Wednesday, November 19, 2008
Fun with math
An e-mail came in asking about predicting file storage requirements and its implications for future analysis. The reader wants to know how the math works. So ... here's a good example.
704 x 480 is a common frame size for digital CCTV. Let's assume that we're dealing with a colour system. With colour, you have 3 channels - red, green, and blue.
704 x 480 x 3 = 1,013,760 bytes = 1,013.76 kb = 1.014 Mb/frame
Then, divide by the compression (remember - compression = loss) factor. Some companies compress at 5:1 - some 60:1 - some 240:1 (ouch).
5:1 = 200 kb / frame - the computer discarded over 800 kb of data.
60:1 = 16.9 kb / frame - the computer discarded almost 1 Mb of data.
Once you've arrived at your frame size, simply multiple the size by the frames per second.
16.9 kb x 30 (fps) x 60 (sec/min) x 60 (min/hour) x 24 (hours/day) = 1 day's storage requirement.
0.0169 x 30 = .507 Mb
.507 x 60 = 30.42 Mb
30.42 x 60 = 1.825 Gb
1.825 Gb x 24 = 43.8 Gb
44 GB / day = 1.3 Tb / month (30 days)
So how do they get more time on the drive? Compress even further, reduce the frames / second, record only on motion, record only during business hours, and so forth.
As far as the implications are concerned, the more that's thrown away combined with the choice of what's discarded will determine the usefulness of the recorded frames for future analysis.