Difference in drop spindle, and drop springs

[Doober]

Thanks I finally found an answer I didn't realize it was such a simple equation though :shock:

http://www.centuryspring.com/pdfs/techfaqs.pdf
"Cutting springs generally decreases the number of active coils. This forces an increase in spring rate. The spring rate is proportional to 1/Na (Na is number of active coils), so reducing the number of active coils by half doubles the spring rate." - bottom of page 3.

 

[Norm Peterson]

Taking that one more step, going back to U = 1/2Kx^2, the spring displacement x is reduced in the same ratio that K increases. But displacement 'x' is squared, so it hits you twice. For a 10% increase in spring rate, about 9.1% less energy is stored:

U = 0.5 * (1.1K) * (x/1.1)^2 = 0.5Kx^2/1.1 = 0.455Kx^2 when you put the comparison in terms of the original K and x.


Norm