加速度计 - 在球滚动的球 [英] Accelerometer - Rolling Ball in Ball

问题描述

我想用手机加速计用于滚动球的球。运动工程 正确的，问题是，当球撞到墙。我怎样才能得到一个平滑的滚动 动画球沿大球的内侧？

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?

这是我目前的code键移动球和检查的交集：

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：）

Thanks for Helping :)

推荐答案

在短短的滑动情况下使用参数圆式

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

• 其中X0，Y0是大圆圈中心
• R = R0，R1
• R0是大圈半径
• R1是小圆半径
• where x0,y0 is the big circle center
• r = R0-R1
• R0 is big circle radius
• R1 is small circle radius

现在的夹角 A

• 最简单的办法是把 A =重力方向这样：
• A = atanxy（acceleration.x，acceleration.y）
• atanxy是ATAN2这是4象限弓tangens
• 如果你没有它用我的 atanxy C ++实现
• 和纠正角度，你的坐标系
• （也许否定和或者增加一些90degree多个）
• the simplest would be to place a=gravity directionso:
• a=atanxy(acceleration.x,acceleration.y)
• atanxy is atan2 which is 4-quadrant arcus tangens
• if you dont have it use mine atanxy C++ implementation
• and correct the angle to your coordinate systems
• (maybe negate and or add some 90degree multiple)

[注]

• 如果您有兼容的协调屏和设备加速计之间的系统
• 然后就扩展加速度矢量大小| R |加（X0，Y0），将它
• 和你有同样的结果withouth的任何测角功能...

进行适当的模拟使用D'兰伯特方程+圆边界

所以2D运动是pretty的方便：

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（|位置big_circle_center |＆GT; big_circle_radius）发生碰撞

• 所以当你不希望任何反弹（所有的能量被吸收），那么：
• 位置 - = big_circle_center;
• 位置* = big_circle_radius / |位置|;
• 位置+ = big_circle_center;
• 现在，你需要删除的径向速度，只剩下切速度
• =正常位置big_circle_center //法线表面
• 正常* =点（速度，正常） //这是正常速度的一部分
• 网​​速度=正常 //现在只是切线速度应留
• 
• 在打完这只是正切（黄色）速度的一部分，仍然...
• 希望我没有忘了什么东西（像化妆单位向量或+/-某处...）
• 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;
• 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
• 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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