# Pose Measurement Method Based on Machine Vision and Novel Directional Target
## 2 Materials  
The imaging process of the new cooperative target under the pinhole imaging model is shown in Figure 1 below. Figure 1 defines four right-angle coordinate systems, of which \(o_{w}-x_{w}y_{w}z_{w}\) is the world coordinate system, the world correspond origin is set at the centre of the target centre circle; \(o_{c}-x_{c}y_{c}z_{c}\) is the camera coordinate system, where the \(z_{c}\) axis coincides with the optical axis of the camera lens. The camera coordinate system can be set at any pose in the world coordinate system, between which the camera's external reference can convert it; \(o_{i}-x_{i}\,y_{i}\) is the image physical coordinate system, with the centre of the image as the origin and the pose of the pixel expressed in physical length; \(o_{p}-x_{p}y_{p}\) is the pixel coordinate system, which is described in pixels. Assuming there exists a point \(p\) on the image, in the pixel coordinate system, its x-axis value is denoted as \(u\), and the y-axis value is denoted as \(v\). Therefore, its pixel coordinates can be represented as \(p^{*}=[u,v]^{T}\), the 3D coordinates of the corresponding spatial point \(P=[X,Y,Z]^{T}\), and the chi-square expressions for both are \(p^{*}=[u,v,1]^{T}\). \(P^{*}=[X,Y,Z,1]^{T}\). Then, the projection relation between the two is shown in Equation (1). \[sp^{*}=K[R|t]P^{*} \tag{1}\]  
where \(s\) is the scale factor; \([R|t]\) is the external reference matrix, which is used to transform between the world coordinate system and the camera coordinate system, where \(R\) is the rotation matrix, \(t\) is the translation vector, and the pose of the target is one-to-one with its external reference matrix. The camera's internal reference matrix, \(K\), is shown in Equation (2). \[K= \begin{bmatrix}a&\gamma&u_{0}\\ 0&\beta&v_{0}\\ 0&0&1\end{bmatrix}  \tag{2}\]  
Figure 1: Schematic diagram of the new cooperative target imaging. where \((u_{0},v_{0})\) represent the coordinates of the image's principal points, \(\alpha\), \(\beta\) represent the pixel \(x_{p}\) axis and \(y_{p}\) focal length values in the direction of the axis, and \(\gamma\) is the tilt factor of the image element axis. Since the world coordinate system is on the target scale now, the \(z_{w}\) axis is perpendicular to the target coordinate, so the z-coordinates of the feature points will all be equal to 0. The three-dimensional coordinate point of \(P\) can be represented as \([X,Y,0]^{T}\) and the corresponding chi-square coordinates are \([X,Y,0,1]^{T}\). Define the first column vector of the rotation matrix \(R\) is \(r_{i}\), which can be expressed by substituting it into Equation (3):  
\[s \begin{bmatrix}u\\ v\\ 1\end{bmatrix} =K \begin{bmatrix}r_{1}&r_{2}&r_{3}&t\end{bmatrix}\begin{bmatrix}X \\ Y\\ 0\\ 1\end{bmatrix} =K \begin{bmatrix}r_{1}&r_{2}&t\end{bmatrix}\begin{bmatrix}X\\ Y\\ 1\end{bmatrix}  \tag{3}\]  
It is important for the feature points of the target itself to be all on the same spatial plane. When the camera captures the feature points, a homography transformation relationship exists between the spatial plane and the image plane, i.e., the spatial coordinates of the feature points can be transformed into pixel coordinates by a homography matrix. Let us define the homography matrix as \(H\). At this point, \(H\) satisfies Equation (4). \[sp^{*}=HP^{*} \tag{4}\]  
Substituting into Equation (3), we obtain  
\[H=\lambda K[r_{1}\;\;r_{2}\;\;t] \tag{5}\]  
which, due to the existence of scale invariance in the chi-square coordinates, yields an arbitrary scalar \(\lambda\). At this point it follows that \(H\) is a \(3\times 3\) matrix defined by \(H=[h_{1}\;\;h_{2}\;\;h_{3}]\), substituting into Equation (5), we have:  
\[[h_{1}\;\;h_{2}\;\;h_{3}]=\lambda K[r_{1}\;\;r_{2}\;\;t] \tag{6}\]  
The rotation matrix \(R\) is an orthogonal matrix, their \(r_{1}\) and \(r_{2}\) are standard orthogonal bases, both mutually perpendicular and of modulus 1, i.e., satisfying \(r_{1}^{T}r_{2}=0\) and \(r_{1}^{T}r_{1}=r_{2}^{T}r_{2}=1\), which can be obtained by combining Equation (6):  
\[\left\{