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Our octagonal roof has a nominal diameter of 10'.

If I choose a latitude of 45° for the roof "bend" I need to know the octagon's Circumcircle of that latitude.

The

__small circle__radius = RCosf . Where f is the__latitude__of the nominal [hemi]sphere containing the double pitched, rotating roof.
Cos 45°= 0.7071

If our octagon's Circumcircle is 10' then R = 5, then the small circle at 45° = 7.07' Ø. [It doesn't matter where in the calculation you multiply or divide by 2 as long as you remember to.]

This small circle marks the

__points__of the octagon. The distance between these points is 7.7' x Pi / 8. The circumference divided by the number of sides. 22.21' /8 = 2.77'. This is the width of the base of the eight triangles forming the upper pitch of the roof. Or the top width of the trapeziums forming the lower pitch of the roof.
The width between the points of the equatorial circumcircle is obviously Pi x D /8.

3.142 x 10' /8 = 3.9'. This is the width of the bottom of the trapeziums.

The height of both the triangular and trapezoid roof panels is 1/8 of the circumference of the circumcircle. So again = 3.9'. We now have the

When I first tried to model my intended roof, to check for the desired overall shape, I was assuming that panels would be straight-sided isosceles triangles. I obviously wasn't thinking straight. Straight sides would only be the case if they formed a straight-tapered roof.

Whereas, bi-pitched roof panels are bent at the junction between the upper triangle and the lower trapezium. The width of this joint must be greater to allow for the depth of the bent form. [Spherical geometry?] This width should, of course, be 2.77 units. Exactly as calculated above. Fingers firmly crossed as I take scissors to cereal packet cardboard in the time-honoured way.

Whoopee! Almost unbelievably, the model appeared

__nominal__sizes of the roof panels.
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When I first tried to model my intended roof, to check for the desired overall shape, I was assuming that panels would be straight-sided isosceles triangles. I obviously wasn't thinking straight. Straight sides would only be the case if they formed a straight-tapered roof.

Whereas, bi-pitched roof panels are bent at the junction between the upper triangle and the lower trapezium. The width of this joint must be greater to allow for the depth of the bent form. [Spherical geometry?] This width should, of course, be 2.77 units. Exactly as calculated above. Fingers firmly crossed as I take scissors to cereal packet cardboard in the time-honoured way.

Whoopee! Almost unbelievably, the model appeared

__exactly__as I was expecting! Next time I'll add double tabs to my four sided units so that I can use clothes pegs on the triangle's inside joints. I'm waiting for the PVA glue to dry on three triangular panel tabs so I can finish the model and take some photos as proof of my calculations.

**Click on any image for an enlargement.**

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