点击一个球体 [英] clicking on a sphere

问题描述

球体可以自由旋转。

如何确定用户点击的球体上的点？

解决方案

鉴于：

• 显示器的高度和宽度
  
• 投影圆的半径，以像素为单位


• 用户点击的坐标

假设左上角是（0,0），随着您向右移动x值会增加，并且随着您向下移动，y值会增加。

将用户的点击点转换为地球坐标空间。 / p>

  userPoint.x  -  = monitor.width / 2 
 userPoint.y  -  = monitor.height / 2 
 userPoint.x / = circleRadius 
 userPoint.y / = circleRadius 
  

找到交点的z坐标。

  //解析z 
 // x ^ 2 + y ^ 2 + z ^ 2 = 1 
 //我们知道x和y，来自userPoint 
如果（x ^ 2 + y ^ 2> 2）满足下面的等式， 1）{
 //用户在球体外点击。翻转
返回-1; 
} 
 
 //负的sqrt比正的更接近屏幕，所以我们更喜欢这一点。 
 z = -sqrt（1  -  x ^ 2  -  y ^ 2）; 
  

现在您知道了（x，y，z）交点，和经度。

假设全球面向用户的中心是0E 0N，

 经度= 90 + toDegrees（atan2（z，x））; 
 lattitude = toDegrees（atan2（y，sqrt（x ^ 2 + z ^ 2）））
  

如果球体被旋转以便0E子午线不直接面对观察者，从经度减去旋转角度。

I have a unit sphere (radius 1) that is drawn centred in orthogonal projection.

The sphere may rotate freely.

How can I determine the point on the sphere that the user clicks on?

解决方案

Given:

• the height and width of the monitor
• the radius of the projected circle, in pixels
• the coordinates of the point the user clicked on

And assuming that the top-left corner is (0,0), the x value increases as you travel to the right, and the y value increases as you travel down.

Translate the user's click point into the coordinate space of the globe.

userPoint.x -= monitor.width/2
userPoint.y -= monitor.height/2
userPoint.x /= circleRadius
userPoint.y /= circleRadius

Find the z coordinate of the point of intersection.

//solve for z
//x^2 + y^2 + z^2 = 1
//we know x and y, from userPoint
//z^2 = 1 - x^2 - y^2
x = userPoint.x
y = userPoint.y

if (x^2 + y^2 > 1){
    //user clicked outside of sphere. flip out
    return -1;
}

//The negative sqrt is closer to the screen than the positive one, so we prefer that.
z = -sqrt(1 - x^2 - y^2);

Now that you know the (x,y,z) point of intersection, you can find the lattitude and longitude.

Assuming that the center of the globe facing the user is 0E 0N,

longitude = 90 + toDegrees(atan2(z, x));
lattitude = toDegrees(atan2(y, sqrt(x^2 + z^2)))

If the sphere is rotated so that the 0E meridian is not directly facing the viewer, subtract the angle of rotation from the longitude.

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