Uda*_*anS 1 python pygame collision-detection collision
我想在 pygame 中制作一个脚本,其中两个球飞向彼此,当它们碰撞时,它们应该彼此弹开,但我不知道该怎么做,所以你能帮助我吗?
要检测两个圆(分别是球)是否发生碰撞,您必须测试圆中心点之间的欧几里得距离是否小于半径之和。我建议使用pygame.math.Vector2/distance_to()进行计算。
下面,1 个圆由中心点(x1, y1)和半径r1定义。第二个圆由(x2, y2)和r2定义:
v1 = pygame.math.Vector2(x1, y1)
v2 = pygame.math.Vector2(x2, y2)
if v1.distance_to(v2) < r1 + r2:
print("hit")
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如果您必须测试圆之间的多次碰撞,您应该比较距离的平方,以避免计算欧几里德距离所需的耗时的平方根计算。distance_squared_to()还有一个用于此目的的函数:
v1 = pygame.math.Vector2(x1, y1)
v2 = pygame.math.Vector2(x2, y2)
if v1.distance_squared_to(v2) < (r1 + r2)**2:
print("hit")
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如果你想让圆反弹,你必须像台球一样将圆的运动向量反映在交点的法向量上。使用 pygame.math.Vector2/reflect_ip()或reflect()计算圆的新方向。
圆圈的运动由(mx1, my1)和 (mx2, my2)给出:
nv = v2 - v1
m1 = pygame.math.Vector2(mx1, my1).reflect(nv)
m2 = pygame.math.Vector2(mx2, my2).reflect(nv)
mx1, my1 = m1.x, m1.y
mx2, my2 = m2.x, m2.y
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最小的例子:
repl.it/@Rabbid76/PyGame-CirclesBounceOff
import pygame
pygame.init()
width, height = 400, 400
window = pygame.display.set_mode((width, height))
clock = pygame.time.Clock()
x1, y1, r1, mx1, my1 = 200, 200, 50, 2, 0.5
x2, y2, r2, mx2, my2 = 300, 200, 50, -1, -1.5
def move(c, v, r, m):
c += v
if c < r: c, v = r, -v
if c > m-r: c, v = m-r, -v
return c, v
hit_count = 0
run = True
while run:
clock.tick(60)
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
x1, mx1 = move(x1, mx1, r1, width)
y1, my1 = move(y1, my1, r1, height)
x2, mx2 = move(x2, mx2, r2, width)
y2, my2 = move(y2, my2, r2, height)
v1 = pygame.math.Vector2(x1, y1)
v2 = pygame.math.Vector2(x2, y2)
if v1.distance_to(v2) < r1 + r2 - 2:
hit_count += 1
print("hit:", hit_count)
nv = v2 - v1
m1 = pygame.math.Vector2(mx1, my1).reflect(nv)
m2 = pygame.math.Vector2(mx2, my2).reflect(nv)
mx1, my1 = m1.x, m1.y
mx2, my2 = m2.x, m2.y
window.fill((127, 127, 127))
pygame.draw.circle(window, (255, 0, 0), (round(x1), round(y1)), r1, 4)
pygame.draw.circle(window, (0, 0, 255), (round(x2), round(y2)), r2, 4)
pygame.display.flip()
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