如何在 pygame 中改进我对 3d 空间的模拟？

Question

至于我之前的问题，我正在尝试在 pygame 中模拟一个 3d 空间。到目前为止，我想出了一个非常简单的想法，它使用第三个坐标作为分母来“压缩”（很确定这里有一些我不知道的术语）屏幕中心周围最远的点并减小它们的大小.

有人可以建议对这个想法进行简单的改进吗？我觉得我可以调整用于投影的分母（请参阅代码）以创建更准确的模拟方式。

如果你运行下面的代码，你会得到一个不错的模拟（比如说）一艘飞船经过一些星星（按 w 或 s）。如果他们走远了，他们就会消失，之后会创建一个新的。但是，如果我应用旋转（a 或 d），很明显模拟效果不佳，因为我并没有真正将 3d 点投影到 2d 屏幕上。

import pygame
import random
import numpy as np

pygame.init()
run=True

#screensize
screensize = (width,height)=(600,600)
center=(int(width/2),int(height/2))
screen = pygame.display.set_mode(screensize)

#delta mov
ds=0.1
do=0.0001

#Stars
points=[]
for i in range(1000):
    n1 = random.randrange(-5000,5000)
    n2 = random.randrange(-5000,5000)
    n3 = random.randrange(-30,30)
    points.append([n1,n2,n3])

while run:
    pygame.time.delay(20)
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            run=False

    ################## keys
    keys=pygame.key.get_pressed()

    if keys[pygame.K_w]:
        for p in points:
            p[2]-=ds
    if keys[pygame.K_s]:
        for p in points:
            p[2]+=ds

    if keys[pygame.K_a] or keys[pygame.K_d]:
        if keys[pygame.K_a]:
            for p in points:
                p[0]=np.cos(-do)*p[0]-np.sin(-do)*p[2]
                p[2]=np.sin(-do)*p[0]+np.cos(-do)*p[2]
        else:
            for p in points:
                p[0]=np.cos(do)*p[0]-np.sin(do)*p[2]
                p[2]=np.sin(do)*p[0]+np.cos(do)*p[2]


    ###############################projection###################

    for p in points:
        #this is to create new stars
        if p[2]<=-30 or p[2]>=30:
            p[0] = random.randrange(-5000,5000)
            p[1] = random.randrange(-5000,5000)
            p[2] =30
        else:
            #this is to ignore stars which are behind the ship
            if p[2]<=0:
                pass
            else:
                try:
                    #THIS IS THE PROJECTION I USE, I TAKE THE RADIUS BECAUSE I GUESS I'LL NEED IT... BUT I DON'T USE IT XD
                    r = ((p[0]**2+p[1]**2+p[2]**2)**(1/2))
                    pygame.draw.circle(screen,(255,255,0),(int(p[0]/p[2]+center[0]),int(p[1]/p[2]+center[1])),int(10/p[2]))
                #this is to prevent division by cero and alike
                except Exception as e:
                    pass

    pygame.display.update()
    screen.fill((0,0,0))


pygame.quit()

Answer 1

一般情况下是通过Homogeneous coordinates 实现的。你的方法很接近。

我推荐操作Cartesian coordinates，其中3个维度具有相同的比例。
绘制点时模拟Perspective projection。
这意味着您必须根据点（例如w = p[2] * 30 / 5000）的深度（z坐标）计算Homogeneous coordinates的w分量，并对x、@987654330执行“透视除法” @ 和z 组件由w 组件，在您绘制点之前。例如：

#delta mov
ds=10
do=0.01

#Stars
points=[]
for i in range(1000):
    n1 = random.randrange(-5000,5000)
    n2 = random.randrange(-5000,5000)
    n3 = random.randrange(-5000,5000)
    points.append([n1,n2,n3])

while run:
    pygame.time.delay(20)
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            run=False

    ################## keys
    keys=pygame.key.get_pressed()

    if keys[pygame.K_w]:
        for p in points:
            p[2]-=ds
    if keys[pygame.K_s]:
        for p in points:
            p[2]+=ds

    if keys[pygame.K_a] or keys[pygame.K_d]:
        if keys[pygame.K_a]:
            for p in points:
                p[0], p[2] = np.cos(-do)*p[0]-np.sin(-do)*p[2], np.sin(-do)*p[0]+np.cos(-do)*p[2]
        else:
            for p in points:
                p[0], p[2] = np.cos(do)*p[0]-np.sin(do)*p[2], np.sin(do)*p[0]+np.cos(do)*p[2]

    ###############################projection###################

    screen.fill((0,0,0))
    for p in points:
        #this is to create new stars
        if p[2]<=-5000 or p[2]>=5000:
            p[0], p[1], p[2] = random.randrange(-5000,5000), random.randrange(-5000,5000), 5000
        else:
            #this is to ignore stars which are behind the ship
            if p[2]<=0:
                pass
            else:
                w = p[2] * 30 / 5000
                pygame.draw.circle(screen,(255,255,0),(int(p[0]/w+center[0]),int(p[1]/w+center[1])),int(10/w))

    pygame.display.update()

---

此外，旋转不正确。当你这样做时

p[0]=np.cos(-do)*p[0]-np.sin(-do)*p[2]
p[2]=np.sin(-do)*p[0]+np.cos(-do)*p[2]

p[0] 在第一行被更改，但在第二行应使用原始值。
做“元组”赋值来解决这个问题：

p[0], p[2] = np.cos(-do)*p[0]-np.sin(-do)*p[2], np.sin(-do)*p[0]+np.cos(-do)*p[2]