从点 (x, y) 以给定角度在边界内绘制一条线 [英] Drawing a line within boundaries from point (x, y) with a given angle

问题描述

我正在尝试在 JavaScript 画布上画一条线.我有两个点 A 和 B(如图所示).

I am trying to draw a line on JavaScript canvas. I have two points A and B (as shown in the picture).

我用这段代码来求这两点之间的角度:

I use this code to find the angle between these two points:

// p1 is point A and p2 is point B    
var theta = Math.atan2(p2.y - p1.y, p2.x - p1.x);

现在我想从点 A 到画布的末端画一条线，我还想得到线结束的点(图片中的点 C).

Now I want to draw a line from point A till the end of the canvas and I also want to get the point where the line ends (point C in the picture).

这可以使用画布的角度和大小(宽度和高度)来实现吗?

Can this be achieved using the angle and size(width & height) of the canvas?

推荐答案

可以不用三角函数来解决这个问题.首先构造给定AB射线的参数表示:

It is possible to solve this problem without trigonometric functions. At first construct parametric representation of given AB ray:

 x = A.X + dx * t
 y = A.Y + dy * t
   where
 dx = B.X - A.X
 dy = B.Y - A.Y

检查哪个边会先相交(t值较小):

The check what edge will be intersected first (with smaller t value):

//potential border positions    
if dx > 0 then
   ex = width
else
   ex = 0

if dy > 0 then
   ey = height
else
   ey = 0

//check for horizontal/vertical lines 
if dx = 0 then
    return cx = A.X,  cy = ey

if dy = 0 then
    return cy = A.Y,  cx = ex


//in general case find parameters of intersection with horizontal and vertical edge
tx = (ex - A.X) / dx
ty = (ey - A.Y) / dy

//and get intersection for smaller parameter value
if tx <= ty then
   cx = ex
   cy = A.Y + tx * dy
else
   cy = ey
   cx = A.X + ty * dx

return cx, cy

宽度 = 400、高度 = 300、固定 A 点和各种 B 点的结果:

Results for Width = 400, Height = 300, fixed A point and various B points:

100:100 - 150:100   400: 100
100:100 - 150:150   300: 300
100:100 - 100:150   100: 300
100:100 - 50:150     0: 200
100:100 - 50:100     0: 100
100:100 - 50:50     0:   0
100:100 - 100:50   100:   0
100:100 - 150:50   200:   0

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