FINDING THE HEIGHT OF THE CENTER OF GRAVITY BY MEASURING WEIGHT CHANGE WHEN TILTING THE VEHICLE:

                                   M
                                    \                  Q'   
                                     \                  \
                                     M                   \                                                                                    M                       Q'  
                          \        c |                 h  \
                       ang X         |                     \
           O-------------------------C----------------------Q
           [---------- a ------------]
           [------------------------- b --------------------]

b = wheelbase or track O-Q from axel to axel
a = length to center of gravity O-C when level
X = angle of tilt   ang QOQ'
h = height of tilt vertically at Q'
c = height of center of gravity above the axel line
M = total weight
W = weight at O

1.	W = M (b-a)/b  = M (1-a/b)
when tilted through angle X, Q is raised by h and more weight is transfered to O 
increasing W to W'
2.	b' = b CosX
3.	a' = a CosX - c SinX
substitute 2 & 3 in 1
4.	W' = M (1 - (aCosX - cSinX)/bCosX) = M( 1 - a/b + (cTanX)/b)
rearranging 1 gives
5.	a = b(1-W/M)
substitute in 4 gives
6.	W' = M( 1 - b(1-W/M)/b + (cTanX)/b = M( W/M + (cTanX)/b)
solving for c
7.	c = (W'/M - W/M) b/TanX = 
8.	      bCotX (W' - W)/M  = 
9.	      b^2(W'-W)/hM

To get the center of gravity above ground add the axel height above ground to c.




Richard Gray    (c) 97/07/26