CCTV Bandwidth & NVR Storage Calculator
Work out the NVR storage, network load and recorder count for an IP camera fleet from its bitrate and retention.
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Worked example: 20 cameras at 6 Mbps for 30 days
Usable storage needed: 38.88 TB (35.36 TiB)
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How it works
CCTV storage and bandwidth formula
CCTV storage is the bitrate written to disk multiplied by the time it is kept. A 1 Mbps stream writes 10⁶ bits each second, so in decimal units each hour of it is 450 MB and each day 10.8 GB; multiply by the cameras and the retention days. Motion and scheduled recording write for part of the day, so storage is sized on the 24-hour average.
Bandwidth is sized at the peak, with every camera sending at once, plus the RTP, UDP, IPv4 and Ethernet headers that carry the video. The bitrate itself is the one you enter, from the camera maker's calculator, the VMS or a measurement: it depends on the encoder settings and the scene, and no standard gives a typical figure for a resolution or codec.
G = Ravg × 3600 × 24 ÷ 8 ÷ 1000
Ravg = R × (1 + oₛ) × h ÷ 24, continuous or scheduled
Ravg = R × (1 + oₛ) × h × (a + i(1 − a)) ÷ 24, motion
S = G × N × D, in GB; S ÷ 1000 in TB; S × 10⁹ ÷ 2⁴⁰ in TiB
B = N × V × (1 + oₙ), with oₙ at least (40 + p + 38) ÷ (1500 − 40 − p)
- G
- storage per camera per day, in decimal gigabytes (10⁹ bytes)
- R, Ravg
- bitrate written to disk per camera, in Mbps (the stream bitrate for each sensor, plus a recorded substream and audio), and its average over 24 hours
- oₛ
- storage overhead for container and VMS metadata on disk, as a fraction; none unless your VMS documents one
- h
- recording window, in hours per day; 24 for continuous recording
- a, i
- motion recording: the share of the window with activity, and the idle rate as a share of the full rate
- N, D, S
- cameras in the group, retention in days, and the group's storage in GB
- B, V
- network load, in Mbps, and the video bitrate per camera, main stream and any substream
- oₙ, p
- network overhead as a fraction, and the RTP payload header in bytes: 2 for H.264, 3 for H.265, 8 for Motion JPEG; 40 bytes are the RTP, UDP and IPv4 headers and 38 the Ethernet framing of a 1500-byte frame
Standards
- IEEE 802.3af, 802.3at and 802.3bt PoE, sized on PSE port reservation
- RFC 3550, RFC 768 and RFC 791: RTP, UDP and IPv4 headers; RFC 6184, RFC 7798 and RFC 2435: H.264, H.265 and JPEG payload formats, for the network overhead floor
- SNIA DDF 1.2: RAID 0, 1, 5, 6 and 10 usable capacity with parity in whole drives, and RAID 1 as a 2 or 3 drive mirror
- IEC 80000-13: decimal and binary prefixes
Worked example
Worked example: 20 cameras at 6 Mbps for 30 days
Twenty IP cameras record H.265 continuously at 6 Mbps each, a bitrate entered from the camera maker's calculator, and the video is kept for 30 days with nothing added for container or VMS metadata.
| Step | Working | Result |
|---|---|---|
| Per camera per hour | 6 × 3600 ÷ 8 ÷ 1000 | 2.7 GB |
| Per camera per day | 2.7 × 24 | 64.8 GB |
| Whole fleet for 30 days | 64.8 × 20 × 30 | 38,880 GB |
| Usable storage needed | 38,880 ÷ 1000, and 38,880 × 10⁹ ÷ 2⁴⁰ | 38.88 TB (35.36 TiB) |
That is usable space: RAID parity, hot spares and any headroom reserve come on top. On the network the same 20 cameras send 126.67 Mbps, which is 120 Mbps of video plus 5.56 per cent, the least the RTP, UDP, IPv4 and Ethernet headers add to H.265 in full 1500-byte frames.
How much storage does a CCTV camera use per day?
Each 1 Mbps recorded around the clock fills 10.8 GB a day: 3600 seconds times 24 hours, divided by 8 bits in a byte, in decimal gigabytes. So a camera recording 4 Mbps continuously fills 43.2 GB a day and 1.30 TB in 30 days. Motion recording fills less, depending on how much of the day it records at the full rate and what it writes in between.
How much bandwidth does a CCTV camera use?
Its stream bitrate plus the packet headers that carry it. The bitrate is set by the encoder settings and the scene, so take it from the camera maker's calculator, the VMS or a measurement. RTP, UDP, IPv4 and Ethernet headers add at least 5.49 per cent for H.264, 5.56 per cent for H.265 and 5.92 per cent for Motion JPEG in full 1500-byte frames, and VLAN tags, RTSP over TCP or a tunnel add more. Twenty cameras at 6 Mbps of H.265 put 126.67 Mbps on the uplink before anyone watches live, and each unicast live viewer pulls its own copy of the streams it watches.
Why is there no bitrate estimate?
Bitrate depends on the encoder settings and the scene, and no standard gives a typical figure: H.264 and H.265 set only a maximum per level. Any estimate from resolution and frame rate would rest on coefficients with no source, so the calculator takes the figure you type and sizes everything from it.
What is the difference between TB and TiB?
A terabyte is 10¹² bytes, which is how drives are sold. A tebibyte is 2⁴⁰ bytes, which is what most operating systems report, so a 10 TB drive shows as about 9.09 TiB. Sizing in one convention and buying in the other leaves a shortfall of about 9 per cent.
How is motion recording modelled?
Motion mode records at the full rate for the active share of the recording window and at an idle rate for the rest, which can model pre and post event buffering. Storage is based on the daily average, while bandwidth is sized at the peak, because the network must carry every camera recording at once.
Why does fitting larger drives not reduce the recorder count?
Because capacity may not be the limiting constraint. A recorder is also limited by its incoming bandwidth and its channel count, and the recorder count is set by whichever is most demanding. If throughput or channels bind, the fleet needs to be split across more recorders or the stream load reduced.
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