pygame旋转一条线

Question

我试图在Pygame中创建一个"雷达".我无法旋转雷达的针头.如何旋转它？

import pygame
from pygame.locals import *
SIZE = 800, 800
pygame.init()
screen = pygame.display.set_mode(SIZE)
FPSCLOCK = pygame.time.Clock()
done = False
screen.fill((0, 0, 0))
degree=0
while not done:
    for e in pygame.event.get():
        if e.type == QUIT or (e.type == KEYDOWN and e.key == K_ESCAPE):
            done = True
            break
    for x in range(1,400,10):
        pygame.draw.circle(screen,(255,255,255),(400,400),x,1)
    line = pygame.draw.line(screen,(10,100,10),(400,400),(400,0),3)
    pygame.transform.rotate(line,degree)
        pygame.display.flip()   
        degree+=5
        FPSCLOCK.tick(40)

这里解决了:

import pygame
import math
from pygame.locals import *
SIZE = 800, 800
pygame.init()
screen = pygame.display.set_mode(SIZE)
FPSCLOCK = pygame.time.Clock()
done = False
screen.fill((0, 0, 0))
degree=0
while not done:
    screen.fill(0)
    for e in pygame.event.get():
        if e.type == QUIT or (e.type == KEYDOWN and e.key == K_ESCAPE):
            done = True
            break
    for x in range(1,400,10):
        pygame.draw.circle(screen,(255,255,255),(400,400),x,1)    
    radar = (400,400)
    radar_len = 400
    x = radar[0] + math.cos(math.radians(degree)) * radar_len
    y = radar[1] + math.sin(math.radians(degree)) * radar_len

    # then render the line radar->(x,y)
    pygame.draw.line(screen, Color("red"), radar, (x,y), 1)
    pygame.display.flip()   
    degree+=5
    FPSCLOCK.tick(40)

使用线条:

x = radar[0] + math.cos(math.radians(degree)) * radar_len
y = radar[1] + math.sin(math.radians(degree)) * radar_len

采取一个angle改变每一帧,以从长度的屏幕中心创建一条线radar_len

Answer 1

您需要计算您的开始和结束位置.

import math
import pygame
from pygame.locals import *

radar = (100,100)
radar_len = 50
x = radar[0] + math.cos(math.radians(angle)) * radar_len
y = radar[1] + math.sin(math.radians(angle)) * radar_len

# then render the line radar->(x,y)
pygame.draw.line(screen, Color("black"), radar, (x,y), 1)