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Re: Healey and car related

To: <pennell@cox.net>
Subject: Re: Healey and car related
From: Roland Wilhelmy <rwil@sbcglobal.net>
Date: Tue, 03 Jan 2006 21:17:55 -0800
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




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