Welcome to
Science a GoGo's
Discussion Forums
Please keep your postings on-topic or they will be moved to a galaxy far, far away.
Your use of this forum indicates your agreement to our terms of use.
So that we remain spam-free, please note that all posts by new users are moderated.


The Forums
General Science Talk        Not-Quite-Science        Climate Change Discussion        Physics Forum        Science Fiction

Who's Online Now
0 members (), 181 guests, and 2 robots.
Key: Admin, Global Mod, Mod
Latest Posts
Top Posters(30 Days)
Previous Thread
Next Thread
Print Thread
Page 7 of 16 1 2 5 6 7 8 9 15 16
Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.



3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
.
Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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.

Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.










3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: May 2010
Posts: 410
I
Senior Member
Offline
Senior Member
I
Joined: May 2010
Posts: 410
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


UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.


3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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.

Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.






3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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

Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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.

Joined: May 2010
Posts: 410
I
Senior Member
Offline
Senior Member
I
Joined: May 2010
Posts: 410
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


UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.


3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.









3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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!

Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.














3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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.

Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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.










3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: May 2010
Posts: 410
I
Senior Member
Offline
Senior Member
I
Joined: May 2010
Posts: 410
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


UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
Joined: Mar 2006
Posts: 4,136
P
Megastar
Offline
Megastar
P
Joined: Mar 2006
Posts: 4,136
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


3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
Joined: May 2010
Posts: 410
I
Senior Member
Offline
Senior Member
I
Joined: May 2010
Posts: 410
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


UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA
Joined: Mar 2010
Posts: 1,100
K
Megastar
Offline
Megastar
K
Joined: Mar 2010
Posts: 1,100
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.

Page 7 of 16 1 2 5 6 7 8 9 15 16

Link Copied to Clipboard
Newest Members
debbieevans, bkhj, jackk, Johnmattison, RacerGT
865 Registered Users
Sponsor

Science a GoGo's Home Page | Terms of Use | Privacy Policy | Contact UsokÂþ»­¾W
Features | News | Books | Physics | Space | Climate Change | Health | Technology | Natural World

Copyright © 1998 - 2016 Science a GoGo and its licensors. All rights reserved.

Powered by UBB.threads™ PHP Forum Software 7.7.5