|A BRIEF DIGRESSION INTO SPIRALS||
The portion of a tree, shown in red at right,
from the base to LLLL... is
Another spiral, the one which motivates this digression, is shown in blue and is formed by the segments on the tip-set
Definition-- Given a ratio r<1 and
an angle theta,
we will define a left-hand spiral by starting with
a base segment and adjoining further segments, each new
length is r times the previous length and each new direction is
to the left of the previous direction.
The definition makes it clear that the spiral is (almost) self-similar (the base segment is missed in the image) by a move involving rotation through angle theta and a shrink by factor r. (We could make it truly self-similar by also adding longer and longer segments at the other end, ie, a doubly infinite construction.)
The center of the rotation is at the tip of the spiral. In the second example above (the blue spiral), the center is the point RLLLL.... If we call the center C and call the segments
[P0, P1], [P1, P2], [P1, P2], … [Pi, Pi+1], …
then the triangles CPiPi+1 are all similar, with equal angle measures theta at the vertex C.
The x-coordinate of the vertex PN on this tip-set spiral is given by the formula
If we extend to the infinite sum, the expression
gives the x-coordinate of the center C of the spiral. We found, by manipulating Fourier series, the function represented by the sum
Let us derive this formula directly from the geometry of the blue spiral.
DIRECT DERIVATION OF THE CLOSED FORM
The picture above has been cut down, rescaled and relabeled for simplicity:
P0 = P, P1 = Q, P2 = R,
segment labels are lengths.
We have similar triangles PQC and QRC, the angles CPQ and CQR both have measure phi, while angles PCQ, QCR, RQA all have measure theta.
CQA has measure phi+theta.
Finally, triangles CAQ and CAP are a right triangles.
We need to express the distance b+1 as a function of r and theta . We see that
The law of sines tells us that
We substitute these into the formula for the sine of a sum of two angles:
Remove the factor to get
We need something to substitute for a2 so we apply the Pythagorean theorem to the two right triangles: .
We subtract, and get .
Then the equation above becomes .
Now we clear the fraction and solve for b:
so and .
The extra factor r2 in the original formula above is required to make the construction fit into one half of the tree.
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