[RndTbl] Re: Hollerith Hijinks

Gilbert E. Detillieux gedetil at cs.umanitoba.ca
Fri May 30 14:43:03 CDT 2003


According to Michael Doob:
> A gigabyte of punch cards?
> 
> I just happen to have a box of cards on my desk (some guys just
> *never* get around to cleaning up). I took a batch and
> really squished them tight: this gave me 60 cards per centimeter.
> That means 60*100*80 =  2^8*3*5^4 = 480000 bytes/meter (I never
> use FORTRAN anymore :-)). So our gigabyte
> of data stands
> 
> 1024*1024*1024/480000 = 2^(22)/(3*5^4) meters = 2^(19)/(3*5^7) km ~ 2.24 km
>
> tall, at least without gravity to compress everything.

Of course, all of you have been basing your calculations on the assumption
that a gigabyte is 1024^3 (or 2^30), which I suppose would make more sense
to us old timers.

However, the new standard is to actually respect the old SI standards, and
use kilo-, mega-, and giga- to represent powers of 1000.  (I.e. 10^3, 10^6,
and 10^9, respectively.)

See the following...

http://physics.nist.gov/cuu/Units/binary.html

This would somewhat reduce the length (or height), volume, and weight
estimates given.  Or, we could restate the question to ask how much a
gibibyte (yuck!) of punch cards would be...

> Now who is the lucky guy who has to retrieve the top one off the stack?

Or who's the poor operator who has to feed them all through the card reader?

-- 
Gilbert E. Detillieux		E-mail:	<gedetil at cs.umanitoba.ca>
Dept. of Computer Science	Web:	http://www.cs.umanitoba.ca/~gedetil/
University of Manitoba		Phone:	(204)474-8161
Winnipeg, MB, CANADA  R3T 2N2	Fax:	(204)474-7609



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