I would like to really understand why when talking about jets we care about thrust but when talking about propeller aircraft we care about power.

(THRUST)="Just a force"
(POWER)= work/time = force x velocity

I know that even though propeller aircraft are called "power producers" and not "thrust producers" as jets, any aircraft needs Thrust because that's what overcomes Drag. The difference is that a prop aircraft uses the Power of the engine to turn the prop while a turbine engine produces Thrust by itself.

I also know that total drag curve comes from induced drag (decreases with airspeed) and parasitic drag (increases with airspeed), so that is exactly what thrust needs to overcome. The total drag you get at a given airspeed is equivalent to the thrust you need to stay level. That makes a lot of sense.

I have no problem with jet theory when thinking about that and the Thrust required / Thrust available curves.

Why is it different for prop aircraft? Why is the power curve important? I mean, aerodynamically. Because I've seen that total drag curve is also the same as Power Required Curve and I don't understand why.

My first thought is: Engine power is important because that is what the prop will use to rotate and generate Thrust. (Propeller efficiency is directly related with HP and other prop aerodynamic factors). However, even with the most powerful engine, if prop efficiency is poor because of the other factors, we won't get too much thrust and that's the only thing that matters. And one more time, thrust is what makes any aircraft fly forward, so total drag is thrust required, not power.

Vx speed is found where the greatest difference between thrust available and thrust required is. Greater excess thrust equals greater angle of climb. (Got it!) that's very logical for me.

But: Vy is where the greatest difference between Power Available and Power Required is found. What?! Why?! How can the power the engine produces relate with airplane's aerodynamic?? It has nothing to do with that. Vy is best rate of climb, the airspeed at which the airplane will get faster to an altitude. I think that the reason should be related with a thrust/drag ratio.

On the other hand, I know that the Power Curve results from multiplying each thrust value times each airspeed value of the thrust curve. That's the fact, just let me explain why it doesn't make any sense to me...

Power=force x velocity, and we can apply that in different ways. If we were talking about electricity we could say Power is measured in Watts, and Watts= Voltage x Current. Because we are talking about prop aircraft, we should use a power unit like HP (for reciprocating engines). HP= Torque X RPM.

Multiplying the Thrust Curves to get Power Curves would mean that Power= Thrust x Airspeed

What kind of power are we getting when multiplying thrust times the airspeed at which the airplane is flying?

Wouldn't that be something like "Aircraft power = thrust x airpeed" or in other words "the amount of power needed to stop that airplane"? Nothing to do with the engine (which is really the reason why a prop airplane is called "power producer")

I'm very confused, even tired of not understanding the exact reasons behind this after reading what should be sufficient. I think I know what power and thrust are, but I just can't understand the way they relate, and to be more specific, how does excess power affect the aircraft flying in real life, aerodynamically. Why to talk about Power instead of Thrust in any case, jets or props. (It's clear there's a good reason not to, and that's what I want to finally understand). Maybe more realistic examples or comparisons instead of graphs and curves would help.

(But those are not a problem either in case you want to try.) I'm a student pilot and I hope I was clear enough with my doubts and explanations.

(Also with English since Spanish is my native language)

  • 2
    $\begingroup$ Here would be a question which addresses how jet thrust changes with speed. $\endgroup$ Commented Oct 21, 2020 at 14:38
  • 2
    $\begingroup$ I'd say the difference between power output of an engine and thrust produced is the efficiency of the propeller or jet efflux plus losses due to friction, accessories etc $\endgroup$
    – Raffles
    Commented Oct 21, 2020 at 20:05
  • $\begingroup$ @quietflyer please can you avoid pseudo-answers in comments. If you want to post an answer please do so. $\endgroup$
    – Jamiec
    Commented Oct 26, 2020 at 8:36
  • $\begingroup$ You've tangled many rather unrelated things together in your question. That makes it fairly difficult to answer comprehensively. $\endgroup$
    – Jan Hudec
    Commented Oct 26, 2020 at 20:18
  • $\begingroup$ Re -- "Because I've seen that total drag curve is also the same as Power Required Curve and I don't understand why." -- this isn't right; they aren't actually the same curves. The question could be improved by fixing this. $\endgroup$ Commented Oct 26, 2020 at 20:31

1 Answer 1


Aircraft power curve

(a.k.a “polar”—a completely illogical name because somebody long ago plotted them in polar coordinates, which makes no sense, but the name stuck)

This is just the power balance of the aircraft: the power provided by the propulsion system minus the power taken away by drag.

You can divide the values by velocity (true airspeed) to get a force (thrust - drag) curve instead. But usually just the power curve is plotted and lines through origin are used to read the interesting force maxima.

Two useful observations can be made about meaning of the power and force here:

  • Excess power is linearly proportional to sustainable climb rate, simply by $P = m g v_v$ (where $P$ is power, $m$ is mass of the aircraft, $g$ is gravity and $v_v$ is vertical velocity).

  • Excess thrust is linearly proportional to sustainable angle of climb. Well, actually there is some tricky trigonometry involved, but at low angles (non-aerobatic aircraft only do very shallow climbs and descents, below about 10°) you can approximate $\sin x \approx x$ and $cos x \approx 1$ and say that $T = m g \gamma$ (where $T$ is thrust, $\gamma$ is flight path angle and $m$ and $g$ as above).

This is true independent of what kind of propulsion, if any, the aircraft has. Power curves work the same for propeller, jet and rocket aircraft and gliders alike. Just the function describing power available for given speed differs by the propulsion type.

Kinds of power

A propulsion system is a device that converts one form of energy to another with some efficiency. So in has input power ($P_{in}$)—which is the rate at which energy comes in—and output power ($P_{out}$)—which is the rate at which it provides the desired energy. They are related by efficiency ($\eta$: $P_{out} = \eta P_{in}$). The rest of the energy—because energy is always conserved—is wasted (often as heat, but in aircraft propulsion the energy given to the reaction mass, the air, is also wasted).

Powered aircraft usually use internal combustion engines consuming hydrocarbon fuel, so the input energy is fuel flow times heating value of the fuel.

The output power of the whole propulsion system is then the energy given to the aircraft (to replace the energy taken away by drag), and this is always thrust times velocity. They are related by the propulsive efficiency, which varies with velocity, and varies differently for different propulsion systems.

When propeller is used, you can split the system in two parts, with their own separate efficiencies: The engine burns fuel, drives the shaft, and wastes the rest of energy as heat of the exhaust gases. It's output power is the torque times shaft speed (the RPM). The propeller then takes this power as input and accelerates some air to produce thrust. But doing so it has to sacrifice some energy to that air (its kinetic energy increases), so its output power is lower than that of the engine.

Note to units

Power=force x velocity, and we can apply that in different ways. If we were talking about electricity we could say Power is measured in Watts, and Watts= Voltage x Current. Because we are talking about prop aircraft, we should use a power unit like HP (for reciprocating engines). HP= Torque X RPM.

Power is always the same dimension, and therefore its basic unit is always Watt. This is a derived unit composed as $\mathrm{W} = \frac{\mathrm{J}}{\mathrm{s}} = \frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{s}^3}$. Power is voltage times current for electricity, and the units work out, because $\mathrm{V}$ is just $\frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{A}\ \mathrm{s}^3}$ (and Ampere is a base unit).

For fuel, power is heat value, in $\frac{\mathrm{J}}{\mathrm{kg}}$, times flow rate, in $\frac{\mathrm{kg}}{\mathrm{s}}$ and we are at Watt again.

And of course torque in $\frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{s}^2}$ times angular velocity in $\frac{1}{\mathrm{s}}$ gives you Watt again. (watch out here; it requires angular velocity, in radians per second, not revolutions per second, but radian works as dimensionless here (metre per metre), so this is prone to error by $2\pi$ that dimensional analysis won't catch).

A horsepower is not an independent unit, just a funny multiple of Watt. Multiple different multiples of Watt, actually, because there are some variants.

Engine power as function of velocity

Piston engine power grows with RPM, then plateaus (at the expense of declining efficiency) and then you hit the maximum permitted RPM.

A constant speed propeller allows running the engine at the optimal RPM, and it's efficiency only decreases slowly with airspeed. Therefore a piston engine with constant-speed propeller produces relatively constant power over the design range of the aircraft, while the thrust declines about inversely proportionally to velocity. Thus it makes most sense to give power as the main figure describing it.

With fixed pitch propeller, you are limited by not being able to reach the optimal RPM at low speeds, and by having to throttle the engine to avoid exceeding the maximum RPM at high speeds, so the power declines faster as you get away from some optimal design speed. That's why fixed propellers are only used on slower planes where it does not matter that much. The power curve is still flatter than the thrust curve (dividing by velocity), so it still makes more sense to give power as the main figure though.

With turbojet engines on the other hand the available thrust remains roughly the same. It first declines with speed somewhat similarly to a propeller, but then the ram pressure starts increasing the effective pressure ratio and the available thrust starts growing again. That's why jet engines are generally rated with thrust, not power. But it isn't really constant over speed, just like the power isn't really constant for a piston engine and propeller.

And this carries over to turbofan engines, that are actually somewhere between. Their thrust decreases with speed, though slower then for propellers, and their efficiency increases. It is an oversimplification to rate them with just static thrust, but it remains the primary value given.

  • $\begingroup$ could we rate "altitude energy use" rate as mg x delta h/s = mv$^2$/s for gliders? Curious as to whether or not we eliminate "1/2" from the steady state energy/time formula. (It graphs as a square instead of a triangle). $\endgroup$ Commented Oct 29, 2020 at 2:42
  • $\begingroup$ @RobertDiGiovanni, which $\frac12$ factor do you mean? Potential energy is $E_p = m h g$, no $\frac12$ factors involved, and power needed for ascent (height grows up, so vertical speed is positive in climb) is simply $P_c = m g v_v$. There is no $t$ factor here, so the differentiation by it does not generate any non-unit factors. Only the kinetic energy gets $\frac12$ factor from going from the first derivative of position (velocity) to the second (acceleration), but the above discussion is for constant speed. $\endgroup$
    – Jan Hudec
    Commented Oct 29, 2020 at 7:19
  • $\begingroup$ thanks for your response. Your section on the interchangability of application for Power is very good. I thought for gliding we could express use of altitude (delta h) per second as mv$^2$/s, which has the same units as fuel consumption J/s that you pointed out. $\endgroup$ Commented Oct 29, 2020 at 9:26
  • $\begingroup$ @RobertDiGiovanni, not just in gliding; always. Gliding is just a special case of engine power output being zero. $\endgroup$
    – Jan Hudec
    Commented Oct 29, 2020 at 9:30

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