Propeller torque & engine torque [Archive]

italia458

26th Mar 2012, 16:00

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 (http://imgur.com/R43ho)

oggers

26th Mar 2012, 16:12

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.

italia458

26th Mar 2012, 16:25

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 (http://www.box.com/s/tpby1m893e6mipiv6sjp)

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

cwatters

26th Mar 2012, 16:31

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.

italia458

26th Mar 2012, 16:59

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.

oggers

27th Mar 2012, 10:34

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:

italia458

27th Mar 2012, 13:14

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.

keith williams

27th Mar 2012, 13:51

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.

oggers

27th Mar 2012, 14:33

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).

?

italia458

27th Mar 2012, 15:42

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.

barit1

27th Mar 2012, 16:01

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).

italia458

28th Mar 2012, 00:47

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

italia458

28th Mar 2012, 03:12

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?

oggers

28th Mar 2012, 08:39

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 a stationary aircraft [producing 200 BHP] 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.”

….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.

if the aircraft isn't moving, it isn't covering distance and so the work = zero.

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

Old Fella

28th Mar 2012, 10:41

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.

rudderrudderrat

28th Mar 2012, 11:15

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)

italia458

28th Mar 2012, 16:20

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.

rudderrudderrat

28th Mar 2012, 17:06

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.

So if we create a lot of thrust but we don't move, we are not creating any THP
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?

oggers

28th Mar 2012, 17:11

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

I will leave the rest for another day :ok:

barit1

28th Mar 2012, 20:59

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!

italia458

29th Mar 2012, 01:28

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.)

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 (http://www.box.com/s/3302d62b8d269bfc96cb)

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).

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 (http://www.box.com/s/86499db811c2ca977456)

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?

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 (http://en.wikipedia.org/wiki/Thrust) -- It also explains what I have been in this thread. Read the "Thrust to propulsive power" section.

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, 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.

rudderrudderrat

29th Mar 2012, 08:23

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.

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

oggers

29th Mar 2012, 11:23

“Oggers... I'm done dealing with this. You're going to have to figure this out on your own.”

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

“Or just be happy with your answer knowing that all these textbooks say that you're wrong.”

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

“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!”

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

“It seems like Harold decided to recant his post”

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

“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.”

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

“Are you going to tell me now that all the forces in a turn are balanced, like so many books and instructors teach it?”

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

“You haven't addressed ANY of the evidence that I've shown you that says you're wrong.”

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

“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.”

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

“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.”

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.

“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.”

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

“When sitting stationary on the runway with full throttle, you're creating tons of thrust. You can ask where that's coming from…”

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

“…and I'd say it's coming from the torque of the engine and the RPM of the crankshaft which is turning the propeller....”

...okaaay...

“..which is power. Since the engine is able to turn the crankshaft, it's creating power...”

...riiiight...

“..if the RPM was zero, it would be zero power, even though there might be lots of torque.”

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

“…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!”

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

“…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!”

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.

“…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!”

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

“…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!”

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.

“…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.”

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.

blackhand

29th Mar 2012, 12:13

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.

italia458

29th Mar 2012, 15:11

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?

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

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.

oggers

29th Mar 2012, 19:29

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:

29th Mar 2012, 20:11

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:

italia458

29th Mar 2012, 20:58

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!

chksix

29th Mar 2012, 21:42

A novel Conveyor Belt topic... :E

oggers

29th Mar 2012, 22:55

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.

blackhand

29th Mar 2012, 23:30

This provides me with the opportunity to compliment italia458 on an excellent analogy with the bolt and the wrench.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.
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.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.

aerobat77

30th Mar 2012, 00:10

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.

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 !;)

italia458

30th Mar 2012, 01:35

Hi blackhand:

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"

Insults eh? All mechanics are good for is turning bolts... don't try to think about physics. My analogy is correct... learn how to spell. Read this: http://en.wikipedia.org/wiki/Work_(physics). You'll see that it does NOT matter what the force applied is... work only happens when you apply a force and move something a certain distance.

What physics courses have you taken? The work equation is taught in high school. To be able to understand what the wikipedia page says about work, you will only require a basic understanding of math. If you know that anything multiplied by zero is equal to zero, you should have no problem understanding the work equation.

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.

That's EXACTLY what I'm saying! And no it's not false. See above.

blackhand

30th Mar 2012, 02:02

work only happens when you apply a force and move something a certain distance.
Is that relatively speaking?

CYHeli

30th Mar 2012, 02:14

A helicopter in a stable hover is producing the right amount of power (lift) to overcome the weight of the helicopter. The engine and blades are doing work because gravity wants to accelerate the aircraft down. Gravity is a constant, so there is always a downwards force, so therefore even if the helicopter is in a stable hover, (vertical - upwards) force is still being applied.

To raise the collective to increase pitch to provide the induced flow and therefore create lift also increases drag on the blades and the blades want to slow down. Engine RPM must be increased to maintain a stable engine RPM and therefore a stable rotor RPM. So as the aircraft lifts, the engine produces more power. In a Robinson R44 for example, the manifold pressure would be below 18 inches at idle and then increase to about 20-21 inches in the hover, but MCP (Max Cont Pwr) might still be about 23.4 inches to allow transition from the hover to forward flight.
Does that add anything to the helicopter theory above?

italia458

30th Mar 2012, 02:23

CYHeli:

A helicopter in a stable hover is producing the right amount of power (lift) to overcome the weight of the helicopter. The engine and blades are doing work because gravity wants to accelerate the aircraft down. Gravity is a constant, so there is always a downwards force, so therefore even if the helicopter is in a stable hover, (vertical - upwards) force is still being applied.

I completely agree. It might be good to specify that the "power (lift)" you mention is BHP(SHP) as we've been talking about THP as well.

blackhand:

Is that relatively speaking?

You're going to have to read the Wikipedia article on work. I'm not going to regurgitate what it says.

CYHeli

30th Mar 2012, 02:44

Got it. From a helicopter point of view we use the expression (vector) of Total Rotor Thrust. We can map out the various forces on a vector diagram, and vertical component of TRT will equal the mass/weight in a stable hover. But the TRT will never be equal to zero unless you sitting on the ground with no pitch applied. But I see where you are coming from.

The helicopter example was probably out of place or just confused the discussion.

The other reason that the helicopter example fails is, what is your point of reference? If you are using the ground then the aircraft has not moved, so you are correct. But the ground is irrelevant.

Use a random parcel of air adjacent to the aircraft in hover as your reference point. This parcel of air is being induced to flow down past the rotor blades by the pitch applied. The parcel of air is travelling at somewhere near 300 feet per second (we call it downwash). The force applied to the air is what keeps the helicopter in a stable hover, so the THP is huge.

Now apply that same reference point to an aircraft sitting on the ground at idle. The aircraft has not moved reference the ground (irrelevant) but measure the THP against the speed (dist/time) of the parcel of air being driven across the prop.

Or is thrust only measured against the distance that the aircraft moves ref the earth?

blackhand

30th Mar 2012, 02:49

You're going to have to read the Wikipedia article on work. I'm not going to regurgitate what it says. Mmmm I'm concerned, that you, as a Physics Guru, missed the implications of "relatively" speaking.
Maybe if you moved on from Newtonian Physics and onto modern physics and explored exchange of energy you wouldn't have to be so rude.
Or is thrust only measured against the distance that the aircraft moves ref the earth? Relatively speaking???

Old Fella

30th Mar 2012, 03:55

Aerobatt77, 13820 Engine RPM which gives 1021 RPM at the Prop via a two-stage reduction gearbox (13.54:1 reduction ratio). The torque meter is mounted between the engine and the input to the gearbox. An outer sleeve which is mounted only at the engine end has a sensor mounted in it which measures the twisting of the driving shaft under load, positive or negative. Of course the measurement accounts for gearbox friction, accessory drive loads and propeller load. As I said, it is a measure of engine torque available and the 19600 in/lbs limiting torque is more about stress on the engine mounting structure than the load on the engine. The three models of the C130 (A-E-H) I operated all had the same Maximum Torque limitation despite different power outputs and propeller installations. At the end of the day, in relation to torque, all I was interested in was whether or not I was getting the torque required for the prevailing conditions. I suspect you know all this stuff anyway. Have a good day.

italia458

30th Mar 2012, 04:23

CYHeli:

Got it. From a helicopter point of view we use the expression (vector) of Total Rotor Thrust. We can map out the various forces on a vector diagram, and vertical component of TRT will equal the mass/weight in a stable hover. But the TRT will never be equal to zero unless you sitting on the ground with no pitch applied. But I see where you are coming from.

Yes, thrust is what is opposing the force of gravity. But THP is different!

The other reason that the helicopter example fails is, what is your point of reference? If you are using the ground then the aircraft has not moved, so you are correct. But the ground is irrelevant.

Use a random parcel of air adjacent to the aircraft in hover as your reference point. This parcel of air is being induced to flow down past the rotor blades by the pitch applied. The parcel of air is travelling at somewhere near 300 feet per second (we call it downwash). The force applied to the air is what keeps the helicopter in a stable hover, so the THP is huge.

Now apply that same reference point to an aircraft sitting on the ground at idle. The aircraft has not moved reference the ground (irrelevant) but measure the THP against the speed (dist/time) of the parcel of air being driven across the prop.

Yes, the ground is the reference and it is not irrelevant! You could hover in one position over the ground all day and you would have done no work on the aircraft. What's your performance, when it takes an infinite amount of time and fuel to get from point A to B? Zero! If something isn't moving, there is no work being done on that something. THP would be zero. Everyone needs to realize that THP is with regard to: the work the thrust does on the aircraft! That's it! Nothing more than that.

Pardon my frustration but I feel like a broken record here. I've explained countless times in this thread what THP is. I've provided numerous resources which all explain it and relate it to performance. I don't know what else to say.

When you're talking about Total Rotor Thrust, that's just thrust. It's a nice way to compare forces. It's similar to talking about what happens to the Total Lift Force in a turn with an airplane. That force (Total Rotor Thrust) comes from the power of the engine (BHP or SHP).

Regarding your parcel of air explanation: as discussed, yes, it requires power to move air. That power comes from the engine. That power is called BHP or SHP. It is not called THP.

Or is thrust only measured against the distance that the aircraft moves ref the earth?

Thrust is related to how much air is displaced. You can see the variables described here: General Thrust Equation (http://www.grc.nasa.gov/WWW/k-12/airplane/thrsteq.html)

THP is related to the distance that the aircraft moves with reference to the earth. Distance/time = velocity which is what is in the THP equation.

I'd be happy to answer any questions you have but please read through this thread because most of them have already been answered.

Cheers! :)

blackhand:

Mmmm I'm concerned, that you, as a Physics Guru, missed the implications of "relatively" speaking.
Maybe if you moved on from Newtonian Physics and onto modern physics and explored exchange of energy you wouldn't have to be so rude.

I'm not a 'physics guru' as you say but I do know a few things about physics. I'm pretty familiar with relativity and what I do realize is that you're being sciolistic. There is no need to complicate this with special or general relativity. None of my references I've used so far have needed to and they've gone into more detail than I've written out on this thread.

CYHeli

30th Mar 2012, 04:52

Thrust is related to how much air is displaced. And you have never heard of downwash? Sorry facetious.

And you are saying that the amount of air displaced (downwash) is irrelevant if the aircraft is not moving relative to the earth. So in the helicopter example, although the blades are moving a whole lot of air and holding the helicopter up against gravity, that the Trust HorsePower is still zero, because we haven't gone anywhere? A force is being applied, but no distance travelled, therefore no work completed?

If that is the case, again I say that the helicopter example is a bad one. Because measuring THP for a helicopter is useless. Again helicopter POH's have very clear performance graphs that display hover in/out of ground effect charts to prove this. (Yes I know that these are a chart to provide a guide as to the helicopter performance based on expected SHP vs weight given certain Dalt)

I think in a scientific calculation you would say that although an engine is producing SHP there is no THP because we haven't gone anywhere, but in reality (a practical sense) you would simply describe that there is insufficient SHP being applied to create enough Trust to provide movement.

italia458

30th Mar 2012, 05:32

And you are saying that the amount of air displaced (downwash) is irrelevant if the aircraft is not moving relative to the earth. So in the helicopter example, although the blades are moving a whole lot of air and holding the helicopter up against gravity, that the Trust HorsePower is still zero, because we haven't gone anywhere? A force is being applied, but no distance travelled, therefore no work completed?

Yup! You got it. Thrust is a force. It's the same as me pushing really hard against the side of a big box. If the box doesn't move, I haven't done any work. I have sure used a lot of energy though! Exactly same with the airplane and the helicopter. The engine is using a ton of energy (fuel) to create that force (thrust)... but since the thrust isn't moving the aircraft, the thrust isn't doing any work. Therefore, there is zero Thrust Horsepower.

If that is the case, again I say that the helicopter example is a bad one. Because measuring THP for a helicopter is useless. Again helicopter POH's have very clear performance graphs that display hover in/out of ground effect charts to prove this. (Yes I know that these are a chart to provide a guide as to the helicopter performance based on expected SHP vs weight given certain Dalt)

How is the helicopter a bad example? I think it's more your 'feeling' that since the engine is producing thrust to keep the aircraft hovering in the air, it must be doing work on the aircraft. I can see how people can think that way - but it isn't correct.

I think in a scientific calculation you would say that although an engine is producing SHP there is no THP because we haven't gone anywhere, but in reality (a practical sense) you would simply describe that there is insufficient SHP being applied to create enough Trust to provide movement.

This is exactly how things get screwed up and taught wrong in the first place! Science IS reality and a theory isn't just some made up thing that a scientist came up with late at night! What happens is someone who doesn't have a physics/engineering background decides that it makes way more sense to explain something their way and so they do. Two problems with that: 1) Have you ever played telephone? If you have, you'd realize that after a few people, the message isn't the same anymore. 2) There is only so much that can be done to simplify something before you actually start 'lying'. And that person doesn't have the physics/engineering background to be able to know which parts could be simplified and which ones can't. The Hydraulic-electric analogy is an example of this: Hydraulic analogy - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Hydraulic_analogy). It works well for explaining the basics to someone because most people are familiar with water, etc. But there is a point when teaching electricity that there aren't any analogies that accurately describe what is going on. You enter an area where all the concepts are new and different in ways you wouldn't imagine. This is a video I think is quite funny... because it's true! Armstrong & Miller Physics Special - YouTube (http://youtu.be/3wHKBavY_h8)

Another example is the forces in a turn. There is a book currently published by Transport Canada that depicts the forces in a turn incorrectly. It shows that all the forces are balanced. A little bit of physics will tell you that in a turn you're constantly accelerating since acceleration is related to velocity and velocity has two components: speed and direction. If you're changing speed, you're accelerating (either positively or negatively). The same goes for direction. In a turn, you're constantly changing direction and, therefore, you're constantly accelerating. Looking at Newton's laws you can see that F=ma. If a mass is accelerating, there must be a net force acting on it. A net force means that the forces CAN NOT be balanced! I had a Class 1 Flight Instructor tell me that we should be teaching it that way because it's "easier to understand", after I explained that it was completely wrong. I never did teach it their way and never had a problem. This Class 1 FI didn't have a good understanding on a number of other topics as well.

oggers

30th Mar 2012, 06:31

@ oggers
Once more into the breach my friend.

It's what I live for blackhand :sad:

italia:

All mechanics are good for is turning bolts... don't try to think about physics.

I wouldn't mention that one around the hangar if I were you.

ft:

What do you make of that me old? Another of italia's 'excellent abstract analogies' that exemplifies 'educated discussion between professionals'? :eek:

:ugh:

tcyandy

30th Mar 2012, 08:27

Dear all,

I am so sorry to reply this late as I was very busy with my school work recently. My sincere apology.

To clarify, in my knowledge, propeller's torque is referring to a drag force while engine torque refers to rotational force generated by the engine.

I have read through all of your replies and I am very thankful for all of your comments.

Now I understand much better on these two parameters.

Basically, now I know that:

At an optimum RPM, the propeller torque is equal to the engine torque, the only difference is its direction. Say, for example, if now the engine torque is in clockwise (viewing from the cockpit), the propeller torque would be in anti-clockwise direction.

Now my questions becomes, what happen if engine torque is greater than propeller torque? I know that the RPM would then become faster, but how would this affect the efficiency of the engine? I don't quite get it. Hope someone would shed me some light on the issue.

rudderrudderrat

30th Mar 2012, 08:53

Hi italia458,
A net force means that the forces CAN NOT be balanced!
You've lost me there with this thread drift.

I'm feeling an acceleration whist standing on the earth (force on my feet) and I'm perfectly balanced. I can do the same in the passenger cabin of an aircraft whilst it performs a turn although I feel the delta g.
Please explain what forces you think aren't balanced.

aerobat77

30th Mar 2012, 09:27

As I said, it is a measure of engine torque available and the 19600 in/lbs limiting torque is more about stress on the engine mounting structure than the load on the engine. The three models of the C130 (A-E-H) I operated all had the same Maximum Torque limitation despite different power outputs and propeller installations

ahoi !

so when the prop rpm was 1021 it would result that the herky had 1021x19600/5252 = 3810 shaft horse power per engine approved due to engine mount limitations- for all allison models.

the allison by itself would , depending on the model, give more but it was not approved, right ?

one more thing , but maybe i understand you not right : the 19600 in/lbs must be the value for the torque on the propeller, not the turbine by itself since a device which spins 13800rpm and produces simultany massive 19600 in/lb of torque would have an asronomic power ( 19600x13800/5252 = 51500 shp!!! )

cheers !

oggers

30th Mar 2012, 10:27

italia: back to the helicopter problem. 'No THP in the hover' you say. The two main arguments you have provided go like this:

'The aircraft doesn't move therefore no work is done on it therefore there is no thrust horsepower, only thrust.'

'Power = Thrust x airspeed. There is no airspeed therefore there can be no thrust horsepower.'

Nobody at all is arguing with the equation you provide, and everyone will recognise the definition of work being a force x distance. They are basics. But it seems you are in denial when it is pointed out that work is being done on the downwash (as CYHeli stated). Your argument is a purely semantic one in which you insist on defining THP as that which moves the aircraft.

Because a helicopter can hover there is an alternative way of determining the efficiency of the rotor system that does not require airspeed of the vehicle. It is called Figure of Merit - not a phrase I was previously familiar with TBH. What is done is the 'power' of the so-called induced airflow is determined. It is the power that results from the thrust. The efficiency of the rotor is then determined by dividing the induced power by the sum of the profile power (from rtr drag) and induced power:

M = Pi/Pi+Po.

None of this requires the aircraft to be moving. It is explained in Basic Helicopter Aerodynamics by Seddon pdf here (http://aerostudents.com/files/rotorcraftMechanicsAndDesign/SeddonBasicHelicopterAerodynamics.pdf). Ch 2.

Markdem

30th Mar 2012, 10:46

italia,

Are you saying that a wind tunnel (helicopter on it side) does no work when YOU are "MOVED" by standing behind it?

Agreed that the aircraft may not move therefore not doing much work ON IT SELF but still moving a hell lot of air behind it.

MD.

Old Fella

30th Mar 2012, 10:50

Aerobat77. The original quesion asked was about the relationship between engine torque and propeller torque. As I am sure you know, the T56 is a constant RPM engine/gearbox/propeller combination. The only indication of engine power available to the crew is torque. For a given atmospheric condition and a given Turbine Inlet Temperature the engine should provide a predetermined torque figure. This can be varied by two things, closing the engine bleed air valves and ram rise as the aircraft accelerates down the runway and becomes airborne. Certainly, in a "hot/high" environment less engine power for a given TIT will be available, thus a lower torque reading results. Conversely, the torque limitation can be reached at a lesser TIT in cold climes. As for all the other theory stated by other posters on this thread, it does not mean anything to the operating crew. The ONLY measure of power available is Torque!!! The T56A-11 was rated at 3750 Equivalent Shaft Horsepower (Shaft Horsepower + jet thrust) The T56A-7 is rated at around 4200 ESHP and the T56A-15 is rated at 4910. It was uncommon in either the C130A (T56A-11) or the C130E (T56A-7) to reach the 19600in/lb limit before reaching the limiting TIT on anything above an ISA day. The C130H (T56A-15) was much more likely to achieve 19600 in/lbs before reaching the limiting TIT, which was higher than on the other engines due to improved turbine design and materials.

aerobat77

30th Mar 2012, 11:28

ahoi !

i think the original question was very well answered- in turboprops the turbine ( single shaft or multishaft free turbine design) provides a high speed low torque power that has to be translated to a low speed high torque power for the prop since the pure rpm,s of a turbine are not suitable for a working propeller.

in pistons without gearing the engine torque is the same as the propeller torque- in geared systems the same like in turboprops but with a significant lower gearing.

the gearbox is nothing other than a torque converter .

a given power can be achieved by low tq high speed or high tq low speed. - also beyond aviation. a ferrari engine may have the same output like a truck engine but the truck engine will have significant bigger torque than the ferrari. the same power output results in the significant higher reving engine of the car.

in aviation you can see the effect nicely on free turbine props like the pt6a. when you without touching the power levers pull back the props and so reduce the speed of the props the prop torque will rise without changing anything in the output - so fuel flow, gas generator speed and ITT will stay the same. you just exchanged torque for prop speed without changing the power delivered.

@ fella : so the allison basicly was not flat rated above isa when you say you will reach ITT limits before the tq limit even on ground when its above ISA?

cheers !

rudderrudderrat

30th Mar 2012, 11:57

Hi Markdem,

Italia has been considering the propeller, engine, airframe combination.

It's easier to consider:
"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%.
THP = SHP * p.e."

p.e. is not zero.
So even if the aircraft is stationary or the helicopter is hovering, there is definitely THP being produced if you are making SHP.

italia458

30th Mar 2012, 13:00

rudderrudderrat:

I'm feeling an acceleration whist standing on the earth (force on my feet) and I'm perfectly balanced.

You are perfectly balanced and that means that there is no net acceleration. Imagine the earth to be stopped (not rotating). It would still have this gravity force as it's related to mass. To really give you a good answer you need to take some general relativity courses. General relativity deals with the theory of gravitation. I will try to explain a little bit about what's going on below.

We're analyzing these problems with reference to the earth and not a point in space so when you're standing on the earth, you are stationary. If we did analyzing the earth as 'it truely is', we would have to triple to quadruple the amount of work so that we compensate for the earth's rotating reference frame.

So, standing on the earth we consider that to be an inertial reference frame. When you are accelerating you're in a non-inertial reference frame. Inertial frame of reference - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Inertial_frame_of_reference)

What you feel when turning in the airplane is the centrifugal force which is a pseudo force. It is only there because YOU are in an accelerating reference frame. You don't actually feel the net increase in the lift force. If you were standing on the earth and saw a plane overhead in a turn, based on where it was going you would have to conclude that there is a net force imbalance that is pulling it into the inside of the turn. That's called centripetal force and it is the horizontal component of lift. There are essentially TWO forces at play when a plane is in a turn. Lift and Weight. At an angle, lift is broken up into its x and y components. The y component essentially needs to offset the weight so that the plane remains at the same altitude and the x component is the centripetal force in this case. There is no other force that is opposing it.

Banked turn - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Banked_turn)
http://selair.selkirk.ca/Training/Aerodynamics/images/lf-turn.gif

oggers:

But it seems you are in denial when it is pointed out that work is being done on the downwash (as CYHeli stated).

You truly are a piece of work!

Here are a couple quotes from myself:

Regarding your parcel of air explanation: as discussed, yes, it requires power to move air. That power comes from the engine. That power is called BHP or SHP. It is not called THP.

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".

That emphasis obviously wasn't enough!

...and now back to a quote from you:

Your argument is a purely semantic one in which you insist on defining THP as that which moves the aircraft.

No, it's not semantics. All the resources I've provided have depicted THP as such. That is what THP is. I have a feeling you REALLY don't like being wrong and are going to continue to fight this point no matter how much I explain that you are in fact not correct on this subject.

Because a helicopter can hover there is an alternative way of determining the efficiency of the rotor system that does not require airspeed of the vehicle.

I highlighted 'alternative' because it is just that. I could analyze efficiency with regard to heat as well... that's called thermal efficiency! But as I've stated quite a number of times - I'm talking about propulsive efficiency.

Markdem:

Are you saying that a wind tunnel (helicopter on it side) does no work when YOU are "MOVED" by standing behind it?

Agreed that the aircraft may not move therefore not doing much work ON IT SELF but still moving a hell lot of air behind it.

Please, read the thread, Markdem. I've stated a few times already, including just above in this post, that you are doing work on the air. But with regard to aircraft performance, moving air is not what is being analyzed!

rudderrudderrat:

p.e. is not zero.
So even if the aircraft is stationary or the helicopter is hovering, there is definitely THP being produced if you are making SHP.

THP has its own equation - if you want to see if there is THP in a situation, you need to use it. As stated numerous times, if the flight velocity is zero, there is zero THP - that is what the equation says, not me.

This is an expanded form of the p.e. equation - http://i.imgur.com/NXLV3.png

rudderrudderrat

30th Mar 2012, 13:27

Hi italia458,

As stated numerous times, if the flight velocity is zero, there is zero THP - that is what the equation says, not me.
That's odd because the relationship "THP = SHP * p.e." is from NAVAVSCOLSCOM-SG-111 page 4 and the equation tells me that I am developing THP if I am converting SHP using a propeller which has an efficiency >0. There is no mention of aircraft speed.

Your formula involving aircraft speed is good for explaining climb performance etc. but is not exclusive to explaining THP.

So, standing on the earth we consider that to be an inertial reference frame.
I disagree - standing on earth is not an inertial frame of reference. Your observation of the "free fall" flight of a golf ball shows you that you are in an accelerating frame of reference.

A "balanced turn" can simply be measured using a spirit level. The bubble has to be in an accelerating reference frame (that's gravity to those of us who haven't studied general relativity) else it wouldn't "float" to the top.

oggers

30th Mar 2012, 14:46

italia:

Regarding your parcel of air explanation: as discussed, yes, it requires power to move air. That power comes from the engine. That power is called BHP or SHP. It is not called THP.

The power that comes from the engine goes through the prop. Rudderrudderrat is correct about propeller efficiency not being zero. If it was zero there would be no thrust.

You can't work out the prop efficiency in the static thrust scenario using the formula you keep giving for propulsive efficiency. I have found an essay by a prop designer that explains why:

Theoretical Prop Efficiency (http://www.jefflewis.net/aviation_theory-theo_prop_eff.html)

"This equation [propulsive efficiency] is very useful for many cases, but you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero. So how do you know how good your prop is doing at low speeds or statically? Well, there's another term that can be used, Figure of Merit, which is Induced Power divided by Power Available, or how much of your power is going into accelerating the air..."

Induced power is the same as thrust horsepower.

italia458

30th Mar 2012, 15:06

That's odd because the relationship "THP = SHP * p.e." is from NAVAVSCOLSCOM-SG-111 page 4 and the equation tells me that I am developing THP if I am converting SHP using a propeller which has an efficiency >0. There is no mention of aircraft speed.

You're correct, except for that last sentence. Look at this picture taken from NAVAVSCOLSCOM-SG-111: http://i.imgur.com/R43ho.png

The speed of the aircraft (to be specific it would be the TAS of the aircraft) is included in the THP equation. It also shows that when at a TAS of 0, the THP is equal to zero.

I already explained this earlier and I used that exactly picture in post #9.

Your formula involving aircraft speed is good for explaining climb performance etc. but is not exclusive to explaining THP.

No, it does explain THP. Read the documents that I've attached in this thread.

As for everyone here: Stop saying what you 'think' is happening and stating it as a fact. If you are going to 'prove' something, please include evidence that proves that point, as I have done. If you have evidence that proves that what I'm saying isn't correct, I'm more than willing to listen. However, none of you are providing evidence that proves that Wikipedia, the U.S. Navy and William Kershner are all wrong. The document that oggers posted does not relate to what I was talking about. I have not read the document in its entirety, but I will eventually as it looks like a fantastic explanation about helicopter aerodynamics.

I disagree - standing on earth is not an inertial frame of reference. Your observation of the free flight of a golf ball shows you that you are in an accelerating frame of reference.

1) I agree that earth is technically not an inertial reference frame, however, forces on earth can be analyzed as being in an inertial reference frame.

2) A reference frame is (courtesy of Wikipedia): "a coordinate system or set of axes within which to measure the position, orientation, and other properties of objects in it, or it may refer to an observational reference frame tied to the state of motion of an observer. It may also refer to both an observational reference frame and an attached coordinate system as a unit." In the case of what we're talking about, it is both an observational reference frame and an attached coordinate system as a unit.

3) In an inertial reference frame, an object can be at rest with zero forces acting on it or at rest with all forces acting on it, balanced. An object can also be in motion with all forces either zero or balanced. An object can also be accelerating in an inertial reference frame due to an imbalance of force.

4) Your golf ball explanation doesn't prove what you're saying. The reason it is falling does NOT prove that it is in an accelerating frame of reference. It shows that there is a force imbalance which is causing the acceleration. Assume that the atmosphere is a vacuum. The wall will be 'fired' up and then will come down. The reason it comes back down to earth is because the force of gravity on the ball is not being opposed! It is accelerating back down to the earth. This is a basic problem that is easily analyzed in an inertial reference frame in ab initio physics courses. You're unnecessarily complicating the problem by analyzing it from a non-inertial reference frame.

5) Just like the golf ball phenomenon can be explained in an inertial reference frame, so too can an aircraft in flight. And that's how it has been done for ages!! If there wasn't lift, the gravity acting on the aircraft would accelerate it back to earth.

italia458

30th Mar 2012, 15:16

oggers:

This equation [propulsive efficiency] is very useful for many cases, but you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero. So how do you know how good your prop is doing at low speeds or statically? Well, there's another term that can be used, Figure of Merit, which is Induced Power divided by Power Available, or how much of your power is going into accelerating the air...

I don't disagree with anything he said! I hope you realize that he also agreed with what I said!

Like he says, you don't know how good your prop is doing at making thrust by analyzing THP. And he's saying that "Figure of Merit" will show you how efficiently the propeller makes thrust. That has nothing to do with THP!

So how does this 'prove' me wrong?

Induced power is the same as thrust horsepower.

NO. IT. IS. NOT!

THP = Thrust x aircraft velocity

Induced power = Thrust x (aircraft velocity + induced velocity at the propeller)

I just taught my nephew the other day (who is 5 years old) that if a word isn't spelled the same, it's not the same! He seemed to understand that right away.

oggers

30th Mar 2012, 15:28

The document that oggers posted does not relate to what I was talking about.

In that case I'm not sure what you think you're talking about but it certainly relates to the power induced by the rotor of a helicopter in the hover. :ugh: Induced power and THP - two names, same concept. Not that the name is important seeing as you already asserted that there was no power at the rotor, only thrust:

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...the "power (lift)" you mention is BHP(SHP)

And the essay (http://www.jefflewis.net/aviation_theory-theo_prop_eff.html) I cited in my last post definitely states that power is induced by a prop. Which didn't come as much of a surprise to me :)

italia: you must try to remain calm.

Now, do I take it from the last post that you now acknowledge there is induced power, and that you get this at the prop in the static thrust scenario? Because you said 'if the aircraft doesn't move no work is done. No work is no THP' Is it just that you didn't think to mention induced power in any of the many pages you have devoted to this so far?

rudderrudderrat

30th Mar 2012, 15:39

Hi italia458,

(1) THP = SHP * p.e. (from NAVAVSCOLSCOM-SG-111 page 4)
The speed of the aircraft (to be specific it would be the TAS of the aircraft) is included in the THP equation.
There is no mention of TAS in equation (1)

The reason it is falling does NOT prove that it is in an accelerating frame of reference.
Correct - but I said "... the "free fall" flight of a golf ball shows you that you are in an accelerating frame of reference."
The ball is in "free fall" during it's flight, the ball is in an inertial frame of reference - not you.

What you feel when turning in the airplane is the centrifugal force which is a pseudo force. It is only there because YOU are in an accelerating reference frame. Just like gravity.

I sincerely hope that you do not confuse your students with the theory of General Relativity, whilst explaining what a simple balanced turn is all about.

italia458

30th Mar 2012, 15:41

italia: you must try to remain calm.

I know... sorry. It gets frustrating having to consistently repeat myself, especially when the recipients will not accept the way things really are.

Feynman sums up what I think about this: you don't WANT to believe what the truth actually is and you won't accept it. It makes much more sense to you to say that THP won't equal zero in a hover and so you will fight everything that says otherwise, disregarding the way it actually is!

You don't like it? Go somewhere else! by Richard Feynman, the QED Lecture at University of Auckland - YouTube (http://youtu.be/iMDTcMD6pOw)

oggers

30th Mar 2012, 15:44

I edited my last post - just so you know.

italia458

30th Mar 2012, 15:53

rudderrudderrat:

There is no mention of TAS in equation (1)

You're going to have to actually listen to what I say. What is the equation for THP? You'll find that TAS is in that equation.

... the ball is in an inertial frame of reference - not you.

As I've said before, being on the earth puts you in an accelerating reference frame (non-inertial reference frame). However, you can analyze a number of phenomena by treating it as an inertial reference frame. Some of these include your golf ball example and an aircraft in flight. Go take a physics course.

I sincerely hope that you do not confuse your students with the theory of General Relativity, whilst explaining what a simple balanced turn is all about.

I don't think you understand the difference between proving and explaining. I do not start by proving what frame of reference we're in and then proving all the forces and so on. You would need to be enrolled in an advanced physics course to begin down that path.

I start by explaining the forces in a turn by drawing something like this: http://selair.selkirk.ca/Training/Aerodynamics/images/lf-turn.gif

This picture is wrong: http://www.aero-mechanic.com/wp-content/uploads/2009/08/4-28.jpg

It shows the forces from an inertial reference frame and a non-inertial reference frame. As I've shown before, if the forces are equal, there is no acceleration. So how does the aircraft turn if the forces are actually balanced, like shown in the second picture?

I usually don't mention how forces in a turn are sometimes depicted incorrectly. But I have the knowledge to be able to answer questions regarding the forces in a turn if the student asks.

italia458

30th Mar 2012, 15:58

oggers:

Is it just that you didn't think to mention induced power in any of the many pages you have devoted to this so far?

Is this a guilt trip now? We were never talking about induced power. Don't blame me for your inadequacies in understanding material. I kept it pretty simple I thought - obviously not. I never said that the propeller wasn't doing any work. You adamantly said that I was wrong about THP being zero when the aircraft is at rest and so this whole discussion has been regarding THP. You were the one to mention induced power and I immediately agreed with it. What else do you want, oggers?

Aviatorjoe

30th Mar 2012, 16:21

As the others said, Brakehorse power and shaft power.

You just have to remember that it is not measured at the propeller. It is measured between the motor and the reduction gear box.

rudderrudderrat

30th Mar 2012, 16:45

@ italia458,

You're going to have to actually listen to what I say. What is the equation for THP? You'll find that TAS is in that equation.
I'm listening - but it only seems to be one way at the moment.

Please explain why you believe THP = SHP * p.e. (from NAVAVSCOLSCOM-SG-111 page 4) is NOT correct.

You repeatedly insist the formula for Engine + Prop + Airframe with regards to Aircraft performance as being the only correct version.

So, standing on the earth we consider that to be an inertial reference frame.
As I've said before, being on the earth puts you in an accelerating reference frame (non-inertial reference frame).
Where do you actually stand on this?

edit
So how does the aircraft turn if the forces are actually balanced, like shown in the second picture?
By a combination of elevator and rudder.

The "centrifugal force" (A) is what you feel (like the force of gravity).
The centripetal acceleration is the reason for (A).

italia458

30th Mar 2012, 17:03

Please explain why you believe THP = SHP * p.e. (from NAVAVSCOLSCOM-SG-111 page 4) is NOT correct.

When did I say that equation wasn't correct? I don't think you realize that the equation in that form is very basic. The terms THP, SHP and p.e. all expand to include a heck of a lot of variables. Like I said before, a physics class would benefit your understanding.

You repeatedly insist the formula for Engine + Prop + Airframe with regards to Aircraft performance as being the only correct version.

No. I. Do. Not! I've only disagreed with everyone incorrect interpretation of what THP is.

Believe what you would like. I think I've had enough with this thread.

Where do you actually stand on this?

When standing on the earth you are technically in an accelerating (rotating) reference frame; which is a special case of a non-inertial reference frame. However, some phenomena can be explained easier from the perspective of an inertial reference frame. I'm pretty sure I said virtually the same thing in a previous post.

italia458

30th Mar 2012, 17:06

By a combination of elevator and rudder.

I do not care what you're doing with the elevator and rudder. If the forces are balanced, the aircraft WILL NOT turn. End of story.

italia458

30th Mar 2012, 17:59

rudderrudderrat:

It would be good to read this.

Inertial frames

In a location such as a steadily moving railway carriage, a dropped ball (as seen by an observer in the carriage) would behave as it would if it were dropped in a stationary carriage. The ball would simply descend vertically. It is possible to ignore the motion of the carriage by defining it as an inertial frame. In a moving but non-accelerating frame, the ball behaves normally because the train and its contents continue to move at a constant velocity. Before being dropped, the ball was traveling with the train at the same speed, and the ball's inertia ensured that it continued to move in the same speed and direction as the train, even while dropping. Note that, here, it is inertia which ensured that, not its mass.

In an inertial frame all the observers in uniform (non-accelerating) motion will observe the same laws of physics. However observers in another inertial frame can make a simple, and intuitively obvious, transformation (the Galilean transformation), to convert their observations. Thus, an observer from outside the moving train could deduce that the dropped ball within the carriage fell vertically downwards.

However, in frames which are experiencing acceleration (non-inertial frames), objects appear to be affected by fictitious forces. For example, if the railway carriage were accelerating, the ball would not fall vertically within the carriage but would appear to an observer to be deflected because the carriage and the ball would not be traveling at the same speed while the ball was falling. Other examples of fictitious forces occur in rotating frames such as the earth. For example, a missile at the North Pole could be aimed directly at a location and fired southwards. An observer would see it apparently deflected away from its target by a force (the Coriolis force) but in reality the southerly target has moved because earth has rotated while the missile is in flight. Because the earth is rotating, a useful inertial frame of reference is defined by the stars, which only move imperceptibly during most observations.The law of inertia is also known as Isaac Newton's first law of motion.

In summary, the principle of inertia is intimately linked with the principles of conservation of energy and conservation of momentum.

Old Age Pilot

30th Mar 2012, 18:07

Read this: http://en.wikipedia.org/wiki/Work_(physics).

:rolleyes:

rudderrudderrat

30th Mar 2012, 18:44

italia458

It would be good to read this.
Yes - I'm very familiar with that - it formed part of my Space, Time and Cosmology Open University course.
And your point?

Like I said before, a physics class would benefit your understanding.My 1960s University degree involved plenty of Maths and Physics.

Thank you for confirming that I completely wasted my time discussing anything with you.

oggers

30th Mar 2012, 20:29

Italia: I’ve been reviewing the thread and would like to respond to some of your previous comments in light of recent developments:

“We were never talking about induced power. Don't blame me for your inadequacies in understanding material. I kept it pretty simple I thought - obviously not. I never said that the propeller wasn't doing any work.”

I’d like to give you the benefit of the doubt but I’m afraid the suggestion that you were sitting on this knowledge about induced power (just when the aircraft was still, obviously, the rest of the time you call it THP) doesn’t quite dovetail with a lot of what you have written:

“It might be good to specify that the "power (lift)" you mention is BHP(SHP) as we've been talking about THP…

The engine is using a ton of energy (fuel) to create that force (thrust)... but since the thrust isn't moving the aircraft, the thrust isn't doing any work. Therefore, there is zero Thrust Horsepower…

When you're talking about Total Rotor Thrust, that's just thrust..

That force (Total Rotor Thrust) comes from the power of the engine (BHP or SHP)…

I believe that when the helicopter is in a fixed position over the ground 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 make a certain amount of THP.”

etc etc. Would've been so much easier just to mention induced power.

“Can you provide evidence that my point is different than what is said in the references I provided?”

Yes. That navy reference you keep using states clearly at the top that it is predicated on the assumption of “equilibrium flight”. That doesn’t include sitting on the ground. For that you need the alternative method of calculating power output from the book and the essay I linked to.

“But the 'Navy stuff' clearly showed that you were in fact creating zero THP when the aircraft was stationary!”

Forgive me for labouring the point as you do seem to rely on that equation, but it assumes equilibrium flight as stated clearly at the top of your reference document. An aircraft on the ground is not in equilibrium flight.

“THP has its own equation - if you want to see if there is THP in a situation, you need to use it. As stated numerous times, if the flight velocity is zero, there is zero THP - that is what the equation says, not me.”

And so.

OTOH, a helicopter in the hover [where you insist there is no THP] is in equilibrium flight. The aircraft itself is not moving but there is a velocity which is the induced airflow. You have acknowledged that there is thrust. Therefore you can apply the formula – weight x [velocity of induced air flow]. The result is THP. It has to be because it’s the power associated with the thrust. Although you insist there is no THP for a helo in the hover.

The same could be said for an aircraft in equilibrium flight into a strong headwind such that groundspeed was zero. You have been very clear:

“I have been very clear in describing that if the 'vehicle' is not moving, then the THP is zero... THP is related to the distance that the aircraft moves with reference to the earth.”

I'd suggest the THP is exactly the same into wind as it would be downwind and the earth has nowt to do with it.

And I really like what you did here:

“There is no need to complicate this with special or general relativity…”

Followed a few hours later by:

“General relativity deals with the theory of gravitation. I will try to explain a little bit about what's going on below.”

italia458

30th Mar 2012, 23:01

oggers:

I’d like to give you the benefit of the doubt but I’m afraid the suggestion that you were sitting on this knowledge about induced power (just when the aircraft was still, obviously, the rest of the time you call it THP) doesn’t quite dovetail with a lot of what you have written:

I don't care what you think. I was not sitting on that information. 'Induced power' had not entered my mine ONCE while in this discussion. Would it have been easier to mention induced power, you ask? No, induced power is how efficient thrust is made. It is not about THP. I do believe it adds another perspective to everything and it's great you introduced it but to accuse me of 'holding out' info is a bunch of bullocks.

Yes. That navy reference you keep using states clearly at the top that it is predicated on the assumption of “equilibrium flight”. That doesn’t include sitting on the ground. For that you need the alternative method of calculating power output from the book and the essay I linked to.

They stated those at the beginning so that they wouldn't have to keep repeating all the different conditions for every specific thing they talk about. It does not affect THP actually. For THP, it doesn't matter and here's why. THP has to do with thrust and velocity. That is it. There is no lift equation in there, lift was not mentioned once, weight was not mentioned once, for all intents and purposes, that aircraft can be assumed to be no different in this condition than in "equilibrium flight". If the TAS of the aircraft is zero, there is zero THP.

And no, it's not just an alternative method of calculating power output. It's a different way. They aren't the same, oggers! I've already proved that with the difference in words and, more importantly, with the difference in equations!

Forgive me for labouring the point as you do seem to rely on that equation, but it assumes equilibrium flight as stated clearly at the top of your reference document. An aircraft on the ground is not in equilibrium flight.

See above.

OTOH, a helicopter in the hover [where you insist there is no THP] is in equilibrium flight. The aircraft itself is not moving but there is a velocity which is the induced airflow. You have acknowledged that there is thrust. Therefore you can apply the formula – weight x [velocity of induced air flow]. The result is THP. It has to be because it’s the power associated with the thrust. Although you insist there is no THP for a helo in the hover.

As stated a number of times, THP = thrust x flight velocity (which is TAS) and it is NOT induced velocity!! You're mixing up all the equations and calling them all the same.

I'd suggest the THP is exactly the same into wind as it would be downwind and the earth has nowt to do with it.

You found a legitimate error, finally, so I would congratulate you for it, but I find it hard to do that because you didn't even realize that it was an error!

In the quote you used, I said THP is relative to the distance the aircraft moves with reference to the earth. That's incorrect. It's TAS, which is flight velocity and explicitly states as so in the Aerodynamics for Naval Aviators text. I made a mistake.

I'd suggest the THP is exactly the same into wind as it would be downwind and the earth has nowt to do with it.

That's how I knew you didn't know it was an error. You just happened to think and suggest something that was actually turned out to be true but, again, with no proof. If you really did know I had made a mistake you'd jump over the opportunity to prove me wrong, something you've been trying really hard to do this whole thread! You could have posted this picture that I've used a few times: http://i.imgur.com/NXLV3.png, and tell me that it's flight velocity and not relative to the ground!

But I really don't care if you knew or not. I'm glad that you did point my mistake out to me. I'm just mentioning this because of you consistently trying to prove me wrong and then accuse me of holding out info. It's just a way of keeping track of what is going on. It's also to prevent you in the end saying "I knew it was wrong...".

Regarding my two, seemingly, different comments separated by a few posts: 1) There is no need to complicate the THP with special or general relativity. That's why I said that! 2) We started getting into more complex examples actually dealing with relativity and not THP anymore, so that's why I said that second comment. Without the context I've provided here, it does seem I'm a bit off my rocker... but I think that's what you were trying to show! Nice try :D

Now, getting back to this induced power and induced velocity stuff. Reference this propulsive efficiency equation: http://i.imgur.com/NXLV3.png

I just realized that the bottom part is actually the formula for the 'induced power' that you provided. In the AfNA document they call it input power. On top is the output power and in this case, it's THP. I think that blows your whole "induced power = THP" out of the water!... again!

Check this out: http://i.imgur.com/1Ev3y.png

That's from the Wikipedia page on Disc loading. You'll notice that they're calculating the power required to hover - which is what you've been trying to figure out all this time. Just like we've been discussing, Power = thrust x velocity. However, in this case, the velocity part is "induced velocity". If you read the AfNA document I posted, immediately before the propulsive equation part they talk about induced velocity. It is 1/2 the total velocity change of the air, and it is measured at the propeller. Since the helicopter is in the hover, the flight velocity is zero so you can take that term out of the induced power equation and you are left with: Induced power = thrust x induced velocity. And that's exactly what the Wikipedia page shows.

I hope now you can see more clearly what THP is and that THP is not induced power.

Not being sarcastic at all, I have to say thanks for introducing the talk about induced power. It's added more clarity regarding all of this for me.

blackhand

30th Mar 2012, 23:20

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.

To which the answer is - YES
To this I don't care what you think. I was not sitting on that information. 'Induced power' had not entered my mine ONCE while in this discussion. Would it have been easier to mention induced power, you ask? No, induced power is how efficient thrust is made. It is not about THP. I do believe it adds another perspective to everything and it's great you introduced it but to accuse me of 'holding out' info is a bunch of bullocks.

The best thread hijack for 2012

italia458

30th Mar 2012, 23:36

oggers:

You accusing me of holding out info and trying to guilt trip me has been bugging me. I shared all the resources I was using throughout this discussion. You and everyone else had access to them on here. I posted the whole AfNA document section on the propellers where it talks about induced power. The reason I hadn't picked up on this earlier and put the two together is because I hadn't read the article word for word and analyzed it. I also hadn't come across the article which you posted about Figure of Merit. Once you did post that, I picked up on it right away and I think it actually backfired on you! It added more evidence to the fact that you don't know what you're talking about. I highlighted that in my last post.

But I should ask the same of you: Have you been holding out this information knowingly as I constantly repeat myself in post after post, trying to explain THP? Have you been sitting on this information? Why didn't you speak up earlier about it? What proves that you haven't just been trolling the whole time, like someone else has mentioned in this thread, and knew all along about this?

I'm not responsible to teach you everything about aerodynamics. You need to take responsibility for your learning. You can't blame me for you not thinking about the efficiency of the prop in creating the thrust. I mentioned that there were other efficiencies that can be analyzed and I agreed that the propeller is doing work on the air. Did I really have to tell you that if anything is doing work, there is a way to analyze the efficiency of the object doing the work? I'm not responsible to think of everything for you.

You finally discovered something that would add to the discussion and then you blame me for not telling you earlier. You really are a piece of work. :ok:

oggers

31st Mar 2012, 09:09

In the quote you used, I said THP is relative to the distance the aircraft moves with reference to the earth. That's incorrect. It's TAS, which is flight velocity and explicitly states as so in the Aerodynamics for Naval Aviators text. I made a mistake.

That's right. You did make a mistake. But not just in that quote (which was actually two quotes anyway). No. You reiterated it across 4 pages:

I have been very clear in describing that if the 'vehicle' is not moving...

.....the distance that the aircraft moves with reference to the earth...”

...when the helicopter is in a fixed position over the ground...

...if the aircraft isn't moving, it isn't covering distance and so the work = zero. When work = zero, Power also = zero.

etc etc.

Who are you trying to kid? You made it central to what you've been prattling on about. Now finally you've dropped it. :D

You accusing me of holding out info and trying to guilt trip me has been bugging me

:{ Well, maybe it will help if you go back and read between the lines. I wasn't really accusing you of holding out info. I was suggesting that it hadn't occured to you.

But let's just be clear about one thing. After 4 pages you have now done a ma-housiv u-turn and decided that the aircraft can stay still after all and still output power from the prop.

oggers

31st Mar 2012, 09:13

blackhand:

The best thread hijack for 2012

Gets my vote :ok:

Old Fella

31st Mar 2012, 10:14

Aerbat77. The Allison is not flat-rated and to clarify my comment on reaching the limiting T.I.T I was simply saying that, especially on the T56A-11 it was common to reach max permissable T.I.T before achieving limiting torque.

italia458

31st Mar 2012, 13:52

oggers:

We had never talked about a wind before and I had assumed that there would be no wind. Therefore, with no wind, TAS is the same as GS.

You hold onto it as if you 'won' or something. If you really were thinking about wind before, you would have made a comment about it the first time I mentioned a position relative to the ground. You also forgot to highlight the times when I referred to the same speed as "TAS". Of course you wouldn't want to provide an unbiased view because that would mean you don't really have a point in this case. I think its quite obvious to everyone that neither of us was thinking about wind.

I should also point out that out of the 4 quotes you posted, only 2 actually relate to what you were talking about. #2 and #3 mention earth or the ground. But #1 and #4 only mention motion.

You never addressed the recent 'issues' I brought up in my last 2 posts about what you said about induced power. It seems as though you actually can't admit being wrong at all. You would never be a good scientist or engineer.

I think the original point of discussion between us was if the aircraft velocity was zero, I said there would be zero THP and you obviously did not agree because you said this:

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.

So, has this discussion changed your answer now, or do you still believe that the THP will equal the BHP because it's 'accelerating a mass of air rearwards'? After this discussion it's quite obvious that I have a better 'grip' on the fundamentals of aerodynamics than you do! :)

EDIT: After 4 pages you have now done a ma-housiv u-turn and decided that the aircraft can stay still after all and still output power from the prop.

Where did I say that the prop does not do work on the air when the aircraft is not moving? I never said that! So, to put it bluntly, you are lying.

barit1

31st Mar 2012, 14:32

And so when a PW4000 runs up to max on a terrestrial test bed, it is actually creating ~70,000 thrust HP? Is that what we surmise? :rolleyes:

barit1

31st Mar 2012, 14:37

Old Fella:...on the T56A-11 it was common to reach max permissable T.I.T before achieving limiting torque.

Hmmm. Since the aircraft performance is dependent on the props being driven at some torque rating, I must ask how you determine your performance-limited TOGW??

italia458

31st Mar 2012, 14:42

barit1:

And so when a PW4000 runs up to max on a terrestrial test bed, it is actually creating ~70,000 thrust HP? Is that what we surmise?

No. THP has to do with aircraft performance. When it's on a testbed, you wouldn't be analyzing THP. The engine would be producing thrust and you could compute how efficient the engine is in making thrust (ie: accelerating air).

EDIT: However, if you were to apply the concept of THP to the testbed scenario, the THP would be zero because the testbed/engine wouldn't be moving.

oggers

31st Mar 2012, 16:33

italia:

So, has this discussion changed your answer now, or do you still believe that the THP will equal the BHP because it's 'accelerating a mass of air rearwards'?

Not changed my pov. BHP x PE = THP. [notwithstanding Tx losses obviously]

After 4 pages you have now done a ma-housiv u-turn and decided that the aircraft can stay still after all and still output power from the prop.

Where did I say that the prop does not do work on the air when the aircraft is not moving? I never said that! So, to put it bluntly, you are lying.

Well, what I wrote and what you wrote underneath it aren't the same. But taking your little statement at face value, you are clearly accepting there that there is a power output from the prop even if the plane is stationary.

An equation for PE (that you have used yourself) is power output divided by power input. An equation for THP is BHP x PE. If you acknowledge there is a power output from the prop you must acknowledge there is THP.

I look forward to reading the wriggling you will do to mitigate this conclusion. Not that I've entirely finished myself.

No. THP has to do with aircraft performance. When it's on a testbed, you wouldn't be analyzing THP

I think most people here, being aviation professionals, would feel the putting of an engine on a tested had quite a lot to do with aircraft performance. Unless you are a glider pilot.

italia458

31st Mar 2012, 16:45

oggers:

You're wrong. End of story.

There is no point explaining it anymore.

Good day!

Old Fella

1st Apr 2012, 02:08

Barit1, I probably did not articulate my point as well as I might. All I was trying to convey was that the 19600"/lbs limiting torque was often not achievable before reaching the Max permissable Turbine Inlet Temperature, especially on the T56A-11 which from memory was limited to 971 degrees C at Take-off power and 927 degrees C Max Continuous. The T56A-15 however with a higher T.I.T limit could readily reach Max Torque below the limiting T.I.T of 1083 degrees C for Take-off. Of course, achievable Torque for the given ambient conditions and Runway length available, surface type etc all are considered to determine Max Take-off weight allowed. Hope this is clearer.

rudderrudderrat

1st Apr 2012, 08:24

Hi,

In my reply (post #24), On RR Tyne with 4 bladed variable pitch props, we had to have both Prop RPM and torque within limits before we called power was set. The engine and gearbox were bolted together so it was not possible to measure their individual torques separately.
Hence the torque sensor (which was bolted between an engine mount and the airframe) simply measured the reaction to the propeller torque - so it indicated actual Prop Torque.

In order to keep passenger comfort during stepped climbs, we kept the prop RPM constant, but varied the fuel flow by "trimming" the (guarded) HP fuel levers. Power changes were called for ("50% trim" etc) and measured by observing the torque.

Since we measured both torque (force) and prop RPM (velocity) one could calculate the propeller horse power being generated (with a known p.e.) - even when stationary on the runway

Since there is no torque meter fitted on a bypass jet engine, the most useful parameters are N1 / EPR which only give an indication of thrust, not power.

Edit for Capt Pit Bull's comments below.
Rolls-Royce Tyne - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Rolls-Royce_Tyne)
"RTy.1 Mk 506
3,259 kW or 4,985 e.s.h.p"

blackhand

1st Apr 2012, 09:24

only give an indication of thrust, not power. mmm mumble mumble walking away.
Is that because there is no electricity?

Old Fella

1st Apr 2012, 10:03

rudderrudderrat, the Allison T56 is essentially a constant speed engine/gearbox/propeller installation with propeller blade angle selectable only in the 'ground' range of throttle lever travel (except feathering or unfeathering of course). Once in the 'flight' range (flight idle to 90 degrees throttle angle) the throttle simply schedules fuel flow and the propeller is governed to 100% RPM (1021 RPM) + or - 2% in basic hydraulic governing mode. The obvious advantage in having such an engine is almost instantaneous response to throttle movement (limited to not moving the throttle from Flight Idle to Max in less than 1 second) Without getting the purists throwing stones the torque value indicated on the C130 is the load imposed on the engine torque shaft ( and thus the engine itself) by the reduction gearbox, accessories and propeller. This whole thread seems to have become the site for "ego boosting" by some, surely not what the original poster was looking for. I am sure you, barit1 and I each would understand that as a crew member our concern would be "Have I got the torque or power I expect within the RPM and T.I.T limitations applicable and, if not, why not?" I am familiar with the use of N1 and EPR (or IEPR on the RB211 engine) as a measure of thrust. Whether we call the motive force Horespower, Thrust, Torque or any other name doesn't really matter. Every aircraft I have operated has had a specific set of parameters which are to be met before and during flight.

Capt Pit Bull

1st Apr 2012, 10:26

Sorry I'm late.

Italia is correct (give or take the odd minor slip / assumed simplification e.g. zero wind etc).

Oggers, you don't understand efficiency.

An engine in a test bed does NOTHING except churn up air thereby producing heat. If the object an engine is attached to does not increase in the total of GPE +KE as a result of the efforts of the engine then no *useful* work is being done. Likewise a person straining on a wrench is just producing heat. In both cases energy is certainly used, but not usefully.

This is my one and only post on the topic. Oggers, you have a long ingrained 'alternative framework'. Unless you are at least prepared to consider that something you think you understand could be wrong I, like italia, see no point in spending any time over it.

pb

oggers

1st Apr 2012, 11:04

Oh yeah nice one Pitbull. I'll chalk that up to being 1st Apr :ok:

oggers

2nd Apr 2012, 13:54

Capt Pit Bull:

The issue I’m disputing is the assertion by italia458 (note, he raised it, not me) that a plane at standstill can knock out 200 BHP and generate static thrust, yet there is zero THP.

“italia is correct.”

That’s a very definite conclusion to draw from the rather subjective argument you offer.

“Oggers, you don't understand efficiency.”

‘Power out over power in’ is pretty uncontroversial.

Unless you are at least prepared to consider that something you think you understand could be wrong I, like italia, see no point in spending any time over it.

THP = BHP x PE. If the thrust is not zero, PE cannot be zero, therefore THP is not zero. There is another equation out there based on airspeed for an arcraft in flight but it does not invalidate this one or vice versa. The trouble is Pit Bull, that italia458 is insisting it does, and you assert that he is correct. I would call that dogma and therefore I have to chuckle at the irony of your post.

“An engine in a test bed does NOTHING except churn up air thereby producing heat."

It develops thrust.

"If the object an engine is attached to does not increase in the total of GPE +KE as a result of the efforts of the engine then no *useful* work is being done..”

Well, hopefullly the engine won't "increase in the total of GPE + KE" during the test because that wouldn't be terribly "useful work". It is on the test-bed for a reason. It is developing thrust and fulfilling a purpose other than moving an aircraft. You are entitled to your opinion - in a certain context I would agree with it - but it does not lead me to the conclusion that the propeller efficiency is zero, or that static thrust = zero THP.

Applying this to the helo scenario, if I want to maintain a steady hover, the work done on the air which provides the thrust to keep me there is 100% useful work. In no sense is it wasted.

aerobat77

2nd Apr 2012, 16:11

oggers ´n italia : do you not have more important things to sort out in your life? :8

oggers

2nd Apr 2012, 19:20

I squeeze it in between the pilates and AA meetings.

2nd Apr 2012, 20:25

If you're not doing any useful work, you can't claim to be efficient. End of story.

As for your uncontroversial proof of understanding, it dictates that if the Joules leaving the system per time unit equals the Joules entering the system per time unit, efficiency is 100%. Any less than 100% efficiency, and less Joules leave the system than enter it.

http://www.htmlforums.com/images/smilies/Burning.gif

That's pretty amazing, and proves Pitbull's point rather nicely. :ok:

blackhand

3rd Apr 2012, 00:49

An engine in a test bed does NOTHING except churn up air thereby producing heat. If the object an engine is attached to does not increase in the total of GPE +KE as a result of the efforts of the engine then no *useful* work is being done. Likewise a person straining on a wrench is just producing heat. In both cases energy is certainly used, but not usefully.Had the dirt track car on the dyno yesterday. Watched very carefully and am sure the car did not move an inch.
Yet the dyno showed 327 rear wheel horsepower at 7200 RPM.
Lots of heat and noise, used 27 litres of fuel.

oggers

3rd Apr 2012, 12:11

Blackhand: rolling road dyno - good example. Makes the point clearly.

3rd Apr 2012, 18:39

oggers,
for once we agree!

That does make the point clearly. Power used, heat and noise generated, car remaining completely motionless - hence, no work done on the car. Efficiency zero (0).

Just as the aforementioned engine on a test stand, or non-moving aircraft during an engine runup. Not achieving anything, other than generating noise and heat.

Obviously not the point you've been failing for make for six pages of this thread, but I'm happy that you've seen the light at long last! :ok:

blackhand

3rd Apr 2012, 19:39

That does make the point clearly. Power used, heat and noise generated, car remaining completely motionless - hence, no work done on the car
You sir, are being "purposely" obtuse.
Work is being done because the rear wheels are turning, as I tried to explain it is "relative".
The same as any stationary engine, test bed or otherwise.

keith williams

3rd Apr 2012, 19:51

The book "The Jet Engine" by Rolls Royce gives the following version of the equation for propulsive efficiency for a jet engine.

Propulsive Efficiency = (Work done on the aircraft) / (Work done on the aircraft + Work done on the exhaust)

We can modify this to apply for a propeller aircraft as follows.

Propulsive Efficiency = (Work done on the aircraft ) / (Work done on the aircraft + Work done on the propwash)

We can also modify it to apply to your car on the dynamometer as follows:

Propulsive Efficiency = (work done on the car) / (Work done on the car + Work done on the dynamometer)

The work done on the jet engine exhaust, the propwash and the dynomometer are all wasted work.

In each case if the aircraft or the car are not moving then the work done on them is zero and the propulsive efficiency is zero.

If we have zero propulsive efficiency this means that we are wasting all of the power that is being generated by our propulsion system. This means that we have no thrust horsepower.

oggers

5th Apr 2012, 12:45

Keith, all reasonable contributions are welcome.

The assumption is we have thrust:

If we have zero propulsive efficiency this means that we are wasting all of the power that is being generated by our propulsion system. This means that we have no thrust horsepower.

I disagree with your conclusion. For a helo in the hover propulsive efficiency is zero. But “the power being generated by our propulsion system” is not being wasted.

Also, propulsive efficiency is not the same as propeller efficiency. If there is thrust, propeller efficiency absolutely cannot be zero. Whereas propulsive efficiency must be zero when the aircraft is stationary, regardless of thrust.

For that reason, any attempt to use propulsive efficiency to prove that THP must be zero in the static thrust scenario, is a logical fallacy. All that is proved is that propulsive efficiency is zero.

THP = BHP x Propeller Efficiency.

If there is thrust PE cannot be zero therefore THP cannot be zero.

“THP is the propeller output or power that is converted to useable thrust by the propeller”. That statement and the equation above are both taken from the document cited by italia458 after he floated this canard.

Before anyone jumps on the word “useable” to suggest the thrust is not useable unless it’s moving the aircraft, consider the hover again. The thrust is not only useable, it is being used.

keith williams

5th Apr 2012, 15:26

Oggers,

My only purpose in contributing to this thread was an attempt to assist blackhand to understand why his car was not producing any THP. I can assure you that it was not in any way meant to convince you of your errors.

The past 6 pages of this thread have shown the following two things quite clearly:

1. You have a long held misunderstanding of the subjects being discussed in this thread.
2. No amount of logical argument by others will convince you that you are wrong.

For these reasons I (and clearly many other members) do not intend to engage in further pointless debate with you.

I suspect that you will come to recognise your mistakes only if you manage to discover them for yourself.

aerobat77

5th Apr 2012, 16:49

you have to split two things here : propulsive efficiency/ power / useful work of the pure engine vs the aircraft as a whole package ( or a car or whatever elese) .

since all we know that any useful work needs a motion by physics a tied up plane or a car on a dyno does not any useful work since it does not move- efficency zero.

BUT

the engine of the car or the plane moves ( spins) , produces a given torque at a given rpm, the prop translates this torque to thrust etc. so the engine by itself of course makes useful work , produces power and has a given efficiency.

the useful work of the pure engine is used to run the dyno,to stress the brakes when they hold the aircraft, to keep our helicopter in hover against gravity.

long story short conclusion : the aircraft as the whole package does not provide any useful work, its engines do with the moment as they start to spin.

cheers !

oggers

5th Apr 2012, 17:07

Keith, I'm not interested in your purpose for contributing. You did contribute.

If you believe that propeller efficiency cannot be other than zero when the aircraft is stationary just ask yourself next time you start the take off run where the aircraft finds the thrust and power to even begin moving. :ok:

blackhand

5th Apr 2012, 20:17

long story short conclusion : the aircraft as the whole package does not provide any useful work, its engines do with the moment as they start to spin.
You must be a mechanic :ok:

rudderrudderrat

5th Apr 2012, 22:31

Hi oggers,

Is this what you are searching for?

"Induced Power
Induced Power is that portion of the power required to produce lift. It is the power required to overcome the portion of rotor drag which is caused by the induced flow tilting the total reaction rearwards.
Induced power is the force required to move a mass of air through the disk at the induced velocity.
If T is the rotor thrust (in a hover equal to weight (Mass * G)), which is a force, and this force moves the air at a velocity Vi , Pi = TVi .

Helicopter Power Required (http://www.griffin-helicopters.co.uk/note/helicopterpower.htm)

oggers

6th Apr 2012, 09:54

RRT,

Thanks that's a nice clear reference.

Pi = T x Vi

THP = T x V

The principle of THP extended to the hover. No need for the aircraft to move.

barit1

7th Apr 2012, 02:34

cwatters

7th Apr 2012, 10:09

Perhaps it would help to remember that aircraft propulsion systems are not "lossless". Not all the power goes into moving the aircraft so you can't simply say the power generated by the engine is

Power = force x velocity

You have to write something like...

Power = (force x velocity) + "losses"

You can define:

force = thrust
velocity = aircraft airspeed

but best leave the definition of "losses" vague for the moment as there are too many things to list and they are NOT constant. Some of the losses are also dependant on velocity.

Now it's clear from the modified equation that the power output is not necessarily zero when the aircraft is stationary. In such a case "Losses" includes throwing a lot of air backwards for no purpose.

In the case of the car on the dyno, "losses" include the load the dyno places on the wheels (eg heating up the brake unit).

Interesting to consider what happens at take off. Lets say you have brakes on, run up the engines and then release the brakes... as the aircraft starts moving through the air the sum of all the "losses" must reduce to balance the equation. (I'll ignore the fact that some engines produce more power when moving).

rudderrudderrat

7th Apr 2012, 11:57

Hi barit1,
How much horsepower does it then take to (ahem) "move" it at zero airspeed?
There may be a clue here...
That depends on the acceleration which the power unit is giving to the aircraft at that very instant.

barit1

7th Apr 2012, 16:05

True, but I believe the unstated assumption here is a steady-state (unaccelerated) vehicle.

rudderrudderrat

7th Apr 2012, 16:23

Are you only considering a shed then?

barit1

7th Apr 2012, 20:13

The conditions I propose:

o Parked, brakes locked (zero GS)

o Zero wind (thus zero IAS)

The questions: How much induced AND profile drag?

and...

From a THP standpoint, how is this any different from running the engine on a test bench?

Keeping in mind the definition of one horsepower (i.e. 33,000 foot-pounds of work per minute), is there any useful work being done?

oggers

11th Apr 2012, 11:17

A few thoughts:

At zero airspeed, how much induced drag does the aeroplane see? How much profile drag?

How much horsepower does it then take to (ahem) "move" it at zero airspeed?

There may be a clue here...

Sure, no power is required if the aircraft is at standstill but that does not mean no power is available. OTOH power is required to accelerate the mass of the vehicle from standstill with or without aerodynamic drag. No power available = no acceleration. THP is the power available. If the aircraft has to move before you get THP then you are in a chicken and egg situation - you can't accelerate the aircraft without power but you can't have power until you move the aircraft.

The question is: if the engine is producing 200 BHP and the prop is providing thrust can we say that the prop is providing THP, even whilst the aircraft is stationary?

THP = BHP x PE

If there is thrust PE is not zero. If PE is not zero THP cannot be zero. THP is not the same as power required, though it may have the same value.

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

RRT,

As an aside, the power required curve in the link you provided should be very familiar to helicopter pilots: the high power required to hover and the ‘hump’ near zero IAS as one transitions to forward flight are important characteristics for rotorheads to consider.

barit1

11th Apr 2012, 12:23

THP is the power available.

Nope. THP is the power DELIVERED TO THE VEHICLE.

THP is not the same as power required, though it may have the same value.

See above.

oggers

11th Apr 2012, 14:09

barit

I said 'THP is the power available'. It wasn't meant as a definitive definition of THP, merely an attempt to use some words to give context to the paragraph.

Nope. THP is the power DELIVERED TO THE VEHICLE.

Now, are you trying to tell me that THP is not the measure of power available? If so, which of the other definitions of power that are out there, that one might call commonplace, eg IHP, BHP, SHP, ESHP do you believe is the right one to use in the context of power output available from a prop? Or do you have one of your own?

barit1

12th Apr 2012, 01:52

To spell it out, using a 10000# thrust engine for example:

At zero TAS, THP = 0

At 375 MPH (i.e. 550 fps) TAS, THP = 10000

At 750 MPH (i.e. 1100 fps) TAS, THP = 20000

...and so on. As you see, the propulsive work delivered to the aeroplane increases proportional to velocity.

In real life, things aren't so tidy, simply because the engine thrust varies a bit with different inlet conditions; but if the machine were adjusted to hold constant thrust, the above would indeed apply.

HINT: Note that the above cases are based on the definition of 1 HP, namely 33000 ft-lb/min, or 550 ft-lb/sec. It helps to think of horsepower in terms of its definition!

oggers

12th Apr 2012, 07:59

barit1

As you see, the propulsive work delivered to the aeroplane increases proportional to velocity

"Propulsive work". Clearly, everyone can see that.

But the question I asked of you was: is THP the measure of power available from the prop or not? Because when I said it was, you responded quite clearly:

Nope. THP is the power DELIVERED TO THE VEHICLE.

I'm asking what you believe is the measure of power available?

keith williams

12th Apr 2012, 10:48

Oggers you are overlooking the power that is being wasted in doing work on the propwash.

When the aircraft is stationary on the ground with engine running all of the propeller power output is being wasted in accelerating the propwash rearwards. So none of the propeller power output is being delivered to the airframe. That means that there is no THP, because as Barit has said "THP is the power delivered to the airframe".

In theory if the aircraft were to fly at its own propwash speed no power would be wasted in the propwash, so all of the propeller power output would be being delivered to the airframe. So it would all be THP. This is of course only true in theory because if the aircraft was flying at its own propwash speed, no air would be being accelerated rearwards by the propeller, so there would be no thrust.

At any speed between the zero and the propwash speed some of the propeller power output would be being wasted in the propwash and the rest would be being delivered to the airframe as THP.

If you think about it this means that the propeller power output is the THP plus the power wasted in the propwash. So we cannot say that propeller power output is THP.

BUT. If you look at any fixed wing diagram of power available against power required you would see that the power available is the THP. It would perhaps be better to say that this is the USEFUL power available.

Your mistake throughout this entire thread has been in assuming that THP is thrust x propwash speed. It is not. Beacuse THP is the power being delivered to the airframe, it is thrust x aircraft speed.

oggers

12th Apr 2012, 11:21

Interesting post Keith. I will give it my full consideration. For now I'll just point out that I most certainly haven't overlooked the propwash, having specifically referred to it as being the place the power went - in my very first post on the topic.

I hope barit1 will answer the simple question of what power is the measure of power available for himself. I think that is totally fair as he insisted in his response to to me that it isn't THP. In capitals no less!

aerobat77

12th Apr 2012, 11:38

you really beat it to death gentlemen :ouch: .

like written above in this particular case you have to split the airframe and the engine. power by definition cannot exist without movement- without movement you can only have a force but not power. so the airframe does not develop any power when the aircraft stands still and no power but anly a force ( e.g the force on the brakes ) is delivered to the airframe by the engine.

but the engine by itself moves ( spins ) and so it of course develops power.

its pretty the same when a diesel locomotive starts to pull a train but its to heavy that it is able to make it move. in this situation the locomotive develops only a force ( on the couplings) but not power.

the diesel engine of the locomotive on the other hand develops power because it spins at a given rpm developing a given torque.

and thats basicly all about this purely theoretic discussion ... :eek:

cheers !

lomapaseo

12th Apr 2012, 12:16

you really beat it to death gentlemen http://images.ibsrv.net/ibsrv/res/src:www.pprune.org/get/images/smilies/shiner.gif

You hit the nail on the head :ok:

Most of what I read is intuition and reason and a lot of that was correct even though there was little agreement :)

It really is all about definitions that allow one to communicate in a common language.

It was fun and we really should all go away satisfied

keith williams

12th Apr 2012, 12:27

Oggers, it was actually your second post in which you said the following:

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

But you were wrong, because you were confusing THP with the power that is being wasted in the propwash. You have then gone on to repeat this mistake thorughout the majority of your posts in this thread.

In the brakes on condition we have:

Propeller power output = power being wasted in the propwash.

In flight we have:

Propeller power output = THP + power being wasted in propwash.

As we accelerate from zero, the THP gradually increases while the power being wasted in the exhaust gradually decreases.

barit1

12th Apr 2012, 12:59

oggers:I'm asking what you believe is the measure of power available?

Power AVAILABLE (not power delivered) is very nicely defined by KW:
In the brakes on condition we have:

Propeller power output = power being wasted in the propwash.

In flight we have:

Propeller power output = THP + power being wasted in propwash.

And I will add: The wasted propwash power is dissipated in heat (friction between air molecules in the turbulent flow).

oggers

12th Apr 2012, 14:32

Okay barit1. I can see you don't want to answer the question. THP is the unit of measurement for power available. I think that is about as close to standard use of terminology as one is going get in this thread, even though you objected to it for reasons that are not clear.

keith williams

12th Apr 2012, 14:39

Oggers you really must try reading instead of just writing!

The POWER AVAILABLE from the PROPELLER is turned into two different types of power.

The first type is THP which is used to push the aircraft forward against the drag.

The second type is the power that is wasted in propelling the propwash rearwards.

So we cannot say that power available is THP.

In the most favourable conditions, that is in forward flight we have:

Power available from propeller = THP + Power wasted in propwash

But on the ground with the brakes on we have:

Power available from the propeller = Power being wasted in propwash

It isn't a difficlut concept, but to get it you need to READ what has been written. Simply waiting a few seconds then writing another post will not do it.

oggers

12th Apr 2012, 15:57

keith, don't worry. I'm reading your last three posts very carefully, and I can see a couple or three things that deserve a considered response :ok:

[edit: it may take til after the weekend though]

rudderrudderrat

12th Apr 2012, 16:25

Hi Keith,

Please could you clear up some doubts in my mind?
The first type is THP which is used to push the aircraft forward against the drag.I thought we simply called that "thrust"?
The second type is the power that is wasted in propelling the propwash rearwards.I thought that is how we generate thrust - by accelerating a mass of air backwards - why would it be considered wasted if it generates something useful?

The only waste I can see is considering all of this with the brakes on, so there is no useful acceleration of the aircraft.

keith williams

12th Apr 2012, 18:57

Hi Keith,

Please could you clear up some doubts in my mind?

Quote:
The first type is THP which is used to push the aircraft forward against the drag.
I thought we simply called that "thrust"?

OK, then I will refine the statement a little. THP is the power that we use to carry out the work of pushing the aircraft forward against the drag at our chosen aircraft speed.The second type is the power that is wasted in propelling the propwash rearwards.

Quote:
The second type is the power that is wasted in propelling the propwash rearwards.
I thought that is how we generate thrust - by accelerating a mass of air backwards - why would it be considered wasted if it generates something useful?

The only waste I can see is considering all of this with the brakes on, so there is no useful acceleration of the aircraft.

We often say that we create thrust by accelerating air rearwards. It would be more true to say that we create thrust by exerting a rearward force on the air. The air then exerts a forward force (the thrust) on our propeller in accordance with Newton's thrid law.

Unfortunately the air is a fluid, so it is not rigid enough to resist the force that we apply to it. This causes it to be accelerated rearwards. This in turn gives it rearward velocity. In giving it rearward velocity we have given it kinetic energy. This represents a waste of energy in that we are never going to get it back nor are we getting any benefit from it.

If you cannot see the difference then let's consider a man in a boat in a shallow canal. He has a long oar with which he can row the boat along or push it along by pushing against the bottom of the canal. The canal bottom and walls are lined with concrete.

If he pushes against the bottom of the canal, the canal bottom will resist this force so it will not move. Only the boat will move, so all of the energy that he uses goes into the boat to give it velocity. None of the energy is being wasted in moving the canal bottom rearward.

But if he chooses to row, then some of the energy will be wasted in accelerating water rearwards and some will go into moving the boat forward. This wasted energy is the equivalent of the energy that we waste when our propeller accelerates air rearwards.

Pushing against the bottom will clearly require less energy (and hence less power) than rowing.

oggers

13th Apr 2012, 12:30

Keith Williams....

Any THP?

....any thrust horsepower or not? Propwash wasted? And propeller efficiency - more than zero or not? Same questions to barit1.

[BTW haven't forgotten to respond to your other stuff :ok:]

barit1

13th Apr 2012, 12:48

The answer is found in the question "how much aerodynamic drag (due to vehicle speed) is there?"

If there's no velocity, and no drag, then there's no work being done to move the plane. And thus there is no thrust horsepower.

No THP = v(ft/sec) x f (thrust) / 550

IF THE ENGINE AND PROP CAN BE REPLACED BY A STATIC BRACKET, THERE IS NO THP.

Yes, the engine is producing rpm and torque (= mechanical hp), and the prop is stirring up the air, but no useful work is being done. Efficiency = zero.

BTW - don't confuse yourself by thinking of the aero drag created by the propwash as useful work!

oggers

13th Apr 2012, 13:08

BTW - don't confuse yourself by thinking of the aero drag created by the propwash as useful work!

That's an odd thing to say barit1. I'm interested, has anybody on this thread suggested such a thing or did the idea arise in your own domain?

Efficiency = zero.

Sorry, is that propeller efficiency = zero? I specifically asked about prop efficiency. Just trying to get clarity, thanks.

keith williams

13th Apr 2012, 14:02

Oggers, some of your posts appear to indicate that you think that if thrust is not zero then propeller efficiency cannot be zero. If you do believe this, then you are wrong (once again).

If you do a search for the definition of propeller efficiency (I've just done one) you will find that the most commonly used definition is as follows:

Propeller Efficiency = (Thrust x Aircraft Velocity) / Propeller power input

If you look at that equation you will see that there are two possible conditions that will give zero propeller efficiency.

These two conditions are, when thrust = zero, and when aircraft velocity = zero.

If either one of these conditions exist then the propeller efficiency will be zero.

In your hovering aeroplane picture the aircraft velocity is zero, so the propeller efficiency is zero.

A number of your arguments appear to be based on the belief that the aircraft velocity is the same as the propwash velocity. It is not.

oggers

13th Apr 2012, 14:51

Keith:

appear to indicate that you think that if thrust is not zero then propeller efficiency cannot be zero. If you do believe this, then you are wrong (once again).

Just to be clear in case I left you in any doubt whatsoever. Yes I DO believe that if thrust is not zero then propeller efficiency is not zero. Thanks for clarifying that you don't though.

I am aware of the equation you mention. I am intensely relaxed about it. It is the obvious equation to use for an aircraft in flight. Earlier in the thread -page 4 #62 - I actually linked to a page from a prop designer who uses it . But, he is careful to point out that it cannot be used to determine the efficiency of a prop in the static thrust situation. It is - according to him - "invalid" in that situation. So, he uses a different equation for that.

keith williams

13th Apr 2012, 15:44

Oggers,

The first point that we should note is that the author of that paper is not a universal authority on these matters. He has his point of view, but this cannot be used as an argument for rewriting all of the texts on these subjects.

The second point to note is that he was using a different equation because he wanted to know how much thrust the prop was producing when the aircraft velocity was zero. The final part of the essay is an equation for thrust produced.

But his equation cannot work out the efficiency of the propeller in converting power input into useful power output. And that is what the definition of propeller efficiency is all about. That is why his equation did not start with the words "Propeller Efficiency = "

But for some bizarre reason you have decided to argue that his equation is the one that should be used to calculate propeller efficiency.

Now if you wish to ignore all of the above then look at the essay a bit more closely. It includes the following equation

Substitute this into the equation for efficiency (1), and simplify to get:
http://www.jefflewis.net/graphics/aviation_theory-prop_eff-eqn_11.gif

The author had earlier defined VA as the aircraft velocity.

When the aircraft velocity is zero, the above equation shows that propeller efficiency is zero.

This entire thread has been about what is and what is not thrust horsepower. So there is nothing to be gained by introducing an equation that works out how much thrust a propeller produces.

If we look again at the basics:

Work is done when a force moves its point of application in the direction of the force.

Work = force x distance.

Power is the rate of doing work

Power = Force x distance / time

Power = force x velocity

From the above we can see that THP = Thrust x aircraft velocity.

Now look at your hovering model aeroplane and give us a logical explanation of why you believe that some THP is being produced.

rudderrudderrat

13th Apr 2012, 19:13

Hi Keith,

Thanks for your lucid explanations. Would you please point out where my logic is failing here.

Ignoring form drag for simplicity:
Consider an aircraft performing continuous rate 1 turns holding over a beacon flying at a speed of V. The wing develops lift and consequently induced drag ID.
Therefore, induced drag power = ID*V.
For sustained flight the thrust power from the propeller must match the induced drag power, therefore THP = ID*V

Consider the model aircraft in a vertical hover. The propeller is now a rotary wing. It is producing lift in a similar way to the aircraft wing, by moving through the air with some velocity Vp and consequently generating induced drag IDp.
The power required to spin the rotary wing (the propeller in this case) = IDp*Vp.

They look very similar to me.

keith williams

13th Apr 2012, 20:15

Happy to oblige.

Ignoring form drag for simplicity:
Consider an aircraft performing continuous rate 1 turns holding over a beacon flying at a speed of V. The wing develops lift and consequently induced drag ID.
Therefore, induced drag power = ID*V.
For sustained flight the thrust power from the propeller must match the induced drag power, therefore THP = ID*V

In the paragraph above you are using the thrust multiplied by the AICRAFT VELOCITY to get the THP. That is entirely correct

Consider the model aircraft in a vertical hover. The propeller is now a rotary wing. It is producing lift in a similar way to the aircraft wing, by moving through the air with some velocity Vp and consequently generating induced drag IDp.
The power required to spin the rotary wing (the propeller in this case) = IDp*Vp.

In this second paragraph the thing that you are calling a "rotary wing" is really just a helicopter rotor or else a propeller. You are then taking the drag that it produces (in the plane of rotation), and multiplying it by the rotational velocity.

That's OK as long as you realise that what you have worked out is the power that is required to keep your prop turning. But this is BHP.

If you want to calculate THP for your rotary wing aircraft you need to multiply the THRUST IN THE DIRETION OF FLIGHT by the AIRCRAFT VELOCITY not by the ROTOR VELOCITY. Neither the thrust nor the aircraft velocity have been included in your description of the schenario.

I don't know if you have studied rotary wing pof very much, but if you look at the power available/power required curves you will find that they are very different from the fixed wing cuves. There is a thing that is called induced power. It is the power that is required to induce the airflow down through the main rotor. Induced power varies markedly with forward speed and it takes into account the varying rotor efficiency (including such things as the tilt of the rotors in forward flight). Because this varying efficiency is taken into account in the induced power curve, the power available curve is shown as a straight(ish) line. This line is really the BHP/SHP. I suspect that you will find few if any references to THP in rotary wing POF.

A number of contributions to this thread have included attempts to draw comparisons between rotary wing pof and fixed wing pof. This is unlikely to be very helpful because the two are very different in many ways. The fact is that few of us know enough about rotary wing pof to make the comparisons very illuminating (I certainly don't)

I have invited oggers to go back to basics and explain why he thinks that an aeroplane that is not moving (brakes on or hovering model) is generating THP. I think that this exercise will be more productive than thinking up imaginary scenarious about circling aircraft and rotating wings. You might like to try it.

barit1

14th Apr 2012, 02:24

oggers:

But, he is careful to point out that it cannot be used to determine the efficiency of a prop in the static thrust situation. It is - according to him - "invalid" in that situation. So, he uses a different equation for that.

Exactly, because it's a different kind of efficiency - i.e. a measure of how much STATIC thrust can be had, compared to some theoretical "perfect" static thrust. That's altogether different from propulsive efficiency, which is

= 100% x (Thrust horsepower) / (engine output hp)

In other words, efficiency = output / input !

keith williams

14th Apr 2012, 08:35

We can calculate the efficiency of any process that we choose to look at.

How efficiently do I digest my food? (not very efficiently, judging by how much I eat and how little exercise I do)

How efficiently do my children use my money? (not at all efficiently)

How efficiently does the government use my taxes (worse than my kids)

If you want to look at how efficiently a propeller produces thrust then that's OK, but it is not propeller efficiency.

Propeller efficiency = THP / BHP

THP and BHP both have very specific definitions, which means that propeller efficiency also has a very specific definition.

THP = Thrust x Aircraft Velocity

So Propeller efficiency = Thrust x Aircraft velocity / BHP

So you will not prove that a propeller is producing THP when the aircraft is not moving.

I (and most of the people reading this thread) am happy to accept the fact that when an aircraft is being held on the brakes, the propeller does a lot of work in propelling the propwash rearwards. But it is doing no work whatsoever in propelling the aeroplane forwards. So it is producing no THP.

As I said to oggers in a pervious post:

If we look again at the basics:

Work is done when a force moves its point of application in the direction of the force.

Work = force x distance.

Power is the rate of doing work

Power = Force x distance / time

Power = force x velocity

From the above we can see that THP = Thrust x aircraft velocity.

Now look at your hovering model aeroplane and give us a logical explanation of why you believe that some THP is being produced.

A Squared

16th Apr 2012, 02:54

one more thing , but maybe i understand you not right : the 19600 in/lbs must be the value for the torque on the propeller, not the turbine by itself since a device which spins 13800rpm and produces simultany massive 19600 in/lb of torque would have an asronomic power ( 19600x13800/5252 = 51500 shp!!! )

Aerobat, the Torque value for the Allison 501 is engine torque. As someone else (Old fella, I think) said, the torque is measured at the engine output/gearbox input shaft. The thing is, the 19600 is *inch* pounds. The equation: torque*rpm/5252 if for *foot*-pounds (or pound-feet) Divide your astronomic power by 12 and you'll get 4300 hp, in round numbers, which is about right for an Allison 501.

barit1

16th Apr 2012, 03:09

A Squared:

Very good breakdown of the SHP problem. I've been stung by that "factor of 12" issue before - it's egg-on-your-face big time!

I will, however, point out 1 or 2 little items: Torque units are properly called "pound-feet" (or "pound-inches") and not "foot-pounds"; the foot-pound is a unit of energy or work.

And the constant 5252 is easy to remember, but it happens to be the quotient of 33000/6.2832 (or 2 pi). I once derived this on my own, many decades ago when I still had sufficient active neurons. :}

oggers

16th Apr 2012, 19:04

Keith Williams and barit1:

“When the aircraft is stationary on the ground with engine running all of the propeller power output is being wasted in accelerating the propwash rearwards. So none of the propeller power output is being delivered to the airframe. That means that there is no THP, because as Barit has said "THP is the power delivered to the airframe".

In theory if the aircraft were to fly at its own propwash speed no power would be wasted in the propwash, so all of the propeller power output would be being delivered to the airframe. So it would all be THP. This is of course only true in theory because if the aircraft was flying at its own propwash speed, no air would be being accelerated rearwards by the propeller, so there would be no thrust.

At any speed between the zero and the propwash speed some of the propeller power output would be being wasted in the propwash and the rest would be being delivered to the airframe as THP.

If you think about it this means that the propeller power output is the THP plus the power wasted in the propwash. So we cannot say that propeller power output is THP.

BUT. If you look at any fixed wing diagram of power available against power required you would see that the power available is the THP. It would perhaps be better to say that this is the USEFUL power available.

Beacuse THP is the power being delivered to the airframe, it is thrust x aircraft speed.”

You say THP is thrust x velocity. I agree. But I say that only works for a specific set of assumptions. Static thrust is not the only scenario that doesn’t conform to the assumptions. Here is another:

Aircraft flying straight and level at speed Va, pilot sets new power to accelerate to Vb. Assume constant propeller efficiency between Va and Vb due variable prop. Final THP = thrust x Vb, we agree that much. Using the hypothesis you outlined above we can draw some conclusions:

The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!

Thrust must have increased, but this didn’t reflect an increase in THP, this was because power was wasted in the propwash?!

It’s all a bit odd, if that’s what you believe. Personally I simply accept that the increase in power results in an increase in THP in accordance with the equation THP = BHP x PE. And that work done on the propwash isn’t wasted if it’s giving thrust.

You say there are two types of power output from the prop. The two types you identify are “wasted propwash power” and “THP”. You both favour this over the common aircraft performance terminology which considers power available to be simply: ‘THP’.

To be clear, nobody is saying that no power is lost in the propwash otherwise THP would simply equal BHP. But that is different from what you are implying – that all the work done on the air is a waste and only work done on the aircraft counts toward THP:

If you think about it this means that the propeller power output is the THP plus the power wasted in the propwash. So we cannot say that propeller power output is THP. BUT. If you look at any fixed wing diagram of power available against power required you would see that the power available is the THP. It would perhaps be better to say that this is the USEFUL power available.

According to Hubert C Smith, Associate Professor Emeritus, Penn State, in his “Illustrated Guide to Aerodynamics”:

“actual power available for thrust is obtained by multiplying BHP times propeller efficiency”

“The [amount of power] available is the thp, which is the bhp times propeller efficiency”

… the same equation you both keep ignoring that appears in numerous sources including the USN training doc already referenced in this thread:

THP = BHP x Propeller Efficiency

…and this quote from the same USN reference: “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.).”

…which directly contradicts the position you both hold:

“you think that if thrust is not zero then propeller efficiency cannot be zero. If you do believe this, then you are wrong (once again).”

In fact, you have both made it clear that you believe a prop with zero propeller efficiency can still produce thrust! And please don’t repeat the equation for propulsive efficiency because it is not valid unless you observe the assumption it depends upon; namely, equilibrium flight. Therefore I have to say the assertions you are making are not credible opinions to hold, but such is the level of cognitive dissonance required to believe in this meme of zero THP for a hovering aircraft. Let’s see it again:

airplane in hover

...zero propeller efficiency, zero thrust horsepower, and zero useful work on propwash. I don't think so.

A Squared

16th Apr 2012, 19:38

A Squared:

Very good breakdown of the SHP problem. I've been stung by that "factor of 12" issue before - it's egg-on-your-face big time!

Yeah, I've made that exact same mistake, publicly. That's why I was able to zero in on it.

I will, however, point out 1 or 2 little items: Torque units are properly called "pound-feet" (or "pound-inches") and not "foot-pounds"; the foot-pound is a unit of energy or work.

Well, saying that it is preferable to use "pound-feet" to avoid confusion with foot-pounds of energy is certainly an *opinion* that has some merit. To say that foot-pounds is not *proper* is a bit of a stretch. You can find a long history of torque being measured in foot pounds. It ain't just something a few ignorant dumbasses do because they don't know any better.

I can't recall ever being confused about whether I was talking about torque, or work and energy

As far as "pound-inches" I have to say, I haven't ever heard that one. I do know that the torque meters on my Herc (and all the official literature associated with it) uses "inch-pounds".

keith williams

16th Apr 2012, 22:09

Oggers,

It's a bit too late for me to reply to your post tonight.

For the moment, would you please state what you believe to be the correct equation for propeller efficiency.

barit1

17th Apr 2012, 03:14

oggers, please re-read post #144. I made it pretty clear that there are two kinds of efficiency, one for static thrust, and one for propulsive efficiency. Please - when you refer to "efficiency" - make it clear which you are talking about.

According to Hubert C Smith, Associate Professor Emeritus, Penn State, in his “Illustrated Guide to Aerodynamics”:

“actual power available for thrust is obtained by multiplying BHP times propeller efficiency”

“The [amount of power] available is the thp, which is the bhp times propeller efficiency”

Smith here is referring to propulsive efficiency - where the plane is being propelled. Don't use this for the stationary example.

keith williams

17th Apr 2012, 10:33

You say THP is thrust x velocity. I agree. But I say that only works for a specific set of assumptions. Static thrust is not the only scenario that doesn’t conform to the assumptions. Here is another:

Aircraft flying straight and level at speed Va, pilot sets new power to accelerate to Vb. Assume constant propeller efficiency between Va and Vb due variable prop. Final THP = thrust x Vb, we agree that much. Using the hypothesis you outlined above we can draw some conclusions:

The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!

Thrust must have increased, but this didn’t reflect an increase in THP, this was because power was wasted in the propwash?!

OK lets’ look at your scenario and see if it really proves your point.

To simplify the situation and to avoid any ambiguities we will assume the following:

1. The aircraft is initially in steady non-accelerated flight. This means that the power required is equal to the THP being generated.

2. The initial speed Va is greater than Vmd so any increase in speed will produce an increase in drag and an increase in power required.

3. There is no reduction gearbox between the engine and the propeller, so the full engine torque is applied to the propeller.

4. The propeller is of the constant speed –variable pitch type.

5. To increase the power, the pilot simply pushes the throttle forward (selecting a higher RPM would just complicate the argument, but not really change the outcome).

6. The pilot has decided to increases airspeed simply by increasing the engine power output (you have already intimated this, but it is best to avoid unnecessary arguments about how best to increase airspeed).

With the aircraft in steady flight at constant RPM the engine torque (causing the propeller to rotate) will be equal and opposite of the propeller torque (the component of propeller total reaction that acts in the plane of rotation to oppose rotation in powered flight). This balance of two opposing torques will maintain a constant propeller RPM.

When the pilot opens the throttle the power output (BHP) of the engine will increase. The increased power will initially take the form of an increase in engine torque. This will cause engine torque to become greater than propeller torque, so the propeller RPM will start to increase.

When the propeller constant speed unit senses this increase in RPM, it will command the pitch change unit to increase the pitch of the propeller blades. This will increase the angle of attack of the blades, thereby increasing propeller total reaction. This in turn will increase both the propeller torque and the thrust. The increased propeller torque will cause the RPM to return to its initial (selected) value. So the overall effects of the pilot opening the throttle will be an increase in thrust, which will cause an immediate increase in THP.

Prior to the pilot opening the throttle, the aircraft was in steady balanced flight, so the thrust was equal to the drag and the power required (drag x TAS) was equal to the THP (thrust x TAS).

The increased thrust caused by the pilot opening the throttle would have the following effects:

1. The thrust would be greater than the drag.

2. The THP being generated would be greater than the power required.

The combination of thrust greater than drag and THP greater than power required would cause the airspeed to increase. As the airspeed increases, it would cause the drag and hence the power required to increase. The acceleration would continue until the increasing drag was once again equal to the thrust and the increasing power required was equal to the THP being generated. The aircraft would then settle at the new into balanced flight at the new higher speed.

Now let’s look again at your comments:

The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!

The situation is actually one in which the THP increase occurred before the airspeed increase and contributed to it. The only thing that you have proved by proposing this scenario, is that you have given insufficient thought to the subject.

Static thrust is not the only scenario that doesn’t conform to the assumptions. Here is another:

I presume that this is a reference to your earlier suggestion that an aircraft attempting to accelerate from brake release was another “impossible chicken and egg” situation. I am quite happy to explain to you why this is not the impossibility that you suggest. But you will probably gain more by working it out for yourself.

work done on the propwash isn’t wasted if it’s giving thrust.

As an aircraft flies through the air it is continuously losing energy due to the drag force. If the airspeed and altitude are to be maintained, this lost energy must be replenished at a rate that is equal to the loss rate. Looked at in this context the purpose of the propulsion system is to take energy from the fuel, convert it into kinetic energy and transfer it to the aircraft. Any energy that is lost to the propwash (or jet efflux) represents a waste of energy.

The enormous improvements in fuel efficiency that have been achieved in turbojets / turbofans over the past few decades have been brought about by reducing the amount of kinetic energy that is wasted in the jet efflux. The old turbojets were quite able to produce thrust, but their very high exhaust velocities meant that they were throwing away enormous amounts of kinetic energy. The fact that modern engines are able produce the same (or even more) thrust, while throwing away much less energy, shows that the energy loss is wasteful.

I am quite happy to respond to the remainder of your post when you have stated the equation that you consider to be correct for propeller efficiency.

rudderrudderrat

17th Apr 2012, 15:54

Hi oggers,

Have a look at:Propeller Aircraft Performance and The Bootstrap Approach: Background (http://www.allstar.fiu.edu/aero/BA-Background.htm)

"Of the four forces acting on the airplane – thrust, drag, lift, and weight – thrust is the most difficult to measure or predict. That is why most books about aircraft performance simply assume that propeller efficiency h is some constant. Commonly cited values are h = 80% and h = 85%. Then thrust T = h P, where P is the engine power. Unfortunately, propeller efficiency is in fact not constant; it varies with air speed and RPM or, more precisely, with the dimensionless ratio of those two variables:
J=V/nd
where J is the "propeller advance ratio." As the propeller rotates through one circle the airplane advances a distance V/n. J is then the ratio of that advance distance to the propeller’s diameter.
Figure 1 is an example of how propeller efficiency varies with advance ratio."
http://www.blackholes.org.uk/PP/e1low.gif
If there is zero advance, then efficiency = 0

keith williams

17th Apr 2012, 16:57

Rudderrudderrat,

If Oggers had wanted to accept the fact that propeller efficiency is zero when forward speed is zero, he just needed to read the Essay to which he referred on page 4 of this thread.

It includes a graph that is pretty much like yours, but it shows propeller efficiency against forward speed.

And it starts at zero-zero! :O :D :ugh:

oggers

15th Jun 2012, 16:58

keith Williams:

I have invited oggers to go back to basics and explain why he thinks that an aeroplane that is not moving (brakes on or hovering model) is generating THP. I think that this exercise will be more productive than thinking up imaginary scenarious about circling aircraft and rotating wings. You might like to try it.

<<Thrust Horsepower: The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill Dictionary of Aviation

Zero thrust horsepower ≡ zero thrust. The model requires thrust to hover therefore the THP is not zero.

Power is required for thrust additional to losses. It is the induced power of the prop/rotor. In the case of the model ‘prop hanging’, that induced power is the thrust times the velocity induced at the prop, along the thrust axis. Induced power is therefore equivalent to thrust horsepower.

The position you are sticking to is that all power going into the slipstream is wasted power, which is true in the context of aircraft velocity but not in the context of thrust. Hence, propulsive efficiency for velocity, and FOM for thrust:

“Since it is the induced power which relates to the useful function of the rotor, the ratio of induced power to total power is a measure of rotor efficiency in the hover. This ratio is called the Figure of Merit” Sedddon: Basic Helicopter Aerodynamics

For a given prop the thrust depends on both induced velocity Vi and the mass flow through the prop: T = ρA(Va + Vi) x ΔV. Va may go to zero but if Vi goes to zero you have no thrust. Statically, the “power converted to thrust” is T x Vi. You will find this is in accordance with simple momentum theory for a prop as well as the definition at the top. It is therefore THP.

Non-statically this is T x (Va + Vi) which eventually becomes ≈ T x Va only. You have said as much but the point that T x Va is actually an approximation (though it is definitive for propulsive power) of ‘power converted to thrust’ seems to have escaped you. In the latter case it is fair to say that THP = T Va and swallow the approximation. But this becomes invalid at low speed as I have been stressing all along. This graph shows where the approximation holds good as the curves run along the top near 100% beginning at about 20m/s for lightly loaded props:

http://www.mh-aerotools.de/airfoils/images/propul1.gif

The efficiency increases as mass flow through the prop picks up at speed.

If Oggers had wanted to accept the fact that propeller efficiency is zero when forward speed is zero, he just needed to read the Essay to which he referred on page 4 of this thread. It includes a graph that is pretty much like yours, but it shows propeller efficiency against forward speed. And it starts at zero-zero! :O:D:ugh:

The graph in the essay is the same as the one above. These are Froude efficiency curves. It does not show that THP goes to zero with Va for a given BHP which is what you think. The speedier end of the curve shows Froude efficiency ≈ 100% of induced efficiency whereas the slower end of the curve shows the opposite. Ergo TVa is a good proxy for induced power at one end of the curve but completely invalid at the other end.

When you add BET you get a graph like the one RRT posted for overall propulsive efficiency. Personally, when I compare the two graphs I see they are “pretty much alike” in the same way that r is pretty much like n (ie different). The r shaped curve is showing that Froude efficiency increases as a proportion of overall efficiency and stays there even after overall efficiency drops off again so TVa as a proportion of “power converted to thrust” stays high. OTOH, as speed goes down the approximation fails. That is the whole point of the essay, the very reason the guy felt the need:

"you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero. So how do you know how good your prop is doing at low speeds or statically?"

..and he spends most of the essay working around the problem.

Looking at the model again, the position that you and barit1 have adopted is that the model has zero THP whilst prop hanging but some THP if it climbs. Therefore it has negative THP if it descends and the prop will drive the engine. In that case you are in autorotation as soon as you start to descend (or shortly after when -THP is sufficient to offset profile power). That would be nice, but it’s not the case. Autorotation does not begin until after RoD exceeds Vi. Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.

By contrast, if you take THP = induced power = T (Va + Vi) then you have THP in the hover, more THP in the climb and less THP in the descent until RoD exceeds Vi and autorotation begins. You will find this is in agreement with momentum theory for a helicopter in descent.

“A number of contributions to this thread have included attempts to draw comparisons between rotary wing pof and fixed wing pof. This is unlikely to be very helpful because the two are very different in many ways. The fact is that few of us know enough about rotary wing pof to make the comparisons very illuminating (I certainly don't)”

I find the comparison to be very helpful because as far as momentum theory goes the rotor and the prop are precisely the same. And we don’t need to go beyond that to determine the induced power required to provide a given thrust.

THP and BHP both have very specific definitions, which means that propeller efficiency also has a very specific definition.

THP = Thrust x Aircraft Velocity

Well, here are 3 credible definitions of THP:

<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation

<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.

<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State

There is absolutely no doubt that the model is developing induced power T x (Va + Vi). That is a given. Induced power is the power converted to thrust for the model or lift for the helicopter. It fits all three of those definitions, whereas T x Va is an approximation that only works when the Froude efficiency is high. Ergo it is invalid statically.

Capt Pit Bull

15th Jun 2012, 18:17

Well, here are 3 credible definitions of THP:

<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation

<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.

<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State

Well as far as I'm concerned none of them are precisely scientifically credible; words like 'equivalent' can be read as meaning 'exactly equal to' but this is not the case, the dimensions of the units involved are wrong. The discrepancy in all 3 cases, to make the units correct, is a velocity term. i.e. TAS. Which takes us right back to square 1.

really, none of those definitions is any more precise than saying 'well, its kind of like this.....'

I bet if you asked Hubert C Smith to give you a formula relating THP and thrust he would say: THP = Thrust x TAS.

pb

italia458

15th Jun 2012, 20:01

Oggers!... You're still arguing the same point I clearly described at the beginning of this thread, eh! :hmm:

keith williams

18th Jun 2012, 13:29

The author of the essay to which you have repeatedly referred did not actually state that the equation is invalid as speed approaches zero. What he actually said was:

While working on a project designing a propeller at work, I wanted to know just how good I was doing. Efficiency is one measure of how well a propeller is performing, but it's not necessarily a good indication of how well the design is performing up to its potential. In aviation, propulsive efficiency is defined as:
Efficiency = T x V / Pavail
where η is efficiency, T is Thrust, V is Velocity, and Pavail is Power Available, or power going into the propeller. Basically, power out divided by power in. This equation is very useful for many cases, but you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero.

The author of the essay had a problem because he was attempting to use the propeller efficiency equation to calculate thrust. He was assuming that the purpose of the propeller is to produce thrust, but this is not true.

The purpose of the propeller is to propel the aircraft forward against the drag force, by producing thrust. The rate at which the propeller is achieving this purpose is equal to the thrust multiplied by the aircraft velocity. This is our old friend the THP.

Because the primary purpose of the propeller is to propel the aircraft forward, it is entirely logical that the overall efficiency of the propeller should be measured in terms of its propulsive efficiency.

Propeller efficiency = (Thrust x Aircraft Velocity/ Brake Horsepower

This can also be stated as (Thrust x TAS) / BHP provided we take care to note that the TAS is velocity of the aircraft relative to the air and is not the propwash velocity.

When the aircraft is running at full throttle on the ground, with the wheel brakes preventing it from moving forward, the propeller is absorbing power but is not achieving any success in propelling the aircraft forward. So the propeller efficiency is zero, which is exactly what the equation yields. The equation is in fact producing the correct result at all airspeeds including zero.

To understand why the author of the essay could not use the equation to calculate thrust when airspeed was zero we need to look at what he was attempting to do.

He was attempting to take the product of (Thrust x zero airspeed) and then divide it by zero airspeed to reveal the thrust.

For any number other than zero and infinity this would be a perfectly reasonable thing to do.

But the product of any number multiplied by zero is zero, and dividing this product by zero simply gives zero / zero = zero.

The simple fact is that the processes of multiplication and division are not reversible when one of the numbers is zero. This fact is not a function of any inaccuracies in the propeller efficiency equation. It is simply the result of one of the properties of the number zero. Any equation involving multiplication and division will encounter exactly the same problem when one of the values goes to zero. But this does not mean that all such equations are invalid.

Looking at the model again, the position that you and barit1 have adopted is that the model has zero THP whilst prop hanging but some THP if it climbs. Therefore it has negative THP if it descends and the prop will drive the engine. In that case you are in autorotation as soon as you start to descend (or shortly after when -THP is sufficient to offset profile power). That would be nice, but it’s not the case. Autorotation does not begin until after RoD exceeds Vi. Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.

Entering autorotation in a helicopter is not simply a matter of throttling back the power until the aircraft starts to descend. If you do this, the blades will stall, the rotor speed will decrease, the blades will fold upwards and the aircraft to crash. To enter autorotation you must lower the collective to get negative pitch (or at least a very low pitch) at which the ROD airflow produces an angle of attack, such that part of the total reaction drives the blades forward. This keeps the rotors turning and keeps generating lift. None of these good things will happen if we simply reduce power in a propeller driven aircraft, because the mechanisms are physically very different. Your use of this scenario simply illustrates your lack of knowledge regarding helicopter POF.

Well, here are 3 credible definitions of THP:

<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation

<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.

<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State

In the second of your definitions the term “horsepower equivalent of the thrust” implies that it is the thrust x TAS. So this definition is quite correct.

In dealing with your first and third definitions I could quite easily find numerous sources of the definition THP = Thrust x TAS. I have at least one example on my office bookshelf. But trading references will prove nothing. Instead let’s see what your definitions produce in terms of a propeller efficiency equation.

Efficiency is simply a ratio of output divided by input, and standard efficiency equations are of the form Efficiency = Output / Input

In your first and third definitions the output is thrust and the input is power.

So we have Efficiency = Thrust output / Power input

If we use imperial units of pounds force for thrust and foot pounds force per minute for power we get:

Efficiency = (lbf) / (ft lbf / min)

Cancelling out the lbf on the top and bottom of this equation we get

Efficiency = 1 / ft/min which is Efficiency = min/ft.

The equation has yielded an efficiency figure that is in units of minutes per foot. What exactly does this mean?

Efficiency is just a ratio of output divided by input, so it has no units.

To get rid of the min/ft in our result we need to multiply by ft/min. But ft/mins are the units of velocity, so to get a proper value for efficiency we need to multiply by velocity. This converts our efficiency equation into Efficiency = (Thrust x TAS) / Power input.

This is of course the propulsive efficiency equation.

The above sequence proves something that we should already have known. That is the fact that power, which is the rate of doing work or expending energy, cannot be converted into a force. Power and force are two totally different things and one cannot be converted into another. So whatever the sources of your first and second definitions, they cannot be correct. It is of course possible that the authors did not intend their definitions to be interpreted literally.

Finally I think that you need to look again at your argument that induced power is the same things at THP. I have drawn the following quotes from various internet sources.

Induced power is the power required to maintain enough lift to overcome the force of gravity.
www.math.usu.edu/~powell/ornlab-html/node7.html (http://www.math.usu.edu/~powell/ornlab-html/node7.html)

Induced Power is that portion of the power required to produce lift.

www.griffin-helicopters.co.uk/note/helicopterpower.htm - 13k

The minimum engine power required to hover is called the "induced power."

scienceworld.wolfram.com/physics/Helicopter.html - 15k

Induced power is what people are referring to when they say helicopters "beat the air into submission." Newton's 2nd law concerning action-reaction applies in this regime where we must force air down to keep the aircraft aloft.

www.navair.navy.mil/safety/documents/Power_Available_vs.doc (http://www.navair.navy.mil/safety/documents/Power_Available_vs.doc)

In each case they show that induced power is exclusively concerned with the process of generating lift. THP is concerned with propelling aircraft through the air, so they are not the same thing. By all means argue that your model aeroplane is generating induced power. But it is most certainly not generating any THP.

barit1

18th Jun 2012, 19:01

keith williams:...The author of the essay had a problem because he was attempting to use the propeller efficiency equation to calculate thrust. He was assuming that the purpose of the propeller is to produce thrust, but this is not true.

The purpose of the propeller is to propel the aircraft forward against the drag force, by producing thrust. The rate at which the propeller is achieving this purpose is equal to the thrust multiplied by the aircraft velocity. This is our old friend the THP.

Well done! :ok:

oggers

21st Jun 2012, 10:49

Entering autorotation in a helicopter is not simply a matter of throttling back the power until the aircraft starts to descend. If you do this, the blades will stall, the rotor speed will decrease, the blades will fold upwards and the aircraft to crash. To enter autorotation you must lower the collective to get negative pitch (or at least a very low pitch) at which the ROD airflow produces an angle of attack, such that part of the total reaction drives the blades forward. This keeps the rotors turning and keeps generating lift. None of these good things will happen if we simply reduce power in a propeller driven aircraft, because the mechanisms are physically very different. Your use of this scenario simply illustrates your lack of knowledge regarding helicopter POF.

Keith, that is a big strawman, and not your first. Nobody has said that entering auto is a simple matter of throttling back the throttle := What I wrote was that your description of what constitutes THP implies you would be in auto as soon as Thrust x RoD (equivalent to T x TAS) exceeded profile power, when what actually happens is that auto begins when RoD exceeds induced velocity by enough to offset profile power. This is a fact.

the mechanisms are physically very different

The actual mechanical arrangement is superfluous, this is about the over-arching physics. Simple momentum theory considers only that a 'disc' accelerates air. This does not tell you if the model is mechanically capable of going into auto, it tells you the theoretical relationship between induced velocity, RoD and thrust.

keith williams

21st Jun 2012, 12:33

when what actually happens is that auto begins when RoD exceeds induced velocity by enough to offset profile power.

As the collective is lowered the aircraft starts to descend and the induced velocity and the induced power both fall to zero. As the aircraft accelerates downwards the direction of the total reaction is tilted in the direction of rotation at some points along the blades and is tilted in the opposite direction at other points. It is only when a certain rate of descent has been achieved that the sum total of the different parts of the total reaction is sufficient to keep the blades rotating in the correct direction and to produce sufficient lift to support the weight of the aircraft. When the sum of lift plus vertical component of drag equals the weight of the aircarft the ROD becomes constant. The actual rate of descent depends upon a number of things include blade pitch angle and aircraft weight.

So even with a helicopter, the autorotative force does not become significant until a certain ROD has been achieved. But none of this confirms your argument that induced power is the same thing as THP

The actual mechanical arrangement is superfluous, this is about the over-arching physics. Simple momentum theory considers only that a 'disc' accelerates air. This does not tell you if the model is mechanically capable of going into auto, it tells you the theoretical relationship between induced velocity, RoD and thrust.

And

Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.

I have known several hundred helicopter pilots over the years, but none of them have flown theoretical discs. They have all flown real helicopters with real rotor mechanisms. You are using the behaviour of real systems to support arguments about theoretical physics, then you are saying that the actual mechanical arrangement of the systems is superfluous. That line of argument is clearly nonsensical because the behaviour of the real systems is determined as much by their physical characteristics as it is by the theoretical physics.

Throughout this thread you have made various assertions, including the following:

1. Propellers continue produce THP when the aircraft velocity is zero.

2. When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.

3. If propellers produced no THP when the aeroplane is standing still on the ground, it will be impossible for the aeroplane to accelerate when the brakes are released.

4. If THP is Thrust x TAS it will be impossible for aircraft to accelerate in flight.

5. THP is the same thing as Induced Power.

6. The propulsive efficiency equation is not the correct equation for propeller efficiency (though interestingly you have never actually stated what the correct equation is).

7. As long as a propeller is producing thrust its efficiency cannot be zero.

In each case your assertions have been incorrect. You have certainly demonstrated tenacity in pursuing these arguments for so long, but you have also demonstrated a very poor understanding of these subjects.

Sillypeoples

21st Jun 2012, 18:29

OP:

Torque is a measurement of power from the engine, at the gear box of the prop/rotor.

The reason for torque indication is the gear box has limitations on how much power it can handle, as it's being resisted on the other side by the prop/rotor, which is being resisted by the drag from the air.

Certainly you can measure torque of the engine, torque to the prop, but usually the gear box is the weakest link, as it's translating the power from the engine to the airfoil...so it usually gets the indicator, with your job keeping it out of the red.

oggers

22nd Jun 2012, 11:56

Keith:

As the collective is lowered the aircraft starts to descend and the induced velocity and the induced power both fall to zero. As the aircraft accelerates downwards the direction of the total reaction is tilted in the direction of rotation at some points along the blades and is tilted in the opposite direction at other points. It is only when a certain rate of descent has been achieved that the sum total of the different parts of the total reaction is sufficient to keep the blades rotating in the correct direction and to produce sufficient lift to support the weight of the aircraft. When the sum of lift plus vertical component of drag equals the weight of the aircarft the ROD becomes constant. The actual rate of descent depends upon a number of things include blade pitch angle and aircraft weight.

Yeah, but irrelevent. :rolleyes:

I have known several hundred helicopter pilots over the years, but none of them have flown theoretical discs. They have all flown real helicopters with real rotor mechanisms.

..and they all imagine their thrust in terms of a man rowing a boat down a canal a la your little analogy? :ok:

You are using the behaviour of real systems to support arguments about theoretical physics, then you are saying that the actual mechanical arrangement of the systems is superfluous. That line of argument is clearly nonsensical because the behaviour of the real systems is determined as much by their physical characteristics as it is by the theoretical physics

I referred to actuator disc theory because it is the right one to tackle this meme of thrust without THP, and I prefer it to your rowing boat story. :)

When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.

I don't recall ever saying that. Feel free to paste a quote though, if you can find one.

keith williams

22nd Jun 2012, 12:27

I thought you might wish to deny one or more of your errors, so I did a search.

1. Propellers continue produce THP when the aircraft velocity is zero.

Page 1 post 8 “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.”

2. When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.

Page 1 post 14 “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. “

3. If propellers produced no THP when the aeroplane is standing still on the ground, it will be impossible for the aeroplane to accelerate when the brakes are released.

Page 6 post 120 “Sure, no power is required if the aircraft is at standstill but that does not mean no power is available. OTOH power is required to accelerate the mass of the vehicle from standstill with or without aerodynamic drag. No power available = no acceleration. THP is the power available. If the aircraft has to move before you get THP then you are in a chicken and egg situation - you can't accelerate the aircraft without power but you can't have power until you move the aircraft.”

4. If THP is Thrust x TAS it will be impossible for aircraft to accelerate in flight.

Page 7 post 148 “Aircraft flying straight and level at speed Va, pilot sets new power to accelerate to Vb. Assume constant propeller efficiency between Va and Vb due variable prop. Final THP = thrust x Vb, we agree that much. Using the hypothesis you outlined above we can draw some conclusions:

The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!

Thrust must have increased, but this didn’t reflect an increase in THP, this was because power was wasted in the propwash?!”

If you look at item number 2 above you will see that

Page 1 post 14 “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. “

implies that

When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.

It really is time for you to face up to the fact that virtually everything that you have written in this thread has been nonsense.