How much storage space is required to store 90-day records?

Hello, Alisa!
Have a nice day to you!
The answer is [0.512Mb/s*3600s/h*24h]/8*90*2142/1024=XX(T)
Thanks!

Thanks for your answer!
Can you tell me more about it? I want to know the exact method so that I can calculate it myself.

Hello, dear!
Please see the link below:

How to calculate the storage space?
https://forum.huawei.com/enterprise/en/how-to-calculate-the-storage-space/thread/561845-887

Hope the answer can help you!

The resolutions of video surveillance images include CIF, D1, 720P, and 1080P. These resolutions have typical bit rates such as 512Kbps, 1 Mbit/s, 2 Mbit/s, 4 Mbit/s, and 8 Mbit/s. In fact, regardless of the analog image or digital image, the video storage space is calculated based on the bit rate, which is not related to the resolution.

The storage capacity is calculated based on the bit rate. The formula for calculating the storage capacity is as follows:

1 hour storage capacity of a camera with the bit rate of n (unit: Mbit/s) =[(n/8) *3600]/1024 GB



Formula parsing:

n Mbps/8 Conversion unit: n MBps

n MBps*3600 Calculate the storage capacity in one hour (unit: MB).

n *3600/1024 The unit is GB.

The following provides an example to describe how to calculate the raw capacity required for storing video images. Assume that there are 100 channels of 720p IPC, the storage bit rate is set to 2 Mbit/s, and all day recording is required. The storage duration is 30 days.

The calculation process is as follows:

(1) First, calculate the raw storage capacity required by a single video every day.

[2Mb/s*3600s/h*24h]/8=21.1GB



(2) Calculate the raw storage capacity required by a single video for 30 days.

[2Mb/s*3600s/h*24h]/8*30=633GB



(3) Finally, calculate the raw storage capacity of 100 channels of videos in 30 days.

[2Mb/s*3600s/h*24h]/8*30*100=63300GB/1024=61.8T

Good answer