How about constructing storage so that there is no actual storage as such, just infinite transmission round a loop?
Of course this idea is not by any means new: In analogue terms, feedback delay lines have been used for 3 decades to store audio data. In digital terms, delay lines have existed for 10 years or so. How about appying the idea to storage?
What got me thinking was a recent article by Siemens and BT. These two companies are just experimenting with 160Gb/sec transmissions and are presently testing over 280 Kilometers sucessfully (in Lab conditions).
‘How much data is in transit at once’, I wondered.
It’s pretty straighforward maths, although I’ve not thought about it like this before.
Lets take it from the beginning:
c, as Einstein would put it, is 299,792,458 Meters per second. (Much as I hate the new money, it’s much easier this way, trust me!).
We’ll go about this in a rather long winded way, primarily because I’m a bit simple: At 10Mbits, such as with our good friend Ethernet, each bit is 29 Meters ‘long’. By long I mean if you could see each bit in transmission it would occupy a length of 29 Meters. Weird concept I know. BTW, Remember that these maths only apply to laser based comms: propogation through cables does not occur at c.
At 100Mbit and 1Gbit, the numbers are pretty easy to work out: 2.9 Meters and 0.29 Meters respectivly.
Okay. In my pretend example, lets say that we have a relay station on Earth and one on the Moon (okay, okay it’s crazy, stick with me a moment). Each relay station simply receives a signal from the other, regenerates it and bounces it back. The Earth station has the ability to pass the signal to an additional receiver and there is a facility for injecting fresh bits onto the stream (presumably replacing anything there already).
In this storage loop, the total capacity of the system and the retrieveal latency are related to the bit ‘length’ and the distance.
Lets keep with the moon example for a moment. The moon is about 402,336,000 meters from the earth, so there is ‘space’ for 13873655 bits along the path in each direction or let’s say around 3 Megabytes in total.
Our retreival time for any given bit is going to be (in the worst case), the round trip time (RTT), which at a distance of 402 million meters is around 2.6 seconds.
All in all, I think you will agree that this is probably the worst storage proposition you’ve ever had: particularly when you think that we have no error or check bits.
Never mind, lets go straight up to Gigabit: Now with each bit taking a mere 29 cm, we can fit a whopping 2,774,731,034
or 346 Megabytes. This is a bit better, now we are merely back in the dark ages.
Next: BT’s Recent Demo was of 160Gbit/Sec, this gives us an effective bit length of a tad under 2mm (although in actual fact, Bt’s work is around multiple parallel transmissions). Lets see what that gives us around 53,788,235,294 bytes in transit at once or around 53 Gigabytes. Much better!
In the real world of course we don’t need to worry about the bit lengths: we can just take the propogation delay in seconds and divide it by the bit rate to get the total data in transit. Not quite as interesting like that though, it it!
Going back to cables: As the propogation happens slower than c the capacity of the system actually increases increases, the tradeoff being the increse in bit recovery time.
All in all, bouncing signals off distant objects may be a really clever way to store high latency friendly data for long periods.
In the next example, We’ll try a Further planetary object like Mars or Jupiter… Watch This space