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Homework 0: Environment construction, simple transformation
Given a point P=(2,1), first rotate the point 45 degrees counterclockwise around the origin, and then translate (1,2), and calculate the
coordinates of the transformed point (homogeneous coordinates are required for calculation).
#include<cmath>
#include<eigen3/Eigen/Core>
#include<eigen3/Eigen/Dense>
#include<iostream>
int main() {
Eigen::Vector3f Point3D(2.0f, 1.0f, 1.0f);
float theta = 45.0f*M_PI/180.0f;
Eigen::Matrix3f RotateMatrix;
RotateMatrix<<cos(theta), -sin(theta), 0, sin(theta), cos(theta), 0, 0, 0, 1;
Eigen::Matrix3f MoveMatrix;
MoveMatrix<<1, 0, 1, 0, 1, 2, 0, 0, 0;
std::cout<<MoveMatrix*RotateMatrix*Point3D;
return 0;
}
Homework 1: Rotation and Projection
Given three points v0(2.0,0.0,−2.0), v1(0.0,2.0,−2.0), v2(−2.0,0.0,−2.0) in three dimensions, you need to transform the coordinates of these three points into screen coordinates and draw the corresponding wireframe triangles on the screen.
Eigen::Matrix4f get_model_matrix(float rotation_angle)
{
Eigen::Matrix4f model = Eigen::Matrix4f::Identity();
rotation_angle = rotation_angle / 180.0 * MY_PI;
Eigen::Matrix4f translate;
translate << cos(rotation_angle), -1 * sin(rotation_angle), 0, 0,
sin(rotation_angle), cos(rotation_angle), 0, 0,
0, 0, 1, 0,
0, 0, 0, 1;
model = translate * model;
return model;
}
Eigen::Matrix4f get_projection_matrix(float eye_fov, float aspect_ratio,
float zNear, float zFar)
{
// eye_fov: 视野的大小
// aspect_ratio: 长宽比
Eigen::Matrix4f projection = Eigen::Matrix4f::Identity();
Eigen::Matrix4f zoomMatrix; // 缩放矩阵
zoomMatrix << zNear, 0, 0, 0,
0, zNear, 0, 0,
0, 0, zNear + zFar, (-1) * zNear * zFar,
0, 0, 1, 0;
float half_eye_angle = eye_fov / 2.0 / 180.0* MY_PI ;
float t = zNear * tan(half_eye_angle); // y轴的最大值
float r = t * aspect_ratio; // x轴的最大值
float l = (-1)*r; // x轴的最小值
float b = (-1)*t; // y轴的最小值
Eigen::Matrix4f moveMatrix; // 平移矩阵
moveMatrix << 1, 0, 0, -(l + r) / 2,
0, 1, 0, -(b + t) / 2,
0, 0, 1, -(zNear + zFar) / 2,
0, 0, 0, 1;
Eigen::Matrix4f zoomMatirx2; // 缩放到[-1,1]范围内
zoomMatirx2 << 2/(r - l), 0, 0, 0,
0, 2/(t - b), 0, 0,
0, 0, 2/(zFar - zNear) , 0,
0, 0, 0, 1;
projection = zoomMatirx2 * moveMatrix * zoomMatrix * projection;
return projection;
}
upside down problem
The problem that the triangle is upside down is caused by the difference in the coordinate axes. The origin in the course is in the lower left corner, while the origin of the opencv library is in the upper left corner. The setting in the course is to look towards the -Z axis, but the zNear and zFar set in the main function are 0.1 and 50 respectively, both of which are positive numbers.
This leads to the fact that if the calculation is still carried out in the previous way, the two axes of the y-axis and the z-axis will be reversed. So there are two changes when modifying the code:
- In the matrix of perspective transformation into orthogonal transformation, the value is positive, and here is the problem of correcting the Z axis.
- When calculating top, it is a negative value, and bottom is a positive value, correcting the problem of the y-axis.
Modify these sections:
float t = -zNear * tan(half_eye_angle); // y轴的最大值
...
zoomMatirx2 << 2/(r - l), 0, 0, 0,
0, 2/(t - b), 0, 0,
0, 0, 2/(zNear - zFar) , 0,
0, 0, 0, 1;
which correctly displays:
Homework 2: Triangles and Z-buffering
Requirements: rasterized triangles,
static bool insideTriangle(float x, float y, const Vector3f *_v) {
Eigen::Vector2f P(x, y);
Eigen::Vector2f A = _v[0].head(2);
Eigen::Vector2f B = _v[1].head(2);
Eigen::Vector2f C = _v[2].head(2);
Eigen::Vector2f AP = P - A;
Eigen::Vector2f BP = P - B;
Eigen::Vector2f CP = P - C;
Eigen::Vector2f AB = B - A;
Eigen::Vector2f BC = C - B;
Eigen::Vector2f CA = A - C;
float res1 = AP[0] * AB[1] - AP[1] * AB[0];
float res2 = BP[0] * BC[1] - BP[1] * BC[0];
float res3 = CP[0] * CA[1] - CP[1] * CA[0];
return (res1 > 0 && res2 > 0 && res3 > 0) || (res1 < 0 && res2 < 0 && res3 < 0);
}
void rst::rasterizer::rasterize_triangle(const Triangle &t) {
auto v = t.toVector4();
// 创建三角形的 2 维 bounding box
float min_x = std::min(v[0][0], std::min(v[1][0], v[2][0]));
float max_x = std::max(v[0][0], std::max(v[1][0], v[2][0]));
float min_y = std::min(v[0][1], std::min(v[1][1], v[2][1]));
float max_y = std::max(v[0][1], std::max(v[1][1], v[2][1]));
// 遍历此 bounding box 内的所有像素(使用其整数索引)。然后，使用像素中心的屏幕空间坐标来检查中心点是否在三角形内。
for (int i = static_cast<int>(floor(min_x)); i <= max_x + 1; i++) {
for (int j = static_cast<int>(floor(min_y)); j <= max_y + 1; j++) {
if (insideTriangle(i + 0.5, j + 0.5, t.v)) {
auto[alpha, beta, gamma] = computeBarycentric2D(i + 0.5, j + 0.5, t.v);
float w_reciprocal = 1.0 / (alpha / v[0].w() + beta / v[1].w() + gamma / v[2].w());
// 计算深度插值
float z_interpolated = alpha * v[0].z() / v[0].w() + beta * v[1].z() / v[1].w() + gamma * v[2].z() / v[2].w();
z_interpolated *= w_reciprocal;
// 更新深度缓冲数组
if (depth_buf[get_index(i, j)] > z_interpolated) {
depth_buf[get_index(i, j)] = z_interpolated;
Eigen::Vector3f color = t.getColor();
Eigen::Vector3f point;
point << i, j, z_interpolated;
set_pixel(point, color); // 设置当前像素的颜色
}
}
}
}
}
Display result: (If there is an upside-down and the depth display is incorrect, please refer to the analysis of the upside-down problem in homework 1)
