计算正多边形顶点的坐标 [英] Calculate coordinates of a regular polygon's vertices

问题描述

我正在编写一个程序，在该程序中我需要绘制任意数量的边的多边形，每个边都由一个动态变化的给定公式转换.涉及一些相当有趣的数学，但我被困在这个问题上.

I am writing a program in which I need to draw polygons of an arbitrary number of sides, each one being translated by a given formula which changes dynamically. There is some rather interesting mathematics involved but I am stuck on this probelm.

如何计算正多边形(所有角度都相等)的顶点坐标，仅给出边数，理想情况下(但不是必须)具有原点在中心?

How can I calculate the coordinates of the vertices of a regular polygon (one in which all angles are equal), given only the number of sides, and ideally (but not neccessarily) having the origin at the centre?

例如:一个六边形可能有以下几点(都是floats):

For example: a hexagon might have the following points (all are floats):

( 1.5  ,  0.5 *Math.Sqrt(3) )
( 0    ,  1   *Math.Sqrt(3) )
(-1.5  ,  0.5 *Math.Sqrt(3) )
(-1.5  , -0.5 *Math.Sqrt(3) )
( 0    , -1   *Math.Sqrt(3) )
( 1.5  , -0.5 *Math.Sqrt(3) )

我的方法是这样的:

void InitPolygonVertexCoords(RegularPolygon poly)

并且需要将坐标添加到此(或类似的东西，如列表):

and the coordinates need to be added to this (or something similar, like a list):

Point[] _polygonVertexPoints;

我主要对这里的算法感兴趣，但 C# 中的示例会很有用.我什至不知道从哪里开始.我应该如何实现它?有可能吗?！

I'm interested mainly in the algorithm here but examples in C# would be useful. I don't even know where to start. How should I implement it? Is it even possible?!

谢谢.

推荐答案

for (i = 0; i < n; i++) {
  printf("%f %f
",r * Math.cos(2 * Math.PI * i / n), r * Math.sin(2 * Math.PI * i / n));
}

其中 r 是外接圆的半径.对不起，错误的语言没有 Habla C#.

where r is the radius of the circumsribing circle. Sorry for the wrong language No Habla C#.

基本上任意两个顶点之间的夹角是2 pi/n，并且所有顶点距离原点r.

Basically the angle between any two vertices is 2 pi / n and all the vertices are at distance r from the origin.

如果你想让中心在原点以外的地方，比如 (x,y)

If you want to have the center somewher other than the origin, say at (x,y)

for (i = 0; i < n; i++) {
  printf("%f %f
",x + r * Math.cos(2 * Math.PI * i / n), y + r * Math.sin(2 * Math.PI * i / n));
}

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