平移坐标，然后在X和Y方向上旋转坐标，使Z‘指向原点

Question

我想建立一个矩阵( 4x4或3x3来转换XYZ->X'Y' Z‘)，它将把我的原点平移{x，y，z}，然后旋转矩阵，使Z’轴指向原点{X=0，Y=0，Z=0}

我试着绕任意轴旋转两次，然后独立平移，但是一旦坐标改变了一次，我的X'Y'Z‘坐标系的结果位置就会出现错误

Answer 1

我解决这个问题的方法是绕任意轴旋转2圈，然后从@0x5453开始计算LookAt矩阵。附件是代码，其中Mv是最终的4x4矩阵：

` `def findWorld2Cam(x，y，z)：

#   EXTRINSIC
camtheta=np.arctan(location[0]/location[2])
T=np.eye(4)
T[0:3,3]=-location
u=[0,1,0]
theta=camtheta
Rtheta=np.array([[np.cos(theta) + u[0]**2* (1 - np.cos(theta)), 
   u[0]* u[1]* (1 - np.cos(theta)) - u[2]* np.sin(theta), 
   u[0] *u[2]* (1 - np.cos(theta)) + u[1]* np.sin(theta), 0],
  [u[1] *u[0]* (1 - np.cos(theta)) + u[2]* np.sin(theta), 
   np.cos(theta) + u[1]**2 *(1 - np.cos(theta)), 
   u[1] *u[2]* (1 - np.cos(theta)) - u[0] *np.sin(theta), 0],
  [u[2] *u[0]* (1 - np.cos(theta)) - u[1] *np.sin(theta), 
   u[2]* u[1] *(1 - np.cos(theta)) + u[0]* np.sin(theta), 
   np.cos(theta) + u[2]**2 *(1 - np.cos(theta)), 0], [0, 0, 0, 1]])
#u=np.dot(Rtheta,[1,0,0,1])
u=[1,0,0]
camphi=np.arctan(location[1]/location[2])
phi=-camphi
Rphi=np.array([[np.cos(phi) + u[0]**2* (1 - np.cos(phi)), 
   u[0]* u[1]* (1 - np.cos(phi)) - u[2]* np.sin(phi), 
   u[0] *u[2]* (1 - np.cos(phi)) + u[1]* np.sin(phi), 0],
  [u[1] *u[0]* (1 - np.cos(phi)) + u[2]* np.sin(phi), 
   np.cos(phi) + u[1]**2 *(1 - np.cos(phi)), 
   u[1] *u[2]* (1 - np.cos(phi)) - u[0] *np.sin(phi), 0],
  [u[2] *u[0]* (1 - np.cos(phi)) - u[1] *np.sin(phi), 
   u[2]* u[1] *(1 - np.cos(phi)) + u[0]* np.sin(phi), 
   np.cos(phi) + u[2]**2 *(1 - np.cos(phi)), 0], [0, 0, 0, 1]])
R=np.dot(Rphi,Rtheta)
up1=np.dot(R,[0,1,0,1])

theta=np.arctan(location[0]/location[2])
phi=np.arctan(location[1]/location[2])
l=[0,0,0]
Mt=np.eye(4)
F=l-location
Mt[0:3,3]=-location
forward=F/np.linalg.norm(F)
up=up1[0:3]/np.linalg.norm(up1[0:3])
left=np.cross(up,forward)
Mr=np.eye(4)
Mr[0,0:3]=left
Mr[1,0:3]=up
Mr[2,0:3]=forward
Mv=np.dot(Mr,Mt)
return Mv
`