Most electricians forget about or ignore this shrinkage on three-point saddles, and as a result, the center of their bend is not centered over the obstruction they are crossing. The difference between B and C is our shrinkage the center of our three-point saddle will move just under 1". What is the conduit shrinkage-that is, the amount by which the center of the bend will be closer to the end of the conduit than the measured length of pipe? We know that A = 4" and angle d = 22.5º.
Assume we need a 4" 3-point saddle, and that we will use 45º as the center bend with 22.5º angle bends on each end. At the same time, we can find the multiplier of a 15º bend by dividing one by the sine of 15º the answer comes back that the multiplier for 15º is 3.86. The calculator tells us that the inverse tangent of. What is the angle to be used? We know that A = 4 and B = 15. Assume we need a 4" offset, and that it must take place in exactly 15". This kind of math is where those multipliers come from! 3846 = 2.6, which is the familiar multiplier for a 22º offset. Note that this can also be written as C = A * (1 / sine(22º)). As an exercise, consider an offset of 12" using two 22º bends. That's a more reasonable length for an offset in 3 1/2" pipe, so it can be used where a 10º offset cannot. That's a little far apart for our bends, so let's try again using a 5º bend. The calculator shows that the sine of a 2-degree angle is. 
Using the sine formulas above, let's try a 2º bend. Normally, this would be impossible using a 10º bend, as two bends cannot be made that close together (12”) in a conduit that large.
Assume that we need a 2" offset in 3 1/2" conduit. 
And make sure that your calculator is set to describe angles in degrees, not in radians radians are useless for the electrician.
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