In a message dated 96-05-11 23:52:34 EDT, khooper@wsp1.wspice.com (Ken
Hooper) writes:

>>So I measure 1.64 feet on my breaker bar
>>and make a mark on the bar and stand with one foot on each side
>>of the bar, and equalize my wieght.
>
>Is this right? 

Not exactly.  He meant to say "one foot on each side of the mark"

>I've wondered this before. Chiefly because the instructions
>to my torque wrench are awfully ticky about having the handle swivel on
>that little fulcrum in there while I'm sweating and grunting. No pulling on
>the bottom of the handle for an extra two inches of leverage! It's hard
>sometimes...

Follow the instructions and don't let the swivel handle tip to either side,
keep it so the force is only transfered through the "axle" that attaches the
handle to the wrench.

>So we want to stand on it, okay. But whatever we put inboard of the
>line--the foot-pound line above--is going to be "lighter" than what we put
>outboard. Less leverage. How do we calculate how much of the mass to put
>outboard? Probably at 390 feet it doesn't matter, but I bet at 4 feet it
>does. Maybe Brother Maher knows, he can do math...

Just center the force at the mark.  1/2 of it 2" left of the mark and 1/2 of
it 2" to the right of the mark will be the same as all of the force at the
mark.

Jim Davis