Please excuse this apparently naive question. no one appears to have the answer.

If a standard packet of data is travelling down a standard fibre, what is the physical time taken between the first digit of data passing a point and the last digit of data. passing the same point, and consequently what is the physical distance of a packet in the conductive material measure from the first unit of data to the last data .

Is it a mile, two miles or 10 feet. please let me have a ball park figure in simple to understand terms.

  • $\begingroup$ the question is inteneded to seek the answer to "How long is a standard "packet" of data. The orginal question is posed wrongly. my appologies $\endgroup$ – Stewart John Lobb Jul 12 '15 at 22:49
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    $\begingroup$ AFAIK, there isn't a "standard" packet of data; the length of the packet is a parameter in the encoded information. Thus, I don't see how there is a way to answer this question. $\endgroup$ – Kyle Kanos Jul 12 '15 at 23:34
  • $\begingroup$ I got the STM-1 datarate wrong: pls see my answer, using the updated data. $\endgroup$ – WetSavannaAnimal Jul 13 '15 at 1:36
  • $\begingroup$ @KyleKanos Actually, in modern optical fiber networks, there is pretty much a standard - SDH: see my answer (although there's not much physics in it). At least for the time being. Optical LANs will use different signalling protocols - like Optical Gigabit Ethernet- from SDH when they become widespread. $\endgroup$ – WetSavannaAnimal Jul 13 '15 at 1:39

https://en.wikipedia.org/wiki/Optical_fiber states that a rule of thumb is that the speed of light in an optical fiber is 200,000 kilometers per second.

https://en.wikipedia.org/wiki/Fibre_Channel states the fibre channel networks run at 2,4,8 and 16 gigabits per second.


states that a fibre channel frame can have up to 2148 bytes.

Assume an average frame is 1024 bytes or 8192 bits.

So it takes .000004096 seconds per frame at 2 gigabits per second.

In that amount of time light in the channel travels .8192 kilometers or 819.2 meters.

For the length of a packet at the other rates just divide by 2.. 4 gigabits/sec yields 409.6 meters etc.


It depends on the packaging: it's highly variable. But by far the most common one is defined by the Synchronous Digital Hierarchy standard or the almost identical American SONET standard. This is pretty much standard on all optical trunk networking. The data for this answer and your question will almost certainly have passed through an SDH link for the long-haul part of its journey.

SDH has as a basic unit the STM-1 (Synchronous Transport Module) frame of $270{\rm columns}\times 9 {\rm rows} = 2430{\rm bytes} =19440{\rm bits}$ at $155.82{\rm MBps}$. The bit length, in an optical fiber of refractive index 1.5 is then $c/(1.5×1.5582×10^8)=1.28{\rm m}$. The STM-1 frame length is thus about 25km.

SDH uses a hierarchy of transport modules, but the bitrate is in proportion to the transport module length, so the frame spatial length is the same for all members: thus, the highest member used is STM-256, whose packets in bits are 256 times as long as STM-1, but they are sent 256 times as fast (40GBps), so these are also 25km long. Above this signalling speed it becomes more economical to wavelength division multiplex STM-256 channels. You can see why. At this colossal signaling speed, the each bit is only 5mm long, so delays in signalling circuitry become very expensive to correct. Not to mention the increased sophistication of hardware to correct the effect of optical dispersion on so wide (in frequency, and thus wavelength) channel. At 1550nm, 40GBps corresponds to a wavelength spread of $\Delta\lambda=-\lambda\Delta\nu/\nu = -\lambda^2\Delta\nu/c = -1.55\times10^{-6}\times 40\times 10^9/3\times 10^8 = 0.032{\rm nm}$, which is actually quite significant when its effect on channel impulse response is calculated.

Outside of the SDH - which is used for most trunk telecommunications, there are a wide variety of standards used, which have different packet lengths. Optical Gigabit Ethernet is one.


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