Centre of gravity of a car

jack action

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

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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)

jddj

Thank you, thats exactly what i was looking for.

grupster2004

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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grupster2004	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.