how do zero pre load shocks work?

[4stroker]

i have heard alot about "zero pre load" shocks, but dont know much about em.

 

why are they good?

 

who makes zero pre load shocks? are they the only way to get ride height adjustment?

 

what does adjusting the preload on a normal (aftermarket) shock do?

 

any info/links greatly appreciated.

 

thanks, mark

[Skidmark]

zero preload shocks use a lighter weight spring that compresses when you sit on the bike. This keeps the weight off the main springs. Its like having a bunch of suspension sag while also having stiffer springs. This lets you have more controll through the little bumps and also be able to hit the big jumps without bottoming. You can adjust the ride height on most shocks, but aftermarket shocks have more height adjustment. Adjusting the preload on a shock gives you a slightly stiffer/softer spring rate. Basically all your doing when you raise the ride height is compressing the spring which makes it stiffer and able to support the weight better.

 

http://www.elkasuspension.com/

http://www.danacreechracing.com/categories/pep-1.html

http://www.customaxisshocks.com/atv.htm

http://www.tcsracing.com/

http://worksperformance.com/

[HotCarl]

Adjusting the preload on a shock gives you a slightly stiffer/softer spring rate.

 

 

Preload does NOT change the spring rate. The rate of a spring is the change in the force it exerts, divided by the change in deflection of the spring. That is, it is the gradient of the force versus deflection curve. Preload is a ride height adjustment only.

 

I'm not saying this to be an ass, I just hate to see people pass along wrong information.

[Skidmark]

Thats what I meant, I just made it easier to understand. If your shock is topped out like most stock shocks are and you up the preload it basically stiffens the spring. I know that may not be exactly what it does but thats the easiest way to explain it. Most people have no idea what it means when you say. "The rate of a spring is the change in the force it exerts, divided by the change in deflection of the spring. That is, it is the gradient of the force versus deflection curve."