您好，我对glm lookAt函数的返回值有疑问。当我以 Debug模式执行时，此时
glm函数的 ... Result[3][2] = dot(f, eye); ...在矩阵的平移z位置中的值错误。值是-2，向我表明前向和眼向 vector 处于相反的位置。我的眼睛，中心 vector 和向上 vector 分别是eye(0,0,2)，center(0,0，-1)和up(0,1,0)。凸轮余弦 vector 为:f(0,0，-1)，s(1,0,0)和u(0,1,0)。用户查看的优势点是(0,0,0)。因此，正确的 View 矩阵应为:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
但我得到这个:
1 -0 -0
-0 1 -0 -0
0 0 0 1-2
0 0 0 0 1
我的代码是:
struct camera {
vec3 position = vec3(0.0f); // position of the camera
vec3 view_direction = vec3(0.0f); // forward vector (orientation)
vec3 side = vec3(0.0f); // right vector (side)
vec3 up = vec3(0.0f, 1.0f, 0.0f); // up vector
float speed = 0.1;
float yaw = 0.0f; // y-rotation
float cam_yaw_speed = 10.0f; // 10 degrees per second
float pitch = 0.0f; // x-rotation
float roll = 0.0f;
...
// calculate the orientation vector (forward)
vec3 getOrientation(vec3 vantage_point) {
// calc the difference and normalize the resulting vector
vec3 result = vantage_point - position;
result = normalize(result);
return result;
}
// calculate the right (side) vector of the camera, by given orientation(forward) and up vectors
mat4 look_at_point(vec3 vantage_point) {
view_direction = getOrientation(vantage_point);
// calculate the lookat matrix
return lookAt(position, position + view_direction, up);
}
};
我试图弄清楚如何解决这个问题，但我仍然不知道。有人能帮我吗？
我正在执行main_cam.look_at_point(vantage_point)函数的主要函数如下所示:
...
GLfloat points[] = {
0.0f, 0.5f, 0.0f,
0.5f, 0.0f, 0.0f,
-0.5f, 0.0f, 0.0f };
float speed = 1.0f; // move at 1 unit per second
float last_position = 0.0f;
// init camera
main_cam.position = vec3(0.0f, 0.0f, 2.0f); // don't start at zero, or will be too close
main_cam.speed = 1.0f; // 1 unit per second
main_cam.cam_yaw_speed = 10.0f; // 10 degrees per second
vec3 vantage_point = vec3(0.0f, 0.0f, 0.0f);
mat4 T = translate(mat4(1.0), main_cam.position);
//mat4 R = rotate(mat4(), -main_cam.yaw, vec3(0.0, 1.0, 0.0));
mat4 R = main_cam.look_at_point(vantage_point);
mat4 view_matrix = R * T;
// input variables
float near = 0.1f; // clipping plane
float far = 100.0f; // clipping plane
float fov = 67.0f * ONE_DEG_IN_RAD; // convert 67 degrees to radians
float aspect = (float)g_gl_width / (float)g_gl_height; // aspect ratio
mat4 proj_matrix = perspective(fov, aspect, near, far);
use_shader_program(shader_program);
set_uniform_matrix4fv(shader_program, "view", 1, GL_FALSE, &view_matrix[0][0]);
set_uniform_matrix4fv(shader_program, "proj", 1, GL_FALSE, &proj_matrix[0][0]);
...
用glm的旋转功能测试的三角形如右图所示。
Triangle shown with the rotate function of glm
最佳答案
我怀疑问题出在这里:
mat4 view_matrix = R * T; // <---
lookAt返回的矩阵已经完成翻译。尝试在三角形内部的(0,0,0)点上手动应用变换。
T会将其转换为(0,0,2)，但现在它与相机重合，因此R会将其发送回(0,0,0)。现在，您将得到射影除以零事故的除法。因此，通过
T删除乘法:mat4 view_matrix = R;
编辑:我想指出的是，从
view_direction计算vantage_point，然后将position + view_direction反馈回lookAt是实现目标的一种相当人为的方式。您在getOrientation函数中执行的操作是lookAt在内部已执行的操作。相反，您可以从view_direction的结果中获取lookAt:mat4 look_at_point(vec3 vantage_point) {
// calculate the lookat matrix
mat4 M = lookAt(position, vantage_point, up);
view_direction = -vec3(M[2][0], M[2][1], M[2][2]);
return M;
}
lookAt at all。