Quote:
Originally Posted by Three Peaks
If A, B, and C are in a straight line to begin with, the amount of movement of point C would equal 1/8" x the ratio of the distances between AB and BC (AB/BC=???? x 1/8")
Bob
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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