我绘制了一个旋转的矩形,我需要检查它是否碰撞。 整个班级:
class Laser:
def __init__(self, player_x, player_y):
self.x = player_x
self.y = player_y
self.original_image = pygame.Surface((2, 1000))
self.original_image.set_colorkey( (0,0,0) )
self.original_image.fill( (255,0,0) )
self.copy_image = self.original_image.copy()
self.copy_image.set_colorkey( (0,0,0) )
self.rect = self.copy_image.get_rect()
self.new_image = pygame.Surface((2, 1000))
self.angle = 0
def continueDrawLaser(self):
if laser_bool:
screen.blit(self.new_image, self.rect)
def rotate(self):
# get rectangle of player and laser, as if the angle would be 0
player_rect = player1.original_player_image.get_rect(topleft=(player1.x, player1.y))
laser_rect = self.original_image.get_rect(midbottom=player_rect.midtop)
self.angle = player1.angle
pivotPos = [player_rect.centerx - laser_rect.x, player_rect.centery - laser_rect.y]
# calcaulate the axis aligned bounding box of the rotated image
w, h = self.original_image.get_size()
box = [pygame.math.Vector2(p) for p in [(0, 0), (w, 0), (w, -h), (0, -h)]]
box_rotate = [p.rotate(self.angle) for p in box]
min_box = (min(box_rotate, key=lambda p: p[0])[0], min(box_rotate, key=lambda p: p[1])[1])
max_box = (max(box_rotate, key=lambda p: p[0])[0], max(box_rotate, key=lambda p: p[1])[1])
# calculate the translation of the pivot
pivot = pygame.math.Vector2(pivotPos[0], -pivotPos[1])
pivot_rotate = pivot.rotate(self.angle)
pivot_move = pivot_rotate - pivot
# calculate the upper left origin of the rotated image
origin = (laser_rect.x + min_box[0] - pivot_move[0], laser_rect.y - max_box[1] + pivot_move[1]) #x,y
# get a rotated image
self.new_image = pygame.transform.rotate(self.original_image, self.angle)
# get new rectangle
self.rect = self.new_image.get_rect(topleft=origin)
这是碰撞函数:
#check if rock collides with laser
def collisionRockLaser(self, laser1):
laser_rect = laser1.rect
rock_rect = pygame.Rect(self.x, self.y, rock_width, rock_height)
if laser_rect.colliderect(rock_rect):
rocks.pop(rocks.index(rock2))
global score
score += 1
这就是我得到的:
我认为这足以通过self.rect,因为它每次都会随着旋转的位置进行更新以检测碰撞,但是似乎我必须使用分离轴定理,您能帮我吗?
答案 0 :(得分:1)
一种选择是创建与直线和矩形相交的算法。
首先创建一个与2个线段相交的算法:
P ... point on the 1. line
R ... normalized direction of the 1. line
Q ... point on the 2. line
S ... normalized direction of the 2. line
alpha ... angle between Q-P and R
beta ... angle between R and S
X ... intersection point
t ... distance between P and X
u ... distance between Q and X
gamma = 180° - alpha - beta
t = | Q - P | * sin(gamma) / sin(beta)
u = | Q - P | * sin(alpha) / sin(beta)
t = dot(Q-P, (S.y, -S.x)) / dot(R, (S.y, -S.x)) = determinant(mat2(Q-P, S)) / determinant(mat2(R, S))
u = dot(Q-P, (R.y, -R.x)) / dot(R, (S.y, -S.x)) = determinant(mat2(Q-P, R)) / determinant(mat2(R, S))
X = P + R * t = Q + S * u
您必须评估距离t
和u
是否为正且小于相应留置线段的长度:
(该算法使用pygame.math.Vector2
)
def collideLineLine(l1_p1, l1_p2, l2_p1, l2_p2):
# normalized direction of the lines and start of the lines
P = pygame.math.Vector2(*l1_p1)
line1_vec = pygame.math.Vector2(*l1_p2) - P
R = line1_vec.normalize()
Q = pygame.math.Vector2(*l2_p1)
line2_vec = pygame.math.Vector2(*l2_p2) - Q
S = line2_vec.normalize()
# normal vectors to the lines
RNV = pygame.math.Vector2(R[1], -R[0])
SNV = pygame.math.Vector2(S[1], -S[0])
# distance to the intersection point
QP = Q - P
t = QP.dot(SNV) / R.dot(SNV)
u = QP.dot(RNV) / R.dot(SNV)
return t > 0 and u > 0 and t*t < line1_vec.magnitude_squared() and u*u < line2_vec.magnitude_squared()
要测试线段是否与矩形相交,必须测试其是否与矩形的4个边相交:
def colideRectLine(rect, p1, p2):
return (collideLineLine(p1, p2, rect.topleft, rect.bottomleft) or
collideLineLine(p1, p2, rect.bottomleft, rect.bottomright) or
collideLineLine(p1, p2, rect.bottomright, rect.topright) or
collideLineLine(p1, p2, rect.topright, rect.topleft))
根据角度在激光上设置一条线:
angle = player1.angle
if angle < 0:
angle += 360
if angle < 90 or (angle > 180 and angle < 270):
laserline = [laser1.rect.topleft, laser1.rect.bottomright]
else:
laserline = [laser1.rect.bottomleft, laser1.rect.topright]
在该行中求值与pygame.Rect
enemy_rect
相交:
collide = colideRectLine(enemy_rect, *laserline)