Chris White

if you are hitting the bump stops its a shock issue not a spring rate issue.

xsboost90

well w/ 600# fronts and stock rears w/ just a big swaybar i would say bump it up to some 700 or so rear coilovers/ some Racer's Edge mounts of course and maybe drop that rear bar down slightly or it'll be oversteer city.

spike951

Thanks to everyone... The advice and insight is greatly welcomed... Well the season is over and I will can not test out the advice... Next step is to remove the rear t-bars, up to 750# rears on a set of konis with the rear bar soften.

vt951

From what I've read on here in other threads... rear coilover spring rates on a 944/951 are only about 56% effective, due to the location and angle of the spring relative to the wheel center. So, 750# rear springs with t-bar delete will result in an effective rate of 420#.

Supposedly, front spring rates are 90% effective, so your 600# front springs have an effective rate of 540#.

Following the same logic, you would need 964# (= 540# / .56) rear springs to match the effective rate of the fronts. Or, go with something like 466# (= 750# * .56 / .90) front springs to match the 750# rears.

I'm no expert, but I would say with 600# front, 750# rears, and no t-bars, you will have an understeer setup.

ehall

(spring rate -tbar rate) /.56 = effective rear spring rate.
though I have seen some debate over the .56 number.

ehall

keep the 30 mm t-bars and add a coil over with 650# springs and you'll be pretty close front to rear.
Otherwise you are looking at a BIG rear spring!

Trucho-951

As I recall, effective spring rate = (motion ratio) ^2 x spring rate

So for example:
Front Effective Rate = (.91)(.91)x 600 lb/in. = 0.828 x 600 = 497 lb/in.

No Torsion Bars
Rear Effective Rate = (.65)(.65) x 750 lb/in. = 0.423 x 750 = 317 lb/in.

Yes, there seems to be some debate about the rear motion ratio.

https://rennlist.com/forums/showpost...0&postcount=45

vt951

??? So if you use a larger t-bar, the effective rate goes down?

ehall

Yes. lets say {550# spring -23.5mm t-bar(126#)}/.56=757.14

or (550-126)/.56=757.14

The actual spring is carrying less load. The larger t-bar is carrying MORE load. So the spring rate, for a set number goes down as the t-bar goes up.

(550-177)/.56=666.xx

same coil spring doing less work.

177# I belive is a 25.5mm t-bar.

ehall

does the .65*.65*spring rate, only apply to a coil over without t-bars?

vt951

Sorry ehall, I usually agree with just about everything you contribute here, but you've got this wrong. Helper springs and t-bars act as springs in parallel. Their spring rates are additive. You won't get a looser suspension by adding bigger t-bars.

vt951

Here's a very good explanation of the motion ratio and effective (wheel) spring rates. I think trucho has got it exactly right. LINK

So, to factor in t-bars, figure out what the effective (wheel) rate is for helper/coilover springs, and figure out what the effective rate is for the t-bar (many threads have this chart), and add them together.

ehall

That's not what you are calculating. You are calculating the effective rate of the spring, itself.
I didn't make up the equation, I just plugged in the numbers. The numbers show an inverse relation between the coil spring and the t-bar( which ofcourse is also a spring, but not in a linear sense.)
That more complicated equation is above my pay grade.

ehall

Like I said, I didn't invent the equation and I'm very open to why it shows an inverse relation. I agree with your point, so I can only surmise that the equation is limited in it's usefulness to finding the coil spring rate one might choose.

ehall

great link. Thanks.