What is the correct center to center length of the diagonal "A" in a one inch pipe offset that uses 45 degree elbows and has an offset distance of 8 inches at "B"?

To determine the correct center to center length of the diagonal "A" in a one-inch pipe offset using 45-degree elbows, it is essential to understand how the geometry of the offset works. When creating an offset with two 45-degree elbows, the diagonal section forms a right triangle. The two legs of this triangle are the offset distance (8 inches in this case) and the length from the elbow pivot point to the outlet.

The total length of the diagonal "A" can be calculated using the Pythagorean theorem. In this case, both legs of the triangle formed by the elbows and the offset represent equal distances due to the symmetrical nature of the 45-degree angles. Thus, both legs need to be considered as equal, making the calculation straightforward:

1. Each leg of the triangle (from the pivot point of the elbow to the vertical offset) is 8 inches.

2. The diagonal length can be calculated as:

[

A = \sqrt{(8^2 + 8^2)} = \sqrt{(64 + 64)} = \sqrt{128} = 8\sqrt{2} \approx 11.31 \text{ inches}.

]

3. When adding