A simple (or simplistic?) view of rotor behavior

[Mark E]

Ed, Chris, Birdy,

You are all absolutely right! - I stand corrected (again!) (I was just having another one of my little mental lapses!) I didn't realize exactly what the correct definition of a rigid rotor was! I had better just stick to making this mental teetering one fly straight.

PS Ed, that sounds ingenious!

 

[Mark E]

But, anyway........

I think I can now visualize Birdy’s leftward roll with increased teeter bearing friction, (which I couldn't a week ago).

…. in S&L flight with the tip planes level at the 3 / 9 o’clock position, there is no movement around the teeter bearing.

As the advancing blade then approaches the 12 position, it is climbing, while the retreating blade is diving, there is a pivoting action on the teeter bearing.

Centrifugal force is moderating these effects and is keeping the blades on one plane.

The only place the friction has an effect is in the vicinity of the 12 / 6 position, when the rotor is trying to teeter on the bearing. (edit: wrong, it is in fact at the limit of it's teeter, and about to reverse it's direction of movement) If the bearing has increased friction, it will resist the rotor attempting to climb towards 12 and dive towards 6. (Edit: See Birdy/Russ below, don't know what I was thinking!, but what I said is wrong, the blades are certainly near level at 3 / 9, but are in the middle of the process of pivoting on the teeter bolt, so the speed of pivot around that bolt must be at it's fastest. At 12 / 6 they are at the limit of their teeter, and are reversing the direction of movement, so at that point are moving more slowly on the pivot.)


The rotor flies across the 12 position at a lower height (or lower lift?), and the 6 at a higher height (compared with the ‘frictionless’ bearing) The upward push at 12 is lessened, so the blade is subsequently lower at 9, and the downward push at 6 is lessened, so the blade is subsequently higher at 3. So the gyro is rolling left. Or at the very least has a lateral movement to the stick.

... or have I just crashed again?

:suspicious:

 

[Jean Claude]

Centrifugal force is moderating these effects and is keeping the blades on one plane.

Grimwat, this is not the centrifugal force that moderates these effects. It is the upward trajectory of the blade, relatively to the plane of drive, which decreases its angle of attack. Thus, the lift balance is restored automatically, with a real plane rotation, different of the drive plane. Imagine two sections of the blade opposed rotating at 200 mph around a center with a pitch of 3°, and the center moving at 50 mph. On one side, section cuts the air at 250 mph , while the other side other section cuts the air at 150 mph. To equalize the lift, it will take an angle of attack of 1.6 ° for the fast stretch, and 4.4 ° for the other. This occurs when the flapping of the blades is 1.4 °.
Jean Claude

 

[birdy]

…. in S&L flight with the tip planes level at the 3 / 9 o’clock position, there is no movement around the teeter bearing.
You got it ass about Mark.
Yes, the blades mat be colse to level wen at 3 n 9, but one is on its way up, and the other is on its way down, in relation to the teeter hinge.
So its at 3 n 9 that the greatest teetering movement is happening.
At 12 n 6, they start to swap sides, so theres buggerall movement there on the teeter hinge.

 

[Russ Hobbs]

Birdy, Do you mean Mark has it 180 out, meaning that the movement about the teetering hinge is greatest at 3-9 and basically neutral at 12-6. If this is the case, in my mind the blade on the right side is swinging upwards in flight pushing the retreating blade down on the left side and at 12-6 they are actually in balance on the teetering hinge point. Which would put the blade at 12 at its highest point in rotation and of course the 6 at it's lowest.

 

[Mark E]

.... not the centrifugal force that moderates these effects. It is the upward trajectory of the blade, relatively to the plane of drive, which decreases its angle of attack. Thus, the lift balance is restored automatically, with a real plane rotation, different of the drive plane. Imagine two sections of the blade opposed rotating at 200 mph around a center with a pitch of 3°, and the center moving at 50 mph. On one side, section cuts the air at 250 mph , while the other side other section cuts the air at 150 mph. To equalize the lift, it will take an angle of attack of 1.6 ° for the fast stretch, and 4.4 ° for the other. This occurs when the flapping of the blades is 1.4 °.
Jean Claude

Thanks, Jean Claude for hitting me with the precise facts of the case! Helps my thinking a lot, and very timely. I read your answer this morning, and thinking about it today, decided I had the rotors behaving the right way, but for some of the wrong reasons. (Edit But I was not trying to say the centrifugal force was dealing with the dissymmetry of lift, but that it was trying to keep the blades on plane in resistance to both that force and the control inputs) But I think I might have to go back and edit a few more of my previous statements!) (I deleted my reply to John/All In ... didn't really make much sense)

hwell:

 

[Mark E]

…. level wen at 3 n 9, but one is on its way up, and the other is on its way down, in relation to the teeter hinge.
So its at 3 n 9 that the greatest teetering movement is happening.
At 12 n 6, they start to swap sides, so theres buggerall movement there on the teeter hinge.

Yep Birdy, Russ, my thinking and what I said is wrong, the blades are certainly near level at 3 / 9, but are in the middle of the process of pivoting on the teeter bolt, so the speed of pivot around that bolt must be at it's fastest.

At 12 / 6 they are at the limit of their teeter, and are reversing the direction of movement, so at that point are moving more slowly on the pivot.

Anyway, makes it hard to explain the sticky bearing, unless it just grabs at the extreme of movement, as it slows and reverses.

 

[birdy]

Russ, yes, assabout is ozzy for backtofrunt.
And
that the movement about the teetering hinge is greatest at 3-9 and basically neutral at 12-6. ,
yes.

in my mind the blade on the right side is swinging upwards in flight pushing the retreating blade down on the left side and at 12-6 they are actually in balance on the teetering hinge point.
Neither blade is pushn anythn anywhere.
If they pushed eachother up n down twice per rev, your hub bar wouldnt last past the TO roll.
They are always blanced in lift, so each will rise and fall equally on either side, no matter the AS difference.
At 12 n 6, they are transitioning from one side to the other, which mearns the indervidual blades AS are swapn. One will see increasing AS, the other decreasen AS, so, from the teeter hinge perspective, the advancing blades climb has stoped and the retreating blades decent has stoped. Movement on the teeter hinge is about to reverse.

unless it just grabs at the extreme of movement, as it slows and reverses.
To feel friction, you need movement. The greater the rate of movement, the greater the friction felt.
So, the greatest movement against the towers thrust washer will be at 3 n 9, and will be tryn to drag the stick with it, left.

 

[Resasi]

Always thought a rigid rotor referred to a gyro rotor which is rigid however have seen that we were talking about a helicopter rigid rotor.

Still following the unfolding thread and although I had grasped the essential elements re advancing retreating and their relative positions in space am still enjoying the unfolding and correction of misconceptions with little added facts that I hadn't yet 'got'.

 

[Mark E]

Russ, yes, assabout is ozzy for backtofrunt.

Ha, yeah I reckoned I got that OK - Russ mentioned 180 out - but I'm really ONLY 90 degrees out (ain't I?) - but then assabout pretty well covered all angles! :lol:


Thanks for the feedback - looks like instead of trying to think of ways to simplify complicated things in my mind I managed to make a simple thing complicated!

Does it just pull the stick left, or do you get a bit of left roll too?

 

[Mark E]

....... am still enjoying the unfolding and correction of misconceptions with little added facts that I hadn't yet 'got'.

Leigh, thanks for the feedback, (and likewise to John/All In on the first page) it is a bit nicer to know that someone else is enjoying the discussion too. (I am, and I appreciate very much the guys who take the time to reply)

I agree with everything Pete/Passin Thru said at the start; and at a first glance it seems improbable the things could fly at all, but it becomes clear that they are truly excellent flying machines. And I marvel at the minds that first theorized and calculated and tweaked and adjusted and made all this actually fly.

 

[Mark E]

Grimwat, this is not the centrifugal force that moderates these effects. It is the upward trajectory of the blade, relatively to the plane of drive, which decreases its angle of attack. Thus, the lift balance is restored automatically, with a real plane rotation, different of the drive plane. Imagine two sections of the blade opposed rotating at 200 mph around a center with a pitch of 3°, and the center moving at 50 mph. On one side, section cuts the air at 250 mph , while the other side other section cuts the air at 150 mph. To equalize the lift, it will take an angle of attack of 1.6 ° for the fast stretch, and 4.4 ° for the other. This occurs when the flapping of the blades is 1.4 °.
Jean Claude

Just bit more on that: Again, Jean Claude, thanks for the feedback, but I wasn't meaning to suggest that the centrifugal force was balancing the dissymmetry of lift, I've pretty well got that sorted in my head now , (I think )- I probably need to go back to that post and make it a bit clearer, but I was making the point that the response to the aerodynamic forces, and to the control inputs is moderated by the centrifugal force (angular momentum), - we can only make it move a relatively small amount each revolution.

And I think it is just the momentum of the blade which dictates it's path across the 12 and 6 positons? - here both blades are seeing equal air speed and so both are (edit .. creating) the same lift - so neutralizing any movement on the teeter, and are not receiving any control inputs. (edit, probably better to say neutral in terms of control inputs)

 

[birdy]

Does it just pull the stick left, or do you get a bit of left roll too?
?????????
Left roll happens wen the stick goes left.
If you let the stick go wen youv got teeter friction, it will move left, so the machine rolls left.

And I think it is just the momentum of the blade which dictates it's path across the 12 and 6 positons?
Cant be just momentum Grim, coz they are still flyn, which means they still have the same amount of airodynamic forces on them.
ASs are near the same but one is in clean air n the other in dirty air, and the 12 oclock one has a greater AOA.
The sole reason theres minimal movement on the teeter at 12 n 6 is coz the teeter action is swapn sides. The advancing blades is entering the retreating side and v/v, so the actual movement on the hinge stops and reverses, twice per rev. [ and wen its stoped, theres no friction.]

 

[Mark E]

Does it just pull the stick left, or do you get a bit of left roll too?
?????????
Left roll happens wen the stick goes left.
If you let the stick go wen youv got teeter friction, it will move left, so the machine rolls left.

And I think it is just the momentum of the blade which dictates it's path across the 12 and 6 positons?
Cant be just momentum Grim, coz they are still flyn, which means they still have the same amount of airodynamic forces on them.
ASs are near the same but one is in clean air n the other in dirty air, and the 12 oclock one has a greater AOA.
The sole reason theres minimal movement on the teeter at 12 n 6 is coz the teeter action is swapn sides. The advancing blades is entering the retreating side and v/v, so the actual movement on the hinge stops and reverses, twice per rev. [ and wen its stoped, theres no friction.]

hmmm stil thinking on this one :suspicious:.. the blade is at this stage (12 / 6), flying parallel to the teeter bolt plane, so the AOA is dictated by that, but there is not the same pressure to adjust AOA as there was at the 3 / 9 position - dirty air aside, they basically see (edit - create) the same lift - so it is not going anywhere aerodynamically, so basically it's momentum carries it (advancing blade) from the control situation at 3 through 12 and on to 9 ? - (the transition from one control stage to the other is gradual, not sudden, )


(a further edit - I think by the time the blades reach 12 / 6 they both have the same AOA - same airspeed, same lift)

 

[Mark E]

...............Left roll happens wen the stick goes left.
If you let the stick go wen youv got teeter friction, it will move left, so the machine rolls left. .................

Yeah,- makes sense - I was just trying to tie it in with CBs saying he just got a lateral movement in the stick (maybe with less 'stickyness'), ie making sure you weren't just indicating that - I didn't think you were ;

But I find this interesting; picturing what the blades are doing - the advancing blade therefore must be flying a lower path to 12, and the retreating a higher path to 6, in effect the friction has reduced the teeter angle. As the advancing blade becomes retreating it will again be taking a lower path than the previous pass - so the left roll will continue unless corrected.

Sorry if I am stating the obvious here, I think it is leading up to another question, as soon as I think a few things out.

:suspicious:

 

[birdy]

so the AOA is dictated by that,
The AOA is dictated by rpm and load.
The teeter bolt only applies a change.

they basically see the same lift
They create the same lift, but theres higher AOA on the 12 Ocl blade, thats why [ as CB explained] the highest point on the disc is just after 12,, its still teetering.

from the control situation at 3 through 12 and on to 9 ? -
?????????
Youv got control over the blades on every point of its parth through 360*.
Wen you control in pitch, your cyclicly activatn lateraly and in roll, longditudinaly on the disc.

As the advancing blade becomes retreating it will again be taking a lower path than the previous pass - so the left roll will continue unless corrected.
Yup.

Dont forget, im only sayn wot i think iv learned.

 

[Mark E]

The AOA is dictated by rpm and load.
The teeter bolt only applies a change.

......................
They create the same lift, but theres higher AOA on the 12 Ocl blade, thats why [ as CB explained] the highest point on the disc is just after 12,, its still teetering.

from the control situation at 3 through 12 and on to 9 ? -
?????????
Youv got control over the blades on every point of its path ...................

This path the blades take and the pitch changes (changes in AOA) are still leaving a few questions in my mind (but I might have to draw something to ask those .....)

But I'm still thinking the AOA at 6 and 12 should be the same. (edit - when they arrive at 12 / 6 blades are then) experiencing the same airspeed and lift, and are traveling parallel to the teeter bolt rotational plane - and true - the pilot has control over this pitch at all points, it is just that in S&L flight, he's neutral in terms of control input ? (edit; on the 6 / 12 line )


edit - I'm just trying to trace how the blades are acting at all 4 points - in pitch and flight path - I think it needs a picture or two - so I'll try to do that when I get time , then ask a few more qestions to get them tracking correctly!

 

[Mark E]

Birdy - if you had private messages activated, I could ask my dumbest questions off record (but one catch, sometimes it's not obvious they're dumb until they get answered!

 

[birdy]

Birdy - if you had private messages activated,
If i wasnt so thick, id know wot your talkn bout.

experiencing the same airspeed and lift,
For the sake of the argument, [simplicity] they are.
[ its only wen you talk to people like CB n JC, the ones who do know, you find out that they aint at quite the same AOA, yet.]

and are traveling parallel to the teeter bolt rotational plane
??????????
Wots that mean?

 

[C. Beaty]

The word “flap” is the source of confusion for most people.

A rotor doesn’t flap in the sense that a bird’s wings flap but viewed at some angle to its real axis of rotation, we can imagine it to flap.

View a spinning bicycle wheel off axis and there could be an illusion of flapping. The valve stem moves nearer and away from the viewer; therefore it’s flapping?

Of course, if the wheel flaps, it must speed up and slow down in order to accommodate those masses that don’t lie on a single plane (say a dished bicycle wheel).

When a rotor is viewed from the rotorhead axis it looks like it’s flapping so we have to use fictitious forces to account for its behavior; Coriolis force to account for the speed up and slow down, also the sum of rotational velocity and flapping velocity to arrive at the angle of relative wind.

Viewed from the real axis of rotation, the tip plane axis, none of the imaginary forces and motions exist. There is only a cyclical variation of pitch if the tip plane and rotorhead axes are not concentric.

The attached sketch illustrates a Cardan joint. Imagine the ring to be a short section of water pipe.

Given a spin, the ring will maintain its orientation in space no matter which way the input shaft (the blue part) is tilted as a result of isolation provided by the Cardan joint (pivot friction will eventually bring the ring axis into alignment with the input shaft).

Extrapolating the sketch to a teetering rotor system, the blue part would be the rotorhead and the red part the rotor. All that tilting the rotorhead can do is rotate the rotorblades about their feathering axes and produce a cyclical variation of pitch as they spin. Aerodynamic force tends to bring the rotor tip plane axis into alignment with the rotorhead axis. During translational flight, the axes are never aligned in order to maintain equilibrium between advancing and retreating blades.

With a 3-blade rotor, an additional set of hinges is required to complete the universal joint.