## Centre of gravity of a car

 Quote by jddj hello i am new to the forum, i came across this thread and have been trying to figure out the derivation to this formula CGH=(Wb x Fwc)/(Tw x Tanθ), but i havent had any luck. could someone please explain where this furmula came from and how to derive it. Thanks

The load portion on the front axle when the car is leveled (Wf/W) is Lr/L [= 1 - Lf/L] (why?)

The load portion on the front axle when the rear axle is lifted (Wfu/W) is Lr'/(L cosA) [= (L cosA - Lf')/(L cosA) = 1 - Lf'/(L cosA)]

The load increase on the front axle (DWf) between the 2 states will be Wfu - Wf.

But: Lf' = Lf cosA - H sinA

So:
Wfu/W = 1 - (Lf cosA - H sinA)/(L cosA)
Wfu/W = 1 - Lf/L + H/L tanA
Wfu/W = Wf/W + H/L tanA
Wfu/W - Wf/W = H/L tanA
(Wfu - Wf)/W = H/L tanA
DWf/W = H/L tanA

Or:

H = (L x DWf)/(W x tanA)
 Thank you, thats exactly what i was looking for.
 The link above helps tremendously http://www.team.net/TR8/tr8cca/wedge...ef/CG_def.html. What if you now have passengers and the vehicle is on level ground? What would the height of the vehicles CG be? I do know the weight of the passengers along with their CG height.
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