Math of convergent assembly
Contents
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.
[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]
To get a continuous scaling law for size scales L we can substitute for i
[math] i = - \log_B(L/s) [/math]
And finally we arrive at:
[math] D(L) = v / (L B^3 C F^4) [/math]
Units are correct: 1/s = (m³/s)/m³
[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 the whole point of the exercise.
- 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.
Beside that:
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 sidenote: 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 poer 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. )
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 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.
