Understanding Spring Rates

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Originally Posted by uninformed

Spring rate check: my coils are 16mm dia wire and according to how "king Springs" count their coils, 8.5 coils. How does that equate to the rate?

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My 240lb/in springs are 16.14mm wire and 6.6 free coils.
16mm and 8.5 free coils would be approx 180 lb/in.

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Originally Posted by Dougal

My 240lb/in springs are 16.14mm wire and 6.6 free coils.
16mm and 8.5 free coils would be approx 180 lb/in.

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"free coils"... I guess it comes down to how they are counted. "king" count from the very start of the coil (tapered end) to the very end (tapered end)

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Originally Posted by uninformed

"free coils"... I guess it comes down to how they are counted. "king" count from the very start of the coil (tapered end) to the very end (tapered end)

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You count all winds and then reduce depending on end type, eg. IIRC closed and ground ends reduce your coil count by 1.75 turns per end (and someone will correct if the memory isn't so hot! :D)

I guess Ill have to add pics.

I pretty certain these coils are 210 lb/in or very close to it because:

A: thats what I custom ordered all those years ago.
B: that 210 lb is still printed on the coil along with the SO(special order) number
C: when the "King" guy came and checked them, he calc'd them at about 215 lb/in.

Quote:

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Originally Posted by uninformed

I guess Ill have to add pics.

I pretty certain these coils are 210 lb/in or very close to it because:

A: thats what I custom ordered all those years ago.
B: that 210 lb is still printed on the coil along with the SO(special order) number
C: when the "King" guy came and checked them, he calc'd them at about 215 lb/in.

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Whenever I rate tested any custom Kings springs (admittedly a very long time ago) I always found the rates exactly as specified.

I count the coils that are free to move when installed.

Over the summer I broke a coil spring on one of my mountainbikes. It was a fox 500lb/in.
I had a rockshox 500lb/in spring to replace it and it's very different. I haven't measured the rate on either, bit late for the original.:angel:

Spring rate is:

k = (G x Dw^4) / (8 x Dm^3 x Na)
where:
k is spring rate in N/mm
G is shear modulus, or modulus of rigidity = 79340 MPa for good carbon and alloy steel springs
Dw is diameter of wire in mm
Dm is mean diameter of coil (OD - Dw) or (ID + Dw) in mm
Na is number of active coils

To convert N/mm to lbs/in, multiply by 5.711

Number of active coils are those coils not in direct contact with the spring seat or an adjacent coil, enabling them to flex and contribute to spring deflection. In other words, the coils that are free to compress under load.

Usually Na is total coils - 2, however when the closed coil at the ends is also tapered, and the spring has sagged, or the precision of the Winding is poor, the number of Inactive Closed coils at either end can be greater, and they must be excluded from the count of the Number of Active Coils.

So look at the spring for where the gap starts and count from there to where the gap closes at the other end.

As you can see the rate increases to the 4th power of wire diameter, so measure it accurately and don't include paint thickness.

Rate is inversely proportional to the 3rd power of mean coil diameter and inversely proportional to number of active coils. I.e. rate reduces if coil diameter, or number of active coils increase.

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Originally Posted by uninformed

Thanks John, as soon as I started reading your latest reply, the light bulb came on and thought, doh, sprung vs unsprung mass.

Lets see If I have this correct for my front end:

Free length: 422mm
static ride height length: 295mm
Rate: 210 lb/in

422-295 = 127

127/25.4 = 5 (to convert to inches as the spring rate is in lb/in, so must be same units)

210 lb/in x 5 in = 1050 lbs.

at the 1.35hz, which is 184.4 lb/in per 1000 lbf, that would be:

184.4 x 1.05 = 193.62 lb/in

If that is correct, my new 190 lb/in springs should be pretty close. Of coarse we are talking about ride and feel more so than handling. I expect with the softer spring to gain a little more body roll and there for, roll steer due to my ride height.

Spring rate check: my coils are 16mm dia wire and according to how "king Springs" count their coils, 8.5 coils. How does that equate to the rate?

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If I guess coil OD = 152 mm and number of active coils = (8.5 - 2) = 6.5 then spring rate comes out at 227 lbs/inch. See spring 3 in this copy from my spreadsheet.
http://www.aulro.com/afvb/attachment...1&d=1390281713

I use 'standard' rules of thumb for the various end types and it's been quite accurate so far.

Don't have easy access to my sheets ATM as I'm in the middle of a move.

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Originally Posted by rick130

I use 'standard' rules of thumb for the various end types and it's been quite accurate so far.

Don't have easy access to my sheets ATM as I'm in the middle of a move.

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In this chart, 'i' is the number of active coils
http://www.aulro.com/afvb/attachment...1&d=1390285968

But it is best to count them because, especially when closed and ground, as all Land Rover coil springs are, the inactive extent of the end coil increases with settlement or poor manufacturing quality.