[ntp:questions] Isolated Network Drift Problem
unruh-spam at physics.ubc.ca
Wed Nov 26 23:45:39 UTC 2008
"Richard B. Gilbert" <rgilbert88 at comcast.net> writes:
>> 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
>"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
1foot/nanosecond is to within about 2% the speed of light in vacuum.
The speed of light in "copper" is the same, depending on how the copper is
arranged. However, the dielectric in most waveguides/lines has a reasonably
high index of refraction at low microwave, and the twisting of the pairs
Cat5e is also important. Looking on the web it seems that a velocity factor
of about .75 for cat5e cable is typical, and impedance of 100ohms.
>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!
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