I'm missing something here and I'm sure someone could enlighten me. Trying to raise my car 1.5 inches for driving in the snow/clearing taller tire. To make this simple we will deal with one corner only. Right front corner weight is 1004 lbs. Stock spring at 180lb. Car sags 3 inches. How does a 1004 lbs only compress the spring 3 in.?

check my math.

150lbs spring would sag 3.6 inches. Add 1.5" for desired height change plus stock free spring length.

Stock spring free length + .6"(difference in rate)+ 1.5"(desired increase)=length to order.

 

Spring rate and wheel rate are not the same thing. Depending on the suspension geometry the spring displacement can be proportionately more or less than the wheel travel. Pick up any suspension or race car setup book. And I think you'll also find that from full droop to static ride height the wheel travel is more than 3 inches.

Dave

 

I understand wheel rate and spring rate are not the same thing. That's not related to the ? I asked.

I relize now, my mistake was not disconnecting the sway-bar. Jacking up only one side, loading the opposite side. That's what I get for being in a hurry.

Macpherson design strut(as the front of an Evo) will always have more wheel rate/travel than spring rate/travel. This is true any time you mount a shock on an A-arm to the chassis. To get less wheel rate you need something that will never show up on a production car, less maybe a Ford GT.

Here's a fine example. My past job experience on a Lights team goes far beyond what books teach you.

 

I'm confused. Are you really looking for an answer? And if so, what is the question? What I can gather seems very rhetorical by nature.

 

djh got the question right. Just wrong answer. Simple answer would have been "hey, did you remember to disconnect your sway-bar?" So I took his comment about picking up a book a bit too personal. In my second post, second paragraph states I found the problem.

 

I'm sorry I didn't correctly discern your level of incompetence.

Dave