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(This question is closely related to this question)

A typical Cessna 172 has a 180 hp engine, with a max. rotational velocity of 2700 RPM. Even small 1 hp electrical motors can achieve that RPM, and therefore, I guess the rest of the power is converted into torque.

What's the use of so much torque, and what happens if there is less torque at the same RPM - i.e.

Can an electric motor perform similar to the Lycoming IO-360 with the same max. RPM, but with just half the torque of the latter?

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    $\begingroup$ Consider: Is it the rotation, or the torque, which ultimately causes the aircraft to move? $\endgroup$ – a CVn Aug 22 '17 at 10:50
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    $\begingroup$ Remember that rpm is a speed, torque is a force. In the absence of any force, speed is constant. add friction/external resisting force and you need to add an internal force to keep the rpm. so you don't want to care only about rpms, but also what you are doing with them. $\endgroup$ – Federico Aug 22 '17 at 11:18
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    $\begingroup$ Then you will need much higher rpm to spend the same horsepower. This could end different ways, e.g. a smaller prop and/or fewer blades, a duct fan or jet fan. Power is torque times rpm times a constant. Less torque same rpm is just less power. The cooling fan in your computer also run in the 2000-4000rpm range but the torque is tiny compared with airplane engine, so is power. . $\endgroup$ – user3528438 Aug 22 '17 at 11:40
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    $\begingroup$ Power = torque X rpm. So the same rpm with half the torque is half the power. $\endgroup$ – Gerry Aug 22 '17 at 11:56
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    $\begingroup$ I think this question is missing any understanding of the mechanics involved. To put it in simple (perhaps simplistic) terms, the horsepower of the engine determines the force (mass * velocity) of the air that the propellor can accelerate backwards, and thus move the airplane forward. Torque and to an extent RPM are really irrelevant in this context: regardless of how much torque your 1 hp electric motor produces, it only accelerates 1 hp worth of air, while a propellor could be designed to work efficiently at 270 rpm, or 27 (think large wind turbines). $\endgroup$ – jamesqf Aug 22 '17 at 17:15
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Even small 1 hp electrical motors can achieve that RPM

Yes, but can it sustain that RPM when something tries to stop it, that is the question. 1 hp combustion engines can get to 2700 rpm as well, but you cannot attach a propeller to it, stick it in the air, and expect it to maintain the RPM. It takes a lot of torque to create aerodynamic thrust.

It is a fact that electric motors are way better at producing torque than combustion engines. An electric motor produces almost maximum torque at the get-go, with the rotor standing still. The combustion engine needs to make RPM in order to make torque. This article compares electric motors with combustion engines for powering a hydraulic pump: a 20 hp electric motor can do the job that a 50 hp combustion engine can do.

Electric motor:

enter image description here

Combustion engine: intermittent torque means torque pulse from combustion.

enter image description here

But wait, is that even possible? Power is power, kilowatt is kilowatt, and it should not matter which engine we use to generate the kilowatts, they should all equate to equal amounts says the first law of thermodynamics, the conservation of energy. But it does matter, since with a combustion engine torque is a function of RPM. So increase RPM, and power increases due to increased RPM and increased torque. At lower RPM the combustion engine may simply lack the torque to speed up, while the electric motor has an almost flat torque curve.

This same happens with outboard motors, when comparing 2-stroke with 4-stroke. 2-stroke has way more torque at lower revs, the 4-stroke may have the same horsepower but may never get to the high revs because it lacks the ability to get the boat planing.

But I digress. To get back to your questions:

What's the use of so much torque, and what happens if there is less torque at the same RPM

To keep the output shaft at this RPM while the propeller is pushing all that air backwards. That requires a serious amount of torque. If there is less torque at the same RPM, there will be less thrust produced and the aeroplane won't fly as fast.

Can an electric motor perform similar to the Lycoming IO-360 with the same max. RPM, but with just half the torque of the latter?

Well, the surprising thing is that a horsepower is not a horsepower. Part of the problem lies in the horsepower output functions, so maybe an electric motor with a lower hp rating can be used. But one thing is very clear: at the RPM of the propeller, a certain torque is required at a certain speed to keep the propeller turning. This amount of torque will always be the same, no matter what kind of engine produces it.

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    $\begingroup$ Long story short, it is simply NOT POSSIBLE to maintain the same RPM when you attach a large propeller (propeller experiences drag thus induces torque). Torque is not just a word. It is the force experienced by the shaft. If you want lower torque you cannot just say "lower torque". Physics won't let you. To lower torque you have to use a smaller prop. Thus torque really is an indirect measure of how large a prop you can attach to the engine. $\endgroup$ – slebetman Aug 22 '17 at 15:10
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    $\begingroup$ @slebetman Or how much pitch the prop has. With a CSP it will alter the pitch to maintain the rpm setting. If you have the power setting too low it will still spin at the set rpm, but at such a low pitch that it won't keep the plane in the air. $\endgroup$ – TomMcW Aug 22 '17 at 16:30
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    $\begingroup$ This is the reason IC engines have starters and flywheels, and why they stall if you try to run them at too slow a speed. IC engines work by converting discrete little "explosions" in the cylinders to continuous rotary motion: if the engine is not turning fast enough, not many of those explosions take place. Electric motors work by continuous magnetic force, which is still strong when the engine is at a standstill. $\endgroup$ – jamesqf Aug 22 '17 at 17:21
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    $\begingroup$ @jamesqf Torque is required, like force is in a rocket. They are one and the same, only the reference frame is rotary instead of linear. $\endgroup$ – Koyovis Aug 25 '17 at 19:52
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    $\begingroup$ @AnandS The engine rotates the propeller, the propeller provides thrust to overcome the aerodynamic resistance. Less resistamce, less thrust required, less torque required from the power plant to drive the propeller. $\endgroup$ – Koyovis Aug 25 '17 at 20:09
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You're ignoring air resistance. A tiny electric motor can accelerate the prop to 2700 rpm in a vacuum. But at 1 bar, the prop moves against the air (effectively pumping air from one place to another), and this requires torque.

An engine with less power won't be able to spin the prop at 2700 rpm.

With a variable-pitch prop, you should be able to observe this. Set the pitch to 0, and the engine can turn the prop with little trouble. As pitch increases, so does the amount of air moved by the prop, and the engine will use more fuel to keep running at the same speed, until you reach the engine's maximum power output.

When you stop applying torque to the propeller while the aircraft is stationary, the propeller will stop turning. If the prop is driven by a piston engine, it will stop abruptly, because the compression stroke takes a lot of power. If you were to decouple the prop from the engine, the prop would stop a bit more gradually, because drag depends on speed (lower speed = less drag).

When you stop applying torque to the propeller while in flight, the aircraft's speed starts pushing the propeller around and the prop acts like a windmill. The drag in this configuration means your aircraft will rapidly lose speed.

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    $\begingroup$ An even simpler observation is that the purpose of a propeller is to convert rotation to linear air movement, and torque to thrust. If there's no torque, there can be no thrust. $\endgroup$ – supercat Aug 22 '17 at 22:27
  • $\begingroup$ So without the necessary torque, the propeller just stops pushing air? What actually happens to the propeller and the associated forces of flight? $\endgroup$ – ClobberXD Aug 26 '17 at 6:52
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    $\begingroup$ @AnandS: No, the propeller does not just stop! It accelerates to the speed where the motor power matches the air resistance. $\endgroup$ – Transistor Aug 28 '17 at 6:57
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It is not RPM that keeps the plane flying. It is the power that does.

As the aircraft moves through the air, it experiences drag. Since drag acts in the (opposite) direction of motion, it does work on the aircraft, taking away energy. This energy must be replaced by the engine. If it is not, the aircraft will either slow down (kinetic energy is depleted) or descend (potential energy is depleted).

Drag comes in two forms: induced and parasite. Parasite drag is caused roughly by “friction” with the air (it's quite complicated actually). Anything moving through fluid experiences it.

More interesting is the induced drag, which is experienced by an airfoil generating lift. To make the air push up on the wing, it must push down on the air due to principle of action and reaction. However, this increases kinetic energy of the air and that must come from somewhere. Due to this, it is not possible to push straight down. Rather a forward component is inevitable that does negative work on the plane to provide this energy. This is the induced drag.

Now a propeller works just like a wing. It moves through the air and pushes it, so it experiences both parasitic and induced drag. The drag must be opposed by the torque, otherwise it will deplete the rotational kinetic energy of the propeller and the propeller will stop.

The engine must have enough power to provide the energy as fast as all the forms of drag deplete it.

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When you put a prop at the end of the shaft of an engine, and wish to turn that prop at a given rpms, the existence of the atmosphere induces a resisting torque that must be compensated by an equal and opposite acting torque supplied by the engine. At any stable rpms, the relation is:

power supplied = angular speed x torque.

Obviously, and for a give angular speed, if the power supplied to the prop is very high, the torque will be very high too...

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    $\begingroup$ Finally, an answer that mentions the key relationship between torque, rpm and shaft power. $\endgroup$ – pericynthion Aug 26 at 15:22
  • $\begingroup$ @pericynthion Please check the answer linked in OP. $\endgroup$ – Koyovis Aug 26 at 19:11
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It all depends on the propeller one can use:

  • The prop should rotate at as much RPM as possible because force is proportional to speed squared and one want a small PropPhiW as possible (the AOA required by the prop-speed / airspeed geometry). Lower PropPhiW means increased efficiency.
  • The prop's tips should rotate below the speed of sound, or else massive additional frictional forces (-> lift/drag goes into the cellar)
  • The prop's AOA0 (the difference betw. set PropAOA - PropPhiW) should be at the max. lift/drag point (most often around 4-6°).

To get that prop rotating one needs torque in the first place, enough to bring the prop into the high efficiency region (low PropPhiW) speedwise and into the maximum lift/drag region (AOA0).

Having a lot of torque makes the setup easier, one has more choices, but one certainly needs an excel-sheet to analyze it fully, operating outside the optimal costs a lot of efficiency.

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    $\begingroup$ Hi and welcome to the site! I see you are trying to employ math terms but these come across as nigh-unreadable. Please see this for a reference on how to use the math notation on this site. also, consider using paragraphs by inserting an empty line in your source text. $\endgroup$ – AEhere supports Monica Aug 26 at 13:43
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NO. Start with the prop. What amount of power will turn that prop at 2700 rpm. 1 hp? Nope.

Let us go back to James Watt once again and see what the horse is doing: lifting weight!

Horsepower and torque have been melded together in modern definitions from applications to internal combustion limits. In order to brag about horsepower, we want rpm high enough to get lots of fuel flowing into the engine, but not too high rpms to burn it up. Friction and heat transfer issues limit internal combustion aircraft engine rpms, jets do a bit better, but still must be careful not to overheat.

So you can compare the torque of your drive shaft/gearing to the drag torque of your prop, but no torque is "wasted", you simply burn as much fuel as is needed to spin the prop at a given rpm.

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  • $\begingroup$ Thanks for answering. If you could reword the third paragraph of your answer to be clearer and more easily understandable, that'd be great. :) $\endgroup$ – ClobberXD Aug 27 at 16:43
  • $\begingroup$ @ClobberXD piston engines are rated torque at a certain rpm. Power will be less at lower rpm and more at higher rpm. Torque is merely forcexleverarm just like a see saw. Your drag torque for the Cessna prop is sitting on the other side. $\endgroup$ – Robert DiGiovanni Aug 27 at 19:14
  • $\begingroup$ Ah, makes more sense now, thanks! $\endgroup$ – ClobberXD Aug 28 at 2:36

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