11 Oct
2005
11 Oct
'05
8:13 p.m.
Patrick Finnegan wrote:
John Foust declared on Tuesday 11 October 2005 07:17
pm:
I was looking for Fortran IV programs and got reading on the History of
FORTRAN.
In 1957 they could not punch out the 2,000 binary cards of FORTRAN II
and had to go tape instead.
Now that is about 72*12 bits per card ... you work out the math.
At 05:55 PM 10/11/2005, Patrick Finnegan wrote:
At 700MB/cd, it'd be about 12 reams per CD,
not 92. 92 reams would
get you about 5.4GB, which is more than a (single-sided, single
layer) DVD-R holds.
My math:
472,500 bits per page
945,000 bits per double-sided page
118,125 bytes per double-sided page
681,574,400 bytes per 650 M CD
5,452,595,200 bits per CD
46,160 pages
92 reams
59,062,500 bytes per ream
^^^
That's bytes per ream, not bits per ream.
681MByte/CD / 59MByte/ream = 11.5 reams/CD
Pat