Latest revision as of 08:37, 5 September 2022

In terms of scaling of volumetric throughput density

Volumetric throughput density refers to:
Processed volume-per-time per volume-of-machinery.

Math (to review):

• v … absolute speed – in m/s – this is kept constant over all layers in this model!
• s … side-length of the product (cube) of the topmost macroscopic assembly chamber
• C … a constant accounting for that the length of robotic motions is not exactly equal to
  the side-length size of the assembly chamber but longer due to curves (and for empty-handed back motions).
  
• B … branching factor, every assembly chamber has B² subchambers -- See: Branching factor
• F … Size ratio of assembly camber size (s*F) to the product size (s) -- See: Chamber to part size ratio
• V … volume of the product
• Vc … Volume of the assembly chamber
• Q … Absolute throughput – in m³/second
• D … Volumetric throughput density – (m³/second)/m³

Time for top assembly chamber (node) to finish one full assembly process.
A full assembly needs to cover the motion distance of B³ motions since the product consists of B³ parts.
Each motion leads through the full chamber size (s*F).
Motions will have curves making the path a bit longer. Accounted for by C (also for empty-handed back-motions x2).
[math] t(0) = (B^3 s F C)/v [/math]
Volume of top nodes final product:
[math] V(0) = s^3 [/math]
Volume of top assembly chamber:
[math] V_C(0) = (s F)^3 [/math]
Absolute throughput of top node:
[math] Q(0) = V(0)/t(0) = (s^2 v) / (B^3 C F) [/math]
Volumetric throughput density of top node:
[math] D(0) = Q(0)/V_C(0) = (v) / (s B^3 C F^4) [/math]

The same for the first sub-layer that has B² assembly chambers that are have a B times smaller sidelength.
[math] t(1) = B^3 (s F C / B^1)/v [/math]
For the volumes we look at the whole sub-layer thus the factor B^2.
[math] V(1) = (s/B^1)^3 (B^2)^1 = s^3 / B^1 [/math]
[math] V_C(1) = (s F/B^1)^3 (B^2)^1 = (s F)^3 / B^1 [/math]
Note: The absolute throughput matches up Q(1) = Q(0) = Q. Just as it needs to. See: Continuity of throughput
[math] Q(1) = V(1)/t(1) = (s^2 v) / (B^3 C F) = Q(0) = Q [/math]
[math] D(1) = Q(1)/V_C(1) = (v B^1) / (s B^3 C F^4) [/math]

Generalizing for arbitrary layers:
[math] t(i) = B^3 (s F C / B^i)/v [/math]
[math] V(i) = (s/B^i)^3 (B^2)^i = s^3 / B^1 [/math]
[math] V_C(i) = (s F/B^i)^3 (B^2)^i = (s F)^3 / B^i [/math]
[math] Q(i) = V(i)/t(i) = (s^2 v) / (B^3 C F) = Q(0) = Q [/math]
[math] D(i) = Q(i)/V_C(i) = (v B^i) / (s B^3 C F^4) [/math]

Be careful! While s is a size it is a constant.
The size s does not carry the information about the scale that we want to consider.
What does carriy the information about the scale we want to consider is the index i.
But i is discrete. We rather want a continuous scaling law for a given continuous sizescale L.
We can do that by calculating L from i like so:
[math] L = sF/B^i[/math]
Now to get a continuous scaling law for size scales L we can substitute B^i for sF/L giving:
And finally we arrive at:
[math] D(L) = v / (L B^3 C F^3) [/math]
Units are correct: 1/s = (m³/s)/m³

We see that:
[math] D \propto L^{-1} [/math]
Halving the size of nanomachinery doubles the volumetric throughput density.
This is the scaling law we wanted to derive and a main point of the exercise
(beside showing continuity of throughput).
See also: Higher throughput of smaller machinery

• Assuming (1kg/s)/m³ as reasonable for a macroscale robot
• then for a 1 000 000 x smaller nano-robotics (1 000 000 kg/s)/m^3 is reasonable

Practically one will just use up to 1 000 000 x less volume of nanomachinery. A thin layer on a chip.

Additional notes to that result

Volumetric throughput density D scales with speed v, robotic-path-curvyness C, and branching factor B as expected.
Interesting is that that the assembly chamber to product size-factor F goes in to the fourth power.

Another side-note: To get the total throughput for a gem-gum factory:
Q = D * namomachineryLayerThickness * chipArea

In terms of replication time

J. Storrs Hall did some analysis on that in the IMM Report 41 (link).
He noticed some discrepancy between the natural scaling for artificial systems and the scaling in natural systems.

Artificial systems: "... a one third power scaling law of replication time to mass – i.e. replication time is proportional to the cube root of the mass."
[math] t_{repl} \propto \sqrt[3]{m_{device}} \propto L^1 [/math]
Natural systems: " ... like animals and plants, have empirically a well-documented one-fourth power scaling law"
[math] t_{repl} \propto \sqrt[4]{m_{device}} [/math]

Math by J. Storrs Hall plus added translations into a more formal setting.

"Working in units of mass of a k-stage system:"
[math] M(k) = \sum_{i=0}^{k} m(k) = m_0 := 1 [/math]
"and working in units of time to replicate a k-stage system"
[math] t_{repl}(k) = M(k)/Q(k) := 1 [/math]

"Let x be the mass of the k+1st stage node:"
[math] m(k+1) := x [/math]
"The mass of the full k+1 stage system is then n+x."
[math] M(k+1) = n+x [/math]
"Thus the mass ratio of a k+1-stage system to a k-stage one is n+x
[math] M(k+1)/M(k) = n M(k) + m(k+1) = n+x [/math]
"The output per unit time of a k+1-stage system ins n,"
[math] Q(k+1) = m/t_{repl}|_{i=k+1} = n Q(k) = n[/math]
"so the time to replicate for a k+1 stage system is (n+x/n)
[math] t_{repl}(k+1) = M(k+1)/Q(k+1) = (n+x)/n [/math]

"Lemma: The mass ratio of k+1-stage node to k-stage node is also n+x"
Note that this is about the ratio of the top nodes this time not the whole system as before.

"The ratio of head node to whole system is x/(n+x),
which equals the ratio of the k-stage node to the whole subsystem it is head of,"
[math] m(k+1)/M(k+1) = x/(n+x) = m(k)/M(k) [/math]
"but since that is 1 by definition, the mass of the k stage node is just x/(n+x)." ( M(k):=1 )
[math] m(k) = x/(n+x) [/math]
"Thus the ratio of k+1-stage node to k-stage node is x/(x/(n+x)), which equals to n+x, qed."
[math] m(k+1)/m(k) = x/(x/(n+x)) = n+x [/math]

(wiki-TODO: Decypher the rest of the math and add it here too. )

Additional notes

Switching to the natural scaling law might be the right choice for the bottom-most processing of matter in gem-gum factories because there resource molecules are taken out of solution phase (or gas phase) and brought into machine phase and viscous flow resistance scales with the fourth power of pipe radius and can get very high.

This is about Poiseuille's Law: [math] Q = (\Delta p \pi R^4) / (8 \mu L') [/math] Where L' is pipe-length (not scale).
(See wikipedia: Hagen–Poiseuille equation)

This means:

• a deviation from the layer topology to more 3D tree like topology
• assembly levels being no longer organized in assembly layers

The the "last mile" transport of resource molecules can be bridged with diffusion transport scaling differently than viscose flow. Some moderate heating can improve on that.

Related

• Convergent assembly
• Scaling law
• Assembly level
• Assembly layer
• Higher throughput of smaller machinery
• Limits to lower friction despite higher bearing area

In Nanosystems

Fig 14.4. "... a hierarchical, convergent assembly process; ..."
ATTENTION! Do not overlook the last sentence in the image description!
"... This structure demonstrates that certain geometrical constraints can be met,
but does not represent a proposed system."
The actually proposed system is not illustrated in the book.
It is illustrated in "Productive Nanosystems From molecules to superproducts".

Table 14.4. (only included in the book not in the dissertation (which was the books prelim version))
Gives example of manufacturing system parameters.

The large number of stages (and thus low branching factor) of stages seems a bit overkill though.
It might interfere too much with the ease of designing products to produce with such manufacturing devices.

Consider 3D printers filling a volume of a cubic centimeter with sort-of-voxels of (0.4x0.4x0.2)mm³ giving 31250 voxels.
Well, granted 3D printers operate in what is equivalent to a streaming parts style of assembly.
When the robotics becomes too small for streaming assembly then just compensate lack of speed with more micro-machinery.
There is enough space for that due to higher throughput of smaller machinery.

With a baseline branching factor of say 32 it would be …

• … 32³ = 32768 voxels to assemble
• … only 4 convergent assembly stages
  from 32nm molecular mill assembly line channels up
  to 32mm macroscopic assembly chambers.
  Add one more and it's meter scale.

Math for cooling systems

In Nanosystems:

• 11.5.1. Murray's law and fractal plumbing – (about the optimal balancing of viscose flow resistance and thermal contact)
• Figure 11.8. ... nearly fractal system of cooling tubes ..
• 11.5.2. Coolant design
• 11.5.3. Cooling capacity in a macroscopic volume

Related

• Higher throughput of smaller machinery
• Deliberate slowdown at the lower assembly levels
• Increasing bearing area to decrease friction
• Optimal sublayernumber for minimal friction & sub-layer
• Level throughput balancing
• Continuity of throughput
• Mesoscale friction

External links

• J. Storrs Hall did some analysis on convergent assembly in the IMM Report 41 (link).
  

Wikipedia:

• Murray's law
• Hagen–Poiseuille equation