Can someone remind me how to locate CG using scales under the wheels of a tricycle gear plane? More to the point, if I know where it SHOULD be, how to calculate the desired % weight on (say) the nosewheel. I remember from a Dave Platt DVD it's about ratios of CG to wheel positions, I've just had a brain fade!

Thanks,

Cam

It's a whole big bunch of math. Not difficult math, really, but a bunch of it. Get ready for the insanely long, but hopefully insightful, explanation.

You can find the CG, to be honest, by measuring weight at ANY two points...landing gear just happen to be convenient ones, since the airplane is sitting on them. If you know their difference from any "0 point", then the formula is:

(A) (Wt1) + (B) (Wt2) = (Wt1 + Wt2) CG

Where A is the distance from your 0 point to the first place you weighed the plane, and Wt1 is its weight at that point.
B is the distance to the second point (from 0) and weight 2 is its weight at that point.

Basically, keep in mind that the whole definition of the CG is that it is the point at which there is an equal moment arm on each side of it. Notice, not equal weight, but an equal moment arm.

Think of any fulcrum....the old standby teeter totter works.

The moment arm is weight x distance from the fulcrum. So, put a 50 lb kid on one side, a 100 lb kid on the other. If the 50 pound kid is 5 feet from the fulcrum, his moment arm is 250 ft-lbs. For the teeter-totter to balance, put the 100 lb kid 2.5 feet from the fulcrum, and viola...balance.

Now...think about this. Let's say you saw this teeter totter suspended in mid-air, and couldn't see the fulcrum...and wanted to know where it was. It would, essentially, be your airplane right now.

Ok, fine. Pick a point "behind' either kid, since it's convenient. Let's say you pick some point 5 feet behind the 100 lb kid. Ok...the moment arm from him to your "zero point" is 500 ft lbs. (5 ft * 100 lbs). The 50 lb kid's moment arm would be 625 ft lbs. (Remember, he's going to be sitting there 7.5 ft from the big kid, plus another 5 to the zero point...so, 12.5 feet * 50 lbs = 625).

Okie doke. guess what...for those two forces to balance, we've got to now "move" our zero-point if you will, right? Move it to the CG, so to speak. How do we do it? The formula above:

(A) (Wt1) + (B) (Wt2) = (Wt1 + Wt2) CG

A is 5 feet (big kid to zero point)
B is 12.5 feet (little kid to zero point)
Wt1 is 100 lbs (big kid)
Wt 2 is 50 lbs (little kid)

500 ft lbs + 625 ft lbs = 150 lbs * CG
1125 ft lbs = 150 lbs * CG
1125 ft lbs / 150 lbs = CG
7.5 ft = CG

Move 7.5 feet back from the zero point, and you have your CG! Sure enough, we look, and there it is.

Just for 'comfort's sake', let's try one from the other side...say, 5' behind the little kid, with little kid being Wt1 and A this time:
250 ft lbs + 1250 ft lbs = 150 lbs * CG
1500 ft lbs = 150 lbs * CG
10 ft = CG
Move 10 feet from the little kid, and sure enough, there's our fulcrum again.

What if we choose a zero point between the two? Say, 2.5 feet "inside" the little kid, who will be A and Wt1 again:
125 ft lbs + 500 ft lbs = 150 lbs * CG
625 ft lbs = 150lbs * CG
625 / 150 =...er...wait...that didn't work so well, now did it?

The math needs to change a bit here. Remember, in this case, the distance from zero point to each kid ISN'T IN THE SAME DIRECTION. So, we need to make one of them negative. Which one? Eh, doesn't really matter, as it happens. *heh*
How about the little kid?
-125 ft lbs + 500 ft lbs = 150 lbs * CG
375 ft lbs = 150 lbs * CG
375/150 = 2.5 Just what we needed.

For that matter, if we'd gone the other way, we'd have -375/150, or -2.5, and had to move that distance from the ero point in the OTHER direction. Either works.

So...to do the scale under the wheel thing, pick a zero point. The shaft of the motor is usually a good spot, since you can hang a plumb bob from there.

Weigh under the main wheels. it's b est to weigh under each one separately, and combine the two. Be sure the plane is sitting dead level, blocking up the other wheels to the height of the scale. This weight will be Wt1, and the distance from the wheels to the plumb bob will be A
Do the same for the nose gear (this'll work on taildraggers too...just weigh at the tail wheel), making its weight Wt2 and its distance to the plumb bob B
Do the math. Of course, in this case, you'll probably be working in oz-in, not ft-lbs, but the math is the same. Whatever distance you come up with, go that far BACK from your zero point, and that's the CG.

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Now, having said all of that...WHY would you EVER do this? Just an exercise in "Gee, what if?" or are you looking for a way to find CG on a "difficult plane"? If the latter, I HIGHLY recommend the "Vanessa CG Machine" http://home.mindspring.com/~the-plum...%20Machine.htm I've used one for everything from a small foamy to a 1/3rd scale Pitts...works like a charm, and beats doing math ANY day.

For what it's worth, in a trike gear airplane, the best "analogy" to the teeter totter example will probably be if you think about the nose wheel as the little kid, and the mains as the big kid, with the CG likely being somewhere between them...thus, this would be like working the second example above...putting the zero point "behind the little kid's back", at the motor shaft.

I agree with gboulton. Go with the Vanessa. It works great and you can make changes "on the fly" (while the plane is suspended) by moving batteries around.

The thing sounds complex when you first read about it but it's very easy to built, even easier to use. And, no math ,...ever. I tweak my giant scale every few weeks just to see the effect on flying.

PS-be sure you use a small level so you have the plane horizontal (level).

Here are a couple of pics of one of my planes suspended on the Vanessa. You can't see it, but I put a small mark on the fuselege where the GC is susposed to be. When the plane is level, the plumb bob points to the GC. Just move around some weight (probably batteries) to get the plumb bob pointing at your mark.

Here's another shot of one on a Vanessa:



Note at the wingtips the piece of string held on with tape. That "drew" a CG line all the way across the wing, as a point for me to aim for.

FWIW, the Vanessa cost me roughly $10 to build...and the rope was a majority of that. *heh* They do, indeed, seem a little "confusing" at first, but once you get the idea, they're really pretty simple.

Thanks guys!

Yeah, I'm balancing a biggish twin, and don't trust my CG machine (ie don't trust it not to collapse!....). Maths doesn't bother me, but the Vanessa is kind of cool, and I not only have a beam right above my island workbench, but a plumb bob and some pulleys sitting idle, so I'm sold.

Again, thanks for your time and help,

Cam