我想要旋转一系列顶点(粉红色),以使顶点图案的一条边与三角形边缘(白色)匹配。
为此,我首先创建两个向量来表示边:floretAB和triangleAB(绿色)。然后我找到两者的交叉积来得到一个轴,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,然后用旋转轴创建一个四元数。最后,我旋转四元数周围的所有顶点。
轮换前
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应该产生什么轮播
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但是,尽管顶点在四元数周围正确旋转,但角度不正确,如下所示:
这是我用来获取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;
}
任何对此的帮助都会非常感激,因为它真的让我烦恼。
谢谢,詹姆斯
编辑:以下是其余代码:
// 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();
}
}
答案 0 :(得分:1)
我认为问题在于你以错误的方式评估角度。 如果我理解你想要实现什么,那么你需要2条绿线之间的角度。您可以使用以下定义正确评估2条绿线之间的点积:
(a, b) = a1*b1 + a2*b2 + a3*b3.
但是dot产品也可以这样评估:
(a, b) = |a|*|b|*cos(theta)
所以你可以评估cos(theta) - 两条绿线之间角度的余弦 - 就像这样:
cos(theta) = (a1*b1 + a2*b2 + a3*b3) / (|a|*|b|)
但我会用另一种方法。我首先将两个向量归一化(即将它们转换为unit-vectors)。你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))然后你将得到以下结果:
(aa, bb) = cos(theta)
其中aa被归一化a和bb被归一化b。
我希望这会有所帮助。
答案 1 :(得分:1)
我认为你的数学过于复杂。
给定两个单位向量(你确实说它们被标准化),那么叉积的大小等于sin(theta)
。不应该需要调用点积或atan2
。
在创建四元数之前,您可能还需要规范化叉积矢量结果 - 这取决于您new Quaternion(x, y, z, theta)
的实现以及是否需要[x, y, z]
进行规范化。
答案 2 :(得分:1)
所述答案对于实数是正确的,但在使用浮点数计算时,在某些角度附近可能会失去准确性。对于arcos(),当角度接近零或PI时,对于arcsin()接近pi / 2和-pi / 2,有效数字的一半可能会丢失。假设输入向量是单位长度,假设输入向量是单位长度,那么一种方法更稳健并且在整个范围内均匀地经历一些舍入误差,并且包括零和包括PI,假设输入向量是单位长度:
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。可以在http://www.cs.berkeley.edu/~wkahan/MathH110/Cross.pdf
找到此参考并归因于Kahan