I have read with interest the post on natescape (http://www.natew.com/rcheli/ under "Setup, Balancing") regarding balde balancing. He asserts that if two blade are statically balanced on a seesaw type of balancer, there is no need to CG match them. In essence, he claims that the physics are such that the centrifugal forces act on the blades in the same manner as gravity. Thus, the lever effect that a CG mismatch has on the seesaw is the same as the increasing amplification of the force asserted by mass as it gets closer outside edge of the spinning rotor disk.

My question is: has anyone tested this theory in practice to see if it is true? In other words, has anyone flown blades that level out on a seesaw balancer (such as the Koll Rotor Pro), but clearly have different spanwise CGs?

I must say that I would love it if all I had to do was one step instead of three when balancing blades. I'm tempted to try the experiment myself, but I would love to have the benefit of prior experience first.

Thanks.

1:take 2 lead weights of equal mass.

2:tie them to two diffrent length strings.

3:spin them at the same RPM.

tell me which one pulls more.

to sum it up.... this person doesnt know what he is talking about.

Unstable: No offense, but you have missed the point. Your proposed experiment does not simulate the situation accurately. A set of blades that balance perfectly on the seesaw but have different spanwise CGs may well be of unequal weight.

Consider this: If you had two blades perfectly the same weight and CG on a seesaw balancer nice and level, and you place one of two identical lead weights on one of the blades at the tip, while placing the other identical lead weight on the other blade at the root, the blades will still have exactly the same weight, but will NOT balance on the seesaw. This is the point that your experiment fails to address.

Actually, after giving this much thought, I think it is entirely possible that Nate is correct. If no one else has tried it, I plan to do so and will report back on the results.

ohhhh.... i did miss the point, but i see what you mean now....

it would be like my above coment but useing diffrent mass weights so that they pull the same while swinging.....
in that case then it "should" work as the force on the head would be even.

but dont take what I say as gospel as I have no degree in physics

maybe I should just learn to keep my mouth shut

let us know what you find out when you try it.

Not at all--All comments are welcome. This is not a very intuitive subject -- you need to do some serious math to prove either conclusion, and I don't have a physics degree either.

Well, I hate to burst someones bubble, but:

>>>>
he claims that the physics are such that the centrifugal forces act on the blades in the same manner as gravity.
>>>>

In physics their is no such thing as centrifugal force unless you read a pre 1930's physics book. Gravity is not fully understood so how can this be applied to rotational motion???


Thanks
Fred

I'm not sure I understand your point. Are you a physics expert? There clearly IS such a thing as centrifugal force in modern physics. To alieviate my inability to word this problem in simple English, I submit the following taken from the Natescape page cited at the beginning of this thread:

************************************************** *
The forces that act on a seesaw are 'moments.' Moment is equal to the product of mass and distance (radius) from the center of rotation:

moment = M x R

The forces that act on a rotating rotor head are centrifugal forces. (some prefer to use the term centripetal, which is a bit more to the point but boils down to effectively same thing). This force is equal to the product of the mass and the square of velocity, divided by the radius of the center of mass:

centri{fugal|petal} force: M x V x V / R

Velocity is distance over time, and in this case distance is the product of two, pi, and radius.

velocity = 2 x pi x R / T

--------------------------------------------------------------------------------

The question then is - can the equation for centrixxxx force be reduced to the equation for moment? The answer is yes. I'm now kicking myself for throwing away the proof, but I do have a counterexample below. Consider a 10 gram blade with a CG 1 centimeter from the root, and a 1 gram blade with a CG 10 centimeters from the root. It's trivial to prove that their moments are the same: 10 x 1 is equal to 1 x 10. But will they exert the same centrifugal force on the rotor head?


F = M x V x V / R

F = M x (D/T) x (D/T) / R

F = M x (2 x pi x R / T) x (2 x pi x R / T) / R

so now we substitute real numbers into each equation...
T has been set to 1 for simplicity.

blade 1: m = 10 grams, r = 1 centimeter, v = 2 x pi x 1
blade 2: m = 1 gram, r = 10 centimeters, v = 2 x pi x 10

F1 = 10g x (2 x pi x 1) x (2 x pi x 1) / 1
F2 = 1g x (2 x pi x 10) x (2 x pi x 10) / 10

divide by (2 x pi) x (2 x pi) to get:

F1 = 10g x 1 x 1 / 1
F2 = 1g x 10 x 10 / 10

multiply to get:

F1 = 10g x 1 / 1cm
F2 = 1g x 100 / 10cm

divide to get:

F1 = 10g / 1cm (note that this looks a lot like the equation for moment)
F2 = 100g / 10cm (that is definitely not a coincidence, I assure you!)

reduce the fraction in the second equation to get:

F1 = 10g / 1cm
F2 = 10g / 1cm

and note that:

F1 = F2

************************************************

Seems like sound reasoning to me

I pay relatively little attention to span wise CG of blades. I never had any vibration because of that (I do not use wood blades though). I check cord wise CG and "seesaw" balance.
As for you question: quote from rays heli manual:
"... if the span wise CG of one blade is out farther than the other, that blades center of mass will essentially make that blade heavier when the disk is at flight RPM. Weight times arm equals movement ... if the CG differs, the total moment of each blade knocks the balance off the center of the main shaft and ... imbalance occurs"
That’s the theory and I believe it is right. In practice it has yet to cause a problem for me. Also a lot of theories relate to a completely stiff system. This is by no means the case in the RC helicopter, blades, shaft, holders, pins and whatever else might flex more or less.
So if you want to be perfect, sorry you still have to do a three step balancing.

I've seen a helicopter fly with one blade and a counter balance in place of the other. Yes, they both weighed the same but as one was a full length blade and the counterweight about 6" as I recall, the CGs were obviously much different.

Also I believe it is common in some racing airplanes to fly with one bladed props with a counterweight. Using only one blade reduces drag.

With that said I have never tried it myself. I always match the CGs of the blades when I balance them.

HelicopterHead,
I was just wondering if you have you tried balancing your blades as per the technique on natescape yet.

brc007: I was going to try it on a new set of Go Quick 550 carbons, but once I added a bit of tape to one blade to get them to balance on the see saw, I checked the CGs and (as luck would have it) they matched perfectly (drat foiled again). So I am going to take all the tape off of my stock woodies and try the experiment with them--but I haven't gotten around to it yet. I'm going to try to get to it this weekend and I'll report back.

I have had a lot of helis that would shake with statically balanced blades. Matching the CGs and then matching the weights always got rid of the shake, period. Most flyers quit balancing blades against each other years ago.

Please don't jump on me for asking this question. I've just started into heli's and haven't tried to balance a set of blades yet, so this may be a dumb question, but ...

Wouldn't matching the CG and then matching the weight be equivalent to balancing blades against each other?

Helicopterhead,
The above post by squirrel is correct in asserting that there is no such thing in physics as centrifugal force. It is merely a construct to explain forces that counteract the centripital force, which is a real force. I do not hold a physics degree, but I did study physics as part of my engineering curriculum.

Brian

The reason you need to CG balance blades on a heli is because of the inertia caused by rotating mass. When a mass is rotating about a fixed point, the inertia is calculated by multiplying the mass by the moment arm distance to the spin center. If the moment arm on one side of the spin center is longer than the other, you will have a larger moment of inertia on the blade where the CG is located farther from the spin center. This will cause the system to oscillate while spinning.

It is sort of like taking a very thin rod with a weight at the end of it and comparing that to a very short but heavy rod. Experience says that you may be able to statically balance them, but the CG is not in the same place for each side. So if you stuck that on your heli, what would happen? You would have a serious vibration problem. The static balance says they are fine, but the CG is off and that is the root cause of the problem.

Brian

Brian- Then how do you explain the above scenario where one blade grip has a blade attached and the other one just has a weight attached to the blade grip. People(not me) have seen it fly without vibrations. I believe that this scenario is exactly what you have described as not going to work. Not being inflamitory or anything, I'm just wondering like everyone else. I must admit to being one of the "sheep" that uses a koll balancer and I do not have any problems with vibration. I don't know the math or science behind why it works I "just do it" and it seems to be fine. I must admit that I don't know why or how my TV,VCR,Car, and computer work either, but I'm a trusting soul.
VD

Brian: I believe my earlier post quoting from natescape adresses the centripital/centrifugal force issue. Construct or not, the point is that there appears (at least to a layman like me) to be some basis in physics for the proposition that differences in CG that cancel each other on the see saw will also cancel each other while in motion. Gravity seems to act upon these forces in the same manner and proportion as centripital/centrifugal force.

As Scotty and Vance's posts point out, there seems to be some practical (real world) examples that prove this point. I know you are not a physics major, but do you have the knowledge to point out what you believe to be the flaw in the above equations? Flmgrip's point is the one that seems the most plausible--i.e., that the equasions do not account for flex, vibration, etc. that are present on a heli in flight. Rather, the equasions seem to assume a perfect world where none of these other factors are present. I'm still not sure I understand why or whether such factors would change the result though. I gues that ultimately the proof will be in the pudding, when I run a practical test this weekend. I will report back. Hard to believe that all you more experienced folks arguing that CG matching is necessary could be wrong, but you never know

Winpattern: I too use the Koll balancer and heretofore have meticulously balanced cord and span CG as well as see saw weight matching (the Koll is truly a great little invention). I have never had any vibration issues doing it this way, and as you can see, this is clearly the conventional wisdom. To answer your question, no. 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). The reason is, a blade with its mass concentrated at the tip will have more leverage (moment) on the see saw than a blade of the same weight with its mass concentrated at the root. Thus, two blades of the same weight will not balance on the see saw unless their CGs are identical. Similarly, blades of different weights may well balance on the see saw if their CGs are such that the lighter blade has more of its mass concentrated at the tip, relative to the heavier blade.

Thus, the "correct" method, is to match the spanwise CGs first, then see saw balance, adding weight to the light blade AT THE CG. Then you will end up with blades that are the same weight AND have the same spanwise CGs. Whew!

Admitidly I am new to the sport....as new as anybody can be. But because of my education in engineering, I do understand the THEORETICAL physics behind these systems. If the system works outside of what the theoretical model says, more power to you. You have beaten the laws of physics!!!

The problem I see with not dynamically testing your blades, is that the static test only tells you how the blades react in a static environment. My dynamics class says that things are more complicated when you set them in motion. But Scotty and Vance are correct, this is not a perfect system. I could spend the 3 or 4 hours and do the math for you on this subject, and tell you the exact outcome of the system, including such things as friction, blade flex, and the numerous other factors involved in rotary flight. But I have tests to study for So good luck with your tests this weekend. Let us all know what turns out.

Brian

Vance,
The reason the above situation works is because that mass that is hanging from the blade grip is designed so its GC is in the same location as the CG for the blade side. It IS possible to get the CG of two completely different shapes to line up in the same position relative to the spin center. But the fact that they AREN'T lined up in the two bladed system is the problem.

Brian

Bwoeb,

I have heard of racers using a one bladed prop with a counterbalance on the other side. This would allow the blade to be balanced, but the CG would definately be different.

They also run at a lot higher RPM then a heli blade does. They also don't get very much (if any) vibration from this.

Wow, there is a lot of supposin'. The examples of the see saw are static examples. If you hang the blades staticaly they will balance all day, but we spin these blades. It has nothing to do with gravity, but rather FORCE.
The definition of center of gravity is the theoretical piont of an object that acts as if the entire mass of the object is centered at that point. Forget about gravity. Gravity has nothing to do with a spinning rotor on a knife egde. It is all about FORCE.

The force of a spinning object (rotor blade in this case) is dependent on three variables: mass (m), speed (v), and distance from the center (r=radius). If the cg of the blade is at a different "r" form the center, it is easy to see that :1- it will be traveling in a bigger circle and therefore at a different speed and, 2- the radius will be different from the other blade.

When you do the math it will come out as mrv^2. If the mass is the same for each blade, you can cancel out the "m" and the equation becones F=rv^2.

For blondes, this means that any change in speed has some drastic changes in force. The speed changes as the cg changes in "r".

You can dynamically balance a prop with a weight on one end, but it will be in balance for only a specific speed. It may be in balance at 20000 rpm but not at 10000 rpm.

The people with great websites are not necessarily experts in helicopters, and there are a lot of posers out there.

Chopper: Not to beat a dead horse (and I still plan to report back on my experiment), but your remarks sound a little condescending. Gravity is also a "force." Gravity is the force acting upon the baldes when they are static, while centrixxxxal is the force acting upon the baldes while in motion.

You make a fundamentally incorrect assumption: that the mass of baldes balancing on a see saw are the same. As my earlier post points out, this will not be the case if the CGs of the blades on the see saw are different. That's why it is at least plausible that CG matching may not be necessary if the blades balance statically on the see saw.

Finally, if your contention is correct, I would expect single bladed racing props to nearly rattle the plane to pieces as they are accelerating/decelerating to their working RPM. I would also expect them to be extremely dangerous because if you have some kind of engine problem that results in a different RPM than expected, the aircraft would nearly come apart from vibration. In short, I kind of doubt that single bladed props would be viable if what you are saying is true.

An easy way to prove this is to put a few wraps of tape about 1/4 of the way out one blade and the same length of tape on the other blade at the tip. My uneducated guess it that it's going to shake despite the fact that both blades weigh the same.

That being said I'd guess the single bladed props have the same "apparent" CG. in other words a heavier weight that is closer to the hub than the blade side.

What I'm trying to say is that the CG can be off as long as the weights of the blades are different to compensate. With both blades weighing the same the CG has to match on them.

I may be all wet here but just my thoughts on the matter.

Next time I go out I'll have to try some things.

I will admit my remarks sound a little condescending. While MY math can certainly be in error, you are blaspheming the name of Sir Isac Newton, and I won't have that! <G>
This is the very basis and difference between static and dynamic balance that we are discussing. There are some posts that keep refering to gravity examples in a dynamic system. Gravity examples work on STATIC systems, while rotational forces (angular momentum) work in dynamic rotor systems.

Do some basic vector analysis, or ask some one to do it , and you will see that if you want to have the forces on each blade the same in all RPM, then the mass has to be equal, and it has to be the same distance from the center. If you graph them, you can make the function curves for dissimilar moments intersect at one or two points, but not for all points. I admit it can be a little confusing.