试图让鼠标拾取工作,但不确定我迷路了
Trying to get mouse picking to work, but not sure where I'm lost
我正在绘制一个 10x10 的正方形网格,深度为 0,并试图突出显示鼠标悬停的那个。我尝试按照此处的教程进行操作:http://antongerdelan.net/opengl/raycasting.html
但我不知道我是否做对了。最后我得到了一个矢量,但我不确定如何处理它。
这是方块的屏幕截图(不确定它有什么帮助..)
http://postimg.org/image/dau330qwt/2
/* Enable attribute index 1 as being used */
glEnableVertexAttribArray(1);
float camera_z = 50;
float camera_x = 0;
float camera_y = 0;
GLuint MatrixID = glGetUniformLocation(program, "MVP");
GLuint ColorID = glGetUniformLocation(program, "input_color");
int mouse_x;
int mouse_y;
while (1) {
int window_width;
int window_height;
SDL_GetWindowSize(win, &window_width, &window_height);
glm::mat4 Projection = glm::perspective(45.0f, ((float)window_width) / window_height, 0.1f, 100.0f);
// printf("Camera at %f %f\n", camera_x, camera_y);
glm::mat4 View = glm::lookAt(glm::vec3(camera_x,camera_y,camera_z), // camera position
glm::vec3(camera_x,camera_y,0), // looking at
glm::vec3(0,1,0)); // up
int map_width = map.width();
int map_height = map.height();
/* Make our background black */
glClearColor(0.0, 0.0, 0.0, 1.0);
glClear(GL_COLOR_BUFFER_BIT);
// go through my 10x10 map and
for (int i = 0; i < map_width; i++) {
for ( int j = 0; j < map_height; j++) {
glm::mat4 Model = glm::translate(glm::mat4(1.0f), glm::vec3(i, j, 0.0f));
glm::mat4 MVP = Projection * View * Model;
glm::vec3 color = random_color();
glUniformMatrix4fv(MatrixID, 1, GL_FALSE, &MVP[0][0]);
glUniform3fv(ColorID, 1, &color[0]);
glDrawArrays(GL_LINE_LOOP, 0, 4);
}
}
/* Swap our buffers to make our changes visible */
SDL_GL_SwapWindow(win);
// printf("Window dimensions %d x %d\n", window_width, window_height);
float normalized_mouse_x = (2.0f * mouse_x) / window_width - 1.0f;
float normalized_mouse_y = 1.0f - (2.0f * mouse_y) / window_height;
printf("Normalized mouse position %f x %f\n", normalized_mouse_x, normalized_mouse_y);
glm::vec3 normalized_mouse_vector = glm::vec3(normalized_mouse_x, normalized_mouse_y, 1.0f);
glm::vec4 ray_clip = glm::vec4 (normalized_mouse_vector.x, normalized_mouse_vector.y, -1.0, 1.0);
glm::vec4 ray_eye = glm::inverse(Projection) * ray_clip;
ray_eye = glm::vec4(ray_eye.xy(), -1.0, 0.0);
glm::vec3 ray_world = (glm::inverse(View) * ray_eye).xyz();
ray_world = glm::normalize(ray_world);
// this prints out values like: World ray: 0.000266, 0.000382, 1.000000
printf("World ray: %f, %f, %f\n", ray_world.x, ray_world.y, ray_world.z);
// l = -(camera_z / ray_world.z)
float l = -(camera_z / ray_world.z);
float mouse_world_x = camera_x + l * ray_world.x;
float mouse_world_y = camera_y + l * ray_world.y;
printf("mouse world %f, %f\n", mouse_world_x, mouse_world_y);
}
更新了 BDL 评论中的代码。我现在得到的输出是:
标准化鼠标位置 0.087500 x 0.145833
世界光线:0.065083、0.081353、499.000000
世界光线:0.000130、0.000163、1.000000
老鼠世界-0.006521, -0.008152
我希望 "mouse world" 行的数字在 1-10 范围内,而不是在 .00x 范围内,不过,根据上面显示 x 和 y 范围的正方形网格的屏幕截图从 0-10.
感谢观看。
从点 C(在本例中为相机位置)开始的给定光线 r 与具有 z=0 的 x/y 平面之间的交点可以计算如下:
C ... Camera position [cx,cy,cz]
r ... ray direction [rx,ry,rz]
We are searching for the point on the ray that has z=0
C + l*r = [x,y,0]
=>
cz + l*rz = 0
l * rz = -cz
l = -(cz / rz)
The xy-coordinates of the intersection are now:
x = cx + l * rx
y = cy + l * ry
剩下要做的是检查这个 (x,y) 坐标位于哪个矩形中。
我正在绘制一个 10x10 的正方形网格,深度为 0,并试图突出显示鼠标悬停的那个。我尝试按照此处的教程进行操作:http://antongerdelan.net/opengl/raycasting.html 但我不知道我是否做对了。最后我得到了一个矢量,但我不确定如何处理它。
这是方块的屏幕截图(不确定它有什么帮助..)
http://postimg.org/image/dau330qwt/2
/* Enable attribute index 1 as being used */
glEnableVertexAttribArray(1);
float camera_z = 50;
float camera_x = 0;
float camera_y = 0;
GLuint MatrixID = glGetUniformLocation(program, "MVP");
GLuint ColorID = glGetUniformLocation(program, "input_color");
int mouse_x;
int mouse_y;
while (1) {
int window_width;
int window_height;
SDL_GetWindowSize(win, &window_width, &window_height);
glm::mat4 Projection = glm::perspective(45.0f, ((float)window_width) / window_height, 0.1f, 100.0f);
// printf("Camera at %f %f\n", camera_x, camera_y);
glm::mat4 View = glm::lookAt(glm::vec3(camera_x,camera_y,camera_z), // camera position
glm::vec3(camera_x,camera_y,0), // looking at
glm::vec3(0,1,0)); // up
int map_width = map.width();
int map_height = map.height();
/* Make our background black */
glClearColor(0.0, 0.0, 0.0, 1.0);
glClear(GL_COLOR_BUFFER_BIT);
// go through my 10x10 map and
for (int i = 0; i < map_width; i++) {
for ( int j = 0; j < map_height; j++) {
glm::mat4 Model = glm::translate(glm::mat4(1.0f), glm::vec3(i, j, 0.0f));
glm::mat4 MVP = Projection * View * Model;
glm::vec3 color = random_color();
glUniformMatrix4fv(MatrixID, 1, GL_FALSE, &MVP[0][0]);
glUniform3fv(ColorID, 1, &color[0]);
glDrawArrays(GL_LINE_LOOP, 0, 4);
}
}
/* Swap our buffers to make our changes visible */
SDL_GL_SwapWindow(win);
// printf("Window dimensions %d x %d\n", window_width, window_height);
float normalized_mouse_x = (2.0f * mouse_x) / window_width - 1.0f;
float normalized_mouse_y = 1.0f - (2.0f * mouse_y) / window_height;
printf("Normalized mouse position %f x %f\n", normalized_mouse_x, normalized_mouse_y);
glm::vec3 normalized_mouse_vector = glm::vec3(normalized_mouse_x, normalized_mouse_y, 1.0f);
glm::vec4 ray_clip = glm::vec4 (normalized_mouse_vector.x, normalized_mouse_vector.y, -1.0, 1.0);
glm::vec4 ray_eye = glm::inverse(Projection) * ray_clip;
ray_eye = glm::vec4(ray_eye.xy(), -1.0, 0.0);
glm::vec3 ray_world = (glm::inverse(View) * ray_eye).xyz();
ray_world = glm::normalize(ray_world);
// this prints out values like: World ray: 0.000266, 0.000382, 1.000000
printf("World ray: %f, %f, %f\n", ray_world.x, ray_world.y, ray_world.z);
// l = -(camera_z / ray_world.z)
float l = -(camera_z / ray_world.z);
float mouse_world_x = camera_x + l * ray_world.x;
float mouse_world_y = camera_y + l * ray_world.y;
printf("mouse world %f, %f\n", mouse_world_x, mouse_world_y);
}
更新了 BDL 评论中的代码。我现在得到的输出是:
标准化鼠标位置 0.087500 x 0.145833 世界光线:0.065083、0.081353、499.000000 世界光线:0.000130、0.000163、1.000000 老鼠世界-0.006521, -0.008152
我希望 "mouse world" 行的数字在 1-10 范围内,而不是在 .00x 范围内,不过,根据上面显示 x 和 y 范围的正方形网格的屏幕截图从 0-10.
感谢观看。
从点 C(在本例中为相机位置)开始的给定光线 r 与具有 z=0 的 x/y 平面之间的交点可以计算如下:
C ... Camera position [cx,cy,cz]
r ... ray direction [rx,ry,rz]
We are searching for the point on the ray that has z=0
C + l*r = [x,y,0]
=>
cz + l*rz = 0
l * rz = -cz
l = -(cz / rz)
The xy-coordinates of the intersection are now:
x = cx + l * rx
y = cy + l * ry
剩下要做的是检查这个 (x,y) 坐标位于哪个矩形中。