# Retrieve yaw, pitch, roll from rvec

I need to retrieve the attitude angles of a camera (using cv2 on Python).

• Yaw being the general orientation of the camera when on an horizontal plane: toward north=0, toward east = 90°, south=180°, west=270°, etc.
• Pitch being the "nose" orientation of the camera: 0° = horitzontal, -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: +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.

I have yet rvec and tvec from a solvepnp().

Then I have computed:
rmat = cv2.Rodrigues(rvec)[0]

If I'm right, camera position in the world coordinates system is given by:
position_camera = -np.matrix(rmat).T * np.matrix(tvec)

But how to retrieve corresponding attitude angles (yaw, pitch and roll as describe above) from the point of view of the observer (thus the camera)?

I have tried implementing this : http://planning.cs.uiuc.edu/node102.h... 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 it gives me some results which are far away from reality on a true dataset (even when applying it to the inverse rotation matrix: rmat.T).
Am I doing something wrong?
And if yes, what?
All informations I've found are incomplete (never saying in which reference frame we are or whatever in a rigorous way).
Thanks.

Update:

Rotation order seems to be of greatest importance.
So; do you know to which of these matrix does the cv2.Rodrigues(rvec) result correspond?:

### Update:

I'm finally done. Here's the solution:

def yawpitchrolldecomposition(R):
import math
import numpy as np
sin_x    = math.sqrt(R[2,0] * R[2,0] +  R[2,1] * R[2,1])
validity  = sin_x < 1e-6
if not singular:
z1    = math.atan2(R[2,0], R[2,1])     # around z1-axis
x      = math.atan2(sin_x,  R[2,2])     # around x-axis
z2    = math.atan2(R[0,2], -R[1,2])    # around z2-axis
else: # gimbal lock
z1    = 0                                         # around z1-axis
x      = math.atan2(sin_x,  R[2,2])     # around x-axis
z2    = 0                                         # around z2-axis

return np.array([[z1], [x], [z2]])

yawpitchroll_angles = -180*yawpitchrolldecomposition(rmat)/math.pi
yawpitchroll_angles[0,0] = (360-yawpitchroll_angles[0,0])%360 # change rotation sense if needed, comment this line otherwise
yawpitchroll_angles[1,0] = yawpitchroll_angles[1,0]+90


That's all folks!

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One thing to note is that solvePnP returns the world coordinates relative to the camera, and you will need to invert the transformation if you need the orientation of the camera with respect to the world.

( 2017-06-20 19:30:50 -0500 )edit

that's what I've already done, see the main post. And thanks to LBerger, I've already read that post many time; it's C++ related and it doesn't help a lot...

( 2017-06-22 17:05:28 -0500 )edit

I wonder; what is the convention used by OpenCV to operate rotation ? In which order ? It's perhaps only a gap between how I imagine these 3 rotations and the reality. For me, with a normal frame, Z pointing upward, first rotation is around Z (yaw), then second rotation is around X axis (pitch) and finally the third one is around the new Z" (roll). But how are the standard camera axis oriented for OpenCV ? And the world reference frame ?

( 2017-06-22 20:25:05 -0500 )edit

The rotation matrix is independent of operation and stores the complete rotational information. When decomposing it to euler angles you choose the order of those rotations.

The camera coordinates are x is right, y is down, z is out into the world. The world coordinates are defined by the points you use solvePnP with.

( 2017-06-22 22:35:31 -0500 )edit

Ok, interesting. Thanks. So if I use a world coordinate system as follow: East=+X, North=+Y, and up=+Z, an horizontal camera looking toward east would yield these angles (in degrees): yaw=0, pitch=270 and roll=0. Am I right?
For the moment, it seems I have to add 90° to my roll angle (as found by my function in first post). I can't figure out why. I would have tell to add this value to the pitch angle as the Z camera axis is tilted 90° from the world Z axis.
And if I want to retrieve azimuth values, I must rotate the yaw values the other way, so: azimuth=360-yaw. This seems to work and I understand why.

( 2017-06-23 05:30:41 -0500 )edit

Hmm, nope. A camera looking east would have a yaw component of magnitude 90. Positive or negative, I'm not sure without coloring my fingers, but I can tell it's needed.

( 2017-06-23 16:08:52 -0500 )edit

My bad; you're right, I was tired. So east=-90° (positive rotation is always given clockwise while looking in the same direction as the axis perpendicular to the plane in which the rotation takes place). Anyway, I need to retrieve azimuth after, so by changing the rotation: azimuth = 360-yaw. This is clear to me.

So it's not the main point.
I would have expected 270 or -90 for the pitch angle for an horizontal camera, and in fact it's almost already near 0°.

And I would have expected a near 0° roll angle for an untilted camera, and there, it's -90° !

I do not understand why.

( 2017-06-25 05:32:42 -0500 )edit

Nope, I just worked it out.

The yaw 90 puts the camera x = -y, camera y = x, camera z = z

Then Roll -90 then rotates around the camera x so you get camera x = -y, camera y = -z, camera z = x, which is what you want. If you did pitch -90, you would be rotating around the y, and would get camera x = -z.

Any yaw and a pitch of -90 gives you your camera x = down, which would mean your camera is turned so the right side of the image is down.

( 2017-06-25 16:30:08 -0500 )edit

But by definition, roll is an intrinsic rotation around y and pitch around x.
I need to figure out the intrinsic rotation order for OpenCV.
I supposed it is: yaw, then pitch, an finally roll (for a "normal" frame, Z pointing up, x to the right and y to the front). Edit: No, I'm wrong with that assumption. Roll is effectively around x and pitch around y. I just figure it out.

( 2017-06-27 16:35:29 -0500 )edit

Ah, no. By definition, pitch is y and roll is x. Always has been. Check the link in your answer.

( 2017-06-27 17:31:19 -0500 )edit

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The rotation matrix is the rotation matrix is the rotation matrix. There is only one. The decomposition into three euler angles is where the rotation order changes. You can go from the rotation matrix to any of those angle orders, you just need to decide which order is what you need.

Remember, the camera is looking up the z axis. So yaw rotates the image, pitch turns the image to the left and right, and roll is up and down in the image. (assuming all the other angles are 0, combining them is less simple)

more

Yes but the rotation matrix given by cv2.Rodrigues(rvec), can only fit one of the matrix presented in my first post. If decomposed an other way, it would yield strange results isn't it?

( 2017-06-30 06:09:35 -0500 )edit

You are mis-understanding what that shows. All of those are valid decompositions and produce valid euler angles, and all start from the same rotation matrix. They merely produce different sets of euler angles. One of those is equivalent to yaw pitch roll, the others are something else. But they all are the same rotation matrix, and the same orientation in space.

( 2017-06-30 16:59:23 -0500 )edit

To be clear, there is no such thing as a "rotation matrix for a x-y'-z" Tait-Bryan sequence". There is only a rotation matrix. By definition, all euler angles describing an orientation produce a particular rotation matrix.

If someone gives you three numbers and says "Here's the euler angles for a rotation", you must ask which order they are, and you'll get a response from that table (or one like it). If someone gives you a rotation matrix, you're done. You already know everything about the rotation.

( 2017-06-30 17:06:26 -0500 )edit

I think (and hope) I'm done with it.

Here's my workaround (if not a final solution) in 6 steps:

0. Imports :

import math, numpy as np, cv2


1. Retrieve rvec and tvec from the cv2.solvePnP() (or with RanSaC) function:

retval, rvec, tvec = cv2.solvePnP(obj_pts, img_pts, cam_mat, dist_coeffs, rvec, tvec, flags=cv2.SOLVEPNP_ITERATIVE)


2. Convert rvec to a 3x3 matrix using cv2.Rodrigues():

rmat = cv2.Rodrigues(rvec)[0]


This matrix is a rotation matrix for a x-y'-z" Tait-Bryan sequence if I'm not wrong (that's what I was searching for days!). So r_total = rz·rỵ·rx (rx occurs first). You can imagine you have first a camera frame (z-axis = through the lens, x=right, y=bottom) which is perfectly superposed to the world frame. You rotate it around x first, then y, finally z. The angles are the Euler angles hereafter. Lowercase axis = camera frame axes, uppercase = world frame axes. Camera frame is firmly attached to the body.

2. If you need, the camera position expressed in the world frame (OXYZ) is given by:

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


4. Create the projection matrix P = [ R | t ]:

P = np.hstack((rmat,tvec))


5. Use cv2.decomposeProjectionMatrix() (or any home-made function) to retrieve Euler angles:

euler_angles_radians = -cv2.decomposeProjectionMatrix(P)[6]


I noticed we have to take the negative values here to be compliant with a conventional rotation (while looking in the same direction as the vector perpendicular to the plane where the rotation occurs; clockwise = positive. It's the conventional sense in mathematics also).
This is due to the fact OpenCV work with a mobile frame around the vector and I was working in my mind with a fix one. Informations here: mathworld.wolfram.com/RotationMatrix.html )

Euler angles are the angles making up the total rotation, but expressed separately after the 3 axis of an oxyz frame firmly attached to the the camera.
Initially, the camera is aligned with the world frame, facing upwards if your world frame Z-axis is pointing upward. X=to the right of the cam. Y=normal to the two others, bottom of the cam.
First rotation is around X, then Y and finally Z. This could be written as r_total = rzryrx (rx occurs first).

Euler angles form a 3x1 vector.
Element [1,0] is the rotation around x, element [2,0] rotation around y, and element [2,0] rotation around z.

6. To retrieve attitude of the camera (just like if it was an airplane) in it's own body-attached frame, here's the magic that seems to work for me yet (one would have to check this with more precise instrument as my eyes...):

eul    = euler_angles_radians
yaw    = 180*eul[1,0]/math.pi # warn: singularity if camera is facing perfectly upward. Value 0 yaw is given by the Y-axis of the world frame.
pitch  = 180*((eul[0,0]+math.pi/2)*math.cos(eul[1,0]))/math.pi
roll ...
more

Thanks. After the solutions, most of them let it go unlike you. Nice

( 2018-08-21 04:57:11 -0500 )edit

I'm trying to understand this line of code:  euler_angles_radians = -cv2.decomposeProjectionMatrix(P)[6]  Doesn't decomposeProjectionMatrix`return the euler angles in degrees? https://docs.opencv.org/4.0.1/d9/d0c/...

( 2019-04-02 09:55:18 -0500 )edit

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