield,


what is a waffle shield , and if it deflects an object while traveling at near c , how much energy is drained from the achieved acceleration that will need to be replaced.

you mentioned using centripetal force for artificial gravity,what if someone walks in the other direction of rotation?
wouldnt this force that they apply for propulsion detract from the rotational velocity?

sure if nothing moves inside the centrifuge of theship , you can have eternal rotation , but if nothing is going to move , then why have the centrifuge?

for every action there is a equal and opposite reaction.

so for every step taken , every item moved within the centrifuge more energy will be required to maintain a precise artificial gravity.

Quote:
Depends on the mass of the vessel, and how close to C you try to get.


apx how much mass are you sudgesting?

toothpaste , porknbeans, french fries, cheesburgers,new underwear,socks,etc....

because they will be moving so they will wear down there clothes.
because we cant suspend life yet.

how many crew members would there be?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 05/31/10 06:46 PM

Quote:
You don't need anything to "allow for" a course change - its all Newton's 3rd law: every reaction has an opposite and equal reaction. So you throw some propellant out with a force of 1N, your ship'll experience an equal force in the opposite direction


why do you need to throw something outside the ship?
why not just throw it out inside the ship.

the same reaction would occur.

and you still have your propellant.

you would only need a compressor and a nozzle that vectors
inside a compartment that has a lower pressure than the compressed propellant.

and you could use nitrogen , its safer you could actually breath it and its innert. so non flamable.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 05/31/10 07:56 PM

Originally Posted By: paul
why do you need to throw something outside the ship?
why not just throw it out inside the ship.

the same reaction would occur.

and you still have your propellant.


Doesn't work that way - material thrown within the ship will encounter resistance with the air, hull, etc. This will generate a force equal to the force of the propellant, thus neutralizing the thrust of the propellant. Its the ol' opposite and equal reaction thingie - the movement of the propellant will "push" on the ship, but the interaction of the propellant with the ship will push back equally. Net effect - zero thrust.

These kinds of internal energy transfer system only work in places where there is friction to counter the unwanted "return" energy of the propellant. Basically, you can thrust in one direction using a lot of force quickly, pushing the object forward. You then recover your propellant slowly, so the force of the propellant moving in the "wrong" direction doesn't exceed the static friction holding you in place.

That doesn't work in space - no friction.

Originally Posted By: paul
you would only need a compressor and a nozzle that vectors
inside a compartment that has a lower pressure than the compressed propellant.


Doesn't work that way, for the reasons mentioned above. In space, the net thrust of this kind of system is zero. And even in places where it is possible, the ISP would suck.

Originally Posted By: paul
and you could use nitrogen , its safer you could actually breath it and its innert. so non flamable.


The propellant used is dictated by the engine driving the spacecraft. Nobel gases are preferred for ion-based engines, due to the relative ease of ionizing them and the lack of intermolecular bonds. If using chemical engines, you need something that burns. If using nuclear, you need a fissile/fussile fuel, etc, etc, etc.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 05/31/10 08:52 PM

I think thats a load !!!

if I have a nozzle at one end of a long pipe and the entire pipe
is at 14.7 psi.

lets say the pipe is 60 inch diameter and 500 ft long.
and its inside your space ship.

and I have a compressor at the other end of the pipe that compresses air to 100 psi.

if I release the compressed air through the nozzle the reaction of the thrust will be felt by the pipe until the pressure in the pipe reaches 100 psi.

and if the compressor can maintain the 14.7 psi then
this can be continuous.

not only that but the friction between the pipe and the compressed air in the pipes that carries the compresed air back
to the nozzle will also be a force in the direction of movement.

according to your way of thinking if there was a hose attached to the place where the nozzle is attached instead of a nozzle and the hose was just laying there on the floor and someone turned on the air valve so that air could come out , the hose would just sit there , and it wouldnt move because you were taught that , and you could just walk up to it and pick it up.

me , I would want the air pressure to be turned off first.
so the hose wouldnt slap me in the head as I approached it.

so if the hose will move then if I held the hose , the ship would move.

the problem here is that when you are driving down
"think street"
and you see a "do not enter" sign you do not enter , instead you turn onto "do not think street" , I dont see any "do not enter" signs.

so I stay on "think street"




Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/01/10 09:20 AM

Watch out Paul, here I come into this thread too! :P

Originally Posted By: paul

so I stay on "think street"


I'm all in favour of thinking too, that's why I like this forum so much.

Just because somebody didn't explain all their reasoning about everything doesn't mean they didn't think it, either right then or way back in the past.

I remember when I was a physics student, a fellow student was spinning on his chair without touching anything else. As we've probably all done, twist your body quickly, then slowly back the other way. You get a net rotation in little fits and starts. He was doing that and said "Hey shouldn't this be impossible?". But then we figured out that the friction is different for different speeds so it does allow angular momentum to not be conserved locally.


The flailing hose example is irellivant because it's obviously losing material, air, that isn't recovered.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/01/10 01:05 PM

Originally Posted By: paul
I think thats a load !!!


Then you think wrong. A space ship is a closed system...but we'll get back to that.

Originally Posted By: paul
if I have a nozzle at one end of a long pipe and the entire pipe is at 14.7 psi.

lets say the pipe is 60 inch diameter and 500 ft long.
and its inside your space ship.

and I have a compressor at the other end of the pipe that compresses air to 100 psi.

if I release the compressed air through the nozzle the reaction of the thrust will be felt by the pipe until the pressure in the pipe reaches 100 psi.

and if the compressor can maintain the 14.7 psi then
this can be continuous.


Absolutely. But you're missing the "closed system" bit.

Originally Posted By: paul
not only that but the friction between the pipe and the compressed air in the pipes that carries the compresed air back to the nozzle will also be a force in the direction of movement.


And here is where you are missing the closed system bit.

Based on your description you have what could be considered a linear air accelerator - air goes in the front, is compressed, and is then forced out the back. Thus giving you a vacuum at the front, providing a forward "pull", and a rear-facing nozzel, giving a "push". On earth this system would give you a great deal of thrust - in fact, add a bit of fire and you've got a jet engine.

Problem is, if you're spraying that air inside of a space ship that thrust disappears - for two reasons. Firstly, nature abhors a vacuum, and your system is generating one at the front of the pump. Hence, there will be an flow of air towards the front of your air pump equal to the air flow entering your pump. Since that air can only come from within your ship, that vacuum will be replaced by ship-board air, and the movement of that air will exert a force equal to the force produced by the vacuum. Newtons laws - opposite and equal reaction.

Secondly, the air coming out of your nozzle has a lot of momentum. That momentum is equal to, but in the opposite direction of, the thrust produced by the nozzle end of your engine. Since your ship is a closed system that air cannot leave the ship, and as such has no option but to transfer that momentum back to the ship. This transfer will occur in the form of impacts between the moving column of air and the walls, etc of your ship. This transfer of momentum back to the ship is in a direction opposite to the thrust of your engine. So just as with the "vacuum thrust", the amount of momentum transfered to the ship will be equal to the momentum of the moving column of air, which in turn is equal to the thrust of your engine. Net effect - zero nozzle thrust.

So:
Vacuum thrust = 0
Nozzle thrust = 0
------------------
Total thrust = 0

Originally Posted By: paul
according to your way of thinking if there was a hose attached to the place where the nozzle is attached instead of a nozzle and the hose was just laying there on the floor and someone turned on the air valve so that air could come out , the hose would just sit there , and it wouldnt move because you were taught that , and you could just walk up to it and pick it up.


Not even close to what I said. The hose will quite efficiently transfer the thrust of the air, resulting in movement of the hose. However, my workshop (equivalent in this case to your spaceship) will stay put as the thrust of that hose is countered because the energy of the air coming out of the hose is transfered to the air and walls of my workshop, producing a net thrust (in the context of my workshop) of zero.

So the hose will flow around and knock the unaware in the head. But people standing outside of my shop are 100% safe from being run over by a moving shop - momentum, in the context of the shop - is zero.

Originally Posted By: paul
me , I would want the air pressure to be turned off first. so the hose wouldnt slap me in the head as I approached it.

so if the hose will move then if I held the hose , the ship would move.


Only if the contents of that hose are free to leave the ship. If they are not, the momentum of the air will be transfered back to the ship, providing a net zero thrust.

Originally Posted By: paul
the problem here is that when you are driving down "think street" and you see a "do not enter" sign you do not enter , instead you turn onto "do not think street" , I dont see any "do not enter" signs.

so I stay on "think street"


LOL, big words for a guy obviously lacking an understanding of basic physics. You cannot generate momentum in a closed system that imparts a net momentum on the closed system itself. You ship is such a closed system.

Open the system though - eject that air out of the ship itself - and you'll move along quite nicely.

Here's the physics you seem to be missing. Read the pages, then come back and we'll talk:

http://en.wikipedia.org/wiki/Closed_system
http://en.wikipedia.org/wiki/Thermodynamic_system
http://en.wikipedia.org/wiki/Newtons_laws
http://en.wikipedia.org/wiki/Momentum

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/01/10 03:19 PM

wow , I cant believe all you wrote just to show how little you
know about physics.

maybe I should say your knowlege of applying physics.

lets forget the compressor , because according to you and your knowledge about physics the only item in question is the fact that this system is a closed system.

and lets put all the compressed air inside a air tank at one end of the ship.

and its still inside the 60 inch diameter pipe and the pipe is still 500 ft long.

the air pressure inside the pipe is at 14.7 psi.

and suppose the air tank holds a mere 1000 cu ft of 100 psi
compressed air inside it.

the volume of 14.7 psi air inside the 500 ft long pipe
is 8817 cu ft.
and the pipe volume capacity is 9817 cu ft.

so the tank holds 1/9.8 the pipes area.

when I release air from the nozzle there is a thrust or a force
that pushes the ship in the opposite direction of the thrust.

this force is felt by the pipe just as if it were being applied
from outside the pipe at the other end of the pipe.

the air inside the pipe at 1 atm presses against every sq inch inside the pipe , not in any particular direction,
and continues to do so the entire time that air is being released from the nozzle.

so all of the pressure inside the pipe itself cancels out.

the only force remaining is the force generated by the thrust
of the nozzle.

so the stuff you argue with = 0 lb force
and
the stuff I argue with = a 100 lb force to a 0 lb force.
if I use a 1 sq inch nozzle.
of course the above does not include the decreasing pressure differential between the 14.7 psi and the final equalized pressure or the initial and final fluid velocity.

which greatly increases the force.

but it alone tells the story.

after the air tank has equalized pressure with the pipe the
thrust will stop , until then

I win the argument / discussion.

or until you can provide some type of solid believable
calculations that support your claim.

if this pipe were in space and you were inside it and the nozzle was pointed directly away from the sun.
and of course your space suit could withstand 100 psi of pressure.

would you still believe that you are right?

I dont believe that we should attempt space travel until
those who learn physics also learn to apply physics
without killing everybody on a space ship because of what they have not been taught.

Quote:
However, my workshop (equivalent in this case to your spaceship)


LOL

your workshop equivalent has a mass of 5.98 × 1024 kg

its attached to the earth.

but put your workshop in space and see what happens.









Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/01/10 04:26 PM

Originally Posted By: paul
wow , I cant believe all you wrote just to show how little you know about physics.


This, coming from the guy promoting a perpetual motion machine in another thread...LOL.

Originally Posted By: paul
lets forget the compressor , because according to you and your knowledge about physics the only item in question is the fact that this system is a closed system.


Because that is the problem with your own model. Your space ship is a closed system; matter can neither enter nor leave. You can exert forces within such a system, but the sum of those forces will always be zero at the level of the system itself.

Originally Posted By: paul
...the air inside the pipe at 1 atm presses against every sq inch inside the pipe , not in any particular direction, and continues to do so the entire time that air is being released from the nozzle.

so all of the pressure inside the pipe itself cancels out.

the only force remaining is the force generated by the thrust of the nozzle.


However, none of the above has anything to do with the forces in your closed system.

The propulsive force generated where the air leaves the pipe is due to the momentum of the moving air; not friction, etc, within the pipe (in fact, friction in the pipe reduces thrust by slowing the air). Its all newtons laws - you use pressure to accelerate a mass of air in one direction, and you get a an equal force on the opposite that'll push on your tank of compressed air.

Momentum, BTW, is a specific physical quantity equal to mass * velocity.

Momentum is also conserved in a closed system - you can neither add it, nor take it away.

Originally Posted By: paul
after the air tank has equalized pressure with the pipe the
thrust will stop , until then

I win the argument / discussion.


Sorry, you don't. Missing half the equation does not equal a valid answer.

I'll walk through this slowly; hopefully you'll see where your error is.

The pressurized air in the tank represents a stored (potential) energy. Upon letting air out of the tank, that potential energy generates a force which accelerates the air down the pipe. Upon leaving the pipe the air will have a momentum, defined by p = mass * velocity. The ship will have an equal momentum in the opposite direction, due to newtons 3rd law (every action has an opposite, but equal reaction). For the sake of ease, lets assume that 1kg of air is moving at a velocity of 100m/s. This will give the air a momentum of:

p(air) = mv
p(air) = 1kg*100m/s
p(air) = 100kg*m/s

Newtons 3rd law dictates our tank (and therefore the ship it is attached to) has the same momentum but in the opposite direction, therefore:

p(tank) = -p(air)
p(tank) = -100kg*m/s

Lets assume our ship/tank combo weighs 100kg. In this case this release of air would impart a velocity of:

p(tank) = -mv, therefore v(tank) = -p/m
v = -100kg*m/s / 100kg
v = -1m/s

The negative sign in this case simply means the ship is moving in the opposite direction of the air expelled from the tank.

In an open system this would lead to the tank going one way and the air going the other - exactly what you propose is happening. In a fully open system (space) that air would continue moving forever - newtons first law (an object in motion stays in motion).

But we're not in a open system. In our closed ship that moving air will eventually hit a bulkhead or hull. When this moving column of air meets the bulkhead/hull that impact creates a force which will be proportional to the momentum of the air (in fact, it will be exactly equal to the force which imparted the air its momentum in the first place). As before, newtons 3rd law dictates this will create an equal, but opposite change in momentum on that bulkhead.

So in this case we have the air imparting a momentum to the bulkhead of:

p(air) = p(bulkhead);
since p(air) = 100kg*m/s; p(bulkhead) also = 100kg*m/s

keeping in mind

p(tank) = -p(air) = -100kg*m/s

Now, if the bulkhead and tank were separate entities, you'd have the tank going one way, and the air would blow the bulkhead in the other. However, the bulkhead and tank are connected as they are both part of the ship. Therefore, the total momentum imparted on the ship will be equal to the sum of these two momentums imparted onto these different parts of the ship:

So we have:
p(ship) = p(tank) + p(bulkhead)
p(ship) = -100kg*m/s + 100kg*m/s
p(ship) = 0kg*m/s

Since velocity is a direct product of momentum, (p=mv)
v(ship) = p/m
v(ship) = 0kg*m/s/100kg
v(ship) = 0m/s

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/01/10 05:32 PM

WOW !!!

so in this closed container ( the pipe ).
air will only move in one direction?
and that direction just happens to be in the direction
of the air comming out of the nozzle...

how convienient for you.

so after all the air has been released out of the tank
there will be a 1000 cu ft block of 100 psi air at the end
furtherest from the nozzle.

I said a believable calculation.

in my world , the air at the other end of the pipe will slowly
compress along with the rest of the air in the pipe.

because pressure is transmitted equally in a closed system.

lets take your assumptions even further.

lets place air pressure guages on every square inch of surface area outside the pipe.

according to your comical assessment the pressure guages will all keep reading 14.7 psi , except the ones located at the end of the pipe.

LOL

the bulkhead is 500 ft away...

Quote:
p(air) = p(bulkhead);


what a load

Originally Posted By: ImageingGeek
Here's the physics you seem to be missing. Read the pages, then come back and we'll talk:





what will we talk about , your fantasies about physics?

Quote:
This, coming from the guy promoting a perpetual motion machine in another thread...LOL.


exactly .. LOL





Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/01/10 07:49 PM

Originally Posted By: paul
WOW !!!

so in this closed container ( the pipe ).
air will only move in one direction?


Ahh, I see - your pipe is the closed system, and doesn't exhaust into the ship?.

If that's the case, it doesn't help you any. Same problem, only confined to a smaller space.

Originally Posted By: paul
so after all the air has been released out of the tank there will be a 1000 cu ft block of 100 psi air at the end furtherest from the nozzle.


I'm not 100% sure I've understood your engine. I am assuming:

1) There is a pipe, sealed on one end
2) There is a pressure tank on the other end
3) You release the air in the pressure tank into the pipe

Regardless, the math is the same. In this situation you'll release the air, and it will flow into the pipe until you reach a pressure equilibrium - i.e. the pipe and tank are all at the same pressure; no pressure gradient.

The initial movement of the air from the tank into the pipe will create a force that will move the ship forward. However, that force will be exactly countered by an equal and opposite force created by the air coming out of the tank interacting with the pipe itself and any air in the pipe.

Net result - no thrust.

Originally Posted By: paul
I said a believable calculation.


That's the science of it. Whether you "believe" science or not isn't my problem. Momentum, kinetic energy, potential energy and force are all well established physical concepts. If you can find some math that allows for those to move a closed system, let us know - that would be nobel-prize kind of work.

Originally Posted By: paul
in my world , the air at the other end of the pipe will slowly compress along with the rest of the air in the pipe.


Nope. The air in the pipe will be compressed equally as air moves in from the tank - gasses naturally fill a void in order to generate an even distribution. You may have a slight imbalance if you release the air quickly enough, but this will only exist for a short period of time.

Originally Posted By: paul

lets take your assumptions even further.

lets place air pressure guages on every square inch of surface area outside the pipe.

according to your comical assessment the pressure guages will all keep reading 14.7 psi , except the ones located at the end of the pipe.


Nope, they wouldn't. I think the issue here is we're sealing up different parts of the system.

I was describing the case where the pipe is open to the inside of this ship. In this case the pressure of the whole ship would increase.

I now think you're describing a case where the pipe is closed, and thus the pressure in the pipe increases, while the remainder of the ship stays the same.

In either case, the net thrust is zero - the initial momentum provided by the movement of the gas from the tank into the pipe or ship will be exactly countered by the interactions of that moving air with the stationary pipe (if the pipe is sealed) or ship (if the pipe is open).

Net thrust, in either case, is zero. Newtons laws - every action has an opposite and equal reaction.

After all, if it worked the way you think it does, NASA would have no trouble sending people to Mars. Just recycle you propellant back and forth.

They don't - strange, isn't it. That the rocket scientists at NASA don't know about this imaginary physics you have created...

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/01/10 08:41 PM

I think you understood the first time.
Originally Posted By: paul

and lets put all the compressed air inside a air tank at one end of the ship.

and its still inside the 60 inch diameter pipe and the pipe is still 500 ft long.


the tank is inside the 500 ft pipe.

there is space between the tank and the pipe.

so that the air inside the pipe presses against the entire inside area of the pipe.

Originally Posted By: ImagingGeek
Regardless, the math is the same.



what math are you refering to.
the
Originally Posted By: ImagingGeek
since p(air) = 100kg*m/s; p(bulkhead) also = 100kg*m/s


If you begin with 100 psi tank pressure
and
14.7 psi external pressure in the pipe.

you dont get an immediate equalization between the two.

you would end up with apx 10 psi inside the pipe after the
pressures have equalized.

its a 1000 cu ft tank inside a apx 10,000 cu ft area so the air
in the tank expands inside the pipe and the pressure reduces as it expands.

the force that you are referring to is fantasy , the same force is applied to both ends and to every square inch of the inside of the pipe.

thereby , no net force from the air acting against the pipe.

and I might add that the force caused by the (air pressure) is multiplied by the entire inside area of the pipe.

not just the end that you choose.

ie..
Originally Posted By: ImagingGeek
p(bulkhead) also = 100kg*m/s


but go ahead and calculate the force that you think is applied to the end of the pipe , it will be the same force that is applied to the opposite end.

thus no net effect.

I agree the pipe will not move for long , it will only move as long as the air has a differential pressure that will cause a net force in the opposite direction of air flow.

exactly like a jet plane would move if it were inside a larger pipe.

and if the jet plane was in the larger pipe in space and it pressed against the end of the pipe , it would transfer its momentum to the pipe , and if it kept on thrusting forward the pipe would continue to move forward.

but I suppose you have a answer to that also...

and it is a no-thinker

zero net force , right?

and in this scenario the air thrust comming out of the tank is exactly the same as thrust comming out of a jet aircraft.










Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/01/10 09:44 PM

lets add up all the forces that apply.
as the air is escaping from the tank.

the ends of the tank

114.7 psi --->
114.7 psi <---

the ends of the pipe

14.7 psi --->
14.7 psi <---

the rear of the nozzle as the air enters the nozzle

114.7 psi --->
14.7 psi <---

the front of the nozzle as the air escapes the nozzle

114.7 psi --->
14.7 psi <---

action = reaction

net result of force to pipe 100 psi <---

nothing is gained inside the pipe , nothing is lost inside the pipe except the energy that was put into compressing the air , but the pipe moves.

I would wager that the exact amount of energy that is is held in the compressed air ( due to its compression ) inside the tank is the exact amount of energy that is applied for propulsion of the pipe.






Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/02/10 12:40 PM

Originally Posted By: paul
lets add up all the forces that apply.
as the air is escaping from the tank.

the ends of the tank

114.7 psi --->
114.7 psi <---


Pressures are potential energy; as in stored energy. You cannot use them directly to determine the force generated upon the release of that pressure. After all, the same 100psi pressure difference with 1000cuft of air has a lot more stored energy than does 1cuft at the same pressure.

Pressure, in this case, is transformed into into momentum thourgh two "conversion" processes. The first is through accelerating the air, giving it kinetic energy. The second is through the expansion of the gas as it leaves a tank - a properly designed nozzle can harness that expansion energy and produce thrust above what the kinetic energy of the air alone provides. The exact amount of force produced is determined by:

F=[m*Ve]+[(Pt-Pa)*Ae], where

m = mass of gas
Ve = exhaust velocity (speed material leaves tank)
(Pt-Pa) = pressure difference between tank and air
Ae = area of nozzle the are is escaping through

You may have noticed that in the above formula you're essentially calculating F(total) = F(from kinetic energy of air) + F(from pressure differential)

In your example we can fill in a few numbers. A cuft of air weights approx 34g, so 1000cuft weights 34,000g (34kg). Your pressure difference is 100PSI, which is SI units is 6895kPa.

Ve is determined by a fairly complex formula:


This becomes more complex when you take into account the pressure decreases as the tank empties, but you get the idea.

Originally Posted By: paul
net result of force to pipe 100 psi <---


Once again, pressure is stored energy and thus cannot be used to determine the net force generated on an object. You need to calculate the force developed by the mass of the gas, combined with the differential pressure between the inside/outside of the tank to determine the force generated.

And once again, we come back to the very simple physical principal you insist on ignoring - newtons 3rd law. As the air leaves the tank, both the tank and that air experience an equal, but opposite force. This force imparts both the tank and air with an equal, but opposite, amount of momentum.

In an open system the momentum behind that air is dissipated into the surroundings.

Where do your propose it goes in your sealed pipe?

The answer, of course, is simple - the momentum of that air is transfered to whatever the air encounters. Namely, the momentum is transfered to the pipe as the air encounters the pipe. Since the momentum of the air is equal, but opposite, to the momentum of the tank, the air will impart and equal, but opposite momentum to the pipe.

One equal, plus an equal but opposite, equals zero.

A + -A = 0

Originally Posted By: paul
nothing is gained inside the pipe , nothing is lost inside the pipe except the energy that was put into compressing the air , but the pipe moves.

I would wager that the exact amount of energy that is is held in the compressed air ( due to its compression ) inside the tank is the exact amount of energy that is applied for propulsion of the pipe.


The amount of ***potential energy*** in the compressed air is the maximum amount which can be "extracted" in the form of thrust. But newtons 3rd law dictates that the force generated will act evenly on both the air coming out of the tank as well as on the tank itself. Since that escaping air is contained in the pipe, it's momentum is transfered to the pipe, thus countering the momentum the tank exerts on the pipe. Giving you a net change in the momentum of the pipe of zero.

But hey, ff you're so confident why don't you build it, patent it, and make yourself rich. After all, you've solved the single biggest problem in sending spacecraft outside of low-earth orbit - and in doing so refuted newton himself.

Or you could accept reality - you are wrong.

Bryan
Posted by: KirbyGillis

Re: Orion, Mission to Alpha Centauri - 06/02/10 03:25 PM

Hi Paul,

One more correction is needed... it would take approximately 16 minutes for light to make the round trip between the Earth and Sun.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/02/10 04:46 PM

Quote:
but you get the idea.


you havent show anything except that you can cut and paste
equations.
and then dump the same ridiculous sob story on me.

what have you shown? nothing.

you go to a great extent to put up a equation and explain all the factors involved (at least you think thats all) and then you dont even supply any result from the equation..

is that the way it works in college these days , the test ask you
which equasion you use to calculate a given set of factors but you are not required to supply a solution.

because that is all you do.

supply a mass of useless words beneath a equation.

lets see if I can solve the problem this way.


The foundational axioms of fluid dynamics are the conservation laws, specifically, conservation of mass, conservation of linear momentum (also known as Newton's Second Law of Motion), and conservation of energy (also known as First Law of Thermodynamics). These are based on classical mechanics and are modified in quantum mechanics and general relativity. They are expressed using the Reynolds Transport Theorem.

In addition to the above, fluids are assumed to obey the continuum assumption. Fluids are composed of molecules that collide with one another and solid objects. However, the continuum assumption considers fluids to be continuous, rather than discrete. Consequently, properties such as density, pressure, temperature, and velocity are taken to be well-defined at infinitesimally small points, and are assumed to vary continuously from one point to another. The fact that the fluid is made up of discrete molecules is ignored.

For fluids which are sufficiently dense to be a continuum, do not contain ionized species, and have velocities small in relation to the speed of light, the momentum equations for Newtonian fluids are the Navier-Stokes equations, which is a non-linear set of differential equations that describes the flow of a fluid whose stress depends linearly on velocity gradients and pressure. The unsimplified equations do not have a general closed-form solution, so they are primarily of use in Computational Fluid Dynamics. The equations can be simplified in a number of ways, all of which make them easier to solve. Some of them allow appropriate fluid dynamics problems to be solved in closed form.

In addition to the mass, momentum, and energy conservation equations, a thermodynamical equation of state giving the pressure as a function of other thermodynamic variables for the fluid is required to completely specify the problem. An example of this would be the perfect gas equation of state:


where p is pressure, &#961; is density, Ru is the gas constant, M is the molar mass and T is temperature.

[edit] Compressible vs incompressible flow
All fluids are compressible to some extent, that is changes in pressure or temperature will result in changes in density. However, in many situations the changes in pressure and temperature are sufficiently small that the changes in density are negligible. In this case the flow can be modeled as an incompressible flow. Otherwise the more general compressible flow equations must be used.

Mathematically, incompressibility is expressed by saying that the density &#961; of a fluid parcel does not change as it moves in the flow field, i.e.,


where D / Dt is the substantial derivative, which is the sum of local and convective derivatives. This additional constraint simplifies the governing equations, especially in the case when the fluid has a uniform density.

For flow of gases, to determine whether to use compressible or incompressible fluid dynamics, the Mach number of the flow is to be evaluated. As a rough guide, compressible effects can be ignored at Mach numbers below approximately 0.3. For liquids, whether the incompressible assumption is valid depends on the fluid properties (specifically the critical pressure and temperature of the fluid) and the flow conditions (how close to the critical pressure the actual flow pressure becomes). Acoustic problems always require allowing compressibility, since sound waves are compression waves involving changes in pressure and density of the medium through which they propagate.

[edit] Viscous vs inviscid flow
Viscous problems are those in which fluid friction has significant effects on the fluid motion.

The Reynolds number, which is a ratio between inertial and viscous forces, can be used to evaluate whether viscous or inviscid equations are appropriate to the problem.

Stokes flow is flow at very low Reynolds numbers, Re &#8810; 1, such that inertial forces can be neglected compared to viscous forces.

On the contrary, high Reynolds numbers indicate that the inertial forces are more significant than the viscous (friction) forces. Therefore, we may assume the flow to be an inviscid flow, an approximation in which we neglect viscosity completely, compared to inertial terms.

This idea can work fairly well when the Reynolds number is high. However, certain problems such as those involving solid boundaries, may require that the viscosity be included. Viscosity often cannot be neglected near solid boundaries because the no-slip condition can generate a thin region of large strain rate (known as Boundary layer) which enhances the effect of even a small amount of viscosity, and thus generating vorticity. Therefore, to calculate net forces on bodies (such as wings) we should use viscous flow equations. As illustrated by d'Alembert's paradox, a body in an inviscid fluid will experience no drag force. The standard equations of inviscid flow are the Euler equations. Another often used model, especially in computational fluid dynamics, is to use the Euler equations away from the body and the boundary layer equations, which incorporates viscosity, in a region close to the body.

The Euler equations can be integrated along a streamline to get Bernoulli's equation. When the flow is everywhere irrotational and inviscid, Bernoulli's equation can be used throughout the flow field. Such flows are called potential flows.

[edit] Steady vs unsteady flow

Hydrodynamics simulation of the Rayleigh–Taylor instability [2]When all the time derivatives of a flow field vanish, the flow is considered to be a steady flow. Steady-state flow refers to the condition where the fluid properties at a point in the system do not change over time. Otherwise, flow is called unsteady. Whether a particular flow is steady or unsteady, can depend on the chosen frame of reference. For instance, laminar flow over a sphere is steady in the frame of reference that is stationary with respect to the sphere. In a frame of reference that is stationary with respect to a background flow, the flow is unsteady.

Turbulent flows are unsteady by definition. A turbulent flow can, however, be statistically stationary. According to Pope:[3]

The random field U(x,t) is statistically stationary if all statistics are invariant under a shift in time.

This roughly means that all statistical properties are constant in time. Often, the mean field is the object of interest, and this is constant too in a statistically stationary flow.

Steady flows are often more tractable than otherwise similar unsteady flows. The governing equations of a steady problem have one dimension less (time) than the governing equations of the same problem without taking advantage of the steadiness of the flow field.




[edit] Laminar vs turbulent flow
Turbulence is flow characterized by recirculation, eddies, and apparent randomness. Flow in which turbulence is not exhibited is called laminar. It should be noted, however, that the presence of eddies or recirculation alone does not necessarily indicate turbulent flow—these phenomena may be present in laminar flow as well. Mathematically, turbulent flow is often represented via a Reynolds decomposition, in which the flow is broken down into the sum of an average component and a perturbation component.

It is believed that turbulent flows can be described well through the use of the Navier–Stokes equations. Direct numerical simulation (DNS), based on the Navier–Stokes equations, makes it possible to simulate turbulent flows at moderate Reynolds numbers. Restrictions depend on the power of the computer used and the efficiency of the solution algorithm. The results of DNS agree with the experimental data.

Most flows of interest have Reynolds numbers much too high for DNS to be a viable option[4], given the state of computational power for the next few decades. Any flight vehicle large enough to carry a human (L > 3 m), moving faster than 72 km/h (20 m/s) is well beyond the limit of DNS simulation (Re = 4 million). Transport aircraft wings (such as on an Airbus A300 or Boeing 747) have Reynolds numbers of 40 million (based on the wing chord). In order to solve these real-life flow problems, turbulence models will be a necessity for the foreseeable future. Reynolds-averaged Navier–Stokes equations (RANS) combined with turbulence modeling provides a model of the effects of the turbulent flow. Such a modeling mainly provides the additional momentum transfer by the Reynolds stresses, although the turbulence also enhances the heat and mass transfer. Another promising methodology is large eddy simulation (LES), especially in the guise of detached eddy simulation (DES)—which is a combination of RANS turbulence modeling and large eddy simulation.

[edit] Newtonian vs non-Newtonian fluids
Sir Isaac Newton showed how stress and the rate of strain are very close to linearly related for many familiar fluids, such as water and air. These Newtonian fluids are modeled by a coefficient called viscosity, which depends on the specific fluid.

However, some of the other materials, such as emulsions and slurries and some visco-elastic materials (e.g. blood, some polymers), have more complicated non-Newtonian stress-strain behaviours. These materials include sticky liquids such as latex, honey, and lubricants which are studied in the sub-discipline of rheology.

[edit] Subsonic vs transonic, supersonic and hypersonic flows
While many terrestrial flows (e.g. flow of water through a pipe) occur at low mach numbers, many flows of practical interest (e.g. in aerodynamics) occur at high fractions of the Mach Number M=1 or in excess of it (supersonic flows). New phenomena occur at these Mach number regimes (e.g. shock waves for supersonic flow, transonic instability in a regime of flows with M nearly equal to 1, non-equilibrium chemical behavior due to ionization in hypersonic flows) and it is necessary to treat each of these flow regimes separately.

[edit] Non-relativistic vs relativistic flows
Classical fluid dynamics is derived based on Newtonian mechanics, which is adequate for most applications. However, at speeds comparable to the speed of light Newtonian mechanics is inaccurate and a relativistic framework has to be used instead.

[edit] Magnetohydrodynamics
Main article: Magnetohydrodynamics
Magnetohydrodynamics is the multi-disciplinary study of the flow of electrically conducting fluids in electromagnetic fields. Examples of such fluids include plasmas, liquid metals, and salt water. The fluid flow equations are solved simultaneously with Maxwell's equations of electromagnetism.

[edit] Other approximations
There are a large number of other possible approximations to fluid dynamic problems. Some of the more commonly used are listed below.

The Boussinesq approximation neglects variations in density except to calculate buoyancy forces. It is often used in free convection problems where density changes are small.
Lubrication theory and Hele-Shaw flow exploits the large aspect ratio of the domain to show that certain terms in the equations are small and so can be neglected.
Slender-body theory is a methodology used in Stokes flow problems to estimate the force on, or flow field around, a long slender object in a viscous fluid.
The shallow-water equations can be used to describe a layer of relatively inviscid fluid with a free surface, in which surface gradients are small.
The Boussinesq equations are applicable to surface waves on thicker layers of fluid and with steeper surface slopes.
Darcy's law is used for flow in porous media, and works with variables averaged over several pore-widths.
In rotating systems, the quasi-geostrophic approximation assumes an almost perfect balance between pressure gradients and the Coriolis force. It is useful in the study of atmospheric dynamics.
[edit] Terminology in fluid dynamics
The concept of pressure is central to the study of both fluid statics and fluid dynamics. A pressure can be identified for every point in a body of fluid, regardless of whether the fluid is in motion or not. Pressure can be measured using an aneroid, Bourdon tube, mercury column, or various other methods.

Some of the terminology that is necessary in the study of fluid dynamics is not found in other similar areas of study. In particular, some of the terminology used in fluid dynamics is not used in fluid statics.

[edit] Terminology in incompressible fluid dynamics
The concepts of total pressure and dynamic pressure arise from Bernoulli's equation and are significant in the study of all fluid flows. (These two pressures are not pressures in the usual sense—they cannot be measured using an aneroid, Bourdon tube or mercury column.) To avoid potential ambiguity when referring to pressure in fluid dynamics, many authors use the term static pressure to distinguish it from total pressure and dynamic pressure. Static pressure is identical to pressure and can be identified for every point in a fluid flow field.

In Aerodynamics, L.J. Clancy writes[5]: To distinguish it from the total and dynamic pressures, the actual pressure of the fluid, which is associated not with its motion but with its state, is often referred to as the static pressure, but where the term pressure alone is used it refers to this static pressure.

A point in a fluid flow where the flow has come to rest (i.e. speed is equal to zero adjacent to some solid body immersed in the fluid flow) is of special significance. It is of such importance that it is given a special name—a stagnation point. The static pressure at the stagnation point is of special significance and is given its own name—stagnation pressure. In incompressible flows, the stagnation pressure at a stagnation point is equal to the total pressure throughout the flow field.

[edit] Terminology in compressible fluid dynamics
In a compressible fluid, such as air, the temperature and density are essential when determining the state of the fluid. In addition to the concept of total pressure (also known as stagnation pressure), the concepts of total (or stagnation) temperature and total (or stagnation) density are also essential in any study of compressible fluid flows. To avoid potential ambiguity when referring to temperature and density, many authors use the terms static temperature and static density. Static temperature is identical to temperature; and static density is identical to density; and both can be identified for every point in a fluid flow field.

The temperature and density at a stagnation point are called stagnation temperature and stagnation density.

A similar approach is also taken with the thermodynamic properties of compressible fluids. Many authors use the terms total (or stagnation) enthalpy and total (or stagnation) entropy. The terms static enthalpy and static entropy appear to be less common, but where they are used they mean nothing more than enthalpy and entropy respectively, and the prefix "static" is being used to avoid ambiguity with their 'total' or 'stagnation' counterparts.

therefore the net force on the pipe = 100 psi <---






Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/02/10 05:04 PM

the 100 psi <---
that I use overcomes the nothing that you have.

we can calculate the mass of the air that travels from point a to point b and this will give us the mass that is being moved within the pipe.
then we can calculate the velocity at which it moves.

then we can calculate the opposition to pipe movement as zero.

simply because there is no opposition to movement.

see if you can understand this one simple thing.

the only place that any force that can be counted as a force
for movement in a direction is where the air is escaping the nozzle.

everything else is canceled out.

what if the tank was just floating inside the pipe
and it is located at the middle of the pipe.

nothing to stop its movement.

then the valve is opened.

are you really going to say it would just sit there and not move?

and why wouldnt it move all the way to the end and apply its momentum to the pipe itself?

because you dont want it to.
because you know physics but you dont know how to apply physics?

because a equation told you it wont move?

knowing physics is like owning a brand new cadilac
with a empty tank of gas.
knowing how to apply physics is like owning all the gas stations in the world.

your cadilac is worthless without the gas.


I sometimes wonder how we ever put a man on the moon given the quality of our college graduates and ability to apply what they have learned or supposedly learned.





Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/02/10 05:23 PM

Originally Posted By: paul
Quote:
but you get the idea.


you havent show anything except that you can cut and paste
equations.
and then dump the same ridiculous sob story on me.

what have you shown? nothing.


Actually, I've shown that physics states your space drive is a physical impossibility.

I'd also point out that in "copying & pasting" equations I've given a mathematical basis for my arguments. Verses you, whose evidence consists of:

a) whining that I provide mathematical evidence of my claims,
b) ignoring basic physical laws such as newtons laws of motion, the conservation of energy, etc, and
c) sticking your fingers in your ears and going "neener-neener-neener" every time you hear something you don't like.

Originally Posted By: paul
you go to a great extent to put up a equation and explain all the factors involved (at least you think thats all) and then you dont even supply any result from the equation..


Because I couldn't - to do that I'd have to know the area of the nozzle on the end of the tank, its divergence angle, the volume of the pipe the tank is sealed in, and so forth. Since you did not provide that info, it is not possible for me to complete the calculation.

But thanx for confirming those mathematics are beyond your ability to comprehend. After all, anyone who bothered to read what the terms meant would realize that several important variables were not accounted for.

Originally Posted By: paul
therefore the net force on the pipe = 100 psi <---


Congratulations, you can cut-and-paste from the wikipedia. I doubt you even understand what you copied - it was mostly irrelevent to the topic at hand, since fluid dynamics will generally describe the flow of the fluid through a pipe, not once it leaves the pipe. And since most of the energy in a system such as yours are derived by the nozzle - i.e. once the air has escaped the pipe - any calculations as to the energy of the moving fluid in the pipe will underestimate the actual amount of thrust provided.

Would this be a bad time to point out I spent a lot of my Phd working out the fluid dynamics of blood flow and how that relates to the distribution of cells within the blood vessel and the forces they experience when interacting with the blood vessels? Even built microfluidic devices that do things like sort cells and mimic our circulatory systems...

I'd also point out that I never once said the pipe wouldn't experience 100psi of pressure - the end closest to the tank would experience that pressure immediately after you open the valve.

However, none of that changes the original issues, namely:

1) Pressure is not a measure of the amount of energy available for work - its merely a measure of force per unit area due to the presence of a fluid. The actual potential energy available is, as I stated earlier, determined by the volume of gas and the pressure it is at. Therefore saying "the pipe experiences 100psi" is meaningless, as that tells you nothing as to the amount of force generated upon release of the air - for that you need to take in account the mass of the air, and the speed it is moving at - using the equations I outlined in my earlier post.

2) You are ignoring what happens once the air leaves the tank. That moving air has kinetic energy. That energy doesn't magically disappear - it has to go somewhere. That somewhere, in a sealed pipe, is the pipe itself. And since the air is moving in opposition to the force on the tank, guess what the effect of that energy transfer is.

Like I said before, build it, show it works, patent it, and show Newton, NASA, etc, to have been wrong.

Bryan

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/02/10 05:54 PM

Originally Posted By: paul
the 100 psi <---
that I use overcomes the nothing that you have.


No, it does not. PSI is a measure of pressure, not the force accelerating the tank.

Originally Posted By: paul
we can calculate the mass of the air that travels from point a to point b and this will give us the mass that is being moved within the pipe.
then we can calculate the velocity at which it moves.


Which was the point of the equasions I provided - the very ones you were whining about.

Originally Posted By: paul
then we can calculate the opposition to pipe movement as zero.


Assuming it is in space, yes.

Originally Posted By: paul
see if you can understand this one simple thing.

[quote=paul]the only place that any force that can be counted as a force for movement in a direction is where the air is escaping the nozzle.


Sorry, that is incorrect. The air leaving the nozzle has kinetic energy (and thus, momentum - the two are closely related). That energy/momentum doesn't magically disappear once the air has moved away from the nozzle - the air will retain that energy/momentum for ever - unless the air encounters something. At which point that energy/momentum will be transfered to the object it encounters.

Originally Posted By: paul
what if the tank was just floating inside the pipe and it is located at the middle of the pipe.

nothing to stop its movement.

then the valve is opened.

are you really going to say it would just sit there and not move?


The tank will move, the pipe it is contained in will not.

For the sake of simplicity, lets assume the tank, when empty, weights 1kg. Lets also assume it had enough air to accelerate it to a speed of 1m/s.

This tank would have:
Kinetic energy = 1/2*m*v^2 = 1/2* 1kg * (1m/s)^2 = 0.5J
Momentum = mv = 1kg * 1m/2^2 = 1kg*m/s

Because of newtons 3rd law, the air that came out of the tank would have the same momentum. Opposite and equal - very important here.

So your tank flies forward, and impacts the front of the pipe. When it does this it will transfer that kinetic energy/momentum to the pipe, pushing the pipe forward.

But at the same time, you have air carrying energy/momentum in the opposite direction. When that air impacts the back of the pipe it'll transfer that energy/momentum to the back of the pipe, pushing the pipe backwards.

Keep in mind - both the tank and the air that came out of it have the same momentum - newtons 3rd law dictates that. So you have an equal amount of momentum being applied to the front and back ends of the pipe.

What happens? It should be self-evident!

Originally Posted By: paul
and why wouldnt it move all the way to the end and apply its momentum to the pipe itself?


I never said it wouldn't. But why would you expect the air expelled from the tank not to do the same thing?

The reason is simple - reality clashes with your preconceptions. so you deny reality.

Originally Posted By: paul

I sometimes wonder how we ever put a man on the moon given the quality of our college graduates and ability to apply what they have learned or supposedly learned.


We put men on the moon by understanding physics. Ever wonder why they didn't use your magical drive to get men there, and instead used big rockets that expel their propellant into space?

It isn't cause NASA is dumb, its because your drive is a physical impossibility.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/02/10 06:01 PM

Quote:
Because I couldn't - to do that I'd have to know the area of the nozzle on the end of the tank, its divergence angle, the volume of the pipe the tank is sealed in, and so forth. Since you did not provide that info, it is not possible for me to complete the calculation.


I said the nozzle area is 1 square inch.
I said the volue of the tank is 1000 sq ft.
I even gave you the pipe volume.
I said the fluid is air so SP = 1.29
the DP is 114.7 - 14.7 = 100

you cant do the math , so you claim that I cant.

ohhhhhhh !!!! a college gadget for sure.

Quote:
Actually, I've shown that physics states your space drive is a physical impossibility.


Actually, you havent SHOWN anything.








Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/02/10 08:34 PM

Originally Posted By: paul
I said the nozzle area is 1 square inch.
I said the volue of the tank is 1000 sq ft.
I even gave you the pipe volume.
I said the fluid is air so SP = 1.29
the DP is 114.7 - 14.7 = 100


You do realize the the exact numbers are meaningless - regardless of how fast or slow the air comes out of the tank, or how much or little of it there there, the momentum of the tank will always be equal, but opposite, to that of the air expelled from the tank.

None-the-less, I did miss your specifications (although even here you didn't give them all). 60" diameter pipe (1.152m), 500' (152.5m) long. You haven't given a divergence angle for the nozzle, so I'll assume it is zero (i.e. there is no cone; just a striaght tube).

Volume of pipe = pi*r^2*h = 3.14*(1.152/2)^2*152.5m = 158.95m^3

I'm going to assume your volume of 1000 sq ft is actually cuft. 1000cuft = 28m^3, which is compressed to 10132kPa (100ATA).

The force generated upon the release of this air is:
F=[m*Ve]+[(Pt-Pa)*Ae]

Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is:
F = (Pt-Pa)*Ae
F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N

As the tank empties and the pipe fills, that force will drop. Thrust ends when Pt = Pa. ASCII doesn't allow me to show my work, integrations and all that, but excel gives a total thrust of just over 60kN*s (kilo-newton seconds).

Originally Posted By: paul
you cant do the math , so you claim that I cant.


Just did the math. And I state again that you cannot; nor do I believe you understand it either. But as I stated at the beginning, the specific numbers do not matter - the force applied to the tank is equal, but opposite to the force applied to the air coming out of the tank. No matter how much force is generated the pipe sits still.

Unless you make enough force to rupture the pipe...

It isn't rocket science - oh wait, in this case it kinda is... smile

And here we are again - exactly where we were at the beginning of our "discussion" - you still do not have the faintest understanding of newtons laws of motion and law of conservation of energy. And once again I've provided yet another layer of mathematics showing the impossibility of your perpetual-motion engine.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/02/10 11:24 PM

Quote:
Volume of pipe = pi*r^2*h = 3.14*(1.152/2)^2*152.5m = 158.95m^3


1 foot = 0.3048 meters


60 inch / 2 = 30 inch
30^2 = 900
900 * 3.14159 = 2827.431

the cross sectional area = 2827.431 sq inch

500 ft * 12 inch = 6000 inch

2827.433 * 6000 = 16964600.329384883487698274269709

there are 1728 cu in in a cu ft

16964600.329384883487698274269709 / 1728 = 9817 cu ft

9817 cu ft = 277 cu meters

somehow you have lost
277 - 158 = 119 cu meters

119 cu meters lost , how did that happen?

And I state again that you cannot


I see what happened now ...

you used 1.152 instead of 1.524 as the diameter

wow , that made a big difference didnt it

Quote:
The force generated upon the release of this air is:
F=[m*Ve]+[(Pt-Pa)*Ae]

Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is:
F = (Pt-Pa)*Ae
F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N


one load after another...

theres something VERY important when calculating force

--------- YOU NEED MASS ---------

F=MA

not

F=whatever you want it to be.

it looks like your using the initial pressure
in the tank (pt)then subtracting the initial pressure
in the pipe (pa?)then multiplying that number by the nozzle area (Ae?).

so where are you going with this ?


Quote:
Because we don't have a cone, m*Ve is zero


so since the mass comes out of a tube as you said , the mass has no velocity?

it just covertly sneeks through the nozzle or perhaps it
is being transported as they do on star trek !!


Quote:
I said the volue of the tank is 1000 sq ft.


yes , after the strain of trying to coax a little common sence out.

but as I recall...................
I did use 1000 cu ft earlier at the begining when I mentioned the tank in the pipe

Originally Posted By: paul
and suppose the air tank holds a mere 1000 cu ft of 100 psi
compressed air inside it.



you know the mass of air.
you know the force , its 100 psi. (believe it or not)

all you need now is the acceleration.

a=f/m
acceleration = force / mass


SP ) standard Pressure dry air has a density of 1.2754 kg/m3.

this will give you the mass and the initial velocity for your formula.


believe me air is light so there wont really be a time to wait as the mass accelerates so we will leave out time for now.


now you have the rest of your formula..

Quote:
And once again I've provided yet another layer of mathematics showing the impossibility of your perpetual-motion engine.


No , you havent provided anything.

but you now have the full formula that you obviously already knew , you just didnt want to allow others to see the outcome of actual numbers , you only want them to believe your twisted math and the endless jargon that flows from your knowlege of what other people said.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/03/10 06:48 AM

Wow this is a new low!!

It's called conservation of momentum. What's so unbelievable about that?

---------------
"The momentum of an isolated system cannot change, ever, for any reason, no matter what happens internally"
---------------

If you're sure it's wrong, then, as ImagingGeek suggested, ACTUALLY BUILD IT AND CLAIM YOUR NOBEL PRIZE!!

As with perpetual motion, this will make you richer than God.

But the advantage here is it's much easier and cheaper. Just get an old PVC pipe, put a pressurized coke bottle inside, rig up a trigger to release the bottle's nozzle, cap the ends, pull the trigger, sit back and wait for the endless stream of money to roll in.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/03/10 06:56 AM

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/03/10 04:00 PM

Now how am I going to prove that paul is wrong ,the only
thing I have is words.
but the words in physics are supposed to be backed up by the math in physics.

AND THE MATH BACKS UP THE WORDS !!!

WHY CANT I USE MATH ?

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/03/10 04:37 PM

reactionless propulsion is even being experimented with today.

heres a good example , the box can be considered a closed system
there is no external forces that apply agaist the box , but the
box moves.

click here to watch video
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/04/10 12:03 AM

Originally Posted By: paul

Quote:
The force generated upon the release of this air is:
F=[m*Ve]+[(Pt-Pa)*Ae]

Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is:
F = (Pt-Pa)*Ae
F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N


one load after another...

theres something VERY important when calculating force

--------- YOU NEED MASS ---------


Try looking up the unites within a pascal, moron. A pascal is a N/m^2. A newton is a kg*m/s^2.

kg, BTW, is the mass...

Funny, you're the one who insists on using pressure as a proxy for force, and are not even aware of what the units of pressure represent...LOL

Originally Posted By: paul

F=MA

not

F=whatever you want it to be.


And, if you plug the units for pressure and area into a formula, you get...exactly what F=mA gives you...

Originally Posted By: paul

it looks like your using the initial pressure
in the tank (pt)then subtracting the initial pressure
in the pipe (pa?)then multiplying that number by the nozzle area (Ae?).

so where are you going with this ?


That's how you calculate the force produced by a fluid moving from a region of higher pressure, to lower pressure.

I though that would be self-evident...

Originally Posted By: paul
Quote:
Because we don't have a cone, m*Ve is zero


so since the mass comes out of a tube as you said , the mass has no velocity?


Nope. The air that comes out of the tank is still at pressure when it leaves the tube connecting the pipe to the outside world. If you have a cone-shaped nozzle you can harness that pressure, and use the pressure to accelerate the gas within the nozzle. As the air expands through the nozzle accelerates it, and the degree of acceleration determines the force generated. Ve is measured relative to the speed of the air before it enters the nozzle, and if that air doesn't accelerate, Ve = 0. Any mass * 0 = 0.

Originally Posted By: paul
it just covertly sneeks through the nozzle or perhaps it is being transported as they do on star trek !!


Nope, it obeys the well understood laws of fluid dynamics - the exact same laws used for designing everything from aromatizers to rockets.

Originally Posted By: paul
you know the mass of air.
you know the force , its 100 psi. (believe it or not)


My god, you are dumb. Pressure is not a measure of the amount of force when the air is released - it is a measure of the force exerted on the tank by the air. The two are not the same - a 1000L tank at 100PSI will generate far more force than a 1L tank at 100PSI.

Strange, isn't it, how these simple concepts elude you!

Originally Posted By: paul
all you need now is the acceleration.

a=f/m
acceleration = force / mass


SP ) standard Pressure dry air has a density of 1.2754 kg/m3.

this will give you the mass and the initial velocity for your formula.


What the hell do you think F=[Pt-Pa]Ae calculates? EXACTLY WHAT YOU JUST WROTE!

What you are missing is the area - pressure is FORCE PER UNIT AREA NOT FORCE.

Originally Posted By: paul

now you have the rest of your formula..


No, you've just shown that you cannot read mathematical formulas, and that you don't understand the difference between pressure and force. Pressure is not force, and more than kinetic energy is momentum.

Originally Posted By: paul

but you now have the full formula that you obviously already knew , you just didnt want to allow others to see the outcome of actual numbers , you only want them to believe your twisted math and the endless jargon that flows from your knowlege of what other people said.


I presented the numbers, for every one to see. You, on the other hand, simply displayed your inability to read a mathematical formula.

LOL

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/04/10 12:17 AM

Originally Posted By: paul
reactionless propulsion is even being experimented with today.

heres a good example , the box can be considered a closed system
there is no external forces that apply agaist the box , but the
box moves.

click here to watch video


did you even watch the video - the tracks the box travels on are clearly attached to the switch.

Having the box on an electrified track is hardly a closed system. Nor, for that matter, is having it on the ground. Ground = friction.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/04/10 01:16 AM

Quote:
F=[m*Ve]+[(Pt-Pa)*Ae]


your formula will not work without the (((((( MASS )))))
of the air.

and your formula will not work without the velocity of
the ((((((( MASS )))))))
---------------------------------
F=[m*Ve]+[(Pt-Pa)*Ae]

Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is:
F = (Pt-Pa)*Ae
F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N
---------------------------------

suppose the mass = 1.29 and the velocity is
5000 m/s

1.29 * 5000 = 6450

we begin with a much larger number than zero ...

6450 + 6471 = 12921 N

twice the amount that you first posted.

however even your math below is incorrect.

F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N

the answer to the above is
6.471 not 6471

and your calling me a moron? LOL

and later you say that I cannot read a formula , when it
is you that seems to lack that ability , not in just
comprehending them but also in performing calculations with them.

they dont just put elements in formulas because they dont have anything else to do.

they put them there because they are a critical part of the formula , otherwise they would have left them out in the first place.


without mass you cannot have a force.
and by assumming that the air has both no mass or velocity
will never render a close solution to the formula.

and how does the formula know how heavy air is?

unless you tell it....

theres no sence in calling people names , it just shows your inability to deliver a inteligent responce therefore you revert to name calling.


Quote:
did you even watch the video - the tracks the box travels on are clearly attached to the switch.



yes I did watch the video , and if the wires are attached to the tracks it only shows that he didnt want people thinking that the box was being pulled by the wires.

it obviously has an electric motor in it , we cant transmit electricity through the air anymore like Tesla did , and maybe the electric motor is too big and power consumming to
have a battery in the box, after all he is just demostrating an reactionless effect.

meaning there is no direct force applied to the box that leaves the box as thrust or a direct push with a force.





Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/04/10 01:49 AM

Originally Posted By: paul
Quote:
F=[m*Ve]+[(Pt-Pa)*Ae]


your formula will not work without the (((((( MASS )))))
of the air.

and your formula will not work without the velocity of
the ((((((( MASS )))))))

and your calling me a moron? LOL

Yes, I am. And you just proved me right.

Mass is an intrinsic value needed for the calculation of pressure. You cannot calculate pressure without first knowing the amount of gas molecules present. pV=nRT, assuming an ideal gas.

Ergo, if you know the pressure, you can calculate force without knowing the mass of the gas itself.

And since it bears repeating:

Pressure is measured as force/area. In Si units that is newtons and meters-squared respectively.

A newton is a kg*m/s^2

You'll notice mass - kilograms - is an intrinsic part of force. And since force is an intrinsic part of pressure, mass is also an intrinsic part of pressure.

Originally Posted By: paul
without mass you cannot have a force.
and by assumming that the air has both no mass or velocity
will never render a close solution to the formula.

Since I assumed no such thing, your point is?

The formula is so simple, I'm amazed you still don't get it. NASA gets it, high schoolers making bottle rockets get it. Heck, even revlon (makers of lipstick, amoung other things) gets it.

Apparently, you don't.

When talking about a compressible fluid emerging from a pipe or other closed structure, the force it produces is equal to the the force generated due to the pressure difference pushing the fluid out of the pipe, PLUS the energy harnessed from the expansion of the gas (if you have a nozzle to take advantage of that expansion).

F=[m*Ve]+[Pt-Pa]Ae

[m*Ve] calculates the energy derived from the nozzle, and is determined by how effectively the nozzle uses the pressure to accelerate the gas. No nozzle and this value is zero.

[Pt-Pa]Ae calculates the force generated due to the pressure difference, taking into account the size of the opening between the high-pressure and low-pressure regions. No mass is necessary, since mass is already accounted for by pressure.

The above IS rocket science, but its still pretty simple science.

Originally Posted By: paul
theres no sence in calling people names , it just shows your inability to deliver a inteligent responce therefore you revert to name calling.

LOL, pot meet kettle...

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/04/10 02:17 AM

F=[m*Ve]+[(Pt-Pa)*Ae]

do you see the (+) sign in the above formula?

lets do this one

number of oranges = 0 + 0

lets put some numbers in there

number of oranges = 8 + 8 = 16

then lets do it your way.

number of oranges = 0 + 8 = 8

all you are calculating is pressure times the nozzle area.

what if the fluid is methane do you think that you will get the same result in you formula if worked the way you say it can be worked?

methane
.717 kg/m^3
liquid methane
415 kg/m3

using 5000 m/s
methane
.717 * 5000 = 3585
liquid methane
415 * 5000 = 2075000

you cant leave this
Quote:
[m*Ve]+

out of this
Quote:
F=[m*Ve]+[(Pt-Pa)*Ae]

or all you have is this
Quote:
F=(Pt-Pa)*Ae


suppose each of the above also have 1000 psi pressure

we get your exact result for each of the above fluids no matter what fluid we use.

and where does the fluid velocity come from?

if you dont include it.

this is rocket science

F = q × Ve + (Pe - Pa) × Ae

where F = Thrust
q = Propellant mass flow rate
Ve = Velocity of exhaust gases
Pe = Pressure at nozzle exit
Pa = Ambient pressure
Ae = Area of nozzle exit

not what you put up.

The product of q*Ve is called the momentum, or velocity, thrust and the product (Pe-Pa)Ae is called the pressure thrust.

you are only including the pressure not the momentum of the mass , because you didnt include mass.

q = the amount of mass that passes through the nozzle.
Ve= the velocity of the mass that exits the nozzle.

you left out the most important part of the formula.








Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/04/10 04:27 AM

OK, you want maths?

Referring to my previous picture

p1 = momentum of plane
p2 = momentum of air pushed back by plane's propeller
p3 = momentum of pipe

total momentum of the whole system ptotal = p1 + p2 + p3

Initially everything's stationary and the plane's in the middle:
p1=0
p2=0
p3=0
ptotal=0+0+0 = 0

Then the plane starts flying:
p1=10
p2=-10
p3=0
ptotal=10-10+0 = 0

Then the air from the propeller hits the pipe
p1=10
p2=0
p3=-10 (notice that the pipe is moving!!!)
ptotal = 10+0-10 = 0

Then the plane hits the end of the pipe
p1=0
p2=0
p3=-10 + 10 = 0 (now the pipe stopped moving)
ptotal = 0+0+0 = 0


TOTAL MOMENTUM REMAINS ZERO AT ALL TIMES!!! You might be confused by the small movement of the pipe, and think that's propulsion, but turn down think street and work out why it's no use for a spaceship.

Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/04/10 12:45 PM

Originally Posted By: paul
F=[m*Ve]+[(Pt-Pa)*Ae]

do you see the (+) sign in the above formula?

Yes, and had you read what I wrote above you'd understand why its there - the force generated as a compressed gas moves from a region of higher pressure to lower pressure is equal to the sum of the force generated by the expansion of the pressurized gas against a nozzle (if there is one) PLUS the force exerted due to the pressure differential.

Strangely enough, when describing sums mathematically, we use '+' signs.

I know, its shocking - using the appropriate mathematical symbol to represent its concordant mathematical procedure.

Originally Posted By: paul
all you are calculating is pressure times the nozzle area.

Which described the force produced given the pressure differential and the area connecting the high-pressure region to the low-pressure one.

If you think that is wrong, certainly you can provide the mathematical evidence it is wrong.

Originally Posted By: paul
what if the fluid is methane do you think that you will get the same result in you formula if worked the way you say it can be worked?

Assuming it is under the same pressure and doesn't go through a phase transition - yes. The force generated due to a pressure differential is always the same for a given pressure difference and cross-sectional area of the opening.

Originally Posted By: paul

you cant leave this
Quote:
[m*Ve]+

out of this
Quote:
F=[m*Ve]+[(Pt-Pa)*Ae]

or all you have is this
Quote:
F=(Pt-Pa)*Ae

But I can leave it out m*Ve specifically deals with the force generated as the gas expands and pushes against the nozzle. Since we don't have a conical nozzle to harness the energy, Ve is zero. m*Ve will always be zero if Ve itself is zero.

Basic math, the fact you are arguing about it says wonders about your understanding of what is quite simple math.

Originally Posted By: paul
suppose each of the above also have 1000 psi pressure
we get your exact result for each of the above fluids no matter what fluid we use.


Exactly - although the amount of gas required to achieve 1000psi will vary gas-to-gas.

Originally Posted By: paul
this is rocket science
F = q × Ve + (Pe - Pa) × Ae

where F = Thrust
q = Propellant mass flow rate
Ve = Velocity of exhaust gases
Pe = Pressure at nozzle exit
Pa = Ambient pressure
Ae = Area of nozzle exit

not what you put up.

LOL, and once again paul shows his total ignorance of math. The formula you just wrote is THE EXACT SAME ONE I HAVE BEEN WRITING ALL ALONG. Substituting one symbol for another - in your case, q for m - doesn't magically make it another formula.

If anything, your good for a morning laugh.

Oh, and by the way F is force, as in newtons (N). You need to integrate to get thrust (which is in newton-seconds [N*s]).

Originally Posted By: paul
The product of q*Ve is called the momentum, or velocity, thrust and the product (Pe-Pa)Ae is called the pressure thrust.
you are only including the pressure not the momentum of the mass , because you didn't include mass.


Nope. q*Ve (m*Ve in the way I wrote it) is the force produced by the nozzle, due to the expansion (and thus acceleration) of the gas by the conical shape of the nozzle. The actual pressure-driven force is accounted for by the [Pt-Pa]*Ae.
Ve is calculated using:


The only real elements of that you need to understand is the ratio: 1-[Pe/Po]

[Pe-Po] is the pressure difference at the entrance verses exit of the nozzle; essentially you are calculating the force generated by the change in pressure as the air expands in the nozzle. In our case there is no expansion, as the nozzle is just a hole. Thus, Pe = Po, ergo:

1-[Pe/Po] = 1-[1] = 0

Since the remainder of the formula is multiplied by that value, the Ve will also end up being zero.

Originally Posted By: paul
you left out the most important part of the formula.

Nope, I understand the formula and thus calculated it correctly. Since Ve is zero, m*Ve (or q*Ve in your case, although its the same bloody thing) is also zero, ergo that part of the math can be ignored.

Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter.

LOL, stupid is funny!

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/04/10 02:36 PM

I used to think that you were just refusing to admit
that the pipe would move , by not putting mass and velocity
into the formula.

now I just think your a complete bone head .... LOL

a force caused by a object that has a mass the same as a car
with a velocity of 5000 m/s

would cause a greater force than a object that has the mass of
your brain , slightly greater if the brain were someone elses that had something useable in it.

you really are a college "gadget".

the only reason that the pipe could move is the momentum
of the mass moving inside the pipe.

ie..
in clasical mechanics momentum
newtons second law
F = (d/dt)*(m*v)

using your boneheaded method by removing the (m*v) in the
above formula will result in
force = distance / distance*time

YOUR WAY PRODUCES NO FORCE , BECAUSE YOU USE NO MASS

if it moves slowly , unconstricted then not much force is produced.
if it moves faster by constriction , then much more force is produced.

Im right and your wrong , your just too boneheaded to admit it.

Quote:
Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter.

LOL, stupid is funny!



no I didnt I included the correct formula ,
and I included all the elements of the formula, if you gadgets
would have learned this stuff or at least that you cant remove
portions of formulas to get a correct answer we wouldnt have
so many buidings and bridges collapsing.

thankfully your just a programmer that only deals in
imagery , so you cant really make things fail.

I did correct your math several times though.
and you know it.


Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/04/10 02:59 PM

Originally Posted By: paul
the only reason that the pipe could move is the momentum of the mass moving inside the pipe.


Exactly - and [Pt-Pa]Ae calculate the portion of that derived from the differential pressure.

Originally Posted By: paul
if it moves slowly , unconstricted then not much force is produced.
if it moves faster by constriction , then much more force is produced.

Im right and your wrong , your just too boneheaded to admit it.


No, you're just too stupid to understand what that formula is calculating. There are two sources of force that move the pipe - the pressure differential and the expansion of the gas once it leaves the pipe (captured by the nozzle).

To calculate the former you need only the differential pressure and cross-sectional area connecting the two regions.

To calculate the later you need the mass and the acceleration provided by the nozzle.

Originally Posted By: paul
Quote:
Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter.

LOL, stupid is funny!



no I didnt I included the correct formula


In that case you can certainly show where your formula varieties from mine.

Of course, we both know you won't, as your formula is the exact same one I have been using since day 1. Swapping symbols doesn't change the formula itself.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/04/10 03:07 PM

I already have shown you , bonehead.

the q*ve = a sum
then you add that sum to the
(pe-pa)*Ae

LOL

F = q × Ve + (Pe - Pa) × Ae

you are leaving out one of the sums.

Quote:
There are two sources of force that move the pipe


true , ..... it just so happens that you are leaving
one of them out !!!

so when you add them together your answer is incorrect
because one of them is zero.

Quote:
the pressure differential and the expansion of the gas once it leaves the pipe (captured by the nozzle).


Wrong ... the pressure differencial and the expansion are
the (Pe-Pa)*Ae side of the formula...

the two forces are

the mass flow rate * velocity (q*Ve)
+
the expanding gasses (Pe-Pa)*Ae

F = (q*Ve)+(Pe-Pa)*Ae

"you're just too stupid to understand "

you have already stated these previously.

Originally Posted By: ImagingGeek
Momentum, BTW, is a specific physical quantity equal to mass * velocity.


so why did you leave out Momentum , mass * velocity

Originally Posted By: imagingGeek
Sorry, you don't. Missing half the equation does not equal a valid answer.


and why did you leave out half the equation.


perhaps to try and twist around a result?
one that would prove you wrong...

Quote:
There are two sources of force that move the pipe

Quote:
There are two sources of force that move the pipe

Quote:
There are two sources of force that move the pipe

Quote:
There are two sources of force that move the pipe

Quote:
There are two sources of force that move the pipe



Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/04/10 06:15 PM

Originally Posted By: paul
I already have shown you , bonehead.

the q*ve = a sum
then you add that sum to the
(pe-pa)*Ae

LOL


And that is different from my formula how? The plus is in there - are you blind, or just stupid?

Originally Posted By: paul
F = q × Ve + (Pe - Pa) × Ae

you are leaving out one of the sums.


No, I am not. m*Ve is zero when you do not have a nozzle which allows for expansion. Ergo m*Ve will always be zero in the case of your perpetual motion machine, as you specified the air was coming out of a tube - as in a non-divergent nozzle.

0 + anything is the same as anything by itself...

Originally Posted By: paul
Quote:
There are two sources of force that move the pipe


true , ..... it just so happens that you are leaving
one of them out !!!

so when you add them together your answer is incorrect
because one of them is zero.


I did not leave one of them out - I provided you with the formula for Ve, and in the case you have described it's value will be zero. Don't take my word for it - do the math for yourself.

Without a divergence cone Po = Pe. No matter how much you try and deny it, having a Po and Pe which are the same means Ve is zero. Basic math.

Originally Posted By: paul
Quote:
the pressure differential and the expansion of the gas once it leaves the pipe (captured by the nozzle).


Wrong ... the pressure differencial and the expansion are
the (Pe-Pa)*Ae side of the formula...


Nope. (Pe-Pa)*Ae refers to the pressure differential between the pressurized tank and the outside environment. Ve is determined by the pressure differential in the stream of gas as it expands once it leaves the tank and enters the nozzle.

A primer on these calcs:
http://www.nakka-rocketry.net/th_thrst.html

More specifically:
"If the pressure ratio (and thus expansion ratio) is 1, then F = 0. The only thrust produced by such a nozzle is the pressure thrust, or Ftotal = (Pe-Pa)Ae. Such a nozzle, of course, would have no divergent portion, since A*/Ae=1, and would be a badly designed rocket nozzle!"

Originally Posted By: paul
Originally Posted By: ImagingGeek
Momentum, BTW, is a specific physical quantity equal to mass * velocity.


so why did you leave out Momentum , mass * velocity


Because that's not what m*Ve is. m (q in your case) is the mass flow rate - as in kg/s, not the mass (as in kg alone). Hence, that portion of the formula calculates the force - not momentum - produced.

That should be self-evident to an "expert" such as yourself - you cannot summate a momentum and force, as the units (and what they describe) are different. The fact you seem to think you can summate different units of measure doesn't exactly help your credibility - not that you have any left that is.

Momentum = mass * velocity = kg*m/s
Force (in simple terms) = mass * acceleration = kg*m/s^2
Force (in m*Ve) = kg/sec * m/sec = kg*m/s^2


Bryan

EDIT: I'm off to the cottage for three days. Maybe use that time to read through Richards pages and learn a little about how rockets generate thrust. Then maybe we can have a conservation that extends beyond me trying to explain simple formulas to you.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/04/10 06:45 PM

Quote:
No, I am not. m*Ve is zero when you do not have a nozzle which allows for expansion.


if its ZERO then NOTHING will come out.

even if there was nothing but a hole in the tank , there
would still be air comming out = m*Ve through the hole , but that might hinder your result.

bonehead.

I dont think theres anything further to discuss as you dont
seem capable of performing math calculations.

but have a nice trip.



Posted by: Amaranth Rose II

Re: Orion, Mission to Alpha Centauri - 06/04/10 06:48 PM

Let's knock off with the insults, please.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/04/10 11:03 PM

Hey, not that I know what that formula's all about but have you two both totally overlooked this really simple miscommunication?

- Paul says Ve is velocity
- Imaginggeek says Ve is zero

Just saying it back and forth to each other doesn't change the fact that you're obviously talking about different quantities!!!!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/06/10 12:21 AM

Originally Posted By: ImagingGeek
F = (10132kPa-101kPa)*0.00064516m^2

1 kPa = .1450377 psi
10132kPa = 1472.959764 psi

we start with 100 psi... not 1472 psi

100 psi / .1450377 = 689.47 kPa
14.7 psi / .1450377 = 101

F = (689.47kPa-101kPa)*0.00064516m^2
F = 588.47kPa * 0.00064516m^2
F = 0.3796573052 N

not your result of

Quote:
F = 6471N


But the above is only the pressure thrust.

you still need to add the momentum or velocity thrust.

the part you left out (m*Ve) or (q*Ve)


F = (m × Ve) + (Pe - Pa) × Ae
air has a mass of 0.0807 lb / cu ft
0.0807 / 1728 = 0.000046701388 lbs / cu in
the area is 1 sq inch
so air at 6.828 atm has a mass of
6.828 X 0.000046701388 = 0.000318877077264 pound mass
0.000318877077264 pounds = 0.000144640209 kilograms
m = 0.000144640209 kilograms
100 pounds = 45.359237 kilograms
a = F/m
45.359237 kg / 0.000144640209 kg = 313600.466382069 m/s^2

F=(0.000144640209 * 313600.466382069 m/s^2)+ (Pe - Pa) × Ae
F=(45.359)+ (Pe - Pa) × Ae
F=(45.359)+ (689.47kPa-101kPa)*0.00064516m^2
F=(45.359)+ 588.47kPa * 0.00064516m^2
F= 45.359 + 0.3796573052
F = 45.738894305199934012421 N

45 kilograms = 99.208018 pounds

the momentum of the air leaving the tank through the nozzle has transfered to the pipe.


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/06/10 03:16 AM

Originally Posted By: paul

you still need to add the momentum or velocity thrust.

the part you left out (m*Ve) or (q*Ve)


Havn't you noticed that ImagingGeek is saying Ve=0 ? Obviously he's not talking about velocity! Did he ever say Ve means velocity? Maybe he's got the wrong formula, why don't you bother to understand what he's thinking instead of doing trivially obvious calculations?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/06/10 03:23 AM

Quote:
Havn't you noticed that ImagingGeek is saying Ve=0 ?


yes I have.
and your point?

NASA Rocket Thrust Equasions



if you will notice in the above formula

the mass flow rate and velocity = the part he left out.

the part that supplies the most thrust.

he insisted that he could leave it out.

that it wouldnt make a difference.

he was wrong.

Quote:
why don't you bother to understand what he's thinking instead of doing trivially obvious calculations?


because what hes thinking is wrong , thats why.

you can believe him if you choose , but I wont because
I think.

and from what I have seen , I think his thinking is wrong.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/06/10 06:33 AM

Yea certainly looks non-zero to me.

He gave another equation for Ve in message 34741, and allowed Po=Pe, but I'm not sure you can do that. For any nozzle shape there'll be more pressure on the inside than the outside - and thus some gas expansion. Without a pressure difference there wouldn't be any flow.

But we'll see better when he returns.


Anyway, how does this show that momentum isn't conserved? Or how does it show that you can get sustained propulsion of a closed cylinder?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/06/10 12:54 PM

yea , he put up a image of a formula to calculate Ve.

I didnt show that momentum isnt conserved , in his way of thinking momentum isnt conserved.

he seems to think that the pressures inside the pipe will
cause the pipe to not move and he is right , but he is claiming that no force will be felt.

he claims that because a force is comming out of the nozzle
that force will be canceled out by the other end of the pipe that is 500 ft away and that is why the pipe wouldnt move.

I say that none of the pressures inside the pipe will cause
the pipe to move , only the momentum of the mass inside
will cause the pipe to move.

the important part here is the mass inside the pipe that is
moving , and the momentum that it transfers to the pipe.

like attaching two gas cylinders together between a valve
and 1 is full of 1500 psi gas and the other is empty.

if you place them in a float , then in a swimming pool
then open the valve the mass inside will move until the
pressure equalizes in the two cylinders , they will transfer momentum to the float and the float will move.


he is so poisoned by things he has heard that he cannot think.

suppose there is a large spinning space station that is supplying
1g to allow people inside the space station to walk around.

and suppose there is a ladder that would allow people to climb up
to the center of rotation.

as people climb up the ladder more and more acceleration is given
to the space station that increases its angular velocity.

the peoples angular velocity decreases because their radius
from the center decreases , therefore they transfer their
angular velocity to the space station.

the result is increased artificial gravity.

those still at the peripery will feel a higher gravity.
the space station has not been pushed from the outside
but its spinning faster.

this is reactionless propulsion.

and the more people that climb up to the center , the faster the space station rotates.

internal momentum changes affect the momentum of the overall system.

I think this is great because this means that we dont
have to pollute the rest of our solar system and beyond by
using gasses that burn and pollute.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/07/10 05:12 AM

Yea you're mostly right. You can cause movement of a closed pipe, or change the speed of a spinning space station by moving things around inside. This is exactly what physicists already know.

BUT! It isn't sustainable, so you can't actaully travel and useful distance. In the spinning space station example, Those people who climbed to the center are now in the center! If they ever go back to the edge the thing will slow down again.

For linear motion, after you've transferred some momentum internally and made the outer pipe move, the mass you moved has to stop, and when it does, it cancels out any momentum it gave the pipe to begin with. So the pipe moves a short distance then stops. You can only repeat the process as long as you have mass at one end of the pipe. If you try to recirculate it in any way, you'll only move the pipe back to where it started from.

This is all clearly obvious by considering that the _total_ momentum of the closed system remains constant. You're just transferring it between the pipe and the internal moving parts.

Science education can't "poision" people because it allows them to think more rigorously, so they have the skills to challenge anything authorities try to tell them. In fact trusting an authority is considered to be non-scientific.

Science isn't just a random collection of facts, it's mostly a tightly interwoven system of many compatible ideas. If one 'fact' was inconsistent with the others, then it'd be easily detected. That's not something that people can fake, especially not when the whole world has access to experiments they can do themselves to discover the truth.

Although I do see your point that when you're immersed in one common set of knowledge it's hard to come up with new, different ideas. Most good new ideas come from young people who don't have such a long lifetime of the same old thing. But equally when you have hardly any knowledge, it's easy to think of millions of ideas that are wrong.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/07/10 02:58 PM

Quote:
For linear motion, after you've transferred some momentum internally and made the outer pipe move, the mass you moved has to stop, and when it does, it cancels out any momentum it gave the pipe to begin



get real now kallog.

it may take several minutes to equalize the pressures in the tank and pipe.

durring this process momentum is being transfered to the pipe giving it
linear acceleration and momentum thus the pipe moves in a linear direction
for a linear distance.

and continues to move in that direction.

any resistive forces that could apply are not stored up inside the pipe then
released when equalization occurs.

so the pipe is still moving , it has momentum , but is no longer accelerating
nor decelerating.


now you also can not decelerate the pipe by re-pumping the
air back into the tank.

because you would have equal and opposite reactions
occuring between the pump and the air tank.

but once youve compressed the tank again you can stop
the pipe by releasing the air in the opposite direction.

Quote:
Although I do see your point that when you're immersed in one common set of knowledge it's hard to come up with new, different ideas. Most good new ideas come from young people who don't have such a long lifetime of the same old thing.


when you are trained to fail its hard to succede.

because you know you cant , youve been taught that
you cant , therefore no one else can either.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/08/10 03:54 AM

Originally Posted By: paul

and continues to move in that direction.


It doesn't continue to move, it stops. The pipe had momentum while it was moving, but it soon transfers that to the internal gasses travelling in the opposite direction, stopping them both.

Please read the post I made earlier listing the states of the system. Tell me where you disagree.

Just saying "it'll keep moving because I was taught in school that moving things keep moving" doesn't work, you were taught wrong, sorry.



Quote:

when you are trained to fail its hard to succede.

When you know nothing, it's hard to succeed either. Notice that the world it full to bursting with people having no science education, but they're not inventing perpetual motion or propellent-free rockets. I wonder why. Some of them are inventing mechanical gadgets, things that just require ingenuity, intelligence, effort and lucky good ideas. But they're not tearing down these solidly established theories.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 04:40 AM

Quote:
The pipe had momentum while it was moving, but it soon transfers that to the internal gasses


the pipe is moving , you say it had momentum then the momentum of the pipe !!!!! transfers to the gasses.

How is the pipe going to transfer momentum to the gasses?

that will be a good one.

the only time that the air comming out of the tank can provide any resistance
to pipe movement is WHILE THE AIR IS COMMING OUT of the tank.

and WHILE THE AIR IS COMMING OUT OF THE TANK , THE PIPE IS MOVING.

and it continues to move.






Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/08/10 08:22 AM

Originally Posted By: paul

How is the pipe going to transfer momentum to the gasses?


When they hit the pipe or indirectly via collisions with other gas molecules which then hit the pipe.

Do you honestly not understand this? It's just a very simple application of the law of conservation of momentum.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 12:32 PM

I myself have a very clear understanding of what happens.

I was questioning your understanding.

you state that the momentum moves the pipe , and you obviously are talking about the momentum of the air in the pipe , and then you state that at the same time the air
is stopping the pipe.

if the air is causing movement or supplying a force for movement then it must be supplying more force than the
resistance for movement that the air is also supplying
according to your assumption.

you cant have it both ways.

the air mass inside the pipe has moved from inside the tank
and distributed to the rest of the pipe.

some of the mass moved over 400 ft.
some moved over 300 ft -->
some moved over 200 ft -->
some moved over 100 ft -->

some small amount moved <--- between the pipe and the tank.

I cant understand your reasoning that the pipe will stop?

if you dont mind could you explain at which point the pipe
looses all momentum and stops?

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/08/10 02:02 PM

Originally Posted By: paul

you state that the momentum moves the pipe , and you obviously are talking about the momentum of the air in the pipe , and then you state that at the same time the air
is stopping the pipe.

Yes, the air both starts it and stops it, not at the same time of course because each air molecule takes some time to slow down.

Quote:

I cant understand your reasoning that the pipe will stop?

if you dont mind could you explain at which point the pipe
looses all momentum and stops?


I'll simplify it to be a gun instead of a rocket, and it's fixed to the front of the pipe, pointing backwards.

You fire the gun, and it gives momentum -p1 to the bullet. At the same time it gives momentum p1 to the gun and pipe. Exactly the same as recoil in a normal gun.

Clearly the pipe's now moving because it has momentum p1.

Eventually the back end of the pipe and the bullet collide, and the bullet becomes embedded in the back of the pipe. They were travelling in opposite directions, with zero total momentum (-p1 + p1 = 0). Now they're a single solid object with the same total momentum - 0.

The pipe has 0 momentum and has therefore stopped moving.


Please let me know if you think:

- Using a bullet analogy is a good enough idealization whether my explanation is correct or not.

- You agree with it but think it differs too much from the air jet case.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 03:59 PM

Originally Posted By: paul
Quote:
No, I am not. m*Ve is zero when you do not have a nozzle which allows for expansion.


if its ZERO then NOTHING will come out.


Wrong, as [Po-Pe](Ae) is non-zero. Ergo you'll have F = 0 + X, where X is equal to whatever [Po-Pe](Ae) equals.

Remember - the formula is the sum, not the multiple of those two terms. A zero on one side of the sum doesn't make the total a zero.

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 04:01 PM

Originally Posted By: kallog
Hey, not that I know what that formula's all about but have you two both totally overlooked this really simple miscommunication?

- Paul says Ve is velocity
- Imaginggeek says Ve is zero

Just saying it back and forth to each other doesn't change the fact that you're obviously talking about different quantities!!!!


Actually, Paul and I agree on this - Ve is a velocity. Where paul is wrong is in how that velocity is calculated - it is the velocity ADDED to the air coming out of the tank, due to the acceleration of the gas caused by a divergent nozzle.

Paul stated the air was coming out of a tube - i.e. there is no divergence. As such Ve (keeping in mind, Ve is the added velocity, not the absolute velocity) will be zero.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 04:59 PM

I have already solved the equation in message #34760
and it is clearly shown what I was telling you.
that BOTH the sums are included for the solution to the equation.

Quote:
F = (m × Ve) + (Pe - Pa) × Ae
air has a mass of 0.0807 lb / cu ft
0.0807 / 1728 = 0.000046701388 lbs / cu in
the area is 1 sq inch
so air at 6.828 atm has a mass of
6.828 X 0.000046701388 = 0.000318877077264 pound mass
0.000318877077264 pounds = 0.000144640209 kilograms
m = 0.000144640209 kilograms
100 pounds = 45.359237 kilograms
a = F/m
45.359237 kg / 0.000144640209 kg = 313600.466382069 m/s^2

F=(0.000144640209 * 313600.466382069 m/s^2)+ (Pe - Pa) × Ae
F=(45.359)+ (Pe - Pa) × Ae
F=(45.359)+ (689.47kPa-101kPa)*0.00064516m^2
F=(45.359)+ 588.47kPa * 0.00064516m^2
F= 45.359 + 0.3796573052
F = 45.738894305199934012421 N

45 kilograms = 99.208018 pounds


F = (m × Ve) + (Pe - Pa) × Ae

any one knowing any math at all would know that
the above equation is basicaly
1 + 1 = ?
which is the what the basic formula is.

you didnt even use the correct pressure you used apx 1400 psi to start with , knowing it was only 100 psi.

Quote:
Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is:
F = (Pt-Pa)*Ae
F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N

As the tank empties and the pipe fills, that force will drop. Thrust ends when Pt = Pa.


100 psi is NOT 10132 Pa !!!!!
1 psi = 6.895 kPa !!!! LOOK IT UP.

and on top of that you botched the results from those two numbers.

ie....F=(10132kPa - 101kPa)* 0.00064516m^2 = 6.47N
6.47N not 6471N

what I think you needed was good ole cheat sheet.

using your atrocity of an result by only using the
0.37N and not adding the 45.35N would of course work in YOUR FAVOR WOULDNT IT?





Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 05:10 PM

Originally Posted By: kallog
Yea certainly looks non-zero to me.
He gave another equation for Ve in message 34741, and allowed Po=Pe, but I'm not sure you can do that. For any nozzle shape there'll be more pressure on the inside than the outside - and thus some gas expansion. Without a pressure difference there wouldn't be any flow.

But we'll see better when he returns.

Yaaaaaayyy - I get to explain this to someone who'll actually try and understand, instead of trying to force the science to fit their preconceptions!

When you have a pressurized gas being released by a tank there are two potential sources of thrust - the pressure pushing the gas out of the tank, and the expansion of that compressed gas once it leaves the tank. The former will be applied onto the tank/rocket/whatever regardless of the nozzle configuration, the later requires a properly designed nozzle to be harnessed.

The former is calculated using [Pt-Pe]Ae, the later using m*Ve. The total force produced is the sum of those two values.

Pressure thrust:
The pressure thrust is determined by Fp = [Pt-Pe]Ae, where:
Fp = force produced by pressure
Pt = pressure in the tank/rocket engine/whatever
Pe = pressure of the environment outside of the tank
Ae = area of the connection between the tank and outside environment

The calculation is fairly obvious (IMO) - pressure is the measure of force per unit area, produced in this case by a gas. So you determine the pressure difference between the tank and the environment, multiply that by the area connecting them, and you'll get the force produced by the pressure differential acting through the opening between the tank and the environment.

Paul complains that this calculation does not take into account mass (and by extension, he should complain it doesn't take into account the exhaust velocity). However, those values are not needed to calculate the pressure force for the following reasons:

1) Pressure is dependent on the amount of gas present (i.e. on its mass, or more specifically, on the moles of gas present). Ergo, by knowing the pressure you already "know the mass", and that quantity is part-and-parcel of the resulting pressure.

2) The movement of air from the tank to the environment is choked - a fancy way of saying it has a maximum possible velocity (the speed of sound). And this is the velocity the gas travels at - with the exception of any slowing due to friction in the nozzle area. Once again, this velocity is part-and-parcel of pressure (its determined solely by gas density).

So despite Pauls complaints to the contrary, [Pt-Pe]Ae calculates the force provided by differential pressure under ideal conditions (i.e. no friction, ideal gas).

Nozzle thrust:
The second source of trust is the stream of gas once it leaves the tank. This stream of gas is under pressure, relative to the environment. This pressure exerts an expansive force, perpendicular to the direction of gas flow. A properly designed nozzle can capture this outwards force, thus "converting" it into additional thrust.

To do so you need a conical nozzle which will allow the gas to expand at a rate close to, but slightly less than, the rate it would expand unencumbered. This will allow the gas to be accelerated to the maximum speed (several times the speed of sound!), at which point the nozzle is impacted by that gas, thus imparting additional force to the nozzle, and thus the tank/rocket.

This value is determined by Fn = m*Ve, where:
Fn = force produced by nozzle
m = mass flow through the nozzle (i.e. kg/s)
Ve = the exit velocity of the gas from the nozzle, relative to the velocity of the gas when it enters the nozzle.

The formula I provided earlier allows for the calculation of Ve, given a known pressure change along the length of the nozzle (determined by the divergence angle), and the characteristics of the gas (pressure, temperature, etc).

In the situation provided by Paul - no divergence, simply a straight tube connecting the tank to the environment - Ve will be zero. This is because the gas is not allowed to expand within the nozzle, thus no additional velocity is given to the gas, and thus no additional thrust is provided. Simply put, in Pauls situation the additional thrust that could be provided is lost in the form of the gas expanding behind the nozzle.

EDIT: I would add at this point that, at least in the case of rocket engines, this expansive force is where the majority of the thrust comes from. In the case of the sugar-potassium nitrate rockets I used to build, 75-80% of the thrust was due to the nozzle, meaning the nozzle "amplified" thrust, on average, 4-5X. People with better machining skills than I often would get increases in thrust of 6X or more, relative to the pressure thrust alone.

The Math:
The total force produced by this tank of pressurized air is equal to the sum of the above two forces:

Ftotal = Fn + Fp = [m*Ve] + [Pt-Pe]Ae

In Pauls case Fn is zero, as there is no divergence of the nozzle and thus Ve is zero. Meanwhile, Fp is some non-zero value, thus:

Ftotal = Fn + Fp
Ftotal = 0 + Fp
Ftotal = Fp


Complications:
There are some complications ignored in the above formula. There is the issue of in-engine friction, which reduces thrust. This can be largely ignored assuming proper engineering, but the true force of such a system will always be slightly less than calculated due to friction. Likewise, those formulas assume an ideal gas, and thus any non-ideal gas (i.e. all of them) will behave a little different than predicted - your total thrust tends to be calc'd right, but the force produced at any one time, the time it takes to empty the tank, etc, will be different than those calculated above.

All that said, Paul accidentally gave us a situation where the biggest complication has been eliminated - notably he has an engine expelling one phase (a gas). Solid rockets, as well as many liquid/gas rockets, produce two-phase flow (i.e. you'll have solid or liquid particulates in the gas stream). In these cases it is not as straight forward as illustrated above, as the force provided by the gaseous fraction components will only produce force due to their acceleration as they leave the engine. This gets quite complicated mathematically speaking, as the particulates can be accelerated twice - first as they go from the combustion chamber into the linear part of the nozzle, and again when they are in the divergent part of the nozzle. That later acceleration is particularly difficult to deal with, as much of the acceleration is perpendicular to the gas stream (and thus does not provide thrust), while at the same time there is some acceleration along the gas stream AND you have particulates impacting the nozzle itself.

Originally Posted By: kallog
Anyway, how does this show that momentum isn't conserved? Or how does it show that you can get sustained propulsion of a closed cylinder?

Exactly - that is where this whole thread started. Momentum is conserved in a closed system. Paul has described a closed system. Ergo, Pauls engine cannot produce a net thrust on his closed system...

...Paul pretends otherwise, by pretending the gas escaping the tank doesn't have momentum of its own, but ignoring reality doesn't change reality.

I realize the above was ridiculously long, but I hope I described things in a way people can understand.

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 05:26 PM

Originally Posted By: paul

100 psi is NOT 10132 Pa !!!!!
1 psi = 6.895 kPa !!!! LOOK IT UP.


My bad, I used ATA in place of PSI for some reason.

Originally Posted By: paul
and on top of that you botched the results from those two numbers.

ie....F=(10132kPa - 101kPa)* 0.00064516m^2 = 6.47N
6.47N not 6471N


Nope, you botched up the calc. A kPa is kilo-Pascals, as in 1000 pascals. You have to multiple kPa by 1000 to get the Pa needed in the formula used above - ergo:

F = (10132kPa - 101kPa)* 0.00064516m^2
F = (10132000Pa - 101000Pa) * 0.00064516m^2
F = 6471N, AKA 6.47kN

Originally Posted By: paul

what I think you needed was good ole cheat sheet.

No, I need to check my work, although I have to say the exact values mean little here; it is the basic physical principals that are important.

Originally Posted By: paul
using your atrocity of an result by only using the 0.37N and not adding the 45.35N would of course work in YOUR FAVOR WOULDNT IT?

Nope, not one iota. Regardless of the magnitude of the force, the physics are the same. 0.37N or 2,000,000,000N, its all the same in the end - the pressurized tank will exert that force forwards on the pipe, while the air coming out the back of the tank will exert that same force rearwards on the tank. Net force is zero either way - whether you've got nano-newtons or giga-newtons (although the later may destroy the pipe completely...)

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 05:31 PM

Quote:
When you have a pressurized gas being released by a tank there are two potential sources of thrust - the pressure pushing the gas out of the tank, and the expansion of that compressed gas once it leaves the tank.


this is not a thrust that is the result of combustible
gasses , it is simply air that is compressed.

I would like to see your results from the following
formula.

F=m*V

and your explaination of it.

although it is a absolute meaningless part of the formula
according to you , do you honestly think that there will be no mass comming out of the tube therefore there will be zero velocity.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 05:45 PM

I didnt botch it up , I worked it exactly as you posted it.
meaning that I need to check all of your formulas plus your math.

F=(10132kPa - 101kPa)* 0.00064516m^2

10132000Pa is still 1469.52 psi...

again you botched it up.
even after you aknowledged the initial mistake.
you still get 6471N
and that is wrong.

it really doesnt matter what the amount of thrust is anyway.

it can be fast or slow but faster is better in this case.

as I have stated before the thrust is not what MOVES the pipe.

the momentum of the mass moving inside the pipe is what moves
the pipe.

do you deny that the air mass moving inside the pipe would
move the pipe?

Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:05 PM

Originally Posted By: paul
this is not a thrust that is the result of combustible gasses , it is simply air that is compressed.

But Paul, it is the same in either case.

Combustion in a confined space produces compressed gas, exactly as you would have in your tank. The physics is the same afterwards - that pressurized air will have potential energy which cam be extracted in the two ways described above.

There are only two ways in which the two systems differ:
1) The pressure dynamics over time are different for a rocket, as the pressure in the "tank" in continually replaced by combustion products, and
2) The two-phase flow issue, as described in my earlier post

Originally Posted By: paul

I would like to see your results from the following
formula.

F=m*V
and your explaination of it.


Already provided, in detail, in my first post today. But to clarify:

If by 'm' you mean simple mass, and V is velocity, than the above formula calculates the momentum of the air exiting your tank. In this case your F should be a P, as F is the symbol for force, not momentum.

If m = mass flow (kg/sec) and V equals the exhaust velocity (i.e. Ve), that it describes the force produced by an divergent nozzle, as described in my last post.

Exhaust velocity is not the same as velocity, but rather is the increase in velocity due to expansion in the nozzle. No expansion, no increase. No increase, Ve = 0.

In simplest terms, Ve = Vx-Vi, where
Ve = exhaust velocity,
Vx = velocity of the gas when it exits the nozzle
Vi = velocity of the gas when it enters the nozzle

For a non-divergent nozzle, as in what you described, there is no acceleration along the length of the nozzle, and ergo Vx = Vi. As such:

Ve = Vx-Vi; Vx=V1, therefore
Ve = Vx-Vx
Ve = 0

The formula I provided earlier for Ve allows for the calculation of Ve without having to actually measure these velocities directly. This is because these velocities will be equal to the speed of sound (Vi) and the speed induced by the pressure, over the length of the nozzle. In the post where I first presented the formula for Ve I went through the critical part of this calc - notably that Ve is a multiple of the fraction of pressure "consumed" by the divergent nozzle. This is strictly dependent on the difference in the pressure of the exhaust when it leaves the tank [Pt] verses when it leaves the nozzle [Pe], as the ratio between these two determines how much of the potential expansion energy in the compressed gas stream is being "captured".

In the case of your straight tube connecting the tank to the outside world there is no divergence. Thus there is no decrease in the pressure of the air stream within your nozzle. Ergo, Pe=Pt, and therefore the part of the formula posted earlier which is:

Fraction of pressure consumed = 1-[Pe/Pt]
= 1-[Pt/Pt]
= 1- [1]
= 0

Since the entirety of the Ve formula is multiplied by that value, Ve itself will be zero.

Originally Posted By: paul

although it is a absolute meaningless part of the formula
according to you , do you honestly think that there will be no mass comming out of the tube therefore there will be zero velocity.


I never said that, and it is clear from this comment you did not read the very post you quote-mined.

The force produced due to the pressure-driven movement of air is determined by [Pt-Pe]Ae. Because pressures are used, both the mass and gas velocity are intrinsic to the measurement, and as such you do not need to calculate them as separate entities.

What you are trying to do is force the m*Ve to your preconceptions, without calculating m*Ve properly. m*Ve is ***not*** the mass flowing through the nozzle times its speed - that doesn't even calculate a force, but rather a momentum. A momentum is not a force, and thus you cannot simply add the momentum you're calculating to the force calc'd by [Pt-Pe]*Ae

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:13 PM

Quote:
the pressurized tank will exert that force forwards on the pipe, while the air coming out the back of the tank will exert that same force rearwards on the tank.


so according to you , there would be no force exerted onto the sides of the pipe?

and dont forget those same exact forces are exerted to the front of the pipe as well , you cant just have it your way.

but the forces from the air as it leaves the tank pressing against the pipe is not what will cause the pipe to move.

it is the mass of air that is moving inside the pipe from one place in front to the rest of the inside of the pipe.

the tube only slows down the time that the mass moves.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:17 PM

Originally Posted By: paul
I didnt botch it up , I worked it exactly as you posted it.
meaning that I need to check all of your formulas plus your math.

F=(10132kPa - 101kPa)* 0.00064516m^2


You did not work it exactly as posted. Had you done that you'd have done the calcs using kilopascals, as indicaterd in what I wrote. You didn't, despite the fact kPa was typed out.

Ergo, my math is right, ignoring the PSI vs ATA foopah...

Originally Posted By: paul

it can be fast or slow but faster is better in this case.

In your case the speed of the gas will always be the same - the speed of sound.

Originally Posted By: paul
do you deny that the air mass moving inside the pipe would move the pipe?


It would. But that change in momentum would be countered by an equal, but opposite momentum due to the momentum of the tank the air was contained in. Thus total change in your pipes momentum would be zero. It may rock back and forth, but it isn't going to go anywhere.

Its that later half you always miss - anytime you give something momentum in a closed system, you get an equal but opposite momentum produced. It doesn't matter which of those you use as a reference point - in the end, the net change in momentum will always be zero.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:21 PM

if there was a piston sitting in the tube , then
your calculations state that the piston would not move.

or would the piston move.

would the pressurized air present a force to the piston?

I already know it would , but for some reason it seems
that you think it would just sit there even though it
only has 14.7 psi on the other end.

Im about to throw my arms up in the air and mark you off as a candidate to use as a tool to deny funding that should apply to work with the idiots at the D.O.E.




Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:26 PM

Originally Posted By: paul
Quote:
the pressurized tank will exert that force forwards on the pipe, while the air coming out the back of the tank will exert that same force rearwards on the tank.


so according to you , there would be no force exerted onto the sides of the pipe?


I never said that; if you're only counter-argument is to twist my words you've already lost.

The increasing pressurization of the pipe has no impact on the numbers what-so-ever, as this pressurization occurs equally in all directions.

However, the air expelled from the tank does have momentum in one direction - opposite to the direction of the momentum of the tank. Momentum is not pressure, and therefor the equalization of pressure in the pipe has nothing to do with where that momentum goes.

Originally Posted By: paul

and dont forget those same exact forces are exerted to the front of the pipe as well , you cant just have it your way.


Wrong, wrong and wrong again. I know repetition doesn't help in your case, but momentum is not pressure
Momentum Is Not Pressure
MOMENTUM IS NOT PRESSURE

Changes of pressure in the pipe have nothing to do with the momentum in this system. Why you'd bring it up is a mystery to me (actually, it isn't, since you've consistently mixed up the differences between force, pressure, momentum and energy).

Originally Posted By: paul
but the forces from the air as it leaves the tank pressing against the pipe is not what will cause the pipe to move.

it is the mass of air that is moving inside the pipe from one place in front to the rest of the inside of the pipe.


So you now think that the momentum of the air in your tank is divorced from the forces which produce that momentum?

Its an interesting world you live in, where things move without forces acting on them.

In the real world things don't work that way - force and momentum are intimately intertwined; you cannot create one without creating/changing the other. Application of forces change momentum, momentums transfer forces.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:30 PM

do you even know anything about momentum.

if the air mass moves from the tank , the air itself is not really moving , the pipe is what moves.

the pipe moves out of the way of the air.

because the pressures inside are equalizing.

the air mass moves inside the pipe because the air pressure is equalizing.




Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 06:45 PM

there really is nothing that your twisting of words and math can accomplish with me that is , I know Im right even
though you have repeatedly re-stated the same old stuff over and over , never once admitting that the mass of air comming out of the nozzle has velocity , simply because the nozzle isnt really a nozzle.

a nozzle only focuses the thrust.
in this case a nozzle is not the reason there is a thrust
the thrust comes from the m*Ve.

but your genius rejects that concept.
so why do they bother putting fuel in jet aircraft all they really need is a nozzle.

paul is just using a tube therefore no velocity , right!!!

if I have a rifle and I cut the barrel off just beyond the
bullet , according to your genius the bullet would just stay there when I fired the riffle.

because the bullet doesnt have a barrel , ohhhh poor poor bullet.

it cant go nowhere because ImageingGeeks math comprehension =ZERO.

and you call people stupid.

now I suppose you will do exactly the same with the momentum of air that moves inside the pipe.

there obviously will be no momentum in the moving air
right?

just like the magic tube.

only now its the magic moving air with no mometum.

just because you write a lot of words and use twisted math
doesnt mean youre saying anything.






Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 07:18 PM

Quote:
Momentum is not pressure, and therefor the equalization of pressure in the pipe has nothing to do with where that momentum goes.


think of all the little air molecule thingies as being
little rockets flying around inside the big old pipe thingie.

some of the rockets fly north , some fly south , some fly east , some fly west , some fly down, the others fly up.

the little rockets that fly up , dont have far to go.
but the little rockets that fly down have a much further trip to take.
most of the litle rockets begin their trip by flying down but they have a harder and harder time finding a place to stop because of all the other rockets getting in the way.

but a few of them get to go all the way down.

and because the little rockets all move inside the pipe the pipe moves , up.
------------------------------------------------
here we go again , the equalization of pressure inside the pipe

has everything to do with the momentum

the pressure is the air , the air mass is what moves , and
the momentum of the moving mass of air

is the momentum.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 07:58 PM

Originally Posted By: paul
there really is nothing that your twisting of words and math can accomplish with me that is , I know Im right

And yet are completely unable to provide one iota of proof. Strange, isn't it - you're right and yet you've been able to convince no one here...LOL

Originally Posted By: paul

even though you have repeatedly re-stated the same old stuff over and over , never once admitting that the mass of air comming out of the nozzle has velocity , simply because the nozzle isnt really a nozzle.


And now you're lying - I've repetitively stated the mass of air has velocity upon exiting the tank. If you re-read my last few posts I've even given the exact speed it will move at - the speed of sound.

Odd, that you have to lie about my claims, in order to make your "point".

Originally Posted By: paul
a nozzle only focuses the thrust.


Nope. And once again the irony - both the page I linked to, as well as the one you linked to, clearly state how nozzles help provide additional thrust. Apparently you don't even read your own sources.

Originally Posted By: paul

in this case a nozzle is not the reason there is a thrust
the thrust comes from the m*Ve.


You're wrong here, as has been pointed out time and time again. Ignoring reality doesn't make it go away.

how m*Ve is calculated is broadly published. You ignore those formula, that doesn't make you right.

Originally Posted By: paul
so why do they bother putting fuel in jet aircraft all they really need is a nozzle.


Wow, the ignorance grows:
1) Jet engines work on a completely different concept. Ergo, not comparable to simple rocket engines

2) The combustion produces pressure. A nozzle is useless without pressure.

Originally Posted By: paul
paul is just using a tube therefore no velocity , right!!!


No increase in exhaust velocity, therefore m*Ve is zero. There will be the velocity intrinsic to [Pt-Pe]Ae.

Originally Posted By: paul
if I have a rifle and I cut the barrel off just beyond the bullet , according to your genius the bullet would just stay there when I fired the riffle.


Nope, and if that is what you think than you completely missed the point. The fact you think guns and rockets work on the same principals is particularity telling...

So there ya'll have it - Pauls argument in a nut shell:
a) ignore basic physics,
b) re-write formulas to suit his purposes,
c) lie about the claims of others, and
d) come up with irrelevant examples

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 08:58 PM

Quote:
And yet are completely unable to provide one iota of proof. Strange, isn't it - you're right and yet you've been able to convince no one here...LOL


I see where kallog also agrees that m*VE cannot be ZERO.
and given that us three are the only participants in this
thread , your wrong.

also who have you convinced?


Quote:
And now you're lying - I've repetitively stated the mass of air has velocity upon exiting the tank. If you re-read my last few posts I've even given the exact speed it will move at - the speed of sound.


the speed of sound is a constant it is the speed at which SOUND travels in air , not the speed limit that air can travel at.

not once have you even tried to calculate or give the exact speed or velocity.

and if you did you just simply stated that it was ZERO.

I calculated the velocity or speed at 313600.466382069 m/s
which is 194.86 miles per second.

Quote:
Nope. And once again the irony - both the page I linked to, as well as the one you linked to, clearly state how
nozzles help provide additional thrust. Apparently you don't even read your own sources.


provide additional thrust , as in the .37N

but you want to neglect the 45N comming from m*Ve

Quote:
You're wrong here, as has been pointed out time and time again. Ignoring reality doesn't make it go away.

how m*Ve is calculated is broadly published. You ignore those formula, that doesn't make you right.


Im not the one that ignores formulas and rearanges them to suit.

Quote:
Wow, the ignorance grows:
1) Jet engines work on a completely different concept. Ergo, not comparable to simple rocket engines

2) The combustion produces pressure. A nozzle is useless without pressure.


so its ignorance that typed the above , you just dont believe
how much you discredited yourself.

you have constantly been saying exactly the opposite of the above.

the combustion produces pressure , in the tank the pressure is
already produced... LOL

so the nozzle is useless without the pressure
did it ever cross your mind? that m*Ve is the result
of the stored pressure.

Quote:

No increase in exhaust velocity, therefore m*Ve is zero. There will be the velocity intrinsic to [Pt-Pe]Ae.


there it is again , the brainwashing must be complete
earlier he attempts to admit that he has given the velocity of air comming out of the tube at the much slower speed , the speed of sound.

So there ya'll have it - Bryan's argument in a nut shell:
a) ignore basic physics,
b) re-write formulas to suit his purposes,
c) lie about the claims of others, and
d) come up with irrelevant examples

BTW , Bryan its a,b,c,d not a,b,b,c

you be the judge.

---------------------------
I have a great idea , can you find an article that tells
how the nozzle itself produces more thrust than the
pressure that is either stored as in the air tank or the pressure that is produced by a jet engine or rocket engine.

meanwhile heres a bottle rocket that just uses air pressure
I noticed how the air came out of the bottle even though it
shouldnt because it doesnt have a space ship type nozzle , just a straight tube.

it just comes out of the bottles neck.

not one shaped like this.


but it sure takes off fast .. ZOOOOOOOM

pressurized air bottle rocket

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 10:01 PM

I was going to wait a while longer but decided to go ahead and
let you know that the side of the equation you use is used
in determining the amount of thrust and duration of flight
in a combustion based rocket engine.

the picture above that I linked to gave it away on this page.

http://www.aerospaceweb.org/design/aerospike/nozzles.shtml

so in reality in my example that does not have a nozzle
your side of the equation is unecessary as it only allows
a area to focus the thrust in a direction and with adjustable
nozzles can be used to adjust thrust amounts.

ie.. less thrust when less thrust is needed.
more thrust when more thrust is needed.

and of course the m*Ve side of the equation is the same as
the momentum equation and is the only side of the equation that we should use , given we have no nozzle.

I was thinking you would eventually figure that out , but
this has dragged on too long and you never did.

F=m*v
or
P=m*v

http://en.wikipedia.org/wiki/Momentum

I was just waiting to see if you could / would ever comprehend it Bryan.




Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 11:08 PM

when Hydrogen is burned in a rocket engine.

the combustion produces a thrust.

then the recombination produces water.

you could burn a Hydrogen rocket engine inside a pipe
and still have all the water you started with.

given you have a big enought pipe.

and a centrifuge to capture and hold the water.
the thrust can spin the centrifuge , and the centrifuge can
seperate the water into H2 and O using membrane technology.

you now have a propulsion system for space travel.



Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 11:16 PM

Originally Posted By: paul
Quote:
And yet are completely unable to provide one iota of proof. Strange, isn't it - you're right and yet you've been able to convince no one here...LOL


I see where kallog also agrees that m*VE cannot be ZERO.


he said no such thing - he said it would appear to be non-zero and would wait for my return to explain why I was saying it was zero. And I did just that, in my first post today.

Originally Posted By: paul
and given that us three are the only participants in this
thread , your wrong.

also who have you convinced?


kallog agrees with me that momentum of a closed system is conserved, and thus your engine is impossible. the closest he's been to agreeing with you is on the m*Ve, where his exact statement was "But we'll see better when he returns."

kirbygillis stopped by to make fun of you, which I assume means (s)he disagrees with you

So all three who've bothered to post vis-a-vis your claims agree that basic physics makes your engine impossible. None were not willing to take your claims vis-a-vis at face value...

Originally Posted By: paul
the speed of sound is a constant it is the speed at which SOUND travels in air , not the speed limit that air can travel at.


Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature. Different mixtures of gasses will have different speeds of sound - for example, the speed of sound in air is highly dependent on the amount of water vapor present. its also dependent on temperature.

The speed of sound is also the maximum speed that a gas can flow at under its own pressure. The speed of sound is determined by a gasses elastic modulus. That same elastic modulus also defines the back pressure of a flowing gas, and thus the degree to which a gas can accelerate itself due to its own pressure.

In simple terms, as a gas accelerates itself it also creates backpressure - resistance to flow generated by the elastic qualities of the gas itself. Gas can only accelerate itself, by its own pressure, to a point where the backpressure equals the gas pressure. Ignoring friction, backpressure will equal gas pressure when the speed of that gas is equal to its speed of sound.

You can get a gas to flow faster than its speed of sound, if you provide it with a either a source of energy other than its own pressure, or with reduced back-pressure. In the case of a nozzle you get the latter; a lowering of back pressure due to expansion of the gas stream.

Originally Posted By: paul
not once have you even tried to calculate or give the exact speed or velocity.


Oh, but I have, several times:

Speed of gasses leaving the tube: speed of sound (assuming room temp air, no humidity, approx 343m/s)

Additional speed of gasses due to acceleration by nozzle (Ve): 0m/s

Originally Posted By: paul
and if you did you just simply stated that it was ZERO.


Another lie - the only thing I stated was zero was Ve. And I provided both the formula and math to show that was the case.

Originally Posted By: paul
I calculated the velocity or speed at 313600.466382069 m/s which is 194.86 miles per second.


Which is wrong - the gas is choked at ~343m/s. Even your own sources described that - maybe you should read your own links...

Originally Posted By: paul
Quote:
Nope. And once again the irony - both the page I linked to, as well as the one you linked to, clearly state how nozzles help provide additional thrust. Apparently you don't even read your own sources.


provide additional thrust , as in the .37N

but you want to neglect the 45N comming from m*Ve


I ignore nothing; you're the one ignoring how those numbers are calculated.

m*Ve produces ZERO newtons of thrust when there is no divergence in the nozzle, as outlined numerous times before. [Pt-Pe]Ae produces the same amount of thrust, regardless of the nozzle design.

Originally Posted By: paul

I have a great idea , can you find an article that tells
how the nozzle itself produces more thrust than the
pressure that is either stored as in the air tank or the pressure that is produced by a jet engine or rocket engine.


It was provided to you two or three pages ago:
http://www.nakka-rocketry.net/th_nozz.html
http://www.nakka-rocketry.net/th_thrst.html

That is written by a fellow named Richard Nakka, a proverbial god in the EX rocketry community for the better part of two decades. On his webpage you'll find everything from basic theory, to engine design, to propellant formulation, to ready-to-go engine designs. He has calculators, applets, and various examples to illustrate every aspect of rocket engines you can imagine.

The relevant quote, in regards to the speed of sound, is:
"The critical point where the flow is at sonic velocity (M=1 at A/A*=1) is seen to exist at the throat of the nozzle. This shows the importance of the nozzle having a diverging section -- without it, the flow could never be greater than sonic velocity!"


Note: underlining provided by me, rest is original.

As for the amount of thrust added by the nozzle:
"The Thrust Coefficient determines the amplification of thrust due to gas expansion in the nozzle as compared to the thrust that would be exerted if the chamber pressure acted over the throat area only . . . The slope of the curve is very steep initially, then begins to flatten out beyond Po/Pe = 5. This is significant, as it indicates that even a nozzle provided with a minimal expansion will be of significant benefit. With such a pressure ratio of 5, the resulting thrust is about 60% of maximum theoretical"[/u]


And here is an applet, and more theory, describing these various factors:
http://www.engapplets.vt.edu/fluids/CDnozzle/cdinfo.html


Originally Posted By: paul
meanwhile heres a bottle rocket that just uses air pressure...


And? the amount of force derived in the case of the bottle rocket filled only with air will described by [Pe-Po]Ae; so that doesn't exactly help your case any. Water complicates things, due to two-phase flow, which I described earlier.

And you may want to look at the source you got the rocket engine picture from:
http://www.aerospaceweb.org/design/aerospike/nozzles.shtml

Even in this page - as in YOUR OWN SOURCE they counter your claim:
"the ultimate purpose of the nozzle is to expand the gases as efficiently as possible so as to maximize the exit velocity (v exit)"

LOL, disproven by your own sources - must hurt.

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/08/10 11:27 PM

Originally Posted By: paul
I was going to wait a while longer but decided to go ahead and let you know that the side of the equation you use is used in determining the amount of thrust and duration of flight in a combustion based rocket engine.

Sorry, that dog doesn't hunt - the properties of gas flow doesn't magically change due to the source of pressure. It doesn't matter if the air is compressed by a fan, piston, chemically (i.e. combustion) or through magic - once compressed, the release of that gas will obey the exact same physical laws.

Originally Posted By: paul
so in reality in my example that does not have a nozzle your side of the equation is unecessary as it only allows a area to focus the thrust in a direction and with adjustable nozzles can be used to adjust thrust amounts.

Wrong again - nozzles harness the expansion of the compressed gas once they leave the tank, regardless of the source of that pressure. The fact you have compressed air, while rockets have fire to compress their air, doesn't change the fact that both will produce pressurized streams of air that can act against that nozzle - or in your example, will be wasted due to the lack of a nozzle.

Originally Posted By: paul
and of course the m*Ve side of the equation is the same as the momentum equation and is the only side of the equation that we should use , given we have no nozzle.

No, it is not, as described by both YOUR and MY links. m is mass flow, not mass. Ergo m*Ve calculates a FORCE NOT MOMENTUM.

Originally Posted By: paul
I was thinking you would eventually figure that out , but this has dragged on too long and you never did.

Figured out what - that you don't read your own sources, or that you don't understand the difference between a force and momentum?

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/08/10 11:53 PM

Quote:
he said it would appear to be non-zero


that doesnt sound like he is agreeing with you , in fact it
sounds alot more like he can read formulas better that you.

Quote:
kirbygillis stopped by to make fun of you, which I assume means (s)he disagrees with you

So all three who've bothered to post vis-a-vis your claims agree that basic physics makes your engine impossible. None were not willing to take your claims vis-a-vis at face value...


kirbygillis was correcting the thing I missed when I used
seconds istead of hours in light speed.

ok , I made 1 mistake , but I corrected it in the next post below it , so anyway that makes it about
50 to 1 in my favor.

Quote:
kallog agrees with me that momentum of a closed system is conserved


I agree that momentum is conserved also , but its not like Im agreeing with you , as that
is something that makes sence.

yes the momentum is conserved in the moving pipe itself.
nothing left the pipe.

Quote:
Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.


so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.

if you will notice everything you are refering to is
about rocket engines , im not using a rocket engine.

so all your links to rocket engines are null.

a rocket engine uses a nozzle to direct the thrust


AFTER IT PASSES INTO THE NOZZLE...

so just like the bottle rocket does not have a nozzle
after the thrust passes the lip of the bottle , my pipe
and tank anology also does not have a nozzle it
just has a 1 inch sq area tube , why do you keep insisting on using nozzles when there isnt a nozzle?

think of it as a smaller pipe that has a hole in it.

Quote:
Wrong again - nozzles harness the expansion of the compressed gas once they leave the tank, regardless of the source of that pressure.


keep reading further down it also says that it directs the gasses in a direction.

Quote:
Wrong again - nozzles harness the expansion of the compressed gas once they leave the tank, regardless of the source of that pressure. The fact you have compressed air, while rockets have fire to compress their air, doesn't change the fact that both will produce pressurized streams of air that can act against that nozzle - or in your example, will be wasted due to the lack of a nozzle.


I never said it had a nozzle , you did.
in fact at first you acknowledged it was just a tube.
so ...WHAT HAPPENED TO THE TUBE?

then you acted as if it had a nozzle from that point
on , but theres no nozzle.

you included a formula to compute the nozzle thrust
and disreguarded the only thing that actually was there
the tube.

theres is no nozzle.

ok , I want to allow the air to expand as it wants...
I dont want the air to be directed towards the rear as you seem to want it to.

I dont want a nozzle , nope no nozzle , none.

OK...no nozzle.

now what , we didnt install a nozzle.

will the pipe move?

is like saying will the bottle rocket go anywhere.

I say it will , what do you say it will do .

go towards the ground instead.

that video was photoshopped wasnt it?
dang conspirators.

they shouldnt promote such realism on the internet.


Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 01:10 AM

Quote:

I agree that momentum is conserved also , but its not like Im agreeing with you , as that is something that makes sence.

so let me get this straight - you agree with the fundamental physical principal which clearly states your "engine" is a physical impossibility, and yet you insist your "engine" will impart momentum...

A little schizoid, are we wink

Quote:
yes the momentum is conserved in the moving pipe itself. nothing left the pipe.

Ergo, the pipe does not move.

Quote:
Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.

so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.

The point is the very part you dishonestly cropped - the speed of sound sets the maximum velocity a gas can accelerate itself to, using its own pressure as the source of that acceleration.

Nor is the speed of sound anywhere near as constant as the gravitational acceleration of earth - the speed of sound changes with the constituents of the gas and temperature.

Quote:
if you will notice everything you are refering to is about rocket engines , im not using a rocket engine.

so all your links to rocket engines are null.

The physical principals are the same, regardless of the source of the pressure. The fact you expect otherwise defies not only the laws of physics,but basic logic as well.

Quote:
a rocket engine uses a nozzle to direct the thrust

Nope, even your own sources refute that claim - the nozzle accelerates the gas, providing thrust beyond that of the pressure differential alone.

Quote:
so just like the bottle rocket does not have a nozzle after the thrust passes the lip of the bottle , my pipe and tank anology also does not have a nozzle it
just has a 1 inch sq area tube , why do you keep insisting on using nozzles when there isnt a nozzle?

But I'm not insisting its a conical nozzle!!!! In fact, I've repetitively stated tat your nozzle is linear - as in a tube, exactly as you describe.

That is, after all, the entirety of the point I've been trying to make the past 10 or so posts - for Ve to have a non-zero value, the nozzle must have a divergent section. No divergent section, no change in the velocity of the gas escaping the "tube", thus Ve = 0.

Quote:
I never said it had a nozzle , you did.
in fact at first you acknowledged it was just a tube.
so ...WHAT HAPPENED TO THE TUBE?

Nothing, I simply used the word "nozzle" to differentiate that part of your contraption from the pipe and the tank. Too many tube-like objects.

As I've stated time and time again, your nozzle is tube-shaped; as in it has no divergent angle.

No divergent angle = no acceleration of the gas within the nozzle

No acceleration of the gas in the nozzle = no Ve

Quote:
you included a formula to compute the nozzle thrust
and disreguarded the only thing that actually was there
the tube.

Wrong again. If you go back and check my calcs you'll see I was using the exact conditions you set in the formula - no divergence in the nozzle (i.e. it is a tube).

So basically you're complaining that I did the math right for the conditions you explained, but chose to use what I thought was a less confusing term for the particular part of your contraption.

Quote:
ok , I want to allow the air to expand as it wants...I dont want the air to be directed towards the rear as you seem to want it to.

Wants and desires have no role in physics - physics obeys set, unalterable laws. In the case of the force generated when a gas moves from a region of higher pressure there are two factors involved - the pressure differential over the exit area; determined mathematically by [Pt-Pe]Ae, and the force generated by the gas as it expands in a divergent nozzle (if there is a divergent nozzle); determined mathematically by m*Ve.

As you said, and as I've agreed since the beginning, you do not have a divergent nozzle; simply a straight pipe. Under that condition, Ve is zero, leaving the sole source of force as the pressure differential - [Pt-Pe]Ae.

As for whether your pipe moves or not, that's a matter of the momentum in the system not the total force generated by the expulsion of air from the tube/nozzle.

Your contraption, nozzleless or nozzle containing, imparts momentum to the tank, which then transfers that momentum to the pipe. However, the same amount of momentum will be transferred to the air coming out of the pipe/nozzle, in the opposite direction. That momentum will be transferred to the opposite end of the pipe, where it will neutralize the momentum induced by the tank.

The nozzle doesn't change the momentum factor - all it does is more effectively transfer the pressure in the tank into momentum. But regardless, that momentum will be equal, but opposite, in the tank verses the air, and thus when the tank and air impart their momentum onto the pipe, the momentum cancels out. The only thing the nozzle does is determine how much momentum you "extract" from the pressurized gas. But regardless of your "conversion efficiency", the momentum imparted onto the tank will always be equal and opposite to the momentum imparted into the air.

Quote:

will the pipe move?

is like saying will the bottle rocket go anywhere.

Completely different situations - the bottle rocket is part of an open system - its exhaust transferes its momentum into the environment, not back into the bottle rocket. If you were to seal that rocket into a pipe, the rocket would move - the pipe would not.

And the reason is simple - the water coming out of the back of the rocket will propel it forward in the pipe. The expelled water will be propelled backwards in the pipe. Both will have the same momentum. When the rocket hits the front of the pipe, it transfers that momentum to the front of the pipe. When the water hits the back of the pipe, it transfers its momentum to the back of the pipe. Since both the rocket, and the water, have the same momentum, the net momentum imparted on the pipe is zero.

Ergo, the pipe will not move.

And since you're stuck on the nozzle issue, bottle rockets are propelled by a liquid - water - not a gas. Because water is incompressible at water-bottle pressures, the water will not expand in volume when it leaves the "tube". Ergo, having a divergent nozzle will not improve performance, as there is no expanding gas to take advantage of.

http://en.wikipedia.org/wiki/Water_rocket#Nozzles

But fill that bottle with compressed air alone, and a nozzle will give you great dividends:
http://en.wikipedia.org/wiki/De_Laval_nozzle

But, and I emphasize this again, regardless of how efficiently or inefficiently you harness the expansion of that gas, the amount of momentum imparted onto the tank will always be equal, but opposite, to the amount of momentum imparted on the gas leaving that tank. And since your system is closed, the net effect will be no net change in momentum of your pipe.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 01:24 AM

Quote:
In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.


momentum , thermodynamics , n such

so you were saying something?

Quote:
although it may be hard to believe, a human sneeze travels at about 120 mph when it exits. the air moves faster, sometimes maybe 200 mph. some people have been able to prove that their sneeze traveled supersonic (breaking the sound barrier). most children sneeze at about 90 mph. adults from 40-50 have the fastest sneezes at sometimes over 800 mph.


the fastest sneezers 800 mph 40-50 yr old people

I guess that what I should get is a bunch of 40-50 year old people because they dont seem to have as hard a time at breaking the sound barrier when they sneeze. LOL

that way we wouldnt need a air tank that is pressurized to 100 psi , they could just sit up front facing backwards and
I could sprinkle pepper over them -- deep space here we come.

----------------------------------
nozzel stuff


Rocket Nozzle Design: Optimizing Expansion for Maximum Thrust
A rocket engine is a device in which propellants are burned in a combustion chamber and the resulting high pressure gases are expanded through a specially shaped nozzle to produce thrust. The function of the nozzle is to convert the chemical-thermal energy generated in the combustion chamber into kinetic energy. The nozzle converts the slow moving, high pressure, high temperature gas in the combustion chamber into high velocity gas of lower pressure and temperature.
------------------ STOP JUST A MINUTE HERE--------------
Gas velocities from 2 to 4.5 kilometers per second can be obtained in rocket nozzles.
.......................................................
note : thats slightly faster than the speed of sound
wouldnt you say?
speed of sound 340 meters per second.

hmmm .... 4500 m/s - 340 m/s = 4160 meters per second faster than the slower speed of sound...
------------------------- continue --------------------
The nozzles which perform this feat are called DeLaval nozzles (after the inventor) and consist of a convergent and divergent section. The minimum flow area between the convergent and divergent section is called the nozzle throat. The flow area at the end of the divergent section is called the nozzle exit area.

Hot exhaust gases expand in the diverging section of the nozzle. The pressure of these gases will decrease as energy is used to accelerate the gas to high velocity. The nozzle is usually made long enough (or the exit area great enough) such that the pressure in the combustion chamber is reduced at the nozzle exit to the pressure existing outside the nozzle. It is under this condition that thrust is maximum and the nozzle is said to be adapted, also called optimum or correct expansion. To understand this we must examine the basic thrust equation:

F = q × Ve + (Pe - Pa) × Ae

where F = Thrust
q = Propellant mass flow rate
Ve = Velocity of exhaust gases
Pe = Pressure at nozzle exit
Pa = Ambient pressure
Ae = Area of nozzle exit

The product qVe is called the momentum, or velocity, thrust and the product (Pe-Pa)Ae is called the pressure thrust.

Let us now consider an example. Assume we have a rocket engine equipped with an extendible nozzle. The engine is test fired in an environment with a constant ambient pressure. During the burn, the nozzle is extended from its fully retracted position to its fully extended position. At some point between fully retracted and fully extended Pe=Pa (see figure below).



Finally, we calculate the thrust,

F = q × Ve + (Pe - Pa) × Ae
F = 100 × 2,832 + (0.05x106 - 0.05x106) × 0.409
F = 283,200 N

notice the above , take into account the
100 * 2832 !!!! because it equals 283,200
you add that amount to the ZERO from the Pe-Pa side and
you get the same number...

amazing

I might add that air can be compressed , combustible gasses
can not be compressed as high as air.

theres a big difference in compressed air and combustible gasses expanding , much higher velocities can be obtained
using compressed air than can be obtained by combustible gasses expanding.

if Uncle Al were here he would put you across his knee and give you a good lashing.



Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 01:42 AM

Quote:
Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.

so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.


As I mentioned earlier, my point was dishonestly cropped by you. But this point is worth revisiting, and as such here is a separate post...

The speed of sound in a given gas is determined by:
S = sqrt(C/d), where
S = speed of sound
C = elastic/bulk modulus of the gas/gas mixture,
1.42×10^5 Pa for air at STP
d = density

The density of a gas is dependent on pressure:
d = p/RT, where:
d = density
p = absolute pressure
R = specific gas constant, for dry air this is
287.058J&#8201;kg&#8722;1&#8201;K&#8722;1
T = temperature, in kelvins

So at 100PSI, at room temp, the density of your gas is:
d100 = p/RT
d100 = 689,475Pa/(287.058J&#8201;kg&#8722;1&#8201;K&#8722;1 * 293K)
d100 = 8.2 kg/m^2

and the speed of sound is:
S100 = sqrt(C/d)
S100 = sqrt(1.42×10^5 Pa/8.2 kg/m^2)
S100 = 131.61 m/s

At atmospheric pressure (14.7PSI):
d14 = 101,352Pa/(287.058J&#8201;kg&#8722;1&#8201;K&#8722;1 * 293K)
d14 = 1.2 kg/m^2

thus the speed of sound is:
S14 = sqrt(1.42×10^5 Pa/1.2 kg/m^2)
S14 = 343.28 m/s
---------------------------------
At this point you're probably asking "WTF is the point?"

The point is simple, and two-fold:
1) Contrary to your claims, the speed of sound is not a constant. In fact, it will be roughly three times SLOWER in your pressurized tank than it will be in your pipe.

2) You've mistakenly tried to calculate the force generated by your system, using m*v where m is the mass of the air, and v is the speed. You're wrong two ways here - m*v in using your measures gives you momentum, not force. And on top of that not-so-insignificant foopah, your v is wrong - its restricted to the speed of sound of the gas, due to back pressure. In the case of your 100PSI tank, that's 131.61 m/s.

So two basic claims of yours - the second of which is te foundation of your "proof" my math is wrong - are in and of themselves 100% wrong.

Kinda like your whole argument, beginning to end...

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 02:03 AM

Originally Posted By: paul
Quote:
In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.


momentum , thermodynamics , n such

so you were saying something?


Yep, that momentum in a closed system cannot change - a point which wikipedia does not contest either in the page you cite, or in its page on momentum:

"Momentum is a conserved quantity, meaning that the total momentum of any closed system (one not affected by external forces) cannot change"

http://en.wikipedia.org/wiki/Momentum

Quote:
...some people have been able to prove that their sneeze traveled supersonic (breaking the sound barrier)


Two point here:

1) I can find no evidence to substantiate your claim. The maximum speeds I've found in various sources - medical and layman is 90-150mph. The fastest speed I've seen claimed by a pseudo-medical source (JFK Health World Museum) is 650mph; about 85% the speed of sound. According to wikianswers, the highest confirmed speed is 166.7km/h; about 103mph.

2) Even if your claim is true, it hardly disproves what I said. Your nose is a divergent structure; the nasal cavity expands in cross section nearly 15X relative to the nasopharinx (where the air enters the nose when you exhale/sneeze). Ergo, your nasal cavity acts exactly like a rocket nozzle would - it allows for acceleration of expired gas, due to the expansion of pressurized gas within the nasal cavity.

In fact, airflow through the nose is a serious area of research, and is very well understood. A few examples:
http://www.springerlink.com/content/x48u517p27517169/
http://www.springerlink.com/content/7869gchy3g05fhut/

Once again, the very physical phenomena I've been describing, in action...well understood, and completely within the laws of physics I have described.

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 02:07 AM

Quote:

------------------ STOP JUST A MINUTE HERE--------------
Gas velocities from 2 to 4.5 kilometers per second can be obtained in rocket nozzles.
.......................................................
note : thats slightly faster than the speed of sound
wouldnt you say?
speed of sound 340 meters per second.


You've added this since my last reply - trying to sneak stuff through, are we?

You need to go back and read what I've written before - it is the nozzle of the rocket that allows for the gas to be accelerated faster than sound. I said that clearly, on several occasions now.

It is that acceleration that creates Ve.

So bascially, you just supported my argument - you need a divergent nozzle to get faster-than-sound traveling air.

Quote:

as you can see the Pe-Pa is just the calculatable part of the moveable nozzle.


And once again paul shows he doesn't even read his own sources. Even your own source confirms Pe-Pa is the pressure force, not the force due to nozzle acceleration.

LOL, your own sources showing you to be wrong. Funny!

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 02:35 AM

I think your loosing it , the speed of sound is not relevant.

it is the speed that air will travel that is relevant.

Quote:
d100 = 689,475Pa/(287.058J&#8201;kg&#8722;1&#8201;K&#8722;1 * 293K)


could you go back to the page where you copied the above
work that someone else did , and insert the correct characters.

so it will be ledgible.




Originally Posted By: ImageingGeek

Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature. Different mixtures of gasses will have different speeds of sound - for example, the speed of sound in air is highly dependent on the amount of water vapor present. its also dependent on temperature.

The speed of sound is also the maximum speed that a gas can flow at under its own pressure. The speed of sound is determined by a gasses elastic modulus. That same elastic modulus also defines the back pressure of a flowing gas, and thus the degree to which a gas can accelerate itself due to its own pressure.



as in the rocket nozzle the gasses are NOT under their own pressure ,,,,, they are COMPRESSED GASSES.

just like in the air tank the air is COMPRESSED not under its own pressure.

so either your right and we have never launched a rocket
or
Im right and we have...

havent you ever seen an aircraft pass the sound barrier?
how do you think it reached that speed , with slower than the speed of sound gasses flowing out of the nozzles?

Quote:
And once again paul shows he doesn't even read his own sources. Even your own source confirms Pe-Pa is the pressure force, not the force due to nozzle acceleration.


and it also states that m*Ve is the momentum and velocity thrust.

and the calculations below show the the m*Ve thrust is much larger than the Pe-Pa thrust.

did you happen to see that part , I have alreasdy covered it for you in a earlier post.


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/09/10 03:29 AM

Hi Paul, sorry to disturb you two's mad fight, but if you find time, could you please respond to my post #34791.

I'm not asking you check the working if you don't want, but at least to identify if that analogy is good enough, because it's a hell of a lot easier than this rocket business!!
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/09/10 06:50 AM

Originally Posted By: ImagingGeek

Ftotal = Fn + Fp = [m*Ve] + [Pt-Pe]Ae


Hi Bryan. Yea I pretty much got the gist of that, cheers. But it still doesn't explain an equation Paul put up, off NASA's website:

Exit Velocity: Ve = Me sqrt(gamma R Te)

According to both you and that same site, for a straight nozzle, Me = 1
gamma is probably on the order of 1
R > 0
Te > 0
So Ve > 0 for a straight nozzle.


You have a different equation for Ve, which gives zero if Pe=Pt. But Pe <> Pt because you need a difference to get any pressure thrust.


Quote:

Ve = the exit velocity of the gas from the nozzle, relative to the velocity of the gas when it enters the nozzle.


Wikipedia: "Ve = Exhaust velocity at nozzle exit"
No mention of being relative to the pre-nozzle velocity. NASA's page equally doesn't say anything about it being an "extra" velocity, they just call it velocity.


So I'm stongly leaning in Paul's direction.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 01:13 PM

Originally Posted By: kallog
Originally Posted By: ImagingGeek

Ftotal = Fn + Fp = [m*Ve] + [Pt-Pe]Ae


Hi Bryan. Yea I pretty much got the gist of that, cheers. But it still doesn't explain an equation Paul put up, off NASA's website:

Exit Velocity: Ve = Me sqrt(gamma R Te)

According to both you and that same site, for a straight nozzle, Me = 1
gamma is probably on the order of 1
R > 0
Te > 0
So Ve > 0 for a straight nozzle.


I assume you mean this page:
http://www.grc.nasa.gov/WWW/K-12/airplane/rktthsum.html

The difference is one of ideal verse non-ideal gases. NASA has the classical (i.e. ideal gas-based) formula on their page, while I provided the one derived for non-ideal gases (the page that my image comes from goes through that entire derivation, if you're interested). basically, paul has the version you'd teach in school to get an approximation of force; I used the one that is used by engineers to actually design and build rocket engines.

The way Mdot is calculated is also different for non-ideal gases, so you cannot really mix. Here is what the non-ideal gas formula looks like, when Mdot (for non-ideal gases) is added into the Ve formula provided earlier:



Everything before the (Pe-Pa)Ae is Mdot*Ve, for non-ideal gases.

I guess I owe paul half an appology - if he were using an ideal gas (the only one we know of is helium, below ~90K) he would have been correct.

Unfortunately, NASA's example math doesn't work (accurately) in the real world.

Originally Posted By: kallog

You have a different equation for Ve, which gives zero if Pe=Pt. But Pe <> Pt because you need a difference to get any pressure thrust.


No, that is not correct. But its my fault - I see I've been mixing pauls and my terms. To be clear:

Pe = pressure of the environment
Pt = pressure of the tank (in pauls case) or throat of an
actual rocket engine
Po = pressure at the end of the nozzle, which is almost
always higher than Pe

Pressure force is [Pt-Pe]Ae - as in the pressure difference between the tank (or wherever pressure is highest in the engine) and the environment.

Ve uses the ratio of Pt/Po - as in the ratio of the pressure difference between the front of the nozzle and the end of the nozzle. In a straight tube Pt will equal Po, ergo [1-(Pt/Po)] will be zero


Quote:

Ve = the exit velocity of the gas from the nozzle, relative to the velocity of the gas when it enters the nozzle.


Quote:
Wikipedia: "Ve = Exhaust velocity at nozzle exit"
No mention of being relative to the pre-nozzle velocity. NASA's page equally doesn't say anything about it being an "extra" velocity, they just call it velocity.


Once again, we have the issue of ideal verses non-ideal gases. Ve, as given in Nasa's formula, is the ideal exit velocity - what you would get if you had a gas that essentially doesn't interact with itself - i.e. no in-flow friction. However, in the real world we don't have ideal gases, so we need to calculate the effective exit velocity:

http://en.wikipedia.org/wiki/Specific_impulse
http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation

For a non-divergent nozzle, with a real gas passing through it, effective Ve is 0.

Bryan

EDIT: Just to add some more sources in regards to theoretical verses real-world calculations of ideal verses effective Ve, thrust calcs, etc in real-world situations:

NASA SP-8039 - derivation of all the formulas I used above, starting from ideal-gas, going to real-world

NASA SP-8115 - formulas for the design of rocket nozzles; covers the nozzle theory and derivation of the equation I provided in this post

NASA TN D-467 - comparisons between the performance of real-world engines and their ideal-gas calculated performance.

Some Considerations on the use of a Pressurized Tank System for a Rocket Engine. R.Sandri, Canadian Aeronautics and Space Journal, Oct.1967 - basically a description of the physics of using pressurized air

All but the latter are available on NASA's webpage.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 02:37 PM

its not a rocket engine Bryan.

its just a tube
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 03:43 PM

Originally Posted By: paul
its not a rocket engine Bryan.

its just a tube


The physics are the same, paul. A rocket engine is, afterall, just a lumpy tube.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 04:24 PM

Quote:
The physics are the same, paul. A rocket engine is, afterall, just a lumpy tube.


then what is the velocity of the air that will come out of the
streamlined tube.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 04:46 PM

Originally Posted By: paul
Quote:
The physics are the same, paul. A rocket engine is, afterall, just a lumpy tube.


then what is the velocity of the air that will come out of the
streamlined tube.



I've answered this many times - whatever the speed of sound is, given the gas mixture in the tube and the temperature/pressure that gas is at.

The effective Ve however is zero, as outlined in my previous posts, as air is not an ideal gas.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 05:54 PM

Quote:
whatever the speed of sound is


well Bryan , since the air is limited to the
"speed of sound"

340.29 m/s

the tube size is not important as the tube can be any size
according to your reasoning it can never travel faster than
the speed of sound.

!!! air is not a ideal gas !!!

fine lets just use the slower speed of sound.

so lets open it up fully to 58 inch diameter.
removing the need for a tube or nozzle.
after all Im only looking for momentum as I have stated.

if I put 1000 cu ft of 1 atm air into the tank that
already has 1000 cu ft of 1 atm of air inside

the pressure doubles.

I now have 2000 cu ft of air inside the tank.
and the pressure is 14.7 * 2 = 29.4 psi

if I put in another 2000 cu ft of 1 atm air inside the
tank the pressure doubles again.

I now have 4000 cu ft of air inside the tank at 58.8 psi

if I put 4000 more cu ft of 1 atm air inside the tank
I now have 8000 cu ft of 1 atm air that has been squeezed into the 1000 cu ft tank.
and the air pressure is doubled again to 117.6 psi
air weights 0.0807 lb / cu ft
8000 cu ft * 0.0807 lb / cu ft = 645 lbs
645 pounds = 292.567079 kilograms

F = (m × V) for force
or
P = (m x V) for momentum

P = 292.567 kg * 340.29 m/s = 99557.62443 kg-m/s

so the momentum is 99,557 kg-m/s

the tank is 54.5 ft long
its diameter is 58 inch

it takes apx 0.16 seconds for the mass of air to leave the tank...


if I use a large or small tube doesnt matter that much
according to your reasoning.

the air will only have a certain speed at that pressure.
and that speed is the speed of sound.

I can increase the time that I allow the air to escape from the tank by using a small orifice or tube.

and this will decrease the momentum that is felt by the pipe each second.

and the ONLY WAY you can counteract this momentum is to supply a force of 99,557 kg-m/s in the opposite direction of the momentum force to make the pipe not move at all.


Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 07:33 PM

Kallog

Quote:
I'll simplify it to be a gun instead of a rocket, and it's fixed to the front of the pipe, pointing backwards.

You fire the gun, and it gives momentum -p1 to the bullet. At the same time it gives momentum p1 to the gun and pipe. Exactly the same as recoil in a normal gun.

Clearly the pipe's now moving because it has momentum p1.

Eventually the back end of the pipe and the bullet collide, and the bullet becomes embedded in the back of the pipe. They were travelling in opposite directions, with zero total momentum (-p1 + p1 = 0). Now they're a single solid object with the same total momentum - 0.

The pipe has 0 momentum and has therefore stopped moving.


the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.

the air moves throughout the pipe , some of the air moves into every cubic inch of area inside the pipe.

because the air pressure itself is equalizing.

because the tank is at one end of the pipe most of the
air in the tank will move toward the other end of the pipe.

and since the pipe is 60 inch diameter and the tank is 58 inch diameter there will basically be just air equalizing at the end where the tank is.

all the other air moves away from the tank.

and it is this movement that moves the pipe.

in my last post I calculated the momentum to be
99,557 kg-m/s

and this movement of mass inside the pipe causes the
pipe to move.

so if you move this mass fast or slow doesnt matter
as the total momentum is the result of P = m*v

in your bullet anology you would still have the reduction
in velocity due to air resistance as the bullet travels the 500 ft distance to the other end.

then F = m* (less velocity)than you started with.

the pipe would still be moving.


Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 07:34 PM

Originally Posted By: paul
Quote:
whatever the speed of sound is


well Bryan , since the air is limited to the
"speed of sound"

340.29 m/s

Paul, your own source says exactly that. From your NASA page:
http://www.grc.nasa.gov/WWW/K-12/airplane/rktthsum.html

The hot exhaust flow is choked at the throat, which means that the Mach number is equal to 1.0 in the throat and the mass flow rate m dot is determined by the throat area

But in your case - 100PSI, the speed of sound is 1/3rd of the value you stated - 131.6m/s (see post #34826 for the calc).

Originally Posted By: paul
the tube size is not important as the tube can be any size according to your reasoning it can never travel faster than the speed of sound.

Right and wrong, depending on what you mean by size.

The radius of your tube will determine the *rate* at which air is released from the tank. And as such, it'll also dictate the force developed. Keep in mind that [Pt-Pe]Ae determines the force derived from pressure; if you double Ae you'll double the force you'll produce.

However, since you have a tank of air the total amount of thrust produced (ISP, measured in newton-seconds) will be the same regardless of the tube's radius - the only difference is a larger diameter tube will produce that thrust in a shorter period of time, while a small-diameter tube will produce that amount of thrust over a long period of time.

Originally Posted By: paul
if I use a large or small tube doesnt matter that much according to your reasoning.

As described above, it depends on what you mean by size. The speed of the air in a tube will be restricted to the speed of sound, regardless of length or diameter. The *mass flow* however will be greater in a larger diameter pipe, but will not be affected by length. Double the cross-sectional area of your pipe, and twice as much mass will flow through it every second. It'll flow at the speed of sound, but there will be more moving

Originally Posted By: paul

I can increase the time that I allow the air to escape from the tank by using a small orifice or tube.

and this will decrease the momentum that is felt by the pipe each second.


Here is where you are missing the implication of choked flow (i.e. flow limited to the speed of sound) - and I think where our entire "conflict" (in regards to your tube) lays.

Forget air for a second, and think of a metal ball in your tube. The 100PSI tank you described earlier produced ~6000N of force through its tube. Pretend for a second that we applied that 6000N of force to the metal ball, but in a way where the source of the force was unconstrained (i.e. not using air confined to the speed of sound). This would accelerate the ball as per the formula F=ma (rearranged to a = F/m to calc the acceleration). The longer the pipe, the more acceleration that ball would experience. Hypothetically speaking (and ignoring relativity) an infinitely long pipe would lead to an infinitely accelerated steel ball.

I believe you will agree with the above description, as that's essentially what you've been saying all along (with the steel ball replaced by a mass of air).

Here is the part you are missing - your flow of air is choked; restricted to a set speed. This is occurring despite the fact that you have ~6000N of force acting on that air. So unlike that steel ball, your air is not continually accelerated - something is holding it back. And, unlike your steel ball, a longer tube won't further accelerate the air - one it gets to the speed of sound it stays there.

The factor that restricts the flow of gas to the speed of sound is the air itself - "real" (as in non-ideal) gases resist their own flow. The faster you push a gas, the more it resists being pushed. When this pushing is pressure-driven, the resistance to flow will equal the force produced by the pressure once the gas reaches the speed of sound. What this means is the momentum of the gas traveling at the speed of sound is perfectly balanced by an opposing force - in this case the backpressure of a non-ideal gas.

So, in your tube you have the situation where your tank initially accelerates the air to the speed of sound - this initial acceleration of the air is generated by the force determined by [Pt-Pe]Ae. At this point you now have a stream of air traveling at the speed of sound, despite the fact the tank's 6000N of force is "trying" to accelerate it further. This lack of acceleration is due to the force provided by the airs resistance to its own flow - and that force is equal, but opposite, to the force provided by the tank.

In other words, once flow is choked your tank is pushing it with 6000N of force, but that force is being countered by 6000N of backpressure.

And keep in mind, just as the tank imparts momentum onto the air due to the 6000N of force available to it, the choked air is in turn pushing back on the tube/tank with 6000N of force. It is this equal, but opposite, forces that prevent the further acceleration of the gas, and it is that opposing force that causes m(dot)Ve to be zero in a tube.

What this means is that your tube-based system experiences a force, and thus gains momentum, from that initial acceleration of the air from rest to the speed of sound - a force determined by [Pt-Pe]Ae. However, the force generated by mass flow - and thus that momentum - is countered by an equal but opposite "choking" force. Ergo m(dot)Ve is zero with a tube.

All of that could be derived by looking at the formulas provided, in a lot fewer words, but the above is the plain-ol-english version.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 08:24 PM

Quote:
What this means is that your tube-based system experiences a force, and thus gains momentum, from that initial acceleration of the air from rest to the speed of sound - a force determined by [Pt-Pe]Ae. However, the force generated by mass flow - and thus that momentum - is countered by an equal but opposite "choking" force. Ergo m(dot)Ve is zero with a tube.


still the mass will come out of the tank given its tube is
58 inch diameter.

and when the mass comes out , and it will come out , the momentum will be mass x velocity.

P=m*V

I dont care how you word it , the pipe will move.

because the mass moves inside.

lets check this...

the pressure differential is

114.7 psi - 14.7 psi = 100 psi

the force is 100 lb/sq inch.

the cross sectional area of the 58 inch tube is
2642.079 sq inch
the air inside the tank weights 0.645 lbs / cu ft
2642.079 cu ft / 1728 cu in = 1.528 cu ft in the cross section

suppose the cross section is 1 inch long.

there is 1.528 cu ft in the cross section.

1.528 cu ft * 0.645 lbs = 0.955 lbs in the cross section

there are 2642.079 sq inches in the cross section

.955 / 2642.079 = 0.0003614577 lb per cu inch of area.

a=f/m

a = 100 psi / 0.0003614577 lbs = 276,657.545 fps

276,657.545 fps / 5280 ft = 52.3972 miles per second.

276 657.545 fps = 84,325.219 meters / second


this incorporates the resistive force and friction is pretty much null in a tube this large especially with air as the fluid.

there is no choking force involved in this situation
there is only fluid flow.

and that restriction is minimal if not non existant
in a pipe this large.

the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s








Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 08:34 PM

Originally Posted By: paul
the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.


Paul, here again you are making a simple mistake. Forces and momentums are vectorial - meaning they have both a magnitude and direction. Pressure is non-vectorial - it has a magnitude, but occurs equally in all directions.

It's not 100% clear what you're trying to claim, but I see two ways to interpret it.

--------------------------------------------

The first interpretation is that you are claiming the pressure-mediated expantion of the air stream will dissipate its momentum, thus leaving only the momentum of the tank, and thus causing your pipe to move. This is wrong - think of the three spatial axis - x, y, z. Imagine your tank is expelling air in the -x direction - this would push the tank in the +x direction. Lets say 100kg*m/s of momentum is given to the tank. This means the tank has 100kg*m/s momentum in the x-axis, but 0 in the y/z axis. Likewise, the air expelled from the tank has -100kg*m/s momentum in the x-axis, but 0 in the y/x axis.

If I understand you correctly, you're saying the pressure of that air stream will counter the momentum of the air stream - as in the expansion of the air stream will dissipate the momentum. This is wrong. Keep in mind pressure is non-directional. So if there is 100PSI of pressure in the air stream, that means there is 100PSI pushing in the +x direction, -x directions, as well as the +/-y and +/-z directions.

So what happens? First, the obvious - the pressure expands the air stream in the +/-y and +/-z directions. In fact, if you were too look at any one air molecule it would now have more momentum, as it would have the initial -100kg*m/s of momentum along the x-axis, plus whatever momentum its now added on the y/z axis. Over all the stream of air has no net gain in momentum, as there will be equal expansion in the +/- directions on the y and z axis. Keep in mind, the expansion on the y/z axis has NO EFFECT on the momentum of the air stream, as this pressure-driven expansion along the y & z axis does not impact momentum on the x-axis.

The big Q is what happens on the x-axis, in regards to pressure. If you pick any one point along the x-axis of the air stream, and look immediately the the + and - of it, there will be essentially no pressure difference. Ergo, the pressure in the stream will equally push in the + and - x directions. Ergo, there is no net pressure-mediated force in the +/- x direction (i.e. expansion will be equal in the + and - x-directions), and thus no net change in momentum along the x-axis will occur.

Ergo, don't expect the pressure mediated expansion of the air stream to magically dissipate the momentum of the air stream - it won't. The net effect of the pressure of the air stream is it will expand it equally in all directions. What this means is any vectorial properties of the air stream (i.e its momentum or forces it applies) will not be affected by its pressure-mediated expansion, as that expansion has no vectorial value (i.e. its even in all directions). When that column of moving air hits the back of the pipe it'll still have a momentum of 100kg*m/s - the only difference is that momentum will be spread over a larger area.

--------------------------------------------

The other way to interpret your claim is that the momentum of the air and tank will equal out (which they do), but the equalization of pressure will provide a net momentum. This interpretation too is wrong - because pressure is non-vectorial, it is incapable of inducing a vectorial force on your pipe. Equalization of pressure occurs equally in all directions; there is no net direction to the force (as air molecules move in a Brownian fashion). Since you have equal momentum of air in all directions the net force - and thus the net momentum - is zero.

Even taking into account simple mass-action, your net is still zero. You start by throwing a mass of air in the -x direction - this will give your pipe an equal momentum in the +x direction. However, the gas will want to reach equilibrium, meaning it'll expand towards the front of the pipe. The force of mass-action will equal (by definition) the amount of force needed to create the initial state. The movement of the air mass forward will be of equal momentum to the initial momentum of the air mass rear-wards. Net effect - pipe doesn't move.

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 08:38 PM

Originally Posted By: paul
Quote:
What this means is that your tube-based system experiences a force, and thus gains momentum, from that initial acceleration of the air from rest to the speed of sound - a force determined by [Pt-Pe]Ae. However, the force generated by mass flow - and thus that momentum - is countered by an equal but opposite "choking" force. Ergo m(dot)Ve is zero with a tube.


still the mass will come out of the tank given its tube is
58 inch diameter.

and when the mass comes out , and it will come out , the momentum will be mass x velocity.

P=m*V

I dont care how you word it , the pipe will move.

because the mass moves inside.


Now you're mixing issues.

The air that leaves the nozzle/tube will have momentum - a momentum given to it by the force [Pt-Pe]Ae.

However, the momentum of the air leaving the nozzle/tube will impart no net momentum on your pipe, for the simple reason it'll transfer that exact same amount of momentum in the opposite direction once it strikes the end of the pipe.

Equal and opposite momentums = your pipe does not move.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 08:40 PM

Quote:
whatever you wrote is no longer important to me


I no longer consider you as someone who should be listened to.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 08:43 PM

Quote:
whatever you wrote is no longer important to me


I no longer consider you as someone who should be listened to.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 08:44 PM

the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 09:10 PM

Originally Posted By: paul
the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s


Absolutely, but this comes back to another common issue with you - you consistently confuse momentum and force.

Lets use simpler quantities to make the math more obvious - your tank is at 1000Pa (1kPa), and contains 1kg of air. For the sake of simplicity, the environment is a vacuum - 0Pa.

With your straight pipe that 1kg of air will be accelerated to the speed of sound - 340m/s. This means that after leaving the tank the air has a momentum of 1kg*340m/s = 340kg*m/2 - regardless of the size of tube you use.

However, the force produced will vary depending on the size of the tube. Take the case where you have a tube with a cross-section of 1m^2 verses 2m^2:

F = [Pt-Pe]Ae
If Ae = 1m^2, then F = [1000Pa-0Pa]1m^s = 1000N
If Ae = 2m^2, then F = [1000Pa-0Pa]2m^s = 2000N

---------------------------------------------
Now what if we add a nozzle that has a divergence angle which doubles the speed of the air (i.e. accelerates it to 2X the speed of sound - 680m/s)?

Now your momentum doubles - 1kg*680m/s = 680kg*m/2 - regardless of the size of tube you use (so long as the nozzle is scaled appropriately).

Your force will also be larger:


The [Pt-Pe]Ae part will remain the same as above - i.e. a 1m^2 will impart 1000N of force. But we've also got the additional acceleration of the air to 2X the speed it would have if there was no nozzle. I've not included enough information here to calculate that, based on the above formula.

But I kinda cheated - the only time you would have a situation where a nozzle would double the speed (and thus double the momentum) of the air passing through it is when you have a nozzle designed such that it produces an amount of force equal to that of [Pt-Pe]Ae.

Ergo, the force produced by our nozzle is:
If Ae = 1m^2, then F = 2000N
If Ae = 2m^2, then F = 4000N

The above nozzle, BTW, is fairly inefficient. Most nozzles accelerate to 3-5X the speed of sound. There is a theoretical maximum you can extract, which depends mostly on the pressure and temperature of the gas exiting the nozzle. For a rocket this usually comes in around mach 6.4. It would be lower in your case, due to the relatively low temperatures involved.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 09:16 PM

click here bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 09:22 PM

wat je schrijft is niet meer belangrijk voor mij

ik je niet meer beschouwen als iemand die moet worden geluisterd naar

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 09:26 PM

Kallog
Denkt u dat dit Bryan is vervelend of wat?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/09/10 09:45 PM

click here bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/09/10 11:36 PM

So I see your arguments are now reduced to childish word games.

Guess we've established the validity of your ideas - they're childish!

LOL
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 01:00 AM

Quote:
quels que soit vous avez écrit n'est plus important pour moi


Je ne vous considère plus comme quelqu'un que l'on devrait écouter.

avec vous il n'y a aucun argument, c'est strickly de cette façon la discussion.

et cette voie est toujours dans la faveur de tyour.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 04:38 AM

Originally Posted By: ImagingGeek

The difference is one of ideal verse non-ideal gases. NASA has the classical (i.e. ideal gas-based) formula on their page, while I provided the one derived for non-ideal gases


Hang on, for a straight nozzle:
With ideal gasses the thrust is pressure thrust + an extra, nonzero term.
With non-ideal gasses the thrust is only pressure thrust, and the other term becomes exactly zero?
What if the gas is nearly ideal but not quite? There's a discontinuity which is really rare in the real world!



Your equation's site says Ve is exit velocity too, just like every other site. None of this "relative to some other speed" or "effective velocity". Actual velocity relative to the rocket.

I think there's a big mix-up with P symbols too. Your equation's site says:
Pe is exit pressure
Po is chamber pressure
Pa is ambient atmospheric pressure

But still we don't get zero exit velocity. I think any ratio of those pressures <> 1 for a tube shaped nozzle.

Why? Because without a pressure difference between the ends of the tube, there's nothing pushing the gas through it.




Quote:

force; I used the one that is used by engineers to actually design and build rocket engines.

Why? This is all just a thought experiment to see if a closed tube can fly isn't it? There's no need for engineering. While we're at it why don't we just use that 90K helium and forget all this non-ideal business? Remember Paul never said it's burning fuel, it's just compressed gas coming out of a tank.

Imagine if Einstein had used equations for the speed of a steam engine in his train thought experiments. He'd have said "In the real world trains are powered by steam engines, and they have friction, etc. These are the (really complicated) equations actually used by engineers." It would be totally pointless. It worked fine with engineless, frictionless, imaginary trains.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 04:53 AM

Originally Posted By: paul

the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.

Air's momentum doesn't magically dissappear because it disperses. It gets transferred to other air molecules and eventually to the walls of the pipe.

[/quote]
then F = m* (less velocity)than you started with.

the pipe would still be moving.
[/quote]

If the force is reduced by friction, then it's being applied over a longer period of time. Either you have high acceleration for a short time, or low acceleration for a long time. Same velocity in the end.

It doesn't matter if it hits the wall suddenly or if the pipe is filled with ballistics jelly, it doesn't matter if the bullet shatters into a million little pieces on its way through. Either way, when the bullet finally becomes stationary relative to the pipe, they becomes like a single object, and that object has the same total momentum that its constituent parts had, which is zero.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 04:55 AM

Paul, if the tube can accelerate itself then do you agree it violate's Newton's 1st law?

"Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it."

Notice it must be an _external_ force. Internal stresses, or thermal motion, or mechanisms, or anything purely internal doesn't count.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 05:18 AM

heres another wording of the first law.

First Law: Every body will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it.

those forces could come from within the pipe , such as the
momentum of the air.

if you were in a pipe in space that has a ladder that runs from one end to the other.
and you climb the ladder , you must place a force in a direction
in order to climb the ladder.

suppose the pipe were very small and short , say only ten meters long , and has the same exact mass as you do.

you apply a 50 lb force to the ladder and pull yourself toward the other end , from your point of view
you are moving up the ladder , but from the point of view of someone outside the pipe , the pipe is moving.

its the same thing , action = reaction.

the same as the mass of the air moving within the pipe.
the only difference is that your mass is not distributed evenly inside the pipe.

action = reaction.

be it a external or a internal force.

which is covered in newtons third law:

Third Law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force -F on the first body. F and -F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and -F the "reaction".

so you climbing the ladder is F
the ladder moving away from you is -F

and

the air mass moving away from the tank is F
the pipe moving away from the air mass is -F

action = reaction

so if you exert 50 lbf only half of that force is used to move
you up the ladder , the other half is used to move the pipe and ladder
in the opposite direction.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 08:15 AM

Originally Posted By: paul

those forces could come from within the pipe , such as the
momentum of the air.

If you consider the "thing" to be the pipe alone, then yes, that's fine, obviously anything inside the pipe is not part of the pipe, so it's classed as an external force. If you consider the "thing" to be the pipe and its contents, then it's the center of mass which maintains constant velocity, even for a normal rocket the center of mass doesn't accelerate - rocket goes one way, exhaust goes the other way, CoM stays at the same velocity as per the 1st law.


Quote:

and you climb the ladder , you must place a force in a direction
in order to climb the ladder.

I'm glad you brought it back to a simple and easy to analyse system.

Yes, I agree with all your reasoning, but you omitted one important part. How do you stop at the top of the ladder? You have to apply a force to the ladder/pipe in the opposite direction. You might do it very slowly so you hardly feel it, but the total impulse (= force * time for constant force) has to be the negative of the impulse you used to accelerate. It decelerates both you and the pipe back to being stationary.

When it's all over, the pipe has moved a short distance then stopped.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 01:34 PM

AH HAAA !!!!

Quote:
When it's all over, the pipe has moved a short distance then stopped.


EXACTLY , thats how it works.

reactionless propulsion , I rest.

the (distance you can travel) depends on the lenght of the ladder and the force you can supply.

the (distance you can travel) depends on the the size of the pipe and the pressure in the tank.

we only used 100 psi , but what if the pressure was
higher , and what if we were using liquid air.

liquid air density 870 kg/m^3

I know it would be alot more technical than just using
compressed air in a non liquid state , but you could go
for quite a trip with 1000 cu ft of liquid air.








Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/10/10 01:52 PM

Originally Posted By: kallog
Originally Posted By: ImagingGeek

The difference is one of ideal verse non-ideal gases. NASA has the classical (i.e. ideal gas-based) formula on their page, while I provided the one derived for non-ideal gases


Hang on, for a straight nozzle:
With ideal gasses the thrust is pressure thrust + an extra, nonzero term.
With non-ideal gasses the thrust is only pressure thrust, and the other term becomes exactly zero?

After seeing your comment I realize I explained things very, very badly. My apologies.

Ideal gases do not exist in the real world - they are a theoretical construct used as an approximation of real gases. For static gases - i.e. ones just sitting around, not moving - ideal gas calculations work very well. Once you start talking about a flowing gas, ideal gases begin to fall flat.

Ideal gases are almost exactly the same as real gases - ideal oxygen would have the same molecular weight as real oxygen, and thus carry the same kinetic energy at a given temperature as oxygen, etc. Where idea gases differ is the atoms/molecules of an ideal gas are are considered to have zero volume (as in they are infinitely small). Real gases of course are made of atoms/molecules, and those atoms/molecules have volume.

Having no volume has two dramatic implications. even at the atomic scale, friction is determined by contact area. By definition, a particle with zero volume also has zero surface area - hence, ideal gases do not experience friction. The second major implication is that atoms/molecules of ideal gases do not interact with each other - if they have no volume, they can never run into each other.

I'd also point out at this point that I stated something very badly - helium is not an ideal gas, but at low temperatures its inter-molecular interactions are so weak that it takes on some of the properties of an ideal gas.

--------------------------------------------
So the Q now is how does the ideal gas differ compared to a real gas in pauls tank/pipe system or in a rocket engine.

The answer is both "no difference" and "big difference". Keep in mind the formula for force: m(dot)Ve + [Pt-Pe]Ae

The later half - [Pt-Pe]Ae - calculates the pressure force. Pressure is simply force per area - if pauls tank was sealed the entirety of the tank would experience the pressure all around. When you cut a hole in it things don't magically change - the pressure (and thus force) remains distributed around the whole tank - the only difference is the force normally applied to the part of the tank where we cut our hole is now applied to the outside environment. This force is the same, whether your talking about a real or ideal gas (ignoring any friction the real gas experiences leaving the tank).

m(dot)Ve however, is another matter. If you push on an ideal gas it moves as per F=ma. In the case of pauls tank/pipe apparatus we have ~6000N of force coming out of the tank due to [Pt-Pe]Ae. Keep in mind its particles have no volume - they don't collide with each other or interact with the tube. So the harder you push, or the longer you push, the faster they will go. Hence, pauls 6000N of force will continually accelerate an ideal gas - there is no opposing force to slow them down. Hence the m(dot)Ve, for an ideal gas without a divergent nozzle, will be non-zero.

Real gases are very different - like an idea gas they too will be accelerated by a force. However, when you push on them, they run into the gas atoms/molecules in front of them, resulting in a force pushing back. The harder you push the stronger this backwards force (termed back-pressure) becomes. Once the gas accelerates to the speed of sound (which is the acceleration imparted by [Pt-Pe]Ae), back-pressure will equal the pressure in the tank, which is why the gas no longer accelerates, and instead holds steady at the speed of sound. Keep in mind what that means - there is 6000N of force coming out of the tank, trying to push that gas forward. Since the gas is not accelerating, that means your back pressure is equal to that 6000N.

So with real gases F = [(m(dot)Ve) - Backpressure] +[Pt-Pe]Ae, and in a straight tube m(dot)Ve and Backpressure are the same, hence the total for that side of the equation is zero.

----------------------------------
So lastly, how a divergent nozzle changes things
Picture the very end of pauls tube, now with a cone attached to it. At the very end of that tube you'll have the gas traveling at the speed of sound. Behind that gas is 6000N of force trying to push it forward, in front of that gas is 6000N of force trying to push it back.

As soon as the gas leaves the tube and enters the nozzle, things change. Backpressure is due to the compaction of gas by its own flow. As soon as you enter the cone, the gas spreads out to fill the cross-section of the cone, thus reducing backpressure. So now you have 6000N of thrust pushing the gas out of the tube, but less than 6000N of backpressure - the gas accelerates, past the speed of sound.

I won't both re-posting the formula I had posted earlier, but if you recall, it calculates Ve based on the change in pressure along the length of the nozzle. What that change in pressure represents is, in essence, the change in backpressure.
Originally Posted By: kallog

What if the gas is nearly ideal but not quite? There's a discontinuity which is really rare in the real world!


As mentioned above, there isn't really such a thing as a gas that acts like an ideal gas. The closest, as I said above, is super-cooled helium. I have no idea how that acts - I used to build my own rocket engines, but the only experience I have with cold helium is its use in cooling electron microscopes.

Originally Posted By: kallog

Your equation's site says Ve is exit velocity too, just like every other site. None of this "relative to some other speed" or "effective velocity". Actual velocity relative to the rocket.


Ve is the accepted term - doesn't mean its an ideal description of what is being measured. But, as outlined in the site I linked to (and the NASA articles I provided), Ve is proportional to the change in pressure through the nozzle.

Originally Posted By: kallog

I think there's a big mix-up with P symbols too. Your equation's site says:
Pe is exit pressure
Po is chamber pressure
Pa is ambient atmospheric pressure


There may be - I was trying to use pauls terms. Its quite possible I mixed them up as I was converting from one to the other.

Originally Posted By: kallog
But still we don't get zero exit velocity. I think any ratio of those pressures <> 1 for a tube shaped nozzle.


If I ever said Ve was zero that was in error. Keep in mind we're talking about net force produced by the engine. When you have a real gas there are three forces created by the engine - the pressure force, the m(dot)Ve force and the backpressure. When a system is choked - as in the gas is flowing at the speed of sound - the force created by m(dot)Ve and backpressure is the same, but in opposite direction. Hence the force due to Ve is zero, as it is countered by backpressure.

Mathematically, there are two ways of handling that. The accurate way is to calculate m(dot)Ve as per the formula I provided. The alternative way is to calculate the effective Ve - basically the amount of velocity not countered by backpressure. With choked flow, effective Ve is zero. I'd also add that the later way of calculating Ve is also what you do for real gases in a non-choked condition.

Quote:
Originally Posted By: kallog

force; I used the one that is used by engineers to actually design and build rocket engines.

Why? This is all just a thought experiment to see if a closed tube can fly isn't it? There's no need for engineering. While we're at it why don't we just use that 90K helium and forget all this non-ideal business? Remember Paul never said it's burning fuel, it's just compressed gas coming out of a tank.


I agree - this started off as a thought experiment. Then I made the "mistake" of actully calcuating the forces involved in pauls device, as he described it. He whined that I had done it "wrong" (silly me, using the air he specified instead of an ideal gas...). And things snowballed to here.

We could do away with the whole force generated by pressurized air, and just accept there is a force. As I said (I think on page 3) - the magnitude of the force produced by the tank is irrelevant. What is relevant is the momentum produced will remain zero so long as that tank stays in a sealed pipe.

I have enjoyed "talking" with you though - so it wasn't entirely wasted.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 02:43 PM

Quote:
He whined that I had done it "wrong"


stating facts is not whinning , bryan.
you were wrong so I said you were wrong.

Quote:
Doesn't work that way - material thrown within the ship will encounter resistance with the air, hull, etc. This will generate a force equal to the force of the propellant, thus neutralizing the thrust of the propellant. Its the ol' opposite and equal reaction thingie - the movement of the propellant will "push" on the ship, but the interaction of the propellant with the ship will push back equally. Net effect - zero thrust.


your still wrong.

Quote:
the magnitude of the force produced by the tank is irrelevant. What is relevant is the momentum produced will remain zero so long as that tank stays in a sealed pipe.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 03:13 PM

Originally Posted By: paul

EXACTLY , thats how it works.

reactionless propulsion , I rest.


Yes, that's what I've been saying all along, but it's not useful propulsion for a spaceship.

If you want to repeat the process and do another step, you have to have another guy waiting at the bottom of the ladder, and ask him to climb it too. If anybody goes back down, they reverse the movement they caused climbing up.

The maximum possible distance you can travel is the length of the pipe. And even that's only achievable with a massless pipe.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 03:25 PM

Quote:
If anybody goes back down, they reverse the movement they caused climbing up.


but hes in zero gravity , he can let go of the ladder after he reaches the end and then apply a tiny force that will send him to the other end.

so he will exert more force climbing than he will returning.

and the tiny force he applies to get to the other end
will cancel out when he reaches the other end.


we could eventually design a human powered spaceship
that has artificial gravity and reactionless propulsion.

as in people at both ends creating artificial gravity by jogging around a circular track.

and people jogging towards one end in the artificial gravity
then climbing up to zero gravity and then propelling themselves
back to the other end , then climb back down and jog to the other end again.

we would really have some healthy astronauts.




Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 03:31 PM

Hi yea I learnt a few things off this. But I still don't like those explanations :P Seems to me a rarified gas would approach being ideal. Also not sure why it really has to go at the speed of sound. I think that's only an upper limit when you have way too much force pushing it. For a bottle rocket I bet there's no speeds of sound happening.

I tend to dislike engineering equations (despite using them every day ;), because of exactly these problems. They're usually simplifications that only apply to special cases. No room for doing your own thing, and very bad for extreme thought experiments.

Perhaps you should tell Paul directly that Ve isn't zero anymore wink
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/10/10 03:36 PM

Hi Paul, I see we've met in real time now wink

Originally Posted By: paul
he will exert more force climbing than he will returning.


If he kicks off with a tiny force, then he spends a long time floating back down. So the pipe keeps on moving backwards slowly for a long time. In the end, the distance (velocity * time) is the same and it's all back where it started.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/10/10 04:05 PM

Originally Posted By: kallog
Hi yea I learnt a few things off this. But I still don't like those explanations :P Seems to me a rarified gas would approach being ideal.

They do, in a way. Because physical interactions are rarer, they flow more readily and experience less back pressure. The end effect of this is an increase in the speed of sound, and thus they don't get choked until much higher velocities.

The speed of sound, BTW, is higher at lower pressures.

Originally Posted By: kallog
Also not sure why it really has to go at the speed of sound. I think that's only an upper limit when you have way too much force pushing it. For a bottle rocket I bet there's no speeds of sound happening.

Absolutely! The speed of sound is simply the maximum speed that a gas will travel at through the narrowest part of whatever it is flowing through. If you have insufficient force, too much friction, etc, you will not get to the speed of sound.

However, in pauls device you will reach the speed of sound quite readily. His ~6000N of pressure force is more than ample to accelerate air to that speed.

Bottle rockets are not a really good comparison though - your expelling water (a liquid) not a gas, so the physics are completely different. That said, flow of a liquid can be choked (and is choked at the speed of sound). However, the speed of sound in water is quite high (~1500m/s!), and I highly doubt a water bottle could support the pressure you would need to achieve that speed.

Originally Posted By: kallog

I tend to dislike engineering equations (despite using them every day ;), because of exactly these problems. They're usually simplifications that only apply to special cases. No room for doing your own thing, and very bad for extreme thought experiments.


I don't think it would be fair to call these equations "engineering equations". They are the physical equations which describe the forces involved in the flow of a non-ideal gas. Engineers use them, not because they are simplified versions of reality, but rather because they are accurate representations of reality. You'll see those equations used throughout the physical sciences as well - not because they are simplified, but rather because they describe realty.

Originally Posted By: kallog
Perhaps you should tell Paul directly that Ve isn't zero anymore wink


I get the feeling he doesn't read my posts anymore...

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 04:11 PM

Quote:
If he kicks off with a tiny force, then he spends a long time floating back down. So the pipe keeps on moving backwards slowly for a long time. In the end, the distance (velocity * time) is the same and it's all back where it started.


yes but if there are other people climbing up the ladder , the
tiny force that he applies will be overcome by the larger forces.

so if you have 1 person stopping at the end and 20 climbing
and 20 floating back.

20 larger forces -------------------->
and 1 tiny force <-
and 1 person stopping<--

so you have the force applied by 20 people
vs the -force of 1 person as he reaches the end plus the tiny force he applies to return , and the tiny force cancels out when he gets there.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/10/10 06:24 PM

the pipe idea kind of reminds me of the cigar shaped ufo's
that are reportedly seen.

and it seems that all other ufo's are saucer shaped , which
could lend itself to possibilities of centrifugal forces in a centrifuge.


none of them seem to be sphere shaped that I have seen photographs of.

even though a sphere shape would seem to be best for space travel.

the German Bell


I believe that according to its shape the Bell was driven by a
hydrogen turbine that powered a centrifuge and its exhaust gasses were trapped in the centrifuge.

it is reported that the Bell had counter rotating barrels one inside the other , this would provide for stability.

but these are just my thoughts on the German Bell and what
purpose it could have served.

I believe that this was a working model for testing a propulsion device that was being developed by the nazis.

that was used in several nazi aircraft.







Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/11/10 03:41 AM

Originally Posted By: paul
so if you have 1 person stopping at the end and 20 climbing
and 20 floating back.


That's not sustainable. If they're climbing faster than they're floating back down then you'll run out of people to climb. They'll have to wait for new ones to finish floating back down. The average rate of people/minute will have to be the same in both directions.

And it still violates the law of conservation of momentum, as well as Newton's 1st law!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/11/10 04:10 AM

I didnt say that , naturally you would have the same number climbing up as is floating down.

heres what I said:
Quote:
yes but if there are other people climbing up the ladder , the
tiny force that he applies will be overcome by the larger forces.

so if you have 1 person stopping at the end and 20 climbing
and 20 floating back.

20 larger forces -------------------->
and 1 tiny force <-
-----
Im adding this part
and 1 tiny force ->
and 1 person stopping<--


it doesnt violate any laws..why would you say it does.

and even if it did violate a law , whats your point.

laws are not binding , they are disposable if proven wrong or
they can be changed to fit.


and if the larger version I put up is used:
and the same number climbing to zero gravity as is climbing down to gravity at the ends of the pipe.


the really neat part about this is that you could use electrically driven weights that are riding on a track or inside a tube driven with electromagnetic forces , the weights could be turned 180 degrees by the tube and they could float back into the tube at the other end.

no people involved just weights and electricity.

thousands of tubes could line the outside of a cylindrical shaped ship.
with a steady flow of weights being moved and creating a force
in the direction of desired movement.

then the tubes all merge in the center where gravity is zero.

so they can float back.

you could build up a enormous acceleration this way in a short time.

depending on the number of weights you use.
and the force that you apply to move the weights.

Im sure there are thousands of ways to get propulsion from
this same principle.












Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/11/10 08:05 AM

Originally Posted By: paul
I didnt say that , naturally you would have the same number climbing up as is floating down.


Please try to think about this and explain it more clearly. I already know it can't work so there's no way I can interpret your vague statements. If you spell it out step by step you'll discover where the fault is yourself.

This is mundanely obvious and doesn't depend on any physical laws, except the law of conservation of people. I suppose nobody magically appears out of nothing or magically evaporates.


Quote:

and even if it did violate a law , whats your point.

My point is what you consistently fail to realise. If you're right you'll be RICHER THAN GOD!!

Put it into perspective. This is a trivially simple idea that millions would have thought through before. Many of those people would be kids who haven't learnt Newton's laws yet. You cling to the arrogant idea that everybody else is wrong simply because they have common ideas. Don't you wonder why you're consistently wrong about every single physical-law-violating idea you've ever had? Don't you notice the mistakes you keep on making?


Quote:

the really neat part about this is that you could use electrically driven weights that are riding on a track or

The really neat part about magic carpets is you could just sit on them and fly around the world.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/11/10 03:58 PM

well , you know talk is cheap , and you can write anything.

but just by writting something doesnt mean anything , it is
only showing that that is what your opinion is on this subject.

you havent shown where any laws have been broken , and
everything that this concept involves has already been
agreed upon by yourself.

not that your agreement would establish if the concept would or would not work.


Quote:
you consistently fail to realise. If you're right you'll be RICHER THAN GOD!!


well from my point of view you consistently fail to realize
that you are basing the feasibility of the concept on laws
that I am showing are not being violated.

yet you wont address these supposed violations that you claim.

and why would God need money?

just because YOU dont want it to work will not cause it to not work.
------------------- think ----------------------

this is a simple thought experiment that is easy to follow.

let us replace the people with weights.

in a single moment the forces are as follows:
lets give the weights a mass and velocity.

if each weight weighs 100 kg
and each weight has a velocity of .5 m/s
to find the force needed to achieve the .5 m/s velocity.

F = m*v
F = 100 kg / .5 m/s = 50 kgf

we want to maintain a constant weight velocity of .5 m/s
so that this experiment will be easier to follow.
so at any given time any weight moving around inside
the pipe will have a velocity of .5 m/s...

when the weight is propelled by a force half of that force is applied to the weight and half is applied to the pipe.
because the pipe has nothing restricting its movement as it
is in space in zero g.

so
25 kg F goes into moving the weight.
and
25 kg F goes into moving the pipe.

result >>>>>>>>>>>> 25 kg >>>>>>>>>>>>> weight

result <<<<<<<<<<<< 25 kg <<<<<<<<<<<<< pipe

20 weights pushing against the pipe.

result = 25 kg * 20 = 500 kg ---------> weight
result = 25 kg * 20 = 500 kg <--------- pipe

lets give the pipe a mass.
the mass of the weights will not be counted as they
are moving inside the pipe independently of the pipe.

suppose we allow the pipe to weigh 500 kg

a = F/m
a = 500 kgf / 500 kg
a = 1 m/s^2

as in physics laws such as newtons 1st,2nd,and 3rd.

everything above is well within the confines of these
3 laws.

---------------
now for the forces that might cause the pipe to not move.

suppose the weights are rideing inside a tube that has
180 degree turn arounds at each end.

when one of the weights reaches this turnaround inside the tube at the end of the pipe , a force must be applied to
turn the weight or reverse its direction 180 degrees.

this can be done easily as in a railgun where magnetic fields
propel a mass in a direction.

as in the below video.

homemade rail gun

the mass or weight is already traveling at .5 m/s
when it enters the turnaround.

so by the time it completes the turnaround it will have lost some of its velocity.

so we will need to supply a force to accelerate the weight to
the required velocity of .5 m/s

for the sake of ease we will say it has lost all of
its velocity and we need to fully accelerate it.

F = m*v
F = 100 kg / .5 m/s = 50 kgf

this force will propel the mass or weight to the other end of
the pipe where it will enter the 180 degree turnaround at
the other end of the pipe.

the mass or weight will still have the same velocity that it
had when it left the tube at the other end and now it enters
the tube again.

so we will need to supply a force to accelerate the weight to
the required velocity of .5 m/s

for the sake of ease we will say it has lost all of
its velocity and we need to fully accelerate it.

F = m*v
F = 100 kg / .5 m/s = 50 kgf

once the weight exits the turnaround it needs no acceleration , its velocity only needs to be maintained.

and this velocity is maintained by the primary rail gun
that is used to accelerate the weights.

at any moment there is a weight inside each turnaround
and the two forces in the turnarounds cancel each other out.

as in newton #3

the total forces involved no matter what you think are as follows.

total force applied to move the weights.
20 weights * 100 kg = 1000 kg

result = 500 kg ----- pushing the -----> weights
result = 500 kg <---- pushing the ----- pipe

acceleration of the pipe is.

a = F/m
a = 500 kgf / 500 kg
a = 1 m/s^2

so the pipe is traveling through space with a acceleration of

1 m/s^2


note: a magic carpet would be nice if they existed.

if you want to achieve a higher pipe velocity you only need to
increase the velocity of the weights.

if you want to slow or stop the pipe , you only need to stop the weights and reverse their direction.

because of this the pipe will need to be long enought to
allow for this , in other words there will need to be a substaintial distance between the weights so that collisions
do not occur.

suppose you have reached your desired velocity and want to stop the weights in order to prevent the weights that are floating from moving around you would need a opposing rail gun that will attract and hold the weights in the stopped
possition or you can store them in another place.

suppose you have traveled to mars and want to decelerate to obtain a orbital velocity around mars.

you would decrease the electricity to the rail gun.
then using the opposing rail gun stop the weights from moving.

then reverse the direction of the weights sligltly.

then remove the opposing rail gun , then slowly accelerate the weights in the opposite direction.

this should cause the pipe to slow down or even stop if desired.

suppose you are traveling to orion you would increase the electricity to the rail gun.

suppose you have accelerated at .5 m/s for a time period and have gained a velocity of 186,000.00 miles per second

to avoid going faster than the speed of light YOU could stop the weights from moving by using the opposing rail gun.

since you are stopping weights moving in both directions
there will be no effect on the pipes momentum.

and you would stop accelerating.

because you probably dont think that anything can
travel faster than 186,000.00 miles per second.

or I could keep on truckin in order to get there in time
for supper.


for those who write in this forum and edit in this forum you
can get a copy of autohotkey and with it you can write a script that will allow you to expand the message to be edited
simply by holding down the mouse button instead of pressing it 50000 times , or maybe the forum can just add a expand message button.








Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/11/10 08:00 PM

Originally Posted By: paul
well , you know talk is cheap , and you can write anything.

but just by writting something doesnt mean anything , it is
only showing that that is what your opinion is on this subject.

you havent shown where any laws have been broken

Actually, kellog and I have been showing exactly which physical law you've been breaking since day 1 - momentum.

Your pipe is a closed system, and momentum is conserved in a closed system.

From wikipedia:

Momentum is a conserved quantity, meaning that the total momentum of any closed system (one not affected by external forces) cannot change. Although originally expressed in Newton's Second Law, it also holds in special relativity

To increase the speed of your pipe - i.e. use it as an engine for a space craft - you need to change its momentum. This is impossible using forces generated within the closed system (i.e. within your pipe).

In order to get a change in momentum of your pipe you must either:

1) Act on it from the outside - i.e. apply a force to the outside of the pipe, or

2) Open the system (i.e. open your pipe) to the outside environment - in other words, make it an open system.

The former requires an outside energy source/source of force. The later turns your device into an overly complex rocket engine.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/11/10 08:11 PM

Quote:
Actually, kellog and I have been showing exactly which physical law you've been breaking since day 1 - momentum.


Actually , I dont recall where you or Kallog have

SHOWED that.

so show it , if you have it to show.

Quote:
1) Act on it from the outside - i.e. apply a force to the outside of the pipe, or


thats stupid , the law states that :

First Law: Every body will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it.

where does it state external forces?
forces impressed on the pipe from within the pipe are still forces impressing on the pipe.

In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.

where does the pipe exchange matter?

Quote:
Momentum is a conserved quantity, meaning that the total momentum of any closed system (one not affected by external forces) cannot change. Although originally expressed in Newton's Second Law, it also holds in special relativity


heres the law , it doesnt say anything about a closed system.

Second Law: A body of mass m subject to a force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma. Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body.


in fact here are all three laws , none of which have the words CLOSED or SYSTEM or EXTERNAL in them.

First Law: Every body will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it.

Second Law: A body of mass m subject to a force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma.

Third Law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force -F on the first body. F and -F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and -F the "reaction".


I used to see those words all the time also , so perhaps
now would be a good time for you to update your word inventory as those words were probably oil influenced
through donations or bribery.



get your laws straight or you could get a ticket.


UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/12/10 01:21 AM

Originally Posted By: paul

Actually , I dont recall where you or Kallog have

SHOWED that.

It was first brought up in post #34667 by kellog, then:
#34673 - by me,
#34804 - by me,
#34827 - by me,
#34908 - by me,

But the best - YOU AGREED IN POST #34823:
I agree that momentum is conserved also

More to the point, from wikipedia:

The law of conservation of linear momentum is a fundamental law of nature, and it states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant. One of the consequences of this is that the center of mass of any system of objects will always continue with the same velocity unless acted on by a force from outside the system.


Originally Posted By: paul

thats stupid , the law states that...

I'm hoping you're joking - you don't honestly believe the 1-sentence rules we use to make it easy for kids to learn the laws are the entirety of the laws? Newtons Principia mathematica - where he described the laws - is over 500 pages long (at least my copy is...).

More to the point, you've cherry-picked your translations to ones which leave out the details you are trying to avoid. Take for example, these definitions of the first law:

-Unless acted upon by a net external force, a body, at rest, will remain at rest and a body, in motion, will remain in motion.

-If net external force on a body is zero, then its velocity remains constant.

And of course, Newtons original:
Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.

The key word in that being impressis, latin for "to act upon" - note its "upon", as in from the outside - not "ab intra" (from within).

The conservation of momentum is a direct consequence of Newtons laws - this is a well accepted fact in physics, and your denials do not change that.

As we've been saying since day 1 - if you think it'll work, build it, patent it, make your millions and get your nobel prize for overturning newtons laws.

Originally Posted By: paul

UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA

that's how the song goes wink

Here's another one for you:
AUUUCUUCUGCUUGUUAAAUCAGCUAACGUAUUGGUCAUACUUAACCUGCUCUUUAAAUCAGCUAAAUGAUUUCUACUGCUAAAGAAAAU

And here's a clue (use compact).

Bryan
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/12/10 03:03 AM

Originally Posted By: paul

just because YOU dont want it to work will not cause it to not work.

I'd love it to work. But it won't, and I don't much like chasing goals that are certain failures. If you want to take that gamble, then go build it with some sewer pipes and skateboards, it's very cheap and easy. Then collect your Nobel prize. This isn't just a curiosity, this is EARTH SHATTERING!!!



Quote:

F = m*v

Oops, don't you mean F=ma? I prefer not to dig through all the rest of the calculations trying to fix up the consequences of this.

It's very handy to use impulse in these things. For a constant force applied over time t:
I = F * t
That happens to be equal to the magnitude of the momentum gained by each object. From momentum you can easily get their velocities.


Quote:

a = 500 kgf / 500 kg

Ouch. This horrible bastardization of units is why it's easy for me to stop bothering to follow. Sure you're probably correct but there's certainly simpler ways to do it. What on earth does a kgf have to do with a machine floating in space? There's no Earth gravity, no weight of a kg up there. Please use consistent units. Even imperial ones, as long as they satisfy equations like F=ma and v=d/t. Your above equation subtly has acceleration in units of g.




Quote:

result = 500 kg ----- pushing the -----> weights
result = 500 kg <---- pushing the ----- pipe



Could you please just step through it with numbered bullet points and simple headings to say what happens when? Is it something like this?

All speeds measured relative to the pipe.
1. Railgun accelerates a 20kg mass up to 1m/s
2-10. Same again for other masses
11. Other railgun decelerates the first mass to 0.
12-20. Same again for other masses
21. Same railgun accelerates the first mass to -1m/s.
22-30. Same again for other masses

???

Railguns have reaction forces pushing the pipe the opposite way. I didn't see that mentioned in your post.




Quote:

because you probably dont think that anything can
travel faster than 186,000.00 miles per second.


Are you trying to start another fight? Yes I do think that, and I also think we can all fly round on magic carpets. I'm right because I say so. You can't prove me wrong without relying on established knowledge that was fed to you by the evil jews who are controlling the world and its oil.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 04:52 AM

Quote:
-Unless acted upon by a net external force, a body, at rest, will remain at rest and a body, in motion, will remain in motion.


the pipe is itself a body.
if you are or are not capable of understanding that is your problem.

I have clearly shown all the possible forces involved and according to the forces
WHICH IS WHAT THE LAWS ARE ABOUT.

the pipe will move just as I described.

your clinging to your STUPID EXCUSSES with missinterpreted laws will do you no good.


BTW , I do agree that momentum is conserved also.

there is clear conservation occurring between the pipe and the weights.

the force that propels the weights is divided between the pipe and the weights.

the momentum is also divided between the pipe and the weights.

I cant help it if you cant understand that , its exactly what would happen in this situation
in zero g.






Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 05:39 AM

All speeds measured relative to the pipe.
1. Railgun accelerates a 20kg mass up to 1m/s
2-10. Same again for other masses
11. Other railgun decelerates the first mass to 0.
12-20. Same again for other masses
21. Same railgun accelerates the first mass to -1m/s.
22-30. Same again for other masses

???

Railguns have reaction forces pushing the pipe the opposite way. I didn't see that mentioned in your post.

well duhhh , thats the reason the pipe moves.

your little 1-30 above is not very well though out.

Quote:
1. Railgun accelerates a 20kg mass up to 1m/s


thats 100 kg and 5 m/s
Im ammased you missed that.
ie....
if each weight weighs 100 kg
and each weight has a velocity of .5 m/s
to find the force needed to achieve the .5 m/s velocity.

F = m*v
F = 100 kg / .5 m/s = 50 kgf
--------------------------------------------------------
Quote:
2-10. Same again for other masses

I suppose your are talking about the weights in the turnarounds.

and if for some stupid reason you are sudgesting that one weight is decelerated
while another is accelerated , you are wrong.
Quote:
11. Other railgun decelerates the first mass to 0.

nope , the other rail gun is only used to stop and hold the weights in place.
and to slightly accelerate the weights.

Im not sure that you read the post , because from your assesment it does not appear that way.
--------------------------------------------------------
Quote:
11. Other railgun decelerates the first mass to 0.

wrong again , you should take a word comprehension refresher course.
I said that the other rail gun is used to stop or hold the weights.
not to decelerate the weights at the same time as the other weights are accelerating.
its pretty easy to see what I wrote.

--------------------------------------------------------
Quote:
12-20. Same again for other masses

nope the opposing weights are floating.

--------------------------------------------------------
Quote:
21. Same railgun accelerates the first mass to -1m/s.

wrong again.

--------------------------------------------------------

its pretty clear to me that neither of you have much mechanical sence
about you.

Quote:
a = 500 kgf / 500 kg

Ouch. This horrible bastardization of units is why it's easy for me to stop bothering to follow. Sure you're probably correct but there's certainly simpler ways to do it. What on earth does a kgf have to do with a machine floating in space? There's no Earth gravity, no weight of a kg up there. Please use consistent units. Even imperial ones, as long as they satisfy equations like F=ma and v=d/t. Your above equation subtly has acceleration in units of g.


that is terrible isnt it , its exactly how it works in zero g.

notice I did not include resistance.

a mass in space is the same as a mass on earth , only you can push it
with less force because there is zero resistance to movement in zero g.

the only difference betweem a mass on earth and a mass in zero g is
gravity.

it still has the same mass.
so it requires the same force.

do you think you can push a 100 kg mass on earth with a 1 kg force.

no because you have to first overcome the inertia of the mass to get it to
start moving.

because of the resistance.

no such bastardization in zero g , sorry mr know it all.

your wrong again.











Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/12/10 07:21 AM

Originally Posted By: paul
your little 1-30 above is not very well though out.

As I said, I didn't bother to follow your complicated working. That was an example of how I'd like you to describe it so it's easy to see. You're asking me to do a lot of work to figure it out, including correcting mistakes along the way. Why not put more effort in yourself? Before guessing an equation, look it up in a text book or on Wikipedia. It's also much easier to do everything with symbols and no numbers.


Quote:

thats 100 kg and 5 m/s
Im ammased you missed that.

As I said, I didn't bother to follow your complicated working.

Quote:

and if for some stupid reason you are sudgesting that one weight is decelerated
while another is accelerated , you are wrong.

As I said, I didn't bother to follow your complicated working.


Quote:

Im not sure that you read the post , because from your assesment it does not appear that way.

As I said, I didn't bother to follow your complicated working.


Quote:

do you think you can push a 100 kg mass on earth with a 1 kg force.

Despite your insults, it's clear you haven't got the faintest idea about mechanics. That's especially easy for a boat floating on calm water, where friction goes towards zero at low speed.


Quote:

no such bastardization in zero g , sorry mr know it all.

your wrong again.

Paul, I've put up with your ongoing insults for a long time. I usually quietly ignore them, but it's getting a bit too much. Please behave more courteously. I'm sure you understand that insults don't make people understand you, rather they polarize people against you.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 08:11 AM

Kallog

you and byran can discuss this between yourselves , Im sure that
between the two of you ,you can arrive at a completely asinine
conclussion.

and as for insults , just reading your post is insulting.

you two come on like your the smartest people in the world but
underneath the facade the only thing I see is two people who
seem inteligent but have a extreme lack of cognitive powers.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 04:29 PM

Originally Posted By: kallog
As I said, I didn't bother to follow your complicated working.


So its normal for you to just simply state your uninformed opinion
on subject matter.
even if you havent read the information presented , you automatically
follow with your opinion of the information that you havent read.

Originally Posted By: paul
do you think you can push a 100 kg mass on earth with a 1 kg force.


Originally Posted By: kallog
Despite your insults, it's clear you haven't got the faintest idea about mechanics. That's especially easy for a boat floating on calm water, where friction goes towards zero at low speed.


given that I was refering to the frictionless environment of zero g
what you posted has pretty much confirmed my assessment of your cognitive abilities
and your mechanical abilities including your knowledge of mechanics.

you used a boat in water -- low friction -- when I was refering to a mass
sitting on the ground -- on earth.

ie...the resistance to movement of a object sitting on earth due to gravity.

Originally Posted By: paul
notice I did not include resistance.

a mass in space is the same as a mass on earth , only you can push it
with less force because there is zero resistance to movement in zero g.

the only difference between a mass on earth and a mass in zero g is
gravity.

it still has the same mass.
so it requires the same force.

do you think you can push a 100 kg mass on earth with a 1 kg force.



I suppose you read the last line of what I posted above and constructed
your anology from that single line.
as you have already informed us that you dont fully read information you
just breeze through it.

Brilliant not even slightly smart , and common among naysayers.










Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 05:21 PM

one more thing that I would like to tell the others that are
following this thread.

at any given time durring acceleration of the pipe.

1) there are 20 masses being accelerated.

being accelerated means that a opposite force is being placed
on the pipe that will give acceleration to the pipe.

2) the opposing 20 masses are not being accelerated.

nor are they decelerating.

3) the forces of the two masses that are being turned around by the tubes (2) 180 degree turnarounds will cancel each other out.

conclusion:

at any moment durring the acceleration of the 20 masses there is a force being placed on the pipe in a direction.

at any moment durring the acceleration of the masses there
are no counter forces being placed on the pipe that could
prevent the pipe from accelerating.

therefore the pipe will accelerate in a direction.

a = F/m

---------------------------------------------------------

thats pretty much as clear as I can put it.
and its simple enought for a grade school student to understand.

if the two seemingly highly intelligent naysayers that have
been involved in this discussion with me dont realize this
then that only shows that their ability to process information
has been flawed or never was there.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/12/10 07:15 PM

Originally Posted By: paul
Quote:
-Unless acted upon by a net external force, a body, at rest, will remain at rest and a body, in motion, will remain in motion.


the pipe is itself a body.
if you are or are not capable of understanding that is your problem.


We all agree it is a body - but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal

Originally Posted By: paul
I have clearly shown all the possible forces involved and according to the forces


No, you've steadfastly ignored newtons laws, in particular newtons 3rd law.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 07:37 PM

Quote:
but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal


it doesnt matter where the force comes from , if it comes from inside the pipe or outside the pipe.
its pretty stupid of you to think that it does.

below I have underlined the word "you" so that
others wont get the opinion that "I" have the same understanding as bryan who just happens to be the "you"
below.

if you were sitting inside a large can on earth
that is completely sealed.

then the large can suddenly was in deep space.

and you stood up inside the can.

according to your understanding of physics the large can would not move as you stand.

I say the large can moves away from you , because of the force that you applied to the large can while standing , and smashes into your head.


if you were outside on the top of the large can and stood up the large can would move according to your understanding of physics.

I say the large can moves away from you , because of the force that you applied to the large can while standing.


you cant just have it your way , bryan.

its basic physics , rock bottom physics , grade school physics.


now I suppose you will reply once again saying that the large can would move as you have repeatedly done in the past , contridicting yourself once again and sinking further into the mire that you created for yourself.


Quote:
No, you've steadfastly ignored newtons laws, in particular newtons 3rd law.



you are the one who is ignoring newtons laws , bryan.

you are ignoring #1 #2 and #3 of newtons laws of motion.

by insisting the pipe would not move.
you try to use newtons laws but they are flawless and
cannot be used to support your missinterpretations of them.

Quote:
but note you need an external force.


no I dont , you do , I know better.

Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/12/10 08:43 PM

Originally Posted By: paul
Quote:
but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal


it doesnt matter where the force comes from , if it comes from inside the pipe or outside the pipe.
its pretty stupid of you to think that it does.


LOL, this from a guy whose continually shown a complete ignorer of newtons laws, and who thinks homeopathy is anything but a scam.

Internal forces are always balanced - equal and opposite as per newtons 3rd law. Ergo, external forces or the ability to impinge a force on the external environment are an absolute requirement for changes in momentum.


Originally Posted By: paul

if you were sitting inside a large trash can on earth
that is completely sealed.

then the trash can suddenly was in deep space.

and you stood up inside the can.

according to your understanding of physics the trash can would not move as you stand.


Nope, that is not what I said, not even close.

A trash can is open - I'd knock the lid off and fly out the top; the trash can would go the other way with equal momentum.

But if the trashcan was sealed, and large enough for me to stand, then when I stand the can would go one way, I the other. When I hit the top of the can the resulting force would cease all of the movement. End effect - can shifted one way, myself shifted the other, but the net center of mass exactly where it was when I started and the total momentum unchanged.

Originally Posted By: paul

I say the can moves away from you , because of the force that you applied to the can while standing.


But in the case you describe, the can is an open system - exactly what kellog and I have been telling you is needed since day 1.

If the lid were welded on - i.e. the trash-can equivalent of your pipe - the can wouldn't go anywhere. No changes in momentum are possible without external forces.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 09:00 PM

you said
Quote:
but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal

I said
Quote:
now I suppose you will reply once again saying that the large can would move as you have repeatedly done in the past , contridicting yourself once again and sinking further into the mire that you created for yourself.

you said
Quote:
But if the trashcan was sealed, and large enough for me to stand, then when I stand the can would go one way,

I said
Quote:
if you were sitting inside a large can on earth
that is completely sealed.


you said
Quote:
But in the case you describe, the can is an open system - exactly what kellog and I have been telling you is needed since day 1.


so completely sealed actually means open , correct.
just like the back of a car is the front of a car , correct.

your really expressing your inteligence , bryan.

so when you are inside the can that means you are outside the can , am I following your pattern of thought corectly , bryan?

ie .. is internal actually external , bryan.


you said

Quote:
but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal


is this how you can have it both ways , bryan.

by simply reversing the meanings of words.

is this what happened to newtons laws , add a little here
subtract a little there , reverse a meaning here and there.

soon you will have it just the way you want it , correct.







Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/12/10 11:24 PM

Originally Posted By: paul
your really expressing your inteligence , bryan.

LOL. This coming from the guy who had to crop my post to make it say waht he wanted to - I described an open and closed garbage can.

Cropping the closed garbage can doesn't change that reality.

And don't think your attempt to re-write what I said made us miss the fact that you were completely unable to refute, or even comment on, the difference between the open verses closed garbage can...

What's the matter paul, cannot make your "point" without altering what people have said? Or is this just what you do when faced with a simple, logical answer that completely demolishes your "hypothesis"?

Originally Posted By: paul

Quote:
but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal


is this how you can have it both ways , bryan.


Nope paul, had you not dishonestly removed part of my post it all would have been consistent. The fact you'd openly try to distort what I wrote - when its all immediately above your post - speaks volumes about your duplicity.

Internal forces in closed systems - i.e. you jumping around in a sealed garbage can - cannot alter momentum of your inertial frame (i.e you + garbage can). To alter the momentum of your inertial frame you need either:

a) an external force, or

b) an open system, so your internal forces can act on the external environment

As I described in my last post...and opposite of how you tried to re-write what I said.

Originally Posted By: paul

is this what happened to newtons laws , add a little here
subtract a little there , reverse a meaning here and there.


Yeah, that's a fair description of what you did. Removed those inconvenient little bits where the laws say things like "external forces".

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/12/10 11:29 PM

Quote:
And don't think your attempt to re-write what I said made us miss the fact that you were completely unable to refute, or even comment on, the difference between the open verses closed garbage can...


why should I refute or comment on something that I didnt say?


Quote:
Nope paul, had you not dishonestly removed part of my post it all would have been consistent. The fact you'd openly try to distort what I wrote - when its all immediately above your post - speaks volumes about your duplicity.


I suppose the below is the "cropping you refer to"


Quote:
We all agree it is a body - but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal



and this is the part that I cropped.


Quote:
but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal


so this is the part I cropped out of the above

We all agree it is a body -

and I cant see where that would make a big difference anyway.

of course , at least I agree , not so sure if you agree because according to newtons laws a body acted upon by a force will move.

and you have constantly dissagreed with that and so Im not
sure that you agree its a body.

you keep saying the pipe wont move , dont you , should I go back and collect every time you said that?











Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/13/10 12:14 AM

#34662

1)

Doesn't work that way - material thrown within the ship will encounter resistance with the air, hull, etc. This will generate a force equal to the force of the propellant, thus neutralizing the thrust of the propellant. Its the ol' opposite and equal reaction thingie - the movement of the propellant will "push" on the ship, but the interaction of the propellant with the ship will push back equally. Net effect - zero thrust.

These kinds of internal energy transfer system only work in places where there is friction to counter the unwanted "return" energy of the propellant. Basically, you can thrust in one direction using a lot of force quickly, pushing the object forward. You then recover your propellant slowly, so the force of the propellant moving in the "wrong" direction doesn't exceed the static friction holding you in place.

That doesn't work in space - no friction.

2)

Doesn't work that way, for the reasons mentioned above. In space, the net thrust of this kind of system is zero. And even in places where it is possible, the ISP would suck.

3)

Only if the contents of that hose are free to leave the ship. If they are not, the momentum of the air will be transfered back to the ship, providing a net zero thrust.

4)

Open the system though - eject that air out of the ship itself - and you'll move along quite nicely.

5)

Because that is the problem with your own model. Your space ship is a closed system; matter can neither enter nor leave. You can exert forces within such a system, but the sum of those forces will always be zero at the level of the system itself

the above are from page 2 and 3

I dont want to spend the rest of the night showing you your insistence that the
pipe wont move.


Im certain I can find plenty in the other 13 pages however.

in my personal opinion , I think , I believe that you need
to learn how to assess a situation , and adapt to it , adapt to the changes.





Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/14/10 11:09 AM

Hello again Paul, I've taken a forced leave of absence due to my failure to appreciate a cultural taboo of western society concerning a specific group of 'chosen' people wink But coincidently it seems all you guys suffered a similar fate, so haha :P


Originally Posted By: paul

at any given time durring acceleration of the pipe.

1) there are 20 masses being accelerated.
...
2) the opposing 20 masses are not being accelerated.
...
3) the forces of the two masses that are being turned around by the tubes (2) 180 degree turnarounds will cancel each other out.


Yep that's much clearer, cheers. I think, let me try to reconstruct it, correct me if I'm wrong:

There are 20 masses being accelerated along the length of the pipe. The reaction to this is what accelerates the pipe.

At the same time 20 other masses are free-floating in the opposite direction.

Both sets of masses reach their ends at the same time, where they're both turned around by a 180deg curved track/etc. The reactions from these two turn-arounds cancel each other out.

Repeat indefinately.



Quote:

if the two seemingly highly intelligent naysayers that have

Cheers! Not everyday that somebody finds I seem highly intelligent smile
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/14/10 11:19 AM

Originally Posted By: ImagingGeek

We all agree it is a body - but note you need an external force. Whether you're talking about a tank of air, a gun, or guys running up & down ladders, all of those forces are internal


Careful there! Nature doesn't care if the ladder's bolted to the inside of the tube or to the outside.

It all depends what you define as the 'body'. If you say it's the pipe, then yes, Paul's right, it can be accelerated by things inside it, they're not the pipe, they're other things. You run into problems if you try to move the pipe further than you move the contents tho.

If you define the 'body' as the collection of the pipe and its contents, then conservation of momentum applies to the center of mass of the collection of things.

Well I'm sure you knew that, but getting careless is no way to sort out a disagreement!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/14/10 02:04 PM

Quote:
There are 20 masses being accelerated along the length of the pipe. The reaction to this is what accelerates the pipe.

Yes , the rail gun is accelerating 20 masses at any given time, and as a result of the force applied to accelerate the 20 masses in one direction , the pipe is accelerating in the opposite direction.
and accelerating 20 masses requires 20 forces.
and these 20 forces cause the pipe to accelerate in the opposite direction according to newtons laws of motion.

Quote:

At the same time 20 other masses are free-floating in the opposite direction.


Yes, because these 20 masses are not being accelerated
they will not counter the acceleration of the 20 masses occuring in the
rail gun.

according to newtons laws of motion.

Quote:

Both sets of masses reach their ends at the same time, where they're both turned around by a 180deg curved track/etc. The reactions from these two turn-arounds cancel each other out.


Since the masses being accelerated will be traveling slightly faster than the masses that are floating the other way , there will be a slight difference in the timming that the individual masses reach the 180 degree turnarounds.

the individual masses that are being accelerated in the rail gun will reach the turnaround slightly faster than the individual masses that are floating to the other turnaround.


Quote:


Repeat indefinately.


Yes , or as long as you need to accelerate the pipe.

so you have a continous force in a single direction
that applies to the pipe causing the pipe to continously accelerate.

according to newtons laws of motion all the above is well within his constraints.

this could be used in a space ship for propulsion.
as long as you have a ample power supply to power the rail guns.
Posted by: Momos

Re: Orion, Mission to Alpha Centauri - 06/14/10 08:53 PM

It took me quite some time to read all your posts smile
In the end I have to say: paul, you are wrong. The explanations of kallog and ImagingGeek are perspicuous. And have been from the very beginning. Maybe you should cool off some time and carefully read them again.


@ImagingGeek
You wrote some interesting stuff I didn't know, especially about the back-pressure and maximum exhaust speed, thanks!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/14/10 10:36 PM

Quote:
In the end I have to say: paul, you are wrong.


well , you've said it.
and everybody has a right to their own opinions.

after all your opinion doesnt matter to me , some may value
it , but I dont.

perhaps you have something perspicuous to base your opinion on.
certainly its not a physical law as the concept adheres to
physical laws.

of course if you dont have any input other than just saying
that I am wrong , then no one will have any proof that I
am wrong , including yourself.

until then , I can just say that you dont exist.
therefore I shouldnt concern myself with what you think.

you may have a different opinion on that , but I can
still say it , think it.

and your opinion or my opinion will not decide if the concept is or is not valid.

but my opinion of your opinion is that you are wrong !!




Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/15/10 03:05 AM

Originally Posted By: paul

the individual masses that are being accelerated in the rail gun will reach the turnaround slightly faster than the individual masses that are floating to the other turnaround.


It's crucial to quantify that in case 'slightly' is actually 'too much'.

- The pipe starts with momentum
pp = 0
- An accelerated mass starts with momentum p1.
- After acceleration at force F for time t it ends up with momentum p1+F*t. The reaction has the opposite effect on the pipe,
pp = 0-F*t
- After turning around 180deg, its momentum in the free-floating stage is -(p1+F*t). This is a change in momentum of -2*(p1+F*t), so the reaction on pipe causes the opposite change in momentum. The pipe gains 2*(p1+F*t),
pp = 0-F*t+2(p1+F*t)
- But at the same time a free-floater is turning around at the other end, imparting momentum -2*p1 onto the pipe.
pp = 0-F*t+2(p1+F*t)-2*p1
Then that just-turned-around free-floater starts to accelerate, imparting -F*t to the pipe during it's travels.
pp = 0-F*t+2(p1+F*t)-2*p1-F*t

There's could also be a slight error because I did everything in the frame of the pipe's initial rest state, but assumed no pipe movement.

This assumption would be valid if the pipe hardly moved at all, say if it had a really huge mass. That's fine, even a huge mass will keep drifting through space given a little momentum.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/15/10 04:32 AM

your right , and its going to require a program to get the slightly just right.

at least I would rather do it that way to be more precise.

because the 20 masses being accelerated (M1 - M20) will
be moving slower at the starting point of acceleration.
and faster at the ending point of acceleration where they enter the turnarounds.

this will set up a gradient between the turnaround exits and entrances.

I think it would be easier to just let a program do all the calculations from pipe P=0 to desired acceleration.

I could make the program build a text file as it runs the calculations.

and then post the text file.


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/15/10 08:39 AM

Go for it.

But if the program says there's motion further than the length of the pipe, what then? I don't want to decipher code to look for bugs.

If the program says there's no long term change in the pipe's velocity, will you still post the results? What will your next step be?
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/15/10 04:09 PM

Originally Posted By: Momos
@ImagingGeek
You wrote some interesting stuff I didn't know, especially about the back-pressure and maximum exhaust speed, thanks!


I only wish I had explained it better in the beginning...

It is an interesting problem when it comes to rocket design. I remember waaaay back in the mid-1990's when I built my first rocket engines, they constantly preformed well below my expectations. I spent the better part of a year thinking the problem was in my design/fabrication. Turns out I had used the ideal gas formulas, and missed out on the choked flow issue.

When I used the correct math, my engine performance was dead-on predicted.

Later designed however did have some design/fabrication flaws - those ones go "whoosh - BOOM", instead of under preforming. Basically expensive confetti = you screwed up...

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/15/10 09:30 PM




Quote:
Go for it.

But if the program says there's motion further than the length of the pipe, what then?


well , I cant think of any reason it wouldnt go further than
the lenght of the pipe.

what then?

NASA perhaps.


Quote:
I don't want to decipher code to look for bugs.


I would use visual basic for something this simple.

however you might get lost in the code that writes the info
to a text file , flat for excell.

Quote:
If the program says there's no long term change in the pipe's velocity, will you still post the results?


sure , why not.


Quote:
What will your next step be?


try to contact someone within NASA I suppose.
what would you do if you were certain it would work?
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/16/10 01:50 AM

Originally Posted By: paul

try to contact someone within NASA I suppose.
what would you do if you were certain it would work?


I'd make sure nobody steals my idea. To do that I'd have to continue the internet discussion in such a way as to make sure nobody believes me. Perhaps by swearing at everyone who replies :P

Then I'd build it myself. I'd pour all my savings into it, I'd borrow money if I had to. I'd learn whatever I needed to design and build it, even if that meant quitting work and enrolling in a course.

Afterwards, I'd either be richer than god, or poorer than an american auto-worker.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/16/10 04:21 PM

Quote:
I'd make sure nobody steals my idea.


how long has magnetism using electricty been around?
surely Im not the first to think of this idea , sadly there
are probably thousands of others.

others who followed the supposedy correct path , keeping the
idea a secret to keep someone from stealing it.

and they most likely ended up with a gag order or even worse.

and just think of the trouble I've had trying to get you and bryan to see its possibilities.
and then think of the trouble I would have trying to get people at
NASA or some other institution to see its possibilities.

because they think the same as bryan does.
book thinking not common sence thinking.

if something seems to violate a law , then it will not work
because of the way they understand the laws or the way they
have been taught to understand the laws.

but I think this would work well in a submarine , do you see
the implications involved there.

you would need no control surfaces or propeller on the submarine
thus extreme silence accompanied by extreme speed.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/16/10 09:45 PM

Originally Posted By: paul
and they most likely ended up with a gag order or even worse.

No, you have no evidence for that occurring. The freedom of sharing information on the internet makes such suppression impossible these days, it's not like the middle ages anymore.


Here's a thought, it violates:
Netwon's 1st law
Newton's 2nd law
Newton's 3rd law
The law of conservation of momentum

That means if you use any of those laws in the calculation, and it results in the behaviour you've predicted, then the result is automatically wrong.

It may mean you've disproven those laws, by showing them to be internally inconsistent. In that case the disproof only needs to be theoretical. There's no need to even build it. Tho of course you won't be able to know if it works or not without building it.

Again, please stop talking about applications for it. Everybody can easily see many amazing applications, just as we can all see amazing applications for magic carpets. It's called counting your chickens.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/16/10 11:27 PM

Originally Posted By: kallog

Here's a thought, it violates:
Netwon's 1st law
Newton's 2nd law
Newton's 3rd law
The law of conservation of momentum

Am I the only one who see's the irony here - paul's little "invention" fall foul of four of the most basic laws of physics. These laws have stood the tests of time, and been validated time and time again - take Noether's proof conservation of momentum.

But kellog and I are the ones who are "crazy" because we don't think his device won't work.

Someone needs to read up on the history of perpetual motion ...

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/16/10 11:56 PM

I give up , you two guys have a nice time dissagreeing with every post that gets posted here until nobody post here anymore.

but I still think you two have a great misunderstanding
of newtons laws , and I dont think that between the both of you you could glue two sticks together

so long everyone , and have a nice life.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/17/10 04:40 AM

So you realised that writing a program won't prove it works, but you still believe it would. Are you going to go and build it in private? Or spend the rest of your life frustrated at other people without actually finding out the truth (whatever it may be) for yourself?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 01:25 PM

Quote:
So you realised that writing a program won't prove it works, but you still believe it would. Are you going to go and build it in private? Or spend the rest of your life frustrated at other people without actually finding out the truth (whatever it may be) for yourself?


No !! I realized that you two guys contridict yourselves to the point that you dont even know what you think.

first you say the pipe will move , then you say it wont.

now you say it will only move the lenght of the pipe.
which is the most ridiculous thing I've ever heard.

theres nothing that will stop the pipe once it starts moving in
the concept that I have put up.

WHY WOULD IT STOP?

you cant give a reason , except that it would break a law.

then you say that it is breaking all three of newtons laws of motion , which is another one of the most ridiculous things I've ever heard.

you've never shown that it would break any of these three laws.
or any other law.

the only reason you or bryan have ever given is that it would break newtons laws , when the fact is that you two guys cant even comprehend newtons laws , and thats a fact , given what you two guys have shown.

I give up because you two dont know what your talking about so
why should I keep trying to explain something so simple to a couple of people that cant comprehend it.

its like talking to a recorded message that never stops spewing out its recorded ignorance.

Originally Posted By: imagegeek
But kellog and I are the ones who are "crazy"


I didnt say that , but Im glad you did , like stupid is stupid does.

ie... I never said you two were "crazy" besides some of
the smartest people in the world were considered "crazy" , I certainly wouldnt accuse you two of being "crazy".








Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/17/10 02:33 PM

Originally Posted By: paul

first you say the pipe will move , then you say it wont.
now you say it will only move the lenght of the pipe.
which is the most ridiculous thing I've ever heard.


Please quote where I said something contradictory. I do have a habit of sometimes understating important points, but I am careful to be precise.




Quote:

WHY WOULD IT STOP?

you cant give a reason , except that it would break a law.


I wrote out the equations showing it would stop in a recent message. If you evaluate the final expression, you'll find it says momentum=0. Feel free to point out any mistake I made.



Quote:

why should I keep trying to explain something so simple to a couple of people that cant comprehend it.


You know somebody else who can comprehend it?


This is a forum for science, not politics. In normal life with real world people you can usually get away with just repeating things and insulting people until somebody agrees. But scientific types demand reasons and evidence.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/17/10 03:35 PM

Originally Posted By: paul
Quote:
So you realised that writing a program won't prove it works, but you still believe it would. Are you going to go and build it in private? Or spend the rest of your life frustrated at other people without actually finding out the truth (whatever it may be) for yourself?

No !! I realized that you two guys contridict yourselves to the point that you dont even know what you think.

Really? Point out one example where we contradict ourselves.

Preferably an example where its a true contradiction, not you mis-interpreting our claims, or lying about them.

Originally Posted By: paul

first you say the pipe will move , then you say it wont.

now you say it will only move the lenght of the pipe.
which is the most ridiculous thing I've ever heard.

But it is the correct answer, paul, and it has been experimentally/functionally verified - the same principal is used by doo-dads on spacecraft (i.e. gyroscopes) to hold or change the orientation of the spacecraft without altering its net momentum.

Let's see if you can follow the bouncing ball:

1) Momentum cannot be changed by internal forces in closed systems.
2) The momentum of a closed system is measured from its center of mass.
3) Internal forces can shift mass around inside of a closed system.
4) The movement in #3 would change the momentum of the closed system, by moving its center of mass, therefore to not violate the conservation of momentum the "equal an opposite" reaction generated by the forces moving the mass creates an equal, but opposite movement of momentum (mass) in the opposite direction.

The net effect in your new system is simple - you move mass forward, and as a consequence the pipe moves backwards due to the equal, but opposite force induced by the movement of the mass forwards. The center of mass of the system is maintained, and thus the law of conservation of momentum is not violated.

Originally Posted By: paul
theres nothing that will stop the pipe once it starts moving in
the concept that I have put up.

laugh
Other than the laws of physics

Quote:
WHY WOULD IT STOP?

you cant give a reason , except that it would break a law.

Because the mass you've moved forward also has to stop. The force stopping the mass moving forward would also stop the movement of the mass moving backwards.

Basic laws of physics, paul - every reaction has an opposite and equal reaction. The ball you throw won't magically stop.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 05:47 PM

are you talking trash again , kallog like the below trash you posted.

Quote:
- The pipe starts with momentum
pp = 0
- An accelerated mass starts with momentum p1.
- After acceleration at force F for time t it ends up with momentum p1+F*t. The reaction has the opposite effect on the pipe,
pp = 0-F*t
- After turning around 180deg, its momentum in the free-floating stage is -(p1+F*t). This is a change in momentum of -2*(p1+F*t), so the reaction on pipe causes the opposite change in momentum. The pipe gains 2*(p1+F*t),
pp = 0-F*t+2(p1+F*t)
- But at the same time a free-floater is turning around at the other end, imparting momentum -2*p1 onto the pipe.
pp = 0-F*t+2(p1+F*t)-2*p1
Then that just-turned-around free-floater starts to accelerate, imparting -F*t to the pipe during it's travels.
pp = 0-F*t+2(p1+F*t)-2*p1-F*t

originally I thought you were only talking about the two
masses that are being turned around at the ends but from your
last post it seems that this must be where you atttempted
to show that the "pipe" would not move.

what a farce of physics...

explaining away 22 masses using only 2 masses.
but normal for your type of genius.

why didnt you add some numbers in the above genius?

you know like 100 kg accelerated at 5 m/s

for a distance of 500 ft then repeat the above

for the other 19 masses being accelerated.

these 20 forces add up to the forces applied to the pipe for movement.

then add up what little negative forces there are from the 2 masses being turned around.


in your poisoned mind you neglect that there are

TWENTY , 20 , MASSES being accelerated at any moment.

NOT JUST ONE --- NOT JUST ONE -- NOT JUST ONE.

but I suppose that is the only way you can show no pipe movement.

ISNT THAT REALLY IT , kallog.

the results are no matter how scrambled your brains are...

at any moment.

there are 20 forces being applied to the pipe that cause
the pipe to move.

and the pipe is not as stupid as you guys are so it dont know it needs to stop just to serve you two guys misinterpretation of newtons laws suddenly as soon as it has traveled the lenght of the pipe. LOL , sickening.

at any moment there are 20 forces that are moving the pipe
and the two forces at the ends would in no way cancel out the 20 forces that move the pipe.

suppose each mass being accelerated applies a force
of 25 kg to the pipe.

that means that each mass can only subtract 25 kg from the pipe.

there are 20 masses at any time supplying a force of 20 kg
so 20 * 25kg = 500kg

now when 1 reaches the end the most it can subtract is

25 kg.

result = 475 kg ---> pipe
25 kg <--- turnaround

so when 2 reach the end
result = 450 kg ---> pipe
50 kg <--- turnaround

see what wonders numbers can do.

and I didnt even use close numbers.

just simple grade school numbers that you two might understand.

when one of the masses moves through the 180 turnaround
do you think it alone will subtract more force than the 20 being accelerated towards the end?

just how dumb are you guys?

no way , the most that the one single mass could possibly subtract is the amount of force that would be required to stop it completely.

no tweedle dumb and tweedle dumber physics dont work that way.

so maybe you can tell us exactly how the two masses turning around will apply more force than the 20 masses being accelerated and keep the pipe from moving.

until you can SHOW SOMETHING OTHER THAN YOUR LIP SERVICE
your wrong and Im right.

kallogs bloopers

Quote:
- After turning around 180deg, its momentum in the free-floating stage is -(p1+F*t).


so according to your brain the mass that has been accelerating for the lenght of the pipe will suddenly
lose all momentum that it has gained...?

that way you have an exact number that just happens to
have be identical number only its a negatve number.
this way you can say +5 added to -5 = 0 movement.

how convenient and ridiculous to say the least.

this way you can say +5 added to -5 = 0 movement.
or as you put it
pp = 0-F*t+2(p1+F*t)-2*p1-F*t


I never said the ends were brakes their just turnarounds.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 06:22 PM

Quote:
Because the mass you've moved forward also has to stop.


only in your mind , even if it you have been shown that it doesnt stop , it only needs to stop in your mind in order for your mind to accept it.

I suppose that when you were a child you didnt use the
little curvy sections of track and your train stopped when it ran out of track , you poor thing , somebody should have put the train track together for you.

Quote:
Basic laws of physics, paul - every reaction has an opposite and equal reaction.


not really , not when its you that is describing action and reaction

according to you and kallog
a 50 kg force can stop a 450 kg force.

even if the two forces are constantly being applied towards each other.

450 kg ---->0 kg<----- 50 kg

they magically cancel each other out.

thats because you two guys are the smartest people on earth.

what I generally refer to as stupid smart people.




Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/17/10 07:26 PM

Originally Posted By: paul
Quote:
Because the mass you've moved forward also has to stop.


only in your mind , even if it you have been shown that it doesnt stop , it only needs to stop in your mind in order for your mind to accept it.

I suppose that when you were a child you didnt use the
little curvy sections of track and your train stopped when it ran out of track , you poor thing , somebody should have put the train track together for you.

I didn't have a train set - I did have an electric car set though. And I apparently paid more attention to my childhood toy than you did to yours.

When a car rocketed around a corner the whole track slid (if you got the cars up fast enough). The reason for this movement is simple - the track produced an inwards force which caused the car to turn, but in turn experienced an opposing force in the opposite direction. If you got things up fast enough that outward force would overcome friction and the track would then slide in the opposite direction of the force on the car.

But, when the car got to the other end of the track and hit the next corner, the track slid back - because once again the track applied a force to the car and thus experienced an opposing force.

Its an imperfect example (due to friction), but does reflect your system somewhat - both the car and the track undergo movement relative to each other, due to the internal forces. But the net product in the end was zero movement - every time the car turned at one end of the oval it created returned the track to its origin.

The major difference in your system is that every acceleration, change of direction, etc, will result in movement of the whole system - due to the absence of friction in space. The net is a sum of zero change in momentum - just as with my childhood toy.

Originally Posted By: paul
Quote:
Basic laws of physics, paul - every reaction has an opposite and equal reaction.


not really , not when its you that is describing action and reaction

according to you and kallog
a 50 kg force can stop a 450 kg force.


Nope, that is not what we are saying. What we are saying is when you move either one of those you get equal forces in both directions. The net effect is zero change in momentum of the total system, since the movement of the mass in one direction is exactly countered by the opposing force which is exerted on the system in the opposing direction.

The problem you are having is quite simple - you're ignoring the "reaction" half of the equation. You forget that when you move one of those masses, you're both exerting the force which moves the mass (the action) as well as exerting a force on the internal environment of your closed system in the direction opposite (the reaction).

Originally Posted By: paul
even if the two forces are constantly being applied towards each other.

450 kg ---->0 kg<----- 50 kg

they magically cancel each other out.


Nope, you've still got it wrong:

(reaction)<--mass-->(action)

<--450kg--> (equal and opposite force/momentum)
<--50kg--> (another equal and opposite force/momentum)

Net force/momentum of <--450kg--> is zero
Net force/momentum of <--50kg--> is zero
Net of the two combined is zero

Like I said earlier, you're missing half the equasion.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 08:05 PM

Quote:
Nope, you've still got it wrong:

(reaction)<--mass-->(action)

<--450kg--> (equal and opposite force/momentum)
<--50kg--> (another equal and opposite force/momentum)

Net force/momentum of <--450kg--> is zero
Net force/momentum of <--50kg--> is zero
Net of the two combined is zero

Like I said earlier, you're missing half the equasion.

Bryan


missing half an equasion is not as bad as missing half a brain , bryan.
but Im not missing half the equasion your missing half the brain , maybe all of the brain , bryam.

so if I push a mass in zero g with 450 kg force and you push the same mass in zero g from the opposite end with 50 kg force then the mass will not move?

according to your understanding of physics , that is.

I think I need to find another name for people such as
yourself.

such as stupid stupid people.







Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 08:16 PM

Quote:

Nope, that is not what we are saying. What we are saying is when you move either one of those you get equal forces in both directions.


just what are you saying bryan?
once again I ask...

1) do you think the pipe will move.

if you think the pipe will move , given that the pipe moves
due to a constant force.

2) do you think the pipe will suddenly stop.


I dont even know why I bother trying to get a straight answer from you two clowns.

neither of you would give an answer that would be truthfull.

if you cant answer the above 2 questions then I dont think there is any reason to reply to any further post that you make , no matter how stupid they are.

Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/17/10 08:32 PM

Originally Posted By: paul
missing half an equasion is not as bad as missing half a brain , bryan.
but Im not missing half the equasion your missing half the brain , maybe all of the brain , bryam.

so if I push a mass in zero g with 450 kg force and you push the same mass in zero g from the opposite end with 50 kg force then the mass will not move?


The insults are really quite ironic, since you're apparently unable to counter my statements without out-and-out lying about them. And I'd point out that repeating the same lie about my claims, which I've now corrected you for 3X, doesn't make that lie true - it just makes you a liar.

Its really simple, as I explained before. You push on the mass with force X. When you push on that mass, the system (i.e. your pipe) experiences a force of -X. Net effect: mass goes one way, the pipe goes the other, the movement of the center of mass (and thus the systems momentum) doesn't change.

So yes, the masses move (as I've said at least 3X now), but so does the pipe they are contained in - net effect, zero change in momentum of the system as a whole.

Same is true when the mass hits the end of the pipe or is otherwise braked. The mass will exert a force on the pipe/brake equal to the force which imparted its movement in the first place. Ergo, the mass stops due to the exertion of force, and that force is also transfered to the pipe. Since the force is exactly the same as what put the mass/pipe in motion to start with, the net effect is the motion of both the mass and the pipe stops.

There it is - so simple even a 2-year old can grasp it.

Now, how are you going to lie about my claims this time? Or perhaps you'll just repeat the same lie you've made the past few posts...

...strange, isn't it - that you can only make your "point" by lying about the claims of others...LOL

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/17/10 08:39 PM

Originally Posted By: paul
Quote:

Nope, that is not what we are saying. What we are saying is when you move either one of those you get equal forces in both directions.


just what are you saying bryan?
once again I ask...

1) do you think the pipe will move.

if you think the pipe will move , given that the pipe moves
due to a constant force.

2) do you think the pipe will suddenly stop.


Answer to 1 depends on what you mean by "move". The pipe will move relative to the mass moving inside of it. The center of mass of the whole system - i.e. pipe + mass - will not move, as observed by an outside observer. In other words, the individual parts will move, but the whole will not.

As for 2, the answer is yes. When the ball hits the end of the pipe, or is turned around, or is otherwise braked, it will exert a force equal to the force that started it moving in the first place. That force will stop the movement of the pipe.

On the other hand, if your pipe were open, so the ball could freely exit into space, than the pipe would continue in its movement unabated, and thus you would have achieved a net change in the momentum of the pipe. I'd point out that is the very essence of how a rocket works.

Originally Posted By: paul
I dont even know why I bother trying to get a straight answer from you two clowns.


Probably because deep down inside you realize we're right.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 09:22 PM

Quote:
The center of mass of the whole system - i.e. pipe + mass - will not move, as observed by an outside observer.


so your saying that if Im standing outside of the pipe I would not notice that it moves , because it will not move.

correct?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/17/10 10:06 PM

imagegeek
I suppose that your understanding of the pipe concept would render a gyroscope used in space in operable as it is inside the space ship , the space ship is a closed system pertaining to the gyroscope.

but thankfully they do use gyroscopes in space ships in space.

the internal gyroscopes do not place a force on the outside of the space ship , it places a force on the inside of the ship.

the gyroscopes keep the ships orentation in a certain position relative to the earth.

otherwise the iss could not maintain its orentation


thus fouling up your entire concept of the laws of newton.
and kallogs entire argument is hereby null.

and heres proof.

the iss and its gyroscopes are used to maintain orentation

meaning that the internal forces are turning the iss !!!

I win unless you can prove otherwise.

remember your entire argument is based on newtons laws as you understand them , so if the pipe cannot move without an external force applied to it , then the iss also cannot move without an external force applied to it.

and if the gyroscope was not turning then the ship would turn therefore the gyroscope is applying a force from inside the ship to keep the iss from turning.

pretty simple stuff.


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/18/10 04:21 AM


You can easily substitute the symbols for numbers, please do. I chose not to because I find it easier that way.

Now I've shown no long-term movement with 2 masses, you can expand the calculation for 20 masses. But why 20? Why not 10? 100? 2?

Quote:

now when 1 reaches the end the most it can subtract is
25 kg.


This might be the crucial mistake.
If a mass has:
25kg and 1m/s. Momentum = 25kgm/s
Then it turns around in a u-bend, it ends up with
25kg and -1m/s. Momentum = -25kgm/s

The momentum has reduced by 50kgm/s!!!! Not 25.



Quote:

I never said the ends were brakes their just turnarounds.


Yep. If an object just turns around, then its moment becomes the negative of what it was before. Bounce a ball against a wall, and it's direction changes to negative, but its speed remains roughly the same. It's lost TWICE the momentum it had.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/18/10 05:21 AM

He's saying the center of mass won't move. Which is what he said. The pipe can still move, just not the center of mass of it and its contents.


Originally Posted By: paul
Quote:
The center of mass of the whole system - i.e. pipe + mass - will not move, as observed by an outside observer.


so your saying that if Im standing outside of the pipe I would not notice that it moves , because it will not move.

correct?
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/18/10 05:42 AM

Originally Posted By: paul

meaning that the internal forces are turning the iss !!!


Correct, but they are not changing the angular momentum of
the whole system (ISS + gyroscopes).

By analogy, moving masses in a pipe are not changing the linear momentum of the whole system (pipe + masses).
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/18/10 12:32 PM

Originally Posted By: paul
the internal gyroscopes do not place a force on the outside of the space ship , it places a force on the inside of the ship.

Yep.

Originally Posted By: paul
the gyroscopes keep the ships orentation in a certain position relative to the earth.

Yep.

Originally Posted By: paul

otherwise the iss could not maintain its orentation


Depends on your reference frame - without a gyroscope the ISS would keep its orientation relative to it own inertial frame, plus whatever changes in orientation were induced by the movement of people/supplies within the ISS itself.

Originally Posted By: paul

thus fouling up your entire concept of the laws of newton.
and kallogs entire argument is hereby null.

Nope, as I pointed out in my post the gyroscopes use newtons laws to maintain the orientation of spacecraft, but cannot be used to change the momentum of the spacecraft.

They disprove your idea; they clearly show that internal forces - i.e. the gyroscopes - are completely unable to alter the momentum of the ship in which they are contained. You can use the gyroscope to reorientate the ship, but you cannot use it to move the center of mass - i.e. fly the ISS to the moon.

Heck, if all you needed a gyroscope once in orbit, why would they have built the Saturn booster rockets to get to the moon? A far simpler way would have been a much smaller rocket + a gyroscope.

The answer is simple - and its not because NASA is dumb. Its because gyroscopes and other internal forces are unable to alter the momentum of an object.

Originally Posted By: paul
and heres proof.
the iss and its gyroscopes are used to maintain orentation

meaning that the internal forces are turning the iss !!!


And? We never once said changes in orientation were not possible - only that changes in momentum - as in what you need for a rocket engine - cannot occur.

Notice in your video the ISS is turning; it is not rocketing off to the moon. Orbit is the same, momentum is the same, its just pointing another way.

The real irony is the turning of the ISS is achieved by the very law you are trying to violate - conservation of momentum. Changes in the spinning of the gyroscope create a change in the angular momentum of that gyroscope. Since momentum is conserved in a closed system, the ISS itself must turn to maintain the systems total angular momentum.

Originally Posted By: paul
I win unless you can prove otherwise.

Win what, the prize for being the worst at grade 10 physics?

Originally Posted By: paul

remember your entire argument is based on newtons laws as you understand them , so if the pipe cannot move without an external force applied to it , then the iss also cannot move without an external force applied to it.

Re-writing our claims again, are we? Kellog and I have been clear since the first post that momentum cannot change. We've also been clear since the first day that shifts in the relative positions of components (i.e. the pipe verses your internal mass) can also occur.

In fact, if you had the vaguest idea of how momentum is determined, you'd realize that shifts in the positioning of the pipe verses the internal mass are necessary to maintain momentum when one object moves within another. If those shifts did not occur, the center of mass would move, and thus the momentum of the system would change.

But what cannot change is the momentum of the whole. A concept clearly demonstrated by your ISS video.

Originally Posted By: paul
and if the gyroscope was not turning then the ship would turn therefore the gyroscope is applying a force from inside the ship to keep the iss from turning.


Actually, you've got that backwards. Relative to its inertial frame, the iss will not spin. It only appears to spin because it orbits the earth. Thus the iss has to be given spin (1 rotation/orbit) so that it always faces the same direction relative to the earth. The gyroscope is used to maintain that spin. The gyroscope is also used to maintain position when mass is moved around inside - other wise, someone walking from one end to the other could alter the iss's orientation in space.

However, no matter what you do with that gyroscope, the ISS stays in the same orbit. The movement of that internal mass, and thus the force created by that internal mass, is completely unable to produce thrust.

Which is the polar opposite of your claim.

Bryan
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/18/10 12:35 PM

Originally Posted By: kallog

Correct, but they are not changing the angular momentum of
the whole system (ISS + gyroscopes).

By analogy, moving masses in a pipe are not changing the linear momentum of the whole system (pipe + masses).


The sad part is that I took ~200 lines to say the same thing frown

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/18/10 08:18 PM

Quote:
This might be the crucial mistake.
If a mass has:
25kg and 1m/s. Momentum = 25kgm/s
Then it turns around in a u-bend, it ends up with
25kg and -1m/s. Momentum = -25kgm/s

The momentum has reduced by 50kgm/s!!!! Not 25.


No , absoloutly not.
I notice you are dreaming again kallog.

but after it leaves the turn round it isnt accelerating !!
its free floating , thus your -25kgm/s is null and void.

and it is the acceleration of the masses that causes pipe momentum.

we are not decelerating the masses yet , other than the slight deceleration as the masses pass through the turnarounds , sure they reverse direction but they do not greatly decelerate.

therefore your -25kgm/s will not affect the motion of the pipe after it leaves the turnaround , because its not in contact with the pipe , and it does not press upon the pipe.

it would only affect the momentum of the pipe if the pipe
were applying a force to accelerate or decelerate it.

and lets not forget that the 2 masses at the ends cancel each other out , they will slightly slow down while passing through the turnarounds however ,and that -force could be used as a negative toward pipe momentum , but what you have posted is not valid according to newtons laws of motion.

so what you end up with is

1) a force is applied for 500 ft to a mass and that mass is accelerated to a given velocity.
2) when that mass reaches the turnaround and passes through the turnaround it will apply a force that would reduce pipe momentum , however that mass is not being stopped only decelerated slightly.


and then that mass free floats for 500 ft back to the other turnaround and passes through the turnaround and it will apply a force that would increase pipe momentum , however that mass is not being stopped only decelerated slightly.

the mass is then accelerated again for a distance of 500 ft

and you didnt mention that there is a constant acceleration of 20 masses as the above is occurring that provides for pipe movement.


Quote:
Yep. If an object just turns around, then its moment becomes the negative of what it was before. Bounce a ball against a wall, and it's direction changes to negative, but its speed remains roughly the same. It's lost TWICE the momentum it had.



but the above would not apply to to pipe momentum , only the mass momentum , I thought that you understood that , I guess I was wrong because you seem to think otherwise or at least you would like for people to think that it would cause the pipe to not move , but as usual you are wrong again.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/18/10 08:36 PM

Quote:
Correct, but they are not changing the angular momentum of
the whole system (ISS + gyroscopes).



but they are changing the momentum of the iss.

otherwise why would they even use it.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/18/10 08:41 PM

Im not sure why Im answering your post

but you seem to think that a gyroscope will not change the momentum of a spacecraft.

corect?

if your spacecraft is spinning and then you stabilize the spacecraft by rotating the gyroscope and the spacecraft stops spinning then the gyroscope has changed the momentum of the spacecraft.

a student spinning by applying a force to a gyroscope

if the above student were inside a can in space he could use the gyroscope to manuver the can , much like space craft use them.

he applies a force to rotate the gyroscope then another to change the axis of he gyroscope in order to move himself and the gyroscope while standing on a swivel.

notice that no one pushes him and causes him to move by applying an external force on him or the swivel or the gyroscope , so if he were in a can in space this internal force he applies would rotate the can he is standing on.

just as newton describes it.
but not as imagegeek describes it.





Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/18/10 09:05 PM

Quote:
He's saying the center of mass won't move. Which is what he said. The pipe can still move, just not the center of mass of it and its contents.


are you also him?
if not why are you replying to questions I ask him?

he has never said that the pipe would move.

remember imagegeek your entire argument is based on newtons laws as you understand them , so if the pipe cannot move without an external force applied to it , then the iss also cannot move without an external force applied to it.

and if the gyroscope was not turning then the ship would turn therefore the gyroscope is applying a force from inside the ship to keep the iss from turning.

pretty simple stuff.


Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/18/10 09:28 PM

Quote:

- An accelerated mass starts with momentum p1.
- After acceleration at force F for time t it ends up with momentum p1+F*t. The reaction has the opposite effect on the pipe,


if I drive a car from new york to california then I apply the brakes and stop the car , does it require as much force to stop the car as it did to get the car to californa from new york?

no it doesnt.

and if there are 19 people driving 19 cars behind me at the same time , and Im only stopping 1 car.

is the force required to stop my 1 car the exact force being used to push the other 19 cars to californa?

no , it isnt.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/18/10 09:49 PM

Quote:
Now I've shown no long-term movement with 2 masses, you can expand the calculation for 20 masses. But why 20? Why not 10? 100? 2?


nope , you didnt show anything except that you have a clear understanding in how to sudgest something will not work using incorrect assumptions.

we are using 20 masses because that is what we started with.

20 masses being accelerated will push the pipe with a greater force than 1 mass being accelerated.

besides I didnt want you two guys to get a even start so
I used 20 to start with.

we could use just 1 mass to prove it will work, and you know it.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/19/10 03:37 AM

Originally Posted By: paul

but they are changing the momentum of the iss.

Yes.

Of course here 'iss' means the rest of the space station, excluding the gyroscopes themselves.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/19/10 03:43 AM

Originally Posted By: paul
besides I didnt want you two guys to get a even start so
I used 20 to start with.


This is just a game to you? In other words you know you're wrong.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/19/10 03:59 AM

Originally Posted By: paul

we are not decelerating the masses yet , other than the slight deceleration as the masses pass through the turnarounds , sure they reverse direction but they do not greatly decelerate.


OK this statement shows that we simply have different interpretations of the meanings of words.

To a physicist, 'acceleration' includes a change in speed or direction. Look up wikipedia to get a more thorough picture. This is the same acceleration in F=ma so it's crucial that we use a consistent definition.




Quote:

the mass is then accelerated again for a distance of 500 ft


Is this how it works?:

1. Accelerate the mass, starts moving the pipe.
2. Turn it around, applies a force the wrong way.
3. Turn around again, applies the same force the opposite way.
4. Repeat.

The two turn-arounds cancel each other out. Leaving only the acceleration to propel the pipe?

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/19/10 04:10 AM

Originally Posted By: paul

the brakes and stop the car , does it require as much force to stop the car as it did to get the car to californa from new york?

no it doesnt.


It may require more or less force. This misuse of terminology can totally lead you in the wrong direction. Of course you can gently accelerate with a tiny force and slowly get up to a high speed. Then crash into a tree with a huge force to stop. Much higher stopping force than starting force.

Equally you can floor it to accelerate with high force, then coast to a stop with a small force.

However the total impulse used to stop the car is the same as the total impulse used to start it. That's symmetric.


On the straight open road the force from the engine balances the force of friction, leaving no net force, and no acceleration - just constant velocity. So that section of the trip can be ignored.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/19/10 12:21 PM

Quote:

but they are changing the momentum of the iss.

Quote:

Yes.

Of course here 'iss' means the rest of the space station, excluding the gyroscopes themselves.



so you unlike imagegeek realize that a internal force can
change the momentum of the iss.

closing his erroneus argument that an external force is required.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/19/10 12:24 PM

Quote:
This is just a game to you? In other words you know you're wrong.


No , this is just a discussion to me.

I know I'm right.

and

I know you guys are wrong.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/19/10 12:36 PM

Quote:
To a physicist, 'acceleration' includes a change in speed or direction. Look up wikipedia to get a more thorough picture. This is the same acceleration in F=ma so it's crucial that we use a consistent definition.



I have a clear understanding of that , and physicist also use decelerate to describe using a force to decelerate a object.

picking at straws are we?

Quote:
Is this how it works?:

1. Accelerate the mass, starts moving the pipe.
2. Turn it around, applies a force the wrong way.
3. Turn around again, applies the same force the opposite way.
4. Repeat.

The two turn-arounds cancel each other out. Leaving only the acceleration to propel the pipe?


but closer to

1. accelerate 20 masses applies 20 positive forces.
2. turn 1 mass around 180 degrees applies a negative force.
3. turn 1 mass around 180 degrees applies a positive force.
4. the 20 opposing masses are free floating applies zero force negative or positive.

5. repeat.

the result of this is constant acceleration of the pipe

even after the pipe has traveled the lenght of the pipe.

or the lenght of a billion pipes or a trilion pipes

as long as you have the electricity to supply an

acceleration to the 20 masses.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/19/10 02:16 PM

Originally Posted By: paul
internal force can change the momentum of the iss.


I can't believe you're having trouble with this. Although most of the ISS changes its angular momentum, the total angular momentum of the ISS and its contents doesn't change. This is analogous to the total linear momentum of the tube and it's contents, which also doesn't change.



Gyro goes one way, rest of ISS goes the other way.

Masses go one way, tube goes the other way.

Gyro keeps spinning indefinately, rest of ISS keeps spinning indefinately.

Masses must reverse direction within the length of the tube, tube reverses direction.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/19/10 02:38 PM


Quote:

1. accelerate 20 masses applies 20 positive forces.

OK, impulse = +20
The pipe's moving forward with momentum +20

Quote:

2. turn 1 mass around 180 degrees applies a negative force.

-2 because it's reversing not just stopping.

Now we have to consider the actual possible layout. I suppose those 20 masses were spaced apart in their accelerator to begin with - or all lumped together. Either way, before any masses get to the 2nd u-bend, which I suppose is at the opposite end of the tube, all the other 19 have reached their first turn-around -

-2 * 19 = -38
Total = +20 - 2 - 2*19 = -20
Now the pipe's going backwards with momentum -20

Quote:

3. turn 1 mass around 180 degrees applies a positive force.

+2

Then the other 19 masses do their 2nd turn-around -

+2 * 19 = +38

Total = +20
Now the pipe's moving forward again with momentum +20


What next?



Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/19/10 03:59 PM

Quote:
What next?


LOL , ok...

Quote:
-2 because it's reversing not just stopping.


were not dealing with a free energy device here.
so you arent allowed to add free energy.

and the turnaround will not add any energy to the mass.

the turnaround does not supply a force to the mass it is like a curve in a road that your car travels on , I cant remember having to press the gas pedal to the floor board just to get through a curve , LOL.

I usually take my foot off the gas. LOL

the mass already has enought momentum to clear the turnaround.

and the most negative momentum you could possibly count
would be the mass * the masses velocity.
which is the masses momentum.

so whatever the single masses momentum is is the most
you can use as a negative momentum.

and the opposite occurs at the other end in the other turnaround , so once again I state that the two masses
momentums in the turnarounds cancel each other out.

the only remaining force is the force used to constantly accelerate the 20 masses.

we havent calculated the force to the pipe due to accelerating the 20 masses "constantly" so we cant use
your +20.

the number will be much greater than that.
to accelerate a 100 kg mass from 0 m/s to 5 m/s
requires more force that to accelerate that same mass
from 0 m/s to 1m/s.

until we know the amount of force applied to the pipe
due to accelerating the "20" masses constantly we will not be able to arrive at a close number.

note: constantly...

and by the time each of the "20 " masses gets to the end of the pipe and enters the turnaround its velocity might be 30-40 m/s faster than when it exited the turnaround at the other end.

so we cant determine the validity of the concept using your
assumptions.











Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/20/10 12:54 AM

Quote:
note: constantly...


constantly :
because
1) there are always 20 masses being accelerated.
2) there are always 20 masses free floating.
3) there are always 2 masses in the 2 turnarounds.

so using your math !!

-2 * 19 = -38
+2 * 19 = +38
+20 * 20000 = +400000
netural 20 * 0 = 0
Total = +400000 -38 +38 = +400000

see I can do the same stuff only in my favor as you do.

LOL
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/20/10 03:21 AM

Originally Posted By: paul
and the turnaround will not add any energy to the mass.


Correct. And of course we're talking about momentum, not energy, which I stated quite clearly.

Suppose a car has momentum of 1000. It crashes into a tree and stops. What's its new momentum? How much did it lose?

This point is crucial to it working or not. So we better clarify it before complicating things with constant accelerations.

Actually we can express it in an even clearer way. Suppose a car has velocity of 1000, it crashes into a tree and stops. What's its new velocity? What's the change in its velocity?

Sounds too simple doesn't it? But I'm curious what your answer will be.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/20/10 04:25 AM

that all depends on your frame of reference and a tree is elastic to a point , then theres the wind resistance as the tree is approached by the car and the wind compresses and pushes back on the car with a momentum of -1000

however since the car is also pushing the compressed air
with a momentuum of +1000 the two momentums cancel each other out , the tree due to the elastisity absorbes the force of the "wind + car" system and the car pushes the tree towards the car at twice the cars momentum.

so after the tree impacts the car the car has a momentum
of car-1000 + tree+2000 = car - 1000.

the car has a momentum of -50,000

then the car will travel at the speed of light in the
opposite direction of its momentum - the doubled tree momentum that is squared and double squared just as the impact occurs due to the right angle acceleration of the wind and its angular velocity.

so
1000^2^2 w.T1-A1= 50,000

but only if the tree is elastic.
if the tree is not elastic then the results would be different.

which results in

-50,000 + +50,000 = -+50,000

depending on your frame of reference.

tree momentum durring collision is

T1(T/A^2) VC1-w = total tree momentum = +2000^3-%@.()

567,hd *(-) + a^3 / AT = wind angular velocity

so 50,000 - 567,hd = -1000

so although it appears that the tree is colliding with the tree , the compressed air between the "tree+car" system
is absorbing the cars momentum , the tree only looks as if the car is hitting the tree because the car is being pushed backwards by the elasticity of the "air+tree" system at the speed of light-c + 1000.





Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/20/10 05:43 AM

Originally Posted By: paul
that all depends on your frame of reference and a tree is elastic to a point , then theres the wind r


I assume all measurements are made from the same reference frame. It stopped from the point of view of whoever measured it as stopping. That same obsersver measured it having p=1000 before collision. And I'm asking for the final and change in momentum from the point of view of that same observer.

The elasticity of tree is also irrellivent because the car stopped. Any transient elastic behaviour was damped out causing the car to end up stopped.

Wind resistance is irrelivent, the car stopped, doesn't matter how.

Do you have an answer or do you give up?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/20/10 11:29 PM

Quote:
Suppose a car has momentum of 1000. It crashes into a tree and stops. What's its new momentum? How much did it lose?


its new momentum is 0 because it lost 1000

unless you want to get technical about it.
because its sitting on the earth , the earth is rotating around the sun at apx 66,000 mph
and who knows how fast our solar system is rotating around our galaxy , and who konws how fast our galaxy is rotating around our universe , etc...

but if its stopped on the earth then in reference to the earth its momentum is zero.
and it lost 1000.

Im not sure where your going with this unless your going to
try again to show that the concept is invalid using only 1 or 2 masses.

because if you stop one of the masses that has been accelerated at 5 m/s then you can subtract all the force that
was applied to accelerate the mass at the end by completely stopping the mass.

but if your going to stop 1 at one end then you must also stop 1 at the other end and this will
cancel out each -momentum with a +momentum.

even if you do that , there are still 20 masses being accelerated constantly at 5 m/s/s applying a force to the pipe.

but try as you may.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/21/10 05:05 AM

Originally Posted By: paul
but if its stopped on the earth then in reference to the earth its momentum is zero.
and it lost 1000.


Yes, same answer I got, funninly enough.

Here's where I'm going. Now the driver manages to restart his car and reverse away from the tree, he speeds up to a mometum of 1000 in the opposite direction and keeps going like that.

Q1: What was his momentum before the collision
Q2: What is his momentum after driving away
Q3: What is the overall change in momentum of the car between the start and end of the story?
Q4: How much momentum did the car lose between the start and end?

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/21/10 12:07 PM

Q1: What was his momentum before the collision
Q2: What is his momentum after driving away
Q3: What is the overall change in momentum of the car between the start and end of the story?
Q4: How much momentum did the car lose between the start and end?

Q1:1000 ; the car was initaly traveling at 1000
Q2:1000 ; the car reaccelerated to 1000 after the collision
Q3:1000 ; after the collision the momentum goes to zero
Q4:1000 ; momentum at start is the same at end only opposite direction.


------ my turn ------

there are a total of 42 mases
20 being accelerated
20 free floating
2 in the turnarounds

Q1: how much force does each 100kg mass apply to the pipe in the opposite direction of the movement of the mass if accelerated for a distance of 500 ft at a constant acceleration of 5 m/s/s if initial velocity is zero.
Q2: how much force will be applied to the pipe in the opposite direction of mass movement by all 20 masses being accelerated at 5 m/s/s before the first mass enters the first turnaround.
Q3: what is the total force applied to the pipe by the 20 free floating masses.
Q4: how much force is required to accelerate 42 100kg
masses at 5m/s/s for a distance of 500 ft.









Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/21/10 01:59 PM

Originally Posted By: paul
Q1: What was his momentum before the collision
Q2: What is his momentum after driving away
Q3: What is the overall change in momentum of the car between the start and end of the story?
Q4: How much momentum did the car lose between the start and end?

Q1:1000 ; the car was initaly traveling at 1000
Q2:1000 ; the car reaccelerated to 1000 after the collision
Q3:1000 ; after the collision the momentum goes to zero
Q4:1000 ; momentum at start is the same at end only opposite direction.


Q1: Yes, 1000
Q2: No, -1000
Q3: Change in momentum = final momentum - initial momentum.
Q4: Negative of the answer to Q3

Do you want to reconsider Q3? That's crucial to the entire moving pipe machine.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/21/10 05:44 PM

why dont you simply tell me what you think the answers are
then see if I agree to them.

meanwhile , I have calculated the braking force of the single mass that is accelerated at 5 m/s/s for a distance
of 152.4 meters = 500 ft.

I get the following , you can check the math for yourself.

500 kg-m/s/s or 500N
100 kg mass
a=f/m
5m/s/s = 500N/100kg

initial velocity = 0m/s
final velocity = 39.0385 m/s
average velocity = 19.51925 m/s
distance the mass traveled in 7.8077 seconds = 152.400448225 meters

so the single mass is traveling into the turnaround at
a velocity of 39.0385 m/s
F=ma
3903.85 N = 100 kg * 39.0385 m/s

the braking force required to stop the single mass is
3903.85 N

at any time durring acceleration the 20 masses being pushed by the 500 kg-m/s/s or 500N are pushing back on the pipe with the same force with a combined force of 500N * 20 = 10000N

we had used 500 kg as the mass of the pipe earlier in the discussion so the 10000N force applying to the 500kg mass
results in a acceleration of 20 m/s/s

a=f/m
20m/s/s = 10000N/500kg

so there is a constant force of 10000N applied to the pipe
at any moment durring acceleration of the 20 100 kg masses.

initial pipe velocity = 0m/s
initial mass velocity = 0m/s

given that all of the masses are distributed evenly
around in the system , by the time a single mass is accelerated between the first turnaround and the next turnaround
the pipe has a final velocity of 156.154 m/s
the pipes average velocity is 78.077 m/s
F=ma

78077N = 500 kg * 156.145


result

78077 N - 3903.85 N = 74173.15 N

we can keep dilly dallying around with examples but using the pipe example would be best.

to an observer outside the pipe the pipe has moved 609.6017929 meters after 7.8077 seconds.

just in case you are curious in 1 hour , 3600 seconds
the pipe will have traveled 129,600,000 meters
or 80,529.706 miles

its final velocity will be 72000 m/s or 44 miles / second

or 2684.323 mph

in 24 hours it will travel 39,158,514.937 miles
or 74649600000 meters

in 1 week 3657830400000 meters or 1,918,767,231.937 miles

and that is by using only 1 set of masses , a space ship could be fitted with hundreds of these tubes and each extra tube adds that much more acceleration , I need to set the program to calculate the speed and time to the nearest star.

what would the fastest acceleration be that humans could withstand?














Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/21/10 10:51 PM

Originally Posted By: paul
why dont you simply tell me what you think the answers are
then see if I agree to them.

I have told you and you don't agree. There really is no point going anywhere beyond this point until you understand how much momentum is transferred from an object that does a U-turn.

The answer to Q3 is -2000
We can verify this with:
Initial momentum + change in momentum = final momentum
1000 + -2000 = -1000


Quote:

we can keep dilly dallying around with examples but using the pipe example would be best.

I agree.


Acceleration of one mass:
500N for 7.8s = impulse of 3900 Ns -> gives the mass a momentum of 3900 kgm/s and the pipe -3900 kgm/s.


Quote:

Turnaround of one mass:
a velocity of 39.0385 m/s
F=ma
3903.85 N = 100 kg * 39.0385 m/s


This is the magnitude of the force that would deccelerate 100kg from 39.0m/s to 0 in 1 second. The mass then has to speed up again from 0 to 39m/s in the other direction, so this force is applied for 2 seconds.

Impulse applied to the mass = -3903.85N * 2s = -7808 Ns
This reduces the mass's momentum by 7808 kgm/s and increases the pipe's by 7808 kgm/s




Quote:

at any time durring acceleration the 20 masses being pushed by the 500 kg-m/s/s or 500N are pushing back on the pipe with the same force with a combined force of 500N * 20 = 10000N


So over 7.8s there's:
10000N pushing one way, impulse = 78000 Ns
and 20 masses doing U-turns, which is 20 * 7808 Ns = 156200Ns in the opposite direction.

The overall force on the pipe is continuously in the opposite direction to what the accelerations should cause. So the pipe's moving backwards.

What about the other U-bend bringing it back to the proper direction? I'm starting to get confused about where all the masses are. While that acceleration was occuring, there were 20 floaters, entering the 2nd u-bend one-by-one? These would go back into the accelerator at some speed - but that's impossible because we specified 20 masses accelerating from 0, not from 39m/s. I'm losing the ability to visualize it. Could you possibly draw a time-sequence of events?

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/21/10 11:11 PM

Originally Posted By: paul

78077 N - 3903.85 N = 74173.15 N


Even if you don't agree with my calculations, you should see a problem here. You've only accounted for 1 of the masses turning around but 20 accelerating. It takes 2s to do a turnaround so there are actually several masses in the u-bend at the same time, and their forces add together.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 01:07 AM

Quote:
This is the magnitude of the force that would deccelerate 100kg from 39.0m/s to 0 in 1 second. The mass then has to speed up again from 0 to 39m/s in the other direction, so this force is applied for 2 seconds.


I never said that the mass would be stopping , that was your
assumption.

Quote:
So over 7.8s there's:
10000N pushing one way, impulse = 78000 Ns
and 20 masses doing U-turns, which is 20 * 7808 Ns = 156200Ns in the opposite direction.


there is (((( ALWAYS )))) 1 mass in 1 turnaround and
there is (((( ALWAYS )))) 1 mass in the other turnaround.
Ive already said that Im not stopping the masses , That would be something a stupid person would do knowing that they would have to re-accelerate the masses using up precious energy.

you always seem to forget the other turnaround so your
Quote:
20 masses doing U-turns, which is 20 * 7808 Ns = 156200Ns in the opposite direction.

also has the other end attached to it.

and the result is :

+156200Ns plus -156200Ns = ZERO

they cancel each other out.

right !!!
yes right , youve agreed to it before already !!!

there is (((( ALWAYS )))) 20 masses being accelerated
and
there is (((( ALWAYS )))) 20 masses free floating

so the final result is: 20 * 3903.85N = 78077N

so at the end of the 7.8 seconds the pipe has a final pipe velocity of 156.154m/s in the opposite direction that the masses were accelerated.

its pretty clear that you are using math that is designed
to work only in your favor , by using the full stop on the masses because it would hopefully prove your point , but the only point you proved today is that you use these left handed
measures to accomplish a goal.

I guess you play around with the numbers until you see an opportunity , LOL , in fact ROFL...




Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 01:23 AM

Quote:
Even if you don't agree with my calculations, you should see a problem here. You've only accounted for 1 of the masses turning around but 20 accelerating.


I counted the braking force of 1 mass because it is the total force applied to 1 mass durring acceleration, the two masses in the turnarounds cancel each other out , why cant you understand that now ...

Originally Posted By: Kallog
Is this how it works?:

1. Accelerate the mass, starts moving the pipe.
2. Turn it around, applies a force the wrong way.3. Turn around again, applies the same force the opposite way.
4. Repeat.

The two turn-arounds cancel each other out. Leaving only the acceleration to propel the pipe?



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/22/10 01:44 AM

Originally Posted By: paul

I never said that the mass would be stopping , that was your
assumption.


A u-bend does stop the mass in 1 dimension, which is all we're considering, and all we need to consider. In order to turn it around. You have to decellerate it down to 0 then back up to full speed. This consumes _NO_ energy because it's an elastic 'collision'. Of course in the real 3D system the _speed_ remains constant throughout the turn, but the component of velocity in the only direction we're considering with must drop to zero and change sign.


Quote:

there is (((( ALWAYS )))) 1 mass in 1 turnaround and


That means we have to change the force it applies when going through:

In 7.8s 20 masses enter the turnaround.
Each mass spends 7.8/20 = 0.39s turning around.
Reversing the direction changes its momentum by 3900*2=7800 Ns
If the turning is done by applying a constant force between the mass and the tube, then that force is:
F = 7800Ns / 0.39s = 20,000N
So this force of magnitude 20,000N is applied continuously at the 1st turnaround.

Just as before, this is double, but opposite the constant 10,000N force applied by the accelerator.

And of course just as before I still havn't mentioned the 2nd turnaround which will help. But I don't know how you want it to work yet. See below..



Quote:

you always seem to forget the other turnaround so your
+156200Ns plus -156200Ns = ZERO
they cancel each other out.

I purposely omitted it because I didn't know what the setup should be. If there are masses going through the 2nd turn at the same speed as all the masses in the 1st turn (39m/s), then they're entering the accelerator already doing 39m/s and will overtake some of the masses starting at 0. But we calculated the forces at the 1st turn assuming all masses were accelerated from 0 to 39, not 39 to 39+39. So the force in the 1st turn now depends on which mass is going through it, and will always be equal or higher in magnitude than at the 2nd turn which takes some stragglers from the previous acceleration boost.

Quote:

there is (((( ALWAYS )))) 20 masses being accelerated
and
there is (((( ALWAYS )))) 20 masses free floating


I'm not sure this is physically possible. Could you produce an animation?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 02:16 AM

the pipe is 152.4 meters long

152 / 20 = 7.6 meters

so the masses begin acceleration seperated by 7.6 meters each.

so we can have a turnaround radius of 4.83 meters.

this means we will need to add 9.66 meters to the lenght of the pipe because of the two turnarounds , however we still accelerate the masses a distance of 152.4 meters.

this way there is 1 mass in the middle of each turnaround when the acceleration begins.

so the turnarounds will need to be used to slightly accelerate the masses in the turnarounds bringing them up to a slight velocity before accelerating them at 5 m/s/s.

if we have to the pipe can always be lenghtened to avoid collisions between the masses.

Im only trying to show the concept not provide you with plans to build anything , so if you are considering collisions forget it.

the 20 masses in the free floating side are held in place by a opposing rail gun accelerator.

the opposing accelerator is removed after the masses are accelerated slightly as I mentioned earlier.

Quote:
A u-bend does stop the mass in 1 dimension, which is all we're considering, and all we need to consider. In order to turn it around. You have to decellerate it down to 0 then back up to full speed. This consumes _NO_ energy because it's an elastic 'collision'. Of course in the real 3D system the _speed_ remains constant throughout the turn, but the component of velocity in the only direction we're considering with must drop to zero and change sign.


No , you dont decelerate the mass .... the mass is only changing direction.

what you are talking about above is simple direction.

Quote:
That means we have to change the force it applies when going through:


No , we do not apply any force while the mass moves through the U turn.

the mass itself applies a force to the U turn and that force is its mass * velocity

100kg * 39.0385 m/s/s = 3903.85 N

which is exactly opposed at the other U turn..........
because its in zero g.

Quote:
Reversing the direction changes its momentum by 3900*2=7800 Ns


changes the momentum of the mass but not the momentum that is
either applied or subtracted from the pipes momentum.


knowing that the masses passing through the turnarounds cancel
each other out , why do you insist on discussing them?

where are you going with this , there is no gain or loss to pipe momentum because the masses pass through the turnarounds.

because we are not including any friction between the masses and the turnarounds.

its like you are arguing about two 5 lb weights tied to a string then flung over a clothesline.

they just sit there because they both present the same force to each other.













Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/22/10 02:59 AM

Originally Posted By: paul
the opposing accelerator is removed after the masses are accelerated slightly as I mentioned earlier.

So it accelerates them just enough to feed them into the main accelerator at the same rate other masses are leaving it, so as to maintain 20 in the main accelerator at all times?


Quote:

No , you dont decelerate the mass .... the mass is only changing direction.


Yes only changing direction, and that requires a force, which can be done just by letting it fly through a U-shaped pipe, or hitting a spring. No need to put any energy in.
Quote:

Originally Posted By: Kallog

To a physicist, 'acceleration' includes a change in speed or direction. Look up wikipedia to get a more thorough picture. This is the same acceleration in F=ma so it's crucial that we use a consistent definition.

I have a clear understanding of that , and physicist also use decelerate to describe using a force to decelerate a object.

picking at straws are we?



Quote:

No , we do not apply any force while the mass moves through the U turn.

We/mass/tube/etc. Somehow there's a force, as you say, and that's what I mean too. We don't need to fire up another rail gun or anything.


Quote:

the mass itself applies a force to the U turn and that force is its mass * velocity

I ignored this before because it didn't really seem to matter, but now we better straighten it out:
F=ma, F<>mv, p=mv.




Quote:

knowing that the masses passing through the turnarounds cancel
each other out , why do you insist on discussing them?

Because they don't. In this new situation with a slight acceleration provided at the turnaround before the accelerator, we have masses going very slowly through that, while _at the same time_ there are masses going at 39m/s through the opposite turnaround. Different speeds, different forces.

Sure the 39m/s masses eventually get to the other turnaround and apply the same but opposite force there. But by the time the last few of them are doing that, there'll be new, even faster masses that've been accelerated twice, going through the 1st turnaround, so they still don't cancel each other out.

This is why I keep asking for an animation. Even tho it's a bit of work to make, it'll clearly show what's happening without having to consider any forces or momentums. I'm sure you have the programming skills to generate such a thing automatically. I don't want to do it because I expect you'll say I've got it wrong and have to do it again.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 03:29 AM

you realize Im sure that the acceleration does not have to be instantly at 5 m/s/s , that the acceleration could take place gradually to avoid collisions , however I assumed you would take that up.

lets drop all the confussing stuff.

we now have 1 mass and thats all.
we dont need a opposing rail gun.
we dont need to control the mass at all.
we just use the instant 5 m/s/s acceleration.

the mass is 100 kg
the pipes mass is 500 kg
the force supplied to the mass is 500kg-m/s/s or 500N
the pipes initial velocity is 0 m/s
the mass initial velocity is 0 m/s

we begin the mass acceleration
its initial velocity is 0 m/s
we accelerate it at 5m/s/s for a distance of 154.2 meters

accelerating this 1 mass results in a force of +3903.85N

to the pipe in the opposite direction.


the mass enters the first turnaround at a
velocity of 39.0385 m/s

it goes through the first turnaround
applies a force of -3903.85 N to the pipe.

it then free floats at a velocity of 39.0385 m/s
to the second turnaround.

it then goes through the second turnaround
and applies a force of +3903.85 N to the pipe.

the forces that apply to the pipe add up as follows

+3903.85 N acceleration
-3903.85 N 1st U turn
+3903.85 N 2nd U turn
-----------
+3903.85 N

its that simple.
the pipes initial velocity was 0 m/s
the pipe has a final velocity of 7.8077 m/s

the process repeats...

the pipes velocity increases every time the mass completes a cycle.










Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/22/10 04:36 AM

Originally Posted By: paul

we now have 1 mass and thats all.

Yay

Quote:

accelerating this 1 mass results in a force of +3903.85N

500N. That's the force supplied to the mass, and equally the pipe. That force is applied for 7.8s

Quote:

it goes through the first turnaround
applies a force of -3903.85 N to the pipe.

OK. Change in momentum = final momentum - initial momentum
= -39m/s*100kg - 39m/s*100kg
= -7800 kg m/s (Ns)
This means the -3900N force was applied for 2 seconds.

Quote:

it then goes through the second turnaround
and applies a force of +3903.85 N to the pipe.

OK. So again the force of +3900N is applied to the pipe for a time of 2s.



It's meaningless to add up forces while ignoring the times they're applied for. I'll include time below...


the forces add up as follows

+500 N acceleration for 7.8s. Impulse = 500*7.8 = 3900Ns
Pipe moves forwards with p=3900 kgm/s, v=7.8m/s

-3903.85 N 1st U turn for 2s. Impulse = -3900*2 = -7800Ns
Pipe reverses direction and has v=-7.8m/s

+3903.85 N 2nd U turn for 2s. Impulse = 3900*2 = 7800Ns
Pipe reverses direction again, v=7.8m/s

I'll do the next cycle. I suppose the accelerator applies the same impulse (change in momentum) each cycle.

So we go on with ..
Impulse = +3900 kg m/s, pipe speeds up to v=15.6m/s
Impulse = -15,600 kg m/s, pipe reverses to v=-15.6m/s
Impulse = +15,600 kg m/s, pipe reverses again to v=+15.6m/s

After 2 complete cycles it's still oscillating back and forth, but going faster and faster.

Maybe it's travelling a longer and longer distance each cycle too, or maybe it isn't. Havn't worked that out.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/22/10 04:41 AM

Where did 15,600 kg m/s come from?

That's the change in momentum of the mass as it goes through the 1st turnaround (for the 2nd time).

Speed on entering turnaround = -39m/s + -39m/s = -78m/s
Momentum entering turnaround = -78m/s * 100kg = -7800 kg m/s
Momentum exiting turnaround = 78m/s * 100kg = 7800 kgm/s
Change in momentum of mass through turnaround = 7800 - -7800 = 15,600 kg m/s
Change in momentum of pipe as mass turns around
= -15,600 kg m/s
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 11:42 AM

Quote:
This means the -3900N force was applied for 2 seconds.


NO !!!!

the mass has a final velocity of 39.038 m/s when it enters the turnaround after being accelerated.

the turnaround would have to have a lenght of
39.038 meters * 2 = 78.076 meters
for the mass to consume 2 entire seconds while passing through it.

so your wrong.

Quote:
It's meaningless to add up forces while ignoring the times they're applied for. I'll include time below...


and its even less meaningless to include erroneous times , Im not sure why your so concerned with time as the time is given in the acceleration of the mass.

ie...5 meters / SECOND / SECOND...

meaning each second..

Quote:
Maybe it's travelling a longer and longer distance each cycle too, or maybe it isn't. Havn't worked that out.


maybe that is where your insertion of 2 seconds will come in !!!

LOL.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/22/10 12:57 PM

Originally Posted By: paul
NO !!!!


Yes. It's your design. You specified 3900N. That requires 2s, and 78m. If you want a shorter turnaround then you need to choose a higher force to get the job done faster. The value of this force is arbitrary, it doesn't change the outcome. That might sound strange, but any change in the force has to be compensated for by applying it for a different length of time so you end up with the same change in momentum.


Quote:

Im not sure why your so concerned with time as the time is given in the acceleration of the mass.


Have a free floating stationary block of 1kg. Push it with 10N for 1s. Then run round the other side and push it in the opposite direction with 10N for 1 hour. The block ends up stationary of course, because the two opposite forces balance, right?



Quote:

maybe that is where your insertion of 2 seconds will come in !!!

As I said before, there's no point going into other details unless you understand what that factor of 2 is for. Without it, the machine will indeed fly. Why don't you try to understand it? Don't have to listen to me, go check google or ask someone else. Or just look at this:
Answer to a physics problem.

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 01:43 PM

Quote:
Yes. It's your design. You specified 3900N. That requires 2s, and 78m. If you want a shorter turnaround then you need to choose a higher force to get the job done faster. The value of this force is arbitrary, it doesn't change the outcome. That might sound strange, but any change in the force has to be compensated for by applying it for a different length of time so you end up with the same change in momentum.



ok , try this experiment out.
get a U shaped pipe , fire a riffle in one end of the pipe.
dont worry you have alot of time to walk out of the way before the bullet comes back towards you.
NO WAIT NOT REALLY , I mean DONT TRY IT !!!

LOL.

I must admit that your logic isnt logical.

the 2 seconds that you are using is a complete falsity.

suppose the mass were attached to a wheel , are you somehow sudgesting that a mass of 100 kg cannot have a velocity of
39.03 m/s given it is attached to the periphery of a wheel with a radius of 4 meters ? or 3 meters? or 2 meters?
or 1 meter?

that somehow the mass cannot travel that fast around a radius?

thats really going a bit to far kallog.

but you knowing that the requirement of 2 seconds in each turnaround would cause the concept to not work , so is that your scientific or your childish way of finding a solution.

to your problem.

when I say Your problem , I mean your inability to find a reason that the concept would not work.

because afterall that is what you are trying to do , you just cant do it because physics keeps getting in the way , spoiling your attempts.










Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 02:08 PM

************** the second cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 39.0385 m/s
its final velocity is 52.193 m/s
its average velocity is 57.930 m/s

the time it takes the mass to accelerate the 152.4 meters is 2.6309 seconds

its braking force is 5219.3 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 7.8077 m/s
final velocity is 10.4386 m/s
average velocity is 11.586088965 m/s

the final pipe velocity is 10.4386 m/s

the pipe has now moved a distance of 60.961 meters

the braking force of the pipe is 5219.3 N

**************** the third cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 52.193 m/s
its final velocity is 63.736 m/s
its average velocity is 66.0178799 m/s

the time it takes the mass to accelerate the 152.4 meters is 2.3086 seconds

its braking force is 6373.6 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 10.4386 m/s
final velocity is 12.7472 m/s
average velocity is 13.203 m/s

the final pipe velocity is 12.7472 m/s

the pipe has now moved a distance of 91.443 meters

the braking force of the pipe is 6373.6 N


with each cycle the mass will be accelerated faster and faster
which in turn will accelerate the pipe faster and faster.

using only 1 mass is stupid but it seems to allow for your understanding
that the pipe will move as I stated , using a constant acceleration of masses
would be the ideal method to accomplish a smoother acceleration.

WOW , just think kallog in a few more cycles the mass will
require less than 2 seconds for acceleration , and your 4 second requirement for the mass to pass through the 2 turnarounds would make the concept , non working...

must be nice to just pick a time requirement out of thin air to use in your favor in a discussion.




Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 08:11 PM

I found a flaw in my calculations , that gave incorrect numbers above.
the flaw is now corrected and I am posting the corrections.

************** the second cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 39.0385 m/s
its final velocity is 55.209 m/s
its average velocity is 47.12375 m/s

the time it takes the mass to accelerate the 152.4 meters is 3.2341 seconds

its braking force is 5520.9 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 7.8077 m/s
final velocity is 11.0418 m/s
average velocity is 9.42475 m/s

the final pipe velocity is 11.0418 m/s

the pipe has now moved a distance of 60.96067362 meters

the braking force of the pipe is 5520.9 N

**************** the third cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 55.209 m/s
its final velocity is 67.617 m/s
its average velocity is 61.413 m/s

the time it takes the mass to accelerate the 152.4 meters is 2.4816 seconds

its braking force is 6761.7 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 11.0418 m/s
final velocity is 13.5234 m/s
average velocity is 12.2826 m/s

the final pipe velocity is 13.5234 m/s

the pipe has now moved a distance of 91.44117378 meters

the braking force of the pipe is 6761.7 N


with each cycle the mass will be accelerated faster and faster
which in turn will accelerate the pipe faster and faster.

using only 1 mass is stupid but it seems to allow for your understanding
that the pipe will move as I stated , using a constant acceleration of masses
would be the ideal method to accomplish a smoother acceleration.

WOW , just think kallog in a few more cycles the mass will
require less than 2 seconds for acceleration , and your 4 second requirement for the mass to pass through the 2 turnarounds would make the concept , non working...

must be nice to just pick a time requirement out of thin air to use in your favor in a discussion.

using 20 sets of tubes with 1 mass in each

in 1 year the pipe could travel a distance of
5.2 trillion miles.

using 40 sets

in 1 year the pipe could travel a distance of
10.43 trillion miles.

hey thats over 330,000 miles /second.

The speed of light is 186,000 miles per second

which can cut the 1 year travel time of
5.2 trillion miles in half to 6 months.













Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 09:52 PM

that is the CHANGE IN MOMENTUM OF THE BALL kallog.

the ball still presents the same force to the wall.
the force that is its mass * velocity.
or your used to using p=mv
the momentum that is its mass * velocity

just like the mass still presents the same force to
the U turns.



its the elastisity of the ball that allows it to bounce
off the wall.

like compressing a spring.
the ball is compressed like a spring because of its
momentum.





Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/22/10 10:19 PM

I must admit that given this new turn of events , I would have to say that I think we could travel to other solar systems using current energy technologies , that is if we
remain too stupid to use free energy sources.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/22/10 11:52 PM

Quote:

the 2 seconds that you are using is a complete falsity.


How long do you think it'll take? Here's a hint. F=3900N, m=100kg, initial speed = 39m/s, F=ma. Any ideas?

Or rather than wasting time, why don't we just change the arbitrary 3900N force that you chose. Make the turnaround force 100,000N and it'll take much less than 2s.



Quote:

because afterall that is what you are trying to do , you just cant do it because physics keeps getting in the way , spoiling your attempts.


I've noticed that every single insult you fire at me is actually describing yourself, not me. Where are your equations? I mean the ones using physics, not just made up. F=mv is just made up.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/23/10 02:33 AM

kallog

Quote:
How long do you think it'll take? Here's a hint. F=3900N, m=100kg, initial speed = 39m/s, F=ma. Any ideas?


if the mass is traveling at a velocity of 39 m/s
and the U turn is 39 meters long , it will take
exactly 1 second.

if the mass is traveling at a velocity of 39 m/s
and the U turn is 19.5 meters long , it will take
exactly 1/2 second.

if the mass is traveling at a velocity of 39 m/s
and the U turn is 9.75 meters long , it will take
exactly 1/4 second.

picture a car on a hill beside a U shaped valley , the car coast down the hill but the car begins to coast up the other hill.

the other hill is the same size and shape of the first hill.

if resistance and friction are zero , the car could coast
up to the top of the other hill.

but will the car need assistance from the hill to coast
up to the other hill?
if so and given that the hill itself will not move and provide a force that would assist the car up the hill
how then can the hill assist the car?

the momentum of the car causes the car to go down the first hill and up the second hill.

the car lost all of its momentum in doing so.

just like the momentum of the mass causes the mass to go half way through the U turn and then through the other half of the U turn.

except the mass has a veocity before it enters the U turn.

now suppose the car was dropped from a
height of 500 ft, the cars momentum would
be enought to send the car down one hill
and up the other hill , and all the way
back up to the altitude it was first dropped from.

Quote:
F=mv is just made up.


nope , its used in hydraulics all the time when describing a fluid pressure that acts against a piston.
there is no time added because the time is itself added

via a=f/m

ie.. apply a force of 10 lb
against a 1 inch hydraulic piston and you get a force
of 100 lb from a opposing 10 sq inch piston.

the F=mv is used to calculate the force that the piston
itself can place on the fluid because of its
mass * velocity.

a lighter piston = less force
a heavier piston = more force

a lighter piston less force required to move the piston itself.
a heavier piston more force required to move the piston itself.

some people work with pistons that weigh several thousand pounds.

thus f=mv


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 03:48 AM

Originally Posted By: paul

its final velocity is 52.193 m/s
its braking force is 5219.3 N


That force is chosen arbitrarily. It need not be 5219.3N. And in fact it cannot be now that you demand <<2s turnaround.

a=F/m
a = 5219.3N / 100kg
a = 52.2 m/s^2
At that acceleration/deceleration, how much time do you think it'll take to stop or turn around?

You must specify a higher braking force.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 04:03 AM

Originally Posted By: paul
that is the CHANGE IN MOMENTUM OF THE BALL

BINGO!


Quote:

the force that is its mass * velocity.

I don't care where you got this equation from or what other applications it's used in. IT IS AUTOMATICALLY WRONG. Why? Because the dimensions are inconsistent:
F = 100kg * 40m/s
F = 4000 kg m/s

Is "kg m/s" a unit of force? Can you convert it to newtons?
You can check with these google searches:
"4000 kg m/s in N"
"4000 kg m/s^2 in N"





Here's another reason it can't be right:
You claim:
F=ma
F=mv and
p=mv

Therefore:
F=p
ma=mv
a=v

Acceleration is always equal to velocity??



Quote:

or your used to using p=mv
the momentum that is its mass * velocity

At least we agree on something.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 04:06 AM

Originally Posted By: paul
remain too stupid to use free energy sources.


No, magic carpets are much better than free energy. Sadly whenever somebody flies a magic carpet, the CIA shoots them. That's why we find bodies dropping out of the sky from time to time.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/23/10 11:34 AM

Quote:

a=F/m
a = 5219.3N / 100kg
a = 52.2 m/s^2

At that acceleration/deceleration, how much time do you think it'll take to stop or turn around?


if the turn around is 52 meters long it will take exactly
1 second to turn around because its velocity is 52m/s.

if the turnaround is shorter it will take less time.

if you apply a force of 5219.3N for 1 second you can stop it in 1 second and 1 meter distance.

applying less force will not stop it as fast and will also require more stopping distance.

applying more force will stop it faster and in less stopping distance.





Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 12:01 PM

Originally Posted By: paul

applying less force will not stop it as fast and will also require more stopping distance.

applying more force will stop it faster and in less stopping distance.


Yea. So if we need <<2s turnaround time we have to use a higher force. 5219N takes 2s and 104m to turn the mass around.

100,000N?

a=F/m
a=-100,000N / 100kg
a=-1000m/s^2

time = (change in v) /a
time to stop = -52.2m/s / (-1000m/s^2)
time to stop = 0.0522s

distance to stop = -1/2 a t^2
distance to stop = 1.36m

Equations from equations of motion
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/23/10 12:06 PM

Quote:
Yea. So if we need <<2s turnaround time we have to use a higher force. 5219N takes 2s and 104m to turn the mass around.


Kallog , its not STOPPING !!!!

why cant you GRASP that ?

no force is REQUIRED to TURN IT AROUND
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 01:23 PM

Suppose a car's coasting towards you at 100km/hr.

You can turn it around using no force. So you can reach out your hand, grab the wing mirror, and the car will spin round and head off in the opposite direction as you let go at the right time.

F=ma. Please look up Wikipedia or a high-school general science textbook instead of saying made up things.


Originally Posted By: paul

Quote:

To a physicist, 'acceleration' includes a change in speed or direction. Look up wikipedia to get a more thorough picture. This is the same acceleration in F=ma so it's crucial that we use a consistent definition.


I have a clear understanding of that , and physicist also use decelerate to describe using a force to decelerate a object.

picking at straws are we?


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 01:50 PM

I think it's starting to come together now.

The recurring problems you have would be consistent with treating velocity and momentum as scalars. Maybe you don't realize they're vectors? F=ma and p=mv are actually a vector equations. They can give wrong results if you only put the magnitudes of the quantities into them.


Can an object accelerate if its speed remains constant?

If an object reverses direction, must its momentum change?

If an object reverses direction, must its velocity change?

If an object reverses direction, must its speed change?

Is it possible to apply an unbalanced force to a moving mass without changing its speed?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/23/10 02:07 PM

Originally Posted By: paul
I have a clear understanding of that , and physicist also use decelerate to describe using a force to decelerate a object.


which object is being decelerated then kallog?

just tell me which object you think is being decelerated and what causes it to decelerate.

ie..where does the force come from that decelerates the object?

at what velocity does the object have the entire time it is decelerating.

what is the objects final velocity when it has finished decelerating?

if you are speaking about the mass as the object that is being decelerated.

1) how can a object declerate without changing its velocity.
2) there is no 2


the change in momentum you are talking about is nothing but
a change in the direction of the object.

that change in direction does not have anything to do with
the momentum of the pipe.


the force that the object presents to the pipe as it presses against the pipe while turning around is the only force that the object can possibly apply to the pipe while turning around.

and it will present a force that is exactly its
mass * its velocity.

How Force is related to Momentum

Momentum measures the 'motion content' of an object, and is based on the product of an object's mass and velocity. Momentum doubles, for example, when velocity doubles. Similarly, if two objects are moving with the same velocity, one with twice the mass of the other also has twice the momentum.

Force, on the other hand, is the push or pull that is applied to an object to CHANGE its momentum. Newton's second law of motion defines force as the product of mass times ACCELERATION (vs. velocity). Since acceleration is the change in velocity divided by time, you can connect the two concepts with the following relationship:

force = mass x (velocity / time) = (mass x velocity) / time = momentum / time

Multiplying both sides of this equation by time:

force x time = momentum



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 02:41 PM

Originally Posted By: paul

just tell me which object you think is being decelerated and what causes it to decelerate.


I wouldn't use the term "decelerate" here because it's a bit vague. I would use "accelerate". The mass going through the u-bend is accelerating all the way through.



Quote:

at what velocity does the object have the entire time it is decelerating.

While it's turning around its velocity is continuously changing from, say 39m/s to the left, through 39m/s upward, till finally reaching 39m/s to the right. Its speed remains the same throughout.

Quote:

1) how can a object declerate without changing its velocity.

It can't. The velocity must change. Therefore, by definition, it is accelerating.


Quote:

the change in momentum you are talking about is nothing but
a change in the direction of the object.

Yes a change in direction. Which is a change in momentum. But not a change in the _magnitude_ of momentum.


Quote:

the force that the object presents to the pipe as it presses against the pipe while turning around is the only force that the object can possibly apply to the pipe while turning around.


Of course. That's the only force I've been talking about all the time, as well as its equal and opposite reaction. What force were you thinking of?


Most of what I just said will appear self-contradictory if you consider 'velocity' and 'speed' to be the same thing.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 02:51 PM

Originally Posted By: paul

and it will present a force that is exactly its
mass * its velocity.

[...]

force = mass x (velocity / time)


Please explain the difference between these two statements.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/23/10 03:01 PM

thank you kallog

now finaly we can connect the dots.

there is no force applied to the pipe because the mass changes direction.

the only forces that apply are the (+) and (-) forces
that the mass applies to the pipe while the mass passes
through the turnarounds.

and the (+)force that is applied to the pipe while the mass
is being accelerated.

these forces are all the same magnatude , therefore
there are 2(+) and 1(-)

the positive forces win , the negative forces loose.

since the total force that the mass can present to the first turnaround is the force that is the product of its mass * its velocity.
and
since the total force that the mass can present to the second turnaround is the force that is the product of its mass * its velocity.

the two forces cancel each other out

the only remaining force is the force that accelerated the mass.

and that force presses against the pipe causing the pipe
to accelerate.

therefore the pipe gains acceleration each time the mass is accelerated and the pipe continues to move in a direction.

with only 1 mass it will accelerate and then jerk back and forth and then be accelerated again then jerk back and forth again etc...etc...etc.

but the jerking back and forth will not stop the pipe from constantly gaining distance away from its initial starting point.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 03:28 PM

That's not connecting the dots, that's just repeating the same things you've been saying over and over again.

Originally Posted By: paul

the two forces cancel each other out


I agree the forces are of equal magnitude and opposite direction. But they're applied at different times. The sequence of turnaround forces to the pipe might look like this. I've ignored the accelerator for now -

+1 1st turn
-1 2nd turn
+2 1st turn
-2 2nd turn
+3 1st turn
-3 2nd turn
...

Since this is a continuous process, there's no reason we have to group 1st and 2nd turns together. We could group each 2nd turn and its following 1st turn together:

-1 2nd turn
+2 1st turn

-2 2nd turn
+3 1st turn

-3 2nd turn
+4 1st turn

Just by putting spaces between the lines, suddenly they no longer cancel each other out!


How about we include the accelerator:

-1 accelerator.
+1 1st turn.
-1 2nd turn.
-1 accelerator.
+2 1st turn.
-2 2nd turn.
-1 accelerator.
+3 1st turn.
-3 2nd turn.

Looks like unbalanced force because each group of three adds up to -1. But write them out in blocks of the same sequence, just grouped into blocks:

-1 2nd turn.
-1 accelerator.
+2 1st turn.

-2 2nd turn.
-1 accelerator.
+3 1st turn.


Clearly each set of 2nd turn, acceleration, 1st turn produces zero total force. They all cancel out!!!

All this is very rough, but it clearly shows that just having the same force at both ends doesn't mean you can ignore them.

Fundamental reason: They don't occur at the same time as each other. Even if you add more masses they can't consistently occur at the same time, try drawing an animation if you don't believe me.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/23/10 03:36 PM

I've actually connected the dots. It's all crystal clear now. (even if McCrystal isn't clear wink

You treat velocity and speed as the same thing. I assumed you knew the difference. I completely understand that my words will be contradictory and ridiculous if you go through all my posts and arbitrarily interchange those two terms.

I'm a bit embarrassed to have not noticed this for so long. But also you should have wondered why I kept changing words for the same thing. Surely not just to make my writing look pretty?

A similar problem exists with momentum. Changing direction means changing momentum.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/23/10 03:40 PM

thank you kallog

now finaly we can connect the dots.

there is no force applied to the pipe because the mass changes direction.

the only forces that apply are the (+) and (-) forces
that the mass applies to the pipe while the mass passes
through the turnarounds.

and the (+)force that is applied to the pipe while the mass
is being accelerated.

these forces are all the same magnatude , therefore
there are 2(+) and 1(-)

the positive forces win , the negative forces loose.

since the total force that the mass can present to the first turnaround is the force that is the product of its mass * its velocity.
and
since the total force that the mass can present to the second turnaround is the force that is the product of its mass * its velocity.

the two forces cancel each other out

the only remaining force is the force that accelerated the mass.

and that force presses against the pipe causing the pipe
to accelerate.

therefore the pipe gains acceleration each time the mass is accelerated and the pipe continues to move in a direction.

with only 1 mass it will accelerate and then jerk back and forth and then be accelerated again then jerk back and forth again etc...etc...etc.

but the jerking back and forth will not stop the pipe from constantly gaining distance away from its initial starting point.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 01:39 AM

Quote:
Looks like unbalanced force because each group of three adds up to -1. But write them out in blocks of the same sequence, just grouped into blocks:

-1 2nd turn.
-1 accelerator.
+2 1st turn.

-2 2nd turn.
-1 accelerator.
+3 1st turn.


Clearly each set of 2nd turn, acceleration, 1st turn produces zero total force. They all cancel out!!!


you are talking about change in momentum which is nothing but the direction the mass is traveling.

I am talking about the forces applied to the pipe.

the process begins with the accelerator.
then the first turn
then the second turn

lets examine the increase in total acceleration of the
pipe as the forces are applied to the pipe , instead of
using the change in momentum of the mass as you do.

to determine if the pipe will move.

+1 accelerator.
-1 1st turn.
+1 2nd turn.

result = +1

+2 accelerator.
-2 1st turn.
+2 2nd turn.

result = +2

total acceleration increase = +3

Im not sure what you hope to accomplish with your constant
usage of CHANGE IN DIRECTION OF THE MASS but you certainly cannot apply
CHANGE IN DIRECTION OF THE MASS as a force that moves the pipe.

Why dont you want to use
the actual forces that apply to the pipe
rather than use the change in momentum of the mass.

Quote:
Most of what I just said will appear self-contradictory if you consider 'velocity' and 'speed' to be the same thing.


velocity = 5 meters per second = 5 m/s
velocity = 5 miles per hour = 5 miles/hour

speed = 5 meters per second = 5m/s
speed = 5 miles per hour = 5 miles / hour

they are both the distance traveled by the object and the time the object traveled.











Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 01:39 AM

thank you kallog

now finaly we can connect the dots.

there is no force applied to the pipe because the mass changes direction.

the only forces that apply are the (+) and (-) forces
that the mass applies to the pipe while the mass passes
through the turnarounds.

and the (+)force that is applied to the pipe while the mass
is being accelerated.

these forces are all the same magnatude , therefore
there are 2(+) and 1(-)

the positive forces win , the negative forces loose.

since the total force that the mass can present to the first turnaround is the force that is the product of its
mass * its velocity.
and
since the total force that the mass can present to the second turnaround is the force that is the product of its
mass * its velocity.

the two forces cancel each other out

the only remaining force is the force that accelerated the mass.

and that force presses against the pipe causing the pipe
to accelerate.

therefore the pipe gains acceleration each time the mass is accelerated and the pipe continues to move in a direction.

with only 1 mass it will accelerate and then jerk back and forth and then be accelerated again then jerk back and forth again etc...etc...etc.

but the jerking back and forth will not stop the pipe from constantly gaining distance away from its initial starting point.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 06:45 AM

Originally Posted By: paul

you are talking about change in momentum which is nothing but the direction the mass is traveling.

I am talking about the forces applied to the pipe.

I'm also talking about the forces applied to the pipe. That's what I said. Please read more carefully.


The accelerator provides twice as much force the 2nd time round? OK that's fine. But it means the turnarounds also provide even more force:

+2 accelerator.
-3 1st turn.
+3 2nd turn.



Move the spaces in the list:


+1 accelerator.
-1 1st turn.

+1 2nd turn.
+2 accelerator.
-3 1st turn.

+3 2nd turn.
+3 accelerator.
-6 1st turn.

+6 2nd turn.
+4 accelerator.
-10 1st turn.


See how each group of 3 forces adds to zero? I've continued your pattern of the accelator's force increasing by 1 each cycle. But you can do that differently if you want, and they still add to zero.









Quote:

velocity = 5 meters per second = 5 m/s
velocity = 5 miles per hour = 5 miles/hour

speed = 5 meters per second = 5m/s
speed = 5 miles per hour = 5 miles / hour

they are both the distance traveled by the object and the time the object traveled.


No they're not the same. I'm really sure this is a big part of your consistent failure to understand anything I write. Until you find out the difference we may as well give up. I'm not going to the trouble of rewriting all my calculations using only speed and no velocity.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 12:03 PM

Quote:
The accelerator provides twice as much force the 2nd time round? OK that's fine. But it means the turnarounds also provide even more force:


No kallog the the total acceleration of THE PIPE increses
each cycle.

the same force is used ie...

2nd cycle the pipe total acceleration increases from

a lower total acceleration to a higher total acceleration.

because the mass accelerates to a higher velocity each cycle.

I dont think I have used the term "speed" so far other than
while trying to convince you that the direction of the mass will not accelerate the pipe , what will accelerate the pipe is the force that is placed on the pipe.

you have never added a direction to velocity , normaly
when you see velocity written down you do not also see
a direction.

ie
50 m/s/north
50 miles per hour/south

so since you insist on using direction from now on
we will attach a direction as we discuss movement.

we will use north and south and left and right
when talking about the moving mass and the moving pipe.

and we will say that the mass is
initialy accelerated north.
then turns left
then travels south without acceleration
then turns right
then is accelerated north again.

this way we can both use a more correct description of velocity.

ie...

accelerated from 0 m/s to 50 m/s/North
50 m/s/Left
50 m/s/South
50 m/s/Right

accelerated from 50 m/s/North to 100 m/s/North
100 m/s/Left
100 m/s/South
100 m/s/Right

etc...etc...etc...etc...



so were both guilty of using velocity the normal way
which is the wrong way , its just that you seem to want to make an issue of it , so in this discussion I will adapt to
the new terms.

if you have used velocity the right way in this or any other discussion in this forum , please post a link so that we can see.

Quote:
I'm really sure this is a big part of your consistent failure to understand anything I write.


if so then it is also a big part of your consistent failure to understand anything I write.

Quote:
The accelerator provides twice as much force the 2nd time round? OK that's fine. But it means the turnarounds also provide even more force:

+2 accelerator.
-3 1st turn.
+3 2nd turn.



according to the above (calculation?) of yours the turnaround accelerates the mass even faster because
you show -3 in turn #1 plus +3 in turn #2
that are the result of the +2 force from acceleration.

even so adding up +2 plus -3 plus +3 shows pipe acceleration. because you end up with +2

but I am curious how you determned that the turnaround adds force , please explain this fantasy of yours , how does the turnaround provide even more force?

and while your explaining please use math.
instead of just picking numbers and forces out of thin air.

but just think kallog if your right then you have discovered another way to produce free energy !! you could capture that extra fantasy force and use it , you still have your original force.


I think you have a extreme lack of the understanding of force and motion




Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 12:53 PM

Originally Posted By: paul

you have never added a direction to velocity , normaly
when you see velocity written down you do not also see
a direction.


All the time I've been treating it as a 1-dimensional system, so there are only 2 directions. I used +/- to indicate these - a 1-dimensional vector is just a number which can be either positive or negative so that's how I wrote them. If you consider it as 3D then it's the same except all the vector quantities are just the x-components of the 3D vectors. You can ignore the y and z components because any forces in those directions do nothing to the motion of the pipe along its length. The geometry can be set up so they cancel out anyway.


Quote:

accelerated from 0 m/s to 50 m/s/North
50 m/s/Left
50 m/s/South
50 m/s/Right


That hides important accelerations:
accelerated from 0 m/s to 50 m/s/North
accelerated from 50m/s/North to 50 m/s/Left
accelerated from 50m/s/Left to 50 m/s/South
accelerated from 50m/s/South to 50 m/s/Right

But it's sufficient to consider only 1 component of velocity for our purposes. That makes it much much simpler.

If you don't like the 1D simplification, then just replace the u-bends with springs. I've been treating them the same as springs all the way through.

Quote:

please explain this fantasy of yours , how does the turnaround provide even more force?


The force at the turnaround is determined by the mass's speed. Each time round it gets faster so applies more force. Even if you switch off the accelerator the mass will keep going round applying high forces at each turnaround, despite the accelerator's force being zero.

I've written out all the equations ages ago. You can go back and find them if you want.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 01:01 PM

Quote:
If you don't like the 1D simplification, then just replace the u-bends with springs. I've been treating them the same as springs all the way through.


if your using springs then how are you getting more acceleration out of a spring.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 01:09 PM

Originally Posted By: paul
Quote:
If you don't like the 1D simplification, then just replace the u-bends with springs. I've been treating them the same as springs all the way through.


if your using springs then how are you getting more acceleration out of a spring.


More than what? Throw a mass against a spring and it applies a force. The average force depends on the mass, speed and spring constant.

That force is exactly the same as the average of the component of force in the longitudinal direction for a mass going through a U-bend that takes the same amount of time to turn around.




Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 01:44 PM

Quote:

+1 accelerator.
-1 1st turn.

+1 2nd turn.
+2 accelerator.
-3 1st turn.

+3 2nd turn.
+3 accelerator.
-6 1st turn.

+6 2nd turn.
+4 accelerator.
-10 1st turn.

See how each group of 3 forces adds to zero? I've continued your pattern of the accelator's force increasing by 1 each cycle. But you can do that differently if you want, and they still add to zero.


you just added up the first two positive numbers then
put that same number in as the negative number , of course
they will always add up to zero if we use your type of
physics , LOL...

lets do this a more correct way.
+1 accelerator.
-1 1st turn.
+1 2nd turn.
------------------------------- = +1
+2 accelerator.
-2 1st turn.
+2 2nd turn.
------------------------------- = +2
+3 accelerator.
-3 1st turn.
+3 2nd turn.
------------------------------- = +3

Total positive force = +12
Total negative force = -6
_____________________________
Total resultant foce = +6

each cycle adds acceleration.

See how the 3 groups of forces adds to +6? I've continued your pattern of the accelator's force increasing by 1 each cycle. but I didnt include any additional fantasy forces as you have , But you can do that differently if you want, and they still add to +6 , unless you use your fantasy forces that is.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 02:47 PM

Doesn't matter. The same set of forces can either "balance" or "not balance", depending on how you group them.


Here's the same numbers non-balanced:

+1 accelerator.
-1 1st turn.
+1 2nd turn.

+2 accelerator.
-3 1st turn.
+3 2nd turn.

+3 accelerator.
-6 1st turn.
+6 2nd turn.

All this just demonstrates the meaninglessness of trying to balance forces that aren't being applied simultaneously.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 02:51 PM

I've done more than my fair share of the leg-work.

How about you just write out your calculations that show the motion of the pipe?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 02:58 PM

Doesnt matter , the positive forces will still add up
to more than the negative forces , so the pipe will move.
no matter how you word it , or change the numbers around.

but if you use fantasy numbers you will always be able to
end up with what ever number you want.

and that is what you do.

+1 accelerator.
-1 1st turn.
+1 2nd turn.
--------------------- = +1
+2 accelerator.
-3 1st turn.
+3 2nd turn.
--------------------- = +2
+3 accelerator.
-6 1st turn.
+6 2nd turn.
--------------------- = +3

Total resultant force = +6

the pipe moves because of the positive force.

as far as I am concerned I have won this discussion because
of my use of non - fictional forces and numbers , you have done nothing but try to use fictional forces and went as far as adding up those fictional forces to arrive at fictional numbers.

I dont think that anything you could post would prove me wrong , so given that you cheat.

I win the discussion and you loose.

reactionless propulsion is a fact.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 03:08 PM

Here's your exact same numbers:

+1 accelerator.
-1 1st turn.
--------------------- = 0
+1 2nd turn.
+2 accelerator.
-3 1st turn.
--------------------- = 0
+3 2nd turn.
+3 accelerator.
-6 1st turn.
--------------------- = 0
+6 2nd turn.
?
?

Total resultant force = 0 + unknown last section which is probably zero too.

the pipe does't move because of the zero force.

as far as I am concerned I have won this discussion because
of my use of non - fictional forces and numbers , you have done nothing but try to use fictional forces and went as far as adding up those fictional forces to arrive at fictional numbers.

I dont think that anything you could post would prove me wrong , so given that you cheat.

I win the discussion and you loose.

reactionless propulsion is a fallacy.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 03:11 PM

Well now that you're right, and you've been proven right, what will you do? Here's some suggestions:

Build it
Patent it
Get a Nobel prize
Become richer than God



Or another possibility:

Do nothing
Feel frustrated
Get nothing
Die poor and forgotten by the world



It's up to you!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 03:17 PM

As I said before I highly doubt that I am the first to
think of this concept for propulsion.

and just as soon as I inform ( the government ) by
applying for a pattent from ( the government ) then
I would be slapped in the face with a gag order or worse.

you apply for a pattent for it , you get rich , LOL.

if you live through that then I will apply for a pattent for the free energy machine to power the accelerator.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 03:30 PM

Quote:
Here's your exact same numbers:

+1 accelerator.
-1 1st turn.
--------------------- = 0
+1 2nd turn.
+2 accelerator.
-3 1st turn.
--------------------- = 0
+3 2nd turn.
+3 accelerator.
-6 1st turn.
--------------------- = 0
+6 2nd turn.
?
?

Total resultant force = 0 + unknown last section which is probably zero too.



your right you just added wrong.

you forgot to add the +6 2nd turn at the end.

3 cycles = 3 forces *3 cycles = 9 forces

you included 9 forces but only added 8

so the Total resultant force = +6

go ahead and add more sections as long as you dont leave out parts the same will occur.

you try to cheat , but always get caught.LOL

so the rest of your post is meaningless other than to
prove that you have a talent for cheating.

Quote:
Total resultant force = 0 + unknown last section which is probably zero too.

the pipe does't move because of the zero force.

as far as I am concerned I have won this discussion because
of my use of non - fictional forces and numbers , you have done nothing but try to use fictional forces and went as far as adding up those fictional forces to arrive at fictional numbers.

I dont think that anything you could post would prove me wrong , so given that you cheat.

I win the discussion and you loose.

reactionless propulsion is a fallacy.



Physics isnt about cheating kallog.

but dont feel bad about not having a talent for physics
because like you , most dont , they just had a good talent for cheating while taking physics courses in college.

your physics skills are a fallacy.

but reactionless propulsion is a fact.






Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/24/10 04:34 PM

I thought your scientific credibility couldn't sink any lower - and then you go and say something like this:
Originally Posted By: paul
but reactionless propulsion is a fact.

Every real physicist on the planet disagrees with you. Reactionless drives are a physical impossibility, defying Newtons laws, conservation of momentum, and general relativity.

You're "model" is properly termed an oscillation thruster. It is well understood to be a physical impossibility - no matter how fast/slow you move your masses, or in whatever configuration they are moved, the total momentum will always balance out to zero.

Some other sources, explaining why your system will not work. You'll notice that their arguments are the same as kellogs and mine:
*Atomic rockets: reactionless drives
*NASA's list of common propulsion errors
*Mathpages description of oscillation drives

Physics: 1,000,000,000,000. Paul: 0.

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 04:43 PM

I dont care what you think imagegeek , I already told you that.

your the one who thinks the pipe wont move at all and
the only proof you have is the misconceptions of newtons laws.

I dont believe that I want to discuss this any further with either of you.

perhaps there are other forums that have inteligent people
in them that can discuss this in an inteligent manner
vs those in this forum who dont even appear inteligent.

only brainwashed.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 04:49 PM

Originally Posted By: paul

your right you just added wrong.

you forgot to add the +6 2nd turn at the end.


Add what? This whole idea of adding up made-up forces applied at different times, and for different durations, is totally meaningless. I'm just showing these additions to emphasise that you can get different results through the same process. Therefore the process is wrong.

I've already detailed my calculations that show the overall motion of the pipe a couple of times. You can go through them and search for faults if you like.

But why don't you produce any actual calculations? I understand if you don't have the skill to do that, but I think you do have a reasonable workshop, and surely the skill to build a prototype of this thing. All you need is some pipes, magnets, maybe some little wheels or an air blower. All cheap easy stuff. That'll be far more powerfully convincing than any explanation.

Keep it secret if you're genuinely scared the CIA will shoot you. Then secretly release a video of it - you know, the way Bin Laden secretly released videos without revealing where he was.

By the way, they can already track you through your posts on here. So it's too late, better get your will written and start winding up your bank accounts and so on.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/24/10 05:00 PM

Originally Posted By: ImagingGeek


Haha, I wish I'd bothered to look it up. Hey Paul somebody else already invented it sorry!

But I realised Paul's device does do something, it spins around, faster and faster, while at the same time oscillating along its length. Would really spill your coffee if you're the poor astronaut tasked with flying it to Alpha Centuri.
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/24/10 06:54 PM

Originally Posted By: paul
I dont care what you think imagegeek , I already told you that.

your the one who thinks the pipe wont move at all and
the only proof you have is the misconceptions of newtons laws.

I dont believe that I want to discuss this any further with either of you.

Paul, I've provided citations from experts in the field - rocket engineers, NASA and mathematicians who specialize in this kind of stuff.

They all say you're full of [censored].

I don't give a crap what you think of my "opinion"; the simple reality is the entierty of the scientific community says you're wrong, the physics support that, and you have not presented one iota of evidence that counters any of those claims - be them mine, or NASA's.

As the saying goes - put up, or shut up1

Bryan

1 By which I mean, provide evidence that counters the evidence/math in the citations I provided
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/24/10 06:55 PM

Originally Posted By: kallog

But I realised Paul's device does do something, it spins around, faster and faster, while at the same time oscillating along its length. Would really spill your coffee if you're the poor astronaut tasked with flying it to Alpha Centuri.


LOL, so paul's basically invented an orbiting clothes washer...or paint mixer...

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 07:27 PM

I dont believe that I want to discuss this any further with either of you.
reactionless propulsion
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 07:32 PM

I dont believe that I want to discuss this any further with either of you.
canoe with reactionless propulsion
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 06/24/10 07:49 PM

I dont believe that I want to discuss this any further with either of you.

Mit potential energy to kinetic energy
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 06/25/10 03:55 AM

Originally Posted By: paul
I dont believe that I want to discuss this any further with either of you.


Translation: "I realize it can't work, but I'm too proud to admit I was wrong."
Posted by: ImagingGeek

Re: Orion, Mission to Alpha Centauri - 06/25/10 03:30 PM

Originally Posted By: kallog
Translation: "I realize it can't work, but I'm too proud to admit I was wrong."



GOOOOOOOOOOOOAAAAAAAAAAAAAAAAALLLLLLLLLLLLLLLLL!
wink

Bryan
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/05/11 06:50 PM

#34780 - June 07, 2010 10:54 PM

Quote:
"It doesn't continue to move, it stops. The pipe had momentum while it was moving, but it soon transfers that to the internal gasses travelling in the opposite direction, stopping them both."


how does the pipe stop?
what causes the pipe to stop?

the air is released from the tank causing the pipe to move due to the mass of air escaping the tank.

the air inside the pipe and tank equalize.

there is no force applied to the pipe that could stop the pipe.

if so then what force and where?



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/06/11 04:36 PM

Hello again Paul smile


Originally Posted By: paul
#34780 - June 07, 2010 10:54 PM
the air inside the pipe and tank equalize.

there is no force applied to the pipe that could stop the pipe.

if so then what force and where?


This all sounds familiar. The 'what force' is the force of the air hitting the pipe. 'where' is all over the inside of the pipe, but more on the back end because that was moving towards the air and crashed into it at a higher relative velocity than the other parts of the pipe.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/06/11 04:57 PM

Quote:
but more on the back end because that was moving towards the air and crashed into it at a higher relative velocity than the other parts of the pipe.


if you release pressurized air into outer space from a space ship do you think the air would travel in a straight line?

no! the air would expand greatly because it is at a higher pressure and is entering a low pressure area.


the force of air inside the pipe on its sides would not apply a force that would stop the pipe.

and remember the end of the pipe is 500 ft away.

so by the time the air reaches the end of the pipe it has diffused into a 50 inch diameter cross section over the entire 500 ft length of the pipe.

so there may be a tiny amount of force applied to the end of the pipe 500 ft away from the nozzle where the air comes out of the tank , but hardly enough to slow the pipe much less stop the pipe.

because as the air mass is moving out of the nozzle
as force (F) which is higher than any (-F) force being
applied at the end of the pipe 500 ft away the force that
causes the pipe to move is greater than any force that
could slow or stop the pipe.

these forces are not applied at seperate times they are applied at the same time.

the greater force overcomes the lesser force.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/07/11 10:12 AM

Originally Posted By: paul
applied at the end of the pipe 500 ft away the force that
causes the pipe to move is greater than any force that
could slow or stop the pipe.


It comes back to the old issue of the time the force is applied for. Now also the area it's applied over. A tiny pressure over a larger area for a long time can cause a lot of acceleration/decelleration.

Also, there would be some longitudinal force applied by the air interacting with the sides of the pipe. If you make the pipe longer to reduce the force on the end cap, you're achieving that by increasing the force on the walls.

Imagine it's millions of ball bearings instead of air. Then what'll happen to them?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/07/11 03:05 PM

thats what I meant when I said the greater force overcomes the lesser force.

the force generated by the air moving through the nozzle is all directed in one direction.

the force applied to the sides would not cause the pipe to move at all because all sides would feel the same force.

the tiny force that is capable of reaching the end will not slow the pipe to any extent.

so the pipe accelerates , and just keeps going.

Quote:
Imagine it's millions of ball bearings instead of air. Then what'll happen to them?


if a ball bearing is ejected through the nozzle it would go all the way to the end and it would impact the end with the same force that was used to propel it.

but ball bearings do not compress and if we were not using compressed air or gasses or something to propel the ball bearings the ball bearings would just sit in the tank and never leave.



Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/07/11 03:22 PM

heres another take on the reactionless propulsion pipe idea.

a steam powered space ship engine...

you could actually boil water in a space ship using solar collectors then direct the steam through a nozzle.

and if you have a nuclear reactor on board that can allow for the production of steam you can travel further from he sun.

but it all boils down to , you dont need to throw something out of a space ship to make the spaceship move , you can throw something inside the ship.


like I said when this all started.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/09/11 03:56 AM

Originally Posted By: paul
the force applied to the sides would not cause the pipe to move at all because all sides would feel the same force.


I didn't mean normal force against the wall, but longitudinal force in the direction of the air's initial motion. ie. parallel to the pipe walls. For a long pipe most of the air's momentum will be transferred to the pipe walls instead of the end cap. But in both cases it's in the same direction.

If you could somehow eliminate the drag on the pipe walls, then even for a long pipe the entire longitudinal momentum of the air when it was released from the tank will be transferred to the cap at the other end of the pipe.

Either way, no motion further than the length of the pipe.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/09/11 04:03 AM

Originally Posted By: paul
heres another take on the reactionless propulsion pipe idea.

Finding ways to use it really isn't an issue. Anyone can easily think of such ideas. But we can also easily think of ways to use magic carpets. Doesn't really do any good.

Quote:

like I said when this all started.

You said many times, but never once bothered to actually demonstrate, either with a correct model based on accepted physics, or a working prototype.

Same as the millions of others who had this same idea. Not one of them could show that it worked either. It's an old and widely known concept, but one which has never ever been built or shown theoretically to work.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/09/11 05:04 AM

Quote:
Either way, no motion further than the length of the pipe.


this is where your thinking is flawed , you somehow think the pipe will stop when it has traveled its lenght.

that doesnt really make much sence.

and you base that on the air that strikes the pipe after leaving the nozzle.

Quote:
For a long pipe most of the air's momentum will be transferred to the pipe walls


this also makes no sence , the air will expand into the entire inside area of the pipe so most of the air will
not "magically" as you like to say , expand out to the
the pipe walls.

I seriously doubt that the air that does rub against the walls of the pipe would cause much resistance to movement.

and the air that actually touches the walls is probably only around .005 percent of the volume of air that leaves the air tank.

so far you havent offered any meaningful opposition.

Quote:
further than the length of the pipe.


and one more thing , if the air is released very slowly
the pipe could actually travel several times its lenght before the air tank empties or equalizes.

have you ever considered that?


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/09/11 10:18 AM

Originally Posted By: paul
and the air that actually touches the walls is probably only around .005 percent of the volume of air that leaves the air tank.

Sure. But I said that was for a long pipe. If you keep making it longer, the wall friction will keep increasing, while the end cap force will decrease by the same amount. Eventually the wall friction will dominate.


Quote:
Quote:
further than the length of the pipe.

and one more thing , if the air is released very slowly
the pipe could actually travel several times its lenght before the air tank empties or equalizes.


However you do it, after the pipe has traveled its length, there's no possibility that any of the air could still be going in the same direction it was when it left the tank, without having turned around at some point. That's where the length-of-pipe limit comes from. Just try to draw a picture of that, it's geometrically impossible!

The slower you release the air, the more mass is left in the tank while you're trying to accelerate it, so the less distance it can go before the released air _has to_ change direction. This reduces the length-of-pipe limit. The length-of-pipe limit can only be approached when the mass of the released air is much higher than that of the pipe and any un-released air.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/09/11 07:18 PM

Quote:
If you keep making it longer, the wall friction will keep increasing, while the end cap force will decrease by the same amount. Eventually the wall friction will dominate.


I didnt make it longer its 500 ft , let me check , brb

---- time passes ----

yep it was 500 ft from the begining

post #34663

Originally Posted By: paul
I think thats a load !!!

if I have a nozzle at one end of a long pipe and the entire pipe
is at 14.7 psi.

lets say the pipe is 60 inch diameter and 500 ft long.


Quote:
Eventually the wall friction will dominate.


the wall friction might dominate between the wall friction and the force felt by the pipe end as a small force that would subtract from the larger force that is generated as the compressed air exits the air tank and passes through the nozzle.

but the resulting force is why the pipe keeps moving.

Quote:
However you do it, after the pipe has traveled its length, there's no possibility that any of the air could still be going in the same direction it was when it left the tank, without having turned around at some point. That's where the length-of-pipe limit comes from. Just try to draw a picture of that, it's geometrically impossible!


that really makes no sence at all.

if i allow a burst of air to excape the tank the pipe would begin to move.

and there is nothing to stop the pipe from moving.

if I wait until the pipe has traversed its lenght and again allow another burst the pipe will accelerate even faster
and I could repeat this until the air in the air tank equalized with the air in the pipe.


Quote:
it's geometrically impossible!


true , which causes the single directional force of the air leaving the nozzle to be the dominant force when determining which force is the greatest force that would give momentum to the pipe.





Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/10/11 02:24 PM

If the pipe travels its length, then obviously none the air in it can remain where it was. Or it would become outside the pipe, left behind floating in space. So any air that was going in the opposite direction to the pipe has to change direction. That imparts a force to the pipe pushing it backwards.

There really is no way to make progress on this without either:
- Using the law of conservation of momentum or
- Working it out in more detail without giving up just before you get to the solution.

Claims of "there's nothing to stop it" have no value, because they're just claims without reasons.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/10/11 11:39 PM

Quote:
If the pipe travels its length, then obviously none the air in it can remain where it was.


I never said that the air would stay where it was.

the majority of the air moves from the air tank to the pipe.

Quote:
So any air that was going in the opposite direction to the pipe has to change direction. That imparts a force to the pipe pushing it backwards.


I agree that the air will be changing direction inside the pipe as the air settles inside the pipe.

but the forces that the air will impart to the pipe during the settling process will not all be in the direction
that would slow the pipe.

the air will apply the same force to the front of the pipe as it will the end of the pipe but most of this force will be to the sides of the pipe.

in other words the negative forces and the positive forces that the air leaving the nozzle applies to the pipe will cancel each other out.

leaving the force generated by the nozzle as the dominate force.

I think there is an easy way to settle this using a experiment.

using pvc pipe and a couple of valves compressed air and water.

and a swimming pool in place of outer space.

releasing the water that is being compressed by air through a pipe that connects to another much larger pipe , should
provide enought visual evidence.














Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/12/11 05:01 AM

Originally Posted By: paul

but the forces that the air will impart to the pipe during the settling process will not all be in the direction
that would slow the pipe.

Sure. But on average they will be in the same direction. It's easy to imagine a random cloud of gas that just disperses in all directions the same. But this isn't like that, the entire cloud of gas starts off moving in one specific direction. It doesn't just forget that it was moving backwards rather than forwards.



Quote:
using pvc pipe and a couple of valves compressed air and water.

and a swimming pool in place of outer space.

Excellent idea, and extremely practical and doable. Go for it. You could use an ordinary PVC toilet pipe as the main pipe and a coke bottle water rocket as the tank. Then the valve can just be a pull-string operated flap in the cap of the coke bottle.

If it doesn't work in a pool, put the same device on a wheels or ice-cubes.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/12/11 05:50 AM

how about just a inflated balloon and a larger un-inflated balloon or a trash bag with a drinking straw between them?

you could pinch off the straw to keep the pressurized air in the inflated balloon.

and if it moves when you release the straw because the air is moving into the bigger balloon or trash bag then that should be good enough.

and really cheap.

we all know that if you blow up a balloon it will fly away if you release it.

without actually doing this simple experiment I can already
say that I believe it will move , and quickly.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/12/11 04:00 PM

Originally Posted By: paul
how about just a inflated balloon and a larger is moving into the bigger balloon or trash bag then that should be good enough.


Believe it or not, I actually did this just now! And it didn't work! I used an emptied pen tube instead of a straw. I tried it with the pen attached to the balloon and it flew along a bit, but much slower than a normal balloon. Then I did it with a very lightweight loose plastic bag on the other end of the pen, and it just dropped to the floor while it was deflating. Absolutely no forward motion!



Quote:

without actually doing this simple experiment I can already
say that I believe it will move , and quickly.


Form your belief after seeing the evidence, not before! If you don't trust my experiment then improve on it. Maybe the mass of the bag stopped it flying away, so tie the screwed up bag to the side of the open straw and see if it also doesn't fly away. Maybe the back pressure from the bag slowed the balloon's deflation. So measure the time compared to doing it with no bag.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/12/11 07:31 PM

all I can say to your experiment is ...

how much did the empty pen tube weigh?

and how large was the smallest portion where air could pass through it.

it doesnt look like you blew the balloon up very much at all judging from the amount of air in the bag...

try the same with a lighter straw and see what happens.
please dont use a coffee stirring straw.

and one more thing you might try using a string attached to the ceiling of your room and also attached to the tube.

and instead of expecting it to move upward against gravity just see if it will move horizontaly.

I bet it does.


Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/13/11 03:35 AM

Originally Posted By: paul
all I can say to your experiment is ...


This experiment should put the matter to rest for good. I won't repeat because I know you'll never trust my results, no matter how many adjustments I make at your request. You have to see it with your own eyes.

If you haven't done it by tomorrow I expect a good reason why not! Don't say it was too rainy to go across town to the balloon shop, just do it! Remember, if this is true then the world stands to gain enormously. You have a social responsibility to make it known.

Also, make sure that before you do the experiment, you know what conclusion you can draw from whatever result it might have. It's no use trying it, and it doesn't work, so you say "oh there's some elusive effect I can't correct for, so no conclusion".

Use controls to eliminate concerns about direction, tube diameter, weight, etc. Repeat it with different balloons and after having blown them up and deflated them several times because they become looser with repeated use. Control every factor that might become an excuse to ignore the result.

By the way, it's quite inconvenient trying to attach the inflated balloon to the tube. See if you can find some other kind of tube that can be closed off but won't collapse under the force of the balloon tightened over it. Maybe use a pen tube with a cap on it, inside the plastic bag. Then you can pop the cap off through the bag.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/13/11 04:28 AM

Im thinking of using one of the long balloons that can be
used to make animal shapes with.

put part of a plastic drinking straw inside the balloon
and then roll 1/4 of the balloon back over the other 3/4 of the balloon then glue the balloon to the straw , allow the glue to dry , and maybe put a little tape around the tube area to keep the balloon from pulling away from the straw when I release the air through the straw , and keeping the straw pinched off I will blow up the end that is only 1/4 the balloons length and twist the balloon where the straw is to keep the air from going through the straw , then tie the balloon off.

then I will stretch the 3/4 of the balloon that is not inflated to limber it up.

then let it fly.

but this might not work because the air will be blowing up the other 3/4 of the balloon which will subtract from the force , so then I will probably try to find a long narrow plastic bag to use that can contain the volume of air that will be inside the inflated balloon.

either way I wont be finished by tomorrow because the stores are mostly closed and I wont walk inside a wallmart.





Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/14/11 06:08 PM

well , found some of the long balloons but they are so cheaply made that they burst before they can be fully inflated.

I will most likely have to try and find some of the thicker material balloons that wont burst so easily.

and since I refuse to be a part of americas destruction by entering a wallmart , I will have to wait until I can find an american made balloon that isnt so cheaply made that will work.

if thats possible.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/14/11 06:40 PM



its amazing that there is still a manufacturer in america.

thank goodness

well the above web site has 3 ft , 6ft even 8 ft balloons.

but I've decided to use 2 empty coleman propane cylinders.

and connect them together and fill one with compressed air
and put a valve in between the two.

they will float on water and they should move because of the air equalizing.




Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/15/11 01:52 PM

Originally Posted By: paul
its amazing that there is still a manufacturer in america.

Hehe, I'm surprised too :P

Quote:

but I've decided to use 2 empty coleman propane cylinders.

and connect them together and fill one with compressed air
and put a valve in between the two.

they will float on water and they should move because of the air equalizing.


Fantastic! The rigid outside shape will make it much more conclusive by preventing any aerodynamic effects of the expanding bag/etc. Do you already have the gear to build it?
Posted by: Amaranth Rose II

Re: Orion, Mission to Alpha Centauri - 03/16/11 02:57 AM

I understand your frustration with the long balloons. Having made a bunch of balloon animals, I discovered that if you stretch them lengthwise several times before you inflate them they don't burst so bad. It exercises the rubber and makes it more flexible so they don't burst. I went through a few balloons myself before I discovered this. Good luck with the propane cylinders, and remember even an empty one still has a volume of propane in it, so be careful.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/16/11 06:46 PM

Rose

I went through several myself and I can imagine what entertainers are going through with the cheaply made chinese balloons , but you get what you pay for and in a republican controlled economy there is no end in sight of cheap.

even the chinese businessmen are moving their businesses to africa where even cheaper labor is available.

and thanks for the warning.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/16/11 10:24 PM

Originally Posted By: paul
Rose
I went through several myself and I can imagine what entertainers are going through with the cheaply made chinese balloons , but you get what you pay for and in a republican


As the only one who did this experiment successfully, without a single balloon failure, I have to say mine was (I only needed one and reused it!) Chinese made, and I got it from Walmart. Maybe there's a reason everybody buys that stuff wink
Posted by: Bill S.

Re: Orion, Mission to Alpha Centauri - 03/17/11 01:05 PM

[quote=K]I got it from Walmart.[quote]

Are you allowed to advertise on this Forum?


BTW, Kallog, I’ve lost track of the thread in which you asked what a libertarian was, so I’m going to slip it in here in deference to the links between politics and sci-fi.

I’m not as politically orientated as my son, so treat my answer with caution. I understand it has to do with supporting unrestricted trade, no subsidies or anything like that, and a belief that we all get what we deserve. If we are poor, it’s because we failed to take the opportunities we should have grabbed, so it’s our own fault.

I was going to say “every man for himself” but I’m sure that's politically incorrect, these days. Somehow, “every person for itself” doesn’t have the same ring to it, though!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/18/11 08:12 PM

Quote:
As the only one who did this experiment successfully


I still dont think that you inflated your balloon enough

its obvious that if you would have inflated the balloon properly the plastic bag would have been bigger or looked as
if it was filled with pressurized air from the balloon.

if you would have used an american made balloon the experiment might have worked because you could have blown it up much more than the cheap chinese balloon you used.




Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/19/11 02:47 AM

Originally Posted By: Bill S.
If we are poor, it’s because we failed to take the opportunities we should have grabbed, so it’s our own fault.


OK yea I guess I am, or rather I'd like a world where that worked. It isn't so good in reality when you have nepotism and all that.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/19/11 02:53 AM

Originally Posted By: paul
I still dont think that you inflated your balloon enough

As I already said, I tried it without the bag. I used the same amount of inflation in both cases. So it was enough, because it did travel without the bag.

But you are kind of right, I was afraid to blow it up too much because it was the last good balloon I had left and I didn't want it to break :P

Anyway! That took me 10 minutes one evening. Why are you still unable to do anything at all? What's causing these endless delays? Have you got the filler valve yet? Have you just got some round balloons like I used? Do it yourself!!!

It doesn't have to be fancy, just get a normal balloon, a plastic bag, a pen tube, some tape, and a rubber band. That's all you need to shatter the world of physics and get a Nobel prize!!! Do it!

You can also use a condom. Those are made to higher quality standards and are more readily available than balloons, and can be inflated further. In fact it would probably be more effective because of the bigger opening.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/20/11 11:59 PM

Quote:
because it did travel without the bag.


well an aircraft will take off as long as its payload is not too heavy.


Quote:
because it was the last good balloon I had left and I didn't want it to break


yea SEE , I went through around 8 of the 12 in the pack
and I used the pump that came with them.

Quote:
t doesn't have to be fancy, just get a normal balloon, a plastic bag, a pen tube, some tape, and a rubber band. That's all you need to shatter the world of physics and get a Nobel prize!!! Do it!


I think that something so earth shattering should be done properly , with proper documentation , thats why.


Quote:
Those are made to higher quality standards and are more readily available than balloons, and can be inflated further. In fact it would probably be more effective because of the bigger opening.


I thought you agreed that a metal tank would be much better
I still think it would and that is why Im going to use one.

of course I could just make a tank out of 12 inch PVC pipe that would hold 420 psi because its minimum bursting pressure is 420 psi.

my compressor will only pump up to 130 psi.

so that should be safe enough.










Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/21/11 06:30 AM

Originally Posted By: paul
I think that something so earth shattering should be done properly , with proper documentation , thats why.

Better to start half-assed then move up, than to have grand plans that never get finished because of too many details.

But yea if the balloon thing worked there'd be a possible excuse that the changing shape somehow pushed the external air around, propelling it. So it can show you're wrong but not right. The rigid tanks can clearly show right or wrong.


Quote:
of course I could just make a tank out of 12 inch PVC pipe that would hold 420 psi because its minimum bursting pressure is 420 psi.

Be careful, that might explode. If the pipe's designed only for liquids, then it won't have the same safety features (like perhaps leak-before-break) as compressed air parts. Also an OSHA violation.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/21/11 01:54 PM

Quote:
Also an OSHA violation.


I expect that there will also be other codes that would need to be followed.

I will have to check to see if I can legally experiment
I wouldnt want to violate anyone.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/21/11 10:40 PM

What's wrong with the two-tanks way? Are you worried that the outlet is too small and won't allow enough flow rate to move it?

Anyway did you get a filling valve?

Hey it could be on an even smaller scale. Get an aerosol can of anything (maybe air), and just attach it to another container.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/29/11 01:25 AM

I have to admit I havent been trying at all , I could
probably build one in a day but the weather changes and I
decide not to do anything.

or mostly I forget to try.

for the last week I have been speculating moving to vb 10
from vb 6 , by attempting to convert some older vb 6 stuff
to vb .net.

Im not sure what Microsoft was thinking when they changed
99% of the language around but from my experience this last week there must have been a extremely large turn over in the vb programming field.

vb 10 seems as if nobody remembered what the real programmers did when they made the vb 6 compiler
so they just scraped off a few layers of the windows api and made a ide from what was left.

Im contemplating assembly again.

but I know I wont.

Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/29/11 12:12 PM

Originally Posted By: paul
decide not to do anything.

or mostly I forget to try.


If you actually believed it would succeed, and recognized the enormity of the discovery, you'd be forgetting to eat or sleep in your unstoppable pursuit of it. I think you know that it won't work, or just don't appreciate the spectacular possibilities it would open up.

Quote:

vb 10 seems as if nobody remembered what the real programmers did when they made the vb 6 compiler

Once you get used to it you'll love it. It really is much more consistent and elegant. But it's like enjoying a delicacy food, you have to be ready to change your way of thinking to appreciate the beauty otherwise it just feels like eating a scorpion. If you use the default migrated code and try to force the old fashioned approach onto the new design it'll be ugly because of bodgy legacy support.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/30/11 05:20 AM

Im sort of getting used to it , but when it comes to using a control array I find that the amount of extra code needed is ridiculous.

dim X as integer

X = index

for i = 1 to 20
commandbutton(X).text = i
commandbutton(X).tag = Y
X = X + 1
next i

turns into

commandbutton1.text = 1
commandbutton2.text = 2
commandbutton3.text = 3
commandbutton4.text = 4
commandbutton5.text = 5
commandbutton6.text = 6
commandbutton7.text = 7
commandbutton8.text = 8
commandbutton9.text = 9
commandbutton10.text = 10
commandbutton11.text = 11
commandbutton12.text = 12
commandbutton13.text = 13
commandbutton14.text = 14
commandbutton15.text = 15
commandbutton16.text = 16
commandbutton17.text = 17
commandbutton18.text = 18
commandbutton19.text = 19
commandbutton20.text = 20

commandbutton1.text = Y
commandbutton2.text = Y
commandbutton3.text = Y
commandbutton4.text = Y
commandbutton5.text = Y
commandbutton6.text = Y
commandbutton7.text = Y
commandbutton8.text = Y
commandbutton9.text = Y
commandbutton10.text = Y
commandbutton11.text = Y
commandbutton12.text = Y
commandbutton13.text = Y
commandbutton14.text = Y
commandbutton15.text = Y
commandbutton16.text = Y
commandbutton17.text = Y
commandbutton18.text = Y
commandbutton19.text = Y
commandbutton20.text = Y

its a ridiculous amount of code that replaces the control array.

and its extremely difficult to manage in code.
Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 03/30/11 11:30 PM

Originally Posted By: paul
Im sort of getting used to it , but when it comes to using a control array I find that the amount of extra code needed is ridiculous.


You can make a control array, unsuprisingly, by using an array of controls, then fill it with CommandButton objects programatically. They can be copied from a single button created in the designer to duplicate all the properties.

It's more elegant to reuse the same general features like arrays that were already there, than to depend on a special one that just serves special cases. Of course you could make your own control array class if you wanted some unique behavior - and you can actually program in unique behavior, unlike VB6.

Certainly duplicating code 20 times is bad practice.
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 03/31/11 01:25 AM

Quote:
They can be copied from a single button created in the designer to duplicate all the propertie


but I dont want all the properties duplicated.

each command button needs at least three assignable properties.

I have found that I can use the
.tag
.assignabledeffinitions

but I need 1 more to use to define the control.

.text is used by the control to represent a item in the array.

I did get the buttons array to work by making a database with buttons using the above three assignable properties.

this entailed using a seperate file for each button.
and a list of button names.

but all of the above could have been done with
this

for i = 1 to 20
button(X).tag = a
button(X).accessabledeffinition = b
button(X).text = c
X = X + 1
next i

but because there is no control array in vb.net
my 28 kb program is already over 800 kb

and Im only half way finished converting it.

in fact I havent touched it today because of the stress.

I have thought today about making a user control that would work exactly like a old style control array.

and the control could have a few more tweaks.






Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 04/13/11 10:40 AM

Originally Posted By: paul
and the control could have a few more tweaks.

That's part of the beauty of .Net. You actually can customize the controls. No need for hideous ActiveX interfaces and all that.

Sounds like you did it the hard way. This code will create an array full of buttons

Code:
  Dim buttons(10) As Button
  For buttonID As Integer = 1 To 10
      buttons(buttonID) = New Button
      buttons(buttonID).Top = buttonID * 30
      buttons(buttonID).Width = 100
      buttons(buttonID).Height = 25
      buttons(buttonID).Text = "Button #" & CStr(buttonID)
      buttons(buttonID).AccessibleDescription = "hey"
      buttons(buttonID).tag = "ho"
      Me.Controls.Add(buttons(buttonID))
  Next


This might not be the most elegant way, but it's certainly easy. Stress comes from craftsmen blaming their tools :P

Posted by: paul

Re: Orion, Mission to Alpha Centauri - 04/14/11 04:17 PM

just using a loop to create buttons is easy.

but can you use the following to change the information
in the properties of any chosen button.

suppose someone clicks on button (0) in your array.

will they be directed to this?
-------------------------------------
private sub button()_click
-------------------------------------

or this

-------------------------------------
private sub button()_text changed
-------------------------------------

the only way I have been able to perform the above is to
use a 0-10 in each buttons tag.

then assign that value to a lable when the button has been clicked.

------------------------------------
private sub button1_click

label1.text = button1.tag
end sub
------------------------------------


then when the label text has changed

--------------------------------------
private sub label1_text changed

the code to change the value in the selected buttons properties.

if button1.tag = label1.text then button1.text = "you clicked B1"
if button2.tag = label1.text then button2.text = "you clicked B2"
if button3.tag = label1.text then button3.text = "you clicked B3"
etc ... 1 for each button in the array

anyway I have the above working well in a program.

but I do like vb.net only because it can build a 64 bit
application.
and can use multi threading.

so I suppose there are better ways to do the above
and make a user control that will perform even better than
the older vb control array.



Posted by: kallog

Re: Orion, Mission to Alpha Centauri - 04/16/11 11:59 AM

Originally Posted By: paul

suppose someone clicks on button (0) in your array.

will they be directed to this?


Yea, you can use .AddHandler to specify a common click event handler for all the buttons. And that can be passed the button that was clicked, so it can identify it by the tag or name property or however.

Well as long as it works. I just hope you don't have to modify it a year later and not have a clue how all those labels and things interdepend on each other!
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 09/05/12 05:50 PM

over a quarter of a million views.

226,228 views

looks like this topic must be an interesting one.
Posted by: Bill S.

Re: Orion, Mission to Alpha Centauri - 09/05/12 08:52 PM

Certainly this is one possible interpretation.

Do you think that the lack of posts between April '11 and Sept '12 might indicate that a lot of those quarter million were disapointed?
Posted by: paul

Re: Orion, Mission to Alpha Centauri - 09/05/12 11:08 PM

looks like this topic must be an interesting one.

Quote:
Certainly this is one possible interpretation.


Quote:
Do you think that the lack of posts between April '11 and Sept '12 might indicate that a lot of those quarter million were disapointed?


well , if nothing else it shows that there arent alot of people who are active posters on this forum!

but it could be that those quarter million were disapointed.

then again there might be something very interesting somewhere inside this topic.

226,463 views right now!

apx 235 hits since I posted 6 hours ago