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#34777 - 06/07/10 02:58 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.

_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34780 - 06/08/10 03:54 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
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.

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#34781 - 06/08/10 04:40 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.






_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34785 - 06/08/10 08:22 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
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.

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#34790 - 06/08/10 12:32 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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?

_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34791 - 06/08/10 02:02 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
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.

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#34800 - 06/08/10 03:59 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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
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#34801 - 06/08/10 04:01 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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
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#34802 - 06/08/10 04:59 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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?





_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34804 - 06/08/10 05:10 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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


Edited by ImagingGeek (06/08/10 05:15 PM)
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#34805 - 06/08/10 05:26 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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


Edited by ImagingGeek (06/08/10 05:32 PM)
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#34806 - 06/08/10 05:31 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34807 - 06/08/10 05:45 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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?

_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34808 - 06/08/10 06:05 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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
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#34809 - 06/08/10 06:13 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34810 - 06/08/10 06:17 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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
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#34811 - 06/08/10 06:21 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.




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3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34812 - 06/08/10 06:26 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
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
_________________________
UAA...CAUGCUAUGAUGGAACGAACAAUUAUGGAA

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#34813 - 06/08/10 06:30 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.




_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34814 - 06/08/10 06:45 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
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.






_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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