Is a hard drive's reliability affected by the amount of data stored on it?
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M$4 Answers
The lifespan of hard drives is measured in working hours, if you keep it always on it will wear off faster.
As it has been already said a very fragmented drive will perform poorly because of the need of physically move the heads all over the platter surface to locate scattered data, but bare in mind that the defragmentation process will also have a wearing effect, do it too often and it will be counter-effective.
You should also consider that the fragmentation issue is not a problem on linux machines as it is on windows.
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M$This could eventually hasten the drive's failure, so it's a good idea to set up a regular defrag schedule on a very full drive I think.
Years of experience installing, troubleshooting and disassembling hard drives in PCs.
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M$Ditto that, simple logic. Piece of machinery: the more the arm has to move, the faster it'll wear out. Keep them files de-fragmented! Plus, stoopid me, on my very old IBM desktop (no longer with me, RIP) I let it get so full there wasn't even enough space left to LET me defrag it. And when I pulled out the hard drive to play Captain Destructo (per safe paranoid protocol) I could even see the tiniest hint of irregularity in the normally mirror-smooth surface.
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M$Short answer: Yes.
Long answer:
Because a hard drive is a physical device, it has a given failure rate. That failure rate says that a certain percentage of the drives will be defective from the manufacturer, and some will work 2 weeks, 4 weeks, 2 years, whatever. It's not as simple as an expiration date, but some devices will "naturally fail" HOWEVER the more use a drive gets, the more likely a failure is to occur. A never used drive has a 0% chance of failure, 0 use but 0% failure. The more you use it, the more likely a failure.
It is a logical step to assume that more data on a drive equals more use, more use equals greater failure rate. This is not to say that you couldn't write the data to the disc once, and then not access it anymore and then wouldn't be really using it much more so the failure rate could decrease.
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M$