我正在尝试确定 UIView 的当前 3D 旋转。我知道要获得 2D 旋转是：
CGFloat radians = atan2f(view.transform.b, view.transform.a);
不确定如何获得当前的 3D 旋转。我希望能够自动反转动画并保留其起点。对象的旋转可能受到用户输入的先前旋转的影响，因此我不能假设它的起始值。我将最后一个用户定义的角度存储在对象的属性上，我认为这在实践中很好，但我很想知道如何获取信息。我对此进行了很多研究，但找不到答案。
【问题讨论】：
标签： ios objective-c core-animation quartz-graphics
在一般情况下，4x4 矩阵可能包含透视、（非均匀）缩放、（非均匀）剪切、旋转（围绕可能不是轴对齐的向量）和平移。如果要在一般情况下提取旋转，则需要将矩阵分解为这些“简单”变换中的每一个的分量。
Graphics Gems II §VII.1 描述了一种算法。附录二提供了算法的代码。代码可以在online here找到。头文件是here。您还需要一些支持文件，例如“C 实用程序”，所以check this link too。
这本书的前言声明代码在公共领域，所以我在下面复制它。
/* unmatrix.c - given a 4x4 matrix, decompose it into standard operations.
*
* Author: Spencer W. Thomas
* University of Michigan
*/
#include <math.h>
#include "GraphicsGems.h"
#include "unmatrix.h"
/* unmatrix - Decompose a non-degenerate 4x4 transformation matrix into
* the sequence of transformations that produced it.
* [Sx][Sy][Sz][Shearx/y][Sx/z][Sz/y][Rx][Ry][Rz][Tx][Ty][Tz][P(x,y,z,w)]
*
* The coefficient of each transformation is returned in the corresponding
* element of the vector tran.
*
* Returns 1 upon success, 0 if the matrix is singular.
*/
int
unmatrix( mat, tran )
Matrix4 *mat;
double tran[16];
{
register int i, j;
Matrix4 locmat;
Matrix4 pmat, invpmat, tinvpmat;
/* Vector4 type and functions need to be added to the common set. */
Vector4 prhs, psol;
Point3 row[3], pdum3;
locmat = *mat;
/* Normalize the matrix. */
if ( locmat.element[3][3] == 0 )
return 0;
for ( i=0; i<4;i++ )
for ( j=0; j<4; j++ )
locmat.element[i][j] /= locmat.element[3][3];
/* pmat is used to solve for perspective, but it also provides
* an easy way to test for singularity of the upper 3x3 component.
*/
pmat = locmat;
for ( i=0; i<3; i++ )
pmat.element[i][3] = 0;
pmat.element[3][3] = 1;
if ( det4x4(&pmat) == 0.0 )
return 0;
/* First, isolate perspective. This is the messiest. */
if ( locmat.element[0][3] != 0 || locmat.element[1][3] != 0 ||
locmat.element[2][3] != 0 ) {
/* prhs is the right hand side of the equation. */
prhs.x = locmat.element[0][3];
prhs.y = locmat.element[1][3];
prhs.z = locmat.element[2][3];
prhs.w = locmat.element[3][3];
/* Solve the equation by inverting pmat and multiplying
* prhs by the inverse. (This is the easiest way, not
* necessarily the best.)
* inverse function (and det4x4, above) from the Matrix
* Inversion gem in the first volume.
*/
inverse( &pmat, &invpmat );
TransposeMatrix4( &invpmat, &tinvpmat );
V4MulPointByMatrix(&prhs, &tinvpmat, &psol);
/* Stuff the answer away. */
tran[U_PERSPX] = psol.x;
tran[U_PERSPY] = psol.y;
tran[U_PERSPZ] = psol.z;
tran[U_PERSPW] = psol.w;
/* Clear the perspective partition. */
locmat.element[0][3] = locmat.element[1][3] =
locmat.element[2][3] = 0;
locmat.element[3][3] = 1;
} else /* No perspective. */
tran[U_PERSPX] = tran[U_PERSPY] = tran[U_PERSPZ] =
tran[U_PERSPW] = 0;
/* Next take care of translation (easy). */
for ( i=0; i<3; i++ ) {
tran[U_TRANSX + i] = locmat.element[3][i];
locmat.element[3][i] = 0;
}
/* Now get scale and shear. */
for ( i=0; i<3; i++ ) {
row[i].x = locmat.element[i][0];
row[i].y = locmat.element[i][1];
row[i].z = locmat.element[i][2];
}
/* Compute X scale factor and normalize first row. */
tran[U_SCALEX] = V3Length(&row[0]);
row[0] = *V3Scale(&row[0], 1.0);
/* Compute XY shear factor and make 2nd row orthogonal to 1st. */
tran[U_SHEARXY] = V3Dot(&row[0], &row[1]);
(void)V3Combine(&row[1], &row[0], &row[1], 1.0, -tran[U_SHEARXY]);
/* Now, compute Y scale and normalize 2nd row. */
tran[U_SCALEY] = V3Length(&row[1]);
V3Scale(&row[1], 1.0);
tran[U_SHEARXY] /= tran[U_SCALEY];
/* Compute XZ and YZ shears, orthogonalize 3rd row. */
tran[U_SHEARXZ] = V3Dot(&row[0], &row[2]);
(void)V3Combine(&row[2], &row[0], &row[2], 1.0, -tran[U_SHEARXZ]);
tran[U_SHEARYZ] = V3Dot(&row[1], &row[2]);
(void)V3Combine(&row[2], &row[1], &row[2], 1.0, -tran[U_SHEARYZ]);
/* Next, get Z scale and normalize 3rd row. */
tran[U_SCALEZ] = V3Length(&row[2]);
V3Scale(&row[2], 1.0);
tran[U_SHEARXZ] /= tran[U_SCALEZ];
tran[U_SHEARYZ] /= tran[U_SCALEZ];
/* At this point, the matrix (in rows[]) is orthonormal.
* Check for a coordinate system flip. If the determinant
* is -1, then negate the matrix and the scaling factors.
*/
if ( V3Dot( &row[0], V3Cross( &row[1], &row[2], &pdum3) ) < 0 )
for ( i = 0; i < 3; i++ ) {
tran[U_SCALEX+i] *= -1;
row[i].x *= -1;
row[i].y *= -1;
row[i].z *= -1;
}
/* Now, get the rotations out, as described in the gem. */
tran[U_ROTATEY] = asin(-row[0].z);
if ( cos(tran[U_ROTATEY]) != 0 ) {
tran[U_ROTATEX] = atan2(row[1].z, row[2].z);
tran[U_ROTATEZ] = atan2(row[0].y, row[0].x);
} else {
tran[U_ROTATEX] = atan2(-row[2].x, row[1].y);
tran[U_ROTATEZ] = 0;
}
/* All done! */
return 1;
}
/* transpose rotation portion of matrix a, return b */
Matrix4 *TransposeMatrix4(a, b)
Matrix4 *a, *b;
{
int i, j;
for (i=0; i<4; i++)
for (j=0; j<4; j++)
b->element[i][j] = a->element[j][i];
return(b);
}
/* multiply a hom. point by a matrix and return the transformed point */
Vector4 *V4MulPointByMatrix(pin, m, pout)
Vector4 *pin, *pout;
Matrix4 *m;
{
pout->x = (pin->x * m->element[0][0]) + (pin->y * m->element[1][0]) +
(pin->z * m->element[2][0]) + (pin->w * m->element[3][0]);
pout->y = (pin->x * m->element[0][1]) + (pin->y * m->element[1][1]) +
(pin->z * m->element[2][1]) + (pin->w * m->element[3][1]);
pout->z = (pin->x * m->element[0][2]) + (pin->y * m->element[1][2]) +
(pin->z * m->element[2][2]) + (pin->w * m->element[3][2]);
pout->w = (pin->x * m->element[0][3]) + (pin->y * m->element[1][3]) +
(pin->z * m->element[2][3]) + (pin->w * m->element[3][3]);
return(pout);
}
Graphics Gems 代码生成欧拉角。有时，您需要 3D 旋转的不同表示。特别是，四元数通常更方便。您可以在TransformationMatrix.cpp 的decompose 函数中找到此分解的WebCore 实现。它为旋转产生一个四元数。
【讨论】：