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Old 08-31-2001, 03:54 PM   #1
Skankin
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Join Date: Dec 1998
Location: London, Ontario, Canada
Posts: 1,349
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Actually the coil-over difference is a factor of 4! (for the front).

Torque = force * distance

Force = (spring constant) * compression

The compression is proportional to the radius. ie. circumference = 2pi*r

So, Torque = k (dist) * compression

Since you're doubling the leverage & the compression, it's equivalent to 4X that of the stock location. (Actually is something like 4.1 because the springs aren't vertical... so they're compressed even more than what the arm movement would indicate).

So a 250# coil-over is similar to a 1000# spring in the stock location. Fortunately, the unsprung weight is reduced and the spring's motion becomes the same as the strut's, so the ride is a little better.

So with coil-overs, you have to take the square of the ratio.


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