3 ms·

His throughput calculation is a bit off: A 2TB drive contains 2,199,023,255,552 bytes, or 17,592,186,044,416 bits. 17,592,186,044,416 / (60sec x 60min x 6hrs)

His throughput calculation is a bit off: A 2TB drive contains 2,199,023,255,552 bytes, or 17,592,186,044,416 bits. 17,592,186,044,416 / (60sec x 60min x 6hrs) = 814,453,058 bits/sec, or roughly 814Mbps.

AaronStanely 14y ago
Incorrect. 814,453,058 bits per second = 776.722963 megabits per second http://www.google.com/search?sugexp=chrome,mod=8&sourceid=chrome&ie=UTF-8&q=814%2C453%2C058+bits%2Fsec#hl=en&sclient=psy-ab&q=814453058+bits+per+second+in+megabits+per+second http://www.google.com/search?sugexp=chrome,mod=8&sourcei...
- Ralith 14y ago
  No, he's perfectly correct. 814.5 megabits per second is 776.7 mebibits per second.
- mmaunder 14y ago
  Yes it's 776 Mbps. Thanks. I forgot to multiply my final result by 2 (forgot I said 2 TB drive and not 1). Fixed.
ChuckMcM 14y ago
Being pedantic here, transmitting one byte (8 bits) of data requires 10 "bit times" (called bauds) because there is a start bit and a stop bit. This is sometimes called 8b/10b encoding since you get 8 bits delivered out of 10 bits sent. In packet systems (like ethernet) it gets a bit weirder since you might have 1500 bytes of data (represented by 15,000 bauds) but you add 36 bytes of network header information (another 360 bauds)) 1500 bytes is now 15,360 bauds [1]. The biggest challenge of shipping drives around though is that hard drives are really really cranky about g-shock so you can easily destroy all of your data if you drop, or even suddenly bang your drive as you are moving it around. I think it was mostly interesting from the perspective of someone who isn't used to dealing with large data sets suddenly having that 'ah ha' moment when confronted with the challenges of managing data with consequences. [1] This is nominally incorrect since the protocol layer overhead isn't 'bauds' like the physical access layer which uses the term strictly for signalling structures, but it works well for this example.

AaronStanely 14y ago

Incorrect. 814,453,058 bits per second = 776.722963 megabits per second http://www.google.com/search?sugexp=chrome,mod=8&sourceid=chrome&ie=UTF-8&q=814%2C453%2C058+bits%2Fsec#hl=en&sclient=psy-ab&q=814453058+bits+per+second+in+megabits+per+second http://www.google.com/search?sugexp=chrome,mod=8&sourcei...

Ralith 14y ago
No, he's perfectly correct. 814.5 megabits per second is 776.7 mebibits per second.
mmaunder 14y ago
Yes it's 776 Mbps. Thanks. I forgot to multiply my final result by 2 (forgot I said 2 TB drive and not 1). Fixed.

ChuckMcM 14y ago

Being pedantic here, transmitting one byte (8 bits) of data requires 10 "bit times" (called bauds) because there is a start bit and a stop bit. This is sometimes called 8b/10b encoding since you get 8 bits delivered out of 10 bits sent. In packet systems (like ethernet) it gets a bit weirder since you might have 1500 bytes of data (represented by 15,000 bauds) but you add 36 bytes of network header information (another 360 bauds)) 1500 bytes is now 15,360 bauds [1]. The biggest challenge of shipping drives around though is that hard drives are really really cranky about g-shock so you can easily destroy all of your data if you drop, or even suddenly bang your drive as you are moving it around. I think it was mostly interesting from the perspective of someone who isn't used to dealing with large data sets suddenly having that 'ah ha' moment when confronted with the challenges of managing data with consequences. [1] This is nominally incorrect since the protocol layer overhead isn't 'bauds' like the physical access layer which uses the term strictly for signalling structures, but it works well for this example.