On 12/9/2016 11:20 AM, John Harmon wrote:
actually said:

You REALLY need to study your high-school math.

This off-topic confusion is all my fault.

I should never have brought toe into this discussion because toe is easily
done at home when you have specs that are in linear dimensions such as
inches but not so easily understood when you have toe specs in angles.
http://i.cubeupload.com/RubZhV.gif
http://i.cubeupload.com/cfaDWp.jpg

Clearly I'm confused how to do the conversion.


Ya, I am to. But first let me say this, The first spec you posted,
0* 14' plus or minus 10', seems this isn't as critical as some posters
are making it.
For toe, it is still a trig problem, but the problem is defining,
side b (a reference point).

http://www.carbidedepot.com/formulas-trigright.asp

I wonder do the shops attach a laser and measure on a wall scale a
defined distance away?

I don't know this, is it a single adjustment that moves both wheels or
do you adjust both wheels separately? (makes a reference even more
important)
Sorry just thinking on the keypad.

You have a trig problem and a measurement problem.
The measurement problem is more difficult.

It is not be hard to convert the 14 minutes to inches using the wheel
diameter as one line.
The angle is how much more is the front of the wheel turned
in more than the rear of the wheel. I'll call the wheel 16"
from front to rear. (just realized this almost the same trig problem for
camber, just rotated 90*)

I'm using the trig calculator above, this time the orientation is correct.
Put the following numbers in, (side c) = 16, (angle A) = .233. The angle
is .233 because 14min/60min = .233.
Your answer is (side a) which is 0.065". So, you want the rear of a 16"
wheel stick out 0.065" more than the front.
Not real easy to measure, But, if you could extend the 16" to 12 ft
(192") with a laser pointer, then (side a) is 0.781".
The laser must be perfectly square with the wheel.
Just some thinking. Hope it makes some sense.
Mikek