Come to think of it, despite my last response,
bits don't occupy space, at least not in theory.

Manchester encoding, used in ethernet, signals a
bit as a one or a zero depending on the
instantaneous change in voltage from +1 to -1
or -1 to +1 volt.  The time spent at a particular
voltage is just that, time spent.  The bit itself
is signal with the voltage change, which in theory
is instantaneous.

Of course, in reality there is no such thing as
a square wave.

-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
whatshakin
Sent: Saturday, October 07, 2000 1:23 PM
To: [EMAIL PROTECTED]
Subject: Re: Ethernet Trivia


Comments inserted.

----- Original Message -----
From: Jay Hennigan <[EMAIL PROTECTED]>
Newsgroups: groupstudy.cisco
To: <[EMAIL PROTECTED]>
Sent: Saturday, October 07, 2000 9:14 AM
Subject: Re: Ethernet Trivia


> On 7 Oct 2000 01:20:43 -0400, whatshakin <[EMAIL PROTECTED]> 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.  Ever see the terms "wavelength" or "short wave" on a radio?

>
> Inversely proportional to the frequency, wave length is the physical
length
> of a signal, based on the distance in free space for one cycle at a given
> frequency.  As the speed of light is slower in media such as twisted pair
> copper and fiber, the length of a bit at a given frequency is shorter than
> it would be in free space.

The physical length of a signal is not inversely proportional to its
frequency.  It differs depending on the line encoding.   Again, a bit is the
term applied to the signal state.  Signal
states occupy line space.
>
> The ones and zeros obviously travel along the wire from the sending to
> the receiving end.  If you could freeze time and take a snapshot, you
> would see a length of wire with a positive voltage, followed by one of
> negative charge, the lengths corresponding to bits.
>
This is quite a good hypothetical scenario, and is indeed correct.

> :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.
>
> And those delay factors are the speed-of-light propagation delay of the
> medium, the delay proportional to the length.  Distance (length on the
wire)
> equals velocity (speed of light in the medium) divided by time (length of
> a bit in fractions of a second).
>
Your formula is correct, however, it does not apply very well to finding
delay propogation over a wire because of the numerous other factors which
need to be applied additionally.  IE: The properties of the wire medium,
EMF, block coding, IFG, protocol overhead...


> :BTW, my calculations for the speed of light resulted in 299,793,100 m/s
>
> Which method did you use?  Laser and a spinning mirror?  :-)

Very observant!  ;-)

>
> --
> Jay Hennigan  -  Network Administration  -  [EMAIL PROTECTED]
> NetLojix Communications, Inc.  NASDAQ: NETX  -  http://www.netlojix.com/
> WestNet:  Connecting you to the planet.  805 884-6323
>
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