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Blade Static & Dynamic Balance - 3/16/2002 2:17:11 AM   
bwoeb


 

Posts: 28
Joined: 3/12/2002
From: Corvallis, OR
Status: offline 		As condescending as chopper may sound, he is correct. You can look at this scenario this way. For the purposes of dynamic balance, you can treat the blade as a point mass. The definition of the CG or CM (center of mass) is if you condensed all of the mass of the object into one infinitely small point, all the forces would be acting on that point. So in effect what you have with two helicopter blades, is two point masses spinning around a the spin center. Now it is not difficult to see what will happen if those point masses are at drastically different distances from the spin center. From the dynamic sense, two things are important.

1) The speeds of the different point masses are different. The math behind this is simple. In order to find the velocity of the mass, multiply the angular velocity (RPM) by the distance to the spin center. You can see with two different distances, you get two different speeds.

2) The "inertia force" of the two different blades is different. That is, the same mass spinning at different distances from the center of the system has different inertia. The math for this is somewhat more complicated. It involves determining the inertia of the blade about itself and then determining the inertia of the blade around the center using the parallel axis theorem. ( won't bore you anymore with technical engineering talk, it bores me too!!)

So you see that there is mathematical reasoning for dynamically balancing your blades. Is it possible that your helicopter will work without balancing your blades? Sure that is possible, but it is based more on luck than anything else. ;-)

I am still interested to know what happens with your test this weekend helicopterhead.

Happy flying.

Brian

(in reply to Adam T)
       Post #: 26

Blade Static & Dynamic Balance - 3/16/2002 2:19:58 AM   
Adam T



Posts: 357
Joined: 1/30/2002
From: Los Angeles, CA, USA
Status: offline 		Hey, Sir Isac Newton is a friend of mine! I will defend his honor to the death!!


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Blade Static & Dynamic Balance - 3/16/2002 5:28:44 AM   
flmgrip



Posts: 1051
Joined: 2/19/2002
From: pasadena, CA, USA
Status: offline 		scotty has a good point and while you guys where discussing all that i was out blowing a gallon of fuel. with blades - for whatever reason - perfectly balanced in flight at any RPM
BTW join my other thread about my strange seesaw problem, all you genius and doctors...


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       Post #: 28

Finally - 4/29/2002 9:39:14 PM   
Adam T



Posts: 357
Joined: 1/30/2002
From: Los Angeles, CA, USA
Status: offline 		At long last, I had the opportunity to test the theory. I have a new set of blades that balance against each other perfectly, but have different span CGs. I bolted them on after double checking the flybar balance, etc., and I did notice a SLIGHT vibration that I don't get with my other blades that are CG matched. My conclusion is that the conventional wisdom is correct--CG matching does matter. There must be a problem with natescape's analysis, although no one has been able to articulate the flaw in his equasions. FYI....


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       Post #: 29

Why he is wrong - 4/30/2002 2:56:32 AM   
bwoeb


 

Posts: 28
Joined: 3/12/2002
From: Corvallis, OR
Status: offline 		Thank you HH for testing the expirament. The reason natescape is wrong in his analysis is that he doesn't take into account the inertia of the blades like I stated before. His math is correct, but it is also incorrect because it doesn't deal with the forces that are actually at work in helicopter blades. He is solving the problem correctly, but he is leaving part of it out.

Brian

(in reply to Adam T)
       Post #: 30

Blade Static & Dynamic Balance - 4/30/2002 3:01:55 AM   
Adam T



Posts: 357
Joined: 1/30/2002
From: Los Angeles, CA, USA
Status: offline 		Bwoeb: I re-read point two in your post and I guess you did articulate it, just not in a way a philosophy major like me can understand! Anyway, just goes to show you how convincing someone can sound while still being wrong. I've been CG matching anyway just to be safe. It's nice to know the work is worth the time.


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       Post #: 31

Blade Static & Dynamic Balance - 4/30/2002 3:17:15 AM   
bwoeb


 

Posts: 28
Joined: 3/12/2002
From: Corvallis, OR
Status: offline 		Don't downplay philosophy. I wish I could think in that much depth.

To be safe is a good idea when talking about a heli that cost you quite a bit if $$$. Most of the blades coming from the factory should be pretty close to balanced, but it is very difficult to have them perfect when you make as many as those places do.

Good luck.

Brian

(in reply to Adam T)
       Post #: 32

Blade Static & Dynamic Balance - 4/30/2002 3:37:54 AM   
Adam T



Posts: 357
Joined: 1/30/2002
From: Los Angeles, CA, USA
Status: offline 		I'm not knocking philosophy, it has served me well. It's just not very helpful when talking to engineers. The only truly perfect blades I have ever bought are V Blades. Every other brand was off enough to require some balancing. Still I have yet to encounter a set of composite blades that was off too far. The stock woodies I got with my kit were so bad that you virtually couldn't get them to balance with tape alone.


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       Post #: 33

Response to Post #7 - 5/2/2002 12:43:17 AM   
Blue Angel


 

Posts: 3
Joined: 12/19/2001
From: Redford, MI, USA
Status: offline 		Hello HelicopterHead: First of all, I want to clear up a few things. There is a "Centrifugal" force and a "Centripetal" force. They are equal and opposite or balancing forces. "Centrifugal" force acts outward on a spinning object, while "Centripetal" force acts inward on a spinning object to maintain equilibrium. Example is the centifugal force of the blade is pulling on the bolt, while the bolt is exerting the same force, centripetal-opposite direction, on the blade.

The second point I want clear up is the math you quote for centrifugal force. We start with
F = M x A, where M equals blade mass and A equals blade acceleration (angular).

However, velocity does not equal 2 x pi x R/T.
V = R x Omega, where Omega is angular velocity. And A = R x Omega x Omega (angular accel.) This is where you went astray in your calculations. I'm sorry, but the keyboard does not have the greek letter "Omega".

Therefor, F in lbs = M x R x Omega x Omega. R is the distance to the blade CG in ft.

Now Omega, Radians/sec. = 2 x pi x N/60, where N equals the RPM of the rotor system.

Therefor, F = M x R x (2 x pi x N/60)x(2xpixN/60).
F = M x R x pi x pi x N x N/900.

Now M = W/g where W = blade weight in lbs. and g = acceleration of gravity in ft/sec/sec.

Therefor, F in lbs = (W/g)(R)(pi x pi)(N x N)/900.
This would be the centrifugal force of the blade acting on the bolt, while the bolt exerts an equal centripetal force on the blade.

No way can a sesaw balanced set of blades be as accurate at minimizing rotor vibration as a set having matched weights and CG's. The bottom line is, keep using the Koll balancer.

Blue Angel

(in reply to Adam T)
       Post #: 34

Blade Static & Dynamic Balance - 5/2/2002 12:47:58 AM   
Adam T



Posts: 357
Joined: 1/30/2002
From: Los Angeles, CA, USA
Status: offline 		Thanks for the informative post, Blue Angel. Interesting reading, but for some reason I have a headache....


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Blade Static & Dynamic Balance - 5/2/2002 1:39:21 AM   
w.pasman


 

Posts: 367
Joined: 4/23/2002
From: ----, NETHERLANDS
Status: offline 		hi,

I dont know what that balancer is, but I assume it works as follows

I assume you align the two blades exactly in a line and have the rotation point of that bold exactly in the middle of the bolt and very close (few mm) from the blades. NOTE a mm of error in this will screw up your balancing. also NOTE that height does not make a difference, as long as both blades are hanging 100% horizontal)
The equation for the torques on left and right blades are:
lefttorque=m r (m=mass of left blade, r=dist to COG of left blade)
righttorque=(m+dm)(r+dr) (dm=mass of right blade minus mass left blade, dr=distance to COG of right blade minus distance to COG left blade)

if the blades balance in the above sketched setup, then lefttorque=righttorque, so
m r == (m+dm)(r+dr)

The equation for dynamic balance is
m (w w) r == (m+dm)(w w)(r+dr)
where w is the rotation speed in rad/s.
(see my webpage [url]www.cg.its.tudelft.nl/~wouter/helipage/tips/balanceforce.html[/url])
As w is the same for left and right blade, the equation for dynamic balance is also
m r == (m+dm)(r+dr)

Equations are the same so the outcome should be the same! Maybe your experiemnt failed because you did not do it accurate enough? 1/10th of a millimeter can make that difference (also on my webpage).

BTW I use this way of balancing as a check after the two-step balancing process. See my page [url]www.cg.its.tudelft.nl/~wouter/helipage/tips/balancing.html[/url]

(in reply to Adam T)
       Post #: 36

Blade Static & Dynamic Balance - 5/2/2002 1:48:26 AM   
flmgrip



Posts: 1051
Joined: 2/19/2002
From: pasadena, CA, USA
Status: offline 		i am glad i just have to fly those things...


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       Post #: 37

Blade Static & Dynamic Balance - 5/2/2002 2:06:12 AM   
w.pasman


 

Posts: 367
Joined: 4/23/2002
From: ----, NETHERLANDS
Status: offline 		Oh and to answer another question:
> You can get a set of blades to balance against one another on the
> see saw, while still having very different spanwise CGs (and even different
> weights).

Yes, even certainly different weights. Exactly that weight difference to make up a dynamic balance :-) A COG and mass inbalance can compensate each other. Again see my webpages.

(in reply to Adam T)
       Post #: 38

Blade Static & Dynamic Balance - 5/2/2002 2:10:22 AM   
flmgrip



Posts: 1051
Joined: 2/19/2002
From: pasadena, CA, USA
Status: offline 		any of you smarties ready to solve my strange problem i have, on the thread called "see saw balance"

thanks


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Please be safe and have fun!

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       Post #: 39

Blade Static & Dynamic Balance - 5/2/2002 7:08:42 AM   
bwoeb


 

Posts: 2