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Re: Re: Propellant fluid dynamics at zero-g



on 1/9/01 6:55 PM, stephen rector at stefano@amug.org wrote:

> Without normal 'adhesion' (by whatever force) to the sidewall, the
> centrifugal force could cease to be present for some of the fluid. Thus,
> zero-g.

I disagree.  You might not have one continuous mass of fluid, but blobs of
it.  Still, ALL of it will be pressed against the outer wall.  It has to by
laws of physics.

It's all based off of Newton's first law -> A body in motion in a given
direction will continue in motion in that direction unless acted upon by
some external force.  So with the spinning motion of the space craft, the
fluid wants to move in a straight line that is tangential to the circle of
rotation.  However, since the entire fuel tank is spinning in a circle, it
won't get to travel in that straight tangential line, but in effect, it will
always be trying to do that - hence the centrifugal force.  It matters not
if the fluid is one constant homogeneous mass or several "globes" or "drops"
of fluid held together by surface tension.

The only place in a spinning craft where there is theoretically no motion
and therefore, no centrifugal force is the axis of rotation.

And in the case of an ignited engine, the inertia of the fuel and the
acceleration of the space craft will create another force that will push the
fuel in the direction opposite of acceleration (Newton's second law).

So there is no way, IMHO, that you could have a blob of zero G fuel in the
middle of a tank in a spinning spacecraft and even more so if the crafts
engines are fired.

Someone also asked the question of what the resulting vector sum of the
centrifugal force and the "acceleration" force (for lack of a better term).
Some suggested that it might be 45 degrees.  It could be, but that would
only be the case if the centrifugal force equaled the force of the engine.
In P3D, we know the engine produces a force of 400 Newtons.  I doubt the
spin of the spacecraft produces anywhere near that much centrifugal force.
So the outlet pipes for the fuel tanks would be much closer to the engine
(or the "bottom of the bird) than 45 degrees.  I think someone calculated
the centrifugal force from rotation.  If you do that, then the vector sum
will be where the fuel pipes come out of the tanks.

73,

Jon
NA9D

-------------------------------------
Jon Ogden
NA9D (ex: KE9NA)

Member:  ARRL, AMSAT, DXCC, NRA

http://www.qsl.net/ke9na

"A life lived in fear is a life half lived."

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