[twobjshelbys]

Originally Posted by AL427SBF

I think you are on the right track but the C movement has to be less than 1/8" so BC/AB x 1/8.

The more convoluted approach ..

A'
.......................................C'

A---------- ----------------C---------------------------------------B

If he is at A looking down the line to B with C in between then ...
Moving to the left 1/8" equates to moving A up 1/8". Now he can see B unobstructed by C.

Q: How far do you have to move C up (C') to be back in line with the new A'B line.

The angle created by A' to B back to A is the same angle as C' to B back to A, call it theta.

Tangent (theta) = (opposite/adjacent) = (1/8")/26" = .00481
Arctan .00481 = 0.00480996291 rad = 0.275590574 degrees

Without knowing where C lies in the original AB line you can't solve for the distance to move C to C' (call it Z). If you did know the distance of CB then ...

Tan (0.275590574) = Z/CB
Z = Tan (0.275590574) x CB

I read that C was the fixed point and B/B' was the midpoint....


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