Hi Wilfred
If BOAC has sorted you please ignore everything after this!
I think your comment about “high speed stall” might have given me a clue as to what is bothering you.
In my book a “high speed stall” is simply one that happens at more than 1g. It can still be wings level (say at the bottom of a loop) or it could be at the normal cruise speed for the type and involve a lot of bank on and pulling hard. It has nothing to do with “high speed” as in compressibility or Mach number effects. It just means the pilot made the aircraft stall at a speed that is higher than the wings level 1g case. Indeed the proper term for what we all refer to as a “high speed stall” is an accelerated stall.
So doing an ordinary stall (wings level with speed reducing in level flight) at say 40K feet does not mean you have experienced a high speed stall even though the Machmeter may well be some way round the dial at the time you stalled.
Just in case the above does not unlock the issue for you, I will spell out my understanding of the subject from the top (forgive the sucking eggs bit but I don’t want to miss out any step just in case that is the one giving problems.)
There are two types of mathematical approach to aerodynamics, one treats the air as if it is an incompressible gas and the other assumes (correctly) that it is compressible. This latter correct approach has the disadvantage that it is much more complicated.
The difference in numerical results between the simple theory and the complex one is very small indeed at low speeds - say below 150 kts - and is usually ignored in normal life because the scatter in results from a whole host of other factors tends to be greater than the compressibility effect. (By these “other factors” I mean inaccuracies in flying, effects of turbulence, instrument and sensor calibrations and errors and so on)
The fact that air IS compressible does effect some aspects of the simple (incompressible) theory more than others. What we are interested in here, namely the loss of the maximum available lift coefficient available from a wing (i.e. the amount of lift we can get before the flow breakdown that we call the “stall” happens) is probably affected as much as anything.
What I am talking about is the breakdown in the simple relationship between stalling speed and applied G. Simple (incompressible) consideration of the lift equation (L = Cl x ½ x density x wing area x speed squared) tells us that if we have a wings level stalling speed of say 100 kts, then at 141.4 kts (or whatever the square root of 2 is) we should be able to pull 2g, ie just manage a 60 deg bank steady level turn at the stall.
But somehow one can never quite achieve this and the shortfall at 173.2 kts is even greater (root 3 when we would expect 3g from simple theory).
Now I know we are taught that a wing always stalls at the same angle of attack (and nobody goes around saying THAT more than me) but it is actually only a true statement if the Mach number at the stall is roughly constant. Numbers for the Harrier in conventional flight are locked in even to my failing memory, so forgive me quoting type specific numbers to make what is a general point:
In a metal wing Harrier at low level the angle of attack you see in a 1g stall wings level is around 12 deg (clean configuration, flaps up) That number will not change even if you double the aircraft weight (although the IAS at which you reach it will be much higher) and equally if you turn too hard you will stall at the same 12 AOA regardless of weight or bank angle in use.
BUT, if you go up to 20K feet and spiral down at say .8 Mach in a high G turn, pulling harder and harder until it stalls, you will probably not see more than 9-10 degrees on the AOA gauge. This reduction in the max lift available has happened because of compressibility effects.
Hope that helps, if not get back to me as there is no such thing as a bad student only a poor instructor (or as my doctor says - there is no such thing as an impotent man only and incompetent woman - advice which I find very comforting)
JF