another terrible ATPL PoF question [Archive]

Arm out the window

13th Oct 2006, 03:43

lifting, height of the tail rotor with respect to the main rotor hub does affect the roll - I agree with AC that the Huey T/R was set high to reduce the effect.
Nose wants to yaw right, and tail left, because of rotor torque.
Tail rotor forces tail right to counter this.
Now you have a side force making helicopter want to move right (tail rotor drift), so we introduce left cyclic to counter it.
If the tail rotor is lower than the main rotor hub, this creates a force couple (left from M/R hub, right from T/R when viewed from the rear) which rolls the fuselage left until an equal and opposite vertical couple formed by lift (up from the M/R hub) and weight (down through the CofG) is reached.
The closer the tail rotor is to the M/R hub, vertically speaking, the less the rolling couple.
The tail rotor on the Huey is high-set, but not as high as the main rotor hub, particularly with an aft CofG, so there is a roll because of it - slight disagreement with AC there - but it's certainly lessened by the higher mounted tail rotor than if it was just on the side of the boom.

[email protected]

13th Oct 2006, 05:28

The point of raising the TR is to give it a better moment arm about the C of G. The horizontal component of the MR thrust is trying to roll the fuselage left about the C of G so mounting the TR high gives it an equivalent lever in the opposite direction.

If you are building a helicopter that will spend most of its working life in the hover, it makes sense raise the TR to produce a more level (laterally) hover attitude which keeps the cabin floor level and stops things like winch cables fouling the fuselage (see the Merlin/Cormorant for this problem).

If you are building a helicopter for high speed cruising then you probably won't bother to raise the TR as it adds weight and another gearbox at the back end.

Battlefield helicopters do benefit from having a higher TR as it is safer both from ground clearance and troop clearance poits of view.

Arm out the window

13th Oct 2006, 06:59

I think we're pretty much in agreeance, lifting.
The part of your post that got me thinking was this:

"Nope, disagree. Height of the t/r has nothing to do with left low.", when all that couple stuff I was blathering on about convinces me that it does.

Anyhow, as you say, I just lift to the hover and hang whichever way the machine wants to.

NickLappos

13th Oct 2006, 15:05

Thank heaven for Crab. The myth that the Tail Rotor's position relative to the Main Rotor Head is a factor in the hover attitude has somehow evolved, much to the surprise of those who actually build these things.

For the record:

the tail rotor makes the left wheel hang low in a hover, due to the sum of two competing factors:
1) the translating tendency (TR thrust tries to move the helo to the right, it takes left bank to stop it)
http://webpages.charter.net/nlappos/attitude1.jpg
2)TR thrust acts to rotate the fuselage, depending on how far above or below the tail rotor is relative to the CG. The higher the tail rotor, the flatter the hover attitude, because the TR roll component cancels the translating tendency roll attitude. the lower the TR, the less cancellation, the more pronounced the roll. (this factor probably lead the aerodynamically challenged to deduce the "theory of rotor heights" explanation.
http://webpages.charter.net/nlappos/attitude2.jpg

HeliComparator

13th Oct 2006, 15:52

Nice picture Nick but a pity you made a fundamental mistake - rotors on proper helicopters go clockwise!

HC

NickLappos

13th Oct 2006, 15:58

Yea, if viewed from the bottom.....;)

HeliComparator

13th Oct 2006, 16:16

Aw shucks - you didn't bite very hard - must be getting soft!

HC

Arm out the window

14th Oct 2006, 00:14

Nick, You said: "The myth that the Tail Rotor's position relative to the Main Rotor Head is a factor in the hover attitude has somehow evolved, much to the surprise of those who actually build these things."
I know you're a vastly experienced test pilot, and maybe I'm not thinking straight about the subject, but can you please shed some light on the following:

In your picture above, the translating and anti-translating forces, as you've labelled them, are vertically displaced. This creates a force couple, making the aircraft roll.
How far does it roll?
Until the opposing couple formed by the lateral displacement between the CofG (from which weight acts straight down) and the rotor hub (from which lift acts straight up) becomes large enough to be equal and opposite to it.
Obviously the greater the roll angle, the greater the displacement between the lateral positions of lift and weight, and so the more the opposing force.

Likewise, the pro-roll couple formed by the translating and anti-translating forces gets stronger the more vertically displaced they are from eachother, so it seems to me that the vertical position of the tail rotor with respect to the main hub is a major factor in how much roll you'll get.

You said: "The myth that the Tail Rotor's position relative to the Main Rotor Head is a factor in the hover attitude has somehow evolved, much to the surprise of those who actually build these things."
To my mind, the vertical displacement between these two determines the strength of the pro-roll couple, and is therefore far from mythical.

To put it another way, say you had a helicopter with a standard main rotor, central CofG, and two antitorque rotors, one on the nose pushing left and one on the tail pushing right.
If the two antitorque rotors were at the same vertical level, there wouldn't be any rolling.
If they were significantly displaced vertically, there'd be lots of roll, increasing with the displacement.
How is this any different to the main rotor / single tail rotor situation?

I rest my case, your honour.:)

NickLappos

14th Oct 2006, 00:49

Arm,
It is unfortunate that we cant just get to a chalk board, it would be so much easier.

Your picture is quite selective, you see the translating component of the rotor thrust, but you don't see the lift portion, which precisely cancels your argument.

One must resolve the forces to the CG to understand their effect of the motions of the body. For the main rotor thrust, for some convenience I show the vector summed at the head. In fact, the main rotor thrust is a single force vector that passes through the aircraft and directly through the CG. Otherwise, the rotor would be rotating the helo (which it does when we maneuver.)

The idea that the height of the rotor is a driver in this roll angle discussion is simply not correct. The rotorhead height DOES help the cyclic control the helo, because the higher the rotorhead, the more lever arm the thrust has when the pilot purposely tilts the lift away from the CG. For teetering rotors, high rotorheads are needed to make the cyclic more powerful. That is why Hueys and Robbies have high masts and Boelkows and Sikorskys have heads tucked down close to the fuselage.

With your dual TR helo (nice idea and a great "thought experiment, Arm!) there would be no translating tendency, because the two would cancel out the lateral force imbalance. Therefore there would be no need for the pilot to tilt the lift to stop the slide, and the helo would hover level. It would hover level no matter how high or low those pesky tail rotors were, as long as they were on the same height. It would hover level no matter how high or low those tail rotors were relative to the main rotor head, as well.

Arm out the window

14th Oct 2006, 02:05

http://i109.photobucket.com/albums/n52/tborella/TRRoll.jpg
Thanks for the discussion, guys.

Sorry, I'm starting to feel a bit thick, but the force diagram as I see it is like this.
The vertical couple between translating force and anti-translating force rolls you.
As the CofG moves laterally with respect to the head, the horizontal couple formed between the two becomes a more powerful restoring force until the two couples balance.
If there was more vertical distance between the tail rotor and the main rotor hub, there'd be a greater rolling force.
Why is this wrong?

NickLappos

14th Oct 2006, 03:01

Arm out the window

14th Oct 2006, 03:29

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.

NickLappos

14th Oct 2006, 03:56

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!

Arm out the window

14th Oct 2006, 04:28

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.

Arm out the window

14th Oct 2006, 05:13

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.

[email protected]

14th Oct 2006, 06:19

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.

yellowbird135

14th Oct 2006, 06:48

Arm,
The picture you show is of a helicopter during a very rapid right rolling maneuver. Note thet the lift vector does not pass thru the CG, so there is one hell of a right rolling moment on the helo.

Nick,
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

Arm out the window

14th Oct 2006, 06:48

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.

[email protected]

14th Oct 2006, 08:41

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.

yellowbird135

14th Oct 2006, 08:44

[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

Ascend Charlie

14th Oct 2006, 23:13

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.

Arm out the window

15th Oct 2006, 00:10

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.

[email protected]

15th Oct 2006, 13:33

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.

Arm out the window

15th Oct 2006, 22:06

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!

NickLappos

15th Oct 2006, 22:33

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.

Stringfellow Dork

15th Oct 2006, 23:13

How's that book coming along, Nick?!

:ok:

Arm out the window

16th Oct 2006, 00:39

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! :)

[email protected]

16th Oct 2006, 04:56

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.

Arm out the window

16th Oct 2006, 07:07

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.

[email protected]

16th Oct 2006, 11:32

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.:)

NickLappos

16th Oct 2006, 12:55

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).

Stringfellow Dork

16th Oct 2006, 13:02

It's more than an urban myth - it's what we are taught by those that set the tests to make sure we reach the required standards! What a joke.

:ugh:

Arm out the window

17th Oct 2006, 01:35

CAUTION - DIAGRAM ALERT!

http://i109.photobucket.com/albums/n52/tborella/BagCGWeb.gif

I talked about this situation in one of my previous posts, so at the risk of being called a gratuitous diagrammist, here's one.

In the top row, if the bag was hanging out to one side and let go, it would swing and come to rest with the CofG aligned with the suspension hook - no surprises there.

In the lower row, left side, a calibrated pull is applied to the bag halfway up.
The bag moves out, and when it gets to as far as it's going to go, two opposing couples are holding it still:
1. The side forces from the suspension hook and the spring balance, trying to move it out, and
2. The vertical forces, up from the hook and weight down from the CofG.

The relative power of these couples comes from two things - the size of the forces, and the arm between them.

Now to the right hand picture - same everything, except the spring balance is now attached to the bottom of the bag. Because the arm is now bigger, the pro-swing couple is greater and so it has to swing further right before the opposing couple (weight and suspension force) is big enough to balance it. Obviously the further it goes, the more of an arm there is for weight to act as a restoring force.

So my point is that this is just like the hovering helicopter, and the distance between the suspension point and where the lower side force is applied DOES matter in how far the thing will roll (or in this case, swing out).

Right, no more diagrams from me now, I promise.

[email protected]

17th Oct 2006, 05:29

Nice diagrams Arm, I still don't know how to get things like that into posts.

However, your bag is fixed at one end and is not therefore a free body - if you could unhook the bag without it falling to the floor and repeat the experiment, the bag would rotate about its c of g until the pull was aligned with it and then the bag would be pulled along.

Because your bag is fixed at one end it can't rotate around its c of g, only around the hook - this might be possible in a helicopter but only if you forget to unplug the ground power set before you get airborne.

I do understand what you are getting at but opposing couples isn't the answer to TR drift and roll. I see your equal and opposite reaction line of thought which produces a couple but one force has created the other which is not the case in the helicopter hovering.

Arm out the window

17th Oct 2006, 06:18

I thought that might be a sticking point - if that's a concern, then replace the ceiling hook with a balloon that just lifts the weight, and a preferably bikini-clad assistant applying the appropriate equal and opposite side force with another spring balance where the bag hangs off it. The equal side forces will stop any lateral motion anyway (as is the case when we apply an appropriate amount of cyclic to counteract tail rotor drift), lift will still equal weight, and the rest of it should work the same.
In that case though, we might as well use the original picture of the helicopter.

As regards whether the helicopter moves around the CofG or not, it doesn't really matter in this case - the basic force diagram should remain the same. Moving the tail rotor further down with respect to the head increases the arm for the lateral couple - same forces and bigger arm equals more torque, therefore more roll.
That in turn will move the head (the suspension point) laterally with respect to the CofG, so creating a wider arm for the vertical couple (lift and weight) and increasing the restoring force provided by it, resulting in a new equilibrium being reached at a greater angle of bank.
I see where you're coming from in saying if the aircraft rotates about the CofG, weight has no arm and therefore won't be a rotating force, but in a complete force diagram, all forces need to be considered, and it's the combination of all that will determine the resulting motion, be it around the CofG or some other point.

Re putting pictures in to posts, I only found out recently, but it's dead set easy - go to photobucket.com, get a log-in name and password, then you can upload pictures to that site.
The site generates a number of references to your picture, and by copying and pasting one of them (the bottom one of three, starts with IMG or something like that, from memory) into your Prune post at the appropriate position, people subsequently reading your post will see your picture, courtesy of the photobucket site.

Cheers

[email protected]

17th Oct 2006, 14:44

Arm, thanks for the photobucket info, I'll give it a go soon.

Back to the question - No! moving the TR down relative to the MR doesn't increase the moment arm - it decreases it because the moment arm is the distance vertically from the TR to the C of G - this is what Nick and I have been trying to get across. However, it does increase the roll angle because the TR force has a shorter lever to affect the c of g.

Try this - if we don't correct for TR drift and allow the aircraft the slide to the right, the height of the TR above or below the aircraft c of g will determine the roll angle - if it is at the same height as the c of g there will be no roll. This happens without any couple. Since the c of g will tend to be lower than the TR in most helicopters the TR will not only provide a translation to the right (presuming we are holding the heading and not allowing the aircraft to yaw) but will also tend to roll the fuselage to the right - the higher the TR above the c of g, the more roll to the right.

Now we apply left cyclic to stop the drift which rolls the fuselage left because the horizontal component of MR thrust is above the c of g. There are two rolling forces, opposing each other and when they are equal the hover attitude is decided. The vertical component of MR thrust may either help or hinder either of the rolling moments depending on the position of the c of g laterally but is a bit part player compared to the main actors and is primarily there to stop the weight pulling it all onto the ground.

Arm out the window

17th Oct 2006, 22:43

I see we're kind of beating around the same bush with slightly different viewpoints.

You say that lowering the tail rotor shortens its arm about the CofG compared to that of the main rotor side force, therefore reducing its effectiveness as a rolling agent - result, more left roll.
I say that lowering the tail rotor increases the arm of the MR side force / TR side force couple, increasing its power and ... creating more left roll.

From the hover, if we level the disc and let the aircraft drift, the left side force from the head goes away. Because the couple mentioned above no longer exists (no MR side force), the body is now free to hang so the CofG is under the head - it's now rolled right compared to how it was in the hover.
As it drifts, drag will come into play, and a rolling couple will be set up, the direction of which will depend on the relative vertical positions of the centre of pressure and the tail rotor side force.

Applying left cyclic to stop restores the original couples, rolling it left again compared to how it was in the drifting state.

imabell

17th Oct 2006, 22:52

crab,

the diagrams are generally saved as .jpg images or gif's like the photos that are posted. right click on arms diagram and look at the properties.

if you construct a diagram and photograph it and email it to me i will put it on the server so you can add it to your post.

there are other ways to do it but you need the programming.

gg

NickLappos

18th Oct 2006, 02:58

Arm,
You still have it quite wrong. The height of the main rotor has NOTHING to do with it, regardless of how many cute diagrams that you draw that make no physical sense. If the MR is 100 feet above the TR, the only thing that determines the degree of roll reduction is the height of the TR above the CG.
THERE IS NO SIDE FORCE STUCK ON THE ROTORHEAD. The main rotor thrust acts as a force vector that runs from the rotor centroid at an angle that runs through the CG (or very very close to the CG). If it does not run through the CG, then the helo will rotate a lot because the pilot has asked for a roll or a pitch rate, and all bets are off. This force can be pictured as a pair of forces AT THE CG that are at right angles, one parallel to the earth, one perpendicular.

It is not easy to picture, but what I am saying has nothing to do with flying experience, it has to do with the simple physics of the situation. If you insist on drawing aircraft that are not in a balanced forse/moment state, or if you insist on drawing forces stuck like arrrows where it seems convenient, you will get (stay?) very confused on this matter.

Arm out the window

18th Oct 2006, 03:28

Nick, I haven't stuck force arrows where it's convenient, I've stuck them where I think they act on the body of the helicopter.
Weight acts down through the CofG.
Rotor thrust, depending on disc tilt, collective setting and rpm makes some force vector that for convenience I split into lift, pointing straight up, and side force. The rotor attaches to the body at the head, and that's where the forces get applied to it - where else? Not the CofG. If you could take the rotor off and apply an equivalent force to the head with a rope, why would that act through the CofG?
Likewise for the tail rotor - it makes a force vector that gets applied to it's hub attachment point.
In the hover, that's pretty much it for forces, unless you count downwash on the upward-facing surfaces.
In translational movement, drag will come into play.

I guess that doesn't convince you, and your view doesn't convince me, so there we are!

Regards

NickLappos

18th Oct 2006, 03:42

Like the drunk said to the frog, "I'm gonna show you one more time..."

Here is what it is like with a REAL diagram of the forces as they REALLY are Note that the MT thrust passes close to the CG, but misses by a bit, so that the MR thrust makes some moment to help cancel the TR moment. If the TR was directly on the CG, the Main thrust would pass right thru the CG. (I used your diagram cause its better than mine, by a long shot!):

http://webpages.charter.net/nlappos/lowTR.jpg

For a High TR, here is what happens:
http://webpages.charter.net/nlappos/highTR.jpg

eagle 86

18th Oct 2006, 03:57

Nick,
Surely, after four pages, we have come back to what I was taught at CFS Tern Hill in 1972 ie designers try to align the thrust of the tail rotor with the thrust of the main rotor to help eliminate "tail rotor roll"?
GAGS E86

Arm out the window

18th Oct 2006, 04:11

Thanks, Nick.
I can follow your line of reasoning, and it's consistent within itself, the thing in it that's got me stuffed is why the force produced by the main rotor doesn't act at the rotor head? I mean, it's a force, and it drags on the fuselage, and that's where it's attached.

Another thing that seems to throw a spanner in the works is the case you mentioned, where the tail rotor is level with the CofG.
If, as you say, the main thrust then passes through the CofG, then according to your view, the helicopter would be sitting wings level, in fixed wing speak, but with tail rotor thrust making it drift right.
Presumably we'd add some left cyclic to stop the drift, tilting the main thrust vector a bit right of the CofG, which would also create a left rolling moment. What then would stop that roll? If the body's rotating around the CofG, it wouldn't be weight shift, and as the tail rotor thrust passes through the CofG, it couldn't be that, either.

In my (possibly confused) world view, it's those varying couples coming to equilibrium that'd do it.

[email protected]

18th Oct 2006, 05:36

Hurray.....one converted, one to go:)

Arm out the window

18th Oct 2006, 07:52

Hey, not so fast there - I ain't caved in yet!
Lifting, sure, forces applied anywhere can be resolved into a force through and a moment about the CofG - happy with that, but isn't a consideration of forces on the helicopter via their size, direction and direct point of application also a legitimate way to go?
Granted, the balloon and spring balance idea wasn't kosher - now, if the bikini girl was in the balloon basket with a little fan to provide the side force, that'd be different, but that's probably taking it a bit far... or perhaps she could be nude?? Now we're talkin'!

Seriously though, let me get this straight about your feather pen scenario - wouldn't the force of the fan act at the centre of pressure, surely over towards the feathery end? And if so, wouldn't that be resolved into a translating force at the CofG plus a moment around it, causing it to turn as well as move away?
If your aim was to get gravity out of the picture, isn't what you said like having a long regular metal beam, say, floating steady in space, putting a small rocket motor on one end of it at right angles to the long axis and firing it? Would you get a simple translation opposite the direction of rocket thrust, or would it turn as well because of the moment around the CofG?

[email protected]

18th Oct 2006, 08:17

OK Arm - another analogy for you...

Sit down on the end of a see-saw - you are applying a force at the end of the plank but the see-saw plank rotates around the fulcrum. The sitting down force equates to the horizontal component of rotor thrust (or the TR thurst), the length of the see-saw on that side of the fulcrum is the moment arm (height of TR or MR above c of g) and the fulcrum is the c of g.

There is no couple, just a single force causing a rotation.

Arm out the window

18th Oct 2006, 09:31

Yes, but the see-saw fulcrum is physically fixed to the spot, not so with the helicopter.

[email protected]

18th Oct 2006, 10:16

Yes, I thought that would be your argument but I thought it might help you visualise that there doesn't need to a be a couple to produce a rotation, just a force acting at a distance from the c of g.

I think we have got to the point where you either believe this is true or continue to convince yourself that opposing couples is the answer - I have run out of analogies to try and illustrate the point.

Maybe one more - imagine a helicopter hovering facing North (no wind, perfect hover), now attach a rope to the nose and pull from a position directly East of the helicopter. First the helicopter will rotate to point East and then (if you are strong enough) translate on an Easterly heading.

Why? Because the initial application of the force was offset from the C of g and produced a rotation - once the force was aligned with the c of g, the rotation stopped and the translation began (no moment arm).

Again, no couple, just one force acting about the c of g.

delta3

18th Oct 2006, 12:28

Nick

nice picture, but if you allow me I would suggest to make a difference between the banking of the rotor and the roll of the heli

the rotor bank opposes the tail rotor trust, a simple static calculation allows one to calculate the required bank (arcus sinus ... etc).

The banking of the heli depends on the moments of the tail rotor trust, the stiffness/excentricity of the rotor (if any) and the position of the CoG. (Not implying you don't know that for the sake of clarity)

d3

NickLappos

18th Oct 2006, 12:41

delta3, your understanding is impeccable. I left out discussions of eccentricity (hinge offset) and the associated head moment for clarity in this discussion.
To all, if the helo has a "rigid" rotor or a high offset articulated, it will require less lateral flapping, and less roll angle, because the rotor will deliver a pure moment in addition to the tilt of its thrust to counter the TR roll effect.
None of this suggests that the height of the MR relative to the TR is of any significance, of course.

thanks Crab and Ummm for diving in, your insights are really helpful to all of us!

Arm, keep asking and pondering, these explanations help me keep in shape, frankly!!

Exo.

18th Oct 2006, 15:15

Nick, in your diagram, you provide the TR Thrust acting out of the tail, but do not set the MR thrust acting from the hub.

I hate to provide another voice of dissent, but I'm also not swayed by this argument. I think I can't get my head around the prospect that the MR thrust acts from the hub, at an arm from the CG, just like the TR Thrust acts from the TR hub at an arm from the CG.

I'm happy that these forces act -about- the CG, but don't see why they act through it. And since you won't get a turning moment if they act through it, only a vector, well, this is where I don't get it I guess.

Thus I lean to Arm's point of view that the two couples - MR Vertical Thrust component and weight acting at CG versus TR thrust and MR thrust Horizontal component - provide the roll, and then balance each other out to provide the angle at which the helicopter naturally falls.

Hence with varying weight, you find it rolls varying amounts.

NickLappos

18th Oct 2006, 22:43

Exo,
The TR has no controlled flapping, its thrust is reasonably assumed to be perpendicular to its disk, and its disk is reasonably assumed to be perpendicular to the helos vertical axis. Thus it is shown differently from the main rotor thrust which is the subject of a great deal of controlled flapping via cyclic.

The roll angle changes with weight because it changes with the power drawn by the MR, which requires more TR thrust to offset the torque. The roll angle for most helos is 1.5 degrees greater in HOGE than in HIGE!

The issue we face here is that we are trying to teach Statics (a branch of mechanics for engineers) to intelligent people who have not had it, and to do so through one (albeit complex) scenario. It is asking an awful lot.

I could try this for you:

No rotorcraft designer believes you. Furthermore, any instructor who said the TR was located to somehow achieve a hover roll angle correction if full of crap, since no designer would spend one pound of weight to appease a pilot! ;)

Arm out the window

19th Oct 2006, 06:02

Lots of points to ponder, Lifting - I was wondering about those links in your posts for a while, then finally decided to click on them - great stuff!

More thinking required for me - even if we do talk about the helicopter rolling around its cg, I'm finding it hard to get past thinking of the two force couples idea - one between lift at the head and weight, and the other between side force at the head and tail rotor thrust.
I can follow the rolling moments around the cg idea OK, but the other way still makes sense to me, too.

Looking back at some of the posts here, it seems that both ways come up with very similar answers - for example, looking at Nick's picture there where the higher tail rotor gets the machine to straighten up and fly right, we could say that tail rotor thrust's further up from the cg creating a bigger moment to make that happen, or alternatively that it's moved closer to the level of main rotor and therefore there's less of a pro-roll couple, as the filthy rotor-height-comparing lobby would say!

onehunglow

19th Oct 2006, 07:49

Been away for very long time. See you are talking about the subject of my moniker.

Important for designers. Boring and ultimately useless and irrelevant for drivers.

Now I know why I went away!

Arm out the window

19th Oct 2006, 09:13

Sure, mate, if you want to categorise yourself and close the door rather than maybe broaden your mind a bit!

delta3

19th Oct 2006, 17:47

Did some calculations for a R44-I by means of my scientifique simulator,
for the amateurs of numbers.

Case 1.

Full fuel, 1 pilot (80kg), total TOW 890 kg ISA sea level, OGE

MR-sideways banking (to left down from pilots perspective) = 2.55 °
Helicopter roll (to left down from pilots perspective) = 1.25°
so relative left banking of MR wrt Body = 1.20°

Interesting data :
vertical forces on body 214 N downward, so total MR trust 8945 N
lateral forces on body 1N to the right (neglectable, also the resulting moment)
TR net trust 395 N
pitch up 3.3°, effectively lowering tail rotor to 1.63 m below MR

Observation: even with pilot on righthand side heli banks to the left

Case 2.

153 lit fuel, 4 POB each 75 kg, total TOW 1080 kg, ISA sea level, OGE

MR-sideways banking (left down from pilots perspective) = 2.63 °
Helicopter roll (to left down from pilots perspective) = 0.73 °,
so relative left banking of MR wrt Body = 1.90°

Interesting data :
vertical forces on body 261 N downward , so total MR trust 10856 N
lateral forces on body 1N to the right (neglectable, also the resulting moment)
TR net trust 499 N
pitch down 1°, effectively putting tail rotor at 1.3 m below MR

Observation : putting more POB, and shifting CoG to the left is overruled,
to a large extent, by much small arm of TR wrt MR, resulting in smaller roll of 0.73° instead of 1.25°.

d3

[email protected]

20th Oct 2006, 05:18

delta3 - observation on 2 - you have raised the TR with respect to the c of g which is why the roll angle is less. You have also ignored the change in the vertical position of the c of g by adding extra pax.

Read Nick's posts, it is NOT the relationship between the heights of main and tail rotors, it IS the distance of main and tail rotors from the c of g.

delta3

20th Oct 2006, 10:53

Crab, you seem to read things in my post that are not there.

SAFETY NOTICE : For math-allergists, skip this post….

1. Model is a full 3-d dynamic model. It is a very extensive mathematical set of differential equations, involving multiple rotating frames.

2. The result of the model is the presented data : so I did not put the tail rotor anywhere, the model did, calculating for instance the rotor blade elements centimeter per centimeter, degree per degree as they rotate until the desired attitude is achieved (so billions of calculations)

3. At the risk of presenting a different opinion, my mathematical experience with respect to the CoG is:

-> for static equilibrium calculation, it does not matter where you put the origin, can be hub, can be CoG. Personally I use the hub, because it is a fixed coordinate, whereas the CoG changes. But one has to be complete in the setup and take all forces and arms into consideration. In that respect all loads where put at the exact 3-D positions (also the passengers, fuel etc). Looking to Helo-practise, I note that different constructors also use different reference points for CoG, including the nose of the Heli.

-> for dynamic equations : the so-called rigid body dynamics around the CoG is the only way to go, because expressing the differential equations not in the CoG become very difficult, read impossible.

4. Some math about the static equilibrium : forces and moments need to be zero

-> Force = 0 => sum(Fi)=0, where Fi are the different 3D-force vectors, such as TR trust, MR trust, aerodynamic forces on body, tail fin, horizontal stabilizer, gravity acting on the heli, fuel and passengers.

-> Moment = 0 => sum(Fi X Li) = 0, where Li are the respective 3D-arms, and X is the cross product of vectors

Changing the reference point to say a position Lo creates the transformation Li = Li' + Lo
sum(Fi X Li) = sum(Fi X Li') + sum(Fi X Lo) = sum(Fi X Li') + sum(Fi) X Lo = sum(Fi X Li'), because at equilibrium sum(Fi)=0. In simple terms : in equilibrium the equations remain quite independent of the choice of the reference point

5. The R44 is theatering, so no moment is acting at the rotor hub. For ridgid/excentiric rotors this moment needs to be put at the rotor hub location. It typically will depend on relative rotor bankings.

d3

[email protected]

20th Oct 2006, 13:52

D3 this was the paragraph I was commenting on

'Observation : putting more POB, and shifting CoG to the left is overruled,
to a large extent, by much small arm of TR wrt MR, resulting in smaller roll of 0.73° instead of 1.25°'

And my point was that you were talking about the postition of the TR with respect to the MR which, if you have read many of the previous posts, you will know has no bearing on the roll angle in the hover.

As for your last post, I'm afraid I didn't understand most of it except your reference to a c of g forward of the nose. It doesn't mean that is where the c of g is - it is a datum point that will always give a positive ie aft of the datum position.

If the datum is under the rotor head it means having positive and negative (fwd of the datum) sums to do when you do your loading calculations which is hardly arduous but in an effort to keep it simple for pilots (which I like) some manufacturers build their RFM loading diagrams about a datum outside the aircraft.

thecontroller

20th Oct 2006, 14:49

blimey, a 'simple' JAA ATPL question, and we're up to 5 pages. what does this say about the JAA principles of flight questions?!

delta3

20th Oct 2006, 15:13

.... Never ask an engineer (let alone a mathematician) ?? ...

That is what I also see in Belgian exams. Many of the questions are ambigous because the full context is either ommitted and/or misjudged.
Never easy to develop good questions. Probably takes a good mix of theoreticians and practitioners.

d3

delta3

21st Oct 2006, 19:25

For illustration I let the simulator generate some plots (rear view). This to have real data and not artistic impressions. Making the vectors visible was not an easy task. I had to fool around with camera viewing angles and alpha transparancy, but I hope it does the job. On my PC I can freely rotate the views, but I don't know how to post that.

1. Pilot case (same data as earlier post)

http://www.portmyfolio.com/prive/heli/Pilot%20only%20rear%20view%20hover.jpg

> Yellow downward arrows are gravity forces of Body,Pilot and 2 fuel tanks
> Red downward arrow is the combined gravity force, acting out of CoG (the starting point of this forces defines the CoG). From the picture one can see that the total gravity forces are to the right of the Body gravity force, illustrating the right shift of gravity by the asymetric loading.
> Both of the above gravity forces are also shifted to the right of the hub because of the counter clockwise roll of the Heli of 1.35° (hub is the reference point in the plot)
> MR is clearly banked (2.55°) and coned
> the offset measured from the top of the MR trust to the vertical (measured horizontally) is equal to the TR-trust (red arrow close to the TR, difficult to see)

Same case before take off

http://www.portmyfolio.com/prive/heli/Pilot%20only%20rear%20view%20weight.jpg

2. 4 POB case

http://www.portmyfolio.com/prive/heli/4POB%20rear%20view%20hover.jpg

One can hopefully distinguish the gravity forces of the pilot and 3 passengers and the resulting total gravity force defining again the CoG (at the root of the red downward arrow). The resulting gravity force is now almost centered because of the symmetric loading (slight offset still possible due to fuel loading).

Discussion of influence of TR height

Approach via Hub reference

http://www.portmyfolio.com/prive/heli/Pilot%20Only%20Forces%20Hub.jpg

Hub = blue cross, same rear view, (sorry for my mediocer free hand drawing)
> MR-trust generates per definition no moment, because it acts in the reference point
> TR generates a counter clockwise moment around Hub
> Gravity generates a clockwise moment around Hub equal to TR moment. Total gravity force is offset to the right of the Hub in part because of assymetric loading, in part because of left roll.

From this viewpoint we can make the following reasoning :
> the deaper the tail rotor, the greater the counter clockwise moment it generates around the Hub.
> the opposing moment by the gravity forces around the Hub must increase to maintain equilibrium, so CoG must shift to the right, equivalent to a left roll.

Approach via CoG reference

http://www.portmyfolio.com/prive/heli/Pilot%20Only%20Forces%20CoG.jpg

CoG = blue cross, same rear view
> MR generates a counter clockwise moment around the CoG
> TR generates a clock wise moment around the CoG
> Gravity by choice of reference point generates no moment

From this viewpoint we can make the following reasoning :
> the deaper the tail rotor, the closer it comes to the CoG, the less the clockwise roll moment it generates around the CoG.
> the opposing moment of the MR-trust generates around the CoG must also decrease to maintain equilibrium.
> This is achieved by shifting the CoG to the right (closer to the extended line of the MR-trust), equivalent to a left roll.

For Nick :
> implicite in the above arguments is that the MR trust does not work through the CoG because of other moments which it has to counter act.
> no aero forces on the tail fin, because in the model interaction with tail rotor is ignored (only net TR-trust needed is calculated in hover)
> no aero forces on stabiliser, because induced velocity is assumed to 'miss' the stabiliser in steady hover.

Hope that helps

d3