Ok All let's see if we can't give a clear explanation of all of the above.

Given a straight rate spring:

It always takes the same amount of weight/additonal weight to compress the spring a given amount regardless of where the spring is within it's range of travel.

The formula for this is: F=kx Where: F=force applied, k=the spring constant in units of force per deflection distance x= spring deflection, in units of distance

As an example, lets's say that we have a spring that has a rate k of 10lbs/in. The formula then is: F=kx F=10x Doing our algebra: x=F/10 In this instance, under a load of 10# the spring deflects 1". If we add another 10# of load, the spring now deflects 2". Add ANOTHER 10# and the spring STILL compresses just an addtional 1" for a total of 3inches @30# of force. Hence adjusting preload does not have the ability to move the spring into a stiffer part of it's range, because the spring rate is constant across its entire travel, up to the point that the coils bind.

Now lets consider the case where the spring rate is not constant. The equation relating force to deflection remains the same, but rate is no longer a constant k, but rather some function of x, the displacement. Let's call the rate R. we can write this as: R=f(x) - this sez: rate is a function of deflection and F=Rx - this sez: force is equal to a function of dispacement, times dispacement let's say that we have a spring where the RATE R = twice the DISPLACEMENT (X). The rate equation will then be: R=2x substituting this into F=Rx we see that: F=2x(x) if we do our algebra and solve for x, we see that x= sqrt(F/2) (sqrt is shorthand for square root) now let's plug in some numbers and see how much force it takes to make the above spring travel through it's range.

If F=8# then: x=sqrt(8/2) x=sqrt(4) x=2 So, The spring traveled 2inches under an 8lb load. if f=18lbs then: x=sqrt(18/2) x=sqrt(9) x=3 So, it took 18lbs to move the spring 3 inches if f=32lbs then: x=sqrt(32/2) x=sqrt(16) x=4 So, it took 32lbs to move the spring 4 inches. let's break this down.

8lbs got you your first 2inches 10lbs more pounds got you the NEXT 1 (8+10=18) 14lbs more pounds got you one more inch for a total of 4 (18+14=32) What this tells us is that in the instance of progressively (not necessarily Progressive-TM) wound springs, preloading either the front or the back results in the movement of the spring into a stiffer portion of it's range.

If you already have the spring compressed 2inches then it will take 10lbs (not the original 8 since it's already in use) to get the spring to move just ONE more inch and an addtional inch on top of that will require 14lbs of force.

This is why your front end feels stiffer when you turn up the preload. Bear in mind that the R the spring rate function in the example above was arbitrarily chosen for simplicity. The rate function for a progressive spring will be different.