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I always see the label and it says 350G's withstandable. What would put this over 350G's? Is it even possible to hit 350Gs of force to a hard drive?

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Is it even possible to hit 350Gs of force to a hard drive?

Sure is. Drop it on the floor.

You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level.

Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in particular directions that will never happen in reality.


If you drop your hard drive from $1~\text{m}$ it will hit the floor at around:

$$ \sqrt{2\times 1~\text{m}\times 1~g}\approx 4.4~\text{m}\cdot\text{s}^{-1} $$

At exactly $350~g$ it would come to a stop in:

$$ \frac{\left(4.4~\text{m}\cdot\text{s}^{-1}\right)^2}{2\times 350~g}=2.8~\text{mm} $$

(Note that due to the way the math works out, the stopping distance when dropping from a height $h$ is just $hg/a$).

Since the actual impact will probably be a varying acceleration and the rigid case of the hard drive will probably deform less than $3~\text{mm}$, the actual peak acceleration can easily exceed $350~g$.

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    $\begingroup$ So let's see here. Say you drop it from a height of 2 m. It will be moving about v = 6 m/s when it hits the floor. At a constant acceleration of 350 g upward, which is about 3500 m/s^2, it would stop in 0.002 s. Over this time it is moving an average of 3 m/s, so it travels 0.006 m, or about 6 mm. I wouldn't think that either the floor or the hard drive case are going to deform by as much as 6 mm for such an impact, so the acceleration must actually be considerably greater than 350 g. $\endgroup$ Jun 10, 2015 at 5:22
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    $\begingroup$ @NateEldredge What the manufacturers guarantee will generally be much less than what it can actually handle. $\endgroup$ Jun 10, 2015 at 6:03
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    $\begingroup$ @Manishearth: Sure. I'm just confirming that dropping it on the floor could plausibly exceed the 350 g rated acceleration. Of course this doesn't mean the drive would necessarily break. $\endgroup$ Jun 10, 2015 at 6:06
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    $\begingroup$ @NateEldredge: The calculation becomes quite a bit harder when you assume it hits the ground corner-first. It's quite likely that the different corners hit the ground over a much longer period than 2 ms. $\endgroup$
    – MSalters
    Jun 10, 2015 at 7:08
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    $\begingroup$ @Superbest The accelerometer is probably saturating. $\endgroup$ Jun 10, 2015 at 13:07
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Here's an application where an ability to withstand high shock is important.

Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research).

We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a known location with the data loggers running. We would then recover the now utterly destroyed computers: their circuit boards shredded into resin and glass fibres. But their log of seismic data was perfectly recoverable from the HDD and we could seismically map the area for minerals prospecting. We would have destroyed thousands of computers in this way. But I don't believe there was a single unreadable HDD, even those within meters of the blast.

At the time, this was actually cheaper than running communications links of the required datarate to all the points in the grid.

As Paul's Answer Points Out, the specification is likely for a parked HDD. Our HDDs were running, but of considerably less data density than HDDs today (they were 160MB drives), which probably explained their ability to withstand such shock whilst working.

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    $\begingroup$ Is it really true that many working physicists love explosions? I am only half joking, as I have come across this line in lots of books, or is it just the type of work many of them themselves in or that there is such a lot to be learned from them? $\endgroup$
    – user81619
    Jun 10, 2015 at 5:23
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    $\begingroup$ @AcidJazz As I said here, the explosion is simply to be an impulse input to the ground: seismic prospecting is simply "measuring the impulse response of the ground". $\endgroup$ Jun 10, 2015 at 6:02
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    $\begingroup$ @AcidJazz Who doesn't love explosions? $\endgroup$
    – cpast
    Jun 10, 2015 at 21:28
  • $\begingroup$ @cpast firemen and bomb disposal dudes......I read George Dyson's book, re project orion and in it, as his propane cooker catches fire on a camping holiday, his father Freeman comes in and says, "yippee, an explosion!" $\endgroup$
    – user81619
    Jun 10, 2015 at 22:12
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You are not the first person to ask this question.

https://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't go around kicking your computer case.

And I just noticed that you also posted the linked question. So technically, you are the first person to ask this question on the SE network.

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    $\begingroup$ I was going to suggest the poster/answerer of that question post here... but the poster DID post here :P $\endgroup$
    – WernerCD
    Jun 10, 2015 at 12:59
  • $\begingroup$ Soccer balls are pretty soft compared to HDDs, so your argument is not very convincing. $\endgroup$ Jun 12, 2015 at 9:52
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A g is a unit of acceleration. I am going to assume the force meant here is 350 times the earth weight of the hard drive. Assuming the drive is 625 grams, that works out to a force of about 450 lbs. This is easy to achieve: just hit it with a hammer.

Edit: as many have pointed out, the rating is given in units of acceleration for good reason. An accelerating reference frame is a better way to picture the problem, so you don't need dynamics here, just kinematics.

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    $\begingroup$ No, they really do mean acceleration. A full-size disk drive should be able to handle 450lbs steady force across it's (solid aluminum) case. $\endgroup$
    – paul
    Jun 10, 2015 at 9:51
  • $\begingroup$ @paul, the questioner asked about a force. $\endgroup$
    – Alan
    Jun 10, 2015 at 17:50
  • $\begingroup$ @cag, I believe he was just mistaken (or loose) on his terminology. By context it's very reasonable to assume he really did mean acceleration. $\endgroup$ Jun 10, 2015 at 20:08
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    $\begingroup$ Indeed, in such problems, what matters is acceleration, though the way it is applied might matter a bit too (and that could be part of the explanation). So your answer is not adequate, but I upvoted you for being the only one to notice that an acceleration is not a force. That kind of loose talk is not acceptable: some guess ok and others are mislead in good faith, as you were. The question and 1st+3rd answers are the guilty party, and you are the victim. To understand the acceleration issue for HD, read Neutron Star- CC @paul $\endgroup$
    – babou
    Jun 15, 2015 at 9:44
  • $\begingroup$ One remarkable aspect of this question is that no one is discussing or questionning the physics of the problem. And the distinction between force and acceleration is very important for that. Since everyone is ignoring the physics of the question, dicussing force or acceleration is a pure matter of arbitrary choice, given the loose talk of almost everyone, that confuses force and acceleration. That is no way to do science. CC @RafaelAlmeida $\endgroup$
    – babou
    Jun 15, 2015 at 9:45

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