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New Data Transmission Speed Record 262

An anonymous reader writes "Gizmag is reporting that a team of German and Japanese scientists have collaborated to shatter the world record for data transmission speed. From the article: "By transmitting a data signal at 2.56 terabits per second over a 160-kilometer link (equivalent to 2,560,000,000,000 bits per second or the contents of 60 DVDs) the researchers bettered the old record of 1.28 terabits per second held by a Japanese group. By comparison, the fastest high-speed links currently carry data at a maximum 40 Gbit/s, or around 50 times slower."
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New Data Transmission Speed Record

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  • Re:2.56 Terabits = ? (Score:3, Informative)

    by bioteq ( 809524 ) <mike AT nanobit DOT net> on Sunday March 26, 2006 @02:44AM (#14996720)
    Not true, actually.

    Most slashdotting happens because of hardware issues, not upstream bandwidth.

    Although, 2tb of bandwidth would be freakin' amazing for some stuff for a botnet to get ahold of..

    Quick! Everyone call up the nearest script kiddie and get to work!
  • by EvanED ( 569694 ) <{evaned} {at} {gmail.com}> on Sunday March 26, 2006 @04:16AM (#14996920)
    You think wrong. Some quotes from the Wikipedia [wikipedia.org] entry:

    In recent years, the use of a byte to mean 8 bits is nearly ubiquitous

    Meaning even today it's not universal.

    A contiguous sequence of binary bits in a serial data stream, such as in modem or satellite communications, or from a disk-drive head, which is the smallest meaningful unit of data. These bytes might include start bits, stop bits, or parity bits, and thus could vary from 7 to 12 bits to contain a single 7-bit ASCII code.

    Here I think is the most revealing definition for the discussion in the present context.

    The eight-bit byte is often called an octet in formal contexts such as industry standards, as well as in networking and telecommunication, in order to avoid any confusion about the number of bits involved.

    Another site [uiowa.edu] says that:

    * Pre-1965, and including the IBM 701, bytes were almost always 6 bits, though they weren't called that much then, but rather characters.

    * 9 bits were sometimes used

    * The PDP-6, PDP-10, and DECsystem 20 all supported changing the byte size with instructions from 1 to 36 bits (probably only some of those)

    The latter reference, looking up the PDP-10 on Wikipedia, gives this quote:

    Some aspects of the instruction set are still considered unsurpassed, most notably the "byte" instructions, which operated on arbitrary sized bit-fields (at that time a byte was not necessarily eight bits)
  • Re:Digg Wins (Score:3, Informative)

    by adolfojp ( 730818 ) on Sunday March 26, 2006 @04:42AM (#14996977)
    Digg is great for learning of new technologies. Slashdot is great because of its discussions. They are complementary websites. I don't see digg replacing slashdot anytime soon, yet, I am glad that digg is there to fill my need of having "as many tech news as possible" available.

    Trying to have good discussions in Digg is futile because of its moderation system. And whenever discussion worthy news are available they are quickly buried by ten articles of what someone somewhere might have said about the color of the new Nintendo console.

    Cheers,
    Adolfo
  • by mrchaotica ( 681592 ) on Sunday March 26, 2006 @04:47AM (#14996989)
    At 100Km/hour, a truck would require 1.6 hours * 60^2 seconds/hour = 5,760 seconds to travel 160 kilometers. At 60 DVDs/second, the truck would have to be carrying 5760*60 = 345,600 DVDs to have equivalent bandwidth. A typical DVD in a case is 14cm wide, 19cm tall, and 1.5cm thick, for a total volume of 399 cm^3 (lets round to 400cm^3). Therefore, the truck would have to have a cargo volume of 400cm^3 * 345,600 = 138,240,000cm^3, or 138.24m^3.

    Now, typical intermodal containers (as used on big rig trucks) are 8.5' by 8.5' by 40', or 2890ft^3. Converted to metric, this is about 82m^3, which is less than the 138.24m^3 required.

    In other words, no, a truck full of DVDs is NOT faster than this connection!*

    *unless you put the DVDs on spindles instead of in cases.
  • Single-channel only (Score:3, Informative)

    by Soft ( 266615 ) on Sunday March 26, 2006 @05:25AM (#14997066)
    To be sure, I believe this is a single-wavelength transmission record. For WDM (multiwavelength), I believe Alcatel's 2002 record of 10 Tbps over 3 x 100 km still hasn't been topped (Frignac et al, OFC 2002 [ieee.org]).
  • by ottffssent ( 18387 ) on Sunday March 26, 2006 @10:24AM (#14997762)
    What are you talking about? U/320 SCSI is 320 megabytes per second. Not bits. In a regular PCI slot on an unloaded bus, a U/320 HBA is limited to 90-110MB/sec, but with PCI-X or PCIe that limitation is removed, and single HBAs can readily sustain 500+MB/sec sequential reads from arrays spanning multiple SCSI channels.

    Of course, in many applications, latency of varying sorts quickly chews that number down to something a bit more sane.

    If you're limited to a hard 20MB/sec over SCSI, the first thing I would suggest is to make sure that you're actually operating at U/320 speeds, and your HBA and drive(s) haven't fallen back to one of the SE modes for some reason (faulty connection, crummy cable, missing / buggy / broken / wrong terminator, etc).
  • by Achoi77 ( 669484 ) on Sunday March 26, 2006 @10:34AM (#14997798)
    2.328306436539 LOC/s

    From Wikipedia:
    1 LOC = 20 tebibytes
    1 tebibyte = 1,099,511,627,776 bytes

    Doesn't sound nearly as impresive, perhaps marketing should stick with 6,984,919,309 nLoc/hr

  • by javaDragon ( 187973 ) on Sunday March 26, 2006 @11:27PM (#15000430) Homepage
    2.5 TB, 60 DVD ?
    Let's see...
    # rpncalc
    rpncalc version 1.35. Copyright (c) 1993-2004 David Frey et. al.
    This is free software with ABSOLUTELY NO WARRANTY.
    For details, type `warranty'.
    Type `quit' to quit and `?' to get a summary.
    2.5 1024 dup dup * * *
    1: 2684354560
    60 /
    1: 4.474e+07
    1024 dup * /
    1: 42.67
    Well, Unless DVD can contain more that 40 GB of data (BluRay ?), there is a possibility of overflow in submitter's calculations.
    My guess is that it's more like the quivalent of 600 DVD per second which has been transmitted.

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