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→Sizing the alarm buffer
*t<sub>0</sub>-t<sub>B</sub> = 8s
*t<sub>A</sub>-t<sub>0</sub> = 3s
*t<sub>BA</sub>-t<sub>AB</sub> = 11s
Once the size of the time window is known, 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.