Hipped roof calculation out of a straight skeleton

[vvoovv]

Suppose the straight skeleton has been calculated for a polygon.
A maximum hipped roof height roofHeight is given. All roof faces have the same pitch α.

So we have
tg(α) = h1/d1 = h2/d2 = hi/di

di is a distance from the straight skeleton vertex till the related edge of the original polygon. di lies in the plane of the original polygon. di is perpendicular to the related edge of the original polygon.

di can be considered as a time when the related straight skeleton event occures.

Since roofHeight defines the maximum roof height and di = hi/tg(α), di is maximal at the vertex of the straight skeleton with roofHeight. Finallly, to get tg(α) one needs to pickup the straight skeleton event that occurs at the very end and take it time di to calculate tg(α) = roofHeight/di.

Once tg(α) is known, the height for every vertex of the straight skeleton is calculated as hi = di * tg(α)