10Base5/Thicknet (was Re: SUN networking problems)

From: Pete Turnbull <pete_at_dunnington.u-net.com>
Date: Sat Mar 24 06:43:49 2001

On Mar 23, 17:42, Bill Gunshannon wrote:
> Two points.
> First, "a piece" of yellow cable won't work. The lengths were part of
> spec and very important. Having to do with reflections and such (anybody
> here still have a TDR??)

I've not seen anything about lengths, except of course for the maximum
length and multiple-of-2.5m tap separation. I'm told the standard
specifically allows for lengths to be joined at intervals which are *not*
multiples of 2.5m, and the overall length does not have to be an exact
multiple either.

> Second, The cable is marked with black stripes. The taps must go on
> stripes. While I have known people who put the transcievers with the N
> type connectors between yellow segments and on the ends of yellow
> I have never known seen it recommended that you could/should cut the
> to insert one of these. I fear that after one or two of them, you would
> move the spacing between tap locations enough to adversely effect your
> network.

I've never been aware of a problem with that. Our old Departmental network
(installed by my predecessors) consisted of several segments, many of which
had lots of N-series transceivers in them. I suppose, though, that when
using N-series transceivers, it might make some sense to chop out a small
piece of coax rather than just cutting it.

I don't have a copy of the standard, and I don't know if it specifies
vampire taps rather than the "taps" with N-connectors (which I dare say are
more properly called "tees") but I do know that it specifically allows a
segment to be made up of sections of cable joined with N-connectors, and
that simple joins do not have to be at multiples of 2.5m. Admittedly, good
practice was to use the same cable (ideally from the same drum) for each
piece, to prevent small impedance mismatches.

To do the maths, I find that the velocity factor for RG8/11U is 0.78, the
standard value for c (velocity of light in vacuo) is 2.998 x 10^8 ms^-2,
and the nominal frequency used for Ethernet is 10MHz, or 10^7 Hz. So one
cycle of the waveform occupies 0.78 x 2.998 x 10 m, which is 23.38
metres[1]. That sounds about right; that would mean about 21 cycles over a
500m length. But there's no obvious relationship between 23.something
metres and the 2.5 metre separation distance, it's about 1/9th. From that,
I can't believe that a small deviation in separation distances is going to
make a significant difference.

Unless someone can correct my assumptions, of course!

[1] Nitpickers might point out that my figure for the velocity factor is
less accurate than the 4 significant figures I used for c. So if I take
 extremes (still within the Ethernet spec) at 0.77 and 0.80, the results
would be 23m and 24m respectively.

> This is, of course, assuming a production network and not two machines.
> But then. if all you had were two or three machines, a couple hundred
> feet of yellow cable seems pretty silly.

Unless they're a long way apart, or you're doing it for
demonstration/nostalgic/"because it's there" purposes :-)

Pete						Peter Turnbull
						Network Manager
						Dept. of Computer Science
						University of York
Received on Sat Mar 24 2001 - 06:43:49 GMT

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