Originally Posted by moon161 (Post 2636339)

Wrong.

Shear does not accumulate like displacement through the length of the spring. If it did, there would be no reaction at one of the supports- either the spring perch or the LCA.

Viewing each coil as a free body is useful, and spring body connects the equal and opposite (in a static case) forces at the spring perch and LCA via shear.

Consider the spring again 10" 10 coils, 100lb/inch. For a 100lb load, each coil sees 100lb, and deflects .1". The spring constant of each coil is the force over the deflection, or 100/.1 = 1000lb/in.

Cut a coil off, 100lb load, deflection is 9x.1" or .9". To achieve 1" deflection, each coil deflects 1/9" = .111", the reaction is 1000*.111=111 lb. The new spring constant is 111 lb/in.

Another way to look at it is each coil being in series w/ the next.

http://en.wikipedia.org/wiki/Hooke%27s_law#Multiple_springs

Springs in series add like resistors in parallel:
If added in series, 1/k = Sum(1/k1 + 1/k2 ....)

10 coils 1/k = 10 * 1/1000 =.01
k = 1/.01 = 100

9 coils 1/k = 9 * 1/1000 = .009
k = 1/.009 = 111

Originally Posted by Craig (Post 2636454)

Sorry, I confused myself trying to do statics in my head. My previous post is incorrect because I didn't recognize that each coil will respond to the full load on the spring, not just a portion of the load. Good catch.

Originally Posted by leathermang (Post 2636458)

Each coil in the spring will respond to its PORTION of the full load on the spring.. even his math agrees with that..
Each coil does not ' see' the 100 lbs... but only ( in the 10 coil example ) one tenth of the 100 lb load...

Originally Posted by leathermang (Post 2636458)

Each coil in the spring will respond to its PORTION of the full load on the spring.. even his math agrees with that..
Each coil does not ' see' the 100 lbs... but only ( in the 10 coil example ) one tenth of the 100 lb load...

Originally Posted by Craig (Post 2636464)

That was my mistake too, if you think of the system as 10 small springs stacked on top of each other, each one has to transmit the full load from the section above to the section below. Each spring section will be displaced a specific amount due to the total load, not just 1/10 of the load. That is consistent with the equation for springs in series (1/k = Sum(1/k1 + 1/k2 ....)), which you can look up anyplace.

Originally Posted by English Bulldog (Post 2636002)

What are the steps to do this.

What is a budget to do this.

Originally Posted by EDBSO (Post 2636503)

Good article here double up sway bars article

I purchased my sway bars from a wrecker for about $50 for both front and rear sway bars. Bushings are extra.

Too cold and too much snow to play with doubling up before spring.

The front end result should look something like this.

http://www.vwbughead.com/412fast/412...le%20sway1.JPG

another full thread

Originally Posted by leathermang (Post 2636473)

HIS MATH shows springs with a deflection of one inch per 100 lbs..
HIS MATH figures that at .1 inch deflection.
The total load ...100 lbs... is spread between the 10 coils ..thus causing each to bend .1 inch... his example set the deflection rate of one inch per coil per 100 lbs load..
Don't give up so fast on your initial judgment..

Originally Posted by Craig (Post 2636700)

The more important point is that making the springs 10% shorter and 10% stiffer isn't going to accomplish anything useful; they will just make the car lower and bottom out easier. If you really want to affect handling, you need higher rate springs anyway.