有没有算法来检测我碰撞过的多边形的哪一边？

Question

如果给出一个n多边形,一条长度线k(at x,y和angle a),是否有一个算法来检测我碰撞过的多边形的哪一边(如果有的话)？到目前为止,我已经尝试测试是否x,y在多边形之外,然后遍历多边形的每个边缘,计算到每个端点的距离.这是一个JS小提琴,展示了我创造的世界.

这是JavaScript(HTML和CSS不值得复制):

var eventLoop,
    maxVelocity = 10,
    agility = 5,
    baseLength = 5,
    degree = ((2*Math.PI)/360),
    world = document.getElementById('world'),
    context = world.getContext("2d"),
    boundry = [[180, 120],[240, 60],[360, 40],[420, 120],[360, 220],[350, 240],[360, 265],[470,360],[450,480],[360,540],[240,550],[140,480],[120,470],[100,360],[120,300],[220,240],[240,220]],
    camera = {
        location: {
            x:300,
            y:90
        },
        angle: 0,
        velocity: 0
    },
    engine = {
        drawWorld: function(shape, context) {
            var point,
                index,
                size = shape.length;

            context.clearRect(0, 0, world.width, world.height);
            context.beginPath();
            for(index = 0; index < size; index++) {
                point = shape[index];
                if(index == 0) {
                    context.moveTo(point[0], point[1]);
                } else {
                    context.lineTo(point[0], point[1]);
                }
            }
            context.closePath();
            context.stroke();
        },
        drawCamera: function(camera, context) {
            var a = camera.location,
                b = this.calcNextPoint(camera, 1);

            context.beginPath();
            context.moveTo(a.x, a.y);
            context.lineTo(b.x, b.y);
            context.stroke();

            context.beginPath();
            context.arc(a.x, a.y, baseLength, 0, Math.PI*2, true); 
            context.closePath();
            context.stroke();
        },
        calcNextPoint: function(camera, moment) {
            return {
                x: camera.location.x + ((camera.velocity*(1/moment))*Math.sin(camera.angle)),
                y: camera.location.y + ((camera.velocity*(1/moment))*(Math.cos(camera.angle)))
            };
        },
        isInside: function(point, shape) {
            var i, j, c = 0;
            for (i = 0, j = shape.length - 1; i < shape.length; j = i++) {
                if (((shape[i][1] > point.y) != (shape[j][1] > point.y)) && (point.x < (shape[j][0] - shape[i][0]) * (point.y - shape[i][1]) / (shape[j][1] - shape[i][1]) + shape[i][0])) {
                     c = !c;
                }
            }
            return c;
        }
    };

document.onkeydown = function(e) {
    e = e || window.event;
    if (e.keyCode == '37') {
        // left arrow
        camera.angle += degree*agility;
    }
    else if (e.keyCode == '39') {
        // right arrow
        camera.angle -= degree*agility;
    }
    else if (e.keyCode == '38') {
        // up arrow
        camera.velocity += 1;
        if(camera.velocity > maxVelocity) {
            camera.velocity = maxVelocity;
        }
    }
    else if (e.keyCode == '40') {
        // down arrow
        camera.velocity -= 1;
        if(camera.velocity < 0) {
            camera.velocity = 0;
        }
    }
}

engine.drawWorld(boundry, context);
engine.drawCamera(camera, context);

eventLoop = setInterval(function() {
    engine.drawWorld(boundry, context);
    engine.drawCamera(camera, context);
    if(engine.isInside(camera.location, boundry)) {
        camera.location = engine.calcNextPoint(camera, 1);
    }
}, 100);

我一直在玩一些JavaScript,它们模拟游戏Flower by ThatGameComapny 的2-Dementional版本,最终我想尝试实现Oculus Rift版本.我要解决的下一个问题是,当玩家与边缘碰撞时,将玩家变回多边形.

Answer 1

所以你基本上想知道多边形的哪些边与给定的线相交,对吧？

这仅仅是非常简单的几个处理线性方程表示的边缘(不等式,更精确).您已经有了很好的inside操作实现,也可以这样做.所有这些算法的共同点是比较线的哪一侧[x1, y1]- [x2, y2]重点[x, y]在于:

compare = function (a, b) { // classical compare function, returns -1, 0, 1
    return a < b ? -1 : (a == b ? 0 : 1);
}
...
_lineSide: function (x, y, x1, y1, x2, y2) {
    return compare(x - x1, (x2 - x1) * (y - y1) / (y2 - y1));
}

如果[x, y]位于线的一侧,则此函数将返回-1 [x1, y1]- [x2, y2]对于另一侧,该函数将返回1,如果它恰好位于该线上,则返回0.这一点并不重要,哪一方是哪一方,只是将它们分开.然而,当y2 - y1零或接近零时,这将不起作用.在这种情况下,你必须xy翻转情况:

lineSide: function (x, y, x1, y1, x2, y2) {
    var eps = 1e-20; // some very small number
    if (Math.abs(y2 - y1) > eps)       // this way we avoid division by small number
        return _lineSide(x, y, x1, y1, x2, y2);
    else if (Math.abs(x2 - x1) > eps)  // flip the situation for horizontal lines
        return _lineSide(y, x, y1, x1, y2, x2);
    else // edge has close-to-zero length!
        throw new this.zeroLengthLineException()
},
zeroLengthLineException: function () {},

现在来测试两条线[x1, y1]- [x2, y2]和[x3, y3]- [x4, y4]相交是否非常容易.只要看看是否[x1, y1]并且[x2, y2]位于[x3, y3]- 的另一侧- [x4, y4]如果[x3, y3]并且[x4, y4]位于[x1, y1]- 的另一侧- [x2, y2].如果是,则线条相交!

// proper: true/false (default false) - if we want proper intersection, i.e. just 
// touching doesn't count
linesIntersect: function (x1, y1, x2, y2, x3, y3, x4, y4, proper) {
    var min_diff = proper ? 2 : 1;
    return Math.abs(this.lineSide(x1, y1, x3, y3, x4, y4) - 
                    this.lineSide(x2, y2, x3, y3, x4, y4)) >= min_diff 
        && Math.abs(this.lineSide(x3, y3, x1, y1, x2, y2) - 
                    this.lineSide(x4, y4, x1, y1, x2, y2)) >= min_diff;
},

最终解决方案 - http://jsfiddle.net/gLTpT/7/

现在,最终解决方案很简单:只需通过调用linesIntersect循环检查给定行与所有多边形边的交集:

http://jsfiddle.net/gLTpT/7/