Pylon Rock
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To: imabell
Nothing was said about a Bell 47 helicopter. Although you didn’t see it, the engine / transmission combination moves up and down to the limits allowed by the two Dynafocal mounts attached to the engine /transmission support cage. The movement is minimal but it is there. There is however something common on the 47 that happens on all of the other Bell 2-blade helicopters and that is the 2-per rev, which gives you a vertical beat.
Also on the 47, the pylon (Mast) and engine will deflect to the limits of the Dynafocal mounts and if the pilot takes it beyond those limits he will come up against the sprague cables on the bottom of the engine. If he goes any further, the mast will bend and cause a growling of the gears in the transmission.
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The Cat
Nothing was said about a Bell 47 helicopter. Although you didn’t see it, the engine / transmission combination moves up and down to the limits allowed by the two Dynafocal mounts attached to the engine /transmission support cage. The movement is minimal but it is there. There is however something common on the 47 that happens on all of the other Bell 2-blade helicopters and that is the 2-per rev, which gives you a vertical beat.
Also on the 47, the pylon (Mast) and engine will deflect to the limits of the Dynafocal mounts and if the pilot takes it beyond those limits he will come up against the sprague cables on the bottom of the engine. If he goes any further, the mast will bend and cause a growling of the gears in the transmission.
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The Cat
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Lu:
It is quite clear that you take this forum very very seriously, and that you absolutely cannot admit that there is something you can learn.
Your loss of lift theory is simply untrue. Period. Would you like me to provide some blade bending data to show that the blades are all bent upward in level flight by the lift they are creating, all subjected to their share of the lift, and that this lift is a relative constant all around the mast?
What exactly will it take to get you to stop saying "I was once told by a Bell engineer..." as some kind of proof for this simply absurd premise?
An open mind is a wonderful thing!
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It is quite clear that you take this forum very very seriously, and that you absolutely cannot admit that there is something you can learn.
Your loss of lift theory is simply untrue. Period. Would you like me to provide some blade bending data to show that the blades are all bent upward in level flight by the lift they are creating, all subjected to their share of the lift, and that this lift is a relative constant all around the mast?
What exactly will it take to get you to stop saying "I was once told by a Bell engineer..." as some kind of proof for this simply absurd premise?
An open mind is a wonderful thing!
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Guest
Posts: n/a
To: Nick Lappos
I do take the forum seriously and I have learned a lot from the participants on the forum. However I will continue to believe the theory of loss of lift until I am set straight by Bell Helicopters. I am also awaiting a copy of a paper delivered to the American Helicopter Society dealing with Fuselage Nodalization. I believe that the answer lies in that paper.
As a test pilot, an engineering graduate and a former Army Snake driver I would like to ask a couple of questions.
1) As a Snake driver did you ever ask yourself or anybody else what causes the 2-per rev bounce and why the frequency of the bounce was two time the rotor speed and not three or four times?
2) As an engineer can you explain in layman’s terms how the energy is developed that would lift and drop a 1000 pound dynamics system by 2+/- inches at a frequency of about 500 times per minute.
I quote,” What exactly will it take to get you to stop saying "I was once told by a Bell engineer..." as some kind of proof for this simply absurd premise?
An open mind is a wonderful thing!
I would like to paraphrase you question. What will it take me to stop saying “I was once told by a Sikorsky engineer…” as some kind of proof that my premise was absurd?
Here is a story about a Sikorsky engineer and his limited understanding of how a helicopter works. I was once involved in a comprehensive 14 month technical training program at Sikorsky. Aside from extensive classroom work I worked in every department of the company from engineering to flight-test. If a part or an assembly was made in an assigned department I built it, I tested it and I flew in it.
On one assignment I was working in the hydraulics test lab. While there I witnessed a hydraulics engineer cobble together a system that was supposed to drop the collective when engine oil pressure was lost. In order to work the collective was raised by a hydraulic piston until a spring was cocked and then the collective dropped. I asked him how the collective was raised and he stated that as hydraulic pressure built up the piston would lift the collective.
I told him that the system would not only not work it could cause serious damage to the airframe. He became extremely belligerent saying that I didn’t understand his device and that it would not damage the helicopter. It seems that his involvement with design was limited to hydraulics and he had no knowledge of theory of flight.
I told him that the hydraulic pump was on the transmission and as the transmission was brought up to speed the blades could not support lift. I further told him that at a slow rotational speed the blades would rise up and stall out falling and hitting the tail cone. The next day the device was gone and two days later, the engineer was gone.
I don’t know what bearing this story has our discussion other than that it takes a lot to change someone’s feelings about something that they hold dear.
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The Cat
I do take the forum seriously and I have learned a lot from the participants on the forum. However I will continue to believe the theory of loss of lift until I am set straight by Bell Helicopters. I am also awaiting a copy of a paper delivered to the American Helicopter Society dealing with Fuselage Nodalization. I believe that the answer lies in that paper.
As a test pilot, an engineering graduate and a former Army Snake driver I would like to ask a couple of questions.
1) As a Snake driver did you ever ask yourself or anybody else what causes the 2-per rev bounce and why the frequency of the bounce was two time the rotor speed and not three or four times?
2) As an engineer can you explain in layman’s terms how the energy is developed that would lift and drop a 1000 pound dynamics system by 2+/- inches at a frequency of about 500 times per minute.
I quote,” What exactly will it take to get you to stop saying "I was once told by a Bell engineer..." as some kind of proof for this simply absurd premise?
An open mind is a wonderful thing!
I would like to paraphrase you question. What will it take me to stop saying “I was once told by a Sikorsky engineer…” as some kind of proof that my premise was absurd?
Here is a story about a Sikorsky engineer and his limited understanding of how a helicopter works. I was once involved in a comprehensive 14 month technical training program at Sikorsky. Aside from extensive classroom work I worked in every department of the company from engineering to flight-test. If a part or an assembly was made in an assigned department I built it, I tested it and I flew in it.
On one assignment I was working in the hydraulics test lab. While there I witnessed a hydraulics engineer cobble together a system that was supposed to drop the collective when engine oil pressure was lost. In order to work the collective was raised by a hydraulic piston until a spring was cocked and then the collective dropped. I asked him how the collective was raised and he stated that as hydraulic pressure built up the piston would lift the collective.
I told him that the system would not only not work it could cause serious damage to the airframe. He became extremely belligerent saying that I didn’t understand his device and that it would not damage the helicopter. It seems that his involvement with design was limited to hydraulics and he had no knowledge of theory of flight.
I told him that the hydraulic pump was on the transmission and as the transmission was brought up to speed the blades could not support lift. I further told him that at a slow rotational speed the blades would rise up and stall out falling and hitting the tail cone. The next day the device was gone and two days later, the engineer was gone.
I don’t know what bearing this story has our discussion other than that it takes a lot to change someone’s feelings about something that they hold dear.
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The Cat
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Lu asked:
1) As a Snake driver did you ever ask yourself or anybody else what causes the 2-per rev bounce and why the frequency of the bounce was two time the rotor speed and not three or four times?
Nick answered:
That is because in the case we discuss, the 2/rev is the mode that tickles the transmission mounts. On some machines, it could be 1/rev or another that drives the limit cycle instability. On some Sikorsky's, it is about 2/3 per rev that makes things hum, and we work to damp this until it is gone. I guess you would believe that a 2/3 per rev mode is excited when the blade looses 2/3 of its lift, huh?
Lu asked:
2) As an engineer can you explain in layman’s terms how the energy is developed that would lift and drop a 1000 pound dynamics system by 2+/- inches at a frequency of about 500 times per minute.
Nick answered:
Lu, you should be able to explain this, as it is fundamental to understanding the nature of dynamic oscillations, a field of study for all engineers. Simply put, the vibration needs almost NO ENERGY, Lu. When a system is in resonance, the natural vibration is at that frequency, and very little energy is needed to make the oscillation keep on going. The trick is to understand that when the transmission is dancing up and down, it gets a down push just as it needs it, so only a little push is needed to keep the dance up.
Remember the analogy of the small boy on a swing. That boy could not lift himself up to head height through normal strength, yet he can do it on a swing by simply adding a little energy when the swing needs it, so his dynamic interaction with the swing is perfectly timed. For a boy to make a swing go from head height in one direction to head height in the other takes almost no energy, as the swing is simply converting the kinetic energy of speed at the bottom into potential energy of lift at the top.
Somehow, you confuse this vast impressive motion with a great need for energy, and you are wrong.
The energy needed to bounce the transmission up and down is small, I estimate that a 1,000 pound force might stretch the mounts an inch or so, so the potential energy would be about 1,000 pounds times 0.083 feet, or 83 foot pounds. This could be applied to the system each second by a small 1/6 horsepower electric motor!
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1) As a Snake driver did you ever ask yourself or anybody else what causes the 2-per rev bounce and why the frequency of the bounce was two time the rotor speed and not three or four times?
Nick answered:
That is because in the case we discuss, the 2/rev is the mode that tickles the transmission mounts. On some machines, it could be 1/rev or another that drives the limit cycle instability. On some Sikorsky's, it is about 2/3 per rev that makes things hum, and we work to damp this until it is gone. I guess you would believe that a 2/3 per rev mode is excited when the blade looses 2/3 of its lift, huh?
Lu asked:
2) As an engineer can you explain in layman’s terms how the energy is developed that would lift and drop a 1000 pound dynamics system by 2+/- inches at a frequency of about 500 times per minute.
Nick answered:
Lu, you should be able to explain this, as it is fundamental to understanding the nature of dynamic oscillations, a field of study for all engineers. Simply put, the vibration needs almost NO ENERGY, Lu. When a system is in resonance, the natural vibration is at that frequency, and very little energy is needed to make the oscillation keep on going. The trick is to understand that when the transmission is dancing up and down, it gets a down push just as it needs it, so only a little push is needed to keep the dance up.
Remember the analogy of the small boy on a swing. That boy could not lift himself up to head height through normal strength, yet he can do it on a swing by simply adding a little energy when the swing needs it, so his dynamic interaction with the swing is perfectly timed. For a boy to make a swing go from head height in one direction to head height in the other takes almost no energy, as the swing is simply converting the kinetic energy of speed at the bottom into potential energy of lift at the top.
Somehow, you confuse this vast impressive motion with a great need for energy, and you are wrong.
The energy needed to bounce the transmission up and down is small, I estimate that a 1,000 pound force might stretch the mounts an inch or so, so the potential energy would be about 1,000 pounds times 0.083 feet, or 83 foot pounds. This could be applied to the system each second by a small 1/6 horsepower electric motor!
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To: Nick Lappos
Thanks for the detailed explanation. Now I want you to carry it one step further. If a 1/6 horsepower motor were strong enough to excite a 1000-pound dynamic system and cause it to bounce up and down I would like for you to define the fractional HP motor as it applies the excitation process. Remove the motor from the equation and substitute the 2 per rev. The motor is powered by electricity. What is the motivational force behind the two per rev and what causes it?
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The Cat
Thanks for the detailed explanation. Now I want you to carry it one step further. If a 1/6 horsepower motor were strong enough to excite a 1000-pound dynamic system and cause it to bounce up and down I would like for you to define the fractional HP motor as it applies the excitation process. Remove the motor from the equation and substitute the 2 per rev. The motor is powered by electricity. What is the motivational force behind the two per rev and what causes it?
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The Cat
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Lu,
The exciting force is most certainly aerodynamic in regards to what we called "verticals" and you most probably mean by 2/rev. However, in the case of that which we called "laterals" and you call pylon rock, the exciting force was mechanical.
Before Mr Chadwick Helmuth confused the issue and we used to make Miss Bell smooth by flag tracking and interpreting a vibration sensitive bum, we would minimise 2/rev verticals by making corrections to the tip path planes of the individual blades. These 2/rev vibes were speed sensitive because the blades responded differently to the forward-speed induced airflow changes - the blades twisted differently, bent differently and had slightly different lift curves because their rigging incidence was different.
If there was a significant split, the 2/rev vertical was obvious in the hover (no forward speed!!!) and we would roll the grips to smooth it out. If the blades flew out of track (the true cause of 2/rev verticals) then we would tab the blades by trial and error to minimise the vibration. I might add that I had the singular pleasure of test flying 5 brand new UH-1Hs from Bell and they were as smooth as many of the 3/4/5 bladers I have flown since - absolutely minimal levels of 2/rev anything.
"Laterals" were different again. Invariably, these vibrations were induced mechanically by mass balancing errors. At near zero lift, very small imbalances resulted in significant rotor swirl/pylon rock. There was no change in vibration level with forward speed in most cases - developing lift did not affect what was essentially a wobbling gyro. They were also characteristically 1-2/rev. If the vibration was of a higher order, say 3 or 4/rev, then the damping characteristics of the transmission mounting system were invariably at fault. Those we would isolate by a pylon rock check in the hover - we would deliberately excite the transmission mounting system and then analyse the time and nature of the damping.
At the risk of being repetitive, I have created or cured most of these characteristic vibrations and I can certainly vouch for the fact that fairly tiny amounts of lead shot in a blade bolt can certainly get a fairly massive and well restrained transmission moving in ways that would water your eyes. I never once worked out how to create extra lift at the front and back to achieve the same thing.
Now for a beautiful '91 Cabernet Sauvignon....
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Stay Alive,
[email protected]
The exciting force is most certainly aerodynamic in regards to what we called "verticals" and you most probably mean by 2/rev. However, in the case of that which we called "laterals" and you call pylon rock, the exciting force was mechanical.
Before Mr Chadwick Helmuth confused the issue and we used to make Miss Bell smooth by flag tracking and interpreting a vibration sensitive bum, we would minimise 2/rev verticals by making corrections to the tip path planes of the individual blades. These 2/rev vibes were speed sensitive because the blades responded differently to the forward-speed induced airflow changes - the blades twisted differently, bent differently and had slightly different lift curves because their rigging incidence was different.
If there was a significant split, the 2/rev vertical was obvious in the hover (no forward speed!!!) and we would roll the grips to smooth it out. If the blades flew out of track (the true cause of 2/rev verticals) then we would tab the blades by trial and error to minimise the vibration. I might add that I had the singular pleasure of test flying 5 brand new UH-1Hs from Bell and they were as smooth as many of the 3/4/5 bladers I have flown since - absolutely minimal levels of 2/rev anything.
"Laterals" were different again. Invariably, these vibrations were induced mechanically by mass balancing errors. At near zero lift, very small imbalances resulted in significant rotor swirl/pylon rock. There was no change in vibration level with forward speed in most cases - developing lift did not affect what was essentially a wobbling gyro. They were also characteristically 1-2/rev. If the vibration was of a higher order, say 3 or 4/rev, then the damping characteristics of the transmission mounting system were invariably at fault. Those we would isolate by a pylon rock check in the hover - we would deliberately excite the transmission mounting system and then analyse the time and nature of the damping.
At the risk of being repetitive, I have created or cured most of these characteristic vibrations and I can certainly vouch for the fact that fairly tiny amounts of lead shot in a blade bolt can certainly get a fairly massive and well restrained transmission moving in ways that would water your eyes. I never once worked out how to create extra lift at the front and back to achieve the same thing.
Now for a beautiful '91 Cabernet Sauvignon....
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Stay Alive,
[email protected]
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Lu asked:
Thanks for the detailed explanation. Now I want you to carry it one step further. If a 1/6 horsepower motor were strong enough to excite a 1000-pound dynamic system and cause it to bounce up and down I would like for you to define the fractional HP motor as it applies the excitation process. Remove the motor from the equation and substitute the 2 per rev. The motor is powered by electricity. What is the motivational force behind the two per rev and what causes it?
Nick answered:
Every rotor produces a natural force wave form at N per rev where N is the number of blades. Basically, there is an oscillation of the lift caused by the natural dynamics of the rotor. I mentioned in a previous post that the magnitude might get as large as 10% of the blade's lift. The frequency is driven by the number of blades, and the amplitude of the oscillation drops very quickly as you get more blades, which is why many-bladed helos are very much smoother than those with few blades.
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[This message has been edited by Nick Lappos (edited 29 June 2001).]
Thanks for the detailed explanation. Now I want you to carry it one step further. If a 1/6 horsepower motor were strong enough to excite a 1000-pound dynamic system and cause it to bounce up and down I would like for you to define the fractional HP motor as it applies the excitation process. Remove the motor from the equation and substitute the 2 per rev. The motor is powered by electricity. What is the motivational force behind the two per rev and what causes it?
Nick answered:
Every rotor produces a natural force wave form at N per rev where N is the number of blades. Basically, there is an oscillation of the lift caused by the natural dynamics of the rotor. I mentioned in a previous post that the magnitude might get as large as 10% of the blade's lift. The frequency is driven by the number of blades, and the amplitude of the oscillation drops very quickly as you get more blades, which is why many-bladed helos are very much smoother than those with few blades.
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[This message has been edited by Nick Lappos (edited 29 June 2001).]
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To: Nick Lappos
I had stated previously that when the fat lady sang (response from Bell) I would either crow about being right or, I would eat crow. This afternoon I received a call form the senior aerodynamicist at Bell. We had a long conversation and I expounded on my views. He gave me a bit of support in stating that there is a variation in generated lift when the blade becomes aligned with the longitudinal axis but that was not the cause of the 2-per rev bounce. He told me that the 2-per rev bounce is generated when the blades are aligned with the lateral axis of the helicopter and it was due to the lift differential across the disc. Now this sounds reasonable but after a bit of thought I became confused. I had always believed that as the advancing blade was moving into the relative wind the pitch had decreased and the opposite was true for the retreating blade as it had higher pitch and was moving with the relative wind. With this scenario in place I was lead to believe that the disc loading was the same on both sides.
I was lead to believe that if there was differential of lift across the disc, the excess lifting forces generated by the advancing blade would result in a perturbing force and through gyroscopic action would result in a flap back (blow back) as this is what happens when you have retreating blade stall. Would you care to comment? Please don’t sprain you arm patting yourself on the back.
I can’t get my money back from the school of engineering as I was an Industrial Design Major.
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The Cat
I had stated previously that when the fat lady sang (response from Bell) I would either crow about being right or, I would eat crow. This afternoon I received a call form the senior aerodynamicist at Bell. We had a long conversation and I expounded on my views. He gave me a bit of support in stating that there is a variation in generated lift when the blade becomes aligned with the longitudinal axis but that was not the cause of the 2-per rev bounce. He told me that the 2-per rev bounce is generated when the blades are aligned with the lateral axis of the helicopter and it was due to the lift differential across the disc. Now this sounds reasonable but after a bit of thought I became confused. I had always believed that as the advancing blade was moving into the relative wind the pitch had decreased and the opposite was true for the retreating blade as it had higher pitch and was moving with the relative wind. With this scenario in place I was lead to believe that the disc loading was the same on both sides.
I was lead to believe that if there was differential of lift across the disc, the excess lifting forces generated by the advancing blade would result in a perturbing force and through gyroscopic action would result in a flap back (blow back) as this is what happens when you have retreating blade stall. Would you care to comment? Please don’t sprain you arm patting yourself on the back.
I can’t get my money back from the school of engineering as I was an Industrial Design Major.
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The Cat
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There are some very knowledgeable people on this thread, so with trepidation, and for the fun of it, the following is humbly submited.
________________________
Lu, could it be that Nick is correct in that the thrust of the rotor disk is quite consistent, irrespective of the location of the blades?
Could it be that in forward flight with a 2-blade rotor the downwash from the blades is striking the fuselage at 2/rev. During hover, this downwash is coming from center of the rotor disk where it is minimal. As the helicopter's forward speed increases, the downwash striking the fuselage will become stronger, since it is now coming from closer to the leading edge of the rotor disk. This will give the fuselage a 2/rev vertical vibration. This segment of the downwash is initiated when a blade is pointing forward but it reaches the fuselage when the blades are out to the sides.
[i.e. The rotor thrust is constant and the parasitic drag oscillates at 2P.]
A continuation of the oscillations, particularly if there is harmonic excitement, will cause both the mast and the fuselage to oscillate. The rotors oscillations will be the greater by far since it has the least mass, but, the rotor's oscillations will be restricted to the mast, hub and inner portion of the disk. The tips will oscillate very little and the thrust will vary very little.
Just a guess.
_________________________
A test for pylon rocking is described on the following web page: http://www.adtdl.army.mil/cgi-bin/at.../1-211/Ch7.htm
[This message has been edited by Dave Jackson (edited 30 June 2001).]
[This message has been edited by Dave Jackson (edited 02 July 2001).]
________________________
Lu, could it be that Nick is correct in that the thrust of the rotor disk is quite consistent, irrespective of the location of the blades?
Could it be that in forward flight with a 2-blade rotor the downwash from the blades is striking the fuselage at 2/rev. During hover, this downwash is coming from center of the rotor disk where it is minimal. As the helicopter's forward speed increases, the downwash striking the fuselage will become stronger, since it is now coming from closer to the leading edge of the rotor disk. This will give the fuselage a 2/rev vertical vibration. This segment of the downwash is initiated when a blade is pointing forward but it reaches the fuselage when the blades are out to the sides.
[i.e. The rotor thrust is constant and the parasitic drag oscillates at 2P.]
A continuation of the oscillations, particularly if there is harmonic excitement, will cause both the mast and the fuselage to oscillate. The rotors oscillations will be the greater by far since it has the least mass, but, the rotor's oscillations will be restricted to the mast, hub and inner portion of the disk. The tips will oscillate very little and the thrust will vary very little.
Just a guess.
_________________________
A test for pylon rocking is described on the following web page: http://www.adtdl.army.mil/cgi-bin/at.../1-211/Ch7.htm
[This message has been edited by Dave Jackson (edited 30 June 2001).]
[This message has been edited by Dave Jackson (edited 02 July 2001).]
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Let me see if I can put to rest the "loss of lift" theory once and for all.
Consider the swept wing jet aircraft. The airflow over the wing does not flow with a velocity vector that is perpendicular to the leading edge of the wing, in fact far from it. Some jet aircraft have had wings with sweep angles up to 45 degrees (the MIG-17 comes to mind), and somehow the wings managed to generate lift. I believe the Concorde has a wing sweep even greater than this.
So if a rotor blade passing across the front of the helo, has a velocity component caused by its rotation and a velocity component caused by the helos forward motion, will the combined velocity vector by straight down the centerline of the blade's airfoil? I don't think so, and therefore the blade will still be generating lift.
The interesting part is that the "sweep angle" of the leading edge relative to this combined velocity vector will be constantly changing from the blade root to the tip, since the velocity vector caused by the blade's rotation will vary constantly from root to tip. Even at the blade root there would still be some airflow from leading edge to trailing edge, however the relative "sweep angle" at the root would be severe.
[This message has been edited by Flight Safety (edited 30 June 2001).]
Consider the swept wing jet aircraft. The airflow over the wing does not flow with a velocity vector that is perpendicular to the leading edge of the wing, in fact far from it. Some jet aircraft have had wings with sweep angles up to 45 degrees (the MIG-17 comes to mind), and somehow the wings managed to generate lift. I believe the Concorde has a wing sweep even greater than this.
So if a rotor blade passing across the front of the helo, has a velocity component caused by its rotation and a velocity component caused by the helos forward motion, will the combined velocity vector by straight down the centerline of the blade's airfoil? I don't think so, and therefore the blade will still be generating lift.
The interesting part is that the "sweep angle" of the leading edge relative to this combined velocity vector will be constantly changing from the blade root to the tip, since the velocity vector caused by the blade's rotation will vary constantly from root to tip. Even at the blade root there would still be some airflow from leading edge to trailing edge, however the relative "sweep angle" at the root would be severe.
[This message has been edited by Flight Safety (edited 30 June 2001).]
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Lu,
You may be looking deeper into it than me at the moment, but could you be meaning to say the 'Angle of Atack' when you are talking about the "pitch Angle" in reference to symetry of lift.
Ie: the leading blade reduces its angle of attack as it flaps up not reduces its pitch angle.
On the point of sweep back, probably not relevent in the rotor blade scenario, but with a span wise flow of air across the blade when it's anywhere in the 12 to 3 oclock position, could the 'fineness' (thickness to cord ratio) be a more efficient producer of lift than when the blade is in the truly perpendicular position ?
It's the big words that make it fun.
You may be looking deeper into it than me at the moment, but could you be meaning to say the 'Angle of Atack' when you are talking about the "pitch Angle" in reference to symetry of lift.
Ie: the leading blade reduces its angle of attack as it flaps up not reduces its pitch angle.
On the point of sweep back, probably not relevent in the rotor blade scenario, but with a span wise flow of air across the blade when it's anywhere in the 12 to 3 oclock position, could the 'fineness' (thickness to cord ratio) be a more efficient producer of lift than when the blade is in the truly perpendicular position ?
It's the big words that make it fun.
Guest
Posts: n/a
To: rotorque
In addressing the pitch angle of the blade I was referring to the fact that the advancing blade in the 3:00 position is at it lowest point of pitch and the retreating blade at 9:00 is at its’ highest point of pitch. In theory the advancing blade with less pitch rotating into the relative wind will have the same degree of lift as the retreating blade which is rotating with the relative wind. In theory this equalizes the lift across the disc. In fact, the advancing blade will flap somewhat and further reduce its’ angle of attack in relation to the relative wind. I believe this flapping manifests itself in blowback or flapback whatever you wish to call it and it is countered with the forward movement of the cyclic stick.
However in order to create an advancing and retreating blade the helicopter must be moving in a given direction. In order to do that, the disc must be tilted in the direction of movement. To accomplish this tilting the lift differential must be greater on the left side (assuming forward flight) and the lift must be less on the right side. This differential of lift across the disc perturbs the rotating disc and the gyroscopic forces cause the disc to tilt up 90-degrees after the upward perturbing force. That is, if you believe in gyroscopic precession as this is not taught in the UK or, I believe in OZ.
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The Cat
[This message has been edited by Lu Zuckerman (edited 01 July 2001).]
In addressing the pitch angle of the blade I was referring to the fact that the advancing blade in the 3:00 position is at it lowest point of pitch and the retreating blade at 9:00 is at its’ highest point of pitch. In theory the advancing blade with less pitch rotating into the relative wind will have the same degree of lift as the retreating blade which is rotating with the relative wind. In theory this equalizes the lift across the disc. In fact, the advancing blade will flap somewhat and further reduce its’ angle of attack in relation to the relative wind. I believe this flapping manifests itself in blowback or flapback whatever you wish to call it and it is countered with the forward movement of the cyclic stick.
However in order to create an advancing and retreating blade the helicopter must be moving in a given direction. In order to do that, the disc must be tilted in the direction of movement. To accomplish this tilting the lift differential must be greater on the left side (assuming forward flight) and the lift must be less on the right side. This differential of lift across the disc perturbs the rotating disc and the gyroscopic forces cause the disc to tilt up 90-degrees after the upward perturbing force. That is, if you believe in gyroscopic precession as this is not taught in the UK or, I believe in OZ.
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The Cat
[This message has been edited by Lu Zuckerman (edited 01 July 2001).]
