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Update faq.rst (#1970) 

The memory math is confusing and incorrect. Original posted added a 5th camera but did not update the final memory size or explain how they arrived at that value.
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jethornton 2017-08-21 10:25:44 -05:00 committed by Isaac Connor
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@ -61,7 +61,7 @@ In *general* a good estimate of memory required would be:
::
Min Memory = 1.2 * ((image-width*image-height*image buffer size*target color space*number of cameras/8/1024/1024 )
Min Bits of Memory = 20% overhead * (image-width*image-height*image buffer size*target color space*number of cameras)
Where:
* image-width and image-height are the width and height of images that your camera is configured for (in my case, 1280x960). This value is in the Source tab for each monitor
@ -69,11 +69,18 @@ Where:
* target color space is the color depth - 8bit, 24bit or 32bit. It's again in the source tab of each monitor
The 1.2 at the start is basically adding 20% on top of the calculation to account for image/stream overheads (this is an estimate)
So let's do the math. If we have 4 cameras running at 1280x960 with 32bit color space and one camera running at 640x480 with 8bit greyscale color space, the system would require:
The math breakdown for 4 cameras running at 1280x960 capture, 50 frame buffer, 24 bit color space:
::
1280*960 = 1,228,800 (bits)
1,228,800 * 24 = 2,359,296,000 (bits)
2,359,296,000 * 50 = 5,898,240,000 (bits)
5,898,240,000 * 4 = 7,077,888,000 (bits)
7,077,888,000 / 8 = 884,736,000 (bytes)
884,736,000 / 1000 = 884,736 (Kilobytes)
884,736 / 1000 = 864 (Megabytes)
864 / 1000 = 0.9 (Gigabyte)
``1.2 * ((1280*960*50*32*4/8/1024/1024 ) + (640 *480 *50*8/8 /1024/1024))``
Or, around 900MB of memory.
Around 900MB of memory.
So if you have 2GB of memory, you should be all set. Right? **Not, really**: