Mark E
Member
- Joined
- Jan 3, 2008
- Messages
- 592
- Location
- Singapore
- Total Flight Time
- A total of about 60 minutes dual hangin onto a stick in Cessna, a glider, and a gyrocopter.
A simple (or simplistic?) view of rotor behavior and control. (doesn't really use the terms flapping or gyroscopic precession)
From https://www.rotaryforum.com/forum/showthread.php?t=22760
Ok, forget the different behavior differnces of longer rotors, then. I need to think out some basics:
Reckon I’ve got the left tilt sorted in my head (maybe!) Much is said throughout these threads in one way or another, but it’s often cryptically explained because the knowledge of the instructor is way ahead of the listener (ie, me). (If I’m way off, and someone screams, I’ll delete, but writing this down helps my thinking a bit, and leaves me open to advice)
Consider CBs thrown 3 bladed rotor, or the boomerang. Thrown, it will arc upwards and away to the left. (edit! Wrong! He actually said it would do a normal inside loop!) If it could continue, it would complete a slow leftward roll. (edit, also wrong!)I think a (somehow balanced for independant flight) two blade rotor really wants to do the same. (edit -not any more I don't! I htink it would also do an inside loop)A rotor would like to keep climbing up (and to the left NOPE) , but a control input stops it. (CB has said this many times, but I am not clear on the input, and have not yet found anyone else’s clearly stated opinion on the matter worded to get to my feeble neurons). (Edit - the control input is important, it is just controlling blade pitch in the 3 /9 axis, thereby sustaining straight and level flight)
That control input is the angle at which the pilot (or trim spring) angles the rotor head in the fore/aft axis (and therefore angles the disk of rotation of the teeter bolt) to the oncoming air stream (ie not the left or right tilt stick tilt, but the forward/aft or N/S stick tilt).
Consider a rotor head on a post in a wind tunnel. Spun up by a prerotator to 300 rrpm. The disk of rotation of the teeter bolt and the disk of rotation of the blade tips are completely parallel to eachother. (this tip path is, my terminology here, “the ideal tip path”).
(and I know the tunnel does not work for the ‘up and away’ action, but it does not get that far)
Lock the rearward tilt of the rotor head so the ideal tip path is tilted upward into the airflow the required amount for flight, then instantly (and theoretically speaking!) turn on an 80 mph (130 km/hr) wind (=forward airspeed).
The advancing blade then wants to set off on a climbing course (and the retreating a diving one, I won’t keep stating this opposite reaction) but is restricted by the disk of rotation of the teeter bolt. (ie the airfoil/blade is compelled to remain in a parallel alignment to the disk of rotation of the teeter bolt)
Which is in turn maintained by the control input of the N/S position of the stick.
Due to the angle that this disk of rotation of the teeter bolt is held, the advancing blade cannot increase it’s AOA in response to a 300 mph (450 Km/hr) wind on its nose and an 80 mph (130 km/hr) wind hitting it’s underside at a 2 (edit 12) degree angle, and to then embark on the spiraling upward journey it wants to take, so has no choice but to climb (levitate) across the “ideal tip path” (and yes, thereby reducing it’s AOA* in relation to the airstream – it is rising say 2 meters (edit - by Chris' later comments, the teeter angle is 2* or less, so this rise is only about 20 cm (8 in) for an * m (24 ft) rotor.......?) in one tenth of a second, so there is an airspeed component at right angles downward to the ideal tip path of about 45 mph (edit- only 4.5 on the recalc??), so there develops a downward airflow vector).
The major point at which the advancing (and retreating) blade want (and fail!) to exert their will on the teeter bolt rotational disk, by changing pitch in the direction they want to, is of course at the 3 and 9 O’clock positions, so the advancing blade is forced on a lower path to 12 oclock than the one it intended to take. Ensuring the disk remains level (and a gyro would fly straight). The same but opposite effect is occurring to the retreating blade.
Now the teeter bolt rotation disk and the actual tip path disk are no longer exactly parallel. This is allowed mainly due to flexing of the blades, (edit, and the teeter, the freedom from the control inputs in the 12/6 position)(and must contribute to the 2/rev shake, two different masses rotating on different disk planes, and this difference varies with control inputs**)
At this point the blade tips are levitating across the ideal tip path at about 3 oclock on the advancing side, reaching a peak at the front (near 12 O’clock), and are doing the opposite on the retreating side, reaching a lowest point at the back (ie 6 oclock)
Gyroscopic forces play a huge part in this, there are tonnes of force involved, the rotor disk does not want to move easily, a (theoretical) weightless rotor would be responding dramatically to the changing airflows. (without getting into a gyroscopic precession discussion, the major forces are in fact being expressed in maximum displacement 90 degrees later)
I still think things are not completely in balance, as the argument between the blades and the teeter bolt (more specifically, the plane of rotation of the teeter bolt) (andother 2/rev contribution?) means an upward tilt force is being applied to the teeter bolt disk plane on one side, and the opposite on the other, and these forces are passed through the controls and airframe. (edit , I now think this is relatively minor, and it would just be felt through the stick)
There is also the issue of the frequently mentioned (again by CB) effect of the flexed up blade at the 12 Oclock position encountering a larger force of air on the underside than the other blade in the 6 oclock position (the built in coning, plus the upward blade flex, and the fact the rearmost blade encounters turbulent air, means that the blade at 6 oclock encounters a lesser upward force)
As the N blade begins it’s retreating path, it ends up flying a little higher at 9 O’clock than it had intended).
*I also understand that exactly what an airfoil does in different airstreams is very much dependant on it’s profile, and it’s chord balance, I just find it easier to think of AOA in stages.
** Without going into intricacies of coning and teeter height, and mass above and below the teeter bolt, which seems to be a very interesting subject.
(without mentioning flapping or precession)
I haven’t put any qualifiers in this (such as 'I think', or, 'perhaps') because it is hard enough to follow anyway, but that's what I'm saying all the way through.
.
From https://www.rotaryforum.com/forum/showthread.php?t=22760
Ok, forget the different behavior differnces of longer rotors, then. I need to think out some basics:
Reckon I’ve got the left tilt sorted in my head (maybe!) Much is said throughout these threads in one way or another, but it’s often cryptically explained because the knowledge of the instructor is way ahead of the listener (ie, me). (If I’m way off, and someone screams, I’ll delete, but writing this down helps my thinking a bit, and leaves me open to advice)
Consider CBs thrown 3 bladed rotor, or the boomerang. Thrown, it will arc upwards and away to the left. (edit! Wrong! He actually said it would do a normal inside loop!) If it could continue, it would complete a slow leftward roll. (edit, also wrong!)I think a (somehow balanced for independant flight) two blade rotor really wants to do the same. (edit -not any more I don't! I htink it would also do an inside loop)A rotor would like to keep climbing up (and to the left NOPE) , but a control input stops it. (CB has said this many times, but I am not clear on the input, and have not yet found anyone else’s clearly stated opinion on the matter worded to get to my feeble neurons). (Edit - the control input is important, it is just controlling blade pitch in the 3 /9 axis, thereby sustaining straight and level flight)
That control input is the angle at which the pilot (or trim spring) angles the rotor head in the fore/aft axis (and therefore angles the disk of rotation of the teeter bolt) to the oncoming air stream (ie not the left or right tilt stick tilt, but the forward/aft or N/S stick tilt).
Consider a rotor head on a post in a wind tunnel. Spun up by a prerotator to 300 rrpm. The disk of rotation of the teeter bolt and the disk of rotation of the blade tips are completely parallel to eachother. (this tip path is, my terminology here, “the ideal tip path”).
(and I know the tunnel does not work for the ‘up and away’ action, but it does not get that far)
Lock the rearward tilt of the rotor head so the ideal tip path is tilted upward into the airflow the required amount for flight, then instantly (and theoretically speaking!) turn on an 80 mph (130 km/hr) wind (=forward airspeed).
The advancing blade then wants to set off on a climbing course (and the retreating a diving one, I won’t keep stating this opposite reaction) but is restricted by the disk of rotation of the teeter bolt. (ie the airfoil/blade is compelled to remain in a parallel alignment to the disk of rotation of the teeter bolt)
Which is in turn maintained by the control input of the N/S position of the stick.
Due to the angle that this disk of rotation of the teeter bolt is held, the advancing blade cannot increase it’s AOA in response to a 300 mph (450 Km/hr) wind on its nose and an 80 mph (130 km/hr) wind hitting it’s underside at a 2 (edit 12) degree angle, and to then embark on the spiraling upward journey it wants to take, so has no choice but to climb (levitate) across the “ideal tip path” (and yes, thereby reducing it’s AOA* in relation to the airstream – it is rising say 2 meters (edit - by Chris' later comments, the teeter angle is 2* or less, so this rise is only about 20 cm (8 in) for an * m (24 ft) rotor.......?) in one tenth of a second, so there is an airspeed component at right angles downward to the ideal tip path of about 45 mph (edit- only 4.5 on the recalc??), so there develops a downward airflow vector).
The major point at which the advancing (and retreating) blade want (and fail!) to exert their will on the teeter bolt rotational disk, by changing pitch in the direction they want to, is of course at the 3 and 9 O’clock positions, so the advancing blade is forced on a lower path to 12 oclock than the one it intended to take. Ensuring the disk remains level (and a gyro would fly straight). The same but opposite effect is occurring to the retreating blade.
Now the teeter bolt rotation disk and the actual tip path disk are no longer exactly parallel. This is allowed mainly due to flexing of the blades, (edit, and the teeter, the freedom from the control inputs in the 12/6 position)(and must contribute to the 2/rev shake, two different masses rotating on different disk planes, and this difference varies with control inputs**)
At this point the blade tips are levitating across the ideal tip path at about 3 oclock on the advancing side, reaching a peak at the front (near 12 O’clock), and are doing the opposite on the retreating side, reaching a lowest point at the back (ie 6 oclock)
Gyroscopic forces play a huge part in this, there are tonnes of force involved, the rotor disk does not want to move easily, a (theoretical) weightless rotor would be responding dramatically to the changing airflows. (without getting into a gyroscopic precession discussion, the major forces are in fact being expressed in maximum displacement 90 degrees later)
I still think things are not completely in balance, as the argument between the blades and the teeter bolt (more specifically, the plane of rotation of the teeter bolt) (andother 2/rev contribution?) means an upward tilt force is being applied to the teeter bolt disk plane on one side, and the opposite on the other, and these forces are passed through the controls and airframe. (edit , I now think this is relatively minor, and it would just be felt through the stick)
There is also the issue of the frequently mentioned (again by CB) effect of the flexed up blade at the 12 Oclock position encountering a larger force of air on the underside than the other blade in the 6 oclock position (the built in coning, plus the upward blade flex, and the fact the rearmost blade encounters turbulent air, means that the blade at 6 oclock encounters a lesser upward force)
As the N blade begins it’s retreating path, it ends up flying a little higher at 9 O’clock than it had intended).
*I also understand that exactly what an airfoil does in different airstreams is very much dependant on it’s profile, and it’s chord balance, I just find it easier to think of AOA in stages.
** Without going into intricacies of coning and teeter height, and mass above and below the teeter bolt, which seems to be a very interesting subject.
(without mentioning flapping or precession)
I haven’t put any qualifiers in this (such as 'I think', or, 'perhaps') because it is hard enough to follow anyway, but that's what I'm saying all the way through.
.
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