Here's my entry:
front of tire moves 1/8 inch left, so we have a triangle about 13.5"
(the tire's radius) on two sides and 1/8" on the third side. arc tan
of .125/13.5 is 0.53 degrees. if the diameter is 27" then one
revolution (the circumference) is about 85 inches (pi * d). 85 times
the tangent of .53 degrees is aout 0.78 inches. So I say the tire and
car move left (in this case) about 3/4 inches.
-Roland
On Tue, 3 Jan 2006 21:52:15 -0500, you wrote:
::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
|