Paul, a shortcut:-

Given the two large triangles are isosceles (so angles 'd' and 'e' are mirrored) and that the three angles of a triangle add up to 180 :-

B + 2d = 180

C + 2e = 180

Adding both sides of the above:-

(B+C)+ 2(d+e) = 360

i.e.

A + 2F = 360

Where

A vertical dropped from the centre of the circle divides both angle A and distance S in half, giving:-

sin(A/2) = (S/2)/(D/2) = S/D

Now, from earlier, A/2 = 180 - F, so:-

sin(180-F) = S/D

(- and remembering sin(180-X) = sin(X))

Cheers

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Given the two large triangles are isosceles (so angles 'd' and 'e' are mirrored) and that the three angles of a triangle add up to 180 :-

B + 2d = 180

C + 2e = 180

Adding both sides of the above:-

(B+C)+ 2(d+e) = 360

i.e.

A + 2F = 360

**F = 180 - A/2****To calculate the required dimensions:-**Where

**'S' is the distance between supports**, and**'D' is the circle diameter**.A vertical dropped from the centre of the circle divides both angle A and distance S in half, giving:-

sin(A/2) = (S/2)/(D/2) = S/D

Now, from earlier, A/2 = 180 - F, so:-

sin(180-F) = S/D

(- and remembering sin(180-X) = sin(X))

**1:**Given support distance and required diameter:-**'Sled angle', F = 180 - arcsin(S/D)****2:**Given 'sled angle' and required diameter:-**Support distance, S = D * sin(F)****3:**Given support distance and 'sled angle'**Resulting diameter, D = S / sin(F)**Cheers

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