quantifying camber change via caster?

Hey guys,

1996 mustang, running max.motorsports c/c plates @ max. caster & -2.5deg camber...

Say the steering is at 1/2 lock, aprox how much more neg camber is the extra caster adding into the equation?

Is there a formula out there?...

Thanks!...

No idea, just didn't want you to feel ignored.

By the way do you think a Volvo 240 has some camber issues at lock

Is there a formula? I'm sure there is, and some 3-dimensional trig would certainly help you find it.

One of the first things you'll need is how many degrees away from straight "1/2 lock" is.

Thinking aloud... I'm conjecturing that if you could turn the wheel 90 degrees, the camber would equal your original caster... I'm further conjecturing that the camber would "advance" smoothly from the static camber value at straight forward to the new camber value at 90 degrees rotation (which, from prior conjecture, is equal to caster). So, if you knew the angle of steering, you could calculate that as a percentage of 90 degrees, multiply that by the difference between camber and caster angles, add that to your static camber, and get your camber at that point in steering.

EDIT: I figured I'd add in some example math...

If the alignment specs were 1 degree camber and 4.5 degrees caster, and assuming we were turning the wheel 20 degrees...

20/90 is .2222, or 22.22%

Multiply that by the difference in values, 3.5, to get .7777

Add that to your original camber value to get 1.77 degrees,

Of course, the above totally disregards any geometry changes that may be happening due to suspension travel... with a mcpherson strut, you're losing camber as the strut compresses, so you'd have to take that into consideration for an outside wheel. It also assumes that your alignment is done at 0 toe.

Again, that's just me postulating, I'm totally accepting of the fact I may be 100% wrong.

talk with John Block at autoware.com he has a strut program.

Hmmm... visualizing the arc the spindle describes during the rotation of the strut, I believe it loses negative camber in both directions. I know my 1st gen RX7 did, which is why the race setups include getting as much negative caster as possible.

Yeah, I'm sure you could model it out and do the math on it. It's just sines, cosines and tangents after that. But, the EASY way to do it is:

Put car in garage. Put HF magnetic digital level on rotor, set to 0 degrees. Turn wheel to 1/2 lock. Read angle off level.

My Locost has 11 degrees caster built in.

Jensenman wrote: Hmmm... visualizing the arc the spindle describes during the rotation of the strut, I believe it loses negative camber in both directions. I know my 1st gen RX7 did, which is why the race setups include getting as much negative caster as possible.

Negative caster ?

Oops, brain fart.

If the caster angle inclines toward the rear of the car, it's positive caster. If it inclines toward the front, it's negative caster.

I've heard a "kinda sorta maybe" rule-of-thumb that 1° of caster is worth 1/2° of camber.

When I align cars I shoot for "as much as I can get" caster.