Helicopter with Main Rotor Below the Cabin

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Main Question or Discussion Point

Overlooking the difficulty in landing such a craft, would a helicopter with the main rotor below the center of mass, differ significantly in stability from normal one with the main rotor above the center of mass?
 

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  • #2
mgb_phys
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Yes - technically speaking the stability would be a right bugger.
 
  • #3
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Think of it like this, it's like standing on a pencil that is stood on it's end.
 
  • #4
berkeman
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At least it makes an ejection seat option easier (you don't need explosive bolts in the rotor blade assembly).
 
  • #5
rcgldr
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It would be less stable, but radio control helicopters flown upside down don't seem significantly less stable because the angular momentum of the main rotor is the dominant factor. The center of mass is set to be very close to the axis line of the main rotor, which reduces any torque reaction related to up thrust or down thrust from the main rotor.
 
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Yep. I remember seeing a documentary on the tv. There was this early model a guy was flying while standing on the craft. It was a public demonstration in a stadium. He was flying so close to the ground it reminded me of a lawnmower. I understand the americans settled for Zarkorzy's? roter over cabin design.
 
  • #7
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Why would it be unstable? There are low wing aircraft...
 
  • #8
sophiecentaur
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boit: Sikorsky was the helicopter guy.

Once airborne, the blades on a full size helicopter curve upwards quite a lot so the centre of effort could be arranged to be above the centre of mass whilst in flight.

BUT one snag with a rotor below the fuselage would, surely, be finding somewhere to put the wheels! :confused:
 
  • #9
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I am wondering if a pendulum effect, in which the center of mass is below the center thrust, helps to stabilize a craft when hovering. At higher velocities the location of the center of pressure becomes more important and at higher accelerations the pendulum effect becomes insignificant. Is this the dominant stabilizing force for a hovering craft and if not, what is?
 
  • #10
sophiecentaur
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Your "pendulum" effect may help a bit bit they are, inherently, not very stable wrt any axis.
I just think they're v. difficult to fly and hovering is the worst.
If the machine is not designed with the cm on the rotor axis, with the fuselage horizontal, then it will tilt until the cm is below the rotor spindle and will need to be compensated for by the blade angle controls if you want to be horizontal.
But it's normal to get the trim of all aircraft (and boats) about right before you take start off, isn't it?
 
  • #11
mgb_phys
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It's very similar to the stability problem of a boat - where you need the center of boyancy above the center of mass.

Imagine the weight at the centre of mass of the cabin pulling down and the lift at the centre of the rotor pushing up.
If the rotor is on top it's like hanging the weight from a rope, if the cabin tilts to the side then it's own weight will pull it back down into line - while the rotors will pull back up and straight.
If the rotor is on the bottom it's trying to increase the tilt, as is the weight - and it will flip over.

It works for model helicopter because the cabin is much lighter (compared to scale) and the rotors are much heavier
 
  • #12
sophiecentaur
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Yes - true but the blades curve upwards* when providing lift (they are flexible) and the cm will be lower, wrt the centre of effort than you might imagine.
*they may curve up much more than they droop down when 'resting' - which is an appreciable amount in pictures of some helicopters I've seen.

BTW, I can't imagine anyone is being really serious about this idea!
 
  • #13
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BTW, I can't imagine anyone is being really serious about this idea!
Of course not.

I was trying to look at the pendulum fallacy from a different perspective. In another thread "Rocket Idea: Bell Shaped Pendulum" posts #1, #3, #4 & #5, the claim is made that positioning the mass below the thruster would not be effective in stabilizing a lunar lander due to the pendulum fallacy. I understand that the pendulum effect (my term) would not be effective when a rocket is traveling through the atmosphere at velocity or at high acceleration, but in the case a hovering lunar lander I do not understand why it would not add stability. I am not suggesting that if the lunar lander is wildly out of control, the pendulum effect is all that would be needed to stabilize it.
 
  • #14
RonL
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Yes - true but the blades curve upwards* when providing lift (they are flexible) and the cm will be lower, wrt the centre of effort than you might imagine.
*they may curve up much more than they droop down when 'resting' - which is an appreciable amount in pictures of some helicopters I've seen.

BTW, I can't imagine anyone is being really serious about this idea!
I think you will find this idea has not only been looked at in a very serious way, but has in fact been quite successful. Look at the "Hiller flying platform" and many other video's on u-tube.
Your comment about the seriousness of an idea, sounds much the same as the wall of resistance that, Sir Frank Whittle, ran into.:rolleyes: there is always a chance old ideas might once again become new.:approve:
 
  • #15
sophiecentaur
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I imagine a "serious" proposal would include a mention of where the wheels go. Please enlighten me. I have tried drawing one and there seems to be a topological problem.
 
  • #16
RonL
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I imagine a "serious" proposal would include a mention of where the wheels go. Please enlighten me. I have tried drawing one and there seems to be a topological problem.
Two links that give some idea of what has been done. It seems that discovery channel may have had u tube remove some videos, as all I could find are poor quality.
The Blakebrough's winged shelf at least gives two views of less quality test flights. If I can find the better discovery clip, I will try to post it.

It seems to me that a larger platform and rotor blades could be a possible design that would give a more stable machine and using lipo batteries attached to the ring, an electric power setup could keep it pretty quite.

http://www.hiller.org/flying-platform.shtml [Broken]

 
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  • #17
sophiecentaur
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The machines shown in the links aren't so much helicopters as hovercraft plus a bit. That actual flying demo seems to require ground effect (it only gets a metre or so high).
To lift a serious payload out of ground effect and to hover, you need larger rotor diameter than that, surely, and then my "where to put the wheels" question would still need to be answered.
With the cm above the rotor, you must also be in a seriously unstable situation - as was the 'flying bedstead' and the Lunar lander. There's no 'pendulum effect' from those arrangements ( no restoring force when perturbed). I think the same problem must exist even with the James Bond 'jet pack'.
With modern control systems this needn't be too much of a problem if there happen to be other advantages. All rockets need stabilising until they are moving through the air fairly fast and that is achieved reasonably well, these days.
But this thread has developed two issues - the stability and the feasibility of such a system. They are getting a bit mixed up, I think.
 
  • #18
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With the cm above the rotor, you must also be in a seriously unstable situation - as was the 'flying bedstead' and the Lunar lander. There's no 'pendulum effect' from those arrangements ( no restoring force when perturbed).
I agree. As I was watching the video I had the feeling the operator was doing a balancing act and the craft was always on the verge of tipping over. I doubt it would be stable with a dead weight on it. With the cm above the center of thrust, any tilt of the craft will tend to push the weight farther from the centerline, exacerbating the unbalance and tilt (positive feedback).

Is there agreement that the pendulum fallacy does not apply to hovering craft?
 
  • #19
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I haven't seen any good arguments here.

Draw the free body diagram; the thrust passes through the center of mass.

There is no restoring torque to keep a helicopter stable, such as in the case of a boat, whether the blades are above or below the center of mass.
 
  • #20
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That being said, I was hoping someone would come up with something to show a difference between the mass being located over vs. under the rotor. After all, paragliders are stable (after a fashion) because the lifting surface is above the center of mass.

A helicopter is a dynamic system.

Imagine a stable paraglider with the pilot proped up above the sail on compressive structural members. I don't think that would work.

Why is a paraglider stable, and an inverted paraglider (apparently) unstable? The same answer should apply to an inverted helicopter where the distance between rotor and center of mass is large.

I don't know the answer to this question.
 
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  • #21
mgb_phys
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There is no restoring torque to keep a helicopter stable, such as in the case of a boat, whether the blades are above or below the center of mass.
Doesn't the lift pass through the centre of the rotor?
In that case it's equivalent to the helicopter being suspended from a rope attached to the hub
 
  • #22
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Yes, except if the rotor is not horizontal, the thrust vector is off vertical. The lift vector still passes through the hub of the rotor but there is an additional vector that causes horizontal movement of the helicopter.

Likewise, as in the above scenario, if the cm is not directly below the center of thrust, there will be a moment tending to return the cm to a position vertical alignment.

If the rotor and center of thrust were below the cm but offset slightly, the moment instead of tending to restore the vertical alignment between the center of thrust and the center of mass, would tend to increase it.
 
  • #23
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Doesn't the lift pass through the centre of the rotor?
In that case it's equivalent to the helicopter being suspended from a rope attached to the hub
Well, yes, but in the equivalently rope model the tension of the rope passes through the center of mass rather than acting on one end of the rotor axle where it could induce a torque.

So the combination of lifting with an off-vertical rotor plus W(=mg) will result in a component of force in the horizontal direction, without a corrective torque to return the helicopter to a vertical orientation.

In the real world, the blades of a helicopter bend upward, or are allowed to swivel upward on a hing. Maybe this effective dihedral lends something to stability without pilot input.

Edit: Skeptics. I hadn't read your post. But in any case there isn't a restoring moment when both forces pass through the center of mass. More, I'm not a helicopter pilot. Can a helicopter fly on it's own, hands-off?
 
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  • #24
sophiecentaur
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That being said, I was hoping someone would come up with something to show a difference between the mass being located over vs. under the rotor. After all, paragliders are stable (after a fashion) because the lifting surface is above the center of mass.

A helicopter is a dynamic system.

Imagine a stable paraglider with the pilot proped up above the sail on compressive structural members. I don't think that would work.

Why is a paraglider stable, and an inverted paraglider (apparently) unstable? The same answer should apply to an inverted helicopter where the distance between rotor and center of mass is large.

I don't know the answer to this question.
One definition of stability is that, for stable equilibrium, a perturbation will increase the potential. For a mass hanging on a string or a mass hanging below whatever is suspending it (i.e. rotor or wing) the potential increases with tilt.
If the cm were above the centre of effort of the blades (if that's all that's involved) then any perturbation will decrease the potential - so it's unstable.

So called Low (fixed) Wing aircraft mostly have dihedral, giving a centre of effort higher than it may appear and a cm which is below where it may appear to be (like double decker buses).
They will mostly be inherently stable (you could hold them up by the wing tips and they would not pitch) but, even if not, when moving through the air, the tail and other surfaces will tend to keep them stable (roll is self corrected).
Dihedral, itself provides more lift on the side that is tilted down, which gives a righting moment so this may also apply to a helicopter rotor, which is tilted up - more lift on the appropriate side to help with stability.

More manouverable aircraft have no dihedral and are much less stable (can be put into a spin, for instance).
 
  • #25
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One definition of stability is that, for stable equilibrium, a perturbation will increase the potential. For a mass hanging on a string or a mass hanging below whatever is suspending it (i.e. rotor or wing) the potential increases with tilt.
If the cm were above the centre of effort of the blades (if that's all that's involved) then any perturbation will decrease the potential - so it's unstable.

So called Low (fixed) Wing aircraft mostly have dihedral, giving a centre of effort higher than it may appear and a cm which is below where it may appear to be (like double decker buses).
They will mostly be inherently stable (you could hold them up by the wing tips and they would not pitch) but, even if not, when moving through the air, the tail and other surfaces will tend to keep them stable (roll is self corrected).
Dihedral, itself provides more lift on the side that is tilted down, which gives a righting moment so this may also apply to a helicopter rotor, which is tilted up - more lift on the appropriate side to help with stability.

More manouverable aircraft have no dihedral and are much less stable (can be put into a spin, for instance).
You reflect my thoughts almost exactly. Rather than speak in general, why would a dihedral winged aircraft be more stable to perturbations with the center of mass below the wing than above?
 

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