- Decide what average bitrate the camera will be set to. This will depend on the resolution, frame rate, image quality, and camera capabilities. As a ballpark figure for a 1080p camera set to 12 FPS, you can use 750 Kbps.
- Decide how many cameras will be used.
- Find out if the internet connection will be shared with other users or will be a dedicated channel for cameras only.
Estimate total bandwidth requirements
To find total bandwidth requirements, multiply average bitrates by the number of cameras, then add any bandwidth required for other users if the connection is not exclusive to cameras.
It is a good practice to add 1 Mbps of overhead for every 5 cameras to account for metadata traffic, spikes in bandwidth consumption, and congestion on the line of your ISP to ensure stable performance. Without this overhead, any instability in any of the components (camera, router/modem, ISP lines) will have a higher chance of causing lost or partial frames, lost event metadata, camera disconnects, etc.
For example, for a project that has 7 cameras set to 1080p and 12 FPS with a dedicated internet channel, the total bandwidth requirement will be:
7 cameras * 750 Kbps + (1+1) Mbps overhead = 5250 Kbps + 2 Mbps = 7.25 Mbps of upload will be enough to ensure stable performance for these 7 cameras, assuming cameras in fact can maintain a 750 Kbps average bitrate. 2 Mbps of overhead is recommended in this example because the number of cameras is greater than 5 but less than 10. But if your internet connection is not very stable, with periods of higher and lower speeds, or if it is shared with other users, a much bigger overhead is recommended.
Approximate data consumption calculations
To calculate approximate storage consumption, simply multiply the number of cameras by the average bitrate and the time for which you want to know the storage requirement of data consumption.
For example, for the above project with 7 cameras configured to 750 Kbps average bitrate, the calculations are as follows:
1. Find the total bitrate for all cameras 7 * 750 = 5250 Kbps
2. Convert Kbps into Kilobytes by dividing by 8, so 5250 / 8 = 656.25 KB is generated every second
3. Convert into MB by dividing by 1024, so 656.25 / 1024 = 0.6408 MB
4. Convert into GB by dividing by 1024 again, so 0.64 / 1024 = 0.0006258 GB
5. Convert into GB per hour by multiplying by 3600, so 0.0006258 * 3600 = 2.25 GB is generated every hour
6. Convert into GB per day by multiplying by 24, so 2.25 * 24 = 54 GB is generated every day
7. Multiply by the number of days to estimate consumption per week, per month, or per year, for example, per month would be 54 * 30 = 1620 GB of data will be generated by these 7 cameras in one month.
For more information on how much bandwidth is needed for cameras, please visit this article↗️