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<DIV> Yes Paul</DIV>
<DIV> I believe this is right. I've always used Nick Gravagnes
formula. The length of the backscale starting at the front bridge pin
Times .026 = 1 1/2 degrees. i.e. a 4 inch rear length times .026 ==.104
thousands of distance bearing. The Only problem with all this extrapolation as I
see it is that the rear foot of the lowell gauge sits behind the rear bridge
pin, which is usually sitting lower than the actual elevation of the
aliquout (or Whatever Del,Terry..grin). WHich means the gauge may actually
read more bearing present than there is. Why? Hmmmm... Because
of bridge slope to the rear & the grasp of the bridge pins. Of course
the opposite could be true.</DIV>
<DIV> </DIV>
<DIV> Dale</DIV>
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<DIV>Terry:</DIV>
<DIV> </DIV>
<DIV>The conversion from the mils measurement (thousandths of an inch) to a
tangent value is straightforward. If I recall correctly, 3 units to the
positive bearing from both sides of the bridge (net) yields 6 x
0.003"=0.018 which is the tangent of 1 degree net bearing over the bridge.
Hmmm, maybe I have to look up my tangent tables to be sure of this, but I
think it's right. 0.026 is the tangent of 1 1/2 degrees, I think. Anyway,
that's the conversion. And someone should make that clear with the
instructions, you're right.</DIV>
<DIV> </DIV>
<DIV>Paul</DIV></FONT></BLOCKQUOTE></DIV>
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