将对象旋转到面部点

时间:2014-05-16 03:16:06

标签: java math 3d rotation

我想将一个物体旋转到一个我有点麻烦的地方。

所以我从一个基数为零的对象开始,并在y轴上对齐。

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我想旋转它以使对象的顶部朝向目的地

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到目前为止我的流程是: 给定轴A

  1. 找到我的位置和我的外观位置之间的距离:D
  2. 创建方向向量:V = D.normalize()
  3. 找到正确的向量:R = A cross D
  4. 找到向上矢量:U = D cross R
  5. 找到向上和方向之间的角度:ANGLE = acos((U点D)/(U.length * D.length))
  6. 按每个轴上的方向缩放的角度旋转
  7. 这是代码表示。我不确定这究竟是什么问题我已经在纸上进行了解决,据我所知,这种方法应该有效,但结果在绘制时完全不正确。如果有人看到任何瑕疵并且可以指出我正确的方向,那就太好了。

        Vector3 distance = new Vector3(from.x, from.y, from.z).sub(to.x, to.y, to.z);
        final Vector3 axis = new Vector3(0, 1, 0);
        final Vector3 direction = distance.clone().normalize();
    
        final Vector3 right = (axis.clone().cross(direction));
        final Vector3 up = (distance.clone().cross(right));
    
        float angle = (float) Math.acos((up.dot(direction)/ (up.length() * direction.length()))); 
        bondObject.rotateLocal(angle, direction.x , direction.y, direction.z);
    

1 个答案:

答案 0 :(得分:11)

这里的基本思路如下。

  • 确定对象面向的方式:directionA
  • 确定对象应面向的方式:directionB
  • 确定这些指示之间的角度:rotationAngle
  • 确定旋转轴:rotationAxis

这是修改后的代码。

Vector3 distance = new Vector3(from.x, from.y, from.z).sub(to.x, to.y, to.z);

if (distance.length() < DISTANCE_EPSILON)
{
    //exit - don't do any rotation
    //distance is too small for rotation to be numerically stable
}

//Don't actually need to call normalize for directionA - just doing it to indicate
//that this vector must be normalized.
final Vector3 directionA = new Vector3(0, 1, 0).normalize();
final Vector3 directionB = distance.clone().normalize();

float rotationAngle = (float)Math.acos(directionA.dot(directionB));

if (Math.abs(rotationAngle) < ANGLE_EPSILON)
{
    //exit - don't do any rotation
    //angle is too small for rotation to be numerically stable
}

final Vector3 rotationAxis = directionA.clone().cross(directionB).normalize();

//rotate object about rotationAxis by rotationAngle