如何计算椭圆圆周上的点？ [英] How do I calculate a point on a ellipse’s circumference?

问题描述

我有X，Y，宽度，高度和t值。

$ b $在J2ME中绘制椭圆曲线中的点b b X 和 Y 是椭圆（根据J2ME）相对于Canvas的位置， t 是相对于椭圆中心的角度（我有问题的图像表示，但不幸的博客不允许插入讨论:)）

  int ePX =（X + width）+（int）（width * Math.cos（Math.toRadians（t）））; 
 int ePY =（Y + height）+（int）（height * -Math.sin（Math.toRadians（t）））; 
  

这个方程是否正确？如果（X，Y）是椭圆的中心，那么我们需要进行更多的计算？

并且宽度和高度是两个轴，那么方程应该是bb
$ b pre code int ePX = X +（int）（width * Math名为.cos（Math.toRadians（T）））;
int ePY = Y +（int）（height * Math.sin（Math.toRadians（t）））;

如果所有的 t都不需要-1与Math.sin的乘法运算绘制整个椭圆。

I want to plot a point in elliptical curve in J2ME

I have values for X,Y,width,height and t.

X and Y are the position of the ellipse(according to J2ME) with respect to Canvas and t is angle with respect to the center of ellipse(I have an image representation of the problem , but unfortunate blog is not allowing to insert into discussion :) )

int ePX = (X + width)+ (int) (width * Math.cos(Math.toRadians(t)));
int ePY = (Y + height)+ (int) (height * -Math.sin(Math.toRadians(t)));

Is this equation correct? or for ellipse do we need to have some more calculations?

解决方案

If ( X, Y ) is the center of the ellipse, and width and height are the two axes, then the equation should be

int ePX = X + (int) (width  * Math.cos(Math.toRadians(t)));
int ePY = Y + (int) (height * Math.sin(Math.toRadians(t)));

The -1 multiplication to Math.sin is unnecessary if you have all t to draw the entire ellipse.

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