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Comment by b11484 | Hacker News Reader

In the article it states that the helium can typically reduce drag by 20%. How much of an effect does that have on the cost of powering the drive? I assume it would at least be less than a 20% improvement overall.

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Yev from Backblaze here -> You'd have to look at the entire cabinet to get a better sense of the power savings. Post author Andy explains a bit more here -> http://disq.us/p/1s8psup.

Maybe this is a stupid question, but why don't manufacturers evacuate the drives (i.e. create a vacuum) instead of filling them with Helium?

It's much easier physically to build something which can hold e.g. +0.1 atmosphere helium, than resist imploding against -1.0 atmosphere. Practically speaking, pumping down to vacuum is a slow process, whereas you can flush with an inert gas. Finally, hard-drive heads actually float on lamina of gas; this is one of the reasons that normal open-circuit hard drives have a maximum operating altitude.

Wait, if it's +0.1 atm helium (for example), what happens if it develops a slight leak, a leak from which helium molecules can escape but air molecules cannot? once the pressure reaches equilibrium, there would be no reason for additional helium to escape, and because air can't enter.... then what?

That isn't how it works though--look up partial pressures of gases. Helium will continue to escape until its partial pressure is at equilibrium with the external environment, i.e. near to zero, and the external gases will enter until they are at equilibrium with the internal environment. At that point the partial pressure of each gas will be equal on both sides.

What happens to these helium hard drives if you repeatedly cycle the pressure, like by taking the hard drive on an airplane?

Since the partial pressure of helium at ground level and flight level is essentially the same (≈0) there should be very little change, providing the seals are intact. Since they are very securely sealed to contain the helium, changes to the external pressure, unless extreme, should not have any significant effect.

He states that they see one drive that is between 94-99%, but that may be a sensor issue. You'd lose the power savings.

Nit: those are not percentages indicating how much helium left, just some undocumented SMART attribute "raw values", there is no direct way to find out their meaning. (This is what the article text implies too).

The raw attribute values could be for example 100 = good 99 = moderate 98 = degraded, or a log scale similar to decibels etc. Or it could be a bitfield that has no meaning as a decimal number. etc

would that still be a problem if you pulled a vacuum outside the drive as well? like put the whole thing in a chamber?

Duskwuff has it right: the air or other gas is a critical component of the system. The spinning platters pull the gas along with them, and the head assembly is designed so that its aerodynamics control the "flying height" [0], the distance between the head and the platter. In a vacuum, there would be no easy way to keep the head from coming into contact with the platter, which is what causes head crashes. (In the era of sealed drives, head crashes are quite rare, but they used to be much more common. In a head crash, the head rips the magnetic coating off the platter, destroying it.)

[0] https://en.wikipedia.org/wiki/Flying_height

You are exactly right. WD explains the same in https://blog.westerndigital.com/rise-helium-drives/ :

«Without air, the heads will crash into the disk»

The head flying height is controlled by air flow (or helium flow, I suppose) under the head. With no atmosphere, there's no way to regulate that gap reliably.

That is definitely a question that a manufacturer should answer :-/ No idea!

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