Roll sensitivity describes how far the car will roll given a certain cornering force applied. Depending on the direction of the calculation, any number of the variables in this can be solved with a combination of knowns and assumptions. This can help not only calculate load on a given tire or suspension point, but allow for measuring (and correcting) suspension geometry change.
If we assume that we are working with “known” spring rates and sway bar rates, we can skip those calculations. Those can also be directly measured off of the car.
So we are going to make a bunch of assumptions here. First of which is that the front and rear roll center heights are the same. This will make things easier in that the neutral roll axis is level with the ground. Effects of suspension geometry are also being neglected here. Tire deflection rates are assumed to be included in the roll rate values. CG and roll centers are in the center of the car, etc.
Thus, we have a moment, M, that is equal to the negative of the center of sprung weight (W) times the distance that sprung weight is above the neutral roll axis (h) times ( lateral acceleration (A) minus the roll angle of the chassis R)
I should probably note that if the neutral roll axis is inclined, we have to find the line perpendicular to the neutral roll axis that intersects the center of the sprun weight and find the length for that line and thats h.
Might be easier to read that in true equation form:
M = -W*h*(A-R)
The negative sign is there so that if A is positive (right hand turn), then the moment on the sprung mass is negative. This means that the roll angle is negative, and the car rolls to the outside of the turn. I’ve seen it written with the negative sign on the other side of the eqn before, but this way makes more sense to me.
Then we have a force F at height h is the same as force F at the origin on the neutral roll axis and a moment M (from above). You can also take that force F and split it amongst the front and rear axle according to the weight on each axle. For our example with a 50/50 weight distribution, the force would be split evenly, which keeps things simple. but far from real world. This force will produce load transfers separate from roll rates.
Back to finding roll angle though. The moment M will produce a roll angle R measured in a plane perpendicular to the neutral roll axis. (This, also, is why we are keeping the height of the front and rear roll centers even.) The magnitude of the roll angle will depend on the sum of the front and rear roll rates. We generally assume the roll to be about the neutral roll axis.
If we equate the Moment M to the roll stiffness moment we get:
Roll rate ( R) / lateral acceleration (A) = (negative of the center of sprung weight (W) times the distance that sprung weight is above the neutral roll axis (h) ) / (front roll stiffness (Kf) + rear roll stiffness (Kr) – center of sprung weight (W) times the distance that sprung weight is above the neutral roll axis (h)) = roll sensitivity (Kroll) Roll sensitivity here is expressed in radians per g of lateral acceleration.
Written as a more easily readable equation that is:
R/A = (-W*h)/(Kf+Kr – W*h) = Kroll
We then can split the moment M to front and rear components. The F gets combined with this too to give load transfer at each axle.
What we are looking for though is a way to express lateral load transfers for the front and rear as related to lateral acceleration.
The equation for the front will be:
Change in weight on axle / A = (W/t) * ( ( (h*Kf’) / (Kf + Kr -W*h) ) + ((l
– a) / (l)) * roll center height of axle) + (unsprung weight of axle / track of axle) * height of roll center of axle
This is for where Kf’ = Kf – (l – a) W*h/l
The equation for the rear will be:
Change in weight on axle / A = (W/t) * ( ( (h*Kr’) / (Kf + Kr -W*h) ) + ((a) / (l)) * roll center height of axle) + (unsprung weight of axle / track of
axle) * height of roll center of axle
This is for where Kr’ = Kr – a W*h/l
l is the distance between the unsprung weight centers.
a is the distance between the center of unsprung weight at that axle and the center of sprung weight for the vehicle.
Notice they are very similar equations. The difference to note is the l-a term for the front and the a term for the rear. L is the distance between the front and rear roll centers. a is the distance from the front roll center to the center of the sprung weight. All on the x axis with plane orientation between those centers not taken into account.
If we make the assumption that this is a single mass system and assume a bunch of other stuff for the sake of making the math not a huge pain, we can get the equations down to:
R/A = (-Wh)/(Kf + Kr) = Kroll
==
Front roll rate / A = (vehicle weight / front track width) *((h*Kf)/(Kf+Kr)+(distance from roll center of axis to center of mass / distance between front and rear roll center)*front roll center height)