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Calculate maximum toppling speed


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#1 staticVoid2   Members   -  Reputation: 218

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Posted 24 April 2014 - 05:17 AM

Hi,

 

I'm trying to calculate the maximum velocity a car body can travel around a curve of radius r before it topples over, given the car base length x(*2), the height of the centre of gravity h and the gravitational force g(9.81m/s).

 

I've spent a while looking for an equation that would let me calculate this value and found this online:

 

Max Velocity = Sqrt( r * g * (x / h) ) [Book: Basic Principles of Physics (3-5.50)]

 

but I don't know if this is correct and/or how it works. I've read multiple sources that state the centre of gravity shifts when applying a centrifugal force to an object and when that centre points falls outside the base of the object the object will topple but it doesn't give any formulas for calculating this.

 

If anyone could clarify or point in the right direction to how to solve this it would be much appreciated.

 

Thanks.

 

 

 



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#2 Buckeye   Crossbones+   -  Reputation: 3805

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Posted 24 April 2014 - 10:41 AM

The formula appears correct IF x = distance from CG to the centerline of the tire.

 

It's a torque equation. Using the diagram:

tipping%20point.png

Just before the car tips, the forces on the car at the outside tire are as shown. Those two forces cause torques at the center of gravity, and, because the car is not tipping, the forces are balanced.

 

Torque due to mg = m*g*x

Torque due to mv2/r = m*v2 * h/r

 

So m*g*x = m*v2*h/r

 

Divide by m and rearrange to get:

 

v2 = g*r*x/h, the square of the equation you posted.


Please don't PM me with questions. Post them in the forums for everyone's benefit, and I can embarrass myself publicly.





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