[ntp:questions] Isolated Network Drift Problem

Joseph Gwinn joegwinn at comcast.net
Thu Nov 27 00:11:04 UTC 2008


In article <gYOdncZ8iqO7PrDUnZ2dnUVZ_ojinZ2d at giganews.com>,
 "Richard B. Gilbert" <rgilbert88 at comcast.net> wrote:

> Unruh wrote:
> > David Woolley <david at ex.djwhome.demon.co.uk.invalid> writes:
> > 
> >> Unruh wrote:
> > 
> >>> With your 100m setup you really want a buffer amp on the line. At 100m, 
> >>> the
> >>> one way trip is .3ms, with reflections every .6ms which might make the
> >>> system a bit weird. To get rid of the reflections you need a 100ohm
> >>> termination if you use cat5e cable to lengthen the 5m wire on the 
> >>> receiver.
> > 
> >> Round trip time for 100m of velocity factor 2/3 cable is approximately:
> > 
> >> 3 * 2 * 100 / (2 * 3E8)
> > 
> >> = 3E2/3E8
> >> = 1 microsecond.
> > 
> > I was taking the speed to be c. And it is only a one way trip-- from the
> > GPS to the computer, not round trip. So I will buy your .5us (having no
> > idea what the velocity factor is of cat 5e cable) but not your
> > round trip ( which is twice that but is irellevant on a properly terminated
> > cable)
> > 
> 
> "c" is not correct.  "c" is the speed of light in vacuum!  Electricity 
> in copper is somewhat slower.  1 nanosecond per foot is a useful rule of 
> thumb!  It's an approximation but a useful one.  If you need something 
> more exact, you might consult the manufacturer.  Cat 5 cable consists of 
> four twisted pairs with two turns to the inch.  The physical and 
> electrical lengths of the cable are slightly different!

Here are the standard rules-of-thumb:

For typical solid dielectrics, propagation is 66% of C.  This also 
applies to light in silica optical fibers.

For typical foam dielectrics, propagation is 75% to 85% of C.

Joe Gwinn




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