zoobyshoe said:
My complaint is that Jeff doesn't ever seem to figure in the low pressure created simply by virtue of the air passing over the front surface of the prop, air which is accelerated, not slowed, and which is shed as vortices at the wing tips, as opposed to being compacted behind the propeller.
You can try to figure all this out, but it is extremely complicated. It is much easier to trust Newton's 3rd law, which says that action = reaction, together with Newton's second law (F = m a), but here in the form: F = d p / d t.
In other words, the trick is first to calculate the entire force that is exerted (by the propeller) on the air, and to find that out, we make the momentum balance: we see what was the momentum of the air "before" and then the momentum of the air "after".
Remember that momentum of something = mass of something x velocity.
Now, here we make an approximation: we consider that there is air that is left "untouched" by the propeller, and an amount of air that is "uniformly displaced" by the propeller. In reality, there will be different categories of air, those that are untouched, there are some that are a bit displaced, there are others that are more displaced etc... so we should sum all these different contributions, but we don't, we simply say that some amount of air is affected in the same way by the propeller, and the rest not at all.
The momentum balance of the unaffected air will of course be 0: it will have the same momentum before and after. So we don't need to take it into account ; what counts is only what is displaced by the propeller. Per unit of time, a certain mass of air is affected by the propeller, and we call that "m". That means that the mass in an amount of time dt is M = m dt. That amount of mass had a velocity v_in before it was affected by the propeller, and is then accelerated and ends up at a velocity v_out (it is the uniformity of v_out which is the approximation we make: not all bits of air are accelerated to the same velocity).
That means that the momentum "after" is M x v_out while the momentum "before" is M x v_in. (for the non-affected air, the momentum in was MMM v_vin and the momentum out is also MMM v_in, where MMM is the big mass of the unaffected air ; so the balance doesn't matter).
So the momentum gained, by the air, in a time dt is given by M x v_out - M x v_in.
That momentum was gained in an amount of time dt, so the force (Newton's first law) that must have been exerted on that air must have been: F = ( M x v_out - M x v_in ) / dt.
Or: F = m x (v_out - v_in). (as M / dt = m).
So, no matter what or how, this must be the force exerted on the air. And as it is the propeller that does this, the opposite force must be the one exerted on the propeller.
So, if we know that a certain amount of air changes velocity, all the rest doesn't actually matter: it is this momentum balance which counts, and it will be the final sum of all the little bits of pressure that work on all the pieces of surface of the propeller. We don't have to go into that detailed description: we have the end result of the balance.