Aerodynamics ~ Coriolis
Iconoclast
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To: Dave Jackson
“The original question was for the fun of it. It was based on a mast rotating with a constant angular velocity, a teetering hub, and a tip path axis that is not aligned with the mast's axis. The tip path axis must be experiencing acceleration and deceleration twice per revolution. The question was if it was best to explain the acceleration/deceleration by the mathematics of a knuckle joint or by the mathematics of cyclical Coriolis (i.e. flapping)”.
I’ll address the second part first. On a two-blade rotorhead that is underslung there is no leading and lagging as that is the reason for the underslinging of the rotorhead. There may be a tendency for spanwise bending but this is taken up by the drag links on the larger Bells and by the blade retention lugs on the Smaller Bells. Spanwise bending is also compensated for to keep it to a minimum by offsetting the pitch change axis slightly ahead of the centerline of the mast. This offsetting is true for most helicopters.
Now, the first part. I must admit to not having read the post that started this but I feel that it is similar to an “engineering text” that I read which addressed a similar if not identical subject. It was very difficult to follow but I think I understood what the author was talking about. First he told the reader to imagine standing directly over the center of a rotor system that was in a hover. Looking down at the blades (4) it could be seen that the blades were equally spaced (90-degrees)and the tip path was a circle. Now the helicopter was put into forward flight and the reader was told that he was standing directly above the and in line with axis of the rotating blades. The reader was told that now the disc is no longer a perfect circle but an oval and that the blades were no longer equally spaced. The advancing blade was forward of the lateral axis of the disc and the retreating blade was also forward of the lateral axis of the rotating disc. The other two blades were disposed over the longitudinal axis of the rotating disc. To put it more plainly the blades formed a peace sign. If you did this experiment on a two-blade rotor the blades are always disposed in the same way and the two rotating discs would always be a circle. It only works with a fully articulated rotorhead that employs flapping and leading and lagging. The Robinson might be a combination of the two.
[ 05 December 2001: Message edited by: Lu Zuckerman ]
“The original question was for the fun of it. It was based on a mast rotating with a constant angular velocity, a teetering hub, and a tip path axis that is not aligned with the mast's axis. The tip path axis must be experiencing acceleration and deceleration twice per revolution. The question was if it was best to explain the acceleration/deceleration by the mathematics of a knuckle joint or by the mathematics of cyclical Coriolis (i.e. flapping)”.
I’ll address the second part first. On a two-blade rotorhead that is underslung there is no leading and lagging as that is the reason for the underslinging of the rotorhead. There may be a tendency for spanwise bending but this is taken up by the drag links on the larger Bells and by the blade retention lugs on the Smaller Bells. Spanwise bending is also compensated for to keep it to a minimum by offsetting the pitch change axis slightly ahead of the centerline of the mast. This offsetting is true for most helicopters.
Now, the first part. I must admit to not having read the post that started this but I feel that it is similar to an “engineering text” that I read which addressed a similar if not identical subject. It was very difficult to follow but I think I understood what the author was talking about. First he told the reader to imagine standing directly over the center of a rotor system that was in a hover. Looking down at the blades (4) it could be seen that the blades were equally spaced (90-degrees)and the tip path was a circle. Now the helicopter was put into forward flight and the reader was told that he was standing directly above the and in line with axis of the rotating blades. The reader was told that now the disc is no longer a perfect circle but an oval and that the blades were no longer equally spaced. The advancing blade was forward of the lateral axis of the disc and the retreating blade was also forward of the lateral axis of the rotating disc. The other two blades were disposed over the longitudinal axis of the rotating disc. To put it more plainly the blades formed a peace sign. If you did this experiment on a two-blade rotor the blades are always disposed in the same way and the two rotating discs would always be a circle. It only works with a fully articulated rotorhead that employs flapping and leading and lagging. The Robinson might be a combination of the two.
[ 05 December 2001: Message edited by: Lu Zuckerman ]
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Kyrillian,
1. What formula are you using?
2. Given that the Robinson is NOT a pure teetering head (3 hinges for a start) I stand by my teetering head comments.
3. Note I said that quoted hinge offsets are generally effective hinge offsets and as such are quoted to describe a fully articulated head of the same characteristics, thus any delta 3 should be already be taken into account. (Should!)
GA
It's all getting a bit academic really innit? Houses get bigger, houses get smaller etc
1. What formula are you using?
2. Given that the Robinson is NOT a pure teetering head (3 hinges for a start) I stand by my teetering head comments.
3. Note I said that quoted hinge offsets are generally effective hinge offsets and as such are quoted to describe a fully articulated head of the same characteristics, thus any delta 3 should be already be taken into account. (Should!)
GA
It's all getting a bit academic really innit? Houses get bigger, houses get smaller etc
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Grey Area,
"1. What formula are you using?"
nu^2 = 1 + (3/2)*(e/(1-e)) the non-dimensionalized natural frequency (omega_n/Omega)
delta_psi = -atan(nu^2 - 1)/(gamma/8)
for small angles and gamma = 8 that turns into -1.5*e (*180/pi to put it into degrees)
In this case the phase lag would be (90 - delta_psi)
"2. Given that the Robinson is NOT a pure teetering head (3 hinges for a start) I stand by my teetering head comments."
The reason I still think of it as a teetering head is that the 'coning hinges' have relatively stiff springs while the teetering hinge has no spring. I would think that the flapping dynamics would be predominantly defined by the teetering hinge, which would make it a teetering head. The fact that the R-44 head has the pitch link points inboard of the coning hinges confirms in my mind that the coning hinges can't be the primary flapping hinges, because if they were the rotor would be unstable and to my knowledge it performs well.
"3. Note I said that quoted hinge offsets are generally effective hinge offsets and as such are quoted to describe a fully articulated head of the same characteristics, thus any delta 3 should be already be taken into account."
I didn't consider this as the definition of effective hinge offsets--I may be incorrect on this tho. I believe, however, that the effective hinge can be determined statically and that it does not take into account anything such as the pitch link/horn attach point. It's the same as the real hinge in articulating heads and it is the location at which a hinge would be located on a hingeless head to get the same deflection. Hinge springs also have an effect (and will decrease the 'phase lag').
"It's all getting a bit academic really innit? Houses get bigger, houses get smaller etc "
Yes, but I'm an engineering student so it _is_ academics to me!
"1. What formula are you using?"
nu^2 = 1 + (3/2)*(e/(1-e)) the non-dimensionalized natural frequency (omega_n/Omega)
delta_psi = -atan(nu^2 - 1)/(gamma/8)
for small angles and gamma = 8 that turns into -1.5*e (*180/pi to put it into degrees)
In this case the phase lag would be (90 - delta_psi)
"2. Given that the Robinson is NOT a pure teetering head (3 hinges for a start) I stand by my teetering head comments."
The reason I still think of it as a teetering head is that the 'coning hinges' have relatively stiff springs while the teetering hinge has no spring. I would think that the flapping dynamics would be predominantly defined by the teetering hinge, which would make it a teetering head. The fact that the R-44 head has the pitch link points inboard of the coning hinges confirms in my mind that the coning hinges can't be the primary flapping hinges, because if they were the rotor would be unstable and to my knowledge it performs well.
"3. Note I said that quoted hinge offsets are generally effective hinge offsets and as such are quoted to describe a fully articulated head of the same characteristics, thus any delta 3 should be already be taken into account."
I didn't consider this as the definition of effective hinge offsets--I may be incorrect on this tho. I believe, however, that the effective hinge can be determined statically and that it does not take into account anything such as the pitch link/horn attach point. It's the same as the real hinge in articulating heads and it is the location at which a hinge would be located on a hingeless head to get the same deflection. Hinge springs also have an effect (and will decrease the 'phase lag').
"It's all getting a bit academic really innit? Houses get bigger, houses get smaller etc "
Yes, but I'm an engineering student so it _is_ academics to me!
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Lu
“A question; how big a problem is this with the Bell”?
Response:
Which problem (not being smart it is just that I did not understand the question).
Response to Response:
Refer to the previous use of the word 'Bell'
“A question; how big a problem is this with the Bell”?
Response:
Which problem (not being smart it is just that I did not understand the question).
Response to Response:
Refer to the previous use of the word 'Bell'
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Kyrilian
Cos φ =
[( ωn / Ω )^2 – 1]
—————————————————————
√ { [( ωn / Ω )^2 – 1] + [4 (c / c crit)^2 ( ωn / Ω )] }
Damping Ratio (c / c crit) =
[(γ / 16)] . [(1 – (e / R))^4] . [ (1+ ⅓ (e / R)) / (1 - (e / R)) ] . [1 / (ωn / Ω )]
Frequency Ratio (ωn / Ω ) =
√ { 1 + [(3/2)(e / R)] / [1 - (e / R)]
γ = Lock Number
(e / R) = effective offset ratio
To Everyone else:
Sorry
[ 05 December 2001: Message edited by: Grey Area ]
Cos φ =
[( ωn / Ω )^2 – 1]
—————————————————————
√ { [( ωn / Ω )^2 – 1] + [4 (c / c crit)^2 ( ωn / Ω )] }
Damping Ratio (c / c crit) =
[(γ / 16)] . [(1 – (e / R))^4] . [ (1+ ⅓ (e / R)) / (1 - (e / R)) ] . [1 / (ωn / Ω )]
Frequency Ratio (ωn / Ω ) =
√ { 1 + [(3/2)(e / R)] / [1 - (e / R)]
γ = Lock Number
(e / R) = effective offset ratio
To Everyone else:
Sorry
[ 05 December 2001: Message edited by: Grey Area ]
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dave
>The tip path axis must be experiencing acceleration and deceleration twice per revolution.
no its not, thats what i ment with the camera showing a circle, the disc or cone is tilted to one side and your looking at it from under the side (showing an oval)
there is no accel or decell 2ce per rev.
>The tip path axis must be experiencing acceleration and deceleration twice per revolution.
no its not, thats what i ment with the camera showing a circle, the disc or cone is tilted to one side and your looking at it from under the side (showing an oval)
there is no accel or decell 2ce per rev.
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vorticey
This subject is quite abstruse, but what the heck.
The following applies to a teetering rotor.
Forgetting the helicopter for a moment, if two shafts bent at angles to each other are connected with a universal joint and the drive shaft moves at constant angular velocity, the driven shaft runs with uneven angular velocity. See: http://klein-gelenkwellen.de/technik/technis5e.htm
Now we equate the drive shaft mentioned above with the helicopter's mast [hub plane], the yoke in the universal joint with the helicopter rotor's [tip path plane], and the driven shaft with the helicopter rotor's [no feathering plane]. If the no feathering plane (and its axis) is tilted, in respect to the hub plane (mast axis) then the mast or the no feathering plane must be experiencing acceleration and deceleration twice per revolution.
If you agree with the above, in that the no feathering plane is experiencing acceleration and deceleration twice per revolution, the following question arises. How does the no feathering plane, and shortly thereafter, the tip path plane handle this acceleration/deceleration?
_____________
The oval has a large and a small 'diameter' measurement. I would suggest that the apparent changing distance between the tips of the two blade represents a cyclical Coriolis.
This subject is quite abstruse, but what the heck.
The following applies to a teetering rotor.
Forgetting the helicopter for a moment, if two shafts bent at angles to each other are connected with a universal joint and the drive shaft moves at constant angular velocity, the driven shaft runs with uneven angular velocity. See: http://klein-gelenkwellen.de/technik/technis5e.htm
Now we equate the drive shaft mentioned above with the helicopter's mast [hub plane], the yoke in the universal joint with the helicopter rotor's [tip path plane], and the driven shaft with the helicopter rotor's [no feathering plane]. If the no feathering plane (and its axis) is tilted, in respect to the hub plane (mast axis) then the mast or the no feathering plane must be experiencing acceleration and deceleration twice per revolution.
If you agree with the above, in that the no feathering plane is experiencing acceleration and deceleration twice per revolution, the following question arises. How does the no feathering plane, and shortly thereafter, the tip path plane handle this acceleration/deceleration?
_____________
The oval has a large and a small 'diameter' measurement. I would suggest that the apparent changing distance between the tips of the two blade represents a cyclical Coriolis.
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Dave, the website you referenced shows that the angular difference if the blades were tilted by 10 degrees would be a maximum of about 1/2 degree. ie drive axis at 90 degrees, driven axis at 89.5 to 90.5 degrees (depending on direction of 10 degree disk tilt).
I think the changes in blade drag and leading/lagging due to flapping over one revolution create much larger accelerations.
I think the changes in blade drag and leading/lagging due to flapping over one revolution create much larger accelerations.
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ok i didnt realise you were talking about the universal joint.
yes it would accelerate and deccelerate the blades causing a vibration as the angle increased. but as heedm said the angle used isnt enough to worrie about.
yes it would accelerate and deccelerate the blades causing a vibration as the angle increased. but as heedm said the angle used isnt enough to worrie about.
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dave
if the speed increases the blade produces lift aswell as centrifugal force.
i think your saying the cone angle would increase and decrease due to the hub spead variations caused by the variable velocity from the universal joint.
i think the most velocity changes might be taken out by drive shaft flex. and the cone angle may not even change due to lift variations due to speed variations
if the speed increases the blade produces lift aswell as centrifugal force.
i think your saying the cone angle would increase and decrease due to the hub spead variations caused by the variable velocity from the universal joint.
i think the most velocity changes might be taken out by drive shaft flex. and the cone angle may not even change due to lift variations due to speed variations
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heedm
True, the variations in drag must add something to the leading and lagging.
A number of people have mentioned that the hub in a teetering rotor must have sufficient strength to handle these in-plane forces. They are then transmitted to and absorbed by the mast.
The concept of a spinning skater raising her arms is used as an example of Coriolis. If this skater was able to raise and lower her arms at twice the rate that she is spining, could this not be considered as a type of intermittent Coriolis?
____________
You've just come up with a project that Lu can work on; if he doesn't centrifugalate. That's Lu speak for heading off on a tangent.
Lu. For you.
It is a given that a teetering rotor hub must have extra strength so that it can handle the in-plane forces of lead-lag and the variations in rotor drag.
You claim that the Robinson's rotor hub is subjected to excessively wear because of these forces.
He's something for you to research. There are two (& maybe more) types of delta-3. One is by 'flap hinge geometry' and the other is by 'control system geometry'. The Robinson uses 'control system geometry'. 'Flap hinge geometry' delta-3 has an in-plane component. In other words when the blade is teetering and pitching it is also lead/lagging. This in-plane component should handle some of the above forces and thereby reduce the hub wear.
You will find some information on http://www.unicopter.com/0941.html#delta3
True, the variations in drag must add something to the leading and lagging.
A number of people have mentioned that the hub in a teetering rotor must have sufficient strength to handle these in-plane forces. They are then transmitted to and absorbed by the mast.
The concept of a spinning skater raising her arms is used as an example of Coriolis. If this skater was able to raise and lower her arms at twice the rate that she is spining, could this not be considered as a type of intermittent Coriolis?
____________
You've just come up with a project that Lu can work on; if he doesn't centrifugalate. That's Lu speak for heading off on a tangent.
Lu. For you.
It is a given that a teetering rotor hub must have extra strength so that it can handle the in-plane forces of lead-lag and the variations in rotor drag.
You claim that the Robinson's rotor hub is subjected to excessively wear because of these forces.
He's something for you to research. There are two (& maybe more) types of delta-3. One is by 'flap hinge geometry' and the other is by 'control system geometry'. The Robinson uses 'control system geometry'. 'Flap hinge geometry' delta-3 has an in-plane component. In other words when the blade is teetering and pitching it is also lead/lagging. This in-plane component should handle some of the above forces and thereby reduce the hub wear.
You will find some information on http://www.unicopter.com/0941.html#delta3
