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Posted by on March 19, 2006, 1:07 pm
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RoyJ wrote:
> You neglected to say that the spring is question has an
> OD of .094"
>
> And this section of the catalog is for continous length which states:
>
> Springs have open ends and can be cut to the length you need. Great for
> manufacturing, utility, and maintenance jobs. To determine the length
> (in inches) to which to cut your spring, take the spring constant and
> divide it by the number of coils per inch. Then take this value and
> divide it by your desired spring rate in lbs./inch.
>
> From what I can tell, this means that the spring constant is NOT the K
> value from your textbook but the spring value for ONE turn. DIVIDE by
> the number of turns to get the spring constant from the textbook.
>
> Don't complain to me, I just read the fine print!!!
Thanks for that, I am inclined to say that they are using the term
"Spring Constant"
in a not so very scientifically correct manner :-|
The spring is no good for me as it has too many turns per inch.
Cheers
>
> 4621vol@volcanomail.com wrote:
> > Hello,
> >
> >
> > Cannot understand Carr Mc Master 'spring constants' in their spring
> > section.
> >
> > They state 'the spring constant is the number of pounds force required
> > to compress
> > the spring one inch'
> >
> > The spring constants being given seem to be far too large.
> > A long 36 inch spring being made out of 0.013 inch wire has a spring
> > constant
> > of 4.8. ie it takes 4.8 pounds to compress the spring one inch.
> >
> > Where am I going wrong.
> >
> >
> > TIA.
> >
> > Jack
> >
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