Determining free length of springs when increasing stiffness?

[rustrocket]

Is there a formula to determine the necesaary free length of the springs when simultaneously lowering your car and increasing stiffness?

 

I'm sure there's one out there, but i just havent found it and i hate math.

 

Thanks.

[74_5.0L_Z]

With springs, the formula that relates deflection to force is as follows:

F=K*x where

 

K is the spring constant (i.e 250 lb / in)

x is the deflection in inches from free length

 

The deflection x is measured from the static position (no force applied)

 

 

As an example,

 

A spring is installed at a corner of the car with a free length of 12 inches, a rate of 250lbs/inch and a loaded spring height of 9 inches. What would be the affect of increasing spring rate to 300 lb/in and decreasing free length?

 

 

First, determine the change by increasing spring rate:

 

Assume the force at the wheel will be unchanged->F=k1*x1=k2*x2

therefore

x2=k1/k2*x1

 

that is x2 = 250/300*3 = 2.5"

 

The deflection of the springs is now only 2.5" rather than 3". If you were to only change spring rate without changing free length, you would raise the car by ~.5"(neglecting the angle of the strut).

 

 

Now, if you want to lower the car while using the stiffer springs. Subtract the amount you wanted to lower the car originally plus the additional lift acquired by going to stiffer springs.

 

Using the same set-up, assume that you wanted to drop the car 1.5" from its original ride height. You would need 300lb/in springs with L=12"-(1.5" + .5)" =10"

 

10 " 300 lb/in springs would lower the car 1.5" as compared to the same car with 12" 250 lb/in springs.

 

All else being equal.

[rustrocket]

Exactly what i was looking for, thanks.