I'm in the process of selecting the spring rate for my coilovers. Clear advice on selection is hard to find so I thought I'd throw my theory out to you all.
Here is my theory:
Assume a front end weight of 1800# or 900# a side.
Assume 6" travel on the shocks so we would want to be somewhere in the middle of the travel @ rest. So, the shocks should compress approx. 3" when loaded. 900# / 3" = 300# spring. If you wanted 2" of compression @ rest you would use a 450# spring, etc., etc..
I realize that there is unsprung and sprung weight and the sides won't weigh exactly the same but I'm thinking these factors can be addressed by the adjustability of the coilovers.
Am I correct in this theory???
Thanks in advance for the advice.

You have part of it right...

What you have described is correct for "wheel rate" or what is happening at the wheel. For the actual spring you have to figure out the "motion rate" because it greatly affects the relationship between "spring rate" and WR. The MR takes into account the leverage difference between spring location and wheel and the travel difference between the two.

For a Chevelle you probably have the spring about halfway between the inner LCA pivot and the wheel centerline or at 0.5. The MR is this squared or 0.25 so you need 4x the desired wheel rate in your spring rate.

Aside from that not so minor detail you are on the right track.

If you really want to get analytic about it you will also need to get into Natural Frequency calculations as well. You see you can get to the same compressed length with quite a number of spring rates by varying the free length. So you have to have some way to choose between these, and that is where NF comes in.

Steve

O.K. so I can visualize the effect of the leverage of the wheel offset from the spring but if I use youor example MR of .25 then in my example I'd use a 300# x4 =1200# spring which can't be right. Can you give me a better discription?

Ok... I have 600 lb/in springs in the front of my Mustang. Let's say the car weighs 3400 lbs and has a 60/40 weight distribution. That is a bit over 2000 lbs over the front end or 1000 lbs per wheel. So at ride height the springs have to balance/provide a 1000 lb push (at the wheel) but where they are located (which is very similiar to a Chevelle) it is really a 4000 lb push (due to the MR). So my springs are compressed something like 6.7" from free length at ride height.

I think this is where you are getting confused: Now I may only have 4" of wheel droop below ride height, or 2" of spring extension. So there is 4.7" of spring compression required just to get the suspension to bolt together. As you go to stiffer springs this particular value decreases and with softer springs it increases. Which is why competition springs are easier to remove than street springs.

BTW I have a front NF of around 1.6 cycles per second if I remember correctly, it has been at least a year since I went through this in detail with real data. I would guesstimate that a stock Chevelle, being the boulevard cruiser that it was, would have a NF closer to 1.2. The general range for NF is 1-2. This is a taste thing but for my butt the 1.6 is about the high end of the range for a daily driver. And from experience it does provide a very nice feel on a mountain road :-)

Steve

Tat's the part I was missing!!There is some compression taking place while the control arms are on their stop in your case 4.7" @ 600#'s is 2820 plus the 2" once the car is on the ground for another 1200# which gets you to the 4000# needed.

Thanks,