Well DAMN! Cut too much coil. suggestions??

I've used them before, and they seem to work just fine. By cutting the coils you've decreased your spring rate. So watch the bumps.

If you put anything between the axle tube and the leaf spring, it will lower the car; pring half, lowering block, 2x4 lumber. If you go down to a metal scrap yard you can probably find some alum of the right thickness to lower the rear however much you want. You can even use multiple layers if you pin them together - 1/2" +

1/4", or whatever. You'll need at least 1/4" of thread extending past the retainer nut on the U bolts. You may have to buy new bolts.

Here's the info you need about the upside down spring trick:

"Latamud's" specification of a 60" link of all-thread must be a typo. You need a piece just long enough to bridge the gap between your existing spring pack and your auxiliary spring. 24" is more than enough. I have not read Latamud's how-to post in a while, but he uses alot of words to describe a fairly simple operation. You take the smallest leaf from a donor spring pack. You remove the center through-bolt from your existing spring pack. You place the donor leaf upside down on top of the main leaf of your existing spring pack. You use the all-thread to draw the two together, until they're close enough to fit in a vise, then you use the vise the rest of the way and to hold the springs together while you put in a new through-bolt. I use a longer than necessary through-bolt and add a spacer or built up washers to the head end to make a registration stud for the axle bracket. The all-thread, I lock two nuts together at one end, put on a fender washer, insert it through the spring pack and the donor spring, put on two fender washers with motor oil between them, then another nut. This top nut is what you use for the tightening process.

It's really easy as pie.

You also need a clamp, for one end of the new spring pack only. You can make your own -- see the pic I added to the linked thread showing how I made mine -- or you can go to a spring shop and have them bend the square U-bolts like you can see on "Opentracker's" car in the linked thread.

You may also need some new axle tube U-bolts, but only if your present U-bolts have less than 1/2" of thread showing past the nuts.

The fender rolling, cut some slits every few inches on the fender lip. Then use a baseball bat or a piece of pipe between the lip and the tire to flatten the lip.

The 1" spring isolators will give you back more than 1" of ride height. They don't cost much. I would try this first, before rolling the fenders. OTPH, you should never have any tire to fender interference, even at full compression.

Also, I assume from now on you will be a vocal dispenser of words of caution to those who cut coil springs. I would never cut a half coil at once, much less 1.5 coils. I have usually seen a 1/4" drop for each

2" of spring removed. And I drive the car around for awhile to let it settle before I remove any more.

Anyway, I'm glad to see you going lower. Unless you're scraping at least once in a while, your Mustang is sitting too tall!

180 Out

wrote

I'll check it out.

I have no interference now, just rubbing with tight turns. without power steering it seems to tun easier at slow speeds when I normally have the HARDEST time getting the wheel around.

No SH-T!!!

No scraping except in the turns and it looks completely badass!!! It's just a little TOO badass for the street. 'Course I had my ass handed to me by a blown 5.0 the other day. My n/a 400W 2.70 open rearend 2" exhaust didn't help, compared to his 10lb blown 5.0 with 4.35's and 4" exhaust.

On Sat, 01 Jul 2006 11:08:35 -0600, ".boB" spewed forth:

Actually he increased the spring rate, ie made the spring stiffer by reducing the number of active coils:

Spring Rate = F/S = Gd4/8ND3 where: F = spring force. S = spring deflection. G = torsional modulus of the material. d = wire diameter. N = number of active coils. D = mean (average) coil diameter.

Very interesting. I have had the intuitive theory that a coil spring and a torsion bar are similar, i.e., that the wire of the coil spring must twist as the coil is compressed. Your formula indicates that the greater the "torsional modulus" of the wire, the higher the spring rate. This seems to verify my intuition.

I have a question, though: why do you have a "4" in the numerator and an "8" and a "3" in the denominator. Why not reduce these three constants to "1/6." Oh, wait, could it be that the formula is Gd^4/8ND^3, where a "^n" indicates an exponent?

Be gentle. I'm only a liberal arts major. : (

180 Out

On 6 Jul 2006 15:03:12 -0700, snipped-for-privacy@hotmail.com spewed forth:

A coil spring and torsion bar work exactly the same way, one is just "wound up" rather than straight.

That would have been a better way to write the formula. I did a cut-n-paste from Paul Haney's webpage, and didn't notice that.

(Confirmed the formula on page 764 of Milliken's book.)