Don Terrill’s Speed-Talk

I am working on a 1970 Monte Carlo. I am working from Moogs specs and can figure how much my car will compress a different spring, but was wondering what that difference will be at the wheel and fender due to suspension geometry. For example, if I install a new spring that will compress .500 more than my current spring, what will that be at my fender when measuring from the ground to my fender lip?

New springs won’t change the suspension geometry. Have someone sit on the finder to compress your current springs .5’’ and check the clearance.

You've probably got this figured out by now, but just in case I made a crude sketch to show off my freehand drawing skills . I don't know where you can look the dimensions up, but a minute with a tape measure will give you a sufficiently accurate answer.
The circle at the right end of the line is the control arm bushings, the coil spring is the, well, coil spring , and the vertical line at the left end is the center of the tire. The distance from the centerline of the control arm bushings to the center of the spring is "C", and the distance from the centerline of the control arm bushings to the center of the tire is "L". Think of this as a lever, so the change in height of the spring will be C/L times the change in height of the tire and the force on the spring will be L/C times the weight on the tire. If C is 8 inches and L is 14 inches and the weight on the front corner is 1000 lbs then the force on the spring will be 1000 lbs * 14"/8" = 1750 lbs. Hope this helps a bit.

ijames wrote:You've probably got this figured out by now, but just in case I made a crude sketch to show off my freehand drawing skills . I don't know where you can look the dimensions up, but a minute with a tape measure will give you a sufficiently accurate answer.

CoilSpring.jpg

The circle at the right end of the line is the control arm bushings, the coil spring is the, well, coil spring , and the vertical line at the left end is the center of the tire. The distance from the centerline of the control arm bushings to the center of the spring is "C", and the distance from the centerline of the control arm bushings to the center of the tire is "L". Think of this as a lever, so the change in height of the spring will be C/L times the change in height of the tire and the force on the spring will be L/C times the weight on the tire. If C is 8 inches and L is 14 inches and the weight on the front corner is 1000 lbs then the force on the spring will be 1000 lbs * 14"/8" = 1750 lbs. Hope this helps a bit.

Thanks, that helps. I was hoping someone could say that an A body spring compressed x equals whatever at the fender.

Moog does not have what I need. I am getting custom springs made. The spring supplier said they can figure what I need from what I have. Thanks again.