The author of the essay to which you have repeatedly referred did not actually state that the equation is invalid as speed approaches zero. What he actually said was:
While working on a project designing a propeller at work, I wanted to know just how good I was doing. Efficiency is one measure of how well a propeller is performing, but it's not necessarily a good indication of how well the design is performing up to its potential. In aviation, propulsive efficiency is defined as:
Efficiency = T x V / Pavail
where η is efficiency, T is Thrust, V is Velocity, and Pavail is Power Available, or power going into the propeller. Basically, power out divided by power in. This equation is very useful for many cases, but you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero.
The author of the essay had a problem because he was attempting to use the propeller efficiency equation to calculate thrust. He was assuming that the purpose of the propeller is to produce thrust, but this is not true.
The purpose of the propeller is to propel the aircraft forward against the drag force, by producing thrust. The rate at which the propeller is achieving this purpose is equal to the thrust multiplied by the aircraft velocity. This is our old friend the THP.
Because the primary purpose of the propeller is to propel the aircraft forward, it is entirely logical that the overall efficiency of the propeller should be measured in terms of its propulsive efficiency.
Propeller efficiency = (Thrust x Aircraft Velocity/ Brake Horsepower
This can also be stated as (Thrust x TAS) / BHP provided we take care to note that the TAS is velocity of the aircraft relative to the air and is not the propwash velocity.
When the aircraft is running at full throttle on the ground, with the wheel brakes preventing it from moving forward, the propeller is absorbing power but is not achieving any success in propelling the aircraft forward. So the propeller efficiency is zero, which is exactly what the equation yields. The equation is in fact producing the correct result at all airspeeds including zero.
To understand why the author of the essay could not use the equation to calculate thrust when airspeed was zero we need to look at what he was attempting to do.
He was attempting to take the product of (Thrust x zero airspeed) and then divide it by zero airspeed to reveal the thrust.
For any number other than zero and infinity this would be a perfectly reasonable thing to do.
But the product of any number multiplied by zero is zero, and dividing this product by zero simply gives zero / zero = zero.
The simple fact is that the processes of multiplication and division are not reversible when one of the numbers is zero. This fact is not a function of any inaccuracies in the propeller efficiency equation. It is simply the result of one of the properties of the number zero. Any equation involving multiplication and division will encounter exactly the same problem when one of the values goes to zero. But this does not mean that all such equations are invalid.
Looking at the model again, the position that you and barit1 have adopted is that the model has zero THP whilst prop hanging but some THP if it climbs. Therefore it has negative THP if it descends and the prop will drive the engine. In that case you are in autorotation as soon as you start to descend (or shortly after when -THP is sufficient to offset profile power). That would be nice, but it’s not the case. Autorotation does not begin until after RoD exceeds Vi. Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.
Entering autorotation in a helicopter is not simply a matter of throttling back the power until the aircraft starts to descend. If you do this, the blades will stall, the rotor speed will decrease, the blades will fold upwards and the aircraft to crash. To enter autorotation you must lower the collective to get negative pitch (or at least a very low pitch) at which the ROD airflow produces an angle of attack, such that part of the total reaction drives the blades forward. This keeps the rotors turning and keeps generating lift. None of these good things will happen if we simply reduce power in a propeller driven aircraft, because the mechanisms are physically very different. Your use of this scenario simply illustrates your lack of knowledge regarding helicopter POF.
Well, here are 3 credible definitions of THP:
<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation
<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.
<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State
In the second of your definitions the term “horsepower equivalent of the thrust” implies that it is the thrust x TAS. So this definition is quite correct.
In dealing with your first and third definitions I could quite easily find numerous sources of the definition THP = Thrust x TAS. I have at least one example on my office bookshelf. But trading references will prove nothing. Instead let’s see what your definitions produce in terms of a propeller efficiency equation.
Efficiency is simply a ratio of output divided by input, and standard efficiency equations are of the form Efficiency = Output / Input
In your first and third definitions the output is thrust and the input is power.
So we have Efficiency = Thrust output / Power input
If we use imperial units of pounds force for thrust and foot pounds force per minute for power we get:
Efficiency = (lbf) / (ft lbf / min)
Cancelling out the lbf on the top and bottom of this equation we get
Efficiency = 1 / ft/min which is Efficiency = min/ft.
The equation has yielded an efficiency figure that is in units of minutes per foot. What exactly does this mean?
Efficiency is just a ratio of output divided by input, so it has no units.
To get rid of the min/ft in our result we need to multiply by ft/min. But ft/mins are the units of velocity, so to get a proper value for efficiency we need to multiply by velocity. This converts our efficiency equation into Efficiency = (Thrust x TAS) / Power input.
This is of course the propulsive efficiency equation.
The above sequence proves something that we should already have known. That is the fact that power, which is the rate of doing work or expending energy, cannot be converted into a force. Power and force are two totally different things and one cannot be converted into another. So whatever the sources of your first and second definitions, they cannot be correct. It is of course possible that the authors did not intend their definitions to be interpreted literally.
Finally I think that you need to look again at your argument that induced power is the same things at THP. I have drawn the following quotes from various internet sources.
Induced power is the power required to maintain enough lift to overcome the force of gravity.
www.math.usu.edu/~powell/ornlab-html/node7.html
Induced Power is that portion of the power required to produce lift.
www.griffin-helicopters.co.uk/note/helicopterpower.htm - 13k
The minimum engine power required to hover is called the "induced power."
scienceworld.wolfram.com/physics/Helicopter.html - 15k
Induced power is what people are referring to when they say helicopters "beat the air into submission." Newton's 2nd law concerning action-reaction applies in this regime where we must force air down to keep the aircraft aloft.
www.navair.navy.mil/safety/documents/Power_Available_vs.doc
In each case they show that induced power is exclusively concerned with the process of generating lift. THP is concerned with propelling aircraft through the air, so they are not the same thing. By all means argue that your model aeroplane is generating induced power. But it is most certainly not generating any THP.