rear engined suspension tuning (spring rates etc)

[juse]

To Tim;
How did you end up with figures like 200/250 lbs on springloads?
I`ve been trying to study these things for a while, and the loadratings seem to variate between 80-150 lbs.
Someone wrote that 80lbs is common with aftermarket struts in U.S., one guy who drives autocross uses 100lbs, one ended on 120lbs and Steve Carter uses 150lbs.
Is it just a matter of opinion or why the scale is so wide?
Yesterday I heard from a Finnish guy who drives rallysprint that he`s not satisfied with the springs he is using now (160lbs), and he tought that he would try 200-220lbs.
Justin

[Tim]

Hi Justin,
Frankly, the figures just come from guesstimation. I assume the stock suspension has 6" travel and 80lb/in spring rate. So 480lb of force will get it bottom out. If I lower the suspension by 4" becoming only 2" travel, at least approx 240lb/in spring will be needed to stop the suspension from bottom out (provided that the vehicle weight is constant). I am not sure whether my assumption is correct or not and that's the reason I post here to ask for opinions.

I would like to obtain a spring rate which I can use as a starting point for my trial and error in the track. FYI, the attached map is the circuit which I will race. Each lap is 4.3km and the longest straight is about 800m. Any inputs for me are appreciated.

Thanks
Tim

[juse]

That sounds like sensible way of figuring out the starting point to springloads, newer thought of it that way myself.
Can`t help you with the track based tuning, since I`ve newer competed in any class myself...yet. I am only planning to try out autoslalom or similar series for roadlegal cars.
I`m very interested of your project and I`m looking forward to learn more from your posts.
Justin

[Tim]

Hello Justin,

I am also a newbie in GL VW and circuit racing although I have owned a standard Beetle for few years and have done some modifications on it. My new project (1303 racer) is still on the sketch. The "donate" car is still sitting in my garage in a kind of "sorry" condition. Currently, I am collecting some used parts, such as Porsche suspension and brake components. I also hope that I can learn from your and others opinions and experiences via this forum. Will keep you posted and will keep asking questions!

Tim

[juse]

Does anyone know the rates of stock rear torsion bars or a formula for calculating the rate?
Allan Staniforth writes in his book "Race and Rally Car Sourcebook: The Guide to Building or Modifying a Competition Car" that the balance between springing frequencies in front and rear is quite important.
So when choosing stiffer springs in front you`ll have to consider the spring rate in rear too.
Justin

[Tim]

Hi Justin,

I've got the book "Baja Bugs and Buggies" which mentioned the torsion-bar spring rate as follows:

For stock torsion bar with 22mm diameter
21-3/4 in bar length - 588in-lb
24-11/16 in bar length - 497in-lb
26-9/16 in bar length - 464in-lb

Please note that this is expressed as torsional force in-lb, not the lb/in as of the spiral spring rate.

Because of leverage, rear suspension stiffness depends on the center-to-center distance from the torsion bar to the axle. In a stock suspension, this distance is about 16-1/2 in. To find the amount of force 1 degree of preload produces in a torsion bar, divide the torsion bar-to-axle distance into the torsion bar spring rate.

e.g. a 21-3/4" stock torsion bar,

588in-lb / 16-1/2in = 35.6lb (change 1 degree)

There are also some torsion bar rate information in the tech article "The perfect suspension" of the GL website.

Hope this helps.

Tim

[juse]

Another comprehensive answer from you Tim, thanks.
But I´m afraid that making those rates comparable to the figures up front, is way too difficult for me. Maths newer was my favourite in school you see...
Justin

[Jeza]

This is something that I've read a bit about following the posts on the forums.

There is one thing that I notice never comes up and that is that Super Beetle front springs are progressive- so the spring rates that Superman posted at Superbeetlesonly.com is for the first inch. The second inch will be stiffer than the first.

This is really important for a car with such a light front end.

I don't know if Steve Carters springs are liner or progressive, but I would imagine that 150 lbs linear springs would be very stiff for a beetle. But then Steve also has a radiator up the front there to.

S1fter has gone for 200 lbs progressive springs, so this is the rating for complete compression, which is about 10 % stiffer than stock. He chose that after testing a stock spring (then carefully filed the info in the bin!!!). But from memory he said that the first inch was about 70lbs, the second was about 150 ish and complete compression was about 180lbs.

Hope this helps - rather than throwing a spanner in the works

Cheers
Jeremy

[Tim]

I think progressive springs are good for street use while the linear springs are more suitable for racing.

Tim

[juse]

As I mentioned before; I own a couple of great books of by Allan Staniforth who writes that when choosing the springrate, the most important thing is wheel frequency.
The formula he gives for this uses sprung weight and CPM(= a factor of wheel frequency).
I toyd with these numbers and got springrates for front from 80lbs to 120lbs for serious street vechiles. I believe this is for linear springs only.
Justin

[Tim]

Hi Justin,

Is it possible to calculate the spring rates for racing? I plan to have my car down to about 800 to 850kg and to have the front and rear weight ratio around 45:55 to 40:60. I have no idea about the wheel frequency for racing though.

Thanks
Tim

[juse]

Tim;
I dont remember the formula, although it isnt any complicated, but I`ll look it up for you as soon as I can.
Justin

[juse]

Here it is Tim, the formula for springrates:
Frequency of suspension in cycles per minute(CPM) for racing cars without wings or ground effect is around 100-125. Figures like 160-200CPM are nearer current trends when moving towards the ground effect area.
A formula for McPherson-setup is without suspension leverage, as there is none of it.

CR = ((WF/187,8) x (WF/187,8)) x SW

CR=coil rate; lbs/inch
WF=wheel frequency; CPM
SW=sprung weight; lbs

Its very hard to write a formula like this...hope you get something out of it.
I also found a site that covers the same matter, but much more complicated way:
http://www.theracesite.com/index.cfm...m_article=1233

Justin