Seeing as how the term project for the Arch 202 quarter would end up being a museum where we would have to pack 20,000 sq. feet of program onto a 5,000 sq. foot lot, it now makes sense that our first exercise the quarter was a space packing exercise. We were to subdivide a volume of given proportions (9-3/8″ x 8-3/4″ x 8-1/2″, the same proportions as the eventual site) using only six given proportional rectangles. Furthermore, no dimension on any of the resulting cells could be less that 1-1/2″ or more that 6″ in any orthogonal direction, and no two cells could have the same dimensions. Sounds confusing, I know. Take a look at the actual assignment handout here.

The simplest way, it seamed to me, was to develop a matrix based on the given, necessary proportions and project it three-dimensionally on to the volume to develop the cells. The process was somewhat similar to the algorithm problem from Arch 201.

The way I went about it was to apply a sort of recursive formula to the object. Each corner of the original vertex would spawn an iteration of the largest of the given rectangle proportions (double square), each of which would spawn from its vertex an iteration of the next largest given rectangle proportions and so on. Using this method I was able to construct the cells out a small number of defining planes, creating a much more simple solution that was arrived at by many of my classmates by pure trial and error.
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