Flappy Bird旋转算法

Question

我正在用c++和DirectX写一个浮华的小鸟克隆。我基本上完成了，除了旋转的算法。我现在有一个(rotation = ((90 * (yVelocity+10) / 25) - 90)/2;)，但它的行为方式与最初的flappy鸟不同。我正在尝试尽可能地复制最初的flappy鸟的旋转，因此任何帮助都将不胜感激。

Answer 1

我会使它等于原始的yVelocity，但将其限制为2个数字。就像这样

rotation = min(topClamp, max(bottomClamp, yVelocity));

你可能想玩一段时间，但这将使旋转依赖于yVelocity，但如果玩家不断向上，那么旋转将被夹在某个数字，而鸟将只是向上看，就像在原始游戏中一样。

Answer 2

    def fall(self):
        # bird falls with accelerated motion
        self.movement += GRAVITY
        self.rect.centery += self.movement

    def rotate(self):
        # factor is positive when falling, therefore rotation is clockwise
        # factor is negative when raising, rotation is counter-clockwise
        factor = -2
        factor *= self.movement

        self.image = pygame.transform.rotozoom(self.temporary, factor, 1)

Answer 3

self.temporary是当前精灵的副本。

""" Bird object """
class Bird(pygame.sprite.Sprite):
    def __init__(self, color):
        super().__init__()

        # sounds
        self.flap = pygame.mixer.Sound('wing.wav')
        self.crash = pygame.mixer.Sound('crash.wav')
        
        # flapping sprites
        self.sprites = []

        for i in [1, 2, 3]:
            temp = pygame.image.load(color + str(i) + ".png").convert_alpha()
            temp = pygame.transform.scale2x(temp)
            self.sprites.append(temp)
            
        # current sprite
        self.image = self.sprites[0]

        # temporary sprite
        self.temporary = self.image

        # current rectangle
        self.rect = self.image.get_rect(center=(BIRD_X, HEIGHT/2))

        # current frame
        self.frame = 0

        # movement amount
        self.movement = 0