Spring rates on the brain



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  • DaveJ

Posted March 05, 2002 - 11:22 PM

#1

Here’s something I realized tonight involving spring rates. I hope I did my math correctly or this is going to be very embarrassing.


A .45KG fork spring means that it takes .45KG or 25.2 lbs of force (weight) to compress this spring 1 inch, (double it for the pair). For 2 inches of compression it takes 50.4 lbs., and an extra 25.2 lbs for each inch. It takes about 302 lbs. to smash one .45 spring.

Going from a .45 to a .47 fork spring is a difference of 4.26%.

Going from a 5.4 to a 5.6 shock spring is a difference of 3.57%.

Without pre-load, if you wanted to compress a pair of forks with a set of .45 springs 3 inches, it would take 151.2 lbs. If you change the springs to .47, it would take 157.92 lbs.

At 7 inches of compression, an additional 15.68 lbs of weight is required to compress the .47 springs to that of the .45s.

Without consideration to damping rates, if you bottom your forks with a .45 spring set, you have applied roughly 604 lbs of force to the front end of the bike. It would take 631 lbs of weight to do the same with .47 springs, (this is probably about the same amount of weight you take to the back of the head when someone decides to use you for a landing pad).

Forks on a 426 will stand, without rider, about 1/8 inch taller when switching from .45 to .47 springs.

Forks on a 426 will stand, with a 200 lb rider (centered) about 1/4 inch taller when switching from .45 to .47 springs.

Every millimeter of pre-load will compress a fork spring with about 1 lb of force. At 7 millimeters of pre-load (on each fork) it will take about 14 lbs. of weight before the forks begin to sag, (the front un-sprung weight of a 426 is about 113 lbs).


Some of these numbers are a little rough, but close enough (I hope).


DaveJ

PS - This is what happens when you drink Red Bull on your way to bed.





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