将对象旋转到面部点

Question

我想旋转一个物体面对一个我有点麻烦的点.

所以我开始使用一个基数为零的对象,并在y轴上对齐.

我想旋转它,使对象的顶部朝向目的地

到目前为止,我的过程是: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. 找到合适的矢量:R = A交叉D.
4. 找到向上矢量:U = D交叉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

这是代码表示.我不确定这究竟是什么问题我已经在纸上解决了问题,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,并指出我正确的方向,那将是伟大的.

    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);

Answer 1

这里的基本思想如下.

• 确定对象面向的方向: directionA
• 确定对象应该面对的方式: directionB
• 确定这些方向之间的角度: 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