How to find out the maximum value of the payload in the ip packet?

Question

linux

A network of two devices is given. I do a ping from one to another, sending 4 packets of 3500 bytes each. I get the result in WebShark.

How to determine the maximum value of payload data in one packet (as I understand it, this is MTU)? You just need to take into account 20 + 8 reserved bytes, that is, there will be 3528, or is it because I get 3610 for each packet during fragmentation and I need to look at it?

Answer 1

each packet, if it is not fragmented, has its own IP header.
each packet, if it alone is part of a fragmented packet, has its own IP header.
the maximum payload value in an IP packet is the maximum ethernet frame size (MTU) minus headers (which are before the data, for ethernet ii without 802.1q - 14 bytes) and trailers (which are after the data, for ethernet ii - 0 bytes) ethernet frame and headers (20 bytes) and trailers (0 bytes) of the IP packet.

Answer 2

MTU size is 1500 bytes, jumboframe has much more

Answer 3

At you it is specified that fragmented ip. On a computer, as a rule, the standard MTU is 1500 bytes, minus 20 bytes of the IP header and 8 bytes of the ICMP header (for echo). 1472 bytes remain - these are the "data" that got inside the ICMP packet.
If you sent pings of 3500 bytes, then 3 IP packets were sent for each such ping. And you can clearly see it on your screen:
First packet - 1472 bytes of data + 8 bytes of ICMP + 20 bytes of IP + 14 bytes of Ethernet = 1514
Second packet - 1472 bytes of data + 8 bytes of ICMP + 20 bytes of IP + 14 bytes of Ethernet = 1514
Third packet - 556 bytes of data (3500-1472-1472) + 8 bytes of ICMP + 20 bytes of IP + 14 bytes of Ethernet = 598
in theory it should turn out like this, it's not entirely clear why the last packet on the screen is 582 bytes.
Yes, there are jumbo frames with MTUs up to 9120 bytes, but they only run between switches and some server NICs.