Now, I realize that the rest of the writeup
s in this node
are meant to be
. 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
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
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.
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
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 :-)