How to calculate the storage capacity of the network camera
For video surveillance solutions to choose the appropriate storage device, you first need to estimate the storage capacity needs.According to the current mainstream compression calculation is as follows:
Formula: stream * time * video because present 8 = N (Byte)
Video streaming: every road transport stream, usually with Kbps, Mbps for unit, said bits per second flow: such as the 720 p streaming is set to 2048 Kbps, namely 2 Mbps
Time: (how many seconds per hour), how many days, months...Video goes: how many roads throughout the storage needs to be completed video data storage.
Unit conversion: 8 bit = 1 byte (bit is bit, byte is byte, one byte is eight bits), 1024 KB = 1 MB and 1024 MB = 1 gb, 1024 g = 1 TB.(h. 264 compression way: D1 clear definition, real-time 25 frames, 500 m/H/road; h. 264 compression way: HAFL D1 resolution, real-time 25 frames, 300 m/H/road; h. 264 compression way: CIF resolution, real-time 25 frames, 200 m/H/road)
Such as: 720 p in the stream of 2 m, 1080 p under the condition of stream 4 m each takes up much capacity per hour?
Formula: stream * time * video because present 8 = N (Byte)
P = 2 MB/S 720 * 3600 * 1 = 900 m8
1080 p = 4 MB/S * 3600 * 1 = 1800 m = 1.8 G 8
Note: (in a single road 720 p hd video camera for example, when the rate is set to: 2 MB/s) single road 720 p storage capacity for a day: (2 MB/s (3600 s) / 8 bit * 24 hours * 1 road = 21 gb;Single way 720 p a month storage capacity is: (2 MB/s * 3600 s) / 8 bit * * 30 days * 24 hours 1 road = 648 gb.
Hint: about the need to communicate with you A few questions: A, use how much compression resolution (CIF, D1, 720 p, 720 p);B, use what size (2 MB/s, 4 MB/s, 8 MB/s, 16 MB/s);C, video system how many road (1 ch, 2 ch...).;How long D, video, save;E, centralized storage or back-end storage before?
