Camera pose from solvePnP

Question

Goal

I need to retrieve the position and attitude angles of a camera (using OpenCV / Python).

Definitions

Attitude angles are defined by:

Yaw being the general orientation of the camera when it lays on an horizontal plane: toward north=0, toward east = 90°, south=180°, west=270°, etc.

Pitch being the "nose" orientation of the camera: 0° = looking horizontally at a point on the horizon, -90° = looking down vertically, +90° = looking up vertically, 45° = looking up at an angle of 45° from the horizon, etc.

Roll being if the camera is tilted left or right when in your hands (so it is always looking at a point on the horizon when this angle is varying): +45° = tilted 45° in a clockwise rotation when you grab the camera, thus +90° (and -90°) would be the angle needed for a portrait picture for example, etc.

World reference frame:

My world reference frame is oriented so:

Toward east = +X
Toward north = +Y
Up toward the sky = +Z

My world objects points are given in that reference frame.

Camera reference frame:

According to the doc, the camera reference frame is oriented like that:

What to achieve

Now, from cv2.solvepnp() over a bunch of images points and their corresponding world coordinates, I have computed both rvec and tvec.
But, according to the doc: http://docs.opencv.org/trunk/d9/d0c/group__calib3d.html#ga549c2075fac14829ff4a58bc931c033d , they are:

rvec ; Output rotation vector (see Rodrigues()) that, together with tvec, brings points from the model coordinate system to the camera coordinate system.
tvec ; Output translation vector.

these vectors are given to go to the camera reference frame.
I need to do the exact inverse operation, thus retrieving camera position and attitude relative to world coordinates.

Camera position:

So I have computed the rotation matrix from rvec with Rodrigues():

rmat = cv2.Rodrigues(rvec)[0]

And if I'm right here, the camera position expressed in the world coordinates system is given by:

camera_position = -np.matrix(rmat).T * np.matrix(tvec)

(src: Camera position in world coordinate from cv::solvePnP )
This looks fairly well.

Camera attitude (yaw, pitch and roll):

But how to retrieve corresponding attitude angles (yaw, pitch and roll as describe above) from the point of view of the camera (as if it was in your hands basically)?

I have tried implementing this: http://planning.cs.uiuc.edu/node102.html#eqn:yprmat in a function:

def rotation_matrix_to_attitude_angles(R):
    import math
    import numpy as np 
    cos_beta = math.sqrt(R[2,1] * R[2,1] + R[2,2] * R[2,2])
    validity = cos_beta < 1e-6
    if not validity:
        alpha = math.atan2(R[1,0], R[0,0])    # yaw   [z]
        beta  = math.atan2(-R[2,0], cos_beta) # pitch [y]
        gamma = math.atan2(R[2,1], R[2,2])    # roll  [x]
    else:
        alpha = math.atan2(R[1,0], R[0,0])    # yaw   [z]
        beta  = math.atan2(-R[2,0], cos_beta) # pitch [y]
        gamma = 0                             # roll  [x]  
    return np.array([alpha, beta, gamma])    

But results are not consistent with what I want. For example, I have a roll angle of ~ -90°, but the camera is horizontal so it should be around 0.

Pitch angle is around 0 so it seems correctly determined but I don't really understand why it's around 0 as the Z-axis of the camera reference frame is horizontal, so it's has already been tilted from 90° from the vertical axis of the world reference frame. I would have expected a value of -90° or +270° here. Anyway.

And yaw seems good. Mainly.

Question

Did I miss something with the roll angle?

Answer 1

I think your transformation is missing a rotation. If I interpret your question correctly, you are asking what the inverse of (rotation by R followed by translation T)

${\hat{R}|\vec{T}}.\vec{r}=\hat{R}.\vec{r}+\vec{T}$

The inverse should return the identity

${\hat{R}|\vec{T}}^{-1}.{\hat{R}|\vec{T}}={\hat{1}|0}$

Working this through yields

${\hat{R}|\vec{T}}^{-1}={\hat{R}^-1|-\hat{R}^-1\cdot \vec{T}}$

As far as I could tell you were using the $-\hat{R}^-1\cdot \vec{T}$ (undoing th translation) part of the answer but leaving out the inverse rotation $\hat{R}^-1$

Rotation+Translation:

${\hat{R}|\vec{T}}\vec{r}=\hat{R}\cdot\vec{r}+\vec{T}$

Inverse of (Rotation+Translation):

${\hat{R}|\vec{T}}^{-1}\vec{r}=\hat{R}^{-1}\cdot\vec{r}-\hat{R}^{-1}\cdot \vec{T}$

Non-latex mode (R^-1*r-R^-1*T) is the inverse of (R.r+T)