Ok this may get a little confusing:
Using your numbers and formula, I get MR = .51. Way back here
, you said you got a motion ratio of .73 for the front, which is much closer to what I'm finding. Can you clear that up for me?
Sorry, my bad- I omitted the SQRT function in the formula I gave above. It should be:
MR=SQRT[(a/b)^2 * (c/d)^2]
The function basically normalizes the two terms. Now if you plug & chug, you get 0.72 not 0.51.
Strictly speaking: MR=SQRT[(a/b)^2 * (c/d)^2]
or more simply: MR=a/b (as in your example)
The answers should be very similar.
Contrary to the name, MR is not WR/SR. You do not multiply MR by SR to get WR. If you look way, way back at this post
I said WR=0.5SR.
What I left out is:
WR = Wheel rate
SR = Spring rate
ACR = Angle correction factor or sin(θ)
Remember, the angle changes slightly with ride height. I found mine to be 75deg.
Alot of folks (me included - and apparently you) call (MR)^2*ACR the Motion Ratio because it is the value that the SR is multiplied by to get WR, but it is not technically accurate - it is just a constant C.
In this case C=(0.72)^2*sin(75)=0.5.
Thatís where I got WR=0.5SR in the earlier post.
So lets do a real world example of of selecting the spring rate for the front of the car:
Letís say weíre going to lower the car to the point where we have 3Ē of suspension travel before hitting the bumpstops. When last on the scales we had LF=844lbs, RF= 848lbs. For simplicity, letís say 850lbs each. I havenít weighed the wheels, tires, knuckles, control arms, strut, spring and CV shaft, but I guessing thereís about 150lbs of unsprung weight at each corner - leaving 700lbs of sprung. Letís assume we are going to be driving on a fairly smooth track (AutoX tends to be much bumpier) so we design for a maximum 1.5g bump. Based on these assumptions, we would be absorbing a 700*1.5=1050lb force over 3Ē, therefore requiring a minimum wheelrate of 1050/3=350lb/in.
Of course WR and SR are not the same, but they a related by a constant that is a function of the control arm linkage ratios, the instant center of the front suspension and angle of the spring.
From above we found WR=0.5SR or SR=WR/0.5
Therefore we, need a 700lb/in spring to handle a 1.5g bump in 3 inches.
The new springs I just installed on my bilsteins are 700lb/in.
Hope this clarifies and I apologize for the confusion. The key is getting the LCA dimensions correct. It is best to remove it from the car.