画布三角形，五角大楼，与彼此的矩形碰撞检测？

Question

我有一些代码绘制一个三角形并返回一个对象，其中包含x1, y1, x2, y2, x3, y3。我还有一个矩形绘图功能，返回x, y, w, h，以及返回x1, y1, x2, y2, x3, y3, x4, y4, x5, y5的五角形绘图功能。我怎样才能做一个碰撞检测功能，可以检测矩形和五边形或三角形和五边形是否碰撞？

Answer 1

测试如果所述三角形的任何边（线段）截获该矩形

如果任何侧截取，那么三角形&矩形发生碰撞的任何一侧。针对每边三角形的

• 测试边＃针对矩形的每侧的三角形的1，
• 测试边＃针对矩形的每侧的三角形的2，
• 测试边＃3
• 如果在测试过程中发现拦截，则可以停止，因为形状确实发生碰撞。

有关“两条线段是否拦截？”的详细信息，请参见下文。

对任何其他多边形进行相同的侧面拦截测试，看看它们是否发生碰撞。

2线测试拦截转载于此：

// point object: {x:, y:} 
// p0 & p1 form one segment, p2 & p3 form the second segment 
// Returns true if lines segments are intercepting 
var lineSegmentsIntercept = (function(){ // function as singleton so that closure can be used 

    var v1, v2, v3, cross, u1, u2; // working variable are closed over so they do not need creation 
           // each time the function is called. This gives a significant performance boost. 
    v1 = {x : null, y : null}; // line p0, p1 as vector 
    v2 = {x : null, y : null}; // line p2, p3 as vector 
    v3 = {x : null, y : null}; // the line from p0 to p2 as vector 

    function lineSegmentsIntercept (p0, p1, p2, p3) { 
     v1.x = p1.x - p0.x; // line p0, p1 as vector 
     v1.y = p1.y - p0.y; 
     v2.x = p3.x - p2.x; // line p2, p3 as vector 
     v2.y = p3.y - p2.y; 
     if((cross = v1.x * v2.y - v1.y * v2.x) === 0){ // cross prod 0 if lines parallel 
      return false; // no intercept 
     } 
     v3 = {x : p0.x - p2.x, y : p0.y - p2.y}; // the line from p0 to p2 as vector 
     u2 = (v1.x * v3.y - v1.y * v3.x)/cross; // get unit distance along line p2 p3 
     // code point B 
     if (u2 >= 0 && u2 <= 1){     // is intercept on line p2, p3 
      u1 = (v2.x * v3.y - v2.y * v3.x)/cross; // get unit distance on line p0, p1; 
      // code point A 
      return (u1 >= 0 && u1 <= 1);   // return true if on line else false. 
      // code point A end 
     } 
     return false; // no intercept; 
     // code point B end 
    } 
    return lineSegmentsIntercept; // return function with closure for optimisation. 
})(); 

---

---

是两条线段拦截？

（归属地user blindman67用于与下面的示例辅助）

在本例中，该函数返回true如果两个线段相交并false如果不是。

的示例设计用于性能，并使用封闭件以保持工作变量

```的Javascript //点对象：{X：Y：} // P0 & P1形成一个段，P2 & P3形成第二分段 //如果线路段被拦截 VAR lineSegmentsIntercept =（函数（）返回真{//功能单以便闭合可用于

var v1, v2, v3, cross, u1, u2; // working variable are closed over so they do not need creation 
           // each time the function is called. This gives a significant performance boost. 
    v1 = {x : null, y : null}; // line p0, p1 as vector 
    v2 = {x : null, y : null}; // line p2, p3 as vector 
    v3 = {x : null, y : null}; // the line from p0 to p2 as vector 

    function lineSegmentsIntercept (p0, p1, p2, p3) { 
     v1.x = p1.x - p0.x; // line p0, p1 as vector 
     v1.y = p1.y - p0.y; 
     v2.x = p3.x - p2.x; // line p2, p3 as vector 
     v2.y = p3.y - p2.y; 
     if((cross = v1.x * v2.y - v1.y * v2.x) === 0){ // cross prod 0 if lines parallel 
      return false; // no intercept 
     } 
     v3 = {x : p0.x - p2.x, y : p0.y - p2.y}; // the line from p0 to p2 as vector 
     u2 = (v1.x * v3.y - v1.y * v3.x)/cross; // get unit distance along line p2 p3 
     // code point B 
     if (u2 >= 0 && u2 <= 1){     // is intercept on line p2, p3 
      u1 = (v2.x * v3.y - v2.y * v3.x)/cross; // get unit distance on line p0, p1; 
      // code point A 
      return (u1 >= 0 && u1 <= 1);   // return true if on line else false. 
      // code point A end 
     } 
     return false; // no intercept; 
     // code point B end 
    } 
    return lineSegmentsIntercept; // return function with closure for optimisation. 
})(); 

```

使用例

var p1 = {x: 100, y: 0}; // line 1 
var p2 = {x: 120, y: 200}; 
var p3 = {x: 0, y: 100}; // line 2 
var p4 = {x: 100, y: 120}; 
var areIntersepting = lineSegmentsIntercept (p1, p2, p3, p4); // true 

该示例容易地修改以返回截距的点。与

if(u1 >= 0 && u1 <= 1){ 
    return { 
     x : p0.x + v1.x * u1, 
     y : p0.y + v1.y * u1, 
    }; 
} 

更换code point A和A end之间的代码或者，如果你想获得对线路的拦截点，而忽略线段开始和结束与替代code point B和B end之间的代码

return { 
    x : p2.x + v2.x * u2, 
    y : p2.y + v2.y * u2, 
}; 

如果没有拦截或返回截获点，这两个修改都会返回false {x : xCoord, y : yCoord}