Need center to center measurement for Tool 2079
George Selby
gselby4x4 at earthlink.net
Fri Mar 28 23:45:34 PDT 2008
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
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