n点圆上的近似位置 [英] Approximate position on circle for n points
问题描述
我正在努力解决以下问题:
我得到了 n
点和一个半径,我必须将它们放在尽可能对称的圆上。
目前,我使用的是这样的:
float theta = 360.0f / n;
int i = 0;
for(Word w:e.getValue()){
double newX = Math.sin(theta * i)* RADIUS + I_OFFSET_X;
double newY = Math.cos(theta * i)* RADIUS + I_OFFSET_Y;
mxCell v2 =(mxCell)graph.insertVertex(parent,null,w.getValue(),newX,newY,OW_WIDTH,OW_HEIGHT,shape = ellipse);
graph.insertEdge(parent,null,,v1,v2);
i ++;
其中 n
是我的点数。
例如,这适用于足够大的n,但对于 n = 3
我得到如下所示:
实际上我喜欢这样的内容:
(坏的绘画技巧不好..)
所以基本上,尽可能对称的东西会很棒。
关于如何解决这个问题的任何提示?
感谢< 3
感谢Jongware,答案很明显。因为我正在处理Java,所以 sin / cos
参数应该以弧度表示。
修正:
double newX = Math.sin(Math.toRadians(theta * i))* RADIUS + I_OFFSET_X ;
double newY = Math.cos(Math.toRadians(theta * i))* RADIUS + I_OFFSET_Y;
像魅力一样工作
I am struggling with the following problem:
I am given n
points and a radius and I have to place them on a circle as symmetrical as possible.
Currently, I used something like this:
float theta = 360.0f / n;
int i = 0;
for (Word w : e.getValue()) {
double newX = Math.sin(theta * i) * RADIUS + I_OFFSET_X;
double newY = Math.cos(theta * i) * RADIUS + I_OFFSET_Y;
mxCell v2 = (mxCell) graph.insertVertex(parent, null, w.getValue(), newX, newY, OW_WIDTH, OW_HEIGHT,"shape=ellipse");
graph.insertEdge(parent, null, "", v1, v2);
i++;
}
where n
is my number of points.
This works fine for a large enough n, but for n=3
for example, I get something like:
I would actually like to have something like:
(bad drawing skills are bad..)
So basically, something as symmetric as possible would be awesome.
Any hints on how to solve this?
Thanks <3
Thanks to Jongware, the answer was quite obvious. Because I'm dealing with Java, all the sin/cos
parameters should be in radians.
Fix:
double newX = Math.sin(Math.toRadians(theta * i)) * RADIUS + I_OFFSET_X;
double newY = Math.cos(Math.toRadians(theta * i)) * RADIUS + I_OFFSET_Y;
Works like a charm
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