ws the progress of the boundary of the dragon curve as generated by the chaos game.
The area within the boundary
We now turn our attention to the area enclosed by the boundary of the dragon curve. To determine the area, let us consider the generation of the dragon curve by iteration as shown in Figure 10.
The area at any level i can approximated by adding up the number of squares that are enclosed by the segments. Because each square is formed by four segments and each segment is shared by two squares, the number of squares is equal to the number of segments multiplied by 2/4 and the total area can be represented by the following equation:
where di is the length of the segment and ni, is the number of segments at level i.
Letting the initial length of the segment at level 0 be equal to a, di and ni can be expressed
From equations (11) and (12) we can compute the area of the dragon curve, represented by S, the limit of Si as i approaches infinity.
Therefore, the area of the dragon curve is a2/2.
An alternate method to compute the area enclosed within the boundary of the dragon curve involves the second IFS boundary structure discussed. Letting b be the length shown in Figure 11 and Si be the area enclosed by the border at any level i, we can calculate the area of the dragon curve by observing the changes in area from one level to the next. Notice that the starting area, S0, is equal to 4b2.
From Figure 11 it can be seen that as the level of generation increases, the boundary of the dragon curve folds in and out, increasing the area at one point and decreasing at another. Upon closer observation we notice from one level to the next, all these areas cancel with the exception of the areas of P1i and P2i. Because P1i increases the area of the dragon curve and P2i decreases it, the total area increases by P1i-P2i from level i to level i+1. Since P1i=b2/2i and P2i=b2/2i+1 the relation between Si+1 and Si is as follows:
From this equation we realize that from level i to level i+1, the area enclosed by the curve increases by .
Using (14), we get that the area enclosed by the boundary of the dragon curve is:
To reconcile this answer for the area with our first solution in equation (13), recall that the first solution uses the length a as shown in Figure 12. From the Pythagorean Theorem we know that a2=9b2+b2 so that b2=a2/10. From here it becomes apparent that the two solutions from equations (13) and (15) are one and the same since S=5b2=a2/2.
In addition we can also calculate the area of each small shape of which the dragon curve is composed. Since it is self-evident that these shapes are similar we can determine the relationship between their size as shown in Figure 13.
Using the relationship between their linear dimension, we can determine the relationship between their areas and we find that .
From Figure 13 it is apparent that
Solving for T0, we get T0=0.4S.
Although at this point in time the dragon curve and its border seem no more than quaint examples of fractals, self-sirnilar structures creating order from chaos, there may come a time when the results from this little mathematical investigation prove to be useful in other areas of science as well. Fractals and chaos are after all, the footprints of nature, appearing here and everywhere and what is science but the quest for knowledge, about nature, this universe that we live in.
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5.Hutchinson, J., Fractal and Self-Similarity, Indiana University Journal of Mathematics 30 , 1981.