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HelicopterMain Discussion › OK, I'm confused now....! Blade balancing question for the engineers out there.....
11-27-2007 08:04 PM  9 years agoPost 141
CK_

rrApprentice

Redondo Beach, CA

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If I'm not mistaken, full size Robinsons and Jet Rangers have free teetering heads with no mechanical flap springs or dampers so no hub moment can be transmitted to the mast. I don't see any reason why this couldn't be done with a freely teetering head. The lift on the blade would cone the blade up and the counterweight down until the lift moment is balanced by the centrifugal flapping moment and the system would be in equilibrium with no vibration.

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11-27-2007 08:17 PM  9 years agoPost 142
rgilmore7

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Mexico City

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Look at this post, it's easy... i used it with good results...
http://www.runryder.com/helicopter/t352560p1/

RaptorRic

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11-27-2007 08:29 PM  9 years agoPost 143
AirWolfRC

rrProfessor

42½ N, 83½ W

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CK,
Tetering head will not eliminate the traveling (rotating) lift vector, it can't. The lift vector is outboard of the shaft center line.

rgilmore7,
That post is exactly what I'm against.
Matching CGs and weights is NOT necessary.
. . . . it works but it's NOT necessary.

The only reason for matching the CGs and weights is if all blades are matched to the same set of numbers and you want to be able to order just one blade and just bolt it on, like in full scale. Still need to do a bit of tracking though.

But that's NOT what happens with these models.

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11-27-2007 08:31 PM  9 years agoPost 144
Hamo

rrVeteran

Ireland

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[quote]There appears to be a slight problem,

106 x 11.25 = 1192.5
156 x 7.5 = 1170

1192.5 <> 1170

These numbers should be equal.

The numbers are not equal because the metal tube is not uniform. Put differently, the weight distribution of the metal tube is not uniform. It has a big metal nut at the end.
Hamo

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11-27-2007 08:33 PM  9 years agoPost 145
AirWolfRC

rrProfessor

42½ N, 83½ W

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. . . . the weight distribution of the metal tube is not uniform. It has a big metal nut at the end.
Which is (should be) accounted for by the stated location of the CG. Uniformity is not a concern here.

The location of the CG from the bolt center line is the number of consequence.

If there is any question, then remove the entire head with the blade(s) and balance on the flybar rod. If you can keep the blade from rotating on the spindle.

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11-27-2007 09:09 PM  9 years agoPost 146
Hamo

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Ireland

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Sorry AirWolf, you are right again. The measurements I gave earlier were just approximate. I didn't know you were going to do maths on them. I checked again a few minutes ago and the CG of the wooden blade is 28.5 cm and the CG of the metal tube is just under 19.5 cm from the root.
If you do the maths now you will see that one works out as 3021 and the other as under 3042 which I think are pretty close. I used a plastic ruler for the measurements.
Hamo

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11-27-2007 09:16 PM  9 years agoPost 147
AirWolfRC

rrProfessor

42½ N, 83½ W

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Either way, the moments need to be equal.

Also, if the blade(s) don't fly with the same lag, same drag, the mass summ of the two "sticks" attached to the hub will not be centered on the main shaft causing an imbalance.

I assure you the 12mm tube has much more drag than the real blade causing greater lag. A shorter, heavier tube will reduce this problem.

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11-27-2007 09:22 PM  9 years agoPost 148
Hamo

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Ireland

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AirWolf, I understand what you are saying. I will make up another counterweight using solid metal rod so that it's thinner and shorter.
I'll let you know how I get on. I really appreciate your expertise and your help. Thank you.
Hamo

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11-28-2007 12:10 AM  9 years agoPost 149
skydude

rrNovice

Gainesville, Florida, USA

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quote

The lift on the blade would cone the blade up and the counterweight down until the lift moment is balanced by the centrifugal flapping moment and the system would be in equilibrium with no vibration.

unquote

The lift will be perpendicular to the blade. As the blade 'cones' up it will be at an angle greater than 90 degrees to the shaft. The lift which should be parallel to the shaft will now be at some angle to the shaft and have a component directed towards it.

The centripedal force of the counterweight will always be perpendicular to the shaft and away from it.

In the proposed configuration both vectors are directed in the same direction. This will have the same result as a single weight spinning on a shaft.

Watch out all you moles!!! (Vae, puto deus fio)

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11-28-2007 12:14 AM  9 years agoPost 150
AirWolfRC

rrProfessor

42½ N, 83½ W

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? ? ? ?

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11-28-2007 12:31 AM  9 years agoPost 151
skydude

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Gainesville, Florida, USA

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The lift force is now directed inboard. There is no offsetting force created = wobble.

--

Watch out all you moles!!! (Vae, puto deus fio)

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11-28-2007 12:42 AM  9 years agoPost 152
AirWolfRC

rrProfessor

42½ N, 83½ W

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I couldn't tell if you were agreeing or disagreeing with the wobble.

I look at it this way,
The summ instantaneous lift vector is never through the center of mass of the heli (because there is only one blade producing lift). Therefore, there will always be some force trying to rotate the heli body as opposed to pure lift. This force trying to rotate the body is changing direction in sync with the head rotation and produces "wobble".

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11-28-2007 06:28 AM  9 years agoPost 153
CK_

rrApprentice

Redondo Beach, CA

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Tetering head will not eliminate the traveling (rotating) lift vector, it can't.
It doesn't have to. All that matters is the reaction at the hub.
The lift will be perpendicular to the blade. As the blade 'cones' up it will be at an angle greater than 90 degrees to the shaft. The lift which should be parallel to the shaft will now be at some angle to the shaft and have a component directed towards it.
Yes that's true. The coning angle is small and the sin of a small angle is a very small number. For example a 10 lb, 5 foot rotor diameter, one blade heli at 1800 rpm with a 3° coning angle would have roughly the equivalent vibration from the tilted lift vector as .1 grams of tracking tape at the blade tip.

I think balancing the drag distribution would be the bigger problem. You can't decouple the drag with a hinge like the lift distribution. If you did, the shaft would turn but the rotor wouldn't go anywhere. Not very useful. That's probably why the counterweight in that photo is a big draggy cylinder. By the time you start adding drag back into the system you're probably negating any efficiency benefit gained by going to a single blade in the first place.

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11-28-2007 08:54 AM  9 years agoPost 154
Hamo

rrVeteran

Ireland

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It appears to me that the real blade will produce variable drag as the pitch changes but the metal tube will have fixed drag, therefore it will not be possible to achieve balance and successfully fly the helicopter.
Does that mean the picture below is a joke ? Also I don't see a flybar.
Or is the rotor head of a special design to allow for this
Hamo

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11-28-2007 11:04 AM  9 years agoPost 155
MILNEI

rrApprentice

Buckingham, England

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I have found an e-mail address for Guenter Knipprath, the chap who designed the heli in the picture and I have asked for some details.

I'll let you know if he replies.......

Anyone got a number for Heli-widows, my wife wants to join!!
Raptor 50 OS Hyper MP2
Mini Titan

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11-28-2007 11:39 AM  9 years agoPost 156
Hamo

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Ireland

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Cool MILNEI. I can't wait to fly with a single blade.
As for heli-widows, spare a thought for this guy's wife and kids.
Hamo

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11-28-2007 03:17 PM  9 years agoPost 157
AirWolfRC

rrProfessor

42½ N, 83½ W

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Hamo,
Know that a round object in a given air flow will have about 10 times the drag as a similar thickness airfoil.

You have a blade that's maybe 8mm thick and a tube that's 12mm thick.
For the same speed, the tube will have about 15 times the drag. Sure the tube is shorter but you get the idea.

That's a lot of lag.

Keep it shorter and thinner.

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11-28-2007 03:25 PM  9 years agoPost 158
skydude

rrNovice

Gainesville, Florida, USA

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quote:

Yes that's true. The coning angle is small and the sin of a small angle is a very small number. For example a 10 lb, 5 foot rotor diameter, one blade heli at 1800 rpm with a 3° coning angle would have roughly the equivalent vibration from the tilted lift vector as .1 grams of tracking tape at the blade tip.

unquote

The sine of 3 degrees is 0.05.

A 10 lb. machine requires 10 lb of lift to hover.

10 lb vectored 3 degrees inboard equal 10 lb x sin 3 deg (0.050) =

0.5 lb

that is a 1/2 pound force directed sideways.

That is actually closer to 0.0806967 grams of tape unbalanced at the end of a 2.5 ft rotor at 1800 rpm.

I never said it was the most or the least of the problems. I suggested that teetering added it's own form of wobble.

I have not run the numbers. Is 3 degrees expected?

And I do not see how the vertical vector approx. 1.675 ft out can ever be balanced to keep from trying to roll the aircraft over. (except maybe with some mechanical wobble gyro)

---

Watch out all you moles!!! (Vae, puto deus fio)

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11-28-2007 03:48 PM  9 years agoPost 159
AirWolfRC

rrProfessor

42½ N, 83½ W

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Where did you get 3º ?

I don't see that in practice. I figure it's more like ½º

As for the wobble, the cone angle has little if anything to do with it.

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11-29-2007 02:30 AM  9 years agoPost 160
skydude

rrNovice

Gainesville, Florida, USA

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I just quoted 3 deg. from another poster.

Calc 4 oz blade, 2.5 ft long spinning @ 1800 generating 10 lb of lift @ approx 1.67 ft out I get 1.6643 degrees. This is with a 1.25 lb weight @ 4 in. out the other side.

As for the cone angle of above # of degrees it has everything to do with the 10lb x sin(cone angle) = 0.29 lb of unbalanced force that will indeed produce wobble.

--

Watch out all you moles!!! (Vae, puto deus fio)

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HelicopterMain Discussion › OK, I'm confused now....! Blade balancing question for the engineers out there.....
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