5 ms·
“Never underestimate the bandwidth of a station wagon filled with magtape, or a 747 filled with CD-ROMs.”
by einarvollset 16y ago
“Never underestimate the bandwidth of a station wagon filled with magtape, or a 747 filled with CD-ROMs.”
- TeHCrAzY 16y agoReminds me of an interesting article re. the bandwidth of a station wagon: http://www.dansdata.com/gz105.htm http://www.dansdata.com/gz105.htm
- crsmith 16y agoOr pigeon http://www.zdnet.com/blog/btl/in-south-africa-carrier-pigeon-faster-than-broadband/24093 http://www.zdnet.com/blog/btl/in-south-africa-carrier-pigeon...
- sjs 16y agoA USB stick? How do they interoperate if they don't follow the spec[1]?!! [1] http://tools.ietf.org/html/rfc1149 http://tools.ietf.org/html/rfc1149
- pjscott 16y agoOr a shipping container filled with hard drives. Conservatively speaking, you could easily get more than 60 PB in a typical shipping container and have plenty of room left over for padding, using 3.5" 2 TB hard drives. Assume it takes a leisurely week to get from, say, one coast of the US to the other; that's a little over 100 Gbit/s sustained throughput. I've been rounding down here; the real bandwidth would be quite a bit higher.
- Retric 16y agoLast I checked 32 micro SD cards had the highest memory density. http://www.sandisk.com/products/mobile-memory-products/sandisk-microsdhc http://www.sandisk.com/products/mobile-memory-products/sandi... 32GB Micro SD = 1 mm * 11 mm * 15 mm = ~6 million / m^3. So 32GB * 6,000,000 ~= 192 PB per m^3. 2TB 3.5 inch HDD = (4 in × 1 in × 5.75 in = ~3000 / m^3. So 3,000 * 2TB = ~6 PB per m^3 PS: Granted that m^3 if SD cards would cost around 1.2 billion and would take a while to upload at the other end.
- spicyj 16y agoI looked incredulously at your dimensions of the 3.5 inch HDD, only to search for the correct ones and found the exact same specs on Wikipedia, then realizing that's probably where you got yours. Looking at hard drive manufacturers' sites, I see that that is in fact the correct dimensions. What does the 3.5 inch figure represent, then? The size of the plates?
- c3o 16y agoThe 3.5 inch figure represents the floppy, remember that? :) A "3.5in HDD" is a hard disk that fits into the bay originally designed to hold a drive for 3.5" floppy disks.
- chancho 16y agoSo then why are laptop hard drives called 2.5in?
- SamReidHughes 16y agoBecause they're smaller.
- plorkyeran 16y agoBecause 2.4" sounds worse. (3.5 * 69.9 / 102 = 2.3985)
- megablast 16y agoI thought it wass due to the fact that the platters are 3.5", just as the 2.5" hd has 2.5" size platters. But no, looked it up, 3.5" have platter sizes of 3.74" apparently, 5.25" had 5.12" platters. Which makes sense when you look at a hdd with the top off.
- spicyj 16y agoOr you could stuff more than one container on a boat. The Emma Mærsk, one of the largest ships, has over 420,000 m^3 of space, which could hold 81 ZB (8.1 million PB) if Retric's (sibling) comment is correct, moving data at just over 1 Eb (exabit) per second. According to Wolfram Alpha, that's big enough to hold the entirety of human knowledge 7000 times.
- rdtsc 16y agoIf you have to copy to the hard drives and copy from the hard drives on the other end, you might be limited to the total IO bandwidth of however many hard drives you can read/write * speed of IO bus. Not too long ago I had realized that my 1Gbps Ethernet network lets me have a higher sustain data throughput than my 5400 rpm laptop hard drive. You would have the same problem with the crate of hard drives depending how many times you have to copy the data...
- gaius 16y agoYep, in practice your HDs would be in NetApp shelves or equivalent, and you would also need to account for the volume of the NetApp heads. That reduces the density by 5-10x. Then there is also the overhead of RAID-DP. Still, that's a shitload of bandwidth.
- crististm 16y agobut what about the latency?