I would expect on a J-3, thinking on it, that tipping it up wouldn't tip it far enough. Unless they're known to stay on the nose in that condition.
The c.g. will be pretty close to the wing in any event.
This makes any suspension somewhat risky, in that it needs to all the plane off the ground, in both attitudes.
I recall seeing a T-6 that tipped up on the nose... when it was pushed back down, the aft fuselage got a very large wrinkle just behind the wing!
When things get this complex, I step back and ask... "Do I NEED that information"?
Many times I find I don't.
Why is the location of the vertical c.g. so important?
Cubs in real life fly solo from the rear seat. The presence of a person there really affects the location of the c.g. of the empty plane, which the model will be, but.. the c.g. of the model will need to be where it is on the single-person full-scale as the aft limit.

Yes, I had thought about that tip issue. Right now I'm pretty sure tipping will work because I've set it up on it's nose several times without it coming over. it rests on exhaust pipes just past balance point. As to why I need it. The science boys are developing modeling software program for an auto pilot system to go in the plane. Vertical CG is just one of a hundred things they need.

Splais,

Please, excuse me for any confusion that my poor English may have caused.

You can use the method described in the drawing above with total confidence that it will give you the spatial location of the CG very accurately.
It is only a suggestion, but I do know it works and I believe it is a very easy way to do what you need.

Now, I understand that you will calculate the actual location of the GC respect to the leading edge of the wing.
Before, I had assumed that you would use the recommended or specified CG location.

That assumption made me think that we should work with the actual location, instead of assuming that the actual CG was where recommended.
Sorry about my assumption, and please disregard my post #22.

I believe you meant post #22.
Since the direction of the weight force acting on the CG is always perpendicular to the ground:

Let’s call X-axis the line made by the ground level.

Let’s call Y-axis the line perpendicular to the ground that crosses the axle of the main wheels.

Please disregard my explanation, as part of post #22.

You are correct; the CG is a unique point in the space, which position respect to the airplane’s body does not change, regardless airplane attitude, as long as the distribution of unitary masses remains unaltered.
However, as we rotate the airplane’s body around the main wheels’ axle, the CG rotates around the axle as well.

No; steps one and two shown in the drawing of post #16, if done carefully, will give you the exact location of the CG, as projected over the side of the fuselage.

First, you find and draw the line that is parallel to the Y-axis and contains the CG, while the airplane is hold in perfectly horizontal attitude.

When you continue rotating the airplane’s body and the CG reaches the position right above the main wheels, the weight force aligns over and coincides with the Y-axis.

At that attitude, there is no torque or moment produced by that weight force, because there is no arm respect to the Y-axis.
All the forces and moments acting over the airplane are in perfect balance (but unstable, since the CG is above the support point).
At that attitude, the actual CG cannot be anywhere else that alongside the Y-axis.

You will have two lines that intercept each other, and your CG will be exactly located at that intersection.

Regarding the tipping issue; I have drawn the airplane at exact scale, and the drawing shows that, as long as the prop is in close to horizontal position, there is enough clearance to go over the balanced attitude.

Regards!

To clarify a point. I know where the Pitch CG is suppose to be (10-12" from leading edge. We have been flying it with the CG near the rear of that range by rough estimate. So yes, I still need to determine the exact CG under two different load conditions; i.e., determine where the equipment needs to go to keep CG where it is now. Secondly for that auto pilot programing I need to measure the Vertical CG. So I am validating one CG and measuring another.

How have you guys estimated that location?
Weight on each wheel at normal stand attitude?

nope, nothing so crude. We use a very scientific method - Lift the plane with the edge of you hand at CG point (level flight) on each side, and measure rough location.

OK, thanks.

Now, could you explain this a little more for me?

"So I am validating one CG and measuring another."

The Pitch CG is spot on, plane flies great. I just need to do a precise measurement of it. But I also still need to measure the distance below the thrust line that the Vertical CG falls on. Your method in the second part of the drawing should do that good enough for now.

I will run a datum line along the fuselage at the thrust line; run another vertical line along fuselage at Pitch CG; tip the plane and mark the Vertical CG, then measure how much it is below thrust line.

I understand now, thank you very much.

The first portion of the drawing shows how to level the plane in order to draw the vertical line (plumb) from the point aft the LE, where the longitudinal CG (what you call Pitch CG) is located at.

I included dimensions between wheels and other, just because I assumed that you would need to estimate the longitudinal location of the CG by weight on each wheel.

Please let us know of your findings.

Splais,

this si getting a little complicated. I have given you the method for finding the Centre of mass from the front of the aircraft and the position of the Centre of mass from the fuselage longuitudinal axis.
You need the height. '
This is easily calculated as well by using the principals of SImple Harmonic Motion. Think of the plane as a pendulum. Its centre of mass is suspended from a string of length L

If you could get the opendulum to swing about the CG position as defined by the distance from the nose of the aircraft you could suspend it or support it under the wing at that point, CG point. This should not be too difficult to do. Yes I know its 50% scale but it just needs to be lifted a few centrimetres so that it can swing unimpeded.
Once the plane is suspended on that linei t will rock back and forth if a force is applied to it. You know the mass of the aircraft I hope. if not the total mass is the sum of reaction forces at the wheels or the sum of the three weights recorded at each wheel when the plane has been <u>levelled.
</u>
Now all you need is a stop watch. the formula is:

L= (T^2 x g) / (4.pi^2)

L = length of your pendulum or the distance of the pivot point to the centre of Mass.
T= is the time period for the plane to rock one complete cycle
g = acceleration due to gravity, assume 9.81 m/sec^2

Once you have the answer, this will please your scientists greatly and they will laud your ingenious application of science. After they reward you you can buy me a special gift from Tower Hobbies next sale

Unfortunately the methods proposed in previous posts that entail tipping the aircraft  and drawing lines will tell you little. but you will have some fantastic racing stripes all over your fuselage.

If you have time, do some reading on simple harmonic motion. You can actually find the cente of mass of a aircraft simply by suspending it and apply a known force and measuring its displacement in 3 axis.
Unfortunately with the advent of CATIA, PRO/E and the like, these techniques are seldom used</p>

I realise I did not take mass into account above.


the equation if taking mass into account is

y = [4(pi^2) I] / M.g.T^2

where I is the moment of inertia calculated from I = m. r^2

Tried it.
Suspending the Kadet was simple.
This one has an 8 foot wing.
Once suspended, any swinging induced also made it rotate around the suspension string.
Ran out of room trying to find the position where it would oscillate in one plane, without rotating.
And there is a "harmonic" with the fuselage bobbing up and down as it oscillates.

Changed the suspension method to move the pivot point closer to the support beam.
Made a video... uploaded to Youtube, rcu has no more room for videos.
.
PendulumCG
http://www.youtube.com/watch?v=BOhMzPDs7tQ
.
Airplane weight: 7 to 7.45 pounds (scale only reads pounds)
Swing time 1:29
Distance from support knot to top of the wing: 17"
Proof of concept left to the student.

Could you explain in simple-eze what you did with those numbers. I guessing this has something to do with TimBles's formula for determining the vertical location of the CG?

I have provided -some- of the numbers.
I haven't timed -an- oscillation.
That would be most accurately achieved by counting video frames, with the NTSC 30 frames per second standard.
Or count the number of oscillations from release to rest.

Allright boys, I think I found the right thread for this question. I overpowered an aircraft making it nose heavy. The obvious and easiest choice is to stick some lead on the tail and make it hang at the spot I want. We all know dead weight is not the best answer. I have plenty of weight already on the aircraft in the form of servos, batteries, RX etc. The battery box is aft of CG as I can manage so it is out of the equation. The rudder and elevator servos are forward of the CG and this is what I want to move back to compensate for the larger engine. Obviously I will have to build a mount for a servos somewhere back on the tail boom to move the servo's location. What I want to avoid is randomly punching holes in my airplane and trying and erring my way to a correct servo location. I understand that I really just need to remove my chosen servo, hang the plane, place the servo in the spot I want it to balance, and then go to work. Really, I am looking for more of an academic exercise.

Does anyone have a link or knowledge on how to take the known balance point, the known weight of the ballast (rudder servo), the future balance point, with X (unknown) being the placement of the ballast to achieve the desired future balance point? I am going to go ahead and get my plane balanced the caveman way (putting the servo on the tail until the plane balances).


Any questions? OK 1,2,3, GO!!

Thanks

It's a simple balance equation. Weight of one thing times distance plus weight of another times distance divided by all up weight gives the location of the balance point. Convention has weights forward of the reference as positive, behind it negative.
Typically for aircraft the datum point is ahead of the nose, this keeps all sign convention simple.
If I were to move the elevator servo back on my FlatFlier AD-5W, the c.g. change would be like this...
Just where it comes out with finger balancing...