Aerodynamics ~ Phase Lag
Iconoclast
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To: Nick Lappos
Quite possibly you have proven my point. In Georgia Tech… How about your design section at Sikorsky. In calculating the loads on the rotorhead do they deal with centripetal force or centrifugal force? Go over to the technical school and ask them how they teach aerodynamics of the helicopter. I’m sure they are still using the blue book and it deals with centrifugal force just as I quoted above. How about when you get into a conversation with your fellow pilots and the subject of rotor dynamics comes up do you address centripetal force or centrifugal force?
I may not have your engineering background but I have over the years worked in engineering departments as a design supervisor and a senior project engineer as well as many other engineering capacities. I asked a lot of questions and kept my eyes and ears open. In almost every case when a new engineering graduate was hired they came in thinking that based on their schooling (UCLA, USC, U of M and many others to include Stamford and MIT) they were capable of designing the entire aircraft, helicopter or spacecraft. However their first job was very menial and they were told that college taught them how to think but at (Douglas, Boeing, Hughes, etc., etc. they would do it their way which did not relate to what the new engineer had learned in college.
That is why I made the comment above about selling their textbooks back to the bookstore or to a new student. They have very little relevance to what goes on in industry. At Douglas we had professors working during the summer and most of them were non-productive because they didn’t understand that Douglas was not the same as the university. The only productive professors were from Long Beach State and that was because their aero department taught the Douglas design handbook and in the drafting classes they taught the Douglas drawing system. Graduates of that program would have a good job waiting for them at Douglas because they did not require indoctrination and training.
I had a summer hire working for me at Douglas and he received a Masters in Aero-Astro Engineering from MIT. He was at that time working on a PHD at Stamford. I asked him to make a few changes on a drawing and he stated that he never had any drafting classes at MIT. At MIT he was told that when he went to work he would have other people to do the drafting. Is that how they taught in the engineering courses at Georgia Tech?
One final question. If a blade separates from the rotorhead what is the force that sends it flying off the rotorhead. What causes the resultant imbalance? Is it an imbalance of centripetal force or is it the result of the imbalance of centrifugal forces. How about when a tipcap fills up with water and the water freezes. Is it centripetal force that causes the imbalance or, is it centrifugal force?
I think you and the others that are deeply involved in engineering studies all think I am an idiot for addressing centrifugal instead of centripetal force. And, that every one else that feels that way is also an idiot because some author of an engineering text says that it is a non-force and you believed him. I think he (the author) would do the engineering student community a good service if he stated that centripetal force is a non-force and that it is the reaction to centrifugal force. It would be much easier to understand the concept.
[ 30 November 2001: Message edited by: Lu Zuckerman ]
Quite possibly you have proven my point. In Georgia Tech… How about your design section at Sikorsky. In calculating the loads on the rotorhead do they deal with centripetal force or centrifugal force? Go over to the technical school and ask them how they teach aerodynamics of the helicopter. I’m sure they are still using the blue book and it deals with centrifugal force just as I quoted above. How about when you get into a conversation with your fellow pilots and the subject of rotor dynamics comes up do you address centripetal force or centrifugal force?
I may not have your engineering background but I have over the years worked in engineering departments as a design supervisor and a senior project engineer as well as many other engineering capacities. I asked a lot of questions and kept my eyes and ears open. In almost every case when a new engineering graduate was hired they came in thinking that based on their schooling (UCLA, USC, U of M and many others to include Stamford and MIT) they were capable of designing the entire aircraft, helicopter or spacecraft. However their first job was very menial and they were told that college taught them how to think but at (Douglas, Boeing, Hughes, etc., etc. they would do it their way which did not relate to what the new engineer had learned in college.
That is why I made the comment above about selling their textbooks back to the bookstore or to a new student. They have very little relevance to what goes on in industry. At Douglas we had professors working during the summer and most of them were non-productive because they didn’t understand that Douglas was not the same as the university. The only productive professors were from Long Beach State and that was because their aero department taught the Douglas design handbook and in the drafting classes they taught the Douglas drawing system. Graduates of that program would have a good job waiting for them at Douglas because they did not require indoctrination and training.
I had a summer hire working for me at Douglas and he received a Masters in Aero-Astro Engineering from MIT. He was at that time working on a PHD at Stamford. I asked him to make a few changes on a drawing and he stated that he never had any drafting classes at MIT. At MIT he was told that when he went to work he would have other people to do the drafting. Is that how they taught in the engineering courses at Georgia Tech?
One final question. If a blade separates from the rotorhead what is the force that sends it flying off the rotorhead. What causes the resultant imbalance? Is it an imbalance of centripetal force or is it the result of the imbalance of centrifugal forces. How about when a tipcap fills up with water and the water freezes. Is it centripetal force that causes the imbalance or, is it centrifugal force?
I think you and the others that are deeply involved in engineering studies all think I am an idiot for addressing centrifugal instead of centripetal force. And, that every one else that feels that way is also an idiot because some author of an engineering text says that it is a non-force and you believed him. I think he (the author) would do the engineering student community a good service if he stated that centripetal force is a non-force and that it is the reaction to centrifugal force. It would be much easier to understand the concept.
[ 30 November 2001: Message edited by: Lu Zuckerman ]
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Lu,
You do make some valid points. Once out of college, there is more learning to be done. I don't buy that there is forgetting to be done. If that were the case then might as well hire high school grads to design airplanes.
What is needed is not a good memory or an ability to quote a book, but an understanding of what is in the book. Apply that understanding to the way things are really done, and you have a useful college grad.
Most of the people posting here fall in that category. Succeeding in whatever they do, but also having an understanding of it that goes beyond company technical training, or the like.
________________
You asked, "If a blade separates from the rotorhead what is the force that sends it flying off the rotorhead."
You must first accept and understand what Newton meant by "An object in motion stays in motion in a straight line unless compelled to move by a force impressed upon it." The blade is in motion and is being compelled to move away from a straight line by the force impressed upon it by the rotor hub, acting towards the center of rotation. Detach the blade suddenly and that force is no longer being impressed upon the blade. Now the blade is free to move in a straight line.
Thus, the force that sends the blade flying off doesn't exist. There is a force that keeps the blade on and it's absence causes the blade to fly off.
________________
You said, "I think you and the others that are deeply involved in engineering studies all think I am an idiot for addressing centrifugal instead of centripetal force."
I sincerely hope that nobody thinks you're an idiot for that reason. Addressing centrifugal force is okay, as long as you understand it. It's not a completely sound method of describing ALL aspects of rotational dynamics, but it is VERY useful in understanding some of them.
Cheers,
[ 30 November 2001: Message edited by: heedm ]
You do make some valid points. Once out of college, there is more learning to be done. I don't buy that there is forgetting to be done. If that were the case then might as well hire high school grads to design airplanes.
What is needed is not a good memory or an ability to quote a book, but an understanding of what is in the book. Apply that understanding to the way things are really done, and you have a useful college grad.
Most of the people posting here fall in that category. Succeeding in whatever they do, but also having an understanding of it that goes beyond company technical training, or the like.
________________
You asked, "If a blade separates from the rotorhead what is the force that sends it flying off the rotorhead."
You must first accept and understand what Newton meant by "An object in motion stays in motion in a straight line unless compelled to move by a force impressed upon it." The blade is in motion and is being compelled to move away from a straight line by the force impressed upon it by the rotor hub, acting towards the center of rotation. Detach the blade suddenly and that force is no longer being impressed upon the blade. Now the blade is free to move in a straight line.
Thus, the force that sends the blade flying off doesn't exist. There is a force that keeps the blade on and it's absence causes the blade to fly off.
________________
You said, "I think you and the others that are deeply involved in engineering studies all think I am an idiot for addressing centrifugal instead of centripetal force."
I sincerely hope that nobody thinks you're an idiot for that reason. Addressing centrifugal force is okay, as long as you understand it. It's not a completely sound method of describing ALL aspects of rotational dynamics, but it is VERY useful in understanding some of them.
Cheers,
[ 30 November 2001: Message edited by: heedm ]
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Lu,
Thats a pretty big call from a guy whos not an engineer, and who holds an unlicensed mechanics certificate.
Every helicopter company I ever worked for always addressed centrifugal forces and the engineers calculated how strong the rotorhead had to be to react those loads. Maybe some wienie in the dynamics department dealt with centripital forces but he most likely was isolated from the real engineers.
Iconoclast
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To: Sling Load
By definition I am a Reliability, Maintainability, Systems Safety Engineer and that is what I am classified as in all of the firms that I have worked for. Granted I do not have an engineering degree as I majored in Industrial Design but my resume (CV) shows all of my experience and that experience is defined by the work that I do and that is RMS Engineering or Product Integrity.
Regarding My A&P license If I remember correctly I got my P license in the 1960s and I got my A license in 1976 and my ticket is still valid with the FAA.
By definition I am a Reliability, Maintainability, Systems Safety Engineer and that is what I am classified as in all of the firms that I have worked for. Granted I do not have an engineering degree as I majored in Industrial Design but my resume (CV) shows all of my experience and that experience is defined by the work that I do and that is RMS Engineering or Product Integrity.
Regarding My A&P license If I remember correctly I got my P license in the 1960s and I got my A license in 1976 and my ticket is still valid with the FAA.
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Lu,
You have a very impressive CV, no question about it, but I don't know how you can call someone a wienie and not a real engineer, and use that as a criticism when you, by definition are not an engineer, and hold no qualifications to counter any aerodynamic theory or that of physics.
I don't see anyone on these forums telling you how to repair a helicopter, im sure youve done plenty of that, so you are in no position to get shirty with professionals in that area.
You have a very impressive CV, no question about it, but I don't know how you can call someone a wienie and not a real engineer, and use that as a criticism when you, by definition are not an engineer, and hold no qualifications to counter any aerodynamic theory or that of physics.
I don't see anyone on these forums telling you how to repair a helicopter, im sure youve done plenty of that, so you are in no position to get shirty with professionals in that area.
Iconoclast
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To: heedm
“Thus, the force that sends the blade flying off doesn't exist. There is a force that keeps the blade on and it's absence causes the blade to fly off”.
Your argument that was made leading up to the above statement is valid on the surface.
My question is, when the designers of a rotorhead do a stress analysis on the basic design to verify structural integrity how do they perform those calculations if there is no such thing as centrifugal force?
How do they calculate centripetal force if they do not know what the applied loads are?
Regarding my comments about selling your engineering texts I meant that the books are of little use if you are working within a structured engineering environment where the company has a design manual that covers any and every type of engineering problem. . I did not imply that you should forget what you learned in college.
“Thus, the force that sends the blade flying off doesn't exist. There is a force that keeps the blade on and it's absence causes the blade to fly off”.
Your argument that was made leading up to the above statement is valid on the surface.
My question is, when the designers of a rotorhead do a stress analysis on the basic design to verify structural integrity how do they perform those calculations if there is no such thing as centrifugal force?
How do they calculate centripetal force if they do not know what the applied loads are?
Regarding my comments about selling your engineering texts I meant that the books are of little use if you are working within a structured engineering environment where the company has a design manual that covers any and every type of engineering problem. . I did not imply that you should forget what you learned in college.
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The centripetal force can be measured by placing a strain gauge between the hub and the blade. Some may think that is measuring centrifugal force. The question is, is the blade pulling away from the root (centrifugal) or is the root pulling the blade towards center (centripetal). Since strain guages normally aren't directional, the instrument won't tell you which is being measured.
The force at the hubs is fairly easy to calculate. It will mostly be centripetal force, and I expect some other aerodynamic stuff could add a bit. The centripetal force is a function of mass distribution and rotational velocity.
Basically, you know how much you have to accelerate the blade to keep it close to the hub, since you know where the mass of the blade is located and the speed it will be spinning, you can work out what force you'll need.
________________
Regarding college learning, I probably didn't come across too clearly, but I essentially agreed with you. Textbook learning provides a foundation. What you actually use is subsequently learned in the real world, but should require that foundation. Not just in knowing, but in understanding.
The force at the hubs is fairly easy to calculate. It will mostly be centripetal force, and I expect some other aerodynamic stuff could add a bit. The centripetal force is a function of mass distribution and rotational velocity.
Basically, you know how much you have to accelerate the blade to keep it close to the hub, since you know where the mass of the blade is located and the speed it will be spinning, you can work out what force you'll need.
________________
Regarding college learning, I probably didn't come across too clearly, but I essentially agreed with you. Textbook learning provides a foundation. What you actually use is subsequently learned in the real world, but should require that foundation. Not just in knowing, but in understanding.
Iconoclast
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To: heedm
Here is a problem:
A helicopter rotor head is five feet in diameter. This diameter scribes the path of the blade cuffs or blade attach points.
The blade is 30 feet long and it weighs 250 pounds.
The rotor is spinning at 250 RPM
The CG of the blade if that is required is at the 45% station measured from the attach point.
Discount aerodynamic effects and any vibratory forces due to a traveling wave down the blade.
Discount any changes in loading due to flapping or leading and lagging.
Now, calculate the centripetal forces at the blade attach point but do not calculate the centrifugal forces developed by the spinning blade.
After making that calculation perform the same analysis on all four blades, which all things being normal will balance itself out so that there is, zero tensile or bending loads applied to the mast.
Now while the rotor system is rotating at 250 RPM detach one blade. As you stated centripetal force is what restrains the blades and one blade breaks off when the centripetal force on that blade drops to zero and it still remains on the other three blades. Will the centripetal forces remain the same on the other blades or will it go up and, why
What part does centrifugal force play in your calculations?
There will be a horrific imbalance before the helicopter breaks apart. What causes the imbalance?
[ 30 November 2001: Message edited by: Lu Zuckerman ]
Here is a problem:
A helicopter rotor head is five feet in diameter. This diameter scribes the path of the blade cuffs or blade attach points.
The blade is 30 feet long and it weighs 250 pounds.
The rotor is spinning at 250 RPM
The CG of the blade if that is required is at the 45% station measured from the attach point.
Discount aerodynamic effects and any vibratory forces due to a traveling wave down the blade.
Discount any changes in loading due to flapping or leading and lagging.
Now, calculate the centripetal forces at the blade attach point but do not calculate the centrifugal forces developed by the spinning blade.
After making that calculation perform the same analysis on all four blades, which all things being normal will balance itself out so that there is, zero tensile or bending loads applied to the mast.
Now while the rotor system is rotating at 250 RPM detach one blade. As you stated centripetal force is what restrains the blades and one blade breaks off when the centripetal force on that blade drops to zero and it still remains on the other three blades. Will the centripetal forces remain the same on the other blades or will it go up and, why
What part does centrifugal force play in your calculations?
There will be a horrific imbalance before the helicopter breaks apart. What causes the imbalance?
[ 30 November 2001: Message edited by: Lu Zuckerman ]
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Heedm:
Your diplomacy is impressive in exposing Lu's lack of understanding. And the basic understanding which you give is so desperately lacking in much of the helicopter world.
You wrote:
"Actually, every action has an equal and opposite reaction.
If you sit on a sled on some ice and throw a brick, the action is the brick moving. The reaction is you moving. The only force involved is the one your muscles created. That force is directed from your hand towards the brick. It is not directed from the brick towards your hand.
On anything that spins, the action is pulling inward to prevent outboard masses from flying off tangentially. The reaction is an equal and opposite pull on the hub………"
So.. when you make the action (which in this case is good to call a force, since not all forces are actions and forces really do have equal and opposite forces, I'm sure he meant Force BTW) on this brick what is the name of the force which squashes an impression of the name of the brick manufacturer on the skin of your hand? It is of course the brick's inertial desire not to be accelerated by virtue of it's mass. That equal and opposite force (back on your hand by the brick) can be found, knowing what it's accn was and knowing it's mass, by : F=ma or alternatively it's just measuring how hard you push it and realising it's the equal and opposite force back on your hand.
You say : "The reaction is an equal and opposite pull on the hub……" … that's what people call centrifugal.
Don't get me wrong I'm really with you on this one, but you can see other people have tried to dilute the Great Man's clear message.:
" Hegel's attack on Newton's theory of the heavenly motions (pp. 97 ff.) is of the same character. Newton explained the heavenly motions by the combination of two separate forces, one an impressed centrifugal, the other an attractive or centripetal force. Hegel's objection to this separation of them is no less great than to Kant's assertion of the distinct independence of repulsion and attraction. Hegel does not deny the convenience of the distinction, but he accuses Newton of mistaking the directions into which the motion is resolved for real and actual forces, independent of each other (p. 102). But these two forces are not different and independent, but identical in the same way as repulsion and attraction, two elements of the total motion which involve each other: they are not combin……"etc.
The importance of the equal and opposite force from the inertial desire of things not to be acctd is being lost…
I'm with Newton on this one…
Your diplomacy is impressive in exposing Lu's lack of understanding. And the basic understanding which you give is so desperately lacking in much of the helicopter world.
You wrote:
"Actually, every action has an equal and opposite reaction.
If you sit on a sled on some ice and throw a brick, the action is the brick moving. The reaction is you moving. The only force involved is the one your muscles created. That force is directed from your hand towards the brick. It is not directed from the brick towards your hand.
On anything that spins, the action is pulling inward to prevent outboard masses from flying off tangentially. The reaction is an equal and opposite pull on the hub………"
So.. when you make the action (which in this case is good to call a force, since not all forces are actions and forces really do have equal and opposite forces, I'm sure he meant Force BTW) on this brick what is the name of the force which squashes an impression of the name of the brick manufacturer on the skin of your hand? It is of course the brick's inertial desire not to be accelerated by virtue of it's mass. That equal and opposite force (back on your hand by the brick) can be found, knowing what it's accn was and knowing it's mass, by : F=ma or alternatively it's just measuring how hard you push it and realising it's the equal and opposite force back on your hand.
You say : "The reaction is an equal and opposite pull on the hub……" … that's what people call centrifugal.
Don't get me wrong I'm really with you on this one, but you can see other people have tried to dilute the Great Man's clear message.:
" Hegel's attack on Newton's theory of the heavenly motions (pp. 97 ff.) is of the same character. Newton explained the heavenly motions by the combination of two separate forces, one an impressed centrifugal, the other an attractive or centripetal force. Hegel's objection to this separation of them is no less great than to Kant's assertion of the distinct independence of repulsion and attraction. Hegel does not deny the convenience of the distinction, but he accuses Newton of mistaking the directions into which the motion is resolved for real and actual forces, independent of each other (p. 102). But these two forces are not different and independent, but identical in the same way as repulsion and attraction, two elements of the total motion which involve each other: they are not combin……"etc.
The importance of the equal and opposite force from the inertial desire of things not to be acctd is being lost…
I'm with Newton on this one…
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Lu, this part is fun. Discussing and learning, asking questions, challenging each other. I don't like the stubborness on other threads, but I do like this.
To calculate the centripetal force in the example you gave, we would need to know how the mass in the blade is distributed. It's not uniform otherwise the center of mass would be in the middle not at the 45% point.
If I assume the mass is linearly distributed the centripetal force works out to 8045.4 Newtons or 1808.6 pounds. It most likely isn't linearly distributed, so that answer is only close.
If I just locate all the mass as a point mass at the center of mass, the centripetal force is an easy calculation, mass times radius times angular speed squared. Works out to 2495.6 pounds. It gives you the order of magnitude.
I wrote out the derivation. If anyone wants it, email me.
___________
You said, "Now, calculate the centripetal forces at the blade attach point but do not calculate the centrifugal forces developed by the spinning blade."
Too late, I already calculated both quantities. Centripetal force is as above, centrifugal force is zero because the blade doesn't develop forces, it is acted upon by a force.
____________
Now we get into losing one blade.
First of all, remember that bit about "For every action there is an equal and opposite reaction."? The action of the centripetal force is to pull the blade towards the hub. The reaction is to pull the hub towards the blade.
With a balanced rotor head, you don't notice the reaction, because it is cancelled by the reactions of the other blade(s). When the rotor becomes imbalanced, it behaves like there is a force pulling the hub away from where there used to be a blade, rotating 250 times per minute. That motion of the hub is the reaction to the centripetal force that the hub exerts onto the remaining blades.
Does that answer your question?
To calculate the centripetal force in the example you gave, we would need to know how the mass in the blade is distributed. It's not uniform otherwise the center of mass would be in the middle not at the 45% point.
If I assume the mass is linearly distributed the centripetal force works out to 8045.4 Newtons or 1808.6 pounds. It most likely isn't linearly distributed, so that answer is only close.
If I just locate all the mass as a point mass at the center of mass, the centripetal force is an easy calculation, mass times radius times angular speed squared. Works out to 2495.6 pounds. It gives you the order of magnitude.
I wrote out the derivation. If anyone wants it, email me.
___________
You said, "Now, calculate the centripetal forces at the blade attach point but do not calculate the centrifugal forces developed by the spinning blade."
Too late, I already calculated both quantities. Centripetal force is as above, centrifugal force is zero because the blade doesn't develop forces, it is acted upon by a force.
____________
Now we get into losing one blade.
First of all, remember that bit about "For every action there is an equal and opposite reaction."? The action of the centripetal force is to pull the blade towards the hub. The reaction is to pull the hub towards the blade.
With a balanced rotor head, you don't notice the reaction, because it is cancelled by the reactions of the other blade(s). When the rotor becomes imbalanced, it behaves like there is a force pulling the hub away from where there used to be a blade, rotating 250 times per minute. That motion of the hub is the reaction to the centripetal force that the hub exerts onto the remaining blades.
Does that answer your question?
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Joepilot,
This is where something that is supposed to be factual and precise appears to become something that is esoteric.
The brick doesn't have muscles. It doesn't have any mechanism that allows it to create a force. Try to move your mouse without any exertion whatsoever. It's not possible. Creating a force requires effort (not a physics term in this case). That is the distinction between a real force and an apparent force.
Look at heavenly bodies. They are attracted to each other. If the bodies orbit each other, then gravity becomes the centripetal force. Anything that gets flung from one of the bodies doesn't get flung due to a mystical flinging force, it gets flung because it's not as attractive. Anyone else got their mind on something other than astronomy?
As far as what Kant has to say about Newton, isn't this the same Kant as in the song, "Emmanual Kant was a real pissant who was rarely barely stable....."?
This is where something that is supposed to be factual and precise appears to become something that is esoteric.
The brick doesn't have muscles. It doesn't have any mechanism that allows it to create a force. Try to move your mouse without any exertion whatsoever. It's not possible. Creating a force requires effort (not a physics term in this case). That is the distinction between a real force and an apparent force.
Look at heavenly bodies. They are attracted to each other. If the bodies orbit each other, then gravity becomes the centripetal force. Anything that gets flung from one of the bodies doesn't get flung due to a mystical flinging force, it gets flung because it's not as attractive. Anyone else got their mind on something other than astronomy?
As far as what Kant has to say about Newton, isn't this the same Kant as in the song, "Emmanual Kant was a real pissant who was rarely barely stable....."?
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To: heedm
“I still have the feeling that he wasn't familiar with the head and assumed it to be like that on the 206 (link-horn connection in-line with teetering hinge). Some of your descriptions make it seem like the swashplate input links are phased wrong, and that this is the sole cause of all the problems”.
Regarding Ray Proutys’ understanding of the Robinson head he wrote an article on it in the AHS technical forum. I also believe he had some involvement in the early design.
Regarding the swashplate input links I have no problem as they cause the swashplate to operate just like on a two blade Bell. Push the cyclic forward and the front end goes down over the nose and the back end goes up over the tail. The similarities between the Robinson and the Bell stop at that point.
On the Bell when the blade is disposed over the lateral axis and cyclic is pushed forward the pitch horn which is 90-degrees ahead of the blade causes the blade angle to be at its’ lowest pitch. Conversely the other blade is at the highest pitch. Precession causes the blades to flap down over the nose and up over the tail.
On the Robinson with the blades disposed in the same manner and forward cyclic is pushed the front of the swashplate will move down just like the Bell. The Robinson blade will not be at its’ lowest pitch like the Bell. The blade will have to travel another 18-degrees until it is at its’ lowest pitch. The Bell is rigged for forward cyclic with the blades disposed as above over the lateral axis. The Robinson blade is advance 18-degrees from that position in order to establish forward cyclic settings but I’m sure you were well aware of that fact. I’m also sure that Ray Prouty is also aware of it as well.
On the infamous Robinson (Cantrell) web site they show a Robinson head and address gyroscopic precession. When you get to the explanation they have an illustration addressing precession on the Robinson but the illustration is for a Bell swashplate. Now if you look above at the positioning of the blades on both helicopters the inputs are wrong for the Robinson because it implies that the Robinson has a 90-degree phase angle. If it does the blade will dip down and to the left with forward cyclic. However Nick sez that pitch flap coupling will cause the blade to respond in 72-degrees as opposed to 90-degrees.
Now I think that if I am correct about the offset the test will prove it, contrary to what you might think.
You have access to a 206 and a Robinson. Perform this test. On both helicopters place the blade over the longitudinal axis. Move the cyclic forward from the rigged neutral position. The Bell blades will not move but the Robinson blades will. Now advance the Robinson blades until the pitch horns are directly above the lateral axis. Move the stick forward and backward from the rigged neutral position and the blades will not move.
Now I don’t have your educational background but I have always been told that with maximum change in cyclic pitch the blades will respond 90-degrees later. If what Nick said and Frank Robinson alluded to by saying it was too technical to explain to non engineers and pilots then I am wrong and all of my instructors and engineers that I have worked with are also wrong.
Based on the weight of the blade I figured that the centripetal force would be between 50,000 and 72,000 pounds, which would cover a helicopter somewhat bigger than an S-58 and up to a CH-37, which has a centripetal force of 72,000 pounds (I use your term not mine). The first figure is quoted in the Blue Book and the Sikorsky service-training department provided the second number.
Lets’ assume the blade is at rest. Due to the mechanical linkage of the blade to the head one point becomes a fulcrum and the weight of the blade places a small amount of stress on the rotorhead. If we placed a strain gage at the point of stress it can be measured. This is the same strain gage that measured the centripetal force yet there is no movement of the blade. If the blade(s) are not rotating can you have centripetal force? Now we start to rotate the blade(s) and the force starts to build until it reaches maximum. The original force was not centripetal and with the blades at speed it is now centripetal. You say tomahto and I say tomato.
It is easy to accept your premise but in order to do that I would have to totally alter my belief system.
Regarding the detached blade, when it happens the blade will fly off. If centrifugal force is zero on your calculation where does the energy come from to propel the rotor off of the rotorhead?
Is it possible that under this condition the centripetal force will drop to zero and an energy transfer takes place with all of the centripetal force being transferred into the blade?
“I still have the feeling that he wasn't familiar with the head and assumed it to be like that on the 206 (link-horn connection in-line with teetering hinge). Some of your descriptions make it seem like the swashplate input links are phased wrong, and that this is the sole cause of all the problems”.
Regarding Ray Proutys’ understanding of the Robinson head he wrote an article on it in the AHS technical forum. I also believe he had some involvement in the early design.
Regarding the swashplate input links I have no problem as they cause the swashplate to operate just like on a two blade Bell. Push the cyclic forward and the front end goes down over the nose and the back end goes up over the tail. The similarities between the Robinson and the Bell stop at that point.
On the Bell when the blade is disposed over the lateral axis and cyclic is pushed forward the pitch horn which is 90-degrees ahead of the blade causes the blade angle to be at its’ lowest pitch. Conversely the other blade is at the highest pitch. Precession causes the blades to flap down over the nose and up over the tail.
On the Robinson with the blades disposed in the same manner and forward cyclic is pushed the front of the swashplate will move down just like the Bell. The Robinson blade will not be at its’ lowest pitch like the Bell. The blade will have to travel another 18-degrees until it is at its’ lowest pitch. The Bell is rigged for forward cyclic with the blades disposed as above over the lateral axis. The Robinson blade is advance 18-degrees from that position in order to establish forward cyclic settings but I’m sure you were well aware of that fact. I’m also sure that Ray Prouty is also aware of it as well.
On the infamous Robinson (Cantrell) web site they show a Robinson head and address gyroscopic precession. When you get to the explanation they have an illustration addressing precession on the Robinson but the illustration is for a Bell swashplate. Now if you look above at the positioning of the blades on both helicopters the inputs are wrong for the Robinson because it implies that the Robinson has a 90-degree phase angle. If it does the blade will dip down and to the left with forward cyclic. However Nick sez that pitch flap coupling will cause the blade to respond in 72-degrees as opposed to 90-degrees.
Now I think that if I am correct about the offset the test will prove it, contrary to what you might think.
You have access to a 206 and a Robinson. Perform this test. On both helicopters place the blade over the longitudinal axis. Move the cyclic forward from the rigged neutral position. The Bell blades will not move but the Robinson blades will. Now advance the Robinson blades until the pitch horns are directly above the lateral axis. Move the stick forward and backward from the rigged neutral position and the blades will not move.
Now I don’t have your educational background but I have always been told that with maximum change in cyclic pitch the blades will respond 90-degrees later. If what Nick said and Frank Robinson alluded to by saying it was too technical to explain to non engineers and pilots then I am wrong and all of my instructors and engineers that I have worked with are also wrong.
Based on the weight of the blade I figured that the centripetal force would be between 50,000 and 72,000 pounds, which would cover a helicopter somewhat bigger than an S-58 and up to a CH-37, which has a centripetal force of 72,000 pounds (I use your term not mine). The first figure is quoted in the Blue Book and the Sikorsky service-training department provided the second number.
Lets’ assume the blade is at rest. Due to the mechanical linkage of the blade to the head one point becomes a fulcrum and the weight of the blade places a small amount of stress on the rotorhead. If we placed a strain gage at the point of stress it can be measured. This is the same strain gage that measured the centripetal force yet there is no movement of the blade. If the blade(s) are not rotating can you have centripetal force? Now we start to rotate the blade(s) and the force starts to build until it reaches maximum. The original force was not centripetal and with the blades at speed it is now centripetal. You say tomahto and I say tomato.
It is easy to accept your premise but in order to do that I would have to totally alter my belief system.
Regarding the detached blade, when it happens the blade will fly off. If centrifugal force is zero on your calculation where does the energy come from to propel the rotor off of the rotorhead?
Is it possible that under this condition the centripetal force will drop to zero and an energy transfer takes place with all of the centripetal force being transferred into the blade?
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Lu confusingly said:
"Regarding the detached blade, when it happens the blade will fly off. If centrifugal force is zero on your calculation where does the energy come from to propel the rotor off of the rotorhead? Is it possible that under this condition the centripetal force will drop to zero and an energy transfer takes place with all of the centripetal force being transferred into the blade?"
Nick sez:
This is an example of how wrong Lu is. He has it all backwards! The blade does not "fly off" and it does not get any "energy". If the hub suddenly broke, Lu, the blade does not fly off. The blade if disconnected simply travels as it would have, in unaccelerated motion in a straight line because the rotorhead can no longer drag it around the circle. The blade does not "fly off" it simply goes straight ahead, in continuous motion, with no force on it.
Any force the blade undergoes must be provided by the hub, because it is the hub which must force the blade to conform to the circle of motion. An unsophisticated fool thinks the blade pulls on the hub, an engineer sees that the hub must jerk the blade away from a straight path to make it move in a circle.
In any engineering department in any manufacturer in the world, your above post would make engineers laugh at its ignorance.
For ppruners who have open minds, try this: whirl a pen on a string above your head. While whirling it, release the string just as the pen points at a wall to your front. Does the pen hit the wall? Does it fly out, away from your fingers? Or does it simply move off in a straight line, tangent to the circle and parallel to the wall it was pointing at?
Lu. don't do this experiment, it will only hurt your brain.
"Regarding the detached blade, when it happens the blade will fly off. If centrifugal force is zero on your calculation where does the energy come from to propel the rotor off of the rotorhead? Is it possible that under this condition the centripetal force will drop to zero and an energy transfer takes place with all of the centripetal force being transferred into the blade?"
Nick sez:
This is an example of how wrong Lu is. He has it all backwards! The blade does not "fly off" and it does not get any "energy". If the hub suddenly broke, Lu, the blade does not fly off. The blade if disconnected simply travels as it would have, in unaccelerated motion in a straight line because the rotorhead can no longer drag it around the circle. The blade does not "fly off" it simply goes straight ahead, in continuous motion, with no force on it.
Any force the blade undergoes must be provided by the hub, because it is the hub which must force the blade to conform to the circle of motion. An unsophisticated fool thinks the blade pulls on the hub, an engineer sees that the hub must jerk the blade away from a straight path to make it move in a circle.
In any engineering department in any manufacturer in the world, your above post would make engineers laugh at its ignorance.
For ppruners who have open minds, try this: whirl a pen on a string above your head. While whirling it, release the string just as the pen points at a wall to your front. Does the pen hit the wall? Does it fly out, away from your fingers? Or does it simply move off in a straight line, tangent to the circle and parallel to the wall it was pointing at?
Lu. don't do this experiment, it will only hurt your brain.
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Thanks Nick, I was looking for another way of saying that.
Lu, the blade that flies off does have energy, but it's nothing to do with centripetal or centrifugal forces. The energy comes from it's rotation.
__________
Dave, sorry this thread didn't stay on topic. Is there any way we can go back to phase lag?
Lu, the blade that flies off does have energy, but it's nothing to do with centripetal or centrifugal forces. The energy comes from it's rotation.
__________
Dave, sorry this thread didn't stay on topic. Is there any way we can go back to phase lag?
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I guess low lifes' and uneducated individuals like myself and this includes many mechanics and some pilots will just have to go through life believing that the blades have centrifugal force acting on them, which balances out the lift on the blades and keeps the blades in almost radial alignment. Since this is what we were taught in tech school with the aids of comic book texts with illustrations that do not relate to those used in engineering texts we just keep believing.
It is difficult to understand, using a CH-37 as an example that the rotorhead must exert 72,000 pounds of measurable force to accelerate a blade in a circle. Is there that much drag and is there that much inertia. Also what makes the blade fly in an almost radial position relative to the center of the rotorhead? What keeps it in alignment if there is no centrifugal force? It is also difficult to understand how a blade that is being dragged through a circular path will not flap up to an extreme if there is no force to balance out the lifting forces. If centripetal force is pulling the blade inward how does this force transmit to the blade to balance the lifting forces and keep the blade in radial alignment (almost) if the blade is isolated from a linear inward pull by the flapping hinge and the lead lag hinge? What force does in actuality allow the blade to fly in an almost radial position and to balance out the lifting forces? Could it be CENTRIFUGAL?
I guess low lifes' and uneducated individuals like myself and this includes many mechanics and some pilots will just have to go through life believing that the blades have centrifugal force acting on them, which balances out the lift on the blades and keeps the blades in almost radial alignment. Since this is what we were taught in tech school with the aids of comic book texts with illustrations that do not relate to those used in engineering texts we just keep believing.
It is difficult to understand, using a CH-37 as an example that the rotorhead must exert 72,000 pounds of measurable force to accelerate a blade in a circle. Is there that much drag and is there that much inertia. Also what makes the blade fly in an almost radial position relative to the center of the rotorhead? What keeps it in alignment if there is no centrifugal force? It is also difficult to understand how a blade that is being dragged through a circular path will not flap up to an extreme if there is no force to balance out the lifting forces. If centripetal force is pulling the blade inward how does this force transmit to the blade to balance the lifting forces and keep the blade in radial alignment (almost) if the blade is isolated from a linear inward pull by the flapping hinge and the lead lag hinge? What force does in actuality allow the blade to fly in an almost radial position and to balance out the lifting forces? Could it be CENTRIFUGAL?
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Lu,
"I guess low lifes' and uneducated individuals like myself...will just have to go through life believing that the blades have centrifugal force acting on them..."
You don't have to go through life believing that, if you prefer you can heed the ramblings of those of us who have studied and used this theory in day to day work. If you prefer you can heed all those online resources I posted. Discussing with some of your friends in the industry is also a good idea. In the end, the decision is up to you whether you want to believe in centrifugal forces or not.
I'm just wondering what to tell my two year old about Santa Claus (Father Christmas). She'll see him at the malls and smell his breath, probably wonder why he's so jolly. Whatever I tell her, the choice is up to her to decide whether to believe in him or not.
_________________
"...what makes the blade fly in an almost radial position relative to the center of the rotorhead?"
Good question. This is one of the places that makes the existance of santarifugal force seem so plausible.
Use the pen that Nick had you tie to a piece of string and then throw across the office. If you can't find it, it should be along the wall somewhere, not stuck in it. 2nd
If you spin the pen around your arm so the string is as close to vertical as you can get it without fouling on your knuckles (problem I have because they're so flattened and calloused from dragging... 3rd). The pen wants to fly off in a straight line. The tension in the string accelerates the pen towards your hand. Add that acceleration to the pen's tangential velocity.
The tension diverts the pens velocity to continue to be tangential but also slightly upwards.
Hmmmmmm. I'm even confusing myself. Take a cone or a dunce cap. Place it on your head. Kidding. Place the cone on your desk, and away from the point, draw a short horizontal arrow. At the arrow head, draw a shorter arrow pointed towards the point of the cone. Now draw the resultant from the feathers of the first arrow to the point of the second. This is the direction that the pen gets deflected.
Of course, you have to throw out that cone and get a wider one because the angle of your string becomes more and more horizontal. Once at the horizontal, the tension in the string only pulls the pen towards your hand in one plane.
__________________
"...difficult to understand how a blade that is being dragged through a circular path will not flap up to an extreme if there is no force to balance out the lifting forces."
I'll let you in on a bit of a lie I've been telling. When the rotors are spinning, coned, producing lift, etc. the hub isn't supplying all of the required centripetal acceleration. That's what is really important, not the centripetal force. The hub pulling on the blades does happen, but once coned, that load is shared by a centripetal component to the blade's lift vector.
Knowing that, we can draw a force diagram. Vertical mast, coned blade. Forces are, load of helicopter: downwards from mast/hinge, weight of blade: downwards from blade CM, Lift of blade: approx normal to the blade, pull of hub: centripetal aligned with blade.
Those are all the forces, you'll notice right away that nothing is pointing outwards. Therefore the blade is not in equilibrium. An acceleration will happen, the net centripetal force will cause that required centripetal acceleration. The vertical forces will cancel if the helicopter is steady.
So, why doesn't the blade cone up to the vertical? Because there is no net force that will cause that motion.
__________________
"...if the blade is isolated from a linear inward pull by the flapping hinge and the lead lag hinge?"
One of the neat things about hinges is they transmit forces that are normal to the hinge axis. That's why your doors don't fall off when you open them. No matter how you align two hinges in sequence there is still at least one line that is normal to both hinge axes. There's no problem transmitting that force. Plus, as I mentioned above the blade lift vector supplies some of that centripetal acceleration.
___________________
Must run. Centra Claus parade is downtown in 15 minutes. 4th
"I guess low lifes' and uneducated individuals like myself...will just have to go through life believing that the blades have centrifugal force acting on them..."
You don't have to go through life believing that, if you prefer you can heed the ramblings of those of us who have studied and used this theory in day to day work. If you prefer you can heed all those online resources I posted. Discussing with some of your friends in the industry is also a good idea. In the end, the decision is up to you whether you want to believe in centrifugal forces or not.
I'm just wondering what to tell my two year old about Santa Claus (Father Christmas). She'll see him at the malls and smell his breath, probably wonder why he's so jolly. Whatever I tell her, the choice is up to her to decide whether to believe in him or not.
_________________
"...what makes the blade fly in an almost radial position relative to the center of the rotorhead?"
Good question. This is one of the places that makes the existance of santarifugal force
seem so plausible.Use the pen that Nick had you tie to a piece of string and then throw across the office. If you can't find it, it should be along the wall somewhere, not stuck in it. 2nd
If you spin the pen around your arm so the string is as close to vertical as you can get it without fouling on your knuckles (problem I have because they're so flattened and calloused from dragging... 3rd
). The pen wants to fly off in a straight line. The tension in the string accelerates the pen towards your hand. Add that acceleration to the pen's tangential velocity. The tension diverts the pens velocity to continue to be tangential but also slightly upwards.
Hmmmmmm. I'm even confusing myself. Take a cone or a dunce cap. Place it on your head. Kidding. Place the cone on your desk, and away from the point, draw a short horizontal arrow. At the arrow head, draw a shorter arrow pointed towards the point of the cone. Now draw the resultant from the feathers of the first arrow to the point of the second. This is the direction that the pen gets deflected.
Of course, you have to throw out that cone and get a wider one because the angle of your string becomes more and more horizontal. Once at the horizontal, the tension in the string only pulls the pen towards your hand in one plane.
__________________
"...difficult to understand how a blade that is being dragged through a circular path will not flap up to an extreme if there is no force to balance out the lifting forces."
I'll let you in on a bit of a lie I've been telling. When the rotors are spinning, coned, producing lift, etc. the hub isn't supplying all of the required centripetal acceleration. That's what is really important, not the centripetal force. The hub pulling on the blades does happen, but once coned, that load is shared by a centripetal component to the blade's lift vector.
Knowing that, we can draw a force diagram. Vertical mast, coned blade. Forces are, load of helicopter: downwards from mast/hinge, weight of blade: downwards from blade CM, Lift of blade: approx normal to the blade, pull of hub: centripetal aligned with blade.
Those are all the forces, you'll notice right away that nothing is pointing outwards. Therefore the blade is not in equilibrium. An acceleration will happen, the net centripetal force will cause that required centripetal acceleration. The vertical forces will cancel if the helicopter is steady.
So, why doesn't the blade cone up to the vertical? Because there is no net force that will cause that motion.
__________________
"...if the blade is isolated from a linear inward pull by the flapping hinge and the lead lag hinge?"
One of the neat things about hinges is they transmit forces that are normal to the hinge axis. That's why your doors don't fall off when you open them. No matter how you align two hinges in sequence there is still at least one line that is normal to both hinge axes. There's no problem transmitting that force. Plus, as I mentioned above the blade lift vector supplies some of that centripetal acceleration.
___________________
Must run. Centra Claus parade is downtown in 15 minutes. 4th
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Joepilot,
I got so excited about Christmas I almost forgot to respond to your post, "When the bloke pushes the tail up he also changes the swash plate attitude - the disc will follow."
I think the bloke was pushing on the helicopter with the unspeakably rigid head. I thought that meant the disc's orientation with respect to the helicopter never changed. If this is so, then the disc won't follow, it will stay with the helicopter.
I got so excited about Christmas I almost forgot to respond to your post, "When the bloke pushes the tail up he also changes the swash plate attitude - the disc will follow."
I think the bloke was pushing on the helicopter with the unspeakably rigid head. I thought that meant the disc's orientation with respect to the helicopter never changed. If this is so, then the disc won't follow, it will stay with the helicopter.
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Lu,
Sir Isaac Newton was a clever chap, he realised that an object will remain in linear motion unless acted upon by a FORCE which would change its velocity. He also realised that a FORCE would be accompanied by an equal and opposite reaction. That reaction is not a force otherwise you would have 2 opposing forces and so no work would be done and nothing would happen. A blade being rotated is experiencing a FORCE which acts toward the centre of the hub - a centripetal force (centripetal meaning acting toward the centre). This radial FORCE causes an acceleration toward the centre of the hub. If that acceleration were not there the blade would fly in a straight line and upset some poor pilots otherwise tranquil day. This is BASIC Newtonian physics, I studied it at school. Now to simplify the understanding of such things non-technical or lower level descriptions of rotating systems call the inertial effects described by Newtonian physics "Centrifugal Force" because is is a difficult concept to grasp. Thankyou Lu for your fine demonstration of someone who can't grasp Newtons Laws of motion. I shall continue in my blind faith in published experts and Nobel laureates over blockheadedness and simplicity.
Sir Isaac Newton was a clever chap, he realised that an object will remain in linear motion unless acted upon by a FORCE which would change its velocity. He also realised that a FORCE would be accompanied by an equal and opposite reaction. That reaction is not a force otherwise you would have 2 opposing forces and so no work would be done and nothing would happen. A blade being rotated is experiencing a FORCE which acts toward the centre of the hub - a centripetal force (centripetal meaning acting toward the centre). This radial FORCE causes an acceleration toward the centre of the hub. If that acceleration were not there the blade would fly in a straight line and upset some poor pilots otherwise tranquil day. This is BASIC Newtonian physics, I studied it at school. Now to simplify the understanding of such things non-technical or lower level descriptions of rotating systems call the inertial effects described by Newtonian physics "Centrifugal Force" because is is a difficult concept to grasp. Thankyou Lu for your fine demonstration of someone who can't grasp Newtons Laws of motion. I shall continue in my blind faith in published experts and Nobel laureates over blockheadedness and simplicity.
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That's why nobody really minds you calling it centrifugal force - it's just easier than explaining it to people like you (Lu).
It's the same when talking about Gyroscpic Precession - It's just easier to explain it like that to people to avoid having to explain what really is going on.
Getting back on topic:
heedm's contribution about a NON-spinning blade in a wind tunnel being commanded to make a sinusoidal flapping motion is a brilliant thought experiment. (maybe wrong - but brilliant)
Also the natural frequency in flap being the same as the rotational frequency. - very nice.
The vacuum is quite good also.(though I think the disc will tilt right if you put the stick forward)
In the NON rotating example at Amplitude zero the upward flapping velocity is max and its accn is zero. If it's Accn is to be zero then neutral pitch would be insuficient implying that the aerodydamic damping (change in relative airflow due to flapping velocity) requires a delay in phase of pitch input. (Heedm:"You stabilize with the blade pitch and flapping position to be 180 degrees out of phase (ie minimum pitch at maximum flap)." So really 160 degrees for example.
-This is why 90deg is not the correct phase delay for a Bell family rotor system
(((Now that's for a blade with Mass but IF it was a blade of Zero(negligable) mass then the phase would be 90deg (neglecting 'damping'))))
If there is a restoring force (spring) of the right magnitude for the mass's oscillation to be in tune with the frequency of the flapping you are demmanding (like there is for a blade) Then you approximately bring the phase around another 90deg..... 160-90=70
Basically a blade tries to flap to the flap position commanded by the pitch and although it is damped in trying to get there. After it has endured 178 degrees of being told to go somewhere it finds it already got there a few degrees before (maybe by 170deg) It is therefore effectively not being commanded to go that way anymore - it's flapped position now being that which one would expect for the pitch.
So -- it's got to be less than 90deg.
And it doesn't really matter that some manufacturers don't incorporate this phenomenon in there designs. Since the pilot sorts out all the disymetries of lift by arranging a disymetry of pitch such that noattitude changes occur - he does this at a higher frequency than Lu grasps these concepts.
Gotta go...
It's the same when talking about Gyroscpic Precession - It's just easier to explain it like that to people to avoid having to explain what really is going on.
Getting back on topic:
heedm's contribution about a NON-spinning blade in a wind tunnel being commanded to make a sinusoidal flapping motion is a brilliant thought experiment. (maybe wrong - but brilliant)
Also the natural frequency in flap being the same as the rotational frequency. - very nice.
The vacuum is quite good also.(though I think the disc will tilt right if you put the stick forward)
In the NON rotating example at Amplitude zero the upward flapping velocity is max and its accn is zero. If it's Accn is to be zero then neutral pitch would be insuficient implying that the aerodydamic damping (change in relative airflow due to flapping velocity) requires a delay in phase of pitch input. (Heedm:"You stabilize with the blade pitch and flapping position to be 180 degrees out of phase (ie minimum pitch at maximum flap)." So really 160 degrees for example.
-This is why 90deg is not the correct phase delay for a Bell family rotor system
(((Now that's for a blade with Mass but IF it was a blade of Zero(negligable) mass then the phase would be 90deg (neglecting 'damping'))))
If there is a restoring force (spring) of the right magnitude for the mass's oscillation to be in tune with the frequency of the flapping you are demmanding (like there is for a blade) Then you approximately bring the phase around another 90deg..... 160-90=70
Basically a blade tries to flap to the flap position commanded by the pitch and although it is damped in trying to get there. After it has endured 178 degrees of being told to go somewhere it finds it already got there a few degrees before (maybe by 170deg) It is therefore effectively not being commanded to go that way anymore - it's flapped position now being that which one would expect for the pitch.
So -- it's got to be less than 90deg.
And it doesn't really matter that some manufacturers don't incorporate this phenomenon in there designs. Since the pilot sorts out all the disymetries of lift by arranging a disymetry of pitch such that noattitude changes occur - he does this at a higher frequency than Lu grasps these concepts.
Gotta go...