javascript - 3 维空间中的轨道倾 Angular

Question

所以我正在从事一个项目，我需要从服务器检索粒子数据(可能会获取多达数千个对象)(以便同步客户端)，并且我正在努力找出轨道如何计算多个粒子的倾 Angular 。

到目前为止，我已经有了轨道总半径、宽度(粒子的起点和终点)

 var asteroidsOrbit = Common.random( 90, 100),
     rotation = Math.PI / 2, // Not sure where to use this yet
     radians = Common.random(0, 360) * Math.PI / 180;

我还通过设置粒子需要的位置来渲染轨道:

     position_x: Math.sin(radians) * asteroidsOrbit,
     position_y: Common.random(-10, 10),
     position_z: Math.cos(radians) * asteroidsOrbit

我注意到默认情况下轨道垂直于 X 平面，表现得理所当然。但是我应该如何对其应用旋转，因为它可能在 x 平面上旋转 20 度？

Answer 1

我用 Python 编写了一些代码，将旋转应用于 3D 空间中的粒子速度，可能会有所帮助:

def make_rotation_matrix(theta=0.,phi=0.,cos_theta=False,cos_phi=False):
    """
    Returns a rotation matrix for given angles.
    Makes the rotation matrix for a given theta angle around the y axis (polar
    angle) followed by another rotation around the z axis by a given phi angle
    (azimuthal angle).

    Parameters
    ----------
    theta : float
        Polar angle, between 0 and pi (or its cosine between -1 and 1).
    phi : float
        Azimuthal angle, between 0 and 2pi.
    cos_theta : boolean
        Wether to consider theta as an angle (False) or its cosine (True).
    cos_phi : boolean
        Wether to consider phi as an angle (False) or its cosine (True).

    Returns
    ----------
    out : np.array
        Rotation matrix (shape == (3,3)).

    Examples
    --------
    >>> make_rotation_matrix(0,10)
    array([[-0.83907153,  0.54402111, -0.        ],
           [-0.54402111, -0.83907153, -0.        ],
           [-0.        ,  0.        ,  1.        ]])
    >>> make_rotation_matrix(10,0)
    array([[-0.83907153, -0.        , -0.54402111],
           [-0.        ,  1.        , -0.        ],
           [ 0.54402111,  0.        , -0.83907153]])
    >>> make_rotation_matrix(0.1,0,True)
    array([[ 0.1       , -0.        ,  0.99498744],
           [ 0.        ,  1.        ,  0.        ],
           [-0.99498744,  0.        ,  0.1       ]])
    """
    if cos_theta:
        cos_theta = theta
        sin_theta = np.sqrt(1. - cos_theta**2)
    else:
        cos_theta = np.cos(theta)
        sin_theta = np.sin(theta)
    if cos_phi:
        cos_phi = phi
        sin_phi = np.sqrt(1. - cos_phi**2)
    else:
        cos_phi = np.cos(phi)
        sin_phi = np.sin(phi)
    return np.array(
    [[cos_theta*cos_phi ,  -sin_phi  ,  sin_theta*cos_phi],
     [cos_theta*sin_phi ,   cos_phi  ,  sin_theta*sin_phi],
     [     -sin_theta   ,      0     ,  cos_theta        ]])


def skew_symmetric_cross_product(vector):
    """
    Returns the skew symmetric cross product of given vector.

    Parameters
    ----------
    vector : iterable
        Iterable object, must have at least 3 elements, corresponding to
        the vectors x, y and z components.

    Returns
    ----------
    out : np.array
        Skew symmetric cross product of the given vector (shape == (3,3)).

    Examples
    --------
    >>> skew_symmetric_cross_product([1,0,0])
    array([[ 0.,  0.,  0.],
           [ 0.,  0., -1.],
           [ 0.,  1.,  0.]])
    >>> skew_symmetric_cross_product([0,1,0])
    array([[ 0.,  0.,  1.],
           [ 0.,  0.,  0.],
           [-1.,  0.,  0.]])
    >>> skew_symmetric_cross_product([0,0,1])
    array([[ 0., -1.,  0.],
           [ 1.,  0.,  0.],
           [ 0.,  0.,  0.]])
    """
    return np.array([[ 0.         , -vector[2] , vector[1]  ],
                     [ vector[2]  , 0.         , -vector[0] ],
                     [ -vector[1] , vector[0]  , 0.         ]])


def rotation_a2b(a,b):
    """
    Returns the rotation matrix that transforms vector a in b.
    From
    http://math.stackexchange.com/questions/180418/

    Parameters
    ----------
    a : iterable
        Iterable object, must have at least 3 elements, corresponding to
        the vectors x, y and z components.

    b : iterable
        Similar to a.

    Returns
    ----------
    out : np.array
        Rotation matrix that transforms a in b (shape == (3,3)).

    Examples
    --------
    >>> rotation_a2b([1,0,0],[1,0,0])
    array([[ 1.,  0.,  0.],
           [ 0.,  1.,  0.],
           [ 0.,  0.,  1.]])
    >>> rotation_a2b([1,0,0],[0,1,0])
    array([[ 0., -1.,  0.],
           [ 1.,  0.,  0.],
           [ 0.,  0.,  1.]])
    >>> rotation_a2b([1,0,0],[0,0,1])
    array([[ 0.,  0., -1.],
           [ 0.,  1.,  0.],
           [ 1.,  0.,  0.]])
    """
    if np.all(a==b):
        return np.identity(3)
    v = np.cross(a,b)
    v_sscp = skew_symmetric_cross_product(v)
    s = np.sqrt(np.dot(v,v))
    c = np.dot(a,b)
    return np.identity(3) + v_sscp +\
           (((1-c)/(s**2))*np.dot(v_sscp,v_sscp))

摘录 self 的报告[编辑]:

由于旋转是围绕粒子速度 (vp) 进行的，因此由 make_rotation_marix 生成的矩阵 R 不能直接应用于 vp - 我们必须首先更改基础以使速度与 (0, 0, 1) 向量共线，应用旋转，然后将基础更改回原始值。由于第一个基础变化的结果已知为 (0, 0, sp)，sp 是粒子的速度(速度大小)，因此跳过第一个基础变化并且 R 应用于 (0, 0, sp)。

为了将基础改回，我们使用矩阵将 (0, 0, 1) 改回初始粒子速度，由函数 Rotation_a2b 提供。矩阵 Ra2b 将向量 a 转换为 b，因此调用时使用 a = (0, 0, 1 ) 和 b 等于归一化粒子的初始速度。事实上，我们对速度进行归一化意味着基数的变化保留了向量的大小 - 或者可以使用 a = (0, 0, sp)。