[Solved] How much does a gigabyte worth of data physically weigh on a hard disk?

What is the physical weight of one gigabyte of memory/storage? Lets say this is on a hard disk.

What is the weight associated with the atoms that are actually storing the data on the disk? How has this value changed as densities of disks have increased?

Enquirer: Callum


Solution #1:

Hard drive density is measured in bits per square inch, the highest of which are currently (5/2013) 750 gigabits per square inch. This means that a gigabyte of data will take up about 6.88 millimeter2. The weight of an area of a platter consists of the substrate (usually glass and ceramic) and the magnetic layer which actually holds the magnetic grains storing the data. The magnetic layer is usually made of a mostly cobalt alloy of 10-20 nm thickness. Assuming 10nm thickness to make the math easier, This gives us about 6.88 * 1013 nm3 of magnetic layer material for one gigabyte.

Given the density of cobalt, this means that we can approximate the weight at 0.612471 micrograms.

I’m not sure how much the substrate weighs, but it’s almost certainly more than that.

2012 Update: This is all about drives that are shipping now – there is a lot of buzz about Seagate getting to 1 terabit per square inch recently, but that is a tech demo and not shipping quite yet.

2013 Update: It looks like the areal density of Hard Drive platters is stagnating, according to an interesting IBM report on the subject of Areal density. TDK says that they can approach the 1.5Tbits/inch2 mark, but they won’t show up in the market until 2014. The Seagate tech touted in last year is supposed to show up in 2014 as well. Next year should be exciting for the weight of gigabytes.

Previously on “How much does a gigabyte weigh on a hard disk?”

  • 2009: Areal Density 400 Gbit/in2 = 1.1518 micrograms (ref)
  • 2010: Areal Density 541.4 GBit/in2 = 0.84817 micrograms (ref)
  • 2011: Areal Density 625 GBit/in2 = 0.734966 micrograms (ref)
  • 2012: Areal Density 744 GBit/in2 = 0.617411 micrograms (ref)
Respondent: jamuraa

Solution #2:

The data kept on a disk does not increase the weight of the disk. The only weight differences in disks would be in the overall size of the disk (example: regular HDDs are larger than laptop HDDs in terms of size and typically mass, and larger sized disks can have more platters to hold data than older ones) itself and in the materials used to make the disk.

Data is stored by switching the magnetic polarity on the disk, not by adding or subtracting something from the actual substance. A full disk will have the same mass and will therefore weigh the same (assuming you don’t move the disk to a location where gravity is stronger or weaker, such as the moon).

Switching the polarity of a hard disk is like turning a magnet around so that the north and south poles are switched. It is not analagous to creating an ion (removing or adding electrons of an atom to give it a positive or negative charage). That could theoretically adjust the mass of the disk, but for all intents and purposes electrons do -not- have mass (so infinitesimally small that it almost appears so at least), so you are back to square one again if the disk did somehow operate in this manner, which it does not.

Respondent: TheTXI

Solution #3:

On the disk an individual bit weighs nothing, it’s just a change in magnetic polarity; see TheTXI’s answer for a more elaborate explanation of this.

In RAM, however, bits are comprised of electrons (or lack thereof) and they do have a mass which is about 9.10938215 × 10?31 kg. So for a GiB of memory, assuming equal distribution for zero and one bits, we get around

4294967296 n × 9.10938215 × 10?31 kg

4294967296 would be the number of one bits in memory (assumed to be 50 %) and n would be the number of electrons that are on average in one bit. I have found one source1 that specified this number at around 105.

So we can give an estimate of how much mass 1 GiB (or 1 GB) of memory would have:

1 GiB, half filled with ones ? 3.91 × 10?16 kg = 391 femtograms

1 GiB, completely filled with ones ? 7.82 × 10-16 kg = 782 femtograms

1 GB, half filled with ones ? 3.64 × 10?16 kg = 364 femtograms

1 GB, completely filled with ones ? 7.29 × 10?16 kg = 729 femtograms

So in general you can assume that weight to be pretty unnoticeable (or, with hard disks to be downright nonexistant).

1 These lecture slides, but they are in German.

Respondent: Joey

Solution #4:

Not-serious answer

It depends on what font size your text is saved in. 24-point font is very heavy, whereas 8-point is quite light. Bold text also increases the weight, and you should avoid saving lots of text in italics, because all the characters lean to the right, which changes the way the disk spins.

Serious answer

Data has no mass.

Respondent: Graeme Perrow

Solution #5:

The correct answer is 0. He didn’t ask how much hard drive does it take to store 1 GB, he asked how much 1 GB weights on a hard disk. As we know it uses magnetic storage and not an electrical charge (which would weigh something), then correct answer is 0.

Respondent: Pyrolistical

Solution #6:

It depends on the data.

Yes, hard drives store data by flipping poles in magnetic domains on the disk–at first glance this means nothing is added or subtracted and thus no weight.

However, that’s not the whole picture. The orientation of those domains matter. There is less total field energy when the domains are 1010101010 than when they are 11111111 or 00000000. I’m sure everyone is familiar with e=mc^2. Putting energy into the domains DOES mean mass, albeit an incredibly small amount of it.

My physics isn’t up to even trying to estimate the mass but I’m sure it’s beyond anything the most sensitive scale could possibly measure.

Respondent: Loren Pechtel

Solution #7:

Depends on where you’re doing the weighing. One of the answers immediately jumps into discussing femtograms, which are not a measure of weight, but instead measure mass.

On the moon things weigh less, on Jupiter they weigh more. In space they weigh nothing.

So, the answer is … depends.

Respondent: Trey Jackson

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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