两个三维向量之间的角度

Question

我有一系列要旋转的顶点（粉红色），以便顶点图案的一条边与三角形的边（白色）匹配

为此，我首先创建两个向量来表示边：floretAB和triangelab（绿色）。然后我找到两个顶点的叉积，得到一个轴，我可以围绕它旋转顶点（红色）

然后，我得到两个向量之间的角度，并用它与旋转轴一起创建一个四元数。最后，我围绕四元数旋转所有顶点

旋转前

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什么样的旋转应该产生

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但是，尽管顶点正确地围绕四元数旋转，但角度并不正确，如下图所示：

这是我用来得到两个向量之间的角度的代码。我不明白我做错了什么：

double[] cross = new double[3];
crossProduct(floretAB.mX, floretAB.mY, floretAB.mZ, triangleAB.mX, triangleAB.mY, triangleAB.mZ, cross);
double dot = dotProduct(floretAB.mX, floretAB.mY, floretAB.mZ, triangleAB.mX, triangleAB.mY, triangleAB.mZ);
double crossMag = Math.sqrt(cross[0]*cross[0] + cross[1]*cross[1] + cross[2]*cross[2]);
double angle = Math.atan2(crossMag, dot);

public static double dotProduct(double vector1X,double vector1Y,double vector1Z,double vector2X,double vector2Y,double vector2Z){

    return vector1X*vector2X + vector1Y*vector2Y + vector1Z*vector2Z;

}

public static void crossProduct(double vector1X,double vector1Y,double vector1Z,double vector2X,double vector2Y,double vector2Z, double[] outputArray){

    outputArray[0] = vector1Y*vector2Z - vector1Z*vector2Y;     
    outputArray[1] = vector1Z*vector2X - vector1X*vector2Z;
    outputArray[2] = vector1X*vector2Y - vector1Y*vector2X;

}

如果您能帮上忙，我将不胜感激，因为这真的让我心烦

谢谢，詹姆斯

编辑：以下是代码的其余部分：

        // get floret p1,p2 vector
    // get triangle p1,p2 vector
    Vector3D floretAB = new Vector3D(florets3D[0], florets3D[7]);
    // get triangle p1,p2 vector
    Vector3D triangleAB = new Vector3D(triangle[0], triangle[1]);

    // get rotation axis (cross) and angle (dot)

    /*
    double[] cross = new double[3];
    crossProduct(floretAB.mX, floretAB.mY, floretAB.mZ, triangleAB.mX, triangleAB.mY, triangleAB.mZ, cross);
    double dotMag = floretAB.getMagnitude() * triangleAB.getMagnitude();
    double dot = dotProduct(floretAB.mX, floretAB.mY, floretAB.mZ, triangleAB.mX, triangleAB.mY, triangleAB.mZ) / dotMag;
    double angle = Math.acos(dot);
    */

    double[] cross = new double[3];
    crossProduct(floretAB.mX, floretAB.mY, floretAB.mZ, triangleAB.mX, triangleAB.mY, triangleAB.mZ, cross);
    double dot = dotProduct(floretAB.mX, floretAB.mY, floretAB.mZ, triangleAB.mX, triangleAB.mY, triangleAB.mZ);
    double crossMag = Math.sqrt(cross[0]*cross[0] + cross[1]*cross[1] + cross[2]*cross[2]);
    double angle = Math.atan2(crossMag, dot);

    // rotate floret so p1,p2 vector matches with triangle p1,p2 vector     
    double[] newVerts = new double[3];
    Quaternion quat = new Quaternion(cross[0], cross[1], cross[2], angle);
    for(int i = 0;i<numfloretVerts;i++){
        Vertex3D vert = florets3D[i];
        quat.RotateVector(vert.getmX(), vert.getmY(), vert.getmZ(), newVerts);
        vert.setmX(newVerts[0]);
        vert.setmY(newVerts[1]);
        vert.setmZ(newVerts[2]);
    }

_

public class Vector3D {

public double mX;
public double mY;
public double mZ;

public Vertex3D point;

/**
 * Constructs a vector from two points. The new vector is normalised
 * 
 * @param point1
 * @param point2
 */
public Vector3D(Vertex3D point1, Vertex3D point2){
    mX = point2.getmX() - point1.getmX();
    mY = point2.getmY() - point1.getmY();
    mZ = point2.getmZ() - point1.getmZ();
    normalise();
    point = point1;
}

/**
 * Normalises the vector
 */
public void normalise(){
    double magnitude = Math.sqrt(mX*mX + mY*mY + mZ*mZ);
    if(magnitude!=0){
        mX /= magnitude;
        mY /= magnitude;
        mZ /= magnitude;
    }
}

/**
 * 
 * @return the magnitude of the vector
 */
public double getMagnitude(){
    return Math.sqrt(mX*mX + mY*mY + mZ*mZ);
}

}

_

public class Quaternion {

private static final double TOLERANCE = 0.00001f;

double w;
double x;
double y;
double z;

public Quaternion(double axisX, double axisY, double axisZ, double angleInRadians){
    setAxisAngle(axisX, axisY, axisZ, angleInRadians);      
}

public void Normalise(){

    // Don't normalize if we don't have to
    double mag2 = w * w + x * x + y * y + z * z;
    if (Math.abs(mag2) > TOLERANCE && Math.abs(mag2 - 1.0f) > TOLERANCE) {
        double mag = (double) Math.sqrt(mag2);
        w /= mag;
        x /= mag;
        y /= mag;
        z /= mag;
    }

}

public void getConjugate(double[] outputArray){

    outputArray[0] = w;
    outputArray[1] = -x;
    outputArray[2] = -y;
    outputArray[3] = -z;

}

public void Multiply(double[] aq, double[] rq, double[] outputArray){

    outputArray[0] = aq[0] * rq[0] - aq[1] * rq[1] - aq[2] * rq[2] - aq[3] * rq[3];
    outputArray[1] = aq[0] * rq[1] + aq[1] * rq[0] + aq[2] * rq[3] - aq[3] * rq[2];
    outputArray[2] = aq[0] * rq[2] + aq[2] * rq[0] + aq[3] * rq[1] - aq[1] * rq[3];
    outputArray[3] = aq[0] * rq[3] + aq[3] * rq[0] + aq[1] * rq[2] - aq[2] * rq[1];

}

private double[] vecQuat = new double[4];
private double[] resQuat = new double[4];
private double[] thisQuat = new double[4];

private double[] conj = new double[4];

/**
 * Rotates a vector (or point) around this axis-angle
 * 
 * @param vectorX the x component of the vector (or point)
 * @param vectorY the y component of the vector (or point)
 * @param vectorZ the z component of the vector (or point)
 * @param outputArray the array in which the results will be stored
 */
public void RotateVector(double vectorX, double vectorY, double vectorZ, double[] outputArray){

    vecQuat[0] = 0.0f;
    vecQuat[1] = vectorX;
    vecQuat[2] = vectorY;
    vecQuat[3] = vectorZ;

    thisQuat[0] = w;
    thisQuat[1] = x;
    thisQuat[2] = y;
    thisQuat[3] = z;

    getConjugate(conj);
    Multiply(vecQuat,conj,resQuat);
    Multiply(thisQuat,resQuat,vecQuat);

    outputArray[0] = vecQuat[1];
    outputArray[1] = vecQuat[2];
    outputArray[2] = vecQuat[3];

}

/**
 * set Quaternion by providing axis-angle form
 */
public void setAxisAngle(double axisX, double axisY, double axisZ, double angleInRadians){
    w = (double) Math.cos( angleInRadians/2);
    x = (double) (axisX * Math.sin( angleInRadians/2 ));
    y = (double) (axisY * Math.sin( angleInRadians/2 ));
    z = (double) (axisZ * Math.sin( angleInRadians/2 ));

    Normalise();
}
}

Answer 1

我认为你的数学太复杂了

给定两个单位向量（你确实说过它们是归一化的），那么叉积的大小等于sin(theta)。应该不需要调用点积或atan2

在创建四元数之前，可能还需要对叉积向量结果进行归一化-这取决于new Quaternion(x, y, z, theta)的实现以及是否需要对[x, y, z]进行归一化

Answer 2

所述答案对于实数是正确的，但在使用浮点数计算时，可能会在某些角度附近失去准确性。对于角度接近零或PI的arcos（），以及接近PI/2和–PI/2的arcsin（），可能会丢失一半的有效数字。假设输入向量为单位长度，则在从零到π的整个范围内（包括零到π的整个范围内）更稳健且仅承受少量均匀舍入误差的方法为：

public double AngleBetween(Vector3D a, Vector3D b)
{
    return 2.0d * Math.atan((a-b).Length/(a+b).Length);
}

注意，这给出了两个向量之间的无定向角度。关于这一点和卡汉的参考资料可在以下网址找到：http://www.cs.berkeley.edu/~wkahan/MathH110/Cross.pdf

Answer 3

我认为问题在于你用错误的方式评估角度。 如果我正确理解了你想要实现的目标，那么你需要两条绿线之间的角度。使用以下定义正确计算两条绿线之间的点积：

(a, b) = a1*b1 + a2*b2 + a3*b3.

但dot产品也可以这样评估：

(a, b) = |a|*|b|*cos(theta)

所以你可以计算cos（θ）-两条绿线之间夹角的余弦-如下所示：

cos(theta) = (a1*b1 + a2*b2 + a3*b3) / (|a|*|b|)

但我会使用另一种方法。首先，我将规范化这两个向量（即，将它们转换为unit-vectors）。您可以通过将每个向量的分量除以向量的长度（sqrt（x1*x1+y1*y1+z1*z1））来完成此操作，然后您将得到以下结果：

(aa, bb) = cos(theta)

其中aa为标准化a，bb为标准化b

我希望这有帮助