FractionalGeometry-Chap2

13 coecient matrix if this matrix is invertible the

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Unformatted text preview: FS for PrxsSol 2.6.1. f 0 -1 1 1 -1 PrxsSol 2.6.2 Not expecting the branches of the tree to be similar to the tree, we take two refertence lines, OA ≈ 4.5cm and BC ≈ 4.8cm. The corresponding lines in the lower left branch are DE ≈ 2.7cm and HI ≈ 2.9cm. Then for this branch, r = 2.9/4.8 = 0.6 and s = 2.7/4.5 = 0.6. The upper left branch also is similar to the tree (to establish this, measure the image of both OA and BC in that branch, but already our diagram is getting too complicated). We measure F G ≈ 2.3, giving r = s = .5. The rotations are ∠EDA = 45◦ and ∠GF A = 20◦ . The right branches are copies of the left branches, rotated clockwise instead of counterclockwise. The trunk follows the construction of the trunk in the tree example, except it is a bit wider in the middle. Assembling all this, we have r 0.6 0.5 0.6 0.5 0.05 0.07 s 0.6 0.5 0.6 0.5 0.6 -0.06 θ ϕ e 45◦ 45◦ 0 20◦ 20◦ 0 −45◦ −45◦ 0 −20◦ −20◦ 0 0 0 0 0 0 0 IFS for PrxsSol 2.6.2. f .6 1 .6 1...
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