another terrible ATPL PoF question
CAUTION - DIAGRAM ALERT!
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.
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.
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.
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.
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
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
Last edited by Arm out the window; 17th Oct 2006 at 06:58. Reason: To add a bit more
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.
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.
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.
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.
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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
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
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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.
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.
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
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
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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!):
For a High TR, here is what happens:
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!):
For a High TR, here is what happens:
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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
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
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.
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.
Hurray.....one converted, one to go
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?
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?
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.
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.
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.
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.
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Bank angle
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
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
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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!!
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!!
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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.
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.