# Does Lowering ONE END of a CAR causes weight
transfertowards this end?

**Chris Pronios
**
hpro@b-link.gr

*Fri, 29 Nov 2002 00:40:19 -0600*

John,
thanks for the thanksgiving workout.
>* It's thanksgiving and the wife is watching Maury... might as well whip out
*>* the calculator. If you were to drop the back end of a classic by 4" and
*>* leave the front at the same height, you would move the CG by .4-.5"
*shifting
>* 10-12 lbs to the rear axle. I based my calculations on 60/40 dist, CG 24"
*>* above the ground, no passenger. If anyone has better data, I'd be happy to
*>* rework the #'s.
*
How exactly did you model the car and come up with the COG displacement? I
am very curious since I cannot fathom HOW the COG "moves" with repsect to
the two axes. In my simplified analysis the COG does not move.
What I mean (rough textual representation:
Car at rest, horizontal ground, horizontal chassis, 60/40 distribution on
axes:
L1=40 L2=60
O========O============O
^ F1 | ^ F2
|* | |
*|* | |
*|* | W
* |
v
Facts:
W=F1+F2
F1*L1=F2*L2 => F1=F2*L2/L1
so
W= F2*L2/L1 + F2 => W=F2(1+L2/L1) +> F2= W/(1+L2/L1)
If we tilt the car a bit, then the forces equilibrium does not change at all
(unless we can actually say that the COG moved to either side and the
relative lengths of the new L1 and L2 changed) and the moments equlibrium
still gives the same result since any sin/cos terms cancel out.
If we tilt by any angle A the formulas become:
W=F1+F2
F1*L1*sinA=F2*L2*sinA => F1=F2*L2/L1
so
W= F2*L2/L1 + F2 => W=F2(1+L2/L1) +> F2= W/(1+L2/L1)
Chris