another terrible ATPL PoF question
The lift vector is acting upward from the hub, being the point where the force from the rotor is transferred to the mast and hence the rest of the machine (plus a little bit from the tilted tail rotor thrust).
The reason the CofG isn't directly under the mast is because of the tail rotor thrust displacing it from there.
The main rotor disc tilt opposes that side force to stop the drift and, in the process creates the roll.
Nick, I must admit I feel a bit of a goat challenging a guy of your experience, but I honestly can't see why this diagram doesn't represent the forces on a teetering head helicopter in a stationary hover, albeit with some exaggerated angles for the sake of clear depiction.
The reason the CofG isn't directly under the mast is because of the tail rotor thrust displacing it from there.
The main rotor disc tilt opposes that side force to stop the drift and, in the process creates the roll.
Nick, I must admit I feel a bit of a goat challenging a guy of your experience, but I honestly can't see why this diagram doesn't represent the forces on a teetering head helicopter in a stationary hover, albeit with some exaggerated angles for the sake of clear depiction.
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Arm,
Please don't worry about jousting with someone with more experience. It is fun, and stimulating for both sides. But what you can't do is to post a drawing with lines at nice places and say 'here it is!" As I said, the lift vector must pass thru the CG, else the craft is accelerating on roll, bigtime.
That is where your diagram makes its most grevious error, and thru that hole passes the entire theory. It is not experience that we draw on here (pun intended!) it is Physics, and the subset of Physics, mechanics.
Considering the size of the lift as 10,000 lbs and the tail thrust at perhaps 400, (for an S76) your drawing shows a helo with a roll acceleration to the right of perhaps 100 degrees per second per second. Try to hold the stick, move the cyclic to the left and get that lift vector back thru the CG, please!
Once you do, the "rotor head height" theory will slide off into space!
Please don't worry about jousting with someone with more experience. It is fun, and stimulating for both sides. But what you can't do is to post a drawing with lines at nice places and say 'here it is!" As I said, the lift vector must pass thru the CG, else the craft is accelerating on roll, bigtime.
That is where your diagram makes its most grevious error, and thru that hole passes the entire theory. It is not experience that we draw on here (pun intended!) it is Physics, and the subset of Physics, mechanics.
Considering the size of the lift as 10,000 lbs and the tail thrust at perhaps 400, (for an S76) your drawing shows a helo with a roll acceleration to the right of perhaps 100 degrees per second per second. Try to hold the stick, move the cyclic to the left and get that lift vector back thru the CG, please!
Once you do, the "rotor head height" theory will slide off into space!
Well, as I said, the angles in the drawing are exaggerated for the sake of clarity - I guess my beef with your argument is that I can't see any particular reason why lift must act through the CofG.
To put it another way, what if we had a helicopter hovering with skids level, for the sake of argument.
If we had a rope tied to the right skid and a bunch of very strong guys pulling the rope directly outwards, they would move the helicopter in their direction.
To stop the movement, the pilot would put in enough left cyclic to oppose that pull exactly.
Now (I think, anyway) we'd have a situation a lot like in my diagram - a force to the right from low down (rope on skids) and one to the left from up high (component of rotor thrust going left), stopping drift but creating roll.
You wind up with the CofG displaced to the right from under the head slightly. The harder those guys pull on the rope, the more the pilot opposes it with cyclic (having to also increase collective because his rotor thrust vector is more tilted too), and the more roll you get, with the CofG moving further out all the time.
The component of rotor thrust going straight up still equals weight, holding the helicopter in the air, but the opposing lateral forces from high and low want to roll it.
It stops rolling when the CofG gets far enough out to generate an equal and opposite rolling couple.
I reckon it's a bit like a punching bag hanging off a hook in the roof - say you pull it sideways at the bottom with a small but constant force. It will move towards you until the opposing force caused by the CofG being displaced away from under the hook is equal to your pull force, and will then stay hanging there stationary with the CofG out to one side because you're holding it there.
May have to call Professor Julius Sumner Miller in for an adjudication.
To put it another way, what if we had a helicopter hovering with skids level, for the sake of argument.
If we had a rope tied to the right skid and a bunch of very strong guys pulling the rope directly outwards, they would move the helicopter in their direction.
To stop the movement, the pilot would put in enough left cyclic to oppose that pull exactly.
Now (I think, anyway) we'd have a situation a lot like in my diagram - a force to the right from low down (rope on skids) and one to the left from up high (component of rotor thrust going left), stopping drift but creating roll.
You wind up with the CofG displaced to the right from under the head slightly. The harder those guys pull on the rope, the more the pilot opposes it with cyclic (having to also increase collective because his rotor thrust vector is more tilted too), and the more roll you get, with the CofG moving further out all the time.
The component of rotor thrust going straight up still equals weight, holding the helicopter in the air, but the opposing lateral forces from high and low want to roll it.
It stops rolling when the CofG gets far enough out to generate an equal and opposite rolling couple.
I reckon it's a bit like a punching bag hanging off a hook in the roof - say you pull it sideways at the bottom with a small but constant force. It will move towards you until the opposing force caused by the CofG being displaced away from under the hook is equal to your pull force, and will then stay hanging there stationary with the CofG out to one side because you're holding it there.
May have to call Professor Julius Sumner Miller in for an adjudication.
It's probably the middle of the night for you guys - sorry to prolong this even more, but your last comment made me think of something else, lifting.
In forward flight, the helicopter's being skull-dragged forward by the mast head.
Drag on the fuselage can be said to act somewhere lower than that - the bottom of the windscreen, say, at a wild guess. Anyhow, it's lower than the mast.
Those two forces make a couple, which tilts the fuselage forward. Why doesn't it keep going? Because as it tilts, the CofG moves backwards with respect to the lifting point (mast head), and that makes another opposing couple.
When the two are equal, you get an equilibrium, and the whole crazy device flies along steadily with rotor thrust giving a component straight up equal to weight and another doing the pulling forwards equal to the drag, plus the CofG isn't directly under the point where lift is acting, it's behind it.
In forward flight, the helicopter's being skull-dragged forward by the mast head.
Drag on the fuselage can be said to act somewhere lower than that - the bottom of the windscreen, say, at a wild guess. Anyhow, it's lower than the mast.
Those two forces make a couple, which tilts the fuselage forward. Why doesn't it keep going? Because as it tilts, the CofG moves backwards with respect to the lifting point (mast head), and that makes another opposing couple.
When the two are equal, you get an equilibrium, and the whole crazy device flies along steadily with rotor thrust giving a component straight up equal to weight and another doing the pulling forwards equal to the drag, plus the CofG isn't directly under the point where lift is acting, it's behind it.
We have postulated at length to I agree (I think) that any force, whether it be weight, thrust, torque or drag acts about the C of G of the object (in this case the airframe). If a force has a moment arm about the c of g then the fuselage will rotate until either a. the force is acting straight through the c of g and the moment is zero or b. ( and the case we are are talking generally about here), another force with an opposing moment arm is generated which equals the original to the point where both forces are in equilibrium.
This is the case with TR drift (or translating tendency) - cyclic rolls the fuselage to the left but it doesn't keep going because the TR thrust creates one opposing couple ( the magnitude of which depends on the height of the TR to a great extent). The other opposing force is the vertical component of MR thrust acting about the c of g, the weight acts through the c of g
If you increase TR thrust (the pulling the skids with a rope analogy) that gives more translating tendency and requires more cyclic to oppose it - this results in more left skid low to hold the hover because the balance of forces is achieved in a new position. If you hover the same helo at a. the lightest weight possible and b. max all up mass you will have significantly different hover attitudes (ignoring the change in the c of g position.
This is the case with TR drift (or translating tendency) - cyclic rolls the fuselage to the left but it doesn't keep going because the TR thrust creates one opposing couple ( the magnitude of which depends on the height of the TR to a great extent). The other opposing force is the vertical component of MR thrust acting about the c of g, the weight acts through the c of g
If you increase TR thrust (the pulling the skids with a rope analogy) that gives more translating tendency and requires more cyclic to oppose it - this results in more left skid low to hold the hover because the balance of forces is achieved in a new position. If you hover the same helo at a. the lightest weight possible and b. max all up mass you will have significantly different hover attitudes (ignoring the change in the c of g position.
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This right rapid roll is not happening because you forgot to mention the counteracting moment of the lateral left vector of the mainrotor thrust to the CoG. In other words, if you position the Total thrust in the centre of the M/R hub, which is, in my opinion, the right place to put the Total Thrust, and it tilts to the left (as in Arm's picture) you have to consider all the forces that can be vectored from this Total Thrust force and and their arms to the CoG
cheers, YB
Yep, lifting, forgot about the down force on the elevator, without which the CofG would have to swing even further out the back until the couple it created with lift acting at the head stopped the motion.
Hi crab. The assumption I think you and Nick are making that I don't agree with is that every force should be thought of in relation to where it's acting with respect to the CofG.
When we change the forces acting on the helicopter, we don't necessarily make it move around the CofG, and there's no reason why it should - turns around the nose, tail or mast are easily done with different combinations of pedal and cyclic, for example.
In the rolling plane, it's the balance of forces as per that picture I drew that will determine which way it rolls and how far it goes.
You guys seem to be considering the CofG as the pivot point around which everything happens, whereas I'm looking at it simply as the point through which the weight of the helicopter can be said to act, and in that regard it's just another point at which force is being applied to the whole combination.
I'm talking about force couples, as um...lifting referred to.
Hi crab. The assumption I think you and Nick are making that I don't agree with is that every force should be thought of in relation to where it's acting with respect to the CofG.
When we change the forces acting on the helicopter, we don't necessarily make it move around the CofG, and there's no reason why it should - turns around the nose, tail or mast are easily done with different combinations of pedal and cyclic, for example.
In the rolling plane, it's the balance of forces as per that picture I drew that will determine which way it rolls and how far it goes.
You guys seem to be considering the CofG as the pivot point around which everything happens, whereas I'm looking at it simply as the point through which the weight of the helicopter can be said to act, and in that regard it's just another point at which force is being applied to the whole combination.
I'm talking about force couples, as um...lifting referred to.
Arm, the reason we are talking about the c of g is because any basic textbook on mechanics or physics will tell you that a force applied to an object will move said object relative to its c of g. If the force is applied directly at the c of g then the object will move in a straight line - if the force is offset ie there is a moment arm, then the object will rotate around that same c of g.
When you yaw the aircraft, you apply a force with the TR which is offset a long way from the c of g (any closer and the TR would have to be bigger to compensate for the loss of moment arm - think see saws). Without any further input the fuselage would rotate around its c of g. If you want to turn about the nose or tail, you must apply different forces using cyclic to modify the effect of the TR push and you will succeed when the mix of forces and moments achieves the required turn. But... all the forces you apply to the fuselage will act about the c of g whether or not the resultant manoeuvre is a pitch, roll, or yaw.
When you yaw the aircraft, you apply a force with the TR which is offset a long way from the c of g (any closer and the TR would have to be bigger to compensate for the loss of moment arm - think see saws). Without any further input the fuselage would rotate around its c of g. If you want to turn about the nose or tail, you must apply different forces using cyclic to modify the effect of the TR push and you will succeed when the mix of forces and moments achieves the required turn. But... all the forces you apply to the fuselage will act about the c of g whether or not the resultant manoeuvre is a pitch, roll, or yaw.
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[quote=Arm out the window;2907760]
You guys seem to be considering the CofG as the pivot point around which everything happens, whereas I'm looking at it simply as the point through which the weight of the helicopter can be said to act, and in that regard it's just another point at which force is being applied to the whole combination.
quote]
Arm, you are right that the A/C CoG is the point at which the total weight of the helicopter acts through, and it also is the pivot around which everything happens. The CoG is not a fixed point but it acts as one due to inertia. But you have to keep in mind that there are quite a lot of 'sub' CoG's. For instance the CoG of the rotor system, or a single blade. It is, however, impossible to consider all the forces acting on a flying helicopter to clarify it in a single two dimensional picture.
Coming back to the inertia around the CoG; consider the following:
Measuring the unbalance of the Main rotor is done with an accelero or veloci meter device. Why do we measure it in the lateral plane? The out of center CoG of the rotor system produces a moment to the A/C CoG. This moment is equally proportional around the mast centre. However the amplitude in the lateral plane is larger than the amplitude in the longitudinal plane, and this is a result from the fact that obviously the inertia of the complete body is higher in the longitudinal plane.
cheers,
YB
You guys seem to be considering the CofG as the pivot point around which everything happens, whereas I'm looking at it simply as the point through which the weight of the helicopter can be said to act, and in that regard it's just another point at which force is being applied to the whole combination.
quote]
Arm, you are right that the A/C CoG is the point at which the total weight of the helicopter acts through, and it also is the pivot around which everything happens. The CoG is not a fixed point but it acts as one due to inertia. But you have to keep in mind that there are quite a lot of 'sub' CoG's. For instance the CoG of the rotor system, or a single blade. It is, however, impossible to consider all the forces acting on a flying helicopter to clarify it in a single two dimensional picture.
Coming back to the inertia around the CoG; consider the following:
Measuring the unbalance of the Main rotor is done with an accelero or veloci meter device. Why do we measure it in the lateral plane? The out of center CoG of the rotor system produces a moment to the A/C CoG. This moment is equally proportional around the mast centre. However the amplitude in the lateral plane is larger than the amplitude in the longitudinal plane, and this is a result from the fact that obviously the inertia of the complete body is higher in the longitudinal plane.
cheers,
YB
AOTW, your statement that the aircraft can pivot about any point other than the CG is a bit questionable.
The "turn about the nose" and "turn about the tail" manoeuvres are co-ordination exercises where the whole aircraft is moved, using all controls, to make the nose or tail stay in one spot. Might as well fly it around a square and then say that the cg was in the centre of the square.
You gotta look at just the forces on the acft. And they do all tend to act about one central point, which can be the CG, or (like aerofoils) the Centre of Pressure, or the rotor head.
Look at the videos of a helicopter gyrating around with loss of T/R thrust. Some forces acting around the CG, others about the rotor head, and when those 2 don't line up, you get roll coupling and it spins towards its side and rolls up big time.
Toss a stick into the air. Apart from the translation caused by the toss, does the stick spin about its end? No, it spins about the cg. Tossing a caber looks like a spin about the end, but it is a combination of a turn about the cg and the movement of the cg in the air.
The "turn about the nose" and "turn about the tail" manoeuvres are co-ordination exercises where the whole aircraft is moved, using all controls, to make the nose or tail stay in one spot. Might as well fly it around a square and then say that the cg was in the centre of the square.
You gotta look at just the forces on the acft. And they do all tend to act about one central point, which can be the CG, or (like aerofoils) the Centre of Pressure, or the rotor head.
Look at the videos of a helicopter gyrating around with loss of T/R thrust. Some forces acting around the CG, others about the rotor head, and when those 2 don't line up, you get roll coupling and it spins towards its side and rolls up big time.
Toss a stick into the air. Apart from the translation caused by the toss, does the stick spin about its end? No, it spins about the cg. Tossing a caber looks like a spin about the end, but it is a combination of a turn about the cg and the movement of the cg in the air.
Morning all (or night as the case may be where you are)
AC, I agree that an object's motion can be broken down into rotation about the CofG and translation, and for a thrown object, that's fine.
For the helicopter, we have forces applied at different parts of the airframe working together to produce a motion.
A turn around the nose or tail isn't a turn around the CofG because we apply these forces in various amounts to achieve the aim and can pivot the helicopter around any point we like.
Nick and Crab seem to be saying that it's the relative moment arm around the CofG of the side forces from the main and tail rotors that's important in determining tail rotor roll.
There's a certain logic in this, but it doesn't tell the full story. Nick's diagram doesn't have weight in it, and he says that lift acts through the CofG.
I don't agree, and I think my diagram has all the appropriate forces in it, and explains tail rotor roll as the balance of two opposing force couples:
1. tail rotor thrust and main rotor side force acting to roll the fuselage left; and
2. lift and weight (lift through the hub and weight through the now laterally displaced CofG) acting to roll the fuselage right.
These are the determining factors, and the size of the left roll is affected by the relative vertical positions of the main and tail rotors, because the force couple has a larger moment arm - doesn't matter where the CofG is or whether these forces are acting through it or not, there's two opposing couples, and the body will roll until they're equal in magnitude.
The upshot of all this is that the relative heights of the main and tail rotors is important in the equation, which Nick in particular seems to be bagging as a theory. The 'vertical position of side forces with respect to the CofG' idea that he and Crab are touting sort of says the same thing, but it doesn't go far enough in consideration of the force couples.
AC, I agree that an object's motion can be broken down into rotation about the CofG and translation, and for a thrown object, that's fine.
For the helicopter, we have forces applied at different parts of the airframe working together to produce a motion.
A turn around the nose or tail isn't a turn around the CofG because we apply these forces in various amounts to achieve the aim and can pivot the helicopter around any point we like.
Nick and Crab seem to be saying that it's the relative moment arm around the CofG of the side forces from the main and tail rotors that's important in determining tail rotor roll.
There's a certain logic in this, but it doesn't tell the full story. Nick's diagram doesn't have weight in it, and he says that lift acts through the CofG.
I don't agree, and I think my diagram has all the appropriate forces in it, and explains tail rotor roll as the balance of two opposing force couples:
1. tail rotor thrust and main rotor side force acting to roll the fuselage left; and
2. lift and weight (lift through the hub and weight through the now laterally displaced CofG) acting to roll the fuselage right.
These are the determining factors, and the size of the left roll is affected by the relative vertical positions of the main and tail rotors, because the force couple has a larger moment arm - doesn't matter where the CofG is or whether these forces are acting through it or not, there's two opposing couples, and the body will roll until they're equal in magnitude.
The upshot of all this is that the relative heights of the main and tail rotors is important in the equation, which Nick in particular seems to be bagging as a theory. The 'vertical position of side forces with respect to the CofG' idea that he and Crab are touting sort of says the same thing, but it doesn't go far enough in consideration of the force couples.
Last edited by Arm out the window; 15th Oct 2006 at 00:12. Reason: To say good morning
OK arm, what happens when the c of g moves (it changes horizontally, vertically and laterally as you add or subtract fuel, crew, pax, internal or external load). If in your diagram, the c of g has inconveniently shifted itself so it is towards the port side of the aircraft then your couple between weight and rotor thrust will keep rolling the aircraft over.
If it is the relative positions of the main and TRs that dictate hover attitude, then why doesn't an R22 (very high rotor mast and low mounted TR) hover 15 degrees left wing low?
Saying that the c of g is irrelevant in the opposing couples is like trying to make a lever work without a fulcrum. Every control input you make creates a force that acts about the c of g to achieve the desired pitch roll, or yaw.
If it is the relative positions of the main and TRs that dictate hover attitude, then why doesn't an R22 (very high rotor mast and low mounted TR) hover 15 degrees left wing low?
Saying that the c of g is irrelevant in the opposing couples is like trying to make a lever work without a fulcrum. Every control input you make creates a force that acts about the c of g to achieve the desired pitch roll, or yaw.
Hi guys.
Crab, I'm not saying the CofG is irrelevant - being one of the points where a force is applied, where it is will make a big difference to the end result.
If, as you say, the CofG was inconveniently to the port side, it would become part of a pro-roll couple. The aircraft would roll until it had gone far enough for the CofG to swing starboard enough to revert back to anti-roll, ie had passed under the mast head and out the other side.
It'd keep going until the lift-weight couple was big enough to equal the pro-roll one.
Whether the CofG would have been within limits in the first place in this scenario is a question for design engineers, which I'm definitely not one of.
The R22, like anything, will hang in equilibrium when all force couples are balanced. Not a great deal of torque to spin that rotor, comparitively speaking, so not heaps of anti-torque required, so my guess there is that the smallish side force from the tail rotor and proportionally small cyclic tilt needed would mean that couple wouldn't be huge in the first place.
Lifting, I see your point about the theoretical vs. reality significance of this effect - and hopefully they'll get Uncle Bert out of the trunk before he starts to stink too much!
Crab, I'm not saying the CofG is irrelevant - being one of the points where a force is applied, where it is will make a big difference to the end result.
If, as you say, the CofG was inconveniently to the port side, it would become part of a pro-roll couple. The aircraft would roll until it had gone far enough for the CofG to swing starboard enough to revert back to anti-roll, ie had passed under the mast head and out the other side.
It'd keep going until the lift-weight couple was big enough to equal the pro-roll one.
Whether the CofG would have been within limits in the first place in this scenario is a question for design engineers, which I'm definitely not one of.
The R22, like anything, will hang in equilibrium when all force couples are balanced. Not a great deal of torque to spin that rotor, comparitively speaking, so not heaps of anti-torque required, so my guess there is that the smallish side force from the tail rotor and proportionally small cyclic tilt needed would mean that couple wouldn't be huge in the first place.
Lifting, I see your point about the theoretical vs. reality significance of this effect - and hopefully they'll get Uncle Bert out of the trunk before he starts to stink too much!
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I haven't done much here the last few days, mostly due to weekend LIFE, guys, but I thought I toss in these little facts:
1) The Comanche hovered 6 degrees left wheel low because the TR (fan actually) was low and almost perfectly lined up with the CG. That was because we didn't want it higher - why add an intermediate GB and why should there be a large yaw to roll coupling in sideslips at high speed flight (where the TR yaw restoring force would roll the aircraft as well as put the nose back straight. - BTW, why does a high TR create MORE roll in forward flight, in spite of that same TR-MR height issue???) The design team agonized about this, and even contemplated tilting the crew seats in a hover to help "fix" the problem.
2) The Fantail demonstrator hovered at 5 degrees left wheel down, vice the S76B that it began life hovering at 3 degrees.
The items we are discussing are not theoretical, and Crab and I are not just supposing, these facts are part of what one does when one designs these things.
Until you guys understand wat a free body diagram is, it is not terribly useful to sketch, debate and sketch again, fun though it might be.
1) The Comanche hovered 6 degrees left wheel low because the TR (fan actually) was low and almost perfectly lined up with the CG. That was because we didn't want it higher - why add an intermediate GB and why should there be a large yaw to roll coupling in sideslips at high speed flight (where the TR yaw restoring force would roll the aircraft as well as put the nose back straight. - BTW, why does a high TR create MORE roll in forward flight, in spite of that same TR-MR height issue???) The design team agonized about this, and even contemplated tilting the crew seats in a hover to help "fix" the problem.
2) The Fantail demonstrator hovered at 5 degrees left wheel down, vice the S76B that it began life hovering at 3 degrees.
The items we are discussing are not theoretical, and Crab and I are not just supposing, these facts are part of what one does when one designs these things.
Until you guys understand wat a free body diagram is, it is not terribly useful to sketch, debate and sketch again, fun though it might be.
Well, OK, Nick - as I said before, your experience speaks for itself and you've been a hell of a long way further in the aviation game than I have, although I've done a fair bit of varied and interesting flying in my time.
I do know a little bit about mechanics though, having a science degree majoring in physics, for what it's worth, and that's why I've been challenging some of the stuff that has come up in this thread.
Check your own diagram back in the early stages of this thread - where's weight? That's the flaw. You are entitled to speak with a great deal of authority, of course, and all respect to your achievements, but that's a fundamental omission that skews the rest of the argument.
It's no biggie in the grand scheme of things, I guess, but when you come in with statements like this:
"Until you guys understand wat a free body diagram is, it is not terribly useful to sketch, debate and sketch again, fun though it might be.", I feel I have to respond - diagrams at dawn at ten paces, perhaps!
I do know a little bit about mechanics though, having a science degree majoring in physics, for what it's worth, and that's why I've been challenging some of the stuff that has come up in this thread.
Check your own diagram back in the early stages of this thread - where's weight? That's the flaw. You are entitled to speak with a great deal of authority, of course, and all respect to your achievements, but that's a fundamental omission that skews the rest of the argument.
It's no biggie in the grand scheme of things, I guess, but when you come in with statements like this:
"Until you guys understand wat a free body diagram is, it is not terribly useful to sketch, debate and sketch again, fun though it might be.", I feel I have to respond - diagrams at dawn at ten paces, perhaps!
Gents, I suggest you get yourself a copy of Ray Prouty's books and read about this and other helicopter issues from an engineer's perspective.
What Nick and I are trying to get over is that the basic P of F from Wagtendonk and others is exactly that - basic. I was taught P of F and I have taught P of F to students but you soon realise it is just a convenient way of explaining what we know happens, rather than empirical proof that it does.
When a guy who was chief test pilot for Sikorsky tells you that these things happen for certain reasons, why don't you believe him? The higher TR provides a rolling couple to the right in fwd flight because it is above the C of G! (that is not rocket science nor secret test pilot stuff). You might not even believe me when I say that TR drift doesn't go away in forward flight either and helicopters sideslip constantly which still has to be overcome with cyclic.
Arm, if you look on Nick's diagrams you will see that the c of g is clearly marked and since the weight of an object is always deemed to act through that c of g why does it need an arrow as well? Just because it doesn't match the P of F book diagram doesn't mean it is wrong.
What Nick and I are trying to get over is that the basic P of F from Wagtendonk and others is exactly that - basic. I was taught P of F and I have taught P of F to students but you soon realise it is just a convenient way of explaining what we know happens, rather than empirical proof that it does.
When a guy who was chief test pilot for Sikorsky tells you that these things happen for certain reasons, why don't you believe him? The higher TR provides a rolling couple to the right in fwd flight because it is above the C of G! (that is not rocket science nor secret test pilot stuff). You might not even believe me when I say that TR drift doesn't go away in forward flight either and helicopters sideslip constantly which still has to be overcome with cyclic.
Arm, if you look on Nick's diagrams you will see that the c of g is clearly marked and since the weight of an object is always deemed to act through that c of g why does it need an arrow as well? Just because it doesn't match the P of F book diagram doesn't mean it is wrong.
Crab, I guess we're not going to come to agreement here.
I'm simply saying that any object, be it helicopter, house or hippopotamus, will move (or not) as determined by the size, position and direction of the forces acting on it.
My diagram shows those forces for a hovering helicopter, and I believe my argument about the force couples stands.
Yes, Nick is a greatly experienced individual and no doubt knows more about flying than I'll ever do if I live to be a hundred, but as he himself said, it's simple mechanics.
Rotor thrust, depending on its size and direction, can be resolved into lift and side force. The extension of that vector downwards may, or may not, pass through the CofG.
Changes in size and direction of any of the forces acting on the helicopter will cause it to move, and not necessarily around the CofG.
Please check my diagram and tell me what's wrong with it, and also with the answers I gave to your points of order before.
Anyway, if we meet at some point, I'll buy you guys a beer for a good robust discussion.
I'm simply saying that any object, be it helicopter, house or hippopotamus, will move (or not) as determined by the size, position and direction of the forces acting on it.
My diagram shows those forces for a hovering helicopter, and I believe my argument about the force couples stands.
Yes, Nick is a greatly experienced individual and no doubt knows more about flying than I'll ever do if I live to be a hundred, but as he himself said, it's simple mechanics.
Rotor thrust, depending on its size and direction, can be resolved into lift and side force. The extension of that vector downwards may, or may not, pass through the CofG.
Changes in size and direction of any of the forces acting on the helicopter will cause it to move, and not necessarily around the CofG.
Please check my diagram and tell me what's wrong with it, and also with the answers I gave to your points of order before.
Anyway, if we meet at some point, I'll buy you guys a beer for a good robust discussion.
Arm, I completely agree with about 90% of your argument and the only area we seem to differ on is how the forces affect the fuselage.
You say that the rotor thrust can be resolved into vertical and horizontal components - I agree! You say that the extension of the vector downwards may or may not pass through the c of g - I agree!
The distance that the vector passes the c of g by whether it be 1mm or 1m is the moment arm and is effectively the size of the lever the force can act on the fuselage. You don't have to have another force pulling in the opposite direction, eg a couple, to create a rotation - you just need one force acting at a distance from the objects c of g. The object will rotate until the moment arm reduces to zero and then the object will translate in the same direction as the force is acting.
The weight will always act through the c of g and therefore has no moment arm so in your diagram it cannot produce a rotation, only a translation which is opposed by the vertical component of rotor thrust. The rotor thrust might act on either side of the c of g and may try to create a rotation either left right but it will generally have a very small moment arm to work with as designers try to keep it within a sensible range as the load, fuel, pax etc change.
The only thing left to combat the rotation caused by the horizontal component of rotor thrust is the TR thrust and its vertical distance above the c of g is what will determine tha final roll angle in the hover. If the c of g was very high or the TR was very low, you would end up with a couple that would keep rotating the fuselage, that's why the c of g is kept low and the TR positioned above it.
Looking forward to the beer.
You say that the rotor thrust can be resolved into vertical and horizontal components - I agree! You say that the extension of the vector downwards may or may not pass through the c of g - I agree!
The distance that the vector passes the c of g by whether it be 1mm or 1m is the moment arm and is effectively the size of the lever the force can act on the fuselage. You don't have to have another force pulling in the opposite direction, eg a couple, to create a rotation - you just need one force acting at a distance from the objects c of g. The object will rotate until the moment arm reduces to zero and then the object will translate in the same direction as the force is acting.
The weight will always act through the c of g and therefore has no moment arm so in your diagram it cannot produce a rotation, only a translation which is opposed by the vertical component of rotor thrust. The rotor thrust might act on either side of the c of g and may try to create a rotation either left right but it will generally have a very small moment arm to work with as designers try to keep it within a sensible range as the load, fuel, pax etc change.
The only thing left to combat the rotation caused by the horizontal component of rotor thrust is the TR thrust and its vertical distance above the c of g is what will determine tha final roll angle in the hover. If the c of g was very high or the TR was very low, you would end up with a couple that would keep rotating the fuselage, that's why the c of g is kept low and the TR positioned above it.
Looking forward to the beer.
Join Date: Apr 2003
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Gents,
Lets not forget what is the point I made:
The tail rotor's position above the CG is the operative term for the contribution it makes to hover roll attitude, and the height of the rotor head relative to the height of the TR is an urban myth regarding same.
I am not sure we agree on that (I am also quite sure the physics does not count our votes).
Lets not forget what is the point I made:
The tail rotor's position above the CG is the operative term for the contribution it makes to hover roll attitude, and the height of the rotor head relative to the height of the TR is an urban myth regarding same.
I am not sure we agree on that (I am also quite sure the physics does not count our votes).