fast377

iTrader: (52)

I have heard that cutting springs increases the spring rate. Any truth to this? I have coil overs on all 4 corners, and once the car is actually together and driveable, I want to be able to "tune" my suspension. I have a few sets of springs that are the correct length for my setup...all having different spring rates. But I have some springs, from different applications. They are semi-weak spring rates, but are considerably longer than needed. If I cut them to length, will the spring rate increase? If so, by how much?

Gearhead1

Yes, the spring rate increases. Think what would happen if you took your spring and unwound it to where it was a straight bar. If you lengthened it, it would be easier to twist. If you shortened it, it would be harder to twist. Same principle as the old Chrysler torsion bar suspensions.

fast377

iTrader: (52)

Good point...I should have thought of that. But how much (per coil cut off) will it increase? Is there a known standard, or does it depend on the particular coil?

Jpr5690

iTrader: (13)

Id Guess Devide The Spring Rate By The Number Of Colils On The Spring As A Estimate?

But That Just A Wild Guess

Actaully I Would Say Jus Tthe Oppsite Though I Mean Wouldent Less Coils Mean Less Material To Hold Up The Car?


But **** If I Really Know

kyles2000z

iTrader: (8)

it makes sine that with less coils, the springs would be weaker ad the car would bottom out easier. just my opinion tho.

fast377

iTrader: (52)

Found this on a motorcycle website: "A spring’s tension rate, in inch lbs per inch, is inversely proportional to the number of coils being used. The fewer the coils the stiffer the spring rate. One way to think of this is it’s easier to bend a piece of rebar that’s 6ft long versus’ one 2ft long. Now you understand why when your 18yr old neighbor kid cut his springs to drop his car lower the car now bounces down the highway. The spring rate has been increased due to less coils but the damping rate remained the same. The springs are now overpowering the damping circuits"

408WS6

iTrader: (1)

Here is the quantitative version of what you just said:

k = d^4 G / (8 D^3 N)

where:

d = diameter of the wire

G = Bulk Modulus (material property)

D = Diameter of the spring

N = Number of turns

k = spring rate

fast377

iTrader: (52)

So what value would I use for G?

trackbird

iTrader: (4)

Completely from memory, I thought it was 178,000, but I might be incorrect. It's the modulus of elasticity of steel, it should be easy to look up (I'm at the office now).

fast377

iTrader: (52)

I'll google it and see what comes up.

1BADAIR

iTrader: (1)

I would say that on a linear spring you most likely won't feel it. But on a progressive spring the change could be drastic

fast377

iTrader: (52)

The springs I am dealing with are linear rate springs. I havn't been fond of progressive rate springs in the past.