In Fig. 5, $C=(C1,C2,C3)T$ are the coordinates of the camera center in the world coordinate system. In this formula, $R$ represents the rotation matrix that rotates the world coordinate axes into the camera coordinates axes. Also, $t=\u2212RC$ can be thought of as a translation vector. In our method, we use a somewhat different approach to computing $P$, which greatly improves the performance of the algorithm. The idea is that in most situations, the camera will point roughly in the direction of $\u2212C$, that is, it will approximately be pointing back to a common position, which we take as the center of the world coordinate system. This has the effect of reducing two of the three angles involved in the rotation matrix to small ranges and essentially removes two degrees of freedom from the search space. To be specific, let $x,y,z$ denote the world coordinate axes, let $x\u2032$, $y\u2032$, $z\u2032$ denote the camera centered coordinate axes as shown in Fig. 5, and let $x^$, $y^$, $z^$ denote the coordinate axes defined as follows. Let $z^=\u2212C$ be the vector pointing from the camera center to the origin of the world coordinate system. The remaining axes $x^$ and $y^$ are arbitrary subject to forming a right-handed orthogonal coordinate system including $x^$. In our case, we took $x^$ to be a vector orthogonal to $z^$, which had 0 third coordinate. We then have $y^=\u2212x^\xd7z^$. We let $R\u2032$ be the rotation matrix that rotates the $(x^,y^,z^)$ coordinate system to the $(x\u2032,y\u2032,z\u2032)$ coordinate system. The matrix $R\u2032$ can be described as follows. Let $\alpha x$, $\alpha y$, $\alpha z$ denote rotation angles about the $x\u2032$, $y\u2032$, $z\u2032$ axes with respect to the camera center. Then Display Formula
$R\u2032=Rz\u2032\xb7Ry\u2032\xb7Rx\u2032,$
where Display Formula$Rx\u2032=(1000cos(\alpha x)\u2212sin(\alpha x)0sin(\alpha x)cos(\alpha x))$
Display Formula$Ry\u2032=(cos(\alpha y)0\u2212sin(\alpha y)010sin(\alpha y)0cos(\alpha y))$
Display Formula$Rz\u2032=(cos(\alpha z)\u2212sin(\alpha z)0sin(\alpha z)cos(\alpha z)0001).$