Hans Franke wrote:
> Now .. let's for now assume a bit hole is about 1 mm in diameter,
> and the gap between is of equal size, thus one byte every 2mm.
> now, assumeing a speed of 40 km/h this gives 40,000,000/2 Bytes/h
> or 20 MByte/h transfer rate (or with around 333 KByte/min or 5 k/s
> almost 300 times the speed you assumed). The problem is still the
> paper transport, not optical sensing, since even an 8 Bit controler
> is be able to shovel awy the data.
> 1*10^5 hours
> or just about 14 Month
Well read this ( almost on topic )
"In order to break the Lorenz codes in a reasonable time the cipher text
had
to be repeatedly scanned at very high speed. This meant at least 5,000
characters per second and in the 1942 this implied hard vacuum photocells
to optically read the holes in the paper tape. The smallest photocells
available
were some developed for proximity fuses in anti aircraft shells. Six
of these in
a row meant an optical projection system to enlarge the image of the paper
tape about 10 times. Dr Arnold Lynch designed the paper tape reader
and used slits cut into black card to form a mask in front of the
photocells."
Quoted from this web page.
http://www.codesandciphers.org.uk/lorenz/colossus.htm
Received on Fri Aug 13 2004 - 15:27:11 BST