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Pigeons' Bandwidth Advantage Quantified 462

An anonymous reader submits "A well documented test took place in the north of Israel, in presence of several dozen Internet geeks and experts. During the test, 3 homing pigeons carried 4 GB (gigabytes) for 100 km distance, achieving, what apparently looks as pigeons' world record in data transfer to a given distance. Bandwidth achieved by the pigeons was 2.27 Mbps...Transferring a similar volume of information through a common uplink of ADSL line would have taken no less than 96 hours..."
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Pigeons' Bandwidth Advantage Quantified

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  • Re:Sure (Score:3, Informative)

    by Mr. Sketch ( 111112 ) * <mister,sketch&gmail,com> on Wednesday March 31, 2004 @07:11PM (#8731014)
    Yes, but is there an RFC [] for IP-over-jet?
  • latency v. bandwidth (Score:4, Informative)

    by Scott Hussey ( 599497 ) <(moc.liamg) (ta) (todhsals+yessuhts)> on Wednesday March 31, 2004 @07:12PM (#8731027)
    Actually, the distance involved matters not. Bandwidth is purely the time to put data on the line. Latency is the time it takes to get from A to B. So the bandwidth would be the same no matter how far they travelled or how fast they flew. A good simile is bandwidth is how many tapes you can load in your trunk per hour. Latency is how fast you can drive those tapes to your destination.
  • Re:Back of envalope (Score:4, Informative)

    by Anm ( 18575 ) on Wednesday March 31, 2004 @07:31PM (#8731233)
    Actually, from this link [], the pigeons flew the 100km in 206, 136, and 233 minutes respectively. That makes 18.12, 27.49, and 16.02 miles per hour. Not sure what motivated pigeon #2.

  • Re:It begins... (Score:3, Informative)

    by fbform ( 723771 ) on Wednesday March 31, 2004 @07:37PM (#8731289)

    Is it already April 1st somewhere?

    That may well be the case, but stranger things [] have happened.

  • by Martin Blank ( 154261 ) on Wednesday March 31, 2004 @07:39PM (#8731319) Homepage Journal
    I dunno. I fit a lot more data on some of the newer DLTs than I do on an equivalent-volume stack of optical media.
  • by LostCluster ( 625375 ) * on Wednesday March 31, 2004 @07:44PM (#8731365)
    When a Clear Channel radio station changes formats and therefore needs a large volume of music on site quickly, they usually send a server that is pre-loaded with the new format worth of music on HDs, and the studio just plugs that into their network. This also gives them the capability to change the format overnight without anybody at the studio complex needing advanced notice, so that soon-to-be-unemployed DJs don't see it coming and therefore leave the station a few days early to ruin the transition... the UPS delivery of the new music comes in a non-descript cardboard box which can be scheduled to be on the site just hours before the changeover happens.
  • Re:Ha! (Score:5, Informative)

    by daveashcroft ( 321122 ) on Wednesday March 31, 2004 @07:51PM (#8731433)
    NOT by GMT...its now 00.51am on 1st April 2004! The world doesnt run on yankee time! ;-)
  • by Dareth ( 47614 ) on Wednesday March 31, 2004 @08:09PM (#8731580)
    I was wondering if homing pigeons were extint.
    This FAQ [] answered that question and many others for me.
  • by irokitt ( 663593 ) <.moc.oohay. .ta. .ruai-setirdnamihcra.> on Wednesday March 31, 2004 @08:11PM (#8731607)
    This is a debate that nobody will ever win. I was taught that bandwidth was the difference between lower/upper frequencies on a wire, i.e.
    "The numerical difference between the upper and lower frequencies of a band of electromagnetic radiation, especially an assigned range of radio frequencies." (thank you Google).
    And under that definition, these pigeons have no bandwidth (unless you're counting the frequency at which they flap their wings ;).

    The Jargon File [] says
    "Used by hackers (in a generalization of its technical meaning) as the volume of information per unit time that a computer, person, or transmission medium can handle. "Those are amazing graphics, but I missed some of the detail -- not enough bandwidth, I guess." Compare low-bandwidth. This generalized usage began to go mainstream after the Internet population explosion of 1993-1994. 2. Attention span. 3. On Usenet, a measure of network capacity that is often wasted by people complaining about how items posted by others are a waste of bandwidth."
  • by Mnemia ( 218659 ) on Wednesday March 31, 2004 @08:11PM (#8731612)

    Incorrect. The bandwidth would remain constant - what would double if the distance doubled is what is usually called a "bandwidth-delay product". This quantity represents the pipe capacity of a given length, or in other words the amount of data in transit at any given moment. This is of course assuming you have an unlimited number of pigeons you can keep sending out.

    You would be correct, using the term bandwidth loosely, if the number of pigeons stayed constant. However, using the strict definition, bandwidth is totally unrelated to line latency/round trip time.

  • but seriously (Score:4, Informative)

    by soricine ( 576909 ) on Wednesday March 31, 2004 @08:17PM (#8731655)
    this actually works. here in nz, a caving tour company uses pigeons to ferry memory sticks back to base so the digital photos can be waiting when the tourists get back. []
  • by Mnemia ( 218659 ) on Wednesday March 31, 2004 @08:21PM (#8731681)
    What you just defined bandwidth by is exactly the same definition I used. Bandwidth is the amount of data that can be transmitted per unit time. The parent poster was only wrong about his statement that bandwidth would be halved if distance (and thus latency) were doubled. *That* is incorrect, because the rate you can transmit data at is not affected by the distance to the receiving station. He would be correct if he said that link utilization would be halved for the same amount of data. But bandwidth is constant.

    I was just saying that what is doubled in this case is the pipe capacity and latency; the bandwidth part stays the same.
  • Re:Um... (Score:2, Informative)

    by Bush Pig ( 175019 ) on Wednesday March 31, 2004 @08:53PM (#8731903)
    Er ... actually, no. It's been April Fools' Day in the Cook Islands for quite a while (I can't be bothered working out how long). It's been April Fools' Day in Adelaide for almost 10 1/2 hours, and in Greenwich (the home of the Greenwich Meridian) for about 50 minutes. I don't think it'll be April Fools' Day in the continental US for some hours, but it's certainly April Fools' Day in China and India.

  • by ( 463190 ) * on Wednesday March 31, 2004 @10:13PM (#8732701) Homepage
    Hmmm - if you want to be pedantic about it then bandwidth is really the defined by the range of frequencies a channel can carry (which, incidentally, determines its information capacity.)

    But even by the more recently accepted definition of bandwidth, you're not quite right. Latency *does* matter when we're talking about packet networking such as pigeon based transport. What if a pidgeon dies in transit? In this case it'll take you up to three hours to learn of his demise, and only then can you send the information again. So high latency + packet loss has reduced your effective bandwidth dramatically - the same happens on non-pidgeon based transports. Of course techniques such as FEC can, and are used to mitigate this. In this case I'm imaginine that pidgeon loss would be quite high, and some sort of RAIP scheme would be desirable on top of a good selective retransmission algorithm.

    Also: how do you know how many pidgeons you can fit in a given amount of airspace? What if you only have ten pidgeons to work with? Here the latency is critical because you need to wait for your pidgeons to return before you can send them again with more data.

    So bandwidth is not the be-all and end-all of total throughput. In many real-world situations, all the bandwidth in the world won't make your connection any work any faster.
  • Re:Ha! (Score:1, Informative)

    by Anonymous Coward on Wednesday March 31, 2004 @10:54PM (#8733077)
    100km = 100000m

    4GB = 4*1024*8 = 32768Mbits

    data transfer rate = 2.27Mbps = 32768/t Mbps

    transfer time = t secs.

    pigeon speed = x m/s = 100km / t secs. = 100000/t m/s

    x : 2.27 :: 100000/t : 32768/t

    x = 100000 / (2.27 * 32768) m/s = 1.3443867 m/s = 4.83979212 km/h
  • Re:Back of envalope (Score:5, Informative)

    by Erik K. Veland ( 574016 ) on Thursday April 01, 2004 @03:24AM (#8734539) Homepage
    It's done [] :


  • RFC1149 (Score:2, Informative)

    by process ( 447778 ) on Thursday April 01, 2004 @05:47AM (#8735013) Homepage
    They compare it to the implementation of RFC1149 [] in Bergen by BLUG [],however this is clearly a breach of 1149.

    From RFC1149:

    Frame Format

    The IP datagram is printed, on a small scroll of paper, in hexadecimal, with each octet separated by whitestuff and blackstuff. The scroll of paper is wrapped around one leg of the avian carrier. A band of duct tape is used to secure the datagram's edges. The bandwidth is limited to the leg length.
    See. One IP datagram, one scroll of paper. The community demands interoperability tests if CPIP is ever to become a standard

  • by El Gringo Loco ( 688126 ) on Thursday April 01, 2004 @12:40PM (#8737743)
    It is well known that Google has used pigeon power for years. []

Loose bits sink chips.