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.


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.