Building the Octagon Pt.62 Octagonal rainproofing.

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An octagonal roof of nominal 10' "diameter" [Circum]circle.

Pi x D = Circumference. 

3.142 x 10' = 31.42'.

31.42' / 8 = 3.9275'

3.9' < 4' = so can be obtained from a normal [4'x8'] sheet  of plywood. A fortunate coincidence. Thin, 4mm [5/32"] Birch plywood is also available in 5'x5' sheets. 1.5m x 1.5m.

A rotating octagon scribes two circles at its "circumference." A smaller one [Incircle] forms Tangents at all the flat sides. While a larger circle [Circumcircle] contains the Points of the angles between the eight sides. This is important because a rotating octagonal roof has a series of [longer] overlapping points and shorter radius tangents. As does the building supporting it.

The octagonal roof's Tangent Incircle must always be larger than the supporting octagonal building's Point Circumcircle. So that rain is always shed outside the building. If the roof's Tangent circle is the same size as the building's, then the roof Tangents will fall inside the building's Point circle at 8 successive angles of rotation. Buy 8 buckets to catch the rain? Or make the roof suitably larger to avoid problems. [P&T are my capitals added here for clarity.]

The radius of the [Incircle] is found by multiplying the Circumcircle by = 0.924.
So a 10' diameter octagon will have a 9.24' diameter Incircle. A 9" difference or 4.5" difference in radius. That 4.5" difference is crucial to making the rain fall outside an octagonal building from an octagonal roof. Some extra overlap is obviously desirable to ensure clearance for flashing attached to the edges [tangents] of the roof where the building's Points pass under it during rotation.

Click on any image for an enlargement.

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