Megabytes, most likely. If they are standard IBM cards, each of them contains a maximum of 80 "bytes" (actually, more like "btwelves", as they consist of 12 bits, not 8). So 62,500 full cards would hold 62,500 * 80 = 5,000,000 bytes, or about 4.77 (binary) Megabytes.
Is that megabytes, or megabits?
ReplyDeleteMegabytes, most likely. If they are standard IBM cards, each of them contains a maximum of 80 "bytes" (actually, more like "btwelves", as they consist of 12 bits, not 8). So 62,500 full cards would hold 62,500 * 80 = 5,000,000 bytes, or about 4.77 (binary) Megabytes.
DeleteAnd generally the first 7 bytes were empty and the last was a continuation character so 72 usable bytes makes 4.5MBytes
DeleteHowever if it's stored as ascii then you dont get a full byte for each character.
If it was stored as binary then indeed one could get the full 4.77 Megabytes.
regardless of how many data units that pile physically represents, you would have a heck of a time making sure that they are read in in order.
ReplyDeleteI-)