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.