I'm sure there is a formula somewhere but this is how I calculate it.
for example : a camera running at 8192 Kbps means it's filling up your drive with 8192 kilobits per second. Storage is measured in BYTES though, not BITS, so you need to do a conversion.
There are 8 bits in a byte so at 8192 Kbps you need 8192/8 KiloBytes for every second of recording. So that would be 1024 KiloBytes per second.
There are 1024 KiloBytes in a MegaByte. so divide that by 1024 to convert KiloBytes to MegaBytes = 1.0 MegaByte per second.
So 8192 kbps (kilo bits per second) equals 1.0 MegaBytes per second.
There are 86400 seconds in a day. So 1.0 x 86400 = 86400 megaBytes of storage. There are 1024 MegaBytes in a GigaByte. 86400 divide by 1024 = 84.374GB
So, my completely unqualified formula is : bitrate in kbps divided by 8, then divide by 1,048,576, then multiply by 86400 gives you your daily storage requirements in GigaBytes.
Please feel free to correct if I've gone wrong. It's been many years since I last used all this stuff
