畫布三角形，五角大樓，與彼此的矩形碰撞檢測？

Question

我有一些代碼繪製一個三角形並返回一個對象，其中包含x1, y1, x2, y2, x3, y3。我還有一個矩形繪圖功能，返回x, y, w, h，以及返回x1, y1, x2, y2, x3, y3, x4, y4, x5, y5的五角形繪圖功能。我怎樣才能做一個碰撞檢測功能，可以檢測矩形和五邊形或三角形和五邊形是否碰撞？

Answer 1

測試如果所述三角形的任何邊（線段）截獲該矩形

如果任何側截取，那麼三角形&矩形發生碰撞的任何一側。針對每邊三角形的

• 測試邊＃針對矩形的每側的三角形的1，
• 測試邊＃針對矩形的每側的三角形的2，
• 測試邊＃3
• 如果在測試過程中發現攔截，則可以停止，因爲形狀確實發生碰撞。

有關「兩條線段是否攔截？」的詳細信息，請參見下文。

對任何其他多邊形進行相同的側面攔截測試，看看它們是否發生碰撞。

2線測試攔截轉載於此：

// 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. 
})(); 

---

---

是兩條線段攔截？

（歸屬地user blindman67用於與下面的示例輔助）

在本例中，該函數返回true如果兩個線段相交併false如果不是。

的示例設計用於性能，並使用封閉件以保持工作變量

```的Javascript //點對象：{X：Y：} // P0 & P1形成一個段，P2 & P3形成第二分段 //如果線路段被攔截 VAR lineSegmentsIntercept =（函數（）返回真{//功能單以便閉合可用於

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 

該示例容易地修改以返回截距的點。與

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

更換code point A和A end之間的代碼或者，如果你想獲得對線路的攔截點，而忽略線段開始和結束與替代code point B和B end之間的代碼

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

如果沒有攔截或返回截獲點，這兩個修改都會返回false {x : xCoord, y : yCoord}