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# RX spring question for spring geeks or engineers...

### #1

Posted 06 May 2005 - 10:58 PM

If I cut one coil off a spring is it stiffer cause it shorter, like a lever? Or, is the longer one that is compressed more stiffer?

I know, I know , I SHOULD know this!

Thanks

### #2

Posted 06 May 2005 - 11:15 PM

the mathamatical formula for the potiential energy a spring uses a spring constant (k) (Newtons per meter), energy (U) (in Joules) and distance compressed (x) (in meters) given that, the formula is: U=(1/2)*k*x^2

as you can see, the length of the spring is irrevalent to the energy absorbed by it - you can play around with the equation to your heart's delight - I am sure it would be quicker than for me to explain it all to you!!

I just finished a physics course on it!!! Simple Harmonic Motion rocks!!!

Good Luck!!

### #3

Posted 06 May 2005 - 11:21 PM

A springs resistance is rated in units of travel per force applied IE lbs/in or KN/mm. So let's look at an example. Let's say that you have a 24in long spring that's rated at 100 lbs/in and you compress it 6 inches, the spring is now 18in long and it's exerting 600 lbs of force. Not let's say you have cut 2 inches from that spring so it's now 22in long and you compress it again to 18in. Since you only compressed it 4 inches this time, it will only exert 400 lbs of force.

Note: This is assuming the spring is wound at a uniform pitch.

### #4

Posted 06 May 2005 - 11:29 PM

### #5

Posted 07 May 2005 - 12:26 AM

### #6

Posted 07 May 2005 - 12:43 AM

But if your stops were at say 16 inches. You could have 6 inches of travel stock at 600 lbs (and still have 2 inches of travel left) OR you could have 6 inches of travel with cut springs and be on the bump stops and be against 600 lbs also.

Am I understanding that right?

So the answer is "it depends"

Note I have a finance degree and NOT an engineering degree, but I love to test hypotheses.

Okay let's see if I can explain this without pulling out one of those big money text books that are collecting dust at the moment.

A springs resistance is rated in units of travel per force applied IE lbs/in or KN/mm. So let's look at an example. Let's say that you have a 24in long spring that's rated at 100 lbs/in and you compress it 6 inches, the spring is now 18in long and it's exerting 600 lbs of force. Not let's say you have cut 2 inches from that spring so it's now 22in long and you compress it again to 18in. Since you only compressed it 4 inches this time, it will only exert 400 lbs of force.

Note: This is assuming the spring is wound at a uniform pitch.

### #7

Posted 07 May 2005 - 09:42 AM

I'm pretty sure the springs on the back of the Rx are not progressive but I understand your logic with a progrssive spring.

I think I get it, my RX spring at stock length, compressed more, would be stiffer than a spring compressed less on the same strut.

Do I understand that the farther it is compressed the more energy it takes to compress it further?

### #8

Posted 07 May 2005 - 02:11 PM

Now I don’t through around my qualifications often but I will mention them here… I am a Mechanical Engineer with 11 years experience in machine design.

A coil spring is just a torsion bar wound up, and its length is a primary factor in its rate. Now when I say length I mean the length of the material used to make the spring, not its height. As long as the material is of constant cross section and the coil diameter is constant the spring’s initial rate will be linear. Only “active” coils count words the length, so wrapped ends and such that don’t move under compression don’t count. That’s how you create a progressive spring rate with the coils closer together on one end, as these coils bottom out the effective length gets shorter and the spring gets stiffer. I don’t have the proper equation at hand (its Saturday by gum), but its not exponential, if you cut the spring in a manner that removes 10% of its active length it will be 10% stiffer.

When cutting coil springs in cars normally the reduction in ride height isn’t proportional to stiffness increase due to the removal of in-active coils that lower the car without adding spring rate.

Captain Smarty Pants

Gary

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### #9

Posted 07 May 2005 - 02:50 PM

It has been 30 years since I last dealt with torsion formulas, so I will leave the proof to the student. But a little practical case might help:

Consider torqueing a bolt with a 3/8" torque wrench to, say, 150ft-lbs. (My apologies to my metric friends .) If you place a 2" extension between the torque wrench and the socket, there will very little torsion "windup" of the extension. If, however, you use a 24" extension there will be considerable torsion "windup of the extension.

If you keep increasing the length of the extension (without altering cross-section or material), at some point the torsional flex will exceed the torque needed to tighten the bolt.

IIRC, the torsional stiffness follows the "inverse square" law: Halve the length and you quadruple the torsional stiffness.

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