# Why is thrust inverse to speed in piston engines?

It is said that piston engines have constant power output, and thus their thrust is inverse to speed (e.g. here), while turbines have relatively constant thrust. I'm looking for an intuitive explanation why is this the case.

I understand that from the energy conservation point of view it makes sense: kinetic energy is proportional to velocity squared, so it is easier to accelerate air from 0 to 100 m/s than from 100 to 200 m/s.

I'd like to better understand what is the direct cause of that. Why can't we simply increase the propeller speed with increasing air speed using a higher gear ratio? It seems to me that if the air comes at us with twice the speed as it did, we should be able to increase the speed of the propeller to match the speed increase. Then, it seems that the propeller blades should interact with the air in the same manner as before, as relative speeds remain the same: the air is twice faster but so is the blade, and it pushes the air with twice the speed. Why is the (physical) work to be done greater?

I'm not sure whether this is the right place to ask this question. Perhaps physics.stackexchange.com would be better?

kinetic energy is proportional to velocity squared, so it is easier to accelerate air from 0 to 100 m/s than from 100 to 200 m/s.

The same is true for the air flowing through the propeller disk. Even if we replace the propeller by a black box, or better a black disk, which simply adds a bit of pressure to the air flowing through it, accelerating air from 0 m/s to 10 m/s is much easier than accelerating it from 100 m/s to 110 m/s. Since engine power is constant, the absolute speed increase will be smaller the higher the entry speed becomes. Thrust is the difference between the impulse of the air flowing towards the propeller disk and the impulse of the air exiting it, so a smaller speed increase means less thrust at higher speed.

The exit speed increase in turbojets is much larger so this effect becomes much smaller in jets. Also, jets benefit from higher entry speed by precompressing the flow ahead of the intake, an effect which raises mass flow through the engine and increases thrust with the square of airspeed. In the subsonic realm both effects roughly cancel each other, so thrust is approximately constant.

• Why it is easier to accelerate from 0 mph to 50 mph rather than from 50 mph to 100 mph? Because more energy is needed? Aug 8 '21 at 10:56
• @Konrad Energy is proportional to speed squared while impulse grows linearly with speed. Aug 8 '21 at 11:15
• So more energy needs to be spent for the same impulse increase? Aug 8 '21 at 12:29
• Thank you for your answer. I was already aware of the greater kinetic energy of the moving air, but I wanted a more low-level analysis. Inspired by your black box propeller argument I made my own analysis. Aug 8 '21 at 13:32
• Another way of looking at it : at higher speed, there is more air (mass) passing the propellor to be accelerated, so you would need more power to give it the same acceleration. Aug 8 '21 at 14:55

Then, it seems that the propeller blades should interact with the air in the same manner as before, as relative speeds remain the same: the air is twice faster but so is the blade, and it pushes the air with twice the speed. Why is the (physical) work to be done greater?

Assuming the physical work to be done per revolution is the same you still need to accomplish it in half the time, needing double the power. But even if the interaction of blade and air results in the same forces acting on the air over the same distance of blade travel (incurring the same work), they interact only half the time and thus convey half the impulse.

I always thought of energy analysis as kind of cheating ;) I know that power input needs to balance kinetic energy increase, but it works like a black box and does not give me true understanding why it is the case.

Inspired by Peter's answer I will try to analyze the situation by modelling the propeller as a disk. But not a black disk, but a moving piston type disk ;)

Let us assume that there is an opaque disk in front of the aircraft and it moves faster than air (by $$Δv$$) towards the aircraft, pushing the air like a piston. After passing the distance $$d$$ it magically teleports to the initial position and repeats the motion, pushing another batch of air volume. We ignore how the air moves behind the disk, it just positions itself so that in the next cycle it can be pushed by the disk. It is a really simplistic approach to modelling a propeller, but let us stick with it for now.

We want to compute the force $$F$$ (and thus work $$W = F\times d\,$$) required to push the air in one cycle.

Case 1: the air is stationary. Within distance $$d$$ we need to accelerate a volume $$V$$ of air from 0 to $$Δv$$. Let us assume constant acceleration $$a$$. $$Δv = a\cdot t$$ $$d = a\cdot\frac{t^2}{2} = \frac{Δv^2}{2a}$$ Thus $$a = \frac{Δv^2}{2d}$$ and $$F = V\cdot\rho\cdot\frac{Δv^2}{2d}$$ Where $$\rho$$ is air density. $$W_1 = V\cdot\rho\cdot\frac{Δv^2}{2}$$

Case 2: the air is moving at speed $$v_0 > 0$$. Within distance $$d$$ we need to accelerate a volume $$V$$ from $$v_0$$ to $$v_0 + Δv$$. $$Δv = a\cdot t$$ $$d = v_0\cdot t + a\cdot \frac{t^2}{2}$$ Now it gets more complicated: $$d = v_0\cdot\frac{Δv}{a} + \frac{Δv^2}{2a} = \frac{2\cdot v_0\cdot Δv + Δv^2}{2a}$$ $$a = \frac{2\cdot v_0\cdot Δv + Δv^2}{2d}$$ Finally, $$W_2 = V\cdot\rho\cdot\left(v_0\cdot Δv + \frac{Δv^2}{2}\right)$$

We can clearly see that $$W_2 > W_1$$. Actually, $$W_2 - W_1 = V\cdot \rho\cdot v_0\cdot Δv$$. Which means it requres more energy to push moving air, than stationary air. This is true, even though our 'piston disk' in Case 2 already moves faster than in Case 1 ($$v_0 + Δv$$ vs just $$Δv$$). The thing is that the faster speed of the propeller is not enough. Relatively, the propeller is faster than air by $$Δv$$ in both cases. However, in Case 2, the propeller together with the air it pushes move faster than in Case 1, and thus traverse the distance $$d$$ faster, and thus there is less time to accelerate the air by the required $$Δv$$. We can compensate this by either (1) larger acceleration, and thus force, and thus expedited energy, OR by (2) longer distance $$d$$, which unfortunately means increase in the work (which is force times distance), so also more expedited energy. That is why pushing moving air (and generally faster air) is more difficult.

• +1 Very illustrating! Would you mind if I sprinkle some MathJax over your answer? Aug 8 '21 at 17:44
• @PeterKämpf please do! I'd like to learn it too. Is there some tutorial somewhere on stackexchange? Sadly, when editing an answer/question there is no toolbox or help button related to math notation. Aug 8 '21 at 20:34
• Yes, SE assumes you speak LaTeX fluently. If you click on the question mark in the header's right corner, via several further clicks you will be guided to this page which gives you an introduction to MathJax. Aug 9 '21 at 5:39
• Please check whether the equations still make sense. I suspect you had the work difference wrong by a factor of 2, but that could as well be my mistake. Aug 9 '21 at 6:55
• @PeterKämpf thank you for correcting my answer! Indeed there was an error in $W_2$. I'm delighted to hear that MathJax is just straight latex! Aug 9 '21 at 13:43

Increasing the rotational speed of a piston engine will increase the number of power strokes per second. This will increase the power output up to a certain point, but marginal efficiency will go down dramatically beyond that. The amount of energy that can be extracted from each milligram of burning fuel will depend upon how far the piston has left to move after it burns. At higher rotational speeds, an increasing portion of the fuel will be burned later in the cycle, when the piston has less remaining distance, thus reducing the amount of energy that can be usefully extracted from each power stroke.

In a road vehicle where the engine and wheels are connected by a rigid transmission, engines will need to be designed to operate usefully at a range of rotational speeds. In aircraft, however, it's common to design engines to operate at one particular rotational speed, and vary the propeller blade pitch so that the load placed on the engine will keep it that speed. If an aircraft had a fixed pitch propeller, flying faster might allow the engine to rotate faster; this could in turn allow it to produce more power if it would otherwise have been below the speed that would produce maximum power output. If, however, a plane is designed to run its propeller at a fixed rotational speed, the power the engine can produce will be essentially independent of the speed at which the plane is flying.

• "the power the engine can produce will be essentially independent of the speed at which the plane is flying" Not quite. There's a limit to how quickly the blades of the engine can go before their tips go supersonic, and that limit will decrease as air speed increases. Aug 10 '21 at 9:35
• @nick012000: Could a prop-driven plane with a constant-RPM propeller travel get anywhere near reaching a speed where blade tips would go supersonic? I suppose maybe in a powered dive that might be possible, but I would think that diving at such speed would cause problems for the rest of the airframe as well. Aug 10 '21 at 14:56
• That depends on how fast the propellers are rotating, and how fast the airplane was designed to go. Some of the WW2 fighter planes were designed to go nearly transonic, for instance, and the main limiting factor for them was the propellers, IIRC. Aug 10 '21 at 15:23
• @nick012000: It's true that if cost were no object and one added more powerful engines in an effort to make a plane fly as fast as possible, propeller tip speed would be a limiting factor, but unless someone were seeking a world record for "fastest piston-driven aircraft" or something I don't such considerations would have been relevant in the decades since other engines have been developed that are more suitable for use at higher speeds. If one considers how much air needs to move through an engine to burn the requisite amount of fuel, and how long the burning fuel-air mixture... Aug 10 '21 at 15:34
• "I don't such considerations would have been relevant in the decades since other engines have been developed" Like I said, WW2-era fighter craft were often designed to push the performance envelopes possible for propeller-driven aircraft, since jet engines had not yet been developed (aside from some experimental stuff built by Nazi Germany late in the war). Aug 10 '21 at 15:37

Because more of what a turbojet does happens inside of a box. What is going on inside the box is (designed to be) isolated to a large degree from conditions outside the box. A turbojet has to compress combustion air in order to work. It does that by converting velocity into pressure (a little at a time). If it gets some free velocity at the inlet, that's nice, but you want the conditions at the combustion chamber largely the same regardless of inlet conditions. And what is going on in the turbine section is also largely isolated from conditions outside the box.

The piston engine also works inside a box. Like the turbojet, the compression, combustion, and power extraction all happen in the box, but unlike the turbojet, the thrust isn't created there.

The propeller does not work in a box. The propeller, unfortunately, is at the mercy of the ambient conditions. As far as flight speed goes, the mass flow through the propeller disc is proportional to speed. $$power = \dot m \times Velocity^2 = density \times Area \times Velocity^3$$

If we look at a before and after picture, and conserve mass $$power = density \times V_{eff} \times Area \times (V2^2-V1^2)$$ For props, $$V_{eff}$$ is just the average of $$V1$$ and $$V2$$, so there is a term that looks like $$\frac{(V2+V1)}{2}(V2^2-V1^2)$$ that has to be constant at constant power. So if the average speed goes up X% the second term has to decrease by X%.

If $$V2 = 150$$ mph and $$V1 = 100$$ mph, the term evaluates to $$1562500$$. If $$V1$$ is increased to $$120$$ mph, the required $$V2$$ is about $$160$$ mph.

$$Force = mass \times acceleration$$ or $$mass flux \times change\ in\ velocity$$. The second is handier. The mass flux changes in proportion to the change in average velocity, from 125 to 140. The delta V changes from 50 mph to 40 mph.

Thrust @ $$V1 = 100$$ mph is proportional to $$125 * 50 = 6250$$.
Thrust @ $$V1 = 120$$ mph is proportional to $$140 * 40 = 5600$$.

The thrust drop is about 11 percent.

The real takeaway here is that it is 90% engineering and 10% physics that leads to this situation. We can control how sensitive the box's internal workings are to external conditions. The more that happens in the box, the better the situation can be managed.