# Fujitsu creates readable, writeable 'nanohole' hard drive

Fujitsu's scheme to produce hard drives that can hold a terabyte of data in each square inch of recording surface has taken a step closer to realisation. It has made a 2.5in disk made of its proposed 'patterned medium' and verified the disc's read/write capability. Fujitsu's approach uses anodised aluminium to create a …

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1. #### So what was the capacity

So what was the capacity / data density of the drive that they created and verified?

Was it greater than currently available drives?

I know that a prototype prrof of concept drive isn't commercially viable, but I'm interested now!

2. #### Re: So what was the capacity

Exactly what I was thinking

3. #### It's roughly 400GB

Using the following assumptions:

* 2.5" diameter platter - roughly 2.2" of usable area

* Single-side

Then

((2.2/2)^2)*Pi) = rough area in inches = ~3.8 sq. in.

13nm density = ~1TB/sq. in.

100nm = ~current density = ~130GB/sq. in.

Giving

130GB/sq.in. * 3.8 sq.in = 494172524409 bytes

or roughly 400GB usable. About the same as the latest high-density 2.5" drives.

Hope this helps...

4. #### about 280-560x more than current technology

Midway down the page ..

http://news.com.com/A+divide+over+the+future+of+hard+drives/2100-1008_3-6108687.html

5. #### Current capacity (lots of math)

1 Inches (in) = 25400000 Nanometers (nm)

so... 1 square inch is 645160000000000 square nm

so now divide that by capacity they're shooting for (1TB - which is 1 followed by a bunch of zeros as far as the hard drive companies are concerned)

so... 645160000000000 / 1000000000000 = 645.16 square nm

now take the square root to find the length that the 13nm pits take up.

this is 25.4nm (which is probably correct since the 13nm pits need some area around that isn't a pit to support the structure and provide that magnetic shielding)

now... do some algebra... 25.4 / 13 = N / 100

13 is the target, 100 is the current working prototype.

N is 195.384615... lots more decimal places... anyhoo...

square this...

we get this...

38175.147928994... again... lots more decimal places... now divide the nm area by this and we have the capacity in bytes.

645160000000000 / 38175.147928994... = 16900000000

16.9 GB. Far cry from 1 TB, but it's getting there.

and hopefully I didn't miss copying a zero somewhere.

6. #### Another advantage not mentioned

Because the magnetic domains are regularly arranged and all very close to the same size, magnetization is more quantized. In traditional drives, magnetic domains vary in size, and so the net magnetization per unit area can be quite variable, resulting in readout noise, which limits density.

Carnegie Mellon University, I believe with funding from IBM mostly, is working on a similar technology to this one, using magnetic nanoparticles, rather than nano-holes. Both are promising.

7. #### So what was the capacity?

I may just be being dense, but I'm still waiting for a figure, e.g. If they released a 3.5" IDE drive (for example's sake) at 13nm, you could store 84TB on your computer, etc.

Am I being dense or is there a figure like that?

8. #### Title

Jesus.

Guys... You know that there's more than 1sq/in on a drive. You know that the stated (new density) is 1TB/sqin. So don't tell me that the capacity will be 16GB.

Also. Brett. Mate. I think you just multiplied the current density, by the area, to get current capacity. I like your thinking though. So lets try again

density * area = capacity

1 (TB per sq in) * 3.8 (sq in) = 3.8 TB. Better I think.

Now - for a multi-platter, multi side arrangement, we might expect four times that.

9. #### RE: Current capacity (lots of math)

Wow! - you made hard work of that. How's this:

(a) they hope to fit 1TB/sq in using 13nm holes. They can currently make 100nm holes.

(b) A 100nm hole is (100/13)^2 ~= 59 times the size of a 13nm hole (forget all the meaningless additional digits which)

(c) If 13nm holes allow 1TB/sq in, we can assume that 100nm holes give roughly 1/59 as much capacity, or 17GB/sq in.

Roughly speaking, a 2.5" platter gives a ring with inner diameter 1.25", outer diameter 2.5" of usable space. The area is thus PI x ((2.5/2)^2-(1.25/2)^2) ~= 3.7 sq in.

For a 3.5" disk this increases to 7.2 sq in per platter-side.

So each 2.5" platter should be able to fit 3.7 x 17 = 63 GB per side at current densities, increasing to 3.7 TB ultimately.

A 5 platter 3.5" drive using both sides should yield 5 (platters) * 2 (sides) * 7.2 (sq in / side) * 1 (TB/sq in) = 72 TB.

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