ANNOUNCEMENT: Live Wireshark University & Allegro Packets online APAC Wireshark Training Session
July 17th, 2024 | 10:00am-11:55am SGT (UTC+8) | Online

Ethereal-users: Re: [Ethereal-users] (Slightly OT) Ethernet length question

Note: This archive is from the project's previous web site, ethereal.com. This list is no longer active.

From: Jaap Keuter <jaap.keuter@xxxxxxxxx>
Date: Wed, 18 Jan 2006 10:54:36 +0100 (CET)
Hi,

The layer protocol model is the best reference you can take in mind.
The minimum ethernet frame length (that is bits on the wire) is 64*8.
Together with the wirespeed this gives a minimum amount of transmission
time. This time is needed for the other endpoint (in a point-to-point
link) or other stations (on a shared medium) to assure recognition of the
frame and collision avoidance. All this is Physical layer stuff.

Then comes the meaning of the bits in the frame. If the ethernet header
says that it's a VLAN tag that is following then the next 2 octets are to
be interpreted that way. From there on the next protocol layer starts.

That is the true meaning of VLAN, a VIRTUAL LAN. It's not a real one, as
seen on the wires, but on a layer above that.

Hope it helps,
Jaap


On Wed, 18 Jan 2006 Ran.Shenhar@xxxxxxxxxxx wrote:

> Hi All,
> I need help with a non-ethereal specific - I know that an ethernet frame is
> of minimum length = 64 bytes, including FCS.
> However, what is the minimum length of a VLAN tagged ethernet frame? Is it
> 64+4, or is it still 64?
>
> I tried looking in IEEE 802.3q 2003 edition, however I was not able to find
> a definite answer.
> If you could also point me to a reference about the answer, it'd be even
> better.
>