Question

以下函数返回具有给定半径的球体上的点。我想添加限制，使得点不能在球体的两极之内绘制。

public static function randomPoint(radius:Number):Number3D
 {

  var inclination:Number = Math.random() * Math.PI*2;
  var azimuth:Number = Math.random() * Math.PI*2;

  var point:Number3D = new Number3D(
   radius * Math.sin(inclination) * Math.cos(azimuth),
   radius * Math.sin(inclination) * Math.sin(azimuth),
   radius * Math.cos(inclination)
  );

  return point;
 }

提前致谢！

Answer 1

听起来你可以限制倾向：

var inclination:Number = (Math.PI/6) + Math.random()*(2*Math.PI-2*Math.PI/6)

随意解决这些常量值，只需保留它们即可显示正常工作。

Answer 2

这就是我到目前为止所做的......这就是我想要的，限制了北极和南极。欢迎任何改进！

var point:Number3D = sphericalPoint(100, inclination, azimuth);

public static function sphericalPoint(r:Number, inc:Number, az:Number):Number3D
{
    var point:Number3D = new Number3D(
        r * Math.sin(inc) * Math.cos(az),
        r * Math.sin(inc) * Math.sin(az),
        r * Math.cos(inc)
    );

    //cheat and use a transform matrix
    var obj:Object3D = new Object3D();
    obj.rotationX = 90;

    point.rotate(point, obj.transform);

    return point;
}

//a number between 0 and 180
protected function get inclination():Number
{
    //default
    //_inclination = Math.random() * Math.PI;

    //number between 60 and 120
    _inclination = Math.random() * (Math.PI*5/6 - Math.PI/6) + Math.PI/6;

    return _inclination;
}

//a number between 0 and 360
protected function get azimuth():Number
{
    //revolve around the Y axis
    _azimuth = Math.random() * Math.PI*2;

    return _azimuth;
}

范围/限制内的球面坐标

2 个答案: