1 jan 2010

Koch pentaflake

Koch pentaflake8.png

The construction of this snowflake is similar to the usual Koch hexaflake.

Start with a pentagon, or the first pentaflake
Koch pentaflake0.png Koch pentaflake1.png
and divide the original edges in four smaller ones, as indicated in the second pentaflake.

The new edges make angles of 108°, 36° and 108° respectively and their length is 0.381966 = *:θ times the original edge length, where 0.618033 = θ =: -: <: %: 5 .

Notice that this division of edges is similar to the division of the (base of the) triangle in Penrose-like Tiling.
Penrose_div_low.png

The Hausdorff dimension is then equal to 1.44042 = - 2 %&^. θ . This is equal to $ {\frac{\log(4)}{\log(2(1+\cos(a)))}} $ with $a$=72°, in list of fractals by Hausdorff dimension.

Koch pentastarflake

is another snowflake with the same construction and the same Hausdorff dimension:
Koch pentastarflake10.png It comes from Koch pentastarflake0.png

RE Boss/J-blog/KochPentaflake (last edited 2010-01-01 17:18:43 by RE Boss)