Four Link Suspension - Drag car

[podo]

Theoretically, how would I calculate the optimum instant center location for the four link suspension used in a drag race car? Car specifics:
NHRA E/altered - competition eliminator
Door car - tube frame
Car weight - 2340 pounds
Rear weight - 1190 pounds (51 percent)
Wheel base - 105 inches
Height of C/G - 17 inches
Manual transmission - 3.25 low gear
Rear gear - 5.83
Motor HP- 800 (470 peak torque)
E.T. - 8.10; 1/4 mile speed - 167 mph
I need to improve, and I need help
Thanks,
Podo

 

[red300zx99x]

I'd run simulations with varing IC's and choose the IC that had the highest average rear wheel load over say the first 60ft or so, or however long you have traction problems. But their are probably other ways

 

[patprimmer]

How do you intend to 167 mph with 5.8 gears.

What is your rolling dia of the tyre.

What RPM are you using.

Is the trans auto or manual.

What is your top gear ratio.

We run about 170mph at over 8000 rpm with 33"tyres and 4.3 diff, and a powerglide.

I can't help with the 4 link as we run ladder bars and set them up about parallel with the ground.

Regards
pat pprimmer@acay.com.au
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[podo]

To red300zx99x;
You described what I think I need - a good simulation package. Do you know of any, and how do I obtain one?
Thanx,
podo

 

[red300zx99x]

Check the forum faqs

 

[BillyShope]

A simple...yet very effective...setup is to keep each link pair parallel (i.e., upper link parallel to lower link), but angled up, from the rear, at an angle with a tangent equal to 17/105 = 0.162 (or 9.2 degrees). This will yield no squat or rise and, at the same time, will not give you any binding problems while cornering. (I'm assuming the links are in the normal "straight ahead" drag race configuration.)

If you wish to cancel out the driveshaft torque (and, of course, you do), add 5 degrees to the right side pair and subtract 5 degrees from the left side. This will get you very close and provide the best 60 foot times.