What is the field of view (FOV) of the default lens?

[Jared Russell]

I can't seem to find any specs for the default CMU cam lens vertical/horizontal field of view. Obviously, I only need one since I know the aspect ratio.

Any ideas? Cheers.

[phrontist]

You could measure it... put some marks on a wall and tell us.

[Joe Ross]

See this thread from last year: http://www.chiefdelphi.com/forums/sh...ght=field+view

I found that thread by searching for field of view in the technical forum and all subforums.

[Astronouth7303]

I did a bit of googling previously, what follows is my calculations for it. (Note that I'm not an optical expert and there's a good chance I'm wrong.)

For rectilinear (normal) lenses, the Field of View (FoV) is calculated as:

2 * arctan (<sensor size>/(<focal length> * 2))

I found that the OV7620 by default has the "Yasan 1/3" Focal Length 6.0mm, F1.8 Glass #604180 LCR1(Coating)" lens and the sensor is 4.86mm x 3.64mm.

After doing the necessary trig, I came up with the answer that the FoV is 44.09589 x 33.74860 degrees. I haven't confirmed this, nor looked at the other thread, yet.

[Rickertsen2]

Originally Posted by Abwehr

I can't seem to find any specs for the default CMU cam lens vertical/horizontal field of view. Obviously, I only need one since I know the aspect ratio.

Any ideas? Cheers.

Needing to know this bit of information jogged some interesting control algorithms in my mind.

[Jared Russell]

Visualization, actually.

I've written an OpenGL application that renders the field in 3D as the camera sees it, based on the tilt, pan, mx, and my values. The FOV is needed to convert mx and my into an angle. Right now I just added in a slider so teams can change it to whatever lens they're using.

The math itself is actually surprisingly complicated (converting to and from 2D screen coordinates to 3D world coordinates), so I don't think I'm going to try to implement something *quite* as precise on the RC. Besides, we don't really need to know the exact range unless the target is centered horizontally in our application, and this constraint turns the equation from matrix math into trig.

It is actually done, and works great - I'm just waiting on our teams' animators to texture the field nicely before I release it on here. I'm a programmer, so my 3D field left much to be desired...

Here's a screen cap from about a week ago - the current build is much prettier.

[Greg Marra]

Originally Posted by Astronouth7303

After doing the necessary trig, I came up with the answer that the FoV is 44.09589 x 33.74860 degrees. I haven't confirmed this, nor looked at the other thread, yet.

From empirical measurements taken last year (which could very well have been incorrect), we found the camera's field of view to be approximately 55 x 35 degrees. These measurements were 'close enough' to accurately find which of the eight possible tetra locations on the floor were the two that had been randomly selected.

It would be wonderful if another team could take these measures and we can see if they line up. On Saturday, I will see if we have the spare manpower to re-do the tests, as I imagine they will be important for this year's camera system as well.

[Jared Russell]

I did it today and got about 34 degrees vertical, which seems in line with the data sheets scattered about the web.