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How to calculate the ride height from spring free length?Helpful Member! 

BillaNichols (Electrical) (OP)
22 Oct 10 18:29
Are these assumptions correct?

I want to change the springs on my 2008 Saturn Sky to increase the natural frequency (NF) and lower the center of gravity (CoG).

The stock front spring has a spring rate of 128 lbf/in, a free length of 15 inches and a compressed length of 10 inches. The spring I've selected has a spring rate of 225 lbf/in and a free length of 12"

The stock spring when assembled in the coilover has a pre-load of 128 lbf/in times 5 inches equals 640 lbf. The new spring will have a pre-load of 225 lbf/in times 2 inches equals 500 lbf. This is because the assembled coilover compresses the spring to a 10 inch length (non-adjustable).

The unsprung corner weight is 655 lbf. At static conditions the spring will compress another 0.12 inch, (655-640)lbf/128 lbf/in = 0.12 inch.

Using the above assumptions the new spring will settle to (655-500)/225= 0.69 inch. This will lower the cars CoG. I'm hoping to lower the car 1 inch.
GregLocock (Automotive)
23 Oct 10 1:59
I think that looks right but you are assuming a 1:1 motion ratio, and you are ignoring the unsprung mass.

 

Cheers

Greg Locock


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BillaNichols (Electrical) (OP)
23 Oct 10 3:41
Thanks for the replay Greg.

Your are correct about the motion Ratio. I'm only considering the shock and spring travel. The MR is .73 for the front end and the coilover is mounted at a 60 degree angle from the horiaontal plane.

I thought that the spring only carries the unsprung weight?

You can buy the springs anywhere but I can't find any info on how to select the free length. Anyhow, any help is appreciated.

Thanks

Biil
Helpful Member!  BrianPetersen (Mechanical)
23 Oct 10 9:02
The motion ratio has to be accounted for because it is also a mechanical leverage that increases the force on the spring over and above what the unsprung weight of that corner is.

655 lb corner weight / 0.73 = 897 lb force on the spring

Stock springs will compress 897 / 128 = 7.00 inches from their free length (i.e. they will measure 8 inches when the car is at rest). This is compressed 2 inches from what you say is the preloaded length, which makes a whole lot more sense, your previous number implied almost no travel available in rebound (going over a dip). This is 2 / 0.73 = 2.75 inches of rebound travel available at the wheel - plausible.

Your proposed springs will compress 897 / 225 = 3.99 inches from their free length, also almost exactly 8 inches when the car is at rest. In other words, there won't be any change in ride height, but there will be more roll stiffness, etc.
BillaNichols (Electrical) (OP)
23 Oct 10 11:01
Cool! :)

Thanks Brain.

So if I use a 10 inch free spring lehgth instead of the 12 inch keeping the same rate of 225 lbf/inch than the compressed spring length will be 6 inches, (free length 10 inches) - (897/225 inch). The ride height will change by [stock compressed spring length ((15 inch) - (897/128)inch)) - new spring compressed lenth ((10 inch - (897/225)inch)] * sin60), ie., (8-6) * sin60 = 1.73 inch.

I think I'll need some adjustable spring perches.

Thanks for you help.

Bill
desertfox (Mechanical)
24 Oct 10 13:59
Hi WildAlan

Just check your figures again because your spring rate of 225lb/in with a compression of 2" yields 450lb force and not 500lbs force.

Regards

desertfox

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