至于我之前的问题，我试图在 pygame 中模拟 3d 空间。到目前为止，我想出了一个非常简单的想法，即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。
如果您运行下面的代码，您将得到一个不错的模拟(比方说)一艘宇宙飞船经过一些恒星(按 w 或 s)。如果它们走得太远，它们就会消失，然后会创建一个新的。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()
请您参考如下方法:
在一般情况下是通过Homogeneous coordinates实现的.您的方法接近于此。
我推荐操作Cartesian coordinates ，其中 3 个维度具有相同的比例。
模拟 Perspective projection当你画点时。
这意味着您必须计算 Homogeneous coordinates 的 w 分量取决于点的深度(z 坐标)(例如 w = p[2] * 30/5000)并执行 x 的“透视划分”， y 和 z 组件由 w 组件绘制点之前。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。当你做的时候
p[0]=np.cos(-do)*p[0]-np.sin(-do)*p[2] p[2]=np.sin(-do)*p[0]+np.cos(-do)*p[2]
第一行的p[0]改变了，但是第二行应该使用原来的值。
做“元组”赋值来解决问题:
p[0], p[2] = np.cos(-do)*p[0]-np.sin(-do)*p[2], np.sin(-do)*p[0]+np.cos(-do)*p[2]
