Wikipedia:Reference desk/Archives/Mathematics/2016 March 30

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March 30

Actual resolution of a clipped fisheye lens?

I have a complex math geometry question. I don't know if I can ask this here. If I should try someplace else please suggest something.

I'm mainly doing this research just for myself but it could possibly be useful in the article Fisheye lens.

Background:

I am researching fisheye security cameras. Some cameras have an entirely circular 360 degree view and it is easy to determine the actual resolution as simply the area of the circle.

Some however have a circular lens that overhangs the edges of the sensor, so the sides of the circle are partially clipped off. This allows more image sensor usage but at the expense of reducing side view, and is best used in long rectangular spaces like hallways.

Non-free image example hosted on imgur.com: http://imgur.com/VuLqFJ5

Product manual:

• Hikvision DS-2CD2942F-IS 4Mp Indoor D/N Network Fisheye Camera
• http://www.hikvision.com/UploadFile/image/2014101611082062758.pdf
• Lens field of view: 186°(horizontal), 106°(vertical)
• Image sensor resolution: 2560 x 1440

I'm not sure how they claim more than 180 degrees field of view, but I believe it's because the lens protrudes out from the base and the base itself sticks out from the mounting surface, so the lens can see slightly into that area beside the camera body.

The question:

How would I calculate the active pixel viewing area for this camera, where the sensor can actually see something, and exclude the black wedges that never detect anything?

I am aware of the imprecision of this calculation, especially around the lens edges where the circular view reflects/refracts and rolls off to infinity. See non-free example: http://imgur.com/4dl9Cqx

Is there a way to determine the usable pixel area calculation directly from the image, vs doing an abstract formula of the resolution and field of view angles?

-- DMahalko (talk) 22:47, 30 March 2016 (UTC)

It looks like you have a circle with a circular segment cut off at the top and bottom. Once you find that net area, you can compare that to the area of the rectangle to figure out what percentage of it is used. StuRat (talk) 04:04, 31 March 2016 (UTC)

The area of the clipped circle is ${\tfrac {1}{2}}wh+{\tfrac {1}{2}}d^{2}\theta$, where $h$ is the height, $w$ is the minimum (top and bottom) width, $d={\sqrt {w^{2}+h^{2}}}$ is the diameter (=maximum width), and $\theta =\tan ^{-1}(h/w)$ is the angle of the diagonal from the horizontal (in radians). You could measure w and h in pixels in a full-resolution sample image and plug them into that to get a total pixel count. -- BenRG (talk) 04:01, 31 March 2016 (UTC)