ssw@cica.cica.indiana.edu (Steve Wallace) (10/30/90)
How far can you run ethernet over fiber before the collision
detecting mechanism breaks down? I came up the following which
seems too far:
Min. frame size = 480 bits
Time to transmit min. frame = frame size / transmission rate
= 480 / 10,000,000 bits per sec.
= .000048 sec.
Distance light travels down the fiber during the
transmission of min frame = .000048 * .8c
= .000048 * 240,000,000 meters/sec
= 11,520 meters
11,520 meters is the round trip time. This divided by 2
(the sender A must hear sender B transmitting) gives a ball park
figure (5,760 meters). I'd give myself a fudge factor of 30
percent and end up with about 4,000 meters. Is this close?
Thanks,
Steven Wallace
wallaces@ucs.indiana.educmilono@netcom.UUCP (Carlo Milono) (10/31/90)
In a perfect media, I believe that you are correct; however, fiber optic cables are *not* perfect, and each batch extruded from the original doped tube has a vintage, much like a wine - with peculiar characteristics. What you see in the industry is the FOIRL standard; you may see a 10BASEF before too long, but I would guess that it would be very much the same as the FOIRL. The practical limits (today) appear to be about half of what your fudged number is: 2km. -- +--------------------------------------------------------------------------+ | Carlo Milono | | Personal: netcom!cmilono@apple.com or apple!netcom!cmilono | |"Sometimes I think the surest sign that intelligent life exists elsewhere | | in the universe is that none of it has tried to contact us." B.Watterson | +--------------------------------------------------------------------------+
pat@hprnd.rose.hp.com (Pat Thaler) (10/31/90)
> > How far can you run ethernet over fiber before the collision > detecting mechanism breaks down? I came up the following which > seems too far: > Min frame size is actually preamble + start of frame delimiter + 512, but the transmitting DTE has to receive the collision signal early enough to send a 32 bit jam and stop transmitting before it has transmitted the 512th bit and to stop transmitting soon enough that collision indication on the media ends before the 512th bit. This means that the round trip from first bit transmitted to collision detected should occur by the 512 bit transmitted including preamble. However, not all of those 512 can go into media delay. Delays in MAUs/transceivers, hubs, AUI/transceiver cables (if any), and DTEs have to come out of that. The amount available for the fiber will vary depending on the type of fiber MAU, presence of AUI cables, etc. It will be less than 480 bit times. FOIRL is spec'ed for up to 1 Km. Some manufacturers support an extended distance, but about the most I've seen is about 2 Km. You have to allow for delay in any repeaters/hubs you use to connect fiber segments together. For instance, in a rough calculation assuming a single repeater, no AUI cables, and 4 MAUs end-to-end, I get about 420 bit times available for the fiber delay. (This was based on delays for a FOIRL (Fiber Optic Inter Repeater Link) MAU which is defined in IEEE 802.3 for connections between repeaters. Using such a MAU to connect to a DTE is not covered by the standard, but a number of manufacturers sell such a product.) Speed of light in fiber is closer to .67c or 2 * 10^8 m/s So, the calculation for fiber distance in that situation would be: 420 BT * 100 ns/BT * 2*10^8 m/s / 2 = 4.2 Km. Now this was a pretty back-of-the-envelope calculation and I may have forgotten something. Such a network would be outside of what is allowed by the current 802.3 standard since the FOIRLs were allowed to connect directly to the DTEs and the fiber distance is more than 1 Km / segment. The standard allows up to five segments and four repeaters end-to-end. Up to three of those can be coax segments of up to 500 m. The rest can be link segments (FOIRL or 10BASE-T) with the FOIRL segments allowed up to 500 m. There will be 10 MAUs and 10 AUI cables of up to 50 m each. So the total distance spanned will be: 5 segments * 500 m/segment + 10 AUIs * 50 m/AUI = 3 Km. Speed of light is actually a bit faster on thick coax (0.77c), but collision propagation is slower. This calculation also had 3 more repeaters and 6 more MAUs. Pat Thaler
gwg@fibercom.COM (Greg Goodchild) (10/31/90)
ssw@cica.cica.indiana.edu (Steve Wallace) asks: >How far can you run Ethernet over fiber before the collision >detecting mechanism breaks down? I came up the following which >seems too far: .... math deleted..... > 11,520 meters is the round trip time. This divided by 2 >(the sender A must hear sender B transmitting) gives a ball park >figure (5,760 meters). I'd give myself a fudge factor of 30 >percent and end up with about 4,000 meters. Is this close? Closer than I thought you would get. You were on the right track in your approach but there is a little more to it which is implementation dependent. In order to avoid what would sound like an advertisement, let me say that there exists a nearly five year old product that implements Ethernet over an active ring. The maximum allowable ring circumference is 8km and therefore, if pulled into a straight line, 4km maximum station to station distance. While not quite answering your question, Carlo Milono brings up a good point when: cmilono@netcom.UUCP (Carlo Milono) writes: >The practical limits (today) appear to be about half of what your fudged >number is: 2km. Knowing that in this sense "practical" means "inexpensive", yes the usual limit between active nodes is 2km. In the above mentioned implementation, neighboring stations on the ring must be within 2km, but because it is a ring architecture (and not a star) you can implement a physically more encompassing network. So, as long as you have at least four stations on your net, two of them can be 4km apart. -- Greg Goodchild INTERNET: gwg@fibercom.com FiberCom, Inc. UUCP: ...!uunet!fibercom!gwg