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桫椤@2005-06-13 21:42

最懒人的方法应该是直接拿现成的...............:o

小狼这个我就非常不客气地收下了.................lol
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小萌有麦@2005-06-18 11:37

原来还有这个东西,真不错啊!
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无雨娃娃@2005-07-16 14:38

我是来谢谢明的HB的>___< 感激ing~~~
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冷火@2005-07-17 00:59

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最初由 无雨娃娃 发布
看起来画出来的花边很PL
函数?貌似使用起来非常麻烦

The following illustrates the Julia set for functions of the form


c sin(z)
where z is a complex number x + iy.

To do this we form the series


zk+1 = c sin(zk)
starting with some initial z0

The behavior of this series determines whether or not the initial z0 is part of the julia set or not. More precisely, if the series tends to infinity then z0 is part of the Julia set, otherwise it isn't. In the examples shown on the right the white regions are in the Julia set, black points are outside the Julia set.

To create images of the Julia set we map pixels in the image onto values of z0 and colour the pixel dependent on the behavior of the series. In the examples on the right the image is mapped onto the range +-2pi in both the real and imaginary axes. The series is tested after 50 terms, it is decided that it tends to infinity if the absolute value of the imaginary part of zk is greater than 50.

Footnote
If you are wondering how to compute the sine of a complex number, you can use the following relationships:

xk+1 = sin(xk) cosh(yk)
yk+1 = cos(xk) sinh(yk)

where zk = xk + i yk



我快要放弃乐 -v-|||


这么恐怖……
某火系数学白痴怎么办??
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