iOS Objective-C: 确定当前UIView的3D旋转
iOS Objective-C: Determine current 3D rotation of a UIView
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先前通过用户输入进行的旋转的影响，因此我无法假设其起始值。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。我对此做了很多研究，但找不到答案。
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（围绕可能不与轴对齐的向量）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每个这些“简单”变换的分量。
Graphics Gems II §VII.1 describes an algorithm for this. Appendix II provides code for the algorithm. The code can be found online here. The header file is here. You'll also need some supporting files like the “C utilities” so check this link too.
这本书的序言声明代码在 public 域中，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。有时您需要不同的 3D 旋转表示。特别是，四元数通常更方便。您可以在 TransformationMatrix.cpp 中的 decompose 函数中找到此分解的 WebCore 实现。它为旋转生成一个四元数。
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先前通过用户输入进行的旋转的影响，因此我无法假设其起始值。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。我对此做了很多研究，但找不到答案。
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（围绕可能不与轴对齐的向量）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每个这些“简单”变换的分量。
Graphics Gems II §VII.1 describes an algorithm for this. Appendix II provides code for the algorithm. The code can be found online here. The header file is here. You'll also need some supporting files like the “C utilities” so check this link too.
这本书的序言声明代码在 public 域中，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。有时您需要不同的 3D 旋转表示。特别是，四元数通常更方便。您可以在 TransformationMatrix.cpp 中的 decompose 函数中找到此分解的 WebCore 实现。它为旋转生成一个四元数。