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Re: spring question



That equation is not the one for the spring rate of a coil
spring. I don't have it memorized, but the stiffness is
_inversely_ proportional to the number of coils and the
(square of the) diameter of the coils (the length of the
wire). The rate is also proportional to the 4th power of the
diameter of the wire.

Imagine a torsion bar (a coil spring is just a coiled
torsion bar) say 1 ft. long (few coils) and one 10 ft (many
coils). Which is easier to twist? Answer: the long one. It
has a lower spring rate.

Bob

Aaron Ness wrote:
> 
> How does a spring get stiffer if you cut it shorter?  F=k*x, where F is the
> force exerted on the spring, k is the spring constant, and x is the
> displacement, or the amount it compresses under the given load F.  The
> spring constant is a function of the metal and its heat treat and geometry.
> By cutting a coil you aren't altering the heat treat (hopefully), and you
> aren't altering the geometry in a manner that affects the spring constant,
> AFAIK.  This means that a given force will result in the same displacement
> as it did before.
> 
> A stiffer spring is represented by a higher value for k, which means the
> same input force would result in less displacement than before.
> 
> Or is there a problem in my reasoning?
> 
> Aaron
> '87 528e
> '82 Scirocco
> '70 Beetle
> 


-- 
Robert Moore
moorer1@home.com

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