It’s not a homework question, I promise!
At my work, we make wire mesh. A customer wants this wire mesh installed 90° turned, so that the openings appear to be diamond shaped.
The outside dimensions of the turned flat piece are 14" x 10.25".
What size of the wire mesh would we have to weave to be able to shear the 90°-turned piece out of it? If you could explain the method, that would be appreciated as I have other sizes to calculate.
I’m pretty sure this is going to involve the Pythagorean theorem? Google wasn’t a lot of help.
Well, in the first place, it’s not 90 degree turned – you’ve rotated your mesh by 45 degrees, if you want it to be squares with the diagonals pointing horizontally and vertically.
In the second place, assuming you want square mesh with equal spacing in both directions, the question you’re asking is how big (on a side) do I want to make a square of mesh so that I can turn it 45 degrees and cut a rectangle with sides 2a by 2b out of it. Call one half the diagonal of your square c. If you place it centered on a coordinate system, the line defining the upper right edge will be y = -x + c, so that at y = 0 you have x = c and at x = 0 you have y = c. Place your rectabgle with the sides 2a long along the x-direction and the side with dimension 2b along the y-direction, with the center at the origin. For the minimum size mesh, the corner of the rectangle, at (x,y) = (a,b) will just touch the line y = -x + c. Inserting (a,b) into that equation we solve for c and get b = -a + c, or c = a + b. The diagonal 2c will be the square root of 2 times the needed side of your square mesh, so the general result will be d > 2(a + b)/SQRT(2), where d is the minimum size of your square mesh. Or d = 2(a+b)/SQRT(2)
In your case, 2a = 14" and 2b = 10.25 so 2d must be at least **24.25/(SQRT(2) = 17.147" **
edited to add:
Your square Calculator gives you the dimensions of the diagonals of your rectangle, but that’s not what you need – you’re asking for how big a square mesh you have to make in order to be able to cut a rectangle of a needed size out of the square rotated by 45 degrees, which is something very different.
I got the same answer as CalMeacham, a square with sides 17.147" (assuming I am understanding the question correctly. Basically, you’re looking for the bounding box of a rectangle turned 45 degrees–something like this, except with a figure turned exactly 45 degrees, so the bounding box would be a square).
If you want to plug it into Wolfram Alpha (actually, you could just put it in a Google search box and it’ll do the same thing, now that I’ve checked) for other dimensions, just use this:
sqrt(10.25^2/2)+sqrt(14^2/2)
Substitute 10.25 and 14 with the dimensions of other objects.
