Does a car’s center of gravity shift with whatever amount of people are on each side of it?
This is in regard to if a car hits a curb or is caught in an angle and opposing the angle is more people on that side of the car, will it shift back due to its center of gravity or another factor?
If you take a car with springs on each side, and no steering or suspension geometry, when the car tilts one side's springs compress and the others lengthen.
This gives a righting moment.
Now if you put more people on one side than the other then the springs on that side compress a bit.
The other side lengthens a bit. This BEFORE you hit the bump.
When you hit the bump the change in the suspension is exactly the same as the first case.
If the four springs have a spring constant k and there is a sideways force of F at a height y above the wheels
And the wheels are r apart with the springs compressing a distance of x
then F*y = 2k x r
ie x = F*y / (2kr) As long as no wheels leave the ground the righting moment requires the same EXTRA compression of the springs no matter were the mass is located.
Only if you are driving around the corner on the outside two wheels, keeping the inside ones off the ground, is the horizontal position of the centre of mass relevant.
The real center of gravity is determined by how the weight is distributed. This also determines righting arm for impacts (the pressure exerted to return to the original status). There are complex formula's used to determine righting arm which while a function of center of gravity, is not the center of gravity.
Answers & Comments
Even without the centre of gravity.
If you take a car with springs on each side, and no steering or suspension geometry, when the car tilts one side's springs compress and the others lengthen.
This gives a righting moment.
Now if you put more people on one side than the other then the springs on that side compress a bit.
The other side lengthens a bit. This BEFORE you hit the bump.
When you hit the bump the change in the suspension is exactly the same as the first case.
If the four springs have a spring constant k and there is a sideways force of F at a height y above the wheels
And the wheels are r apart with the springs compressing a distance of x
then F*y = 2k x r
ie x = F*y / (2kr) As long as no wheels leave the ground the righting moment requires the same EXTRA compression of the springs no matter were the mass is located.
Only if you are driving around the corner on the outside two wheels, keeping the inside ones off the ground, is the horizontal position of the centre of mass relevant.
The real center of gravity is determined by how the weight is distributed. This also determines righting arm for impacts (the pressure exerted to return to the original status). There are complex formula's used to determine righting arm which while a function of center of gravity, is not the center of gravity.