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.

Loads of thanks!!

Andy

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.

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.

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

Ah yes, that might be what he's referring to.

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.

italia:

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.

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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.

So when he said "propeller torque" and "engine torque".... he really meant "thrust horsepower" and "brake horsepower"? :ugh:

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.

italia:

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.


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:

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

I'll respond to that tomorrow if nobody picks it up in the meantime :ok: BTW I added some more to my last post - just so you know.

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

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.

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.

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. :ugh:

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.

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).

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.

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.

tcyandy

What are you studying that has made you ask this question?

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.

:\ 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:

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

?

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?

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.

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).

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

Harold... that's a repeat of what was already said on the thread. It also caused a decent amount of confusion. Can you be more specific about the propeller torque you mention? What specific force and what specific lever arm/axis?

italia you are still waffling. I know what thrust, power and torque are. Before becoming a pilot I was a transmission design engineer.

I have made two simple points: the first was that the [3 now!] posters who have said exactly what Harold did at #20 are unlikely to be talking about the inertia thing you assumed.

The second point is that you are wrong with this:

….yes please do explain how you think this is the case because the standard blurb about power required and power available to maintain level flight don't apply to an aircraft being used as a big stationary fan on the ground. But that doesn't mean a fan has no thrust and no thrust horsepower.

Do you believe a helicopter in a hover is producing any thrust horsepower? Or if the helo is 'wheels light': any THP?

Torque
 

Obviously GM-Allison don't give a fig about differentiating between Engine Torque and Propeller Torque in the C130. One indication only, Engine Torque, displayed in in/lbs and measured between the core engine and the gearbox by the Torque Meter Shaft assembly. On all up to and including the 'H' model a limit of 19600 in/lbs applied, this being an engine mounting limit rather than an engine limitation. Don't know about the J model, no seat for a F/E.

Hi Old Fella,

VC9 and Tyne turbo props was similar.

"Take Off Power" was called when we had 15,250 N1 rpm and an indication of 430 psi on an engine "torque meter" (no idea about the pivot radius, nor the piston area to obtain the correct dimensions).

It didn't matter if the aircraft was stationary or not - it still produced that amount of power to accelerate the air mass backwards and with some relative velocity. (Mass * Acceleration * relative Velocity)

Oggers... I'm done dealing with this. You're going to have to figure this out on your own. Or just be happy with your answer knowing that all these textbooks say that you're wrong. I could derive the equations to show you how they got to Power = thrust x airspeed but it's not worth it because you'll just say that I'm wrong!

It seems like Harold decided to recant his post and I couldn't care if 10000 people had the same point of view, it doesn't mean that they're correct!

You're just the same as people who are told one thing in a somewhat convincing manner and believe it like a religion even in the face of clear evidence showing that what they were initial taught is wrong. Are you going to tell me now that all the forces in a turn are balanced, like so many books and instructors teach it? You haven't addressed ANY of the evidence that I've shown you that says you're wrong. So why don't you take the time now to write a little blurb and tell the US Navy that they don't know what they're talking about and write to William Kershner and tell him he doesn't know what he's talking about. :rolleyes:

I'd be happy to go into detail on these matters for someone who was interested in learning about why they say Power = thrust x airspeed. But considering our interaction so far, I think I'll pass.

Edit: This is the nice teacher in me coming out: You said you were a transmission design engineer. I think your misunderstanding of this is coming from that. When sitting stationary on the runway with full throttle, you're creating tons of thrust. You can ask where that's coming from and I'd say it's coming from the torque of the engine and the RPM of the crankshaft which is turning the propeller.... which is power. Since the engine is able to turn the crankshaft, it's creating power... if the RPM was zero, it would be zero power, even though there might be lots of torque. Imagine yourself turning a bolt with a wrench, when you apply a lot of torque to the wrench and it doesn't move, there is no work being done. When the bolt finally starts to move, then you start doing work and your boss will agree with you! Even though you will tell your boss that you've been 'working' on the bolt all day and it hasn't moved, your boss will yell at you and correctly tell you that you've done zero work! Back to the airplane... so to get that thrust we need power from somewhere and its coming from the engine. That power is termed BHP or SHP. But I can create 10000 BHP on my engine and go nowhere! Even though I've actually been moving pistons up and down in the engine and doing lots of work there, I've done no work on the vehicle because it hasn't moved! And that's what we're talking about when we say THP. We're talking about the 'work' that the thrust does and that relates to aircraft performance. So if we create a lot of thrust but we don't move, we are not creating any THP. Thrust/THP and Torque/BHP(SHP) are separate concepts.

Hi italia458,

We all agree that if a force doesn't move then no useful work has been done.

The graphs of Power V Speed in post #20 aren't very useful. They've plotted Power (Thrust * Speed) v Speed.
The gradient of the graph merely gives Thrust (which is more useful.)

Rolls Royce "the Jet Engine" has used the same graph but labelled the Y axis as "Propulsive Efficiency" % (work done on aircraft / Energy imparted to engine airflow).

I much prefer the RR explanation and mathematics.

We agree that we are not creating any useful THP - but we are most certainly accelerating a mass of air and have given it a relative velocity - thus we are developing THP. Otherwise where is the energy of the fuel we are using going to?

italia: one word: hover. One question: any thrust horsepower or not?

I will leave the rest for another day :ok:

Permit me, please, another thought problem:

A C-130 with a failed #4 starter lines up behind another C-130 for a "buddy start". The "mother ship" runs up a couple engines while standing on the brakes. 100% RPM, rated torque, zero airspeed: No thrust horsepower relative to this stationary ship.

But the recipient ship feels the propwash, and its #4 prop begins to windmill, finally permitting it to reach self-sustaining speed.

Obviously there is useful horsepower available in the propwash, even though it's not being used by the mother ship.

So my conclusion is: Horsepower is in the frame of reference of the beholder!

Yes, the slope of the line = thrust. What's your point? The reason the slope equals thrust is because of the equation: Power = thrust x speed. The slope of any line is rise over run. And that equals thrust in this case. But I don't think you understand where that's coming from. The graph might make more sense to you if you look at the thrust graph. http://i.imgur.com/FLRfH.png

Power is a function of thrust and airspeed (Power = thrust x airspeed). So, when you plug in the thrust at a certain airspeed, the resultant is THP. So, when the aircraft is not moving, the airspeed is zero and the THP is zero.

As a note, I'll use BHP or SHP in this post but for all intents and purposes, they are the same thing.

And you'll ask where does that thrust come from? It comes from the power of the engine (BHP). And that power is equal to the torque multiplied by the RPM. And that torque enables the RPM. And the burning of fuel inside the cylinder creates a force which acts at a certain distance from the rotational axis of the crankshaft, at a certain range of angles, which creates the torque. The thrust is directly related to the BHP of the engine. If you increase the BHP generated by the engine, you will increase the thrust.

As I have said before, THP has to do with aircraft performance. Yes, you are creating lots of thrust and that requires power to do that and that power comes from the engine which is the BHP. No physics laws have been violated!

Aircraft performance is with regard to climbing, descending, gliding, turning, accelerating, decelerating, etc. I have a question for you - if your airspeed is zero, what is your climbing performance? What is your turning performance? It's quite obvious that it is zero! When you increase your speed to the point where Power available and Power required are equal, you will be able to maintain level, non-accelerated flight. If you increase your speed to a point where PA is less than PR, you will have negative performance. If you increase to a speed where PA is greater than PR, you will either accelerate or you will climb - or do both, until you reach a point where they will equal each other. When PA and PR are equal, TA and TR will be equal as well. You've probably heard that angle of climb is dependent on excess thrust and rate of climb is dependent on excess power. That is correct! The airspeed for maximum excess thrust will be different than the airspeed for maximum excess power. That difference is not only related to the difference between PA and TA, but the difference between TR and PR. TR is equal to drag and PR is equal to TR x velocity, which is equal to drag x velocity and so on. Just like TR is equal to the addition of parasite drag and induced drag, the PR curve is equal to the addition of two velocity curves: V cubed and 1/V - which are parasite power required and induced power required, respectively. It is a somewhat hard concept to understand because the difference isn't entirely obvious. But these graphs and equations will explain and prove that to be the case, as long as you understand where all the data comes from and what the data means. If you understand the purpose of these graphs and material I think you'll be able to understand better what THP really means.

I uploaded a section from the US Navy document about Thrust and Power: US Navy - Thrust and Power.pdf - File Shared from Box - Free Online File Storage

If you could provide a picture of the graph that would be great. I should point out that it is not the same graph. It might have a similar shape but it is not the same. Also, with jet engines the efficiency and amount of energy imparted for propulsive work is related to thrust, whereas with propeller aircraft it's generally related to power (BHP). A jet engine makes thrust a propeller engine makes power. Fuel flow is generally directly related to thrust in a jet engine and in a propeller aircraft is directly related to power. There are some variances but that's what they're based on. An example of variances are: a jet engine at 50 knots is creating a heck of a lot less power than at 500 knots assuming that in both cases the engines are producing 75% thrust. The fuel flow at 75% thrust in the case of the slower speed will be higher than in the case of the faster speed due to differences in efficiency, which is backwards to what most people think.

The document above will illustrate propeller (propulsive) efficiency as THP = SHP x pe (propeller efficiency). And working with the equation you get:

pe = THP/SHP = Thrust x velocity/SHP = Thrust x (distance/time) / Torque x RPM x 2pi

As the document says, gearbox inefficiencies and propeller drag will decrease pe from being 100%.

Note the document below (Aerodynamics for Naval Aviators) which also explains propeller efficiency.

Aerodynamics for Naval Aviators.pdf - File Shared from Box - Free Online File Storage

I understand what you're trying to say but calling it THP isn't correct in this case. To really get a good understanding of how this THP actually works with regard to aircraft performance you have to start at the very basics and work up. The material to cover is at least the amount in a full university course and depending on how much detail you want to go into, it could cover multiple courses.

This is a good wiki article on Thrust: Thrust - Wikipedia, the free encyclopedia -- It also explains what I have been in this thread. Read the "Thrust to propulsive power" section.

Yes, that's definitely a relativity scenario! There is no difference - from the frame of reference of the #4 engine - between an airstart and your scenario.

Hi italia458,

Thanks for the concise explanation.

NAVAVSCOLSCOM-SG-111 Page 4 at the top: (my bolding.)
"Thrust horsepower (THP) is propeller output, or the power that is converted to usable thrust by the propeller."
We all agree.

Please explain oggers' helicopter hover (when ias = zero) using your concept of the thrust horsepower of the blades (behaving like a big propeller).

It's okay. I figured it out in 1981 with a little help from the grown-ups at Farnborough.

Not disagreed with a single reference you’ve given. Just pointed out they don’t support your point.

No need, I already passed the exams. But it’s irrelevant here. It doesn’t support your point.

Well, it's gone. Maybe because it was a carbon copy of two previous ones and as such breached the guidelines?

Actually I like clear evidence. What you've posted though, is more like a load of red-herrings, strawmen and general verbosity...

...case in point. Where did that come from?

There is no evidence. There is only you stamping your feet and insisting on your own personal definition of what constitutes THP.


Because – as I pointed out before - that Navy stuff is correct. But it doesn’t support your point, it is redundant in that context.

You said that twice in the same post. It’s not required but I see you went ahead and laboured the point anyway in your next post.

Because transmission designers are known for having trouble understanding thrust, torque and power?

Thanks, I.…oh wait you’re telling me anyway:

...okaaay...

...riiiight...

Whaaaaaat?!!! Stop the record. An engine at zero RPM and lots of torque? Better have a good handbrake!

What if I don’t have a boss - is work still being done?

To be honest mate, if I can’t get a bolt off quickly by the simple application of torque I don’t spend all day pulling on the wrench. But if that's your method, fair enough.

Go nowhere sure but can you make 10 000 BHP and not generate so much as a thrust pony?

Are you telling me that when I was releasing bolts you were in the engine? BTW everyone knows that 'no work is being done on the vehicle'. But: work...is...being...done...on...the...air.

Nice story, thanks for sharing. But who is this ‘we’? Is there someone in there with you? If so why not ask them if a helicopter is producing any THP in the hover.

Torque is measured from the torque meter sensing torque on the propellor shaft.
So one would say that the engine torque and prop torque are essentially the same.

Hi rudderrudderrat:

Yes I agree with the quote you provide from the Navy document. Do you understand that the Navy document does not mean what you had previously said about THP? The following sentence is not the same: "We agree that we are not creating any useful THP - but we are most certainly accelerating a mass of air and have given it a relative velocity - thus we are developing THP. Otherwise where is the energy of the fuel we are using going to?

The engine is producing BHP using the energy of the fuel. "THP" means "usable thrust", as the Navy document said - not "useful (usable) THP".

Does that make sense?

Re: oggers' helicopter hover situation - I haven't studied any aerodynamic stuff regarding helicopters, however, I believe that when the helicopter is in a fixed position over the ground (in a hover), the THP will be zero. The engine will definitely be creating lots of BHP(SHP) and burning lots of fuel to produce the thrust that is keeping the helicopter in the hover - but I believe THP will be zero. Work will be done to lift the helicopter off the ground into the hover position which will obviously take a certain amount of THP.

This is my reasoning. First of all, I'm assuming that the same concept of THP (ie: with regard to the airplane's performance) is applied to the helicopter scenario. As I've stated, THP is related to the performance of the aircraft. First assume that the helicopter is a small cardboard box that is empty. It is resting on 4 poles situated at the four corners of the box so that you are free to place your hand underneath the box and lift it. When the box is resting on the 4 pole platform I think we can agree that there is no work being done on the box. Then you place your hand underneath it and apply a 10 Newton force in the direction of the normal (vertical) axis. Let's assume it takes 100 Newtons of force to overcome the force of gravity on the box. So increasing the force of your hand on the box all the way to 100 Newtons will essentially transfer the weight (mass x gravity) of the box to your hand. At the point where you are applying 100 Newtons of force, the box still hasn't moved. Work = force x distance. Therefore, no work has been done on the box. If no work has been done on the box, then power equals zero. Power = work / time.

Then you apply a 150 Newton force for 1 second before reducing the force to 100 Newtons. That accelerates the box upwards and then brings the box to rest again at a position that is a certain distance from the original position (resting on the 4 pole platform). Since you changed the position of the box you did work on the box. And since you did work on the box, power was used to move that box. But once it comes to rest again, it goes back to the same condition - zero work and zero power on the box.

It's the same with the helicopter. I'll disregard ground effect for this thought experiment. The pilot increases the thrust to get it off the ground and then reduces it so that he comes to a hover a certain distance off the ground. He still needs some force that will oppose the mass x gravity of the helicopter and the thrust provides that. The engine is doing lots of work to generate that thrust - but the thrust is not doing any work on the helicopter when it's in the hover! No work = no power!

-------------------------

oggers.. why don't you prove me wrong?! If you've taken all these exams and gotten educated on these topics, do you have any references that you could provide that proves me wrong? So far, you've been very adamant in saying I'm wrong but you've provided no counter argument to the references that I've provided... which leads me to my next point...

You say you don't disagree with any of the references I've provided. That's progress. But you say that they don't support my point. Can you provide evidence that my point is different than what is said in the references I provided?

Well this is what you said: "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..."

But you actually don't understand what that 'prop efficiency' really is! Here is the equation for propeller (propulsive) efficiency: http://i.imgur.com/NXLV3.png

And you say that's just semantics... haha :=

Then you said: "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."

But the 'Navy stuff' clearly showed that you were in fact creating zero THP when the aircraft was stationary! So you obviously don't agree with the 'Navy stuff'. You're still not getting that we're talking about aircraft performance. I'm not sure how many times I should repeat "aircraft performance" before it sinks in. Of course you're doing 'work' on the air because you're moving it backwards. And I'd like to add emphasis to the part where I said: "doing 'work' on the air". You are NOT doing any work on the airplane!

What's interesting though is that you said in your most recent post: "Are you telling me that when I was releasing bolts you were in the engine? BTW everyone knows that 'no work is being done on the vehicle'. But: work...is...being...done...on...the...air.

So you do understand that no work is being done on the 'vehicle' and yet when I say that no work is being done on the airplane when it's stopped (therefore there is no power), you have told me that I'm wrong. Hmm... haha :ugh:

I've never disagreed that the engine is doing work (BHP) but I have been very clear in describing that if the 'vehicle' is not moving, then the THP is zero.

italia:

What if the helicopter was to climb at a very small rate, say, 1 ft/min. Does this thrust without power you speak of suddenly become THP after all? Or just a tinsy winsy bit THP? What if the helo is winching pax up whilst maintaining a steady hover: the AUM increases with every pax. but the aircraft hasn't moved. This requires a change of thrust and engine power of course but according to you THP remains at zero throughout :*

What if said helo were to descend at a very small rate. Is it using more THP because it is now moving than it did in the hover. According to you the answer is yes because you say there was zero THP in the hover! Or would you now say it is negative THP?! :confused:

What of that old carrier footage where the planes used their prop wash to aid in coming alongside? The aircraft are lashed to the deck. Work is being done on the carrier, THP is being generated, you would probably agree. If during this maneouvre the carrier comes briefly under the influence of a current that halts its progress toward the jetty, does THP magically go to zero and then come back when the current abates? According to you it goes to zero :rolleyes:

A little understanding is indeed a dangerous thing.

Force times velocity has been iterated ad nausea throughout this thread.

It's multiplication.

What happens when you multiply by a very small number, as compared to when you multiply by a larger number? And as compared to when you multiply by zero?

Now, exactly which were those exams you were making a big deal out of having passed? Elementary algebra included?

(Vector, direction, magnitudes - those terms are left for googling by the interested reader.)

Now, I really, really do not care to argue the point, or any point, with you. It seems pointless, so to speak - a state of full duplex communication doesn't appear to ever be reached - and you are not exactly going out of your way to exhibit a charming online personality either.

Besides, you simply have to be trolling. This is a behaviour I detest, as it severely detracts from the usability of online forums for those who actually seek knowledge and understanding. Please stop, and let this forum be a place of educated discussion between professionals.



I do want to thank you for giving me a chance to respond on-topic (or at least in-line with the current direction the topic is taking) in this thread though.

This provides me with the opportunity to compliment italia458 on an excellent analogy with the bolt and the wrench. Well done, good sir! A good teacher is one who can bring an abstract concept into the every-day world most people can grasp without prior education on the subject. :ok:

oggers... As ft said, you're not contributing to anything here. You consistently keep ignoring my questions asking you to prove your point and instead try to find any way that you can prove me wrong. As for your ridiculous questions - you already know the answer. Just ask yourself if the aircraft is moving or not and you will know the answer!

Thanks for the comment ft! I was seriously doubting the intellect of the human race - but I guess there is always a bad apple in every bunch!

A novel Conveyor Belt topic... :E

ft

There's really no need to get so upset. If what you genuinely want is to engage in an 'educated discussion between professionals', then I don't understand why you didn't:

a) Address the OP

b) State clearly whether you agree with italia's hypothesis that a helo in the hover is not generating any thrust horse power.

Hopefully you are being satirical, as a mechanic I can tell you that his analagy is wrong.
The amount of "force" needed to overcome the friction and torque to "break" the bolt is considerable, perhaps why limp wristed theorists don't actually "work"
@ oggers
Once more into the breach my friend.
Are you saying that if a force is applied to an object that is less than that required to move it, then no work has been done. You must see intuitively that this is a false argument.

you mixed something up fella. the 19600 in/lbs are surely the value for the propshaft- not the engine. since we know that the simple formula for calculating power from torque and rpm,s is torque ( in in/lb) x rpm / 5252 you can calculate the power.

the prop of the herky spins about 1000 rpm and the engine about 13800rpm , right ?

soo... 19600x1000/5252 which results in a little more than 3730hp.

imagine the engine by itself would have this torque .... 19600x13800/5252... :sad:

the turbine is a high speed low torque thing and the prop / via a gearbox a high torque low speed device. at turboprops the propshaft torque is measured.

but not sure if the thread opener talked turboprops...

one more word to the usuful thrust/power when the engine are pulling but the aircraft is hold by the brakes... the aircraft as a whole thing does not develop any useful work, but the engines of course do - their use in our case is just to strenght the brakes.

cheers gents !;)