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How much memory is 1 terabyte harddrive

Last response: in Windows 7
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August 13, 2012 9:18:27 AM

say i have like a thousand or close to thousand videos saved on my computer and they all 1 gb. one i get to 1000 of those videos, sayin they the same size, my hard drive will be full. is it like ipod memory, cause i like to fill up my computer with videos, movies
a b 8 Security
a c 99 } Memory
a c 395 $ Windows 7
August 13, 2012 12:43:15 PM

1024GB = 1TB

A 1 terabyte hdd actually is only 931 gb.
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a b } Memory
August 13, 2012 1:03:07 PM

SR-71 Blackbird said:
1024GB = 1TB

A 1 terabyte hdd actually is only 931 gb when formatted.

It has nothing to do with formatting, it's different units. Hard drive manufacturers use SI, memory vendors and Windows use binary. An SI terabyte equals approximately 931 binary gigabytes, or 1000 SI gigabytes.
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a b } Memory
a b $ Windows 7
August 13, 2012 1:28:46 PM

I wouldn't do that if I were you. I hate the last 1/4 of the HDD because the speed is noticeably slower. Well, at least noticeable slower than the edge, not sure if you are as inpatient as I am. So I tried not to fill it up and buy new HDD if I need more space before HDD is full.
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August 13, 2012 1:40:35 PM

I currently have 2 x 1TB hard drives to the brink full with movies about less 20gb on each, they both still copy at about average 90-100MB/s to and from any of my other hard drives. and now getting close to filling my third with 200GB left and about 600GB left on my fourth and 300GB left on my 500GB hdd, they all run really fast even when almost completely full
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August 13, 2012 1:41:19 PM

Pyree said:
I wouldn't do that if I were you. I hate the last 1/4 of the HDD because the speed is noticeably slower. Well, at least noticeable slower than the edge, not sure if you are as inpatient as I am. So I tried not to fill it up and buy new HDD if I need more space before HDD is full.


Hard drive's don't access in a linear plane/fashion, it sounds a defrag is in order.
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a b } Memory
a b $ Windows 7
August 13, 2012 1:48:45 PM

Na, it is all defragged. Also, the drive that mirrors it (not RAID1, just mirrored with software) is written in a sequential manner so there is no fragmentation. It's just how mechanical drive is. The inner edge is slower than the outer edge.
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a b } Memory
a b $ Windows 7
August 13, 2012 1:49:12 PM

There is a noticable decrease in thoughput when comparing the 20 % outer to the inner most 20%. Fagmented drive Only makes matters worse.

Transfer speed is based on RPMs (fixed for a given drive), Media Density (also fixed for a given drive) and Distance tha the media occupies verse time to transverse (angular velocity vs distance). Simplified - Takes Much longer to read a cluster at the inner circumfornace than if the Cluster is on the outer edge.
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August 13, 2012 2:54:16 PM

Pyree said:
Na, it is all defragged. Also, the drive that mirrors it (not RAID1, just mirrored with software) is written in a sequential manner so there is no fragmentation. It's just how mechanical drive is. The inner edge is slower than the outer edge.


You are correct, I'm sorry for misunderstanding. I thought you had meant that it took longer because it each sector/cluster as it goes up in size it takes longer to access, as if it was on a straight linear plane. Since there are multiple sectors/clusters that are equidistant to the read/write arm, there should be multiple instances where the access time is exactly the same and shouldn't be noticable to the user, but in regards to inner/outer portions of the disk you are 100% correct.

I still find it strange that there is a large noticable difference though, I use a 2TB 7200RPM standard mechanical disk that is roughly 80-85% of capacity, and I rarely notice a difference in access times despite it's location. Perhaps I'M too slow to notice the difference :pt1cable: 
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a b } Memory
a b $ Windows 7
August 13, 2012 4:15:57 PM

ie 7200 RPM drive. and say the data to be read is contained in one revoluion of the drive when the data is at the outermost track (& sequencially placed). That data would be read 139 microSec. If the data where on the intermost track it would probably take about 833 microsec (about 6 X longer). EX based on intermost track at .5" and outer track at 3" or Outer throuput of 140 mb/s then inner throulput would ONLY be about 23 mb/s.

If inner track was 1/4 In and outer track 3.5" then differnce would be considerably more (ie 14 X longer to read same data on inner track vs Outer track- or if Throughput was 140 mb/s fo outer most track then throughput for the innermost track would be ONLY 10 mb/s).
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