why minimum Ethernet payload size is 46 bytes ?

轉載於 http://www.itechtalk.com/thread212.html

shows Ethernet Node A and Ethernet Node B at the farthest ends of a 5-4-3 network
using 10Base5 cabling.
When Node A begins transmitting, the signal must propagate the network length. In the worst-case collision scenario, Node B begins to transmit just before the signal for Node A’s frame reaches it. The collision signal of Node A and Node B’s frame must travel back to Node A for Node A to detect that a collision has occurred.
The time it takes for a signal to propagate from one end of the network to the other is known as the propagation delay. In this worst-case collision scenario, the time that it takes for Node A to detect that its frame has been collided with is twice the propagation delay. Node A’s frame must travel all the way to Node B, and then the collision signal must travel all the way from Node B back to Node A. This time is known as the slot time. An Ethernet node must be transmitting a frame for the slot time for a collision with that frame to be detected. This is the reason for the minimum Ethernet frame size. The propagation delay for this maximum-extent Ethernet network is 28.8 μs. Therefore, the slot time is 57.6 μs. To transmit for 57.6 μs with a 10 Mbps bit rate, an Ethernet node must transmit 576 bits. Therefore, the entire Ethernet frame, including the Preamble field, must be a minimum size of 576 bits, or 72 bytes long. Subtracting the Preamble (8 bytes), Source Address (6 bytes), Destination Address (6 bytes), EtherType (2 bytes), and FCS (4 bytes) fields, the minimum Ethernet payload size is 46 bytes.