The emergence of network video systems calls for increased use of hard disk storage. This gives rise to a number of questions, ranging from how much hard disk is needed to how to build fail-safe hard disk storage.
Required hard disk space
Factors to consider when calculating storage needs:
Number of cameras
Number of hours per day the camera will be recording
How long the data must be stored
Motion detection (Event) only or constant recording
Other parameters such as frame rate, compression, image quality and complexity
Note that the calculation examples below are examples only and do not take into consideration any overhead or other technical issue that may result in a higher file size than mentioned below.
The calculation examples below do not consider storage space for the operating system or video management software.
JPEG or Motion JPEG
For JPEG/Motion JPEG where single files are received, storage requirements vary by changing the frame rate, resolution and compression: Cameras 1, 2 and 3 in the table below have different storage requirements according to their fps (frames per second) and resolution settings.
Calculation:
Image size x frames per second x 3600s = KB per hour /
1,000 = MB per hour
MB per hour x hours of operation per day/1000 = GB per day
GB per day x requested period of storage = Storage need
| Camera | Resolution | Image size (KB) | Frames per second | MB/hour | Hours of operation | GB/day |
| No. 1 | CIF | 13 | 5 | 234 | 8 | 1.9 |
| No. 2 | CIF | 13 | 15 | 702 | 8 | 5.6 |
| No. 3 | 4CIF | 40 | 15 | 2160 | 12 | 26 |
Total for the 3 cameras and 30 days of storage = 1002 GB.
MPEG-4
In MPEG-4, the images are received in a continuous data stream, i.e not individual files. It is the bit rate - measuring the amount of video data transmitted - that determines the corresponding storage requirements. The bit rate is a result of specific frame rate, resolution and compression, as well as the level of motion in the scene.
Calculation:
Bit rate / 8 (bits in a byte) x 3600s = KB per hour / 1,000 = MB per hour
MB per hour x hours of operation per day / 1000 = GB per day
GB per day x requested period of storage = Storage need
| Camera | Resolution | Bit rate (Kbit/s) | Frame per second | MB/hour | Hours of operation | GB/day |
| No. 1 | CIF | 170 | 5 | 76.5 | 8 | 0.6 |
| No. 2 | CIF | 400 | 15 | 180 | 8 | 1.4 |
| No. 3 | 4CIF | 880 | 15 | 396 | 12 | 5 |
Total for the 3 cameras and 30 days of storage = 204 GB
H.264
Calculation:
Bit rate / 8 (bits in a byte) x 3600s = KB per hour / 1,000 = MB per hour
MB per hour x hours of operation per day / 1000 = GB per day
GB per day x requested period of storage = Storage need
| Camera | Resolution | Bit rate (Kbit/s) | Frame per second | MB/hour | Hours of operation | GB/day |
| No. 1 | CIF | 110 | 5 | 49.5 | 8 | 0.4 |
| No. 2 | CIF | 250 | 15 | 112.5 | 8 | 0.9 |
| No. 3 | 4CIF | 600 | 15 | 270 | 12 | 3.2 |
Total for the 3 cameras and 30 days of storage = 135 GB
Note: The formula does not take into account the amount of motion, which is an important factor that can influence the size of
storage required.
See also: Storage considerations