加速度计 - 滚球中的球 [英] Accelerometer - Rolling Ball in Ball

问题描述

我想使用手机加速度计来滚动球中的球.运动有效正确的，问题是当球击中墙时.我怎样才能顺利滚动球沿着大球内侧滑动的动画?

i want to use the phones accelerometer for rolling a Ball in a Ball. The movement works correctly, the problem is that when the ball hits the Wall. How can i get a smooth rolling animation that the ball slides along the inner side of the bigger ball?

这是我当前移动球和检查交叉点的代码:

This is my current code to move the Ball and check the intersection:

    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];
    }
};

感谢您的帮助:)

推荐答案

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 是大圆心
• r = R0-R1
• R0 是大圆半径
• R1 是小圆半径

• x0,y0 is the big circle center
• r = R0-R1
• R0 is big circle radius
• R1 is small circle radius

现在角度a

最简单的方法是放置a=gravity direction所以:

the simplest would be to place a=gravity directionso:

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

atanxy 是 atan2，它是 4 象限的弧线.如果你没有它，请使用我的

the atanxy is atan2 which is 4-quadrant arcus tangens. If you don't have it use mine

• atanxy C++ 实现

并校正坐标系的角度(可能否定和或添加一些 90 度倍数)

and correct the angle to your coordinate systems (maybe negate and or add some 90degree multiple)

[注释]

如果您在屏幕和设备加速度计之间有兼容的坐标系，那么只需将加速度矢量缩放到 |r| 的大小，然后添加 (x0,y0) 即可没有任何测角函数的相同结果......

If you have compatible coordinate systems between screen and device accelerometer then just scale acceleration vector to size |r| and add (x0,y0) to it and you have the same result without any goniometric function ...

为了正确模拟，请使用 D'ALember 方程 + 圆边界

所以 2D 运动非常简单:

so the 2D movement is pretty easy:

// 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 如果 (|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)然后:

now if (|position-big_circle_center|>big_circle_radius) an collision occurred so when you do not want any bounce (all energy was absorbed) then:

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

现在您需要移除径向速度并保留切线速度:

Now you need remove the radial speed and left just tangent speed:

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

所以在这之后速度的切线(黄色)部分仍然存在......希望我没有忘记一些东西(比如制作单位向量或 +/- 某处......)

so after this just tangent (Yellow) part of velocity remains ... Hope I did not forget something (like make unit vector or +/- somewhere...)

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