A writer recently used Thrust = $\eta \frac {P}{V}$ here to determine the maximum forward airspeed of an aircraft for a given set of conditions.

While efforts made to answer were very impressive and reflected a lot of effort, there was a nagging question that seems difficult to reconcile:

does thrust always decrease when airspeed increases?

Wouldn't a better expression of prop thrust be:

RPM × AoA$^1$ to the relative wind?

Before the mass flow argument comes in, let's consider that specific impulse of props are in the thousands while the best (mass flow) rockets barely top 400. Props pull by making lift, and loss of $\Delta$ mass flow velocity will have far less effect on thrust than non-optimal AoA at higher airspeeds.

The drag of the prop is consuming the engine power, not the planes forward speed!

Which leads us to:

jets are constant thrust, props are constant power?

Really? Isn't engine power strictly a function of fuel flow?

As an aircraft gains airspeed, prop pitch must coarsen to maintain optimal AoA. Thrust loss is only cosine of this angle, in other words, not much at all at lower angles, especially if prop RPM is high, which brings us to this, courtesy of Stipa and Caproni from the 1930s.

I know, too slow, not enough power, too much drag. But if the propeller were recessed into the duct where air was slower and higher pressure, might this not be an analog of today's fan jets? Could ram pressure at increasing airspeeds eliminate the need for variable pitch?

$^1$ as part of a prop "lift equation"

  • $\begingroup$ The same author has also linked this answer which better explains how that equation is derived an its limitations. If it's anyway not clear maybe you can ask to expand that answer. $\endgroup$
    – sophit
    Jan 11, 2023 at 15:48
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    $\begingroup$ (Prop rpm)$^2$ × prop lift equation at optimal AoA × cosine deviation angle from thrust line (coarsening to maintain AoA at higher speeds) × a $\Delta$ Mass flow factor (small but significant) seems a bit better than a fudgy $\eta$. But the "authors" number seem OK. Just curious. Power output of the engine seems to get confused with the kinetic energy state of the moving aircraft F×V or mV$^2$/s. Engine power only turns the prop. $\endgroup$ Jan 11, 2023 at 17:47
  • $\begingroup$ Well, if a simple $\eta$ is enough, there's no need to overcomplicate it 😉 Also because the next less-simple mathematical model to represent a propeller would be the Blade Element Theory that, albeit simple in its hypothesis, is quite lengthy in its results. $\endgroup$
    – sophit
    Jan 11, 2023 at 18:26

3 Answers 3


The drag of the prop is consuming the engine power, not the planes forward speed!

Power output of the engine seems to get confused with the kinetic energy state of the moving aircraft F×V or mV2/s. Engine power only turns the prop.

There's a bit of misunderstanding here.

Most, if not all, of the engines just make something rotate: a car's engine makes its wheels rotate; a drill's engine makes a drill bit rotate; a propeller's engine makes the propeller rotate and so on.

But these rotations are not connected to any power until they create useful work. An engine of a car which is on a perfectly iced surface is not using any power even if it's spinning at its max rpm, except for the power needed to win its own internal mechanical friction. An engine of a drill which is just turning freely without doing any hole is not using any power except the one needed to win its own internal electro-mechanical friction. An engine of an airplane with a propeller that is not producing any thrust, is not using any power if not the one needed to win its own internal aero-mechanical friction.

Only when these rotations create a work then there's a power associated to them. The car's engine produces power when the wheels make the car travel at some speed $V$. The drill's engine makes the electric meter spin very fast only if it's drilling a hole in a reinforced concrete wall. The propeller draws power only when it pulls the airplane at a speed $V$.

So, connecting airplane's speed with propeller's thrust and with engine's power does actually make sense and it's physically correct. And since we do not live in a perfect universe, some power get lost in the process and we model this lost via a simple efficiency factor $\eta$.

Wouldn't a better expression of prop thrust be: RPM × AoA to the relative wind?

Indeed, $\eta$ depends on the blade pitch (AoA) as measured at ¾ of the span and on the advance ratio $J=\frac{V}{nd}$, where $n$ is the rotating speed [rev/s].


By including the efficiency factor $\eta$, the author gave the expression $T=\eta\cdot\frac{P}{v}$ the needed flexibility to be correct for a propeller and piston engine driven airplane. Efficiency is not constant, and especially at high speed might suffer from too small angles of attack (for fixed pitch propellers) or Mach effects (for all propellers).

Just look at the propeller efficiency for different pitch settings, plotted over advance ratio J (below).

propeller efficiency for different pitch settings

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    $\begingroup$ Thanks Peter. Does it not seem the "efficiency" is pitch related? Does an aircraft doubling airspeed always mean half thrust for the same engine power output? (Not to be confused with the aircraft "power state" which is Force × Velocity). $\endgroup$ Jan 11, 2023 at 15:55
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    $\begingroup$ @RobertDiGiovanni Yes if efficiency does not change. Since engine power is constant and power is force times speed by definition, the only reason why thrust won't be half at twice the speed is a change in efficiency. $\endgroup$ Jan 11, 2023 at 19:47

the propeller draws power only when it pulls the airplane at a speed V

Let's look not at thrust loss at higher speed, but on the runway at full throttle with brakes set.

According to the formula Thrust = $\eta\frac{P}{V}$ when the brakes are released thrust is nearly infinite. Is this actually true?

Another Cessna 172 rolls onto the runway at 3 m/s (around 6 knots), which one gets off the ground in the shortest runway distance?

Both accelerate at 3 m/s$^2$. Take off speed is 24 m/s. Easy one, rotate after around 8 seconds. But the rolling start bought me 1 second of acceleration. The last second speed increases from 21 m/s to 24 m/s, averaging 22.5 m/s.

22.5 meters, or about 74 feet shorter takeoff run.

Will the plane sitting at full throttle have infinite thrust when brakes are released? Will acceleration markedly drop as it rolls forward?

It seems, as theoretically and mathematically correct $\eta\frac{P}{V}$ is, the proof would be to watch takeoff rolls of jets and props. Increasing drag should begin to reduce acceleration, but at low speeds force = mass × acceleration, at higher speeds acceleration is 0 (steady state) and force = drag.

It may be the prop acceleration curve from start to rotation is straighter than we think, even for fixed pitch aircraft$^1$.

$^1$ increasing efficiency may actually increase rate of acceleration during takeoff roll.

  • $\begingroup$ "According to the formula when the brakes are released thrust is nearly infinite. Is this actually true?" No it's not and I have already pointed you to this other answer of mine where I explain its limitations and what to use instead. $\endgroup$
    – sophit
    Jan 13, 2023 at 6:10
  • $\begingroup$ Oh, BTW HP force (could) = mg weight + m$\Delta$V/t weight + m$\Delta$V/t Horse (+ Drag Horse and weight + pulley friction). Last 2 are minimal. No mechanical advantage single pulley, though I did consider adding a gear for one poney (or having 2 ponies). $\endgroup$ Jan 13, 2023 at 10:33
  • $\begingroup$ But once the Horse got going (to steady state) HP required = mg weight (+ Drag Horse and weight + pulley friction). I got a real kick out of comparing these thrust formulae to the thrust curves of Estes model rocket engines. Big spike to get started, then a plateau until burnout. Back in the 1700's, oats and hay were more easily had than rocket fuel ;-). $\endgroup$ Jan 13, 2023 at 10:44
  • $\begingroup$ "Some of the formulas tried didn't scale well." which one? I'd be glad to explain (or correct) it better. Have you double checked the units? And btw what's your design? $\endgroup$
    – sophit
    Jan 13, 2023 at 14:05
  • $\begingroup$ @sophit I'll put it in a new question. But try this one: footpounds of torque (at a given rpm) ÷ feet 3/4 prop arc radius = pounds drag. Pounds drag × 4 ($\eta$ factor) = pounds thrust. Way less L/D than glider wing but props are in the 300-400 mph range. (I'll go metric eventually). (Should work for 2 or 3 bladed props). Good day. $\endgroup$ Jan 13, 2023 at 14:55

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