Unwrapping the Onion
In the comments to the part 2, Chris Hall pointed out a problem with the inner surfaces of the near and far roof slabs: they do not intersect the left and right slabs at the ridge lines (in plan view). I gave the rafters at H and G the same width as the principal rafters at E and F. This does not affect our exercises for finding the footprints for these rafters, but it messes up another aspect of the layout. Consider the solid made by the inner surfaces of the rafters. If this is congruent to the shape of the main roof, then not only do many aspects of the layout become much neater, but ideals of symmetry (and, I suspect, various spiritual ideals as well) are satisfied. In a real roof there are several such solids formed by the inner and outer surfaces of the common rafters, purlins, and principal rafters, and they all should be congruent. The desired configuration in our model looks like this:Notice the similar shapes of the inner and outer surfaces, with all the rafters lying between them. In solid geometry terms, the inner surface is the same as the outer, scaled uniformly about a point in the ground plane lying directly below the peak of the roof. The inner solid produced by my original layout, which I'm not illustrating here, would be an oddly skewed version of the main roof surface.
How do we layout the correct inner surface of the plan? We said that the inner surface has been scaled around a point underneath the peak, so the inner peak must also be directly under the outer peak. Therefore, in each elevation view that cuts through the two peaks, the distance between the two will be the same. The dimensions and layout of the principal rafters are given, so the widths of the rafters in the others surfaces will be determined by this principle. Here's a corrected version of the plan:
Back to Footprints
With that out of the way, we return to laying out rafters. The next rafter will have a completely irregular cross section with one side lying against the near surface of the roof. We start as we did for the rafter at D by drawing an elevation view of the rafter and folding a perpendicular cross section plane down to the ground:and then find,in the cross section, the edge that lies against the near surface. This is the same construction we did last time, running a line from point 2 to the intersection of the folding line and the gutter line:
Next we draw the rest of the cross section:
At this point in 3d, we can place the rafter in the model and see what we have:
Next, we find where the points 4 and 6 end up in the footprint:
Point 9 is found by extending a parallel to the elevation cross section -- the ridge line -- through point 6 to the edge of the roof. To find point 11, we do the same construction to transfer a surface intersection from the cross section view to the plan, only in reverse. The line between 2 and 4 is extended to intersect the fold line at point 8; the line from point A to 8 is then the left edge (and dévers de pas line) of the footprint. Point 11 is located on it with the same parallel method used to find point 9.
To find the last vertex of the footprint, we transfer the last two edges:
In 3D, we see how the lines in the cross section come down to give us the edges of the footprint:
Now, the plane of the elevation view at H passes through point K (on the ground) and is perpendicular to the roof surface. The cross section plane of rafter A also contains K. It is perpendicular to the roof surface because by definition it is perpendicular to the upper edge of the rafter, which lies in that roof surface. So, the rafter cross section plane must intersect the roof elevation at the normal line from K to the roof surface:
Therefore, the distance from K to the roof plane will be the same in the two planes. We run a line from the intersection of the fold line and roof gutter tangent to the arc we drew back to the ridge line, which gives use point 2 and the first edge.
I don't see that this is easier than drawing the elevation, but the geometry that supports this construction is interesting.
The next rafter will have an equilateral triangle as a cross section. Mazerolle says that "The rafter B ... is obtained in the same manner as that which came before." That's not exactly true in the case of the development in the book, and we'll also add an additional twist that forces us to proceed differently: the bottom of the rafter will be level. We do start by constructing an elevation view of the rafter at B:
We will give the triangle a size such that the footprint of the rafter will just touch the intersection of the inner surfaces of the left and far roofs. This implies that the rafter will "run up" the edge formed by those surfaces; therefore, in the cross section view, the lower edge of the triangle will also touch that intersection. We know how to find those surfaces and intersections in the cross section view:
The footprint is established with the same methods used before:
The two outer edges pass through the intersections of those edges in the cross section and the fold line. The end points of the third edge are found with parallels through the cross section view. As expected, the footprint just touches the intersection of the inner ground lines. The footprint agrees with the 3D view of the rafter:
The last rafter, erected at C, is a hexagon that lies against the far roof surface. The cross section and footprint are constructed using the same methods we've used already:
As we've come to expect, the 3D view shows the rafter sitting nicely on its footprint:
As a justification for fooling with such an exotic footprint, Mazerolle makes the point that if the rafter cross section was circular, then the inscribed hexagon could be used to find the elliptical footprint, with the help of a pistolet or French curve:
The draftsman would run the French curve through points of the footprint to find the shape that is known to be an ellipse. I don't have a set of French curves handy, and they would be awkward to use with Blender, so I made the ellipse by rotating and scaling a circle. That was tricky, because none of the lines between vertices of the hexagon lie on the major or minor axes of the ellipse. We could probably change the orientation of the hexagon a bit to make this method more practical.
We are now equipped to find the devers de pas footprints, and the corresponding dévers de pas surface lines, in just about any situation. While Mazerolle's carpentry drawings usually use the first technique from part 2, where we used a trait carré normal line from an existing elevation view, we should now be able to handle whatever he throws at us in the world of devers de pas.