Date:         Mon, 13 May 1996 09:53:15 -0700
Sender:       Vanagon Mailing List <vanagon@vanagon.com>
From:         wabbott@mtest.teradyne.com (William Abbott)
Subject:      Re: torque, rear brakes

	Yep, that's why they put that little pivot on some kinds of
torque wrenches- as long as the pivot is free, you're are pushing
AT the point of the pivot, and the wrench is calibrated.
	Consider if you and your pal Pat wiegh 390lb together-
if the two of you stand 1 foot away from the center of the drive, that's
390 foot pounds of torque. However, neither of you wants to stand on
the other's foot, so you agree to stand on either side of a line 1 foot
from the drive center. Further assume your shoes are 4 inches wide.
	You're exerting 390 lb of force, *centered* 1 foot from the
socket drive. If you take each of you separately, one is exerting 195lb
of force centered 10 inches (12 - (4/2)) from the socket center, the other
is exerting 195lb of force centered 14 inches (12 + (4/2) from the socket
center. That's 162.5 foot pounds and 227.5 foot pounds, 390 total.
	(note that each of you is standing on one foot for this demo).

	You can continue subdividing the width over which the wieght is
applied, multiply the wieght of each slice by the distance of its center
from the socket center and then add the results up and you'll always get
390 foot pounds.

	*NOTE* that this depends on the wrench being horizontal as you
balance on it- if the handle of the wrench is at an angle, the force
applied is decreased to sine of the angle.

	Maher isn't the only guy who can do math, but he is better than I-
Steve, have I got this right? Does the mass integrate smoothly?

	Bill