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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" 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. |
work only happens when you apply a force and move something a certain distance. |
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? |
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. blackhand: Is that relatively speaking? |
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? |
You're going to have to read the Wikipedia article on work. I'm not going to regurgitate what it says. 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? |
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.
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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. 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. 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? 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. |
Thrust is related to how much air is displaced. 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. |
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. 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 Once more into the breach my friend. italia: All mechanics are good for is turning bolts... don't try to think about physics. ft: What do you make of that me old? Another of italia's 'excellent abstract analogies' that exemplifies 'educated discussion between professionals'? :eek: :ugh: |
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. |
Hi italia458,
A net force means that the forces CAN NOT be balanced! 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. |
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 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 ! |
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. |
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. |
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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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 ! |
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. |
rudderrudderrat:
I'm feeling an acceleration whist standing on the earth (force on my feet) and I'm perfectly balanced. 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). 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". ...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. 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. 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. 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. This is an expanded form of the p.e. equation - http://i.imgur.com/NXLV3.png |
Hi italia458,
As stated numerous times, if the flight velocity is zero, there is zero THP - that is what the equation says, not me. 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. 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. |
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. 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. |
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. 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. 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. 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. |
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... 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. 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. |
The document that oggers posted does not relate to what I was talking about. 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) 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? |
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. The reason it is falling does NOT prove that it is 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. 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. |
italia: you must try to remain calm. 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 |
I edited my last post - just so you know.
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rudderrudderrat:
There is no mention of TAS in equation (1) ... the ball is in an inertial frame of reference - not you. 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 start by explaining the forces in a turn by drawing something like this: http://selair.selkirk.ca/Training/Ae...es/lf-turn.gif This picture is wrong: http://www.aero-mechanic.com/wp-cont...09/08/4-28.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. |
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? |
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. |
@ 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. 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). edit So how does the aircraft turn if the forces are actually balanced, like shown in the second picture? The "centrifugal force" (A) is what you feel (like the force of gravity). The centripetal acceleration is the reason for (A). |
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. Believe what you would like. I think I've had enough with this thread. Where do you actually stand on this? |
By a combination of elevator and rudder. |
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. |
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italia458
It would be good to read this. And your point? Like I said before, a physics class would benefit your understanding. Thank you for confirming that I completely wasted my time discussing anything with you. |
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.” “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.” “Can you provide evidence that my point is different than what is said in the references I provided?” “But the 'Navy stuff' clearly showed that you were in fact creating zero THP when the aircraft was stationary!” “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.” 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.” And I really like what you did here: “There is no need to complicate this with special or general relativity…” “General relativity deals with the theory of gravitation. I will try to explain a little bit about what's going on below.” |
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: 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. 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. 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. 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. 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. 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. |
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 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 |
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