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ArishMell · 70-79, M
IG = 1 X 10^3 M, and 46.5Gb = 4.65 X 10 Gb.
So
(4.65 X 10^10) / (1.4 X 10^6) = 3.321 X 10^[10-6] = 3.321 X 10^4 = 33.21 X 10^3.
You can't have a fractional disc, so you'd need 34 000 floppy-discs.
I think that's correct though I suspect I've been too liberal with the indices' amplitudes - please, mathematicians among you, do verify or correct as necessary.
And lots of loading time.
.....
I recall seeing one computer at work that used 8-inch floppy-discs, but by then 5.25" was becoming the norm.
So
(4.65 X 10^10) / (1.4 X 10^6) = 3.321 X 10^[10-6] = 3.321 X 10^4 = 33.21 X 10^3.
You can't have a fractional disc, so you'd need 34 000 floppy-discs.
I think that's correct though I suspect I've been too liberal with the indices' amplitudes - please, mathematicians among you, do verify or correct as necessary.
And lots of loading time.
.....
I recall seeing one computer at work that used 8-inch floppy-discs, but by then 5.25" was becoming the norm.
GuyWithOpinions · 31-35, M
@ArishMell nice!