如何修复光线投射器中的扭曲透视?

问题描述

我正在使用 SDL 的 C API 编写光线投射器。我花了数周时间试图修复臭名昭著的鱼眼效应,但无济于事。根据 this 来源,我可以将计算出的距离乘以一半 FOV 的余弦值来修复它。这对我不起作用。尽管如此,我的代码中仍然有余弦校正。

这里有两张图片展示了失真:

Image 1

Image 2

我认为我的代码一个核心问题可能是我的角度增量是恒定的,而增量应该更小,因为我更接近屏幕边界。不幸的是,我不知道如何开始实施。

如果可能的话,有人可以看看我的代码并给我一个关于如何去除鱼眼的提示吗?要向任何方向移动,请使用箭头键。使用“a”和“s”键分别向左和向右转。

这就是我的编译方式:clang `pkg-config --cflags --libs sdl2` raycaster.c

#include <SDL2/SDL.h>
#include <math.h>

typedef struct {
    float x,y,prev_x,prev_y,angle,fov;
} Player;

enum {
    map_width = 12,map_height = 15,screen_width = 800,screen_height = 500
};

const float
    move_speed_decr = 0.08,angle_turn = 2.0,ray_theta_step = 0.4,ray_dist_step = 0.8,darkening = 1.8,width_ratio = (float) screen_width / map_width,height_ratio = (float) screen_height / map_height;

const unsigned char map[map_height][map_width] = {
    {1,1,1},{1,4,3,2,1}
};

SDL_Window* window;
SDL_Renderer* renderer;

float to_radians(float degrees) {
    return degrees * M_PI / 180;
}

float distance(float x0,float y0,float x1,float y1) {
    return sqrt(((x1 - x0) * (x1 - x0)) + ((y1 - y0) * (y1 - y0)));
}

void shade(int* color,int darkener) {
    int darkened = *color - darkener;
    *color = darkened < 0 ? 0 : darkened;
}

void draw_rectangle(SDL_Rect rectangle,int r,int g,int b) {
    SDL_SetRenderDrawColor(renderer,r,g,b,SDL_ALPHA_OPAQUE);
    SDL_RenderFillRect(renderer,&rectangle);
    SDL_RenderDrawRect(renderer,&rectangle);
}

void raycast(Player player) {
    float relative_x = player.x * width_ratio;
    float relative_y = player.y * height_ratio;
    float half_fov = player.fov / 2;
    float width_fov_ratio = (screen_width / player.fov) / 2;
    float distort_adjust = cos(to_radians(half_fov));

    // the core of my problem may be in the constant increment of my angle
    for (float theta = player.angle - half_fov,screen_x = 0;
        theta < player.angle + half_fov;
        theta += ray_theta_step,screen_x += width_fov_ratio) {

        float radian_theta = to_radians(theta);
        float cos_theta = cos(radian_theta),sin_theta = sin(radian_theta);

        float d = 0,new_x,new_y;
        while (d += ray_dist_step) {
            new_x = cos_theta * d + relative_x;
            new_y = sin_theta * d + relative_y;

            int map_x = new_x / width_ratio,map_y = new_y / height_ratio;
            int map_point = map[map_y][map_x];

            if (map_point) {
                int dist_wall = distance(relative_x,relative_y,new_y) * distort_adjust;
                int twice_dist_wall = 2 * dist_wall;
                if (twice_dist_wall >= screen_height) break;
                else if (map_point) { // succeeds when a wall is present
                    int r,b;
                    switch (map_point) {
                        case 1: r = 255,g = 255,b = 0; break;
                        case 2: r = 0,g = 128,b = 128; break;
                        case 3: r = 255,g = 165,b = 0; break;
                        case 4: r = 255,g = 0,b = 0; break;
                    }

                    int color_decr = dist_wall / darkening;
                    shade(&r,color_decr);
                    shade(&g,color_decr);
                    shade(&b,color_decr);

                    SDL_Rect vertical_line = {
                        screen_x,dist_wall,width_fov_ratio + 1,screen_height - twice_dist_wall
                    };

                    draw_rectangle(vertical_line,b);
                    break;
                }
            }
        }
    }
}

void handle_input(const Uint8* keys,Player* player) {
    SDL_Event event;

    while (SDL_PollEvent(&event)) {
        if (event.type == SDL_QUIT) {
            SDL_DestroyWindow(window);
            SDL_DestroyRenderer(renderer);
            exit(0);
        }

        else if (event.type == SDL_KEYDOWN) {
            float radian_theta = to_radians(player -> angle);
            float move_x = cos(radian_theta) * move_speed_decr,move_y = sin(radian_theta) * move_speed_decr;

            // handle arrow keys
            if (keys[SDL_SCANCODE_UP]) player -> x += move_x,player -> y += move_y;
            if (keys[SDL_SCANCODE_DOWN]) player -> x -= move_x,player -> y -= move_y;
            if (keys[SDL_SCANCODE_LEFT]) player -> x += move_y,player -> y -= move_x;
            if (keys[SDL_SCANCODE_RIGHT]) player -> x -= move_y,player -> y += move_x;

            // handle 'a' and 's' for angle changes
            if (keys[SDL_SCANCODE_A]) player -> angle -= angle_turn;
            if (keys[SDL_SCANCODE_S]) player -> angle += angle_turn;

            // safeguards for invalid positions and angles
            if (player -> x < 0) player -> x = 0;
            else if (player -> x > screen_width) player -> x = screen_width;

            if (player -> y < 0) player -> y = 0;
            else if (player -> y > screen_height) player -> y = screen_height;

            // move the player to their prevIoUs coordinate if they're in a wall
            if (map[(int) player -> y][(int) player -> x])
                player -> y = player -> prev_y,player -> x = player -> prev_x;

            if (player -> angle > 360) player -> angle = 0;
            else if (player -> angle < 0) player -> angle = 360;

            player -> prev_y = player -> y,player -> prev_x = player -> x;
        }
    }
}

int main() {
    SDL_CreateWindowAndRenderer(screen_width,screen_height,&window,&renderer);
    SDL_SetwindowTitle(window,"Raycaster");    

    Player player = {5,5,60};
    SDL_Rect the_ceiling = {0,screen_width,screen_height / 2};
    SDL_Rect the_floor = {0,screen_height / 2,screen_height};
    const Uint8* keys = SDL_GetKeyboardState(NULL);

    while (1) {
        handle_input(keys,&player);

        draw_rectangle(the_ceiling,96,96);
        draw_rectangle(the_floor,210,180,140);

        raycast(player);

        SDL_RenderPresent(renderer);
        SDL_UpdateWindowSurface(window);
    }
}

在 RandomDavis 的帮助下,失真有所减轻。这是新的结果。尽管如此,仍然存在一些翘曲:

new 1

new 2

new 3

注意: 对于仍在为这个问题苦苦挣扎的人,我在这里解决了它: How do I fix warped walls in my raycaster?

解决方法

好的,我找到了一个 guide 讨论了这个确切的问题。

在画墙之前,有一个问题必须要注意。这个问题被称为“鱼缸效应”。鱼缸效应的发生是因为光线投射实现将极坐标和笛卡尔坐标混合在一起。因此,在不直接位于观察者面前的墙壁切片上使用上述公式将提供更长的距离。这不是我们想要的,因为它会导致如下图所示的观看失真。 Blockquote

enter image description here

因此,为了消除视觉失真,从图 17 中的等式获得的所得距离必须乘以 cos(BETA);其中 BETA 是投射的光线相对于视角的角度。在上图中,视角 (ALPHA) 为 90 度,因为玩家面朝上。因为我们有 60 度的视野,所以最左边的光线的 BETA 是 30 度,最右边的光线是 -30 度。

相关问答

Selenium Web驱动程序和Java。元素在(x,y)点处不可单击。其...
Python-如何使用点“。” 访问字典成员?
Java 字符串是不可变的。到底是什么意思?
Java中的“ final”关键字如何工作?(我仍然可以修改对象。...
“loop:”在Java代码中。这是什么,为什么要编译?
java.lang.ClassNotFoundException:sun.jdbc.odbc.JdbcOdbc...