Construction
Animation
Begin with a line segment. For the first iteration, replace this segment with two segments, the first scaled by r and the second scaled by r2, where r is equal to (1/φ)(1/φ) with φ being the golden ratio, i.e. \(\varphi = \frac{{1 + \sqrt 5 }}{2}\). The new segments are positioned in such a way that the original segment would have been the third side of the triangle formed by the new segments. Following along the original segment, we place the two new segments to the left. For the second iteration, replace the first segment with two new segments scaled in order by r and r2 and placed to the left. Replace the second segment from the first iteration with two new segments scaled in order by r2 and r and placed to the right. Continue this construction, always alternating the new segments between left and right along the segments of the previous iteration, and alternating the scaling order between (r, r2) and (r2, r). This generates the "golden dragon curve". The following figure shows the first three iterations for this construction.
Remember that \(r = {\left( {\dfrac{1}{\varphi }} \right)^{\frac{1}{\varphi }}} = 0.74274\) where \(\varphi = \dfrac{{1 + \sqrt 5 }}{2} = 1.618033988\).
We can use the Law of Cosines to determine the value of the angle A.
\[{r^4} = 1 + r^2 - 2r\cos A \Rightarrow \cos A = \frac{{1 + r^2 - {r^4}}}{{2r}} = 0.839678466\]Therefore A = cos−1(0.839678466) = 32.893818°. A similar calculation with the Law of Cosines for angle B shows that
\[B = {\cos ^{ - 1}}\left( {\frac{{1 + {r^4} - {r^2}}}{{2{r^2}}}} \right) = {46.98598225^ \circ }\]
Because of the need to alternate left and right when constructing
the dragon, we need to think of constructing the second line segment
by a rotation through 180°−B = 133.0140178° followed by a translation of the origin
to the point (1,0). This yields the following IFS
\({f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{0.62367} & { - 0.40337} \\
{0.40337} & {0.62367} \\
\end{array}} \right]{\bf{x}}\) |
scale by r, rotate by 32.893818° |
\({f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ - 0.37633} & { -0.40337} \\
{0.40337} & { -0.37633} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
1 \\
0 \\
\end{array}} \right]\) |
scale by r2, rotate by 133.0140178° |
The attractor of this IFS will be the golden dragon curve. The golden dragon consists of two self-similar pieces corresponding to the two functions in the IFS.