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## Caster

 To: tigers@Autox.Team.Net Caster Anita Barrett Sat, 30 Aug 1997 12:09:34 -0400
 ```At 11:50 AM 8/30/97 -0400, you wrote: >Jim Barrett wrote: >> >> I hope I am calculating correctly on caster. My theory is that, >> (assuming zero camber to start), if one could turn the spindle 90 degrees >> then the "camber" would be the caster angle. Therefore if I turn the >> spindle 22 1/2 degrees the camber angle would be 1/4 of the caster. >> That is why I did the 4x calculation. >> I have a very expensive alignment set... >*********************** >Jim, > >It's clear that you've spared no expense with your alignment set. I'm >impressed. Think that your 4X rule might stand a bit of refinement >tho. I suspect that coupling between camber and caster is not a linear >function of steering angle but varies with the Sine of steering angle. >Mathematically it would look something like this: > > Caster = Measured_camber X sin(Steering_angle) > >If this is correct, then, for a steering angle of 22.5 degrees, > > Caster = Measured_camber X sin(22.5) > = Measured_camber X 0.38 > >The 4X rule would then become a 1/0.38 = 2.6X rule. Note also that in >the limit, for an angle of 90 degrees, Sin(90) = 1 and camber = caster. > >Don't take this as gospel. I haven't been through the analysis to say >that its correct or otherwise. It just looks right to me intuitively. >-- >Intuitively yours, Will Seay - B382001570 - wseay@sprynet.com > Will, I should have known better, I don't know if you are exactly right but I know I am completely wrong. I have trouble thinking in multiple dimensions sometime when I am setting there staring at a greasy front end and sweat is running into my eyes. I think that maby Florida has fried my brain. At least one can determine if both sides are about equal caster angles by turning the tires a fixed angle on both sides. I fixed my pull to the right by moving a 3/32" shim from back to front on the left side still have no idea what the caster is set to, but it seems to now be equal. ```
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