加速度计 - 球滚球

Question

我想用手机加速度计来滚动球。运动正常，问题是当球撞到墙上时。我怎样才能得到一个平滑的滚动动画，球在大球的内侧滑动？

这是我当前移动球并检查交叉点的代码：

    onSuccess: function(acceleration) {
        var xPos = this.xPos + (-1 * (acceleration.x * 0.5));
        var yPos = this.yPos + (acceleration.y * 0.5);

        var intersect = this.intersection(xPos + 32, 
                                          yPos + 32, 
                                          32, 
                                          self.canvas.width * 0.5, 
                                          self.canvas.height * 0.5, 
                                          self.canvas.width * 0.5);
        if (!intersect) {
            this.yPos = yPos;
            this.xPos = xPos;
        }

        this.cnv.clearRect(0.0, 0.0, this.canvas.width, this.canvas.height);    
        this.cnv.drawImage(this.target, this.xPos, this.yPos);
    },

    intersection: function(x0, y0, r0, x1, y1, r1) {

        var a, dx, dy, d, h, rx, ry;
        var x2, y2;

        /* dx and dy are the vertical and horizontal distances between
         * the circle centers.
         */
        dx = x1 - x0;
        dy = y1 - y0;

        /* Determine the straight-line distance between the centers. */
        d = Math.sqrt((dy*dy) + (dx*dx));

        /* Check for solvability. */
        if (d > (r0 + r1)) {
            /* no solution. circles do not intersect. */
            return false;
        }
        if (d < Math.abs(r0 - r1)) {
            /* no solution. one circle is contained in the other */
            return false;
        }

        /* 'point 2' is the point where the line through the circle
         * intersection points crosses the line between the circle
         * centers.  
         */

        /* Determine the distance from point 0 to point 2. */
        a = ((r0*r0) - (r1*r1) + (d*d)) / (2.0 * d) ;

        /* Determine the coordinates of point 2. */
        x2 = x0 + (dx * a/d);
        y2 = y0 + (dy * a/d);

        /* Determine the distance from point 2 to either of the
         * intersection points.
         */
        h = Math.sqrt((r0*r0) - (a*a));

        /* Now determine the offsets of the intersection points from
         * point 2.
         */
        rx = -dy * (h/d);
        ry = dx * (h/d);

        /* Determine the absolute intersection points. */
        var xi = x2 + rx;
        var xi_prime = x2 - rx;
        var yi = y2 + ry;
        var yi_prime = y2 - ry;

        return [xi, xi_prime, yi, yi_prime];
    }
};

谢谢你的帮助:)

Answer 1

在滑动情况下使用参数圆方程

x=x0+r*cos(a)
y=y0+r*sin(a)

哪里：

• x0,y0是大圆圈中心
• r = R0-R1
• R0是大圆半径
• R1是小圆半径

现在角度a

最简单的是放置a=gravity directionso：

a=atanxy(acceleration.x,acceleration.y)

atanxy是atan2，是四象限的arcus tangens。如果你没有它使用我的

• atanxy C++ implementation

并纠正你的坐标系的角度（也许否定和或添加一些90度的多重）

[笔记]

如果屏幕和设备加速度计之间有相容的坐标系，那么只需将加速度矢量缩放到|r|大小并添加(x0,y0)就可以得到相同的结果而没有任何测角功能...

为了正确模拟，使用D'ALembert方程+圆边界

所以2D运动非常简单：

// in some timer with interval dt [sec]
velocity.x+=acceleration.x*dt;
velocity.y+=acceleration.y*dt;
position.x+=velocity.x*dt;
position.y+=velocity.y*dt;

现在if (|position-big_circle_center|>big_circle_radius)发生碰撞，所以当你不想要任何反弹（所有能量都被吸收）时：

position-=big_circle_center;
position*=big_circle_radius/|position|;
position+=big_circle_center;

现在你需要移除径向速度并保持切线速度：

normal=position-big_circle_center; // normal vector to surface
normal*=dot(velocity,normal);      // this is the normal speed part
velocity-=normal;                  // now just tangential speed should be left

所以在此之后，只有切线（黄色）部分的速度仍然存在...希望我没有忘记一些东西（比如制作单位矢量或+/-某处...）