keith Williams:
I have invited oggers to go back to basics and explain why he thinks that an aeroplane that is not moving (brakes on or hovering model) is generating THP. I think that this exercise will be more productive than thinking up imaginary scenarious about circling aircraft and rotating wings. You might like to try it
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<<Thrust Horsepower: The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill Dictionary of Aviation
Zero thrust horsepower ≡ zero thrust. The model requires thrust to hover therefore the THP is not zero.
Power is required for thrust additional to losses. It is the induced power of the prop/rotor. In the case of the model ‘prop hanging’, that induced power is the thrust times the velocity induced at the prop, along the thrust axis. Induced power is therefore equivalent to thrust horsepower.
The position you are sticking to is that all power going into the slipstream is wasted power, which is true in the context of aircraft velocity but not in the context of thrust. Hence, propulsive efficiency for velocity, and FOM for thrust:
“Since it is the induced power which relates to the useful function of the rotor, the ratio of induced power to total power is a measure of rotor efficiency in the hover. This ratio is called the Figure of Merit” Sedddon: Basic Helicopter Aerodynamics
For a given prop the thrust depends on both induced velocity Vi and the mass flow through the prop: T = ρA(Va + Vi) x ΔV. Va may go to zero but if Vi goes to zero you have no thrust. Statically, the “power converted to thrust” is T x Vi. You will find this is in accordance with simple momentum theory for a prop as well as the definition at the top. It is therefore THP.
Non-statically this is T x (Va + Vi) which eventually becomes ≈ T x Va only. You have said as much but the point that T x Va is actually an approximation (though it is definitive for propulsive power) of ‘power converted to thrust’ seems to have escaped you. In the latter case it is fair to say that THP = T Va and swallow the approximation. But this becomes invalid at low speed as I have been stressing all along. This graph shows where the approximation holds good as the curves run along the top near 100% beginning at about 20m/s for lightly loaded props:
The efficiency increases as mass flow through the prop picks up at speed.
The graph in the essay is the same as the one above. These are Froude efficiency curves. It does not show that THP goes to zero with Va for a given BHP which is what you think. The speedier end of the curve shows Froude efficiency ≈ 100% of induced efficiency whereas the slower end of the curve shows the opposite. Ergo TVa is a good proxy for induced power at one end of the curve but completely invalid at the other end.
When you add BET you get a graph like the one RRT posted for overall propulsive efficiency. Personally, when I compare the two graphs I see they are “pretty much alike” in the same way that r is pretty much like n (ie different). The r shaped curve is showing that Froude efficiency increases as a proportion of overall efficiency and stays there even after overall efficiency drops off again so TVa as a proportion of “power converted to thrust” stays high. OTOH, as speed goes down the approximation fails. That is the whole point of the essay, the very reason the guy felt the need:
"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. So how do you know how good your prop is doing at low speeds or statically?"
..and he spends most of the essay working around the problem.
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.
By contrast, if you take THP = induced power = T (Va + Vi) then you have THP in the hover, more THP in the climb and less THP in the descent until RoD exceeds Vi and autorotation begins. You will find this is in agreement with momentum theory for a helicopter in descent.
“A number of contributions to this thread have included attempts to draw comparisons between rotary wing pof and fixed wing pof. This is unlikely to be very helpful because the two are very different in many ways. The fact is that few of us know enough about rotary wing pof to make the comparisons very illuminating (I certainly don't)”
I find the comparison to be very helpful because as far as momentum theory goes the rotor and the prop are precisely the same. And we don’t need to go beyond that to determine the induced power required to provide a given thrust.
THP and BHP both have very specific definitions, which means that propeller efficiency also has a very specific definition.
THP = Thrust x Aircraft Velocity
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
There is absolutely no doubt that the model is developing induced power T x (Va + Vi). That is a given. Induced power is the power converted to thrust for the model or lift for the helicopter. It fits all three of those definitions, whereas T x Va is an approximation that only works when the Froude efficiency is high. Ergo it is invalid statically.