Need center to center measurement for Tool 2079

[George Selby]

At 09:02 PM 3/27/2008, you wrote:
>I'm making my own version of Tool 2079 - which will be 35" long. Then
>I can use my Craftsman torque wrench with a 129 ft-lb setting to give
>me 258 ft-lbs at the crank bolt.
>
>The archive info says that applied torque for the crankbolt is:
>With tool 2079 = 258 ft-lbs
>Without tool 2079 = 332 ft-lbs
>
>My calculation indicates the distance for tool 2079 should be 15.5
>inches. A reference in the archives says 11 3/4 inches.

A little confused here, is it going to be 35" long, or 15.5" long (or 11.75"?)

Furthermore, I think the net result is you need 332 ft-lbs at the 
crank bolt, which means calculations attempting to give 258 ft-lbs at 
the crank would be wrong.  Tool 2079 gives some extra leverage which 
means 258 ft-lbs at the tool will result in 332 ft-lbs at the crank 
bolt.  I can't remember enough physics to remember how to multiply 
torque with an extension.

What I'm thinking that you're thinking right now is:  Tool 2079 
assumes you can get a 258 ft-lb torque wrench.  You want to know how 
much longer you need to make your 2079 tool (let's call it 2079a) so 
that you can use your lower capacity torque wrench.

Doing some simple math here (proportions) gives me the following 
rough guess at a solution to your problem:

If a 11.75" (assuming that's correct) extension gets you from 332 to 
258, that means each inch of length reduces the amount to torque at 
the wrench by ~6.3 ft-lbs.  You need to drop the torque requirement 
by 203 ft-lbs, which means you would need an extension ~32.25" long (203/6.3)

End result: a 35" tool 2079a would probably be a good length.

George Selby