Here are my calculations without any knowledge of 'arc tangents', etc. :
Diameter of tire = 24"
Radius = 12"
1/8" or .125" at 12"
Circumference = Pi d
3.141259 times 24" = 75.39816"
75.39816 divided by 12 = 6.28
6.28 times .125" = 0.785 "
or, as Roland said, about 3/4 inch
(The Other) Len
Vacaville, CA
1967 3000 MKIII HBJ8L39031
> [Original Message]
> From: <pennell@cox.net>
> To: <healeys@autox.team.net>
> Date: 1/3/2006 6:53:36 PM
> Subject: Healey and car related
>
> Listers,
>
> On a recent trip alone to NC I had plenty of time to think about stuff.
Most of it not too productive - just staring down the highway.
>
> For some reason the following came to me. Suppose you take a car in good
condition and the front end aligned perfectly. With both front tires
pointing straight imagine a line running along the pavement straight ahead
from the center of both front tires.
>
> Now the steering is changed such that the front of both front tires are
1/8 inch left (or right) of exactly straight. After one revolution of the
tire how far to the left (or right) of the centerline is the center of the
tire now displaced? For simplicity, assume the tire has a diameter of 24
inches.
>
> Keith Pennell
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