What does hard drive shock up to 200G mean?

[xenolith]

The Western Digital Caviar BB/JB hard drives are rated to withstand up to 200Gs of non-operating shock and 65Gs of operating shock for a period of 2ms. In comparison, Maxtor Plus 9 hard drives are rated up to 300/60Gs for 2ms respectively.

Can anybody explain (or have a link to someplace that can explain) what these ratings compare to in layman's terms?

Thanks

 

[Utterman]

Pretty much if you were to dropped a hard drive off from a desk to the group, there is a certain amount of G's the hard drive faces as it falls to the groud. It is saying that if you were to drop it, it is better to drop it when it's not running compared to when it is running. I wouldn't buy a hard drive just because it can stand a higher amount of G's. I would look at the performance because it is not everyday that you drop a hard drive from a desk.

 

[xenolith]

Thanks utterman. I was just wondering if anybody (maybe a tech site) has done a test on this subject. Like finding out that a hard drive would have to fall at a such-and-such distance onto a such-and-such surface to produce 200Gs for 2ms. And possibly even expanding on the subject with the drive surrounded by packing materials to produce 200Gs.

I just think it would be interesting to know how durable these drives are physically.

 

[dszd0g]

This is somewhat important since hard drives are shipped. If they had no tollerance for shock they would all die in shipping. I would bet most hard drives would fail by dropping them from standing height onto concrete.

Quantum used to do a demonstration at some of their events where they would take a Quantum SCSI drive (as I recall SCA), yank it out of a running computer, drop it on the carpetted floor, and then plug it back into the still running computer and show that no data was lost. Kind of impressive, but not very useful from the practical point of view. I wouldn't make my purchasing decisions based on it.

This figure is probably much more important for laptop drives than desktop. However, I have not researched whether it has any real significance (and has actually been verified) or whether it is pure marketing.

200G = 1961m/s^2

 

[xenolith]

Originally posted by: dszd0g
This is somewhat important since hard drives are shipped. If they had no tollerance for shock they would all die in shipping. I would bet most hard drives would fail by dropping them from standing height onto concrete.

Exactly. I believe this would be important to know for that very reason too since I've received some OEM drives from unscrupulous vendors that don't pack drives as I would like. Sometimes they would ship it with only the drive rattling around inside a box with the invoice .

I've also slipped-up by dropping my fair share of drives onto a hardwood bench, but only at relatively short distances ( few inches). Again, I just think it would be nice to know what Gs these actions produce on a drive, and what role it plays on a drive's possibility of failure down the road.

 

[dszd0g]

Originally posted by: xenolith

I've also slipped-up by dropping my fair share of drives onto a hardwood bench, but only at relatively short distances ( few inches). Again, I just think it would be nice to know what Gs these actions produce on a drive, and what role it plays on a drive's possibility of failure down the road.

What, if you knew it didn't hurt your drive you wouldn't worry about dropping it? Or the opposite, if you knew that it would be bad you would have less accidents?

Seriously though, of course dropping drives isn't good for them. Chances are if it's a short fall on the desk it isn't going to hurt them. If you drop them onto a hard floor they may be toast.

The more I think about it I would think a force like Newtons would be a more useful measure than acceleration like G's. But I really don't enjoy Physics enough to whip out my Physics books to work out some Kinetics equations.

 

[RaynorWolfcastle]

the high figure comes from the fact that when you drop something on the floor, the floor doesn't "give" much. eg, assume that you drop a hard disk from a distance of 1m, it will have speed
v(f)^2 = v(i)^2 + 2a*d
v(f) ~= 5 m/s

now assume that the floor "gives" 2mm or 0.002m
then by the same eqn
0 = 25 + 2*a*0.002

a = -6250m/s^2 or ~625G :Q ... Of course, this acceleration only lasts for a fraction of a second, hence the 2ms period

Keep in mind that I pulled the "give" figure out of my @ss so it may be off by a significant margin. Nevertheless, these calculation give you an idea of what kind of shock these drives can withstand

 

[dszd0g]

Icecool, I've never liked that technique even though the math is easy. I haven't seen a good way to measure the give, which makes it pretty useless. I would think that the give would be much less than 1mm.

I would think it would be much more useful to treat it as an inelastic collision and measure the force exerted on the hard drive. Or actually on most surfaces the hard drive does bounce, so it would have to be an elastic collision.

 

[xenolith]

Originally posted by: dszd0g

What, if you knew it didn't hurt your drive you wouldn't worry about dropping it? Or the opposite, if you knew that it would be bad you would have less accidents?

Yeah, I know, only wish I wasn't so clumsy.

The more I think about it I would think a force like Newtons would be a more useful measure than acceleration like G's. But I really don't enjoy Physics enough to whip out my Physics books to work out some Kinetics equations.

Me too, in fact it's past my bed time anyway.

 

[RaynorWolfcastle]

Originally posted by: dszd0g
Icecool, I've never liked that technique even though the math is easy. I haven't seen a good way to measure the give, which makes it pretty useless. I would think that the give would be much less than 1mm.

I would think it would be much more useful to treat it as an inelastic collision and measure the force exerted on the hard drive. Or actually on most surfaces the hard drive does bounce, so it would have to be an elastic collision.

I really wasn't going for a very accurate result, I was trying to use quick and dirty calculations to give a qualitative idea of what's happening

 

[Pugchucker]

It basically means that you could fire it out of a cannon directly into a brick wall and then pop it into your computer and it would run fine. Its been a while since i took college physics buit if i remember correctly a "G" is a measure of acceleration and 1g is equivalent to the pull of earth's gravity...so 200G is like the force of gravity on Jupiter. So you could like drop the hard drive on Jupiter and itwould be ok...just in case you ever go there...and bring your computer.

Either that or what icecool said