Riding Tips - Preloading

**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. **

**Let's define some addtional terms: **

**Ride Height--The height of the front or rear end with NO
rider on it. The suspension is TOPPED OUT (as if you were flying the bike--fully
extended). Ride height by this definition can not be changed by setting preload.
**

**Static Sag--(measured at each end separately)The amount of
change between topped out and the bike at it's resting point with NO rider.
**

**Race Sag- The difference between TOPPED OUT and suspension
position with rider in full gear and in riding position on the bike. It is true
that changing the preload will change static sag, unless static sag is zero
(like on REALLY light bikes). The spring does, indeed, compress and then the
fork extends a little bit in response but the net total of the adjustment (screw
in --fork expand) still leaves the screw having gone in further then the fork
expanded. Hence you have upped the preload on the spring AND slighty changed
your static and race sag, but not the ride height. (read the definition above
again). **

**Ride height is changed only by moving the forks in the trees,
or by changing rear shock length or linkages, or by changing tire size, or wheel
diameter. **

**Hope this helps. Cowritten by Kevin Glick and John Dahl MS
BE **