Computer vision 根据其他两台摄像机的校准输出，计算第三台摄像机中的3D点

我从三个不同的角度记录了一个心理治疗课程：

cam 1聚焦于患者

cam 2的重点是治疗师

凸轮3从侧面捕获两个凸轮

我使用Matlab校准工具校准cam1和cam2，并使用cam1和cam2对cam3和cam3进行外部立体校准，我拥有所有会话中的所有旋转向量和固有矩阵

现在，给定2个点（来自cam1/2），我感兴趣的是计算“真实世界”cam3中的3D点（使用我之前找到的校准矩阵从两个摄像机的每个点绘制的向量之间的交点）

我用python编写了以下代码来查找3D点：

'''
External calibration output between 2 cameras:
  ---------------------------------------------
  R - Rotation Matrix of camera one relative to camera two
  O - Translation vector of camera one relative to camera two
  
  Anotations:
  ---------
  org - of camera which represents the real world (with origin [0,0])
  rel - of relative camera with translation vector
  d - direction vector of each camera, i.e,
      np.linalg.inv(K).dot([u, v, 1])
      [u, v] - pixel in the image
      K - Calibration matrix (the intrinsicMatrix)
      
  Output:
  ------
  3D point of each camera
'''
def find3DPoints (d_org, O_org, d_rel, R_rel, O_rel):
    dist = 1000
    alpha=0
    beta=0
    X_org=0
    X_rel=0
    
    d_rel = np.matmul(np.linalg.inv(R_rel), d_rel)
    
    while (dist > 10) :
        alpha = np.random.randint(200)
        beta = np.random.randint(200)
        
        #print("alpha: ", alpha)
        #print("beta: ", beta)
        
        X_org = alpha*d_org + O_org
        X_rel = beta*d_rel + O_rel
        
        dist = np.linalg.norm(X_org-X_rel)
        #print("distance: ", dist)
        
    print("distance between 2 closest points: ", dist)
    print("alpha: ", alpha)
    print("beta: ", beta)
    print("X Origin: ", X_org)
    print("X Relative: ", X_rel)
    return X_org, X_rel

我得到的结果令人满意：

distance between 2 closest points:  8.663262014946904
alpha:  140
beta:  81
X Origin:  [-41.49543898  11.50037104 140.        ]
X Relative:  [-41.69268956   3.02469031 141.78214397]

然后，我尝试编写一个更好的解决方案，使用线性代数，使用最小二乘法（Ax=b）求α和β：

我考虑它是因为：

X=O_组织+阿尔法*d_组织

X=O_rel+beta*d_rel

==>

[d_rel，d_org]*[-beta，alpha]=O_rel-O_org

[d_rel，d_org]=>3x2矩阵

O_rel-O_org=>3x1向量

def find3DPointByLeastSquare(d_org, O_org, d_rel, O_rel):
   A=np.array([d_org, d_rel])
   #print("A=",A)

   b=O_rel-O_org
   #b=np.array(O_org)-O_rel
   #print("b=",y)

   # Ax=b
   alpha, beta = np.linalg.lstsq(A.T, b, rcond=None)[0]
   #beta=-beta
   print("alpha: ", alpha)
   print("beta: ", beta)

   X_org = O_org + alpha*d_org
   print("X Origin=", X_org)

   X_rel = O_rel + -beta*d_rel
   print("X Relative=", X_rel)

   return X_org, X_rel

但是我得到的结果与第一个解决方案不同！！！ 我也希望得到同样的结果。原因可能是什么

alpha:  199.83226356812804
beta:  -204.61307546220422
X Origin= [ 27.31677265  16.63486267 199.83226357]
X Relative= [ 67.03447507 -28.86206184 198.19026467]