First off, this is not an aero question but I felt that the aerodynamicists here would be the most likely to have an answer. I used to know how to do this! I was playing around checking the CG of a new plane. It was off by 3/4". I know I could just add weight and keep checking till I get it right but I thought I'd do a simple calc (or so I thought) that would tell me how much to add without going through the trial and error method.
I started with a basic balance beam equation based on nose moment/tail moment at current CG. I used 36 as tail moment 16.75 for nose moment and a total weight of 122oz. This calc'd to a force of 38.7 oz at the tail and 83.2 on the nose. This is where I get stuck. How would you calc to find the weight needed to shorten the tail moment by 3/4"?
I know it'd be much simpler to just add weight till it's right but I'd really want to know how to calculate it. Like many things, it's just good to know how!
Thanks,,,basmntdweller

You need to get out of the basement more.

While it doesn't account for the mass distribution exactly, you can do something like this (assuming my physics isn't that rusty):

Assumption = aircraft balanced around the moment arms and forces you gave: 83.2 * 16.75 = 38.7 * 36

Then shift the fulcrum .75" forward while holding the forces the same:

(83.2 + x) * 16 = 38.7 * 36.75
x = (38.7 * 36.75) / 16 - 83.2
x = 5.7 oz (or so for a starting point)

Actually, it'll be more since you won't be able to add the weight at the nose end of the balance beam (at the spinner). You could add a second moment arm on the left side of the equation to make a better guess like this:

(83.2 * 16) + (x * 10) = 38.7 * 36.75


x = (1422 - 1331) / 10 = 9.1 oz in the fuel tank area

I'd just add modelling clay until it sits level...

But.....

OK, we have an out of balance 80 oz model. We want to move the balance point 1 inch forward of where it is now. The distance from the exising balance point to the area we can add our lead lump is 11 inches. So if we look forward in time to when it's balanced we will have 80 oz of model sitting one inch behind where the new point will be. That is 80 oz-in of torque. Our lead lump is sitting 10 inches in front of the new balance location (now you see why I chose 11 inches... ). So we need 80 oz-inches of torque in the nose. 80oz-in / 10 in = 8 oz.

Now plug in your own numbers.

Howzzat.

Thanks thud,
That was where I wanted to go! I am assuming that you can't account for mass distribution because the mass is not evenly distributed on each side of the fulcrum, correct? I calced to move the fulcrum back .75" and got 1.067oz. I'll set up my balancer and check it out tomorrow.
Thanks,,,basmntdweller

Steve, I'd love to get out of the basement more but I'm a wimp when it comes to flying in 20mph winds at 20 degrees. I stay where it's nice and warm.....

Hmm, I get 2.6 oz using my method

That's 122 x .75= 91.5 in-oz. 91.5 / 35.25 = 2.6 added to the tail to shift the CG back 3/4 inch.

I think you are trying to remember how to "sum the moments". Here's the formula after all is said and done:

X = (W * D1)/D2

X= Amount of weight to add
W=Original weight of plane
D1=Distance from orig. CG to desired CG
D2=Distance from desired CG to the position of the added weight.

Okay, now I'm confused! Bmatthews theory looks and sounds familiar to me. As I said, I have done this before but it's one of those use it or lose it deals. I hadn't used it for a long time. Interpreting Bruce's logic, I arrive at a requirement of approx 2.6 oz. I am wanting to move the CG aft .75". Maybe you guys can compare my answers to what you come up with.
Thanks again,,,basmntdweller

My intuition tells me that you can't use the total weight of each side as though it's at the end of the fulcrum. But my physics is rusty too.

It's going to be interesting to see just how much weight you end up with on the tail.

HEY, THAT'S MY ANSWER.....

Just nicely put in a box with decorative cellophane wrapping.....

And just how long has it been since your physics classes hauckf?

Summing the moments is a phrase I also recall. hauckf's answer is the same as Bmatthews except in direct algabraic form versus "word problem". Thud's answer made sense for the most part but the 1.067oz just didn't seem like near enough to do the job. The 2.6oz seems closer.
I will confirm tomorrow!
basmntdweller

Oops, I missed the "shorten the tail moment" part.

For point masses, the selected moment arms in this case should have been the mid points of the nose/tail lengths since we don't know the exact weights and moment arms.

hauckf's method is better.

Bmatthews: How long? Hmmm, lets see now.... 42 years (1961).

Draw a line representing the model, with the distances and directions of the forces shown on it. Choose a convenient point on it (try to remove one of the variables by choosing a point that puts it at the fulcrum, but the point can be anywhere).

Then; clockwise moments = anticlockwise moments.

(a moment is the weight times the distance from the chosen point; the same as torque)


David C.

Well, your memory is better than mine. I last did this sort of Physics in about 1972 or so....

This forum is sure a lot better than crossword puzzles...

Here's another little brain-teaser: Given the plane's weight and CG, the battery weight and it's location with respect to the CG, how do you calculate how far to move the battery to move the CG a given distance?

Almost the same, but different

I'm way rusty, but think you are correct. It seems if using the mass behind the CG as a point mass it would have to have a moment arm length equal to the distance to it's center of mass, not to the tail.

See my other message. It works for any number of weights.

David C

I setup this evening to try and be as accurate as possible with my measurements. After rechecking the existing CG, I found that in order to get where I think I want it to balance, I have to shift the CG .91" to the rear. I reweighed the plane on a better set of scales and still weighed out at 122oz. It's good to know my cheap scales are fairly accurate too. So, summing the moments, I calculated (122*.91)/35.25=3.149oz to shift it the desired amount. I set up my plane on the balance stand at the desired CG. I added one stick of the stick-on type weight to the location I figured for. It balanced perfect with that one stick. I figured that getting that close was pretty decent. The calculation was only off by .15oz. I was getting ready to write this and then wondered how accurate that stick-on weight was. I got out my Ohaus gram scales and the stick of lead weighed 89.12 grams. Converting to oz I get 3.143oz. Pretty freakin close to the predicted 3.149 isn't it!!!
Thanks for the help and the physics refresher. I feel really bad as you guys have much better memories than I as my physics class was about ten years ago!
basmntdweller

basmntdweller: Glad it worked out like we thought it would; kinda renews one's faith in math etc. Also, I really liked the part about checking the actual weight of the stick-on when it didn't come out exactly as you thought it would. You'd make a good engineer!
David Cutler: You are right, of course, about the solution to the moving the battery question, but setting the problem up is a litttle tricky because the battery weight is included in the total weight of the plane. Try it. Remember, the sum of the forces must be zero as well as the sum of the moments. I think the answer is:

Xf = (W*Xcg + B*Xo)/B

Where:

W = Weight of plane (incl. battery)
B = Weight of battery
Xcg= Distance form orig CG to Desired CG
Xo = Distance from orig CG to orig. battery position
Xf = Distance form orig CG to final battery position

Note: Distances measured from the orig. CG forward are positive. Distances measured from the orig. CG aft are negative. Example: If the orig battery location is 3" behind the orig CG, Xo = -3 etc.

Only two things to remember:-

1/ Clockwise moments = anticlockwise moments.

2/ Every couple has its moment.

I'll just have to take the second one on trust.

-David C.

p.s. a 'moment' is the weight times the distance from any point.