Talk: Digital Domain Core

Intro

I think the most likely design we can estimate for the core is basically a quantum dot cellular automaton (QDCA) based logic circuits. Its molecular QDCA logic circuits might achieve switching speed that is within the gigahertz range, while consuming a lot less energy.

The cells are each comprised of four quantum dots that can encode at least two states: 0 and 1 depends on the positioning of electrons and quantum dots. Then those cells can be arranged to form logic gates in a grid with size of about 9x9, and there might be anywhere between one to three such groups for each gate. Note that cells are much farther apart from the dots within each cells.

What is out possible cell construct? From our previous note we get that the interior of the core is:

comprised of foamed carbon nanotube constructs filled with beryllium-copper-platinum-iridium (BCPI) enriched buckyballs, acting as programmable arrays of quantum dots.

There's a concept called endohedral fullorenes, or endofullorenes, where each one of such molecules are basically fullorenes that have additional atoms, ions, or clusters enclosed within their inner spheres.

Maybe it's related to the BCPI compounds in that manner, each is enclosed in such structures. We can consider that the quantum dots are in fact made of such structures: endofullorenes with BCPI cores.

According to the Wikipedia page on fullorenes:

The van der Waals diameter of a buckminsterfullerene molecule is about 1.1 nanometers (nm). The nucleus to nucleus diameter of a buckminsterfullerene molecule is about 0.71 nm.

Which describes a ` C_{60} `.

On endofullorenes page, there they described possible electron transfers from the interior molecules or molecules attached on the outside. Which might be the basis of our quantum dots.

If we assume the stacking is done in a manner that creates cubic configuration of fullorenes, we get that each balls occupied a space of about 1.331 nm cubic. It's one ` C_{60} ` per 1.331e-27 meter cubic.

QDCA Arrays

Maybe each cell is then comprised of a scaffold of an inner cell of 9x9 to contain the 4 dots, enclosed in another layer of normal fullorenes gap. So each cell is a 10x10 plane. Maybe they're further protected with a layer above and below, so 10x10x3 sheet can be formed. It's a sheet comprised of 300 balls each. They will occupy a volume of 3.993e-25 m³ each. To make a simple gate, they are arranged into 9x9 grids. A gate might comprise of 1-3 of such grids. So the mean is maybe 2 grids per gate.

Each grid is then comprised of 81 cells (24.3k balls). Each cell is comprised of 4 dots. The amount of dotted cells apparently range from 3 to 7 for each grid. Suppose the average is 4 dotted cells per grid. So each grid in average has 32 dots, encoding 4 bits.

So each gate is comprised of 2 grids (162 cells, 48.6k balls), then 64 dots for 8 bits. Assume an operation in a gate basically manipulate those bits, we can use Landauer principle to estimate our computing energy costs. At T=300k, it's 2.87e-21 J/bit. Ideally, each gate consumes 2.3e-20 joules per operation.

Assume we have to excite all 48.6k balls to change the states of just 8 bits (therefore those 8 bits are encoded in 48.6kilobits state information), we get: 1.4e-16 joules per operation. It makes an erasure of just 8 bits costs that much. We get that the cost is then 1.74e-17 J/bits.

Meanwhile information density is 6075 balls per bit. It's actually ` C_{60} `, so it's about 364500 atoms per bit.

We can ignore the trapped particles inside ` C_{60} ` as normally we probably have just about 100 of such atoms in each gate. It's dwarfed by all of those carbon atoms.

6075 balls per bit means each bit is encoded in a space of 8.09e-24 m³. It's 364,500 carbon atoms for such volume, with each carbon atom has a mass of 12*1.66053906660e-27 kg, we get a carbon mass of about 12*6.05e-22 kg (it is 7.26E-21 kg/bit, or 7.26E-24 g/bit). The density is then, ~900 kg/m³? Add some metal structures within and maybe we can get something like ~1200 kg/m³, with 70% be the computing substrates and 30% for support structures at 2000 kg/m³.

Actually, let's just consider the 6075 balls. Its volume is 8.09e-24 m³. It is comprised of 364.5k carbon atoms, therefore the mass is 7.2631978773084E−21 kg of QDCA for each bit. Specific information is 7.2631978773084E−21 kg/bit.

Its information density is 1 bit per 8.09e-24 m³ * 70% = 1.77e23 bit/m³. Is it?

QDCA Computing Speed

Its computation cost is 1.74e-17 J/bits. Then its computational cost to erase all bits per m³ is ~3 MJ. How fast is it? At computing switching of about 10 GHz, we can do it like 10^10 times per second, and it's, um, 3.07e16 Watts. Too crazy.

Let's scale them down to cgs units.

Now we get:

1. Information density of 1 bit per 8.09e-18 cc * 70% = 1.77e17 bits/cc.
2. At 10 GHz switching rate, the computational power requirement is 1.74e-17 J/bits * 1.77e17 bits/cc * 10^10 Hz = 31 GW/cc.
3. Material density is about ~1.2 g/cm³

Realistically, we can't have 31 GW/cc, as we're limited to blackbody radiation or our device would just shatter. What's then, our theoretical upper limit for energy intake? Well, it depends on the size of the orb.

If we limit it to about 30W, we would need sophisticated cooling mechanism, perhaps even water cooling. The structure would then need to maximize surface area deep to the core to remove heat as quick as possible. It would have grooves and maybe comprised of foamed structures instead. Which is why the support structures, primarily beryllium bronze alloys, can attain density of about 2 g/cm³.

Our 30W budget under computing cost of 1.74e-17 J/bits gave us computing speed of 1.72E18 bps. Comparable to human brains.

Structure

Copper beryllium density is 8.25 g/cm³ btw. Adjust the figure accordingly:

The main computing structure is comprised of 67% quantum dots cellular automaton (QDCA) as its main logic circuits and 34% beryllium-bronze alloy meshes that function as inter-connective circuitry support structures. The QDCA units are comprised of diamondoid structures infused with platinum-iridium enriched endofullorene quantum dots. The beryllium-bronze alloy meshes are structured into an acyclic tree from the center to the periphery, with connections between branches facilitated with clusters of specialized QDCA bridge circuits of its neighboring branches. The branches of the beryllium-bronze meshes that aren't connected to bridge-circuits end in a dense cluster general-purpose QDCA logic units that forms the logic circuits. The beryllium bronze branches are not the actual wiring, as the wiring is apparently facilitated with internal lightways comprised of diamondoid fibers that transport data packets in form of modulated light pulses of various wavelengths.

The configuration allows information to be transferred efficiently from one logic circuits to the next by travelling away from or toward the root branch. Alternately inter-branch connections are possible by travelling sideways across the bridge circuits. It allows the system to to not get congested by downstream (rootward) or upstream (tipward) data transfers, as information packets can be routed through the bridge-circuits up to the target branch.

The main computing circuitry has a density of 3.41 g/cm³. However, the tree structure has open gaps in between branches that allows circulation of water coolants to get rid of waste heat. Therefore the computing circuitry actual density (30% empty volume included) is 2.39 g/cm³. As its specific information is 7.2631978773084E−24 grams per bit, and the QDCA is basically 67% by volume with composite rho of 0.9g/cc * 67% = 0.6g/cc, we get that its information density is its composite rho times its specific information = 0.6g/cc * 7.2631978773084E−24 g/bit = 8.3E+22 bits/cc.

However, the computing core is further foamed with about 30% of empty space, so we also multiply the information density with 70% = 5.81E+22 bits/cc.

Now we information density for the computing core, we can consider the geometry of the DDC. Basically it's comprised of the shell and the computing core. The core has a diameter of 6.7 cm. The Shell is made of perforated beryllium bronze layer covered in 1 mm thick BCN-based light transceivers on the inner side and the outer side. The total shell thickness is 15 mm.

As it is perforated, its density would also be multiplied with 70%. Given that it is comprised of 2 mm thick BCN emitter arrays with density of 2 g/cc, and 13mm thick beryllium bronze shell with density of 8.25 g/cc, and that it is situated from r=3.35cm to r=3.5cm, its aggregate density is 7.42 g/cc. With its perforation, we multiply that with 70% and we get 5.19 g/cc.

Our DCC has a diameter of 7 cm, or radius of 3.5 cm: 1. Its computing core (r=0cm) extends from the center up to r=3.35cm. Its density is 2.39 g/cc, and its volume is 472.44 cc, therefore its mass is 1129.13 grams. 2. With information density of 5.81E+22 bits/cc, the core can hold up to 2.74E+25 bits of information. 3. It consumes 30 watts of power and its computing power can reach up to 1.72E+18 bps. 4. Its shell with density of 5.19 g/cc and occupy the region between r=3.35cm to 4=3.5cm has a volume of 66.35 cc, therefore its mass is 344.36 grams. 5. The DCC has a total mass of 1473.49 grams.

Data Transfer

Now the other fun part: its data bandwidth. Given that it is a spherical construct, we actually had something similar to calculate its theoretical upper limit of data transfer. However it was originally designed to calculate the bandwidth of a Hayward class wormhole. Plugging the diameter of our DCC (7 cm) and its power consumption (30W), we get a transfer rate of 1.37E+22 bps. At that rate, its entire memory content can be downloaded or uploaded within just 2000 seconds or about 33 minutes.

Obviously it wouldn't use the power used for computation, in fact I expect that the emitters are powered individually, probably about 20 watts in general. Therefore the total energy consumption of the DCC can be up to 50 watts during its operation, with data transfer of (20 watts) 1.01E+22 bps top. With that transfer rate, its entire memory can be uploaded or downloaded within just 2700 seconds or 45 minutes.

In practice, under normal operation we can expect it to just use about 30 watts of power, and data transfer rate is more or less equal to its computing capacity of about 10^18 bps.

Data Storage

Given its memory and its computing power, consider how long is it before the memory space is exhausted? Assuming its memory space is filled up at the same rate as its computing speed, it would take up to around 16 million seconds for the storage space to run out. It is 4425 hours or 184 days. Of course, practically it would employ some sort of compression, and a lot of data are redundant so can be skipped for efficiency.

From a quick Google search, i got:

1 Hour of 4K at 24FPS takes up around 15.8GB for H264 and 7.9GB for HEVC/H265. 1 Hour of 4K at 30FPS consumes about 20.5GB for H264 and 10.3GB for HEVC. 1 Hour of 4K at 60FPS hogs up space approximately 23.4GB for HEVC.

Source: https://www.macxdvd.com/mac-video-converter-pro/4k-video-file-size.htm

Say for 1 hour of 4k at 60 fps we get 23.4GB, it's 2.34E10 bit per hour. Then the storage space can store about 1.17E15 hours of footages, or 4.89E13 days, or 133 billion years. That's absurdly large amount of storage. Of course video formats and a kuinite robot memory is not comparable. However with this quick analysis, we can safely and reasonably assume that the memory can serve the robot for its entire service life plus some (a lot of) extra data.

Conclusion

The Digital Domain Core is a spherical core with a diameter of 7 cm and mass of about 1.5 kg. Its outer surface is a perforated beryllium bronze shell covered in BCN-based light emitters. The interior is composed primarily of platinum-iridium endofullorenes programmable quantum dots arrays suspended in diamondoid matrices covering fine branching beryllium-bronze wires that converges at the center of the structure.

On average it consumes 30 watts of power with maximum computing speed of 1.72E+18 bps. Its memory can store up to 2.74E+25 bits of information.