26 Oct
1998
26 Oct
'98
9:51 p.m.
> The size isn't the real problem - you already
get 16 Gig in less
> than 320 cm^3 (using hard disk technology) which is more than
> 100 Meg per cm^3, which gives us 100x100x100x100 Meg or 100 Tera
> per m^3 (Only heat will be a problem, but if we assume that this
> will shrink by the factor 2 within the next few years, we get
> enough space for cooling without developing a new technology).
> 100 Tera are 100x2^40 Bytes so, for 16 exabytes you need
> 10x2^18 m^3 or 64x64x64x10 m^3 - just the size of a ordinary
> 160 store skyscraper. Nothing real big - isn't it? - and especialy
> not a mountain. and if we assume a increasing density by 10 within
> the next years, it is less than a warehouse.
> (I just left the disc acces time out of
calculation, but acording
> to any information availabel from disk manufacturers the internal
> caches will eliminate this almost to zero :)
Not so...
We can have 1 Terabytes now.
Seagate just announced their new drives in 3.5" I
think in 50GB.
Use 20 of them and still fit in 1 minicomputer box.
Just anounced ... hmm ok, will be availabel next year. So
lets see 1 Tera in a minicomputer box ? Lets take the size of my
minitower beside my desk (latest series SIEMENS Pro M7) its
20x46x44cm or 40.480 cm^3 (I cut off some plastic parts and
the sidedes to lower the size and we ignore the need of power
cables) lets say 40.000 cm^3 that gives us 25 units per m^3
(1 m^3 = 1.000.000 cm^3) - still four times more tan in my
last experiment. Remember I never said 1 Tera is out of reach
the mainframe I'm working on has a bit more than just a Tera
as disk storrage. It's just about how to get 16 Exas of address
space filled with minimum size requirements.
Ok, but we can continue to use the Barracuda 50 as a known
base for calculation. It will be a full height 3 1/2" drive
thats 42x102x147 mm^3 or 629,748 cm^3. since this drive still
radiates some 20W of heat, we still need cooling, but as in
my first example we just double the space to add high eficient
cooling (and cables and montage space) so we get some 1260 cm^3
now again packing it to the m^3 gives us 793 units - lets say
800 for easy calculation which gives us exactly 50x800 = 40.000 Meg
or 40 Tera. I think we have still two years or so to wait before
geting our 100 Tera per m^3...
And for the price, the Baracuda 50 is anounced at USD 2.100,
so 800 drives are just USD 1.620.000. Well I think we could
get som off from Segate, if we sell more of our
Storrage Meters ((c) ClassicCmp)
:)))
And can be run
off the 15A 115VAC or two plus the computer iteslf. :)
Didn't I say anything about power requirements ?
And please just show me one drive (from today!)
that is just connected to 115V (or 230V). We need
a 5/12V DC supply with about 20W per drive (~6W
at 5V and 14W at 12V with 30W peak). So 800 drives
are somewhat like 16 kW so lets get 20 kW and regarding
this, we just cant use 115 V even 400 V 3L~ will be
kind of stressed :). And at our 12V line we get a
current up to 1800 A peak or 700 A idle wich leads
us to a copper rail of some 20 cm^2 to get a load of
less than 100A per cm^2 (for US imaginations thats
3x1" solid copper) for the main supply line. And a
second one will be needed for 5V and ground. Maybe
we take silver (real cheap right now) so we could
halve the size :)
Oh, and a power supply of this capacity would be
almost the same size than our Storage Meter (c),
so we have to half the capacity, or just don't
tell te customers that we need dubble the proposed
space just to fire it up.
All calculations ar rough estimation based on actual
10 Gig drives. Real proportions are not used to
simplify calcualtion.
Servus
hans
P.S.: it feels good to think of copper in cm^2 instead
of fine line wire...
--
Ich denke, also bin ich, also gut
HRK