# Milliken Equation

### #1

Posted 14 November 2011 - 16:56

The equation is found in Chapter 16, Page 589, I have just modified it to assume that all units used are the same (N and m):

K₵f = (Krf * t^2)/2

₵ is supposed to be fi, where K₵f is front roll rate

Krf is front ride rate

and t is track

I have tried using trig for a certain example I am studying to give a value for how many Nm are required to roll the chassis 1 degree. This method makes sense to me geometrically, but gives a different value to Milliken's as above.

Shouldn't suspension geometry be considered?

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### #2

Posted 14 November 2011 - 18:09

(b)Shouldn't suspension geometry be considered?

Lets move 1 wheel up by one mm, and one wheel down by 1mm. The roll angle is then 2/t/1000 radians.

The torque across the vehicle is kf*1/1000*t N m

So the roll stiffness is kf/2*t^2 Nm/rad. Perhaps he wasn't calculating what you thought he was calculating

(b) no

**Edited by Greg Locock, 14 November 2011 - 18:53.**

### #3

Posted 15 November 2011 - 10:17

For the torque across the front axle, why are you using the whole track, t but only using the 1mm of bump movement of the outer wheel? Shouldn't you use either half track with the movement of one wheel, or the total movement of both wheel for the whole track as you have done when calculating the roll angle in rads?

### #4

Posted 15 November 2011 - 22:01

because a torque is the force at one end mulitplied by the distance between the two forces.Thanks for the reply.

For the torque across the front axle, why are you using the whole track, t but only using the 1mm of bump movement of the outer wheel? Shouldn't you use either half track with the movement of one wheel, or the total movement of both wheel for the whole track as you have done when calculating the roll angle in rads?

### #5

Posted 22 November 2011 - 03:38

### #6

Posted 22 November 2011 - 23:21

. . . or a moment.Take a minute . . .