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Let t<sub>0</sub> be the time associated with the occurrence of an alarm. t<sub>0</sub>-t<sub>B</sub> is the time before the alarm to be recorded. t<sub>A</sub>-t<sub>0</sub> is the time after the alarm to be recorded. Consequently, t<sub>B</sub>-t<sub>A</sub> is the size of the entire window to be recorderrecorded. For example, for the the PoC here described:
*t<sub>0</sub>-t<sub>B</sub> = 8s
*t<sub>A</sub>-t<sub>0</sub> = 3s
*t<sub>B</sub>-t<sub>A</sub> = 11s
Once the size of the time window is knowknknown, it is straightforward to determine how much RAM is required for the buffer. In the case under discussion, the size of a frame is approximately 1280x720x8bpp=921600 byte. One second of recording is thus 921600x300≈264 MByte. In conclusion, the buffer has to be at least 264x11=2904 MByte to contain all the required frames.
===Alarm recordings===
The processing platform is equipped with an SSD that is used to store alarm buffers only. Of course, its size should be determined depending on the maximum number of alarms to be stored at the same time.
==Future work==