1 | initial version |

What you are looking for is called Homogeneous transformation matrix.

To get the 3D coordinate of the camera in the poster frame, assuming a known camera pose, you just have to compute what is called the inverse of the homogenous matrix (see this course, page 72).

2 | No.2 Revision |

What you are looking for is called Homogeneous transformation matrix.

To get the 3D coordinate of the camera in the poster frame, assuming a known camera pose, you just have to compute what is called the inverse of the homogenous matrix (see this course, page 72~~).~~, Ridig Body Motion – Homogeneous Transformations by Claudio Melchiorri).

**I strongly recommend to read a course about homogeneous transformation** (the one linked should be good).

I recall the equations in case the link is no more accessible.

3 | No.3 Revision |

What you are looking for is called Homogeneous transformation matrix.

To get the 3D coordinate of the camera in the poster frame, assuming a known camera pose, you just have to compute what is called the inverse of the homogenous matrix (see this course, page 72, Ridig Body Motion – Homogeneous Transformations by Claudio Melchiorri).

**I strongly recommend to read a course about homogeneous transformation** (the one linked should be good).

I recall the equations in case the link is no more accessible.

The superscript `t`

in `cRp^t`

simply means the transpose of matrix.

Note:

`solvePnP()`

returning a rotation vector and a translation vector, you will have to use `Rodrigues()`

to compute the rotation matrix from the rotation vector.

Copyright OpenCV foundation, 2012-2018. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.