>Date: Tue, 2 Nov 1999 08:09:58 -0800 (PST) >From: Tom Eastep <[EMAIL PROTECTED]> >> Now to the ques-- why, the minimum ethernet datasize is 46 bytes? Why >The minimum size of an ethernet frame is 64 bytes; this is to ensure that >the first byte of the frame reaches the receiver before the last byte is >transmitted. Given that there are 18 bytes of header and framce check >sequence in an ethernet frame, the minimum data size if 46 bytes. I suppose the reason is a bit different. Maximum length of cable is 2600m (5*500+2*50), and the time signal needs to travel it is 13.5us, which is enough to send 135 bits, providing no delay on transceivers or repeaters. Minimal packet has 512 bits plus preamble (usually 56 bits) = 568 bits, over 4 times the time. So, why packets need to be long? If two stations start sending at almost the same time, it is required for both stations know about collision. It means first station must still be sending when getting collision signal from the second. I suppose twice the network delay would be sufficient if there would be no more delays than this on cable, but there might be, and it is assumed sum of all delays - including time needed by the collising detection circuitry (it must be at least 1us) - is small enough for these 600 bits to be sufficient, including some "margin of safety" - to allow error called "late collision" if time when it is detected is larger than can be expected from allowed network size (sorry, I don't remember and cannot find the time after which collision is late - maybe 320 bits is reasonable). The calculation of 46 data bytes is correct (14 bytes are used for two addresses and length/type field, 4 for CRC). Jerzy - To unsubscribe from this list: send the line "unsubscribe linux-net" in the body of a message to [EMAIL PROTECTED]
