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I'd like to know how I can go about calculating the angle of some pixel in a photo relative to the webcam that I'm using. I'm new to this sort of thing and I'm using a webcam. Essentially, I take a photo, process it, and I end up with a pixel value in the image that is what I'm looking for. I then need to somehow turn that pixel value into some meaningful quantity---I need to find a line/vector that passes through the pixel and the camera. I don't need magnitude, just phase.

How does one go about doing this? Is camera calibration necessary? I've been reading a bit about it but am unsure.

Thanks

Jean-Luc
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    To be able to do this, you'd need to know the distance of the object corresponding to the pixel, and the field-of-view of the camera. – Oliver Charlesworth Jul 06 '13 at 03:33
  • Okay, do you have any recommendations as to how I can solve my issue. I essentially am looking for a particular colour object. I find the pixel coordinate(s) which best matches the colour I am looking for, then I need to be able to have some way of saying which direction the object is... Is this not possible without depth? I would've thought I'd be able to get some bearing from the pixel coordinates... – Jean-Luc Jul 06 '13 at 03:55
  • Draw a diagram of the situation. You have a right-angled triangle, whose corners are (A) the object, (B) the camera, and (C) a point at the centre of the image but at the depth plane of the object. How are you going to compute the angle ABC? – Oliver Charlesworth Jul 06 '13 at 03:58
  • Note that in my first comment, that "and" should have been an "or". – Oliver Charlesworth Jul 06 '13 at 03:59

2 Answers2

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You don't need to know the distance to the object, only the resolution and angle of view of the camera.

Computing the angle requires only simple linear interpolation. For example, let's assume a camera with a resolution of 1920x1080 that covers a 45 degree angle of view across the diagonal.

In this case, sqrt(19202 + 10802) gives 2292.19 pixels along the diagonal. That means each pixel represents 45/2292.19 = .0153994 degrees.

So, compute the distance from the center (in pixels), multiply by .0153994, and you have its angle from the center (for that camera -- for yours, you'll obviously have to use its resolution and angle of view).

Of course, this is somewhat approximate -- its accuracy will depend on how much distortion the lens has. With a zoom lens (especially wider angle) you can generally count on that being fairly high. With a fixed focal length lens (especially if it doesn't cover an angle wider than 90 degrees or so) it'll usually be pretty low.

If you want to improve accuracy, you can start by taking a picture of a flat rectangle with straight lines just inside the angle of view of the camera, then compute the distortion based on the deviation from perfectly straight in the resulting picture. If you're working with an extremely wide angle lens, this may be nearly essential. With a lens covering a narrower angle of view (especially, as already mentioned, if it's fixed focal length) it's rarely likely to be worthwhile (such lenses often have only a fraction of a percent of distortion).

Jerry Coffin
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  • Thanks you, this is very helpful. Do you think 'camera calibration' would be necessary/worth it? – Jean-Luc Jul 06 '13 at 13:24
  • @user968243: I believe OpenCV's camera calibration is intended to measure distortion, so it's pretty much what the last paragraph of the answer talks about. Its utility will depend on the lens' distortion level, which I don't know. – Jerry Coffin Jul 06 '13 at 15:22
  • Answer is only a first order approximation. You can easily get a much more accurate answer by calibrating the camera. This is particularly true for wider angle lenses (>90 deg FOV), which usually exhibit significant nonlinear distortion. – Francesco Callari Jul 06 '13 at 16:38
  • @JerryCoffin, you wrote: "I believe OpenCV's camera calibration is intended to measure distortion". Your belief is demonstrably false. Good luck with your rep. – Francesco Callari Jul 06 '13 at 17:00
  • opencv's camera calibration will also give you the focal length of the lens in pixels (as well as distortion) – Martin Beckett Jul 06 '13 at 17:02
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    @FrancoCallari: ...and yet the OpenCV [documentation](http://docs.opencv.org/doc/tutorials/calib3d/camera_calibration/camera_calibration.html) says it does precisely that. – Jerry Coffin Jul 06 '13 at 17:04
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    @JerryCoffin. Kindly invent an algorithm that is capable of estimating the non-linear distortion of a lens and does NOT also use or produce an estimation of the linear part. Then publish it and apply for next year's CVPR'14 Best Paper award ;-) – Francesco Callari Jul 06 '13 at 17:48
  • @JerryCoffin, thanks for the answer. This is great way of getting approx angle with having to worry about calibration. if I want to know only the horizontal angle of each pixel, should I just divide the FoV by width in pixels? not sure what do you mean by diagonal FoV. – mdasari Jun 30 '21 at 19:14
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    Yes, that's basically it. Diagonal FoV is the FoV covered by a line from one corner to the opposite corner--traditionally, when lenses are rated for FoV is what they're talking about. – Jerry Coffin Jul 02 '21 at 03:55
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Recipe:

1 - Calibrate the camera, obtaining the camera matrix K and distortion parameters D. In OpenCV this is done as described in this tutorial.

2 - Remove the nonlinear distortion from the pixel positions of interest. In OpenCV is done using the undistortPoints routine, without passing arguments R and P.

3 - Back-project the pixels of interest into rays (unit vectors with the tail at the camera center) in camera 3D coordinates, by multiplying their pixel positions in homogeneous coordinates times the inverse of the camera matrix.

4 - The angle you want is the angle between the above vectors and (0, 0, 1), the vector associated to the camera's focal axis.

Francesco Callari
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  • Hi, I would like to arrive to have from the pixel P position the bearing vector. Which is a unit vector that goes from the camera center to the object identified by the pixel P. To do that I should arrive just to the point 3 of your description. The camera_matrix is the 3x3 matrix that for example comes out of a calibration of the camera. Is all that right? Thanks in advance – desmond13 Dec 19 '16 at 14:50