I'm looking at some of the study notes for CNX from www.optimized.com and I
can answer most situations but there are two that confuse me:
a) Significantly more than 8 bytes of "55" or "AA" hexadecimal data
appended to the end.
Four bytes get appended by the initial host (for local collisions) and 4
more by the repeater (for remote echoing). Would this be the second machine
noticing the collision (later) and appending it's four bytes too? And how
about the repeater coming back (for a total of 16 bytes)?
B) Streaming "00" or "FF" patterns of hexadecimal data appended to the end.
I have no clue. The closest I could come up with was that one book stated
that the the jam signal was not defined as a AAAA/5555 sequence and could be
any sequence ie FFFF or 0000.
I've read two Ethernet books and they tend to skim over top collisions and
don't go into any depth.
Thanx
--
Kevin L. Kultgen
"Priscilla Oppenheimer" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> At 10:20 PM 10/6/00, whatshakin wrote:
> >This makes it sound like there is actually something tangible being put
on
> >the wire. Bits are merely ones and zeros which are signaled by different
> >voltages etc in the line encoding.
> >
> >Bits do not occupy line space.
>
> Sure they do. Wasn't it Einstein that said time and distance are related?
> (Don't quote me on that.) Seriously, the discussion of how much space a
bit
> takes on the wire has been going on since Ethernet was invented. See
> Optimized Engineering's Technical Compendium for a discussion that brings
> up the space issue.
>
> http://www.optimized.com/COMPENDI/EN-Propa.htm
>
> And here's some more related trivia. Do you know why a 32-bit jam is sent
> when a station detects a collision? It's to increase the time and distance
> of the collision event. It's to avoid the case where the collision happens
> right next to you and you've stopped transmitting by the time the
collision
> event hits a repeater that could be 500 meters away.
>
> Why didn't they use a 16-bit jam? Because on thick coax Ethernet cable the
> signal travels at 231,000 kilometers per second. This means a bit occupies
> 23.1 meters on thick Ethernet. 16 x 23.1 wouldn't have been enough. 32 x
> 23.1 is 739 meters. An extension of 32 bits allows the sender to busy out
a
> maximum 500-meter segment. This explains both the jam size and the fact
> that a repeater extends a received fragment by at least 32 bits.
>
> I know this is a strange way of looking at things, but it is one way that
> engineers, including the inventors of Ethernet, looked at the Ethernet
> parameters.
>
> Priscilla
>
>
> >Measurements of how fast data can be moved over a wire are the time it
takes
> >for a signal at one end to be heard at the other. The amount of data
> >(signals) which can be moved across a wire is ascertained by the line
> >encoding method, and how many signals the encoding system can be made to
> >produce in a second. Minus the delay factors between point A and B of
> >course.
> >
> >I seem to recall reading some papers from folks at the US Berkley
computer
> >science dept a few years back that researched the various line encoding
> >techniques etc that were quite interesting. I cannot find them now that
I
> >need them though!!
> >
> >BTW, my calculations for the speed of light resulted in 299,793,100 m/s
> >
> >
> >----- Original Message -----
> >From: Leigh Anne Chisholm <[EMAIL PROTECTED]>
> >To: Kevin L. Kultgen <[EMAIL PROTECTED]>; Tim O'Brien
> ><[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
> >Sent: Friday, October 06, 2000 4:15 PM
> >Subject: RE: Ethernet Trivia
> >
> >
> > > Ahh Kevin, your post reminds me of some research I did when I was
putting
> >together my paper on LAN Switching for CertificationZone. I was looking
at
> >how to calculate the round-trip propagation delay for 10BaseT networks.
> >Here's a few technical numbers for you you (and possibly other Groupstudy
> >members) might find interesting.
> > >
> > > --- Beginning of Calculations ---
> > >
> > > Electrical signals travel in a copper wire travel (propagate) at
> >approximately two-thirds the speed of light. Remembering that the speed
of
> >10 Mbps Ethernet is 10,000,000 bits/second, we can determine the length
of
> >wire that one bit occupies, by using the following calculation:
> > >
> > > Speed of Light in a Vacuum = 300,000,000 meters/second
> > >
> > > Speed of Electricity in a Copper Cable = 200,000,000 meters/second
> > >
> > > 20,000,000 meters/second / 10,000,000 bits/second = 20 meters per
bit
> > >
> > > The minimum size Ethernet frame consisting of 64 bytes (512 bits)
occupies
> >10,240 meters of cable.
> > >
> > > --- End ---
> > >
> > >
> > > -- Leigh Anne
> > >
> > >
>
>
> ________________________
>
> Priscilla Oppenheimer
> http://www.priscilla.com
>
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