A question came up on the ABC Self-Service-Science Forum, asking for a solution to find the dimensions of a diagonal brace looking like the diagram below:
The solution is not as simple as it might look because the angle of the brace is not equal to the angle of the diagonal. The solution I came up with is shown below:
The length of the diagonal of the brace is equal to the diagonal of the opening it fits in. The side length of the brace, before cutting is then found by pythagoras:
(Diagonal^2 – Brace Width^2)^0.5
The end of the brace is then at the intersection of a circle centred at the bottom left corner, with radius equal to the brace length, and a circle centred at the top right corner, with radius equal to the brace width (green lines in the diagram). I already had a User-Defined Function (UDF) to find the intersection point of two circles, so it was a simple matter to copy the code into a new spreadsheet, and set up the required input and length calculations.
Having found the coordinates of this point the position of the skew cuts can be found by simple proportion.
The spreadsheet (including open source code) can be downloaded from FrameBrace.xlsb, and the UDF to find the circles intersection point (along with many other geometrical and interpolation functions) can be downloaded from IP2.Zip
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