666K cards at 80 bytes per is 53,333,280.
A IBM 350 for a RAMAC 305 weighs over a ton and hold 5 million 6 bit
characters 3.5MB.
So you would need 15 of those.
Or a PC hard disk drive from 1994 of 60MB.
Or about 30 3.5 floppy disks.
Or a small flash drive of 64M to 2G.
Or a micro SD card of 2G.



On Tue, Dec 1, 2015 at 2:41 PM, Barry Merrill <[email protected]> wrote:
> I think a box of 2000 IBM cards is on the order of 6 pounds,
> so a TON of JCL cards would be 333 boxes, or about 666,666
> card images.
>
> But, the useful weight is zero, since we only use the holes.
>
> Barry
>
>
> Herbert W. "Barry" Merrill, PhD
> President-Programmer
> MXG Software
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> -----Original Message-----
> From: IBM Mainframe Discussion List [mailto:[email protected]] On
> Behalf Of Farley, Peter x23353
> Sent: Tuesday, December 01, 2015 1:59 PM
> To: [email protected]
> Subject: What's a "ton" of JCL? [was:RE: Straightforward way to determine
> hardware architecture level?]
>
> Re: "ton" of JCL, at least one large shop of my prior acquaintance (20 or so
> years ago) had over 250,000 members in the production applications JCL
> libraries.
>
> Not sure how much of that was obsolete at the time, but the batch operations
> control product they used had vast quantities of data as well.
>
> I think that counts as a "ton" or 2 . . . :)
>
> Peter
>
> -----Original Message-----
> From: IBM Mainframe Discussion List [mailto:[email protected]] On
> Behalf Of Peter Relson
> Sent: Tuesday, December 01, 2015 9:30 AM
> To: [email protected]
> Subject: Re: Straightforward way to determine hardware architecture level?
>
> <Snipped>
>
> . . . migrating from Cobol 4 to Cobol 5 without changing a ton of JCL (how
> much JCL is a "ton" anyway?).
>
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-- 
Mike A Schwab, Springfield IL USA
Where do Forest Rangers go to get away from it all?

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