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Propeller torque & engine torque

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Propeller torque & engine torque

Old 30th Mar 2012, 02:23
  #41 (permalink)  
 
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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.
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Old 30th Mar 2012, 02:44
  #42 (permalink)  
 
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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?

Last edited by CYHeli; 30th Mar 2012 at 02:54.
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Old 30th Mar 2012, 03:55
  #43 (permalink)  
 
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Engine Torque

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.

Last edited by Old Fella; 30th Mar 2012 at 04:13.
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Old 30th Mar 2012, 04:23
  #44 (permalink)  
 
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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

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.
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Old 30th Mar 2012, 04:52
  #45 (permalink)  
 
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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.
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Old 30th Mar 2012, 05:32
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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. 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

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.
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Old 30th Mar 2012, 06:31
  #47 (permalink)  
 
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@ oggers
Once more into the breach my friend.
It's what I live for blackhand

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'?

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Old 30th Mar 2012, 08:27
  #48 (permalink)  
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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.
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Old 30th Mar 2012, 08:53
  #49 (permalink)  
 
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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.
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Old 30th Mar 2012, 09:27
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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 !
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Old 30th Mar 2012, 10:27
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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. Ch 2.

Last edited by oggers; 30th Mar 2012 at 12:45.
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Old 30th Mar 2012, 10:46
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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.
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Old 30th Mar 2012, 10:50
  #53 (permalink)  
 
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Approved Torque

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.
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Old 30th Mar 2012, 11:28
  #54 (permalink)  
 
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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 !
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Old 30th Mar 2012, 11:57
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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.
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Old 30th Mar 2012, 13:00
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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

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://selair.selkirk.ca/Training/Ae...es/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
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Old 30th Mar 2012, 13:27
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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.

Last edited by rudderrudderrat; 30th Mar 2012 at 14:28. Reason: spelling
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Old 30th Mar 2012, 14:46
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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

"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.
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Old 30th Mar 2012, 15:06
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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.
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Old 30th Mar 2012, 15:16
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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.
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