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# Propeller torque & engine torque

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# Propeller torque & engine torque

26th Mar 2012, 02:50

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Propeller torque & engine torque

Hi guys,

Just a very quick question, what is the relationship between the propeller torque and the engine torque? Are they actually referring to the same rotational force but only different in teminology?

Hope you could shed me some lights.

Andy
26th Mar 2012, 02:56

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One might expect the engine torque to be that measured at the engine, and the propellor torque that measured at the propellor (i.e. they are measured on either side of the gearbox/transmission system).

That would be analagous to brake horsepower and shaft horsepower.
26th Mar 2012, 04:58

Join Date: Jun 2010
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Like Mad Scientist said, torque is torque. In the case you mentioned, one is measured at the propeller and one at the engine (crankshaft). Propeller torque will be less than engine torque, mainly due to friction.
26th Mar 2012, 05:08

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propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller. If the propeller is at a constant RPM then these are equal and opposite
26th Mar 2012, 06:01

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propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller.
Ah yes, that might be what he's referring to.

If the propeller is at a constant RPM then these are equal and opposite
Yes and no. When the engine is rotating at any RPM, constant or not, it is creating a turning moment (torque) on the airplane. This is due to the propeller drag. If the propeller is rotating clockwise, it wants to rotate the airplane counter-clockwise.

What you're talking about is engine inertia which only produces a turning moment (torque) when the engine is accelerating (positive or negative).

Roll-Wise Torque Budget [Ch. 9 of See How It Flies]

Paragraph 9.5 and 9.6 talk about that.
26th Mar 2012, 13:55

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italia:

What you're talking about is engine inertia which only produces a turning moment
No I really don't think that is what morrisman means. He is talking about the fact that if the RPM is stable, then that is because thrust horsepower equals brake horsepower at that moment in time.

-------

tcyandy:

What I would add to the answers above is that the two torques are different due to the presence of the reduction gearbox. The slower turning shaft - ie the prop - will have a higher torque than on the engine side. The power will be the same though as the relationship is: power = torque x rpm

As an aside, this relationship of shaft speed, torque and power transmitted is illustrated by the spindly drive shafts you can see on the tail boom of a helo. The rpms are stepped up to reduce torque and enable a smaller shaft diameter, then reduced again at the tail assembly.

Last edited by oggers; 26th Mar 2012 at 14:18.
26th Mar 2012, 14:46

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No I really don't think that is what morrisman means.
So when he said "propeller torque" and "engine torque".... he really meant "thrust horsepower" and "brake horsepower"?

He is talking about the fact that if the RPM is stable, then that is because thrust horsepower equals brake horsepower at that moment in time.
Thrust horsepower is equal to thrust multiplied by velocity (TAS). You could be creating 200 BHP, at a steady RPM, while stopped on the ground and your thrust horsepower would be zero.
26th Mar 2012, 15:43

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italia:

So when he said "propeller torque" and "engine torque".... he really meant "thrust horsepower" and "brake horsepower"?
No, you don't get it. Torque of shaft x revs of shaft = power, so it's the same as THP and BHP. I very much doubt he was talking about inertia, the way you asserted.

You could be creating 200 BHP, at a steady RPM, while stopped on the ground and your thrust horsepower would be zero.
Not that myth again. It's disappointing to have to point out to one who calls himself an instructor that in your 'brakes on' scenario the aircraft is still producing 200 THP as well as 200 BHP because it is accelerating a mass of air rearwards in a futile attempt to turn the earth and the atmosphere in opposite directions. Come on italia, pull your socks up. These are fundamentals that instructors should have a grip on.

Last edited by oggers; 26th Mar 2012 at 16:08.
26th Mar 2012, 16:00

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Not that myth again. It's disappointing to have to point out to one who calls himself an instructor that in your 'brakes on' scenario the aircraft is still producing 200 THP as well as 200 BHP because it is accelerating a mass of air rearwards in a futile attempt to turn the earth and the atmosphere in opposite directions. Come on italia, pull your socks up. These are fundamentals that instructors should have a grip on.
Very quick to throw the insults out. Yes, I am a flight instructor. Here is today's lesson:

When talking about performance, the paragraph below is appropriate.

"Power is the rate of doing work, and work is a force times a distance. Power required (PR) is the amount of power that is required to produce thrust required. PR is the product of TR and velocity (V). If V is expressed in knots, then the product of TR and V must be divided by 325 to give power in units of horsepower. Thus, thrust horsepower only depends on thrust and velocity."

If you're talking about propeller efficiency, the paragraph below is appropriate.

"In a turboprop, power available is determined by the performance of the engine/propeller combination. Engine output is called shaft horsepower (SHP). Thrust horsepower (THP) is propeller output, or the power that is converted to usable thrust by the propeller. The ability of the propeller to turn engine output into thrust is given by its propeller efficiency (p.e.). Under ideal conditions, SHP would equal THP, but due to friction in the gearbox and propeller drag, THP is always less than SHP. Propeller efficiency is always less than 100%."

So, with regard to your statements:

He is talking about the fact that if the RPM is stable, then that is because thrust horsepower equals brake horsepower at that moment in time.
...the aircraft is still producing 200 THP as well as 200 BHP...
THP will never equal SHP or BHP.

Those two paragraph quotes were taken directly from the "Fundamentals of Aerodynamics" which was prepared by the U.S. Naval Aviation Schools Command.

EDIT: Here is a picture taken from the same textbook. imgur: the simple image sharer
26th Mar 2012, 16:12

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I'll respond to that tomorrow if nobody picks it up in the meantime BTW I added some more to my last post - just so you know.
26th Mar 2012, 16:25

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No, you don't get it. Torque of shaft x revs of shaft = power, so it's the same as THP and BHP. I very much doubt he was talking about inertia, the way you asserted.
I understand it very well. We should probably wait until he responds to see what he meant!

I'd recommend checking out this article to see the difference between power and torque. If you realized the difference, you would see that they aren't interchangeable.

Power and Torque.pdf - File Shared from Box - Free Online File Storage

It will also help you understand why THP is equal to zero when stopped on the ground. Power = Force x velocity

Last edited by italia458; 26th Mar 2012 at 17:10.
26th Mar 2012, 16:31

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propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller. If the propeller is at a constant RPM then these are equal and opposite
That depends on how you define engine torque, eg before or after any gearbox.

If you define it as after the gearbox you are correct (otherwise the aircraft would roll).

If define it as before the gearbox (eg between crankshaft and airframe) then it's normally less than the prop torque due to use of reduction gearbox.
26th Mar 2012, 16:59

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If you define it as after the gearbox you are correct (otherwise the aircraft would roll).
Propeller torque is measured after the gearbox. So measuring 'engine torque after the gearbox' is propeller torque! Both would be measured in the same place so they would obviously be the same.

I think some of the confusion is related to the ambiguous term, "torque". Torque is defined as the tendency of a force to rotate an object about an axis, fulcrum, or pivot. That means ANY force and there are lots of forces in play when an airplane is in flight. As I mentioned above, propeller drag constitutes a torque and is sometimes understood as propeller torque, mostly to people who have only flown direct drive propeller engines. I think tcyandy needs to clarify exactly what he is asking so that he can get a clear answer.
27th Mar 2012, 10:34

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Italia:

For the sake of clarity I agree that THP will not equal BHP because of prop efficiency (I should have been more careful to specify SHP in front of a pedant) and frictional losses in the intermediate gearing. But that is just semantics, beside the principle morrissman was getting at, which was I believe:

At constant RPM the power required at the prop shaft = power provided by the crankshaft.

The torque, were it to be measured on each shaft, would be different however. Really quite simple. He wasn’t talking about the inertia tangent you went off on.

Now, about this stopped aircraft knocking out 200 BHP but no THP because it doesn’t have any velocity nonsense. What about a helicopter in a hover? Any thrust there (and you can include frictional losses and rotor efficiency in your answer if it will make you happy)? The answer is pretty self-evident. The aircraft doesn’t have to move to produce thrust. Do you, as a flight instructor, agree with that or not?

Because when you say “Today’s lesson is…” you quote an extract from a credible text on aerodynamics. But you have taken the text completely out of context. Of course we can work out power required by taking drag and multiplying by airspeed! But that does not mean a stationary aircraft can do no work and therefore produce no THP!!! The work is done by moving the mass of air one way and the mass of the Earth a tiny imperceptible amount the other way.
27th Mar 2012, 13:14

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At constant RPM the power required at the prop shaft = power provided by the crankshaft.
I can see that that might be exactly what he wanted to know. To me it seems obvious as Fnet=ma. When there is no acceleration (ie: steady RPM), there has to be no net force. This whole thing is ridiculous because from the beginning, the OP didn't ask a clear question. If what you wrote was what the OP asked if it was true, this thread would be done after someone wrote Fnet=ma.

Now, about this stopped aircraft knocking out 200 BHP but no THP because it doesn’t have any velocity nonsense. What about a helicopter in a hover? Any thrust there (and you can include frictional losses and rotor efficiency in your answer if it will make you happy)? The answer is pretty self-evident. The aircraft doesn’t have to move to produce thrust. Do you, as a flight instructor, agree with that or not?
I understand what you're trying to say but I can tell you that with regard to aircraft performance, what I said is correct. If you would like I can send you a copy of that U.S. Naval aerodynamics text. You're getting tied up on the word thrust and you've used it extensively throughout that paragraph and I agree with everything you said. There is lots of thrust when the helicopter is in the hover! And I agree the aircraft doesn't have to move to produce thrust. I would never teach a student otherwise.

However, you're not understanding what THP actually is. Thrust Horsepower is POWER! It is NOT thrust. Re-read my post with the paragraph explaining that and then look at the picture I included. Yes, the engine will be producing a certain SHP that will go to the propeller and produce a certain amount of thrust. But, THP is only related to the work the thrust does on the aircraft. Work = force x distance, therefore, if the aircraft isn't moving, it isn't covering distance and so the work = zero. When work = zero, Power also = zero. In the document I included, "Power and Torque", I broke down those equations so you should be able to see exactly how power relates to thrust(torque).

Because when you say “Today’s lesson is…” you quote an extract from a credible text on aerodynamics. But you have taken the text completely out of context.
How do you know the text is credible? You obviously don't have the text because you say that I quoted it out of context.

Of course we can work out power required by taking drag and multiplying by airspeed! But that does not mean a stationary aircraft can do no work and therefore produce no THP!!!
Oh you were so close! I thought you had it. Yes a stationary aircraft is doing zero work and it is producing no THP! I'm usually pretty careful to not state flat out that I know I'm correct but in this case I think I could come out and say that. I'm not going to continue to argue if it's true or not but if you would like to understand it, I have no problem going into detail and explaining it.
27th Mar 2012, 13:51

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tcyandy

If you are studying for the JAA/EASA ATPL then the answer they are looking for is.

Propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller. If the propeller is at a constant RPM then these are equal and opposite.

If you are studying something more technical then some of the other posts in this thread are more relevant.
27th Mar 2012, 14:33

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How do you know the text is credible?
Because you told me it was "Fundamentals of Aerodynamics" produced by the US Naval Aviation Schools Command. That and the fact that drag x airspeed = power required is hardly controversial.

But you are waffling again. The bottom line is you believe:

You could be creating 200 BHP, at a steady RPM, while stopped on the ground and your thrust horsepower would be zero.
This is incorrect. If you are creating 200 BHP it's because you are moving a mass of air over a distance, and at a rate equivalent to 200 BHP x [efficiency of prop and transmission].

BTW I see Keith Williams has now posted exactly the same thing as morrissman. Would you make the same statement now to Keith as you did to morrissman? ie

What you're talking about is engine inertia which only produces a turning moment (torque) when the engine is accelerating (positive or negative).
?
27th Mar 2012, 15:42

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Because you told me it was "Fundamentals of Aerodynamics" produced by the US Naval Aviation Schools Command. That and the fact that drag x airspeed = power required is hardly controversial.
See, that's the difference between you and I. You seem to think that credibility is in a name (in general and in particular, in another thread). I believe credibility is in the truth or accuracy of the material. No one will ever convince me to believe in a name.

You seem to understand that Power required = drag x airspeed but you don't believe that Power available = thrust x airspeed?

This is incorrect. If you are creating 200 BHP it's because you are moving a mass of air over a distance, and at a rate equivalent to 200 BHP x [efficiency of prop and transmission].
After telling my that my statement about THP was incorrect, you provide one sentence that doesn't mention THP once! Did you really accomplish anything?

You are consistently highlighting your lack of understanding of the matter and yet you are forceful in your opinion that myself and all aeronautical engineers are wrong. Just because something seems correct, doesn't mean it's correct. I have maths and physics that prove that you are incorrect and yet you still try to convince me otherwise. Unless you can prove something wrong with the maths and physics used to illustrate THP (which I would love to see), you aren't proving anything worthwhile.
27th Mar 2012, 16:01

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IF tcyandy is talking about a direct-drive, ungeared engine, then there is ONLY ONE TORQUE to consider. It is the torque the engine crankshaft delivers to the prop.

You may call it prop torque, or engine torque, as you like - but they are THE SAME THING.

If there's a gearbox in the system, then there are an input torque (i.e. crankshaft) and an output torque (i.e. prop). If the gear ratio is 2:1, then rpm is halved, and torque is doubled (well, almost, minus a very small heat loss).
28th Mar 2012, 00:47

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oggers... To clarify further:

You seem to understand that Power required = drag x airspeed.

Drag is the same as Thrust required for level flight; it's basically the force that needs to be offset with thrust so the airplane can fly.... therefore,

Power required = thrust required x airspeed

And Power available is,

Power available = thrust available x airspeed

This picture shows it: http://i.imgur.com/R43ho.png

What might be confusing is the Thrust Horsepower term. It is still power and is talking about the power generated by the thrust that propels the aircraft forward. If the picture was showing the BHP or SHP as well, it would be virtually a straight line near the top of the graph.

Here is a picture I took from another textbook of mine that shows the THP available and required curves. It's not 100% clear but I think it shows enough: http://i.imgur.com/3e4Ru.jpg?1