The myth of Gyroscopic Precession.
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The myth of Gyroscopic Precession.
A very simple experiment can be done to disprove the belief that there is a strong relationship between gyroscopic precession and the helicopter's rotor-disk. Lu can even do it, without having to ask others.
Take the proverbial bicycle wheel and spin it horizontally on its axis. Then move the axle in an attempt to *roll* this rotating disk. Feel the desire for the disk to *pitch*. Think ' Wow; gyroscopic precession'.
Take the same bicycle wheel and remove its rigid rim. Locate weights at the outer end of every spoke. Now spin this new 'disk' also. Then move the axle in an attempt to *roll* this rotating disk. Watch the disk *roll*. Think 'What the ****; where did the gyroscopic precession go?'.
Then finally think about whether the rigid rim or the loosely connected spokes best represents the helicopter's hub and blades.
______________
heedm
Would you like to comment as to whether there is no gyroscopic precession in the second example, or whether there is, but it is too small to be relevant.
[ 02 December 2001: Message edited by: Dave Jackson ]
Take the proverbial bicycle wheel and spin it horizontally on its axis. Then move the axle in an attempt to *roll* this rotating disk. Feel the desire for the disk to *pitch*. Think ' Wow; gyroscopic precession'.
Take the same bicycle wheel and remove its rigid rim. Locate weights at the outer end of every spoke. Now spin this new 'disk' also. Then move the axle in an attempt to *roll* this rotating disk. Watch the disk *roll*. Think 'What the ****; where did the gyroscopic precession go?'.
Then finally think about whether the rigid rim or the loosely connected spokes best represents the helicopter's hub and blades.
______________
heedm
Would you like to comment as to whether there is no gyroscopic precession in the second example, or whether there is, but it is too small to be relevant.
[ 02 December 2001: Message edited by: Dave Jackson ]
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To: Dave jackson
As I indicated previously the book I read at Boeing indicated that the system as you described would in effect exhibit gyroscopic precession.
"Then finally think about whether the rigid rim or the loosely connected spokes best represents the helicopter's hub and blades".
The answer is the loosely connected spokes.
If you could scribe the blade position in relation to each other especially when the disc is tilted and lead lag is present and the blades flap individually it is obvious that the blades can move in relation to each other and are not connected by a rim as is the bicycle wheel.
______________
[ 02 December 2001: Message edited by: Lu Zuckerman ]
As I indicated previously the book I read at Boeing indicated that the system as you described would in effect exhibit gyroscopic precession.
"Then finally think about whether the rigid rim or the loosely connected spokes best represents the helicopter's hub and blades".
The answer is the loosely connected spokes.
If you could scribe the blade position in relation to each other especially when the disc is tilted and lead lag is present and the blades flap individually it is obvious that the blades can move in relation to each other and are not connected by a rim as is the bicycle wheel.
______________
[ 02 December 2001: Message edited by: Lu Zuckerman ]
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Dave, I thought you'd know me better by now. Of course I want to comment! 
In the second example, I'm assuming at rest the spokes are hanging. In helospeak they now have flapping hinges.
With that disk, when you move the axle, you impart no torque to the disk. The result of the original angular momentum plus the zero torque you added, equals the original angular momentum. So the disk doesn't tilt like the first one.
So why does the disk roll? Because the tension in the spoke pulls the masses into a plane that is perpendicular to the axis of rotation.
Also, because the second wheel is not rigid, it's not right to call it a gyroscope.
[ 02 December 2001: Message edited by: heedm ]
In the second example, I'm assuming at rest the spokes are hanging. In helospeak they now have flapping hinges.
With that disk, when you move the axle, you impart no torque to the disk. The result of the original angular momentum plus the zero torque you added, equals the original angular momentum. So the disk doesn't tilt like the first one.
So why does the disk roll? Because the tension in the spoke pulls the masses into a plane that is perpendicular to the axis of rotation.
Also, because the second wheel is not rigid, it's not right to call it a gyroscope.
[ 02 December 2001: Message edited by: heedm ]
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Lu:
Forget Boeing, forget books, forget lead-lag. We are talking about a bicycle wheel.
We are talking about one small aspect of the helicopter's rotor.
Lu think about the tree, not the whole damn forest.
__________
heedm
Will get back to you. The dog has to go so badly, that he has his back legs crossed.
Forget Boeing, forget books, forget lead-lag. We are talking about a bicycle wheel.
We are talking about one small aspect of the helicopter's rotor.
Lu think about the tree, not the whole damn forest.
__________
heedm
Will get back to you. The dog has to go so badly, that he has his back legs crossed.
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heedm
Thanks for the very clear explanation. To perhaps bring the subject closer to a conclusion, does the following sound correct?
The small hub of the bicycle is being subjected to gyroscopic precession. Let's say we increase the diameter of this hub and reduce the length of the spokes proportionately. The total assemblage will be subjected to more gyroscopic precession, but the spokes will never be subjected to gyroscopic precession.
If this is true, then the blades of a helicopter are never subjected to gyroscopic precession. This is true of a teetering rotor and of a rotor with flapping hinge offset.
____________
An interesting adjunct;
The blade's tip generates far more lift then the blade's root. It might be fair to say that aerodynamic precession relocates the blade's tip and this causes a centripetal force to relocate the blade's root.
[ 02 December 2001: Message edited by: Dave Jackson ]
Thanks for the very clear explanation. To perhaps bring the subject closer to a conclusion, does the following sound correct?
The small hub of the bicycle is being subjected to gyroscopic precession. Let's say we increase the diameter of this hub and reduce the length of the spokes proportionately. The total assemblage will be subjected to more gyroscopic precession, but the spokes will never be subjected to gyroscopic precession.
If this is true, then the blades of a helicopter are never subjected to gyroscopic precession. This is true of a teetering rotor and of a rotor with flapping hinge offset.
____________
An interesting adjunct;
The blade's tip generates far more lift then the blade's root. It might be fair to say that aerodynamic precession relocates the blade's tip and this causes a centripetal force to relocate the blade's root.
[ 02 December 2001: Message edited by: Dave Jackson ]
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Dave,
The reason the spoke and mass wheel doesn't lag is because no force is being applied to the "disk" (masses in this case). The action of you tilting the axle doesn't get transmitted. Initially the masses will want to continue to spin in the plane that they are already spinning in, but shortly after that they will realign themselves with the plane of rotation.
When a helicopter blade generates lift, a force is being applied to the "disk".
The reason the spoke and mass wheel doesn't lag is because no force is being applied to the "disk" (masses in this case). The action of you tilting the axle doesn't get transmitted. Initially the masses will want to continue to spin in the plane that they are already spinning in, but shortly after that they will realign themselves with the plane of rotation.
When a helicopter blade generates lift, a force is being applied to the "disk".
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heedm
>" The reason the spoke and mass wheel doesn't lag is because no force is being applied to the "disk" (masses in this case). "<
Don't you think that the weights will be subjected to a small amount of torque and lag, in the primary plane of rotation, because of aerodynamic drag?
>" The action of you tilting the axle doesn't get transmitted. Initially the masses will want to continue to spin in the plane that they are already spinning in, but shortly after that they will realign themselves with the plane of rotation. "<
I think that tilting the hub will impose an out-of-plane force on the rotating weights. This out-of-plane or realignment force will be the sine of the angle of misalignment times the centripetal force.
>" The reason the spoke and mass wheel doesn't lag is because no force is being applied to the "disk" (masses in this case). "<
Don't you think that the weights will be subjected to a small amount of torque and lag, in the primary plane of rotation, because of aerodynamic drag?
>" The action of you tilting the axle doesn't get transmitted. Initially the masses will want to continue to spin in the plane that they are already spinning in, but shortly after that they will realign themselves with the plane of rotation. "<
I think that tilting the hub will impose an out-of-plane force on the rotating weights. This out-of-plane or realignment force will be the sine of the angle of misalignment times the centripetal force.
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To: Dave Jackson
In the experiment you are describing you can place the spinning system in a vacuum and it would still work. It would be difficult to make the experiment work properly so you would have to suspend the hub and the weights in a gimbaled system. It is not necessary to misalign the shaft all you have to do is to put a perturbing force on the gimbal ring supporting the gimbal ring that supports the hub and weights and the gyroscopic forces will realign the hub and weights in another plane 90-degrees later in the direction of rotation..
[ 02 December 2001: Message edited by: Lu Zuckerman ]
In the experiment you are describing you can place the spinning system in a vacuum and it would still work. It would be difficult to make the experiment work properly so you would have to suspend the hub and the weights in a gimbaled system. It is not necessary to misalign the shaft all you have to do is to put a perturbing force on the gimbal ring supporting the gimbal ring that supports the hub and weights and the gyroscopic forces will realign the hub and weights in another plane 90-degrees later in the direction of rotation..
[ 02 December 2001: Message edited by: Lu Zuckerman ]
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Dave,
I should have said, "The reason the spoke and mass wheel doesn't lag is because no force resulting from tilting the axle is being applied to the "disk" (masses in this case)."
You're right that forces are being impressed upon the masses, I was just referring to a specific force. Didn't make that clear, sorry.
You said, "I think that tilting the hub will impose an out-of-plane force on the rotating weights. This out-of-plane or realignment force will be the sine of the angle of misalignment times the centripetal force."
I agree.
I should have said, "The reason the spoke and mass wheel doesn't lag is because no force resulting from tilting the axle is being applied to the "disk" (masses in this case)."
You're right that forces are being impressed upon the masses, I was just referring to a specific force. Didn't make that clear, sorry.
You said, "I think that tilting the hub will impose an out-of-plane force on the rotating weights. This out-of-plane or realignment force will be the sine of the angle of misalignment times the centripetal force."
I agree.
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Lu
I am trying to keep this thread very simple.
The bicycle wheel (with its rigid rim) has been used to represent gyroscopic precession in a helicopter's rotor.
The bicycle wheel is therefor an excellent candidate to dispute this belief.
I believe that;[list=1][*]a bicycle wheel with individual weights is a better representation of a helicopter rotor.[*]a bicycle wheel with individual weights does not exhibit gyroscopic precession.[/list=a]
You have agreed with line 1.
heedm, with a lot more knowledge in dynamic then either of us, has agreed with both lines.
___________
Re line 2.
Vacuums and gimbals complicate the discussion, so lets keep it simple and only consider the bicycle wheel.
The following two question are asked in all seriousness.[list=a][*]Assume that the spokes are all attached (hinged) at the very center of the bicycle's axle. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane.[*]Assume that the hub now has a very large diameter, which is 90% of the 'wheels' diameter, and that spokes are all attached (hinged) to the rim of this hub. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane.[/list=a]
[ 03 December 2001: Message edited by: Dave Jackson ]
I am trying to keep this thread very simple.
The bicycle wheel (with its rigid rim) has been used to represent gyroscopic precession in a helicopter's rotor.
The bicycle wheel is therefor an excellent candidate to dispute this belief.
I believe that;[list=1][*]a bicycle wheel with individual weights is a better representation of a helicopter rotor.[*]a bicycle wheel with individual weights does not exhibit gyroscopic precession.[/list=a]
You have agreed with line 1.
heedm, with a lot more knowledge in dynamic then either of us, has agreed with both lines.
___________
Re line 2.
Vacuums and gimbals complicate the discussion, so lets keep it simple and only consider the bicycle wheel.
The following two question are asked in all seriousness.[list=a][*]Assume that the spokes are all attached (hinged) at the very center of the bicycle's axle. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane.[*]Assume that the hub now has a very large diameter, which is 90% of the 'wheels' diameter, and that spokes are all attached (hinged) to the rim of this hub. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane.[/list=a]
[ 03 December 2001: Message edited by: Dave Jackson ]
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"Assume that the spokes are all attached (hinged) at the very center of the bicycle's axle. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane"
Depends on moment of inertia, angular velocity, and friction in the flapping hinge. If there is no friction in the flapping direction (ie not hinge, no air resistance, etc.) then the weight will never realign.
The weight acts like a pendulum. The restoring force (force that brings pendulum to it's median position) comes from the rotational motion. In a frictionless environment, pendulums would continue to swing forever.
"Assume that the hub now has a very large diameter, which is 90% of the 'wheels' diameter, and that spokes are all attached (hinged) to the rim of this hub. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane."
Same answer, it depends on.....
In A, the natural frequency of the weight would be the same as the angular frequency of the wheel. Thus, the weight would go through one full cycle as the wheel turns once.
In B, the natural frequency of the weight would be higher than that of the wheel, so it would go through more than one cycle for one turn of the wheel. To get an idea of how much of a difference it makes, if we neglect the weight of the spoke and simplify the system, the weight will do 3.16 cycles for one turn of the wheel.
______________
I had to convince myself that this was accurate, so I built a teetering pencil, spun it on my cordless scredriver. I saw the exact same motion that I expected, and for a sudden change of the axle by about 30 degrees, it took about 7 revolutions for the teetering pencil to return to perpendicular to the axle.
Pictures available by email if interested.
Depends on moment of inertia, angular velocity, and friction in the flapping hinge. If there is no friction in the flapping direction (ie not hinge, no air resistance, etc.) then the weight will never realign.
The weight acts like a pendulum. The restoring force (force that brings pendulum to it's median position) comes from the rotational motion. In a frictionless environment, pendulums would continue to swing forever.
"Assume that the hub now has a very large diameter, which is 90% of the 'wheels' diameter, and that spokes are all attached (hinged) to the rim of this hub. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane."
Same answer, it depends on.....
In A, the natural frequency of the weight would be the same as the angular frequency of the wheel. Thus, the weight would go through one full cycle as the wheel turns once.
In B, the natural frequency of the weight would be higher than that of the wheel, so it would go through more than one cycle for one turn of the wheel. To get an idea of how much of a difference it makes, if we neglect the weight of the spoke and simplify the system, the weight will do 3.16 cycles for one turn of the wheel.
______________
I had to convince myself that this was accurate, so I built a teetering pencil, spun it on my cordless scredriver. I saw the exact same motion that I expected, and for a sudden change of the axle by about 30 degrees, it took about 7 revolutions for the teetering pencil to return to perpendicular to the axle.
Pictures available by email if interested.
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If the spokes are hinged in flap AND lead lag at zero distance then it is not possible to rotate them at all. If they were already spinning then a change in axis of rotation would have no effect.
If the hinges (still at zero distance) were only free to flap and rigid in lead and lag then of course you could spin it. If you then changed the axis of rotation of the hub it would be equivalent of the NON free lead-lag hinge imparting a force which is in the original flapping direction. So for the 90deg
either side of the phase position which corresponds to the axis about which you have attemted to rotate the wheel's axle you have been urging the point mass at the end of the spoke to rise (or fall) - NATURALLY by the end of that 180degs the point mass will have achieved the maximum displacement from the original plane of rotation.
This is just an explanation of what people refer to gyroscopic procession. I don't think it helps people to understand it to 'hide' behind the term GP.
NOTHING really happens 90deg AFTER anything even with gyros. You spend 180deg pushing something up it gets to be high by the end of that pushing up process.
It certainly doesn't help to just tell people to use the mathematical function of taking the CROSS product of the INERTIAL MOMENTUM and the APPLIED MOMENT vectors - the result of which, by using the RIGHT HAND RULE, predicts something happening 90deg from somewhere else. CALLING this gyroscopic precession and pumping it out to the science student body and leaving them to figure a way to explain this APPARENTLY mysterious phenomenon to folk like LU just clouds the issue.
Sure there are still a few sin functions floating around - but the only reason to use them here is to bamboozle old ladies. To conclude - still '90deg later'.
If the hinges (still at zero distance) were only free to flap and rigid in lead and lag then of course you could spin it. If you then changed the axis of rotation of the hub it would be equivalent of the NON free lead-lag hinge imparting a force which is in the original flapping direction. So for the 90deg
either side of the phase position which corresponds to the axis about which you have attemted to rotate the wheel's axle you have been urging the point mass at the end of the spoke to rise (or fall) - NATURALLY by the end of that 180degs the point mass will have achieved the maximum displacement from the original plane of rotation.
This is just an explanation of what people refer to gyroscopic procession. I don't think it helps people to understand it to 'hide' behind the term GP.
NOTHING really happens 90deg AFTER anything even with gyros. You spend 180deg pushing something up it gets to be high by the end of that pushing up process.
It certainly doesn't help to just tell people to use the mathematical function of taking the CROSS product of the INERTIAL MOMENTUM and the APPLIED MOMENT vectors - the result of which, by using the RIGHT HAND RULE, predicts something happening 90deg from somewhere else. CALLING this gyroscopic precession and pumping it out to the science student body and leaving them to figure a way to explain this APPARENTLY mysterious phenomenon to folk like LU just clouds the issue.
Sure there are still a few sin functions floating around - but the only reason to use them here is to bamboozle old ladies. To conclude - still '90deg later'.
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Joepilot said, "If the spokes are hinged in flap AND lead lag at zero distance then it is not possible to rotate them at all. If they were already spinning then a change in axis of rotation would have no effect."
If the flap hinge is offset, then there will be an effect. The weights will flap at a frequency higher than that of the wheel.
Also with an offset hinge, since all the weights aren't flapping about the same point, then when you tilt the axis you add a translation at the flapping hinge to at least some of the weights. This really confuses the dynamics.
If the flap hinge is offset, then there will be an effect. The weights will flap at a frequency higher than that of the wheel.
Also with an offset hinge, since all the weights aren't flapping about the same point, then when you tilt the axis you add a translation at the flapping hinge to at least some of the weights. This really confuses the dynamics.
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Lu
My previous two questions A. & B. were intended to seek simple responses.
A. will take forever, or say, 1,000s and 1,000s of degrees.
B. will only take, say, 30 or 40 degrees.
The intent was to simply show that the phase angle can be more than 90-degrees or it can be less than 90-degrees.
Bottom line; 90-degrees is valid for a gyroscope, but not for a helicopter's rotor.
GP was a convenient way to describe the old teetering rotors, gimbal rotors and gyrocopter rotors
____________
heedm
For the fun of it, you might want to use your teetering pencil to look at the 'knuckle joint' v.s. 'cyclical Coriolis'. I built the same thing a while back to help envision the effects that are mentioned in the lead posting on the 'Aerodynamic - Coriolis' thread.
With the teetering pencil lying on the desktop and the 'mast' being rotated and not normal to the desktop, is the acceleration/deceleration caused by the knuckle joint or by a cyclical type of Coriolis?
______
JoePilot
>"If the hinges (still at zero distance) were only free to flap and rigid in lead and lag then of course you could spin it. If you then changed the axis of rotation of the hub ..... "<
This is a description of the gyrocopter's rotor head.
>".......it would be equivalent of the NON free lead-lag hinge imparting a force which is in the original flapping direction. "<
This force, which is caused by tilting the hub, is because this tilting has pitched one blade down and pitched the other blade up. The blades now fly to position, hence 'aerodynamic precession'
[ 03 December 2001: Message edited by: Dave Jackson ]
My previous two questions A. & B. were intended to seek simple responses.
A. will take forever, or say, 1,000s and 1,000s of degrees.
B. will only take, say, 30 or 40 degrees.
The intent was to simply show that the phase angle can be more than 90-degrees or it can be less than 90-degrees.
Bottom line; 90-degrees is valid for a gyroscope, but not for a helicopter's rotor.
GP was a convenient way to describe the old teetering rotors, gimbal rotors and gyrocopter rotors
____________
heedm
For the fun of it, you might want to use your teetering pencil to look at the 'knuckle joint' v.s. 'cyclical Coriolis'. I built the same thing a while back to help envision the effects that are mentioned in the lead posting on the 'Aerodynamic - Coriolis' thread.
With the teetering pencil lying on the desktop and the 'mast' being rotated and not normal to the desktop, is the acceleration/deceleration caused by the knuckle joint or by a cyclical type of Coriolis?
______
JoePilot
>"If the hinges (still at zero distance) were only free to flap and rigid in lead and lag then of course you could spin it. If you then changed the axis of rotation of the hub ..... "<
This is a description of the gyrocopter's rotor head.
>".......it would be equivalent of the NON free lead-lag hinge imparting a force which is in the original flapping direction. "<
This force, which is caused by tilting the hub, is because this tilting has pitched one blade down and pitched the other blade up. The blades now fly to position, hence 'aerodynamic precession'
[ 03 December 2001: Message edited by: Dave Jackson ]
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Yes that just clouds the issue.
I think your higher frequency arguement for an offset flapping hinge is very good.
I thought my explanation was good for a flapping hinge at NO distance though. (I would tho'
)
I thought it answered Dave's point - and gives some insight into why Gyroscopic Precession doesn't really happen (even in Gyros)...it's just a way of explaning things.
I think your higher frequency arguement for an offset flapping hinge is very good.
I thought my explanation was good for a flapping hinge at NO distance though. (I would tho'
)I thought it answered Dave's point - and gives some insight into why Gyroscopic Precession doesn't really happen (even in Gyros)...it's just a way of explaning things.
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Dave:"This force, which is caused by tilting the hub, is because this tilting has pitched one blade down and pitched the other blade up. The blades now fly to position, hence 'aerodynamic precession'"
I thought these were suposed to be spokes without a rim. - thats what I was describing - and the inclined flapping hing would push the blade(spoke) up.
It would still be at it's highest point after 180 deg of being pushed up - No?
good night
I thought these were suposed to be spokes without a rim. - thats what I was describing - and the inclined flapping hing would push the blade(spoke) up.
It would still be at it's highest point after 180 deg of being pushed up - No?
good night
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the disc would'nt align itself strait away, after the spinning axis is moved the disc would move in a slightly advanced changing axis(gyroscopic force or reaction) eventualy coming back to perpendicular with the axis
in A the spokes would align the weights fairly quickly and in B the alignment would take a while longer because the leverage in B is much less (the arm from the axes)
if your tetering pencil was a true teater joint (no arm) and no friction in the pin then the disc would not have to align itself with the axes at all, only offset hinges provide a throwing effect (centripital) on the blade.
both a & b have offset hinges
take the bike and lift the front wheel of the ground, spin the wheel up.now turn the handle bars, the bike feels a rolling motion but is held by you. just as the rotorblades are feeling a slight gyroscopic proccesion but are held by you on the cyclic.
in A the spokes would align the weights fairly quickly and in B the alignment would take a while longer because the leverage in B is much less (the arm from the axes)
if your tetering pencil was a true teater joint (no arm) and no friction in the pin then the disc would not have to align itself with the axes at all, only offset hinges provide a throwing effect (centripital) on the blade.
both a & b have offset hinges
take the bike and lift the front wheel of the ground, spin the wheel up.now turn the handle bars, the bike feels a rolling motion but is held by you. just as the rotorblades are feeling a slight gyroscopic proccesion but are held by you on the cyclic.
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JoePilot & vorticey
A. had no offset whereas B. had a humoungeous offset. It was just an attempt to show Lu that phase angles can be greater or less than 90-degree.
______________-
My current belief is that
Gyroscopic precession explains the activity inside the flapping/teetering hinges.
Aerodynamic precession, including non-gyroscopic dynamics, explains the activity outside the flapping/teetering hinges.
Therefor;
Gyroscopic precession has nothing to do with teetering rotors.
Gyroscopic precession has next to nothing to do with all other rotors.
A. had no offset whereas B. had a humoungeous offset. It was just an attempt to show Lu that phase angles can be greater or less than 90-degree.
______________-
My current belief is that
Gyroscopic precession explains the activity inside the flapping/teetering hinges.
Aerodynamic precession, including non-gyroscopic dynamics, explains the activity outside the flapping/teetering hinges.
Therefor;
Gyroscopic precession has nothing to do with teetering rotors.
Gyroscopic precession has next to nothing to do with all other rotors.
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Dave: Agreed
VORcity: "if your tetering pencil was a true teater joint (no arm) and no friction in the pin then the disc would not have to align itself with the axes at all, only offset hinges provide a throwing effect (centripital) on the blade."
- No because the teetering hinge has (even at no arm) a 'non-freely flapping' component if you incline it. So the free teetering spokes MUST still align themselves.
GP is a red herring (- just a useful term/concept.)
VORcity: "if your tetering pencil was a true teater joint (no arm) and no friction in the pin then the disc would not have to align itself with the axes at all, only offset hinges provide a throwing effect (centripital) on the blade."
- No because the teetering hinge has (even at no arm) a 'non-freely flapping' component if you incline it. So the free teetering spokes MUST still align themselves.
GP is a red herring (- just a useful term/concept.)
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Using the example of a hub mounted on an axis and attached to that hub were a series of weights attached to arms that were attached to the hub with horizontal hinges. This means that the weight on the arm is free to “flap” but not “lead / lag”.
Now, spin the shaft until the weights are extended outward at a speed sufficient to create “gyroscopic rigidity”. If you had some means of perturbing the individual weights and arms you would experience precession and the weights would move downward on one side of the disc and upward on the opposite side of the disc. If the spinning shaft is rigid and can’t be displaced as a part of the “gyroscopic rotor” the arms and weights would eventually come back to their original position which is extended outward radially from the hub. Now, in this example there is no swashplate, which would keep the arms, and weights in the commanded position and that is why the arms and weights would return to the radial position.
Now lets talk about the CH-53, which has a lot of blades just like the example above has a lot of weights and arms. It also has a swashplate and the helicopter is a free body in space. If you perturb the blades via the swashplate then the blades will move to the commanded position (gyroscopic or aerodynamic precession) and because of the high level of interlock between the blades and the head (centripetal or, centrifugal force) the head will align itself with the blades. The example above could not do that because the shaft although capable of spinning could not be displaced from its' position.
Now, spin the shaft until the weights are extended outward at a speed sufficient to create “gyroscopic rigidity”. If you had some means of perturbing the individual weights and arms you would experience precession and the weights would move downward on one side of the disc and upward on the opposite side of the disc. If the spinning shaft is rigid and can’t be displaced as a part of the “gyroscopic rotor” the arms and weights would eventually come back to their original position which is extended outward radially from the hub. Now, in this example there is no swashplate, which would keep the arms, and weights in the commanded position and that is why the arms and weights would return to the radial position.
Now lets talk about the CH-53, which has a lot of blades just like the example above has a lot of weights and arms. It also has a swashplate and the helicopter is a free body in space. If you perturb the blades via the swashplate then the blades will move to the commanded position (gyroscopic or aerodynamic precession) and because of the high level of interlock between the blades and the head (centripetal or, centrifugal force) the head will align itself with the blades. The example above could not do that because the shaft although capable of spinning could not be displaced from its' position.