OK, my math skills have slipped in the decades since last used . . . .

If I have a spindle with a steering arm radius (center line through the ball joints to the center line of the outer tie rod stud hole) is approximately 4.72" and the rack is 3.5 turns lock to lock, how much will I need to shorten the arm to end up with 3.0 turns lock to lock? 2.5 turns?

Professor Neil says "Please show your work".

Please keep your eyes on your own paper and you may begin . . . . . . NOW.

[This message has been edited by sspeedstreet (edited 09-20-2004).]

trick question, depends on if you are currently hitting bump stops - without any resistance such as bumpstops your stearing wheel would remain 3.5 turns lock to lock and simply reduce your turning radius

edit: if you are serious about finding the answer its a lot easier if you measure the linear travel of your rack from lock to lock.

[This message has been edited by Kohburn (edited 09-20-2004).]

The relationship is fairly linear so your equation will look like this:

(new travel / current travel) * current steering arm length = new steering arm length.

for 3 turns lock to lock you'll have (3.0/3.5) * 4.72 = 4.04
for 2.5 turns you'll get 3.37.

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Doug Chase
Chase Race
Custom roll cage and exhaust fabrication

quote Originally posted by Doug Chase: The relationship is fairly linear so your equation will look like this: (new travel / current travel) * current steering arm length = new steering arm length. for 3 turns lock to lock you'll have (3.0/3.5) * 4.72 = 4.04 for 2.5 turns you'll get 3.37.

Do you really think it could be assumed linear on a radius this short?

quote Originally posted by sspeedstreet: Do you really think it could be assumed linear on a radius this short?

I think so.

The only thing that makes it non-linear is that the shorter steering arm is going to cause more angle of the tie rod. This will will increase your rack travel slightly but my gut feel is that this amount will be negligible.

Doug

Doug, I went through the longer calcs using the different circumferences and you are right. Thanks.