A simple (or simplistic?) view of rotor behavior

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

Birdy the private message system is part of the facilities in the Rotary Wing Forum where individuals can discuss something in private and not through a particular thread, perhaps clear a particular point, or apologize for insulting them...or not, as the case may be.

If you click on a persons Name/Avatar/picture/logo a drop down menu will give some options. second one down is Send a private message to ...

Click that and you will get a box similar to the one you use for posting on the thread except it will go only to that person.

To see if you have any, top right of the opening page where it says welcome Birdy it will also tell you if you have any.

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

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 ....................

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 ............................

Thanks Chuck, you may have answered one of my upcoming questions in regard to the tip path plane -
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. I can think of the forces involved better if I think of the blades flying to a certain position, ie fly higher or lower, but at the same time I'm aware that the application on a force at a certain point would result in the movement of a gyroscope 90 degrees later. I was going to ask whether the rotor tips followed a perfect path, or whether they did in fact vary slightly from that path (perhaps only a few mm)... ie as per my mental picture ... which I will continue to see anyway!

Yes, the term flap has sent my mind off on a few tangents for sure, (and I'd banish it! ... the word, not my mind.)

Aerodynamic force tends to bring the rotor tip plane axis into alignment with the rotorhead axis.

The above phrase worries me a little, I picture the aerodynamic forces as balanced at 12 / 6 (ie equal lift) but at 3 / 9 the aerodynamic forces are trying to change the path - it is the control input to fly level that actually ensures the lift is equalized, reducing (moderating is a better word?) pitch on the advancing blade and increasing pitch on the retreating blade, so the rotor disk plane does not keep tilting upwards, the effect of "flapping" actually ensures the rotor tip disk plane is not aligned with the teeter bolt rotational axis, so applying a pitching force ... (in fact, this has been the crux of my mental approach )

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.........................

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

Sorry, I have to apologize for creating my own terminology! (it's the way my mind works :twitch - I'm thinking of the plane the teeter bolt rotates in as a disk (a bit like a heli swash plate?) because it is the difference in the angle that this bolt is on, and the path the rotor tip is traveling that applies a force affecting the pitch of the blade (?) - when the advancing blade is at 3 oclock (according to JC in a post above, 'flapping angle of 1.4* ') the difference in the angle that teeter bolt is at (ie the plane it is rotating in), and the path the blade tip is taking is 1.4*. (usually shown as the angle between the spindle bolt and the vertical axis of the rotor plane tip path - much easier to say, but for me, more confusing as to what is happening).

But when the blades are at 12 / 6 the teeter bolt is in alignment to the blade tip path, with the blade at 12 having teetered upward, and vise versa for the blade at 6.

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Aerodynamic force tends to bring the rotor tip plane axis into alignment with the rotorhead axis.
I should have qualified that as without horizontal translation, i.e., hovering or vertical descent.

Because of coning and undersling, the tips of a seesaw don’t follow a perfect circular path except without translation. But the deviation is small enough to be ignored in the context here.

I once made up a training aid for Bensen Days a number of years ago that used a scale model rotor and an old starter ring gear to define the tip plane path. The rotor blade tips had pins at ¼ chord that rode on the starter ring gear. The starter ring gear was pivoted and tiltable to alter the tip plane axis, providing a clear visual example of the topic herein.

Someone asked if he could have it for the PRA museum to which I agreed. I have no idea whatever became of it. Perhaps I’ll make another one for the forthcoming Bensen Days.

To see if you have any, top right of the opening page where it says welcome Birdy it will also tell you if you have any.
orrite, ill hava look.

To see if you have any,
No messages, but theres a coupla things there bout sumone wantn to be my frend?!?!

[Sum people must be more lonely n i thought.]

Nice picture JC.

OK Birdy I see you are online. I also see by clicking on your name that you apparently do not have that facility switched on.

I'll have a look and see where/how that is done.

Yup I guess some people need a friend if only a virtual one.

Go to your control panel. Go to edit options. Scroll down till you get to Messaging and Notification.

Scroll down to Private messaging then click on the button that says enable private messaging.

That should do it.

Now see I took so long Birdy is off-line. Oh well maybe get it later.

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To see if you have any,
No messages, but theres a coupla things there bout sumone wantn to be my frend?!?!

[Sum people must be more lonely n i thought.]
With 6,000 cows for companions, who needs cyberfriends?

............................................................................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.......................

Thanks once again Mr Beaty, I appreciate your patience, and the diagram is very descriptive. The model sounds worthwhile.

A sketch is worth a thousand words.
Jean Claude
....................................

Thanks also JC for a very detailed diagram. I wanted to clarify one thing with a drawing of my own, but haven't got the skills or the time right now ................ maybe in a couple of weeks when I get back home again.

............... theres a coupla things there bout sumone wantn to be my friend?!?!

Wasn't me, mate! I can find better looking friends than you, and anyway you've never even bought me a beer (or a bundy).

................

Go to your control panel. Go to edit options. Scroll down till you get to Messaging and Notification.

Scroll down to Private messaging then click on the button that says enable private messaging...................................

Well sorted Leigh - I had a look at that too, but gave up too quickly. I didn't remember seeing the options when I signed on!

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Scroll down to Private messaging then click on the button that says enable private messaging.
Wot give you the impression i wanted any messages?...........................

Ill hava ganda.

Far as i can figure, its on.

private messages

private messages

Far as i can figure, its on.

I finally sussed it out Birdy! - I think you have to be a paid subscriber ... I must have had special privileges before!

Anyway, I don't have any secret messages for you at present, but this place has one the best atmospheres I have seen on any forum - great information, great patience on the part of those who do respond. Definitely worth the price, and now - it's easy to pay directly with credit card! (note, extra stars are a bonus!):yo:

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Sorry Birdy, guess as CB says, it's just the moos then.

repost showing figs 1 to 5

repost showing figs 1 to 5

Note - reposted #56 ... have put the pics in here linked to an online album, in text. The previous post #56 used the my rotary forum album. (feedback appreciated)

Sorry if I am being too pedantic here, but I always had trouble visualizing the two blades at 3 / 9 simply 'swapping ends... ie teetering' quickly enough to cope with the dissymmetry of lift, it is usually just stated that the teetering deals with the lift issue;

Hopefully the last of the 'basic' questions for a while (please excuse the scratchy hand drawings - best I can do right now!): for a gyro set up as per Jean Claude's diagram in post #46 (fig 1) with 10* mast tilt ..... the set up is such that spindle bolt it in line with the mast in normal S & L cruise, and the tip disc plane is inclined rearward another 2 degrees. (correct?) Ignoring coning and teeter block height/undersling for simplicity.

[fig 1

Therefore the teeter bolt, in rotation, is forming an imaginary disc on a plane with 2* less inclination than the imaginary disc the rotor tip plane path forms. (fig 1 and 2) Viewed from the side at 9 oclock, the teeter bolt is therefore increasing the pitch on the retreating blade at this point. The blade has less AOA before it reaches this point, (coming from 12) then aligns itself with the teeter bolt plane at 9, increasing pitch and AOL, then again it's AOA increases as it swings through to 6, flying parallel to the teeter bolt plane ... etc (side view fig 2)

1. This 'control input point' is where JC's calculated retreating and advancing AOA (post #23) come in to play? (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 ° ). Would the actual flapping angle therefore be greater overall? (probably relates to Q 4 also)

2. So is it therefore the pilot adjusting the blade pitch at this point and therefore the pitch AOA? ... or is the pitch of the blade that is tilting the teeter bolt rotational plane?

3. If left to it's own devices, would the rotor keep attempting to adjust for the dissymmetry of lift until it hit the stops? The teeter bolt rotational plane has to be held to that position, or the rotor disc will keep pulling it back? (by trim spring or pilot?

https://i.pbase.com/o4/48/591148/1/118911691.bd6Bwx0F.fig2.jpg
fig 2 (front and side view of dic and teeter bolt rotational plane)

Below in Fig 3 is the 'teeter bolt rotational plane' defined, and also showing the different pitch at point A and C (ie 12 and 9 oclock)

(fig 3)

Fig 4 shows CB's cylinder with the blade tip path traced on it ... and 180* of the path (from 12 to 6) if we lay the pipe out flat**:

Q 4. I always find it easiest to think of the blades flying to a lower or higher point ... but do they simply follow a precise path, being pitched as required around that path, therefore creating differing amounts of lift/force? ie - so the teetering angle would not vary with different maneuvers... I find that difficult to imagine?

Fig 4

Q6. Or do they fly a sigmoid path as in Fig 5? (badly drawn ...)

fig 5

** I started to draw the full 360* path of the blade tips laid out flat, and realized I could only draw it as a curve, but wanting to draw the tracing of a flat disc. Then I realized we actually need to project it onto the inside of a sphere, not a cylinder (as per Mercator map projections) .. this makes my head hurt.

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Jean Claude

Mark despite flippant excursions I have certainly gained from your efforts, and continue to do so despite contributing little.

no pictures?

no pictures?

Jean Claude

Dang it! Can no one else see them (fig 1 to 5) too?

I uploaded to my gallery here and then linked to the post. I'm on a different computer now so have not got the originals.

(but I can see the pics from here - will see what I can do)

edit, update - I have posted the whole thing again, but this time used the attachment function to upload from my computer (after downloading them from the post below! - previously I put them in my online rotary forum album and linked to that - I thought it would be more reliable and it is easier to see the pics in the text;:noidea:

(brainwave; maybe my album is private ! - will change setting - (any feedback appreciated, I'd like to know the best way to show pics, thanks.)

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Note - reposted #56 below ... there I have put the pics in here linked to an online album, within the text. The previous post #56 used the my rotary forum album. This post uses the upload, but puts em in a box at the bottom ;( (feedback appreciated)

Sorry if I am being too pedantic here, but I always had trouble visualizing the two blades at 3 / 9 simply 'swapping ends... ie teetering' quickly enough to cope with the dissymmetry of lift, it is usually just stated that the teetering deals with the lift issue;

Hopefully the last of the 'basic' questions for a while (please excuse the scratchy hand drawings - best I can do right now!): for a gyro set up as per Jean Claude's diagram in post #46 (fig 1) with 10* mast tilt ..... the set up is such that spindle bolt it in line with the mast in normal S & L cruise, and the tip disc plane is inclined rearward another 2 degrees. (correct?) Ignoring coning and teeter block height/undersling for simplicity.

fig 1

Therefore the teeter bolt, in rotation, is forming an imaginary disc on a plane with 2* less inclination than the imaginary disc the rotor tip plane path forms. (fig 1 and 2) Viewed from the side at 9 oclock, the teeter bolt is therefore increasing the pitch on the retreating blade at this point. The blade has less AOA before it reaches this point, (coming from 12) then aligns itself with the teeter bolt plane at 9, increasing pitch and AOL, then again it's AOA increases as it swings through to 6, flying parallel to the teeter bolt plane ... etc (side view fig 2)

1. This 'control input point' is where JC's calculated retreating and advancing AOA (post #23) come in to play? (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 ° ). Would the actual flapping angle therefore be greater overall? (probably relates to Q 4 also)

2. So is it therefore the pilot adjusting the blade pitch at this point and therefore the pitch AOA? ... or is the pitch of the blade that is tilting the teeter bolt rotational plane?

3. If left to it's own devices, would the rotor keep attempting to adjust for the dissymmetry of lift until it hit the stops? The teeter bolt rotational plane has to be held to that position, or the rotor disc will keep pulling it back? (by trim spring or pilot?

fig 2 (front and side view of dic and teeter bolt rotational plane)

Below in Fig 3 is the 'teeter bolt rotational plane' defined, and also showing the different pitch at point A and C (ie 12 and 9 oclock)

(fig 3)

Fig 4 shows CB's cylinder with the blade tip path traced on it ... and 180* of the path (from 12 to 6) if we lay the pipe out flat**:

Q 4. I always find it easiest to think of the blades flying to a lower or higher point ... but do they simply follow a precise path, being pitched as required around that path, therefore creating differing amounts of lift/force? ie - so the teetering angle would not vary with different maneuvers... I find that difficult to imagine?

fig 4

Q6. Or do they fly a sigmoid path as in Fig 5? (badly drawn ...)

fig 5

** I started to draw the full 360* path of the blade tips laid out flat, and realized I could only draw it as a curve, but wanting to draw the tracing of a flat disc. Then I realized we actually need to project it onto the inside of a sphere, not a cylinder (as per Mercator map projections) .. this makes my head hurt.

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