I have to be careful that I've answered this right... Let me see if I can rescue this.
There are two factors at work, displacement and force:
Let's take a simple example, a 100 lb/in spring located directly over the wheel at the outer end of an A-arm. The installation ratio is 1:1, and the wheel travel will be whatever the shock is, let's say it's 1".
If you want to double the wheel travel, the shock must be moved inward along the lower A-arm until 2" of wheel travel equals 1" of shock travel. One inch of wheel travel results in 1/2" of spring travel.
What happens are two separate things:
1. The spring is now compressed half as much (1/2"), so the spring rate must be doubled to balance it.
2. The leverage of the spring is now half; requiring double the spring rate to balance it.
Each factor require a doubling of the spring rate in order to maintain the original 100 lb/in force at the wheel. This is where the squared term comes from in the installation ratio.
So for a bike shock, you'll have to work out the installation ratio to see what you end up with, then decide if the resulting rate is acceptable.
yea thats what I meant. I knew the equations just never simplified it to see the square function. lol. the exchange of knowledge is allways welcome. to me atleast.
with this being said, what does everyone run for wheel travel? street? track?
1" droop, 3" compression, plus another inch for a bump rubber.
does this mean you design for a 4g bump? thinking about it, it makes since?