使用JS计算两个矩形节点之间的线相交偏移

时间:2018-08-31 23:49:00

标签: javascript math trigonometry

我不想问像这样的问题的其他人,但是我不是专家,在计算线相交的某些位置时我需要一点帮助。

使用Javascript和一组具有一定宽度和高度的已知矩形节点(节点的x和y位置代表每个节点的中心),我想计算该行的起点和终点的线偏移量,以便这条线只接触节点的边缘。

[_____]---->[_____]

上面的ascii艺术显示了2个矩形节点,中间有一个箭头。箭头在起点或终点均不与节点相交。假设直线始终是水平或垂直,这很容易计算,但是如果直线成一定角度,我仍然希望直线仅接触边缘。

假设下面的ascii艺术中有一个箭头指向目标节点(由管道符号表示)

[_____]
    \
     \|
   [______]

已知变量:

  • 来源(1):{x,y,w,h}
  • 目标(2):{x,y,w,h}
  • 线的theta(Math.atan(y2-y1,x2-x1))

所需的输出:

  • 源节点上的交点
  • 目标节点上的交点

示例:

{
    sourceIntersect: {x, y}, 
    targetIntersect: {x, y}
}

1 个答案:

答案 0 :(得分:0)

我发现有一个可以计算两条线的交点的工具。如果我将矩形视为一组4条线,则可以使用line-intersect npm项目获取这4条线之一上的交点。在下面的SO问题中找到了解决方案。

calculating the point of intersection of two lines

这是我的代码的示例:

// first, get the sizes of the element.
// here we use getBoundingClientRect for an SVG
const clientRect = svgRectElement.getBoundingClientRect();

// extract the width and height
const w = clientRect.width;
const h = clientRect.height;

// trg represents the target point from the element above
// the x and y for trg relate to the center of the element
const top = trg.y - h / 2;
const bottom = trg.y + h / 2;
const left = trg.x - w / 2;
const right = trg.x + w / 2;

// a line extends from src{x,y} to trg{x,y} at the center of both rectangles

// another line extends from left to right at the top of the rectangle   
const topIntersect = lineIntersect.checkIntersection(src.x, src.y, trg.x, trg.y, left, top, right, top);

// another line extends from top to bottom at the right of the rectangle
const rightIntersect = lineIntersect.checkIntersection(src.x, src.y, trg.x, trg.y, right, top, right, bottom);

// another line extends from left to right at the bottom of the rectangle
const bottomIntersect = lineIntersect.checkIntersection(src.x, src.y, trg.x, trg.y, left, bottom, right, bottom);

// another line extends from top to bottom at the left of the rectangle
const leftIntersect = lineIntersect.checkIntersection(src.x, src.y, trg.x, trg.y, left, top, left, bottom);

// only one of the intersect variables above will have a value
if (topIntersect.type !== 'none' && topIntersect.point != null) {
  // topIntersect.point is the x,y of the line intersection with the top of the rectangle
} else if (rightIntersect.type !== 'none' && rightIntersect.point != null) {
  // rightIntersect.point is the x,y of the line intersection with the right of the rectangle
} else if (bottomIntersect.type !== 'none' && bottomIntersect.point != null) {
  // bottomIntersect.point is the x,y of the line intersection with the bottom of the rectangle
} else if (leftIntersect.type !== 'none' && leftIntersect.point != null) {
  // leftIntersect.point is the x,y of the line intersection with the left of the rectangle
}

要获得源节点的交集,我会将其作为trg参数传递给我的函数,而目标节点将代表src参数,从本质上诱使系统认为绘制了线条向后。这将提供源节点的交集,该交集在上面的变量trg中表示。