我想我明白为什么调用glRotate(#,0,0,0)会导致被零除。旋转矢量a被归一化:a'= a / | a | = a / 0
这是glRotate唯一可能导致零除的情况吗?是的,我知道glRotate已被弃用。是的,我知道矩阵在OpenGL manual上。不,我不知道线性代数足以自信地回答矩阵中的问题。是的,我认为这会有所帮助。是的,我已经在#opengl中问了这个问题(你能说出来吗?)。不,我没有得到答案。
答案 0 :(得分:2)
我会说是的。我会说你对标准化步骤也是正确的。 OpenGL manual中显示的矩阵仅由乘法组成。并且将矢量相乘会产生相同的结果。当然,如果你得到(0,0,0)
的向量,那会很奇怪。 OpenGL在|x,y,z|=1
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1
中表示glRotate
(或OpenGL将规范化)。
因此,如果它不能正常化,你最终会得到一个非常空的矩阵:
{{1}}
哪会以最奇怪的方式破坏你的对象。因此,请勿使用零向量调用此函数。如果您愿意,请告诉我原因。
如果某些简单的{{1}}过于复杂,我建议您使用像manual这样的库进行矩阵计算。
答案 1 :(得分:2)
为什么在check for that?时可以除以零:
/**
* Generate a 4x4 transformation matrix from glRotate parameters, and
* post-multiply the input matrix by it.
*
* \author
* This function was contributed by Erich Boleyn (erich@uruk.org).
* Optimizations contributed by Rudolf Opalla (rudi@khm.de).
*/
void
_math_matrix_rotate( GLmatrix *mat,
GLfloat angle, GLfloat x, GLfloat y, GLfloat z )
{
GLfloat xx, yy, zz, xy, yz, zx, xs, ys, zs, one_c, s, c;
GLfloat m[16];
GLboolean optimized;
s = (GLfloat) sin( angle * DEG2RAD );
c = (GLfloat) cos( angle * DEG2RAD );
memcpy(m, Identity, sizeof(GLfloat)*16);
optimized = GL_FALSE;
#define M(row,col) m[col*4+row]
if (x == 0.0F) {
if (y == 0.0F) {
if (z != 0.0F) {
optimized = GL_TRUE;
/* rotate only around z-axis */
M(0,0) = c;
M(1,1) = c;
if (z < 0.0F) {
M(0,1) = s;
M(1,0) = -s;
}
else {
M(0,1) = -s;
M(1,0) = s;
}
}
}
else if (z == 0.0F) {
optimized = GL_TRUE;
/* rotate only around y-axis */
M(0,0) = c;
M(2,2) = c;
if (y < 0.0F) {
M(0,2) = -s;
M(2,0) = s;
}
else {
M(0,2) = s;
M(2,0) = -s;
}
}
}
else if (y == 0.0F) {
if (z == 0.0F) {
optimized = GL_TRUE;
/* rotate only around x-axis */
M(1,1) = c;
M(2,2) = c;
if (x < 0.0F) {
M(1,2) = s;
M(2,1) = -s;
}
else {
M(1,2) = -s;
M(2,1) = s;
}
}
}
if (!optimized) {
const GLfloat mag = SQRTF(x * x + y * y + z * z);
if (mag <= 1.0e-4) {
/* no rotation, leave mat as-is */
return;
}
x /= mag;
y /= mag;
z /= mag;
/*
* Arbitrary axis rotation matrix.
*
* This is composed of 5 matrices, Rz, Ry, T, Ry', Rz', multiplied
* like so: Rz * Ry * T * Ry' * Rz'. T is the final rotation
* (which is about the X-axis), and the two composite transforms
* Ry' * Rz' and Rz * Ry are (respectively) the rotations necessary
* from the arbitrary axis to the X-axis then back. They are
* all elementary rotations.
*
* Rz' is a rotation about the Z-axis, to bring the axis vector
* into the x-z plane. Then Ry' is applied, rotating about the
* Y-axis to bring the axis vector parallel with the X-axis. The
* rotation about the X-axis is then performed. Ry and Rz are
* simply the respective inverse transforms to bring the arbitrary
* axis back to its original orientation. The first transforms
* Rz' and Ry' are considered inverses, since the data from the
* arbitrary axis gives you info on how to get to it, not how
* to get away from it, and an inverse must be applied.
*
* The basic calculation used is to recognize that the arbitrary
* axis vector (x, y, z), since it is of unit length, actually
* represents the sines and cosines of the angles to rotate the
* X-axis to the same orientation, with theta being the angle about
* Z and phi the angle about Y (in the order described above)
* as follows:
*
* cos ( theta ) = x / sqrt ( 1 - z^2 )
* sin ( theta ) = y / sqrt ( 1 - z^2 )
*
* cos ( phi ) = sqrt ( 1 - z^2 )
* sin ( phi ) = z
*
* Note that cos ( phi ) can further be inserted to the above
* formulas:
*
* cos ( theta ) = x / cos ( phi )
* sin ( theta ) = y / sin ( phi )
*
* ...etc. Because of those relations and the standard trigonometric
* relations, it is pssible to reduce the transforms down to what
* is used below. It may be that any primary axis chosen will give the
* same results (modulo a sign convention) using thie method.
*
* Particularly nice is to notice that all divisions that might
* have caused trouble when parallel to certain planes or
* axis go away with care paid to reducing the expressions.
* After checking, it does perform correctly under all cases, since
* in all the cases of division where the denominator would have
* been zero, the numerator would have been zero as well, giving
* the expected result.
*/
xx = x * x;
yy = y * y;
zz = z * z;
xy = x * y;
yz = y * z;
zx = z * x;
xs = x * s;
ys = y * s;
zs = z * s;
one_c = 1.0F - c;
/* We already hold the identity-matrix so we can skip some statements */
M(0,0) = (one_c * xx) + c;
M(0,1) = (one_c * xy) - zs;
M(0,2) = (one_c * zx) + ys;
/* M(0,3) = 0.0F; */
M(1,0) = (one_c * xy) + zs;
M(1,1) = (one_c * yy) + c;
M(1,2) = (one_c * yz) - xs;
/* M(1,3) = 0.0F; */
M(2,0) = (one_c * zx) - ys;
M(2,1) = (one_c * yz) + xs;
M(2,2) = (one_c * zz) + c;
/* M(2,3) = 0.0F; */
/*
M(3,0) = 0.0F;
M(3,1) = 0.0F;
M(3,2) = 0.0F;
M(3,3) = 1.0F;
*/
}
#undef M
matrix_multf( mat, m, MAT_FLAG_ROTATION );
}