Thread: Parabolic leaf springs

Originally Posted by stirlsilver

Knowing the spring rate of your packs is good info to have. But the thing that has me hung up is the actual dead weight that the springs have to carry and working that out... Otherwise it will be difficult to tell what height the car will settle to when you put in your newly tuned leaves that have been reset to a custom free camber. It's really something that has to be measured, but for that you need some real big scales! Or do you...?

Actually I found the free camber values the easiest to work out.

While weighing each corner would be the best option, you can use the free camber values in the S3 manual to extrapolate what your custom springs need for a free camber. There is a free camber equation in the SAE design manual.

I did the rear first, and they sagged/settled a bit, but I think that was because I had made the upper stage slightly too soft first time around, AND the spring place didn't temper the springs after resetting. Still, they worked fine for a year or so antil I could be bothered pulling them out (did the fronts at the same time). The fronts worked finr first go, and have been in the car since 1998, same goes for the 2nd iteration of the rears.

When I got the fronts done (and the 2nd iteration of rears done) I added 1" to my calculated free camber values, to allow for any slight settling which might occur.

I would imagine one problem with the amount of free camber that you use is that if you go too far you may find that you will have a problem with shackle inversion? Maybe not so much a problem with the front but It looks like it could happen on the rear.

I'll have to go back to the library and photo copy some pages out of the book. The book did have some pretty heavy duty equations in there, and I didn't have time to sift through which ones were the most relevant. Also, the drawings which were supposed to illustrate the axle path was nuts!

Originally Posted by stirlsilver

I would imagine one problem with the amount of free camber that you use is that if you go too far you may find that you will have a problem with shackle inversion? Maybe not so much a problem with the front but It looks like it could happen on the rear.

I'll have to go back to the library and photo copy some pages out of the book. The book did have some pretty heavy duty equations in there, and I didn't have time to sift through which ones were the most relevant. Also, the drawings which were supposed to illustrate the axle path was nuts!

I didn't have any problems with shackle inversion. I think landies aren't prone to this - unlike cruisers!!!

My driver's side front spring had so much free camber when I got it reset, that I had to use a chain and hi-lift off the back axle to flatten it ouit enough to connect the shackle up (no load on spring). Once the weight was on it everything was fine though. It never inverted on me. I have never seen an inverted rear either.

I like how this all sounds. Especially since the camber helps out with how far the axle is able to move down. Great info you are providing ben! I'll probably use your free camber measurements then since they seem to have worked quite well for you.

Ben, is that equation you provided correct? I just tried to use it and it is producing very small values. I just checked the equation and the units don't really work out.

Assuming:
r = spring rate (lb/in)
w = width of leaves (in)
n = number of leaves (dimensionless)
t = thickness of leaves (in)
L = length of main leaf (in)
SF = stiffness factor (dimensionless)

You get:
r = (w.n/12) * (t.1x10^3/L^3) * SF
= (in) * (in/in^3)
= 1/in

Is there a material stiffness value that is supposed to go in or something (to introduce lb into the units??)

Anyway, I measured my leaf packs and this is what I have:

Front (right and left)
6.5mm x 2 (main leaves)
4.5mm x 6

Rear (right and left)
7mm x 7

What are your thoughts? I'm looking to remove maybe one more leaf from the rear and two from the front, cut them down for even stepping, chamfer the ends and have them reset and tempered. Obviously I'll do some calcs to check it all (which I was hoping to do with the equation you provided)

Sorry - you are right - should have been:

r = (w.n/12) * (t.1x10^3/L)^3 * SF

Actually, forget that equation, it is oversimplified. I found the proper equation - which I used to design my springs.

it is:

r = 8.E.n.w.t^3 / 3.L^3 * SF

E = young's modulus. Which for steel is:
3.05x10^7 lb.in^-2 or 2.1x10^11 N.m^-2

the equation above is dimensionally correct.


For anyone who is intimidated by the equations, forget about them. You can either design them by trial and error or ask me.

EDIT - for anyone interested, if you substitute in the numerical value for E (lb/in2), you can rearrange the equation to become:
r = 0.25.w.n/3 * 1x10^9.t^3/L^3 - which can be further rearranged into the first equation.

OK... that makes sence................