Category talk:Custom board
Add topicNecessary Custom Board Dimensions for Board Makers[edit source]
To satisfy my curiosity, I'll give the computation here that confirms that the minimum diameter is 1 31/32" to circumscribe a Large mid. Here's my calculation.
A Large has height 1 3/4". If it's lying flat, the isosceles triangle that needs to fit the board has base 1" and height H, where
(1) H = ((1 3/4)^2 + (1/2)^2)^(1/2) = 1.82".
We can write H = x + R, where R is the radius of the circumscribed circle, and x is the distance from the center of the circle, to the middle of the base. Both x and R are unknown to us. In addition to H = x + R, we get this equation to solve:
(R^2 - x^2) = (1/2)^2, because the base of the Large is still 1/2.
Hence, if we divide this by H = R + x, we are left with R - x, thus
(2) (R - x) = (R^2 - x^2) / (R + x) = (R^2 - x^2) / H = (1/2)^2 / 1.82 = 0.137".
Adding (1) and (2) we find 2R (the diameter of the circle):
(3) 2R = (H + (R - x)) = (1.82 + .137) = 1.957" ~ 1 31/32" - rest, with rest < 1/85".