hi..i have some problems in calculating the storage needs in a cctv system.
i know that the factors that i have to considered when calculating storage needs are:
-number of cameras
-number of hours per day, that the camera will be recording
-how long the data must be stored
- frame rate , compression, image quality and complexity

for the bellow exemple :
if i have 1 camera whit:
resolution= 640 * 480
frames per second=25
period of storage= 8 hours
compression = MJPEG (50)

i know that i have to use this formula (FOR 1 CAMERA):
required storage space=image size* frames per second * period of request

my principal question is how i can calculating the image size?
how resolution and compression influences the image size?
can anybody help me? :?
best regards

please check the time recording table of the DVR. if you don't have it, you can have an actual test. connect one camera to the dvr and select you desired setting. record for at least 2 mins and note down the quality setting and fps setting, from there you can estimate the space the will be occupied by the video for a certain setting.

i don´t have the time recording table of the DVR. what you are saying is that is impossible to know the storage needs even if i knew the resolution,frame rate , compression mode(Mpeg4 or mjpeg), number of cameras, etc....unless i do a pratical test. so, when i need install a diferents cameras or DVR´s, whit a diferents resolutaion , i have to do new tests? thanks for your help, i am a newbie in cctv, and all information are welcome.

what dvr are you using?

i'm going use the samsung SHR-2040

check this link http://www.samsung-security.com/Portals/0/UserManuals/SHR-2040-2041-2042-ENG-F.pdf

visit this site for samsung SHR-2040 storage calculator http://www.gviss.com/default.aspx?tabID=435