So the British couple are flying down The Five in an 86 Country Squire with about a hundred hastily-labeled quarter-inch tapes, 4 gallons of Jim Beam, a case of Old Crow, and an Exabyte tape robot in the back, and the british driver turns to the inexplicably leather-clad female next to him and says 'Ricochet me'.

She laughs maniacally and throws the bigass knife switch mounted on the ceiling and the first of many tapes is jacked into the tape drive.

meanwhile, at the colo site across town...

A SparcSTOREDGE array clatters to life, spinning and whirring, slamming it's glut of ASCII porn down the wire.

Never underestimate the bandwidth of a station wagon full of quarter-inch tapes.

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The SETI@Home project produces about 35 Gigabytes of data a day. While this does not sound like that much to someone with a nice connection (a T1 is 1.544 Megabits/second - or about 0.2 Megabytes/second), even on a T1 this data would take 2 days to transmit a single day's worth of data. However, The Arecibo telescope's (in Puerto Rico) bandwidth is more similar to that of DSL connection or a dialup which is used for much less bandwidth hungry projects.

Instead, these tapes are sent on 35 Gigabyte DLT tapes are sent out by postal mail to Berkeley.

While in 1981 a station wagon filled with QIC tapes was a fair bit - consider today the bandwidth of a 747 filled with DVD-ROMs. One thing to consider though - the length of time it takes to read the data from the media.

Make certain you have a large timeout on any packets sent via station wagon.

Now, I realize that the rest of the writeups in this node are meant to be humorous. And, to a certain extent, they are. It's a good yarn, there are a couple of nice touches, there's a fair bit of history, a little poetry, it's basically a well-polished node. There's just this one, teeny-tiny, itsy-bitsy detail they forgot to include, a nugget of information which got left out when the rest of the authors crafted this little gem. See if you, too, can guess what it is.

Okay. For the rest of you who get just a little ticked off when strung along for three writeups only to be left hanging, the bandwidth of a station wagon full of quarter-inch tapes is approximately:

13 Petabytes per second.

For comparison purposes, this is equivalent to about 650 strands of perfectly saturated, single-mode fiber optic cable.


This figure will, of course, vary depending on a number of factors. In order to compensate for your own rate of travel and storage media, simply fill in the blanks below to get your tally! It's fun for kids of all ages!

BW = (( WV / (TW * TL * TH ) ) * TC * WS / WL) , where

BW = bandwidth in bytes / second
WV = the volume of your station wagon, in cubic meters
TW = the width of each individual quarter-inch tape, in meters
TL = the length of each individual quarter-inch tape, in meters
TH = the height of each individual quarter-inch tape, in meters
TC = the capacity of each individual quarter-inch tape, in bytes
WS = the speed of your station wagon, in meters/sec
WL = the length of your station wagon, in meters

This figure assumes average instantaneous bandwidth down the length of the wagon; in reality, I would assume that the bulk of the data transfer would occur in the region nearest the trunk.

To get my figure, simply plug in: WV = 2.72, TW = 0.054, TL = 0.073, TH = 0.0105, TC = 35.0 * 10 ^ 9, WS = 26.8, WL = 4.75. These numbers are meant to describe a stuffed 2001 Subaru Outback doing 60MPH using 35GiB tapes of this form factor.


Hmmmm. It seems there are a few engineers in the audience. Go figure. At least two people commented that ' it's not the length of the wagon that counts, it's the distance between endpoints, or the length from tape drive to tape drive, that determines bandwidth'. Well, I suppose this calls for clarification.

I'm told that the term 'bandwidth' applies to a communications channel. As such, a station wagon hardly counts -- it'd be like asking for the bandwidth of an IP packet< /u>. It wouldn't make sense. Similarly, it's not so much the bandwidth of the wagon as the bandwidth of the channel along which the wagon travels. With this in mind, walk with me through the following justifications.

When it's said that a SCSI bus (for example) is sustaining 20 million bits per second ( for example), what's implied is that a) if one observes the output of the bus, during every second in time, 20 million valid bits appear, on average, and b) if one observes the input of the bus, every second 20 million valid bits are being shoved onto the bus. What's not being said is how long it takes for a given bit to go from being shoved into the bus inlet to being taken out of the bus outlet. This number's usually called latency, I'm told. Regardless, in this case, the bus (channel of interest) is sustaining a bandwidth of 20 million bits per second. On average. The length of the bus is irrelevant as far as bandwidth is concerned: doubling the length of the physical bus will not change the fact that 20 million bits per second are coming out of it / going into it (at steady state), it will merely double the time it takes for a given bit to go in and then come out the other end.

So, to be proper, it should be mentioned that by 'bandwidth of a station wagon' I have computed the 'bandwidth of a one-lane road of indefinite length packed bumper-to-bumper with station wagons, each carrying quarter-inch tapes'. After all, it's the road that is really the communications channel in question, the wagon is simply the data packet.

However, if you view the communications medium as 'a road of given length with exactly one station wagon on it, carrying quarter inch tapes', then it is vital to know the length of the road in order to compute the time-averaged throughput attainable on this communications medium. Some would say that this is closer to what is assumed by the node title. I guess it depends on your perspective. What a great way to say we're both right :-)

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