Difference between revisions of "Higher throughput of smaller machinery"
(Moved basic math section to the top -- and added some improvements - subchapters) |
(added section: == Intuition ==) |
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Line 7: | Line 7: | ||
* n = 1 ... one robotic unit – this is just here to not mix it up with other dimensionless numbers | * n = 1 ... one robotic unit – this is just here to not mix it up with other dimensionless numbers | ||
− | * s ... sidelength of the blocks that the robotic unit | + | * s ... sidelength of the product blocks that the robotic unit assembles <br>s^3 ... volume <br>rho*s^3 ... mass |
− | * f ... frequency of operation | + | * f ... frequency of operation that is how often per unit of time a block assembly gets finished |
* v ... constant speed | * v ... constant speed | ||
Line 45: | Line 45: | ||
f = v/(constant*s) ~ v/s <br> | f = v/(constant*s) ~ v/s <br> | ||
f' ~ v/s' ~ v/(s/2) ~ 2f | f' ~ v/s' ~ v/(s/2) ~ 2f | ||
+ | |||
+ | == Intuition == | ||
+ | |||
+ | How the heck can throughput rise when the very first assumption was <br> | ||
+ | that speed of robotic manipulation will be kept constants over all size scales?! | ||
+ | |||
+ | The answer: | ||
+ | * [[In place assembly]] is assumed here. | ||
+ | * Removal of the product from the manufacturing structure (the scaffold) is not included. | ||
+ | |||
+ | That is not a problem in the case of basic nanofactories where: <br> | ||
+ | (1) a stack of monolayers (only one layer per assembly level) is the baseline for the approach which | ||
+ | * makes productivity constant over all size scales and | ||
+ | * makes part removal correspond to pusing products up in the next assembly level (where streaming further up continues) | ||
+ | (2) the bottom assembly level where there is a thicker stack of monolayers of the same assembly level has much lower assembly than transport speeds | ||
+ | |||
+ | {{speculativity warning}} <br> | ||
+ | Higher degrees of filling space with nanomachinery than just thin monolayers may not be completely impossible. <br> | ||
+ | It would certainly be way beyond everyday practical. <br> | ||
+ | Transport speeds for product removal would "just" need to be faster than what is robotically possible. <br> | ||
+ | Imaginable would e.g. be a [[ultra high throughput microcomponent recomposer unit]] that is a block full of nanomachinery that has lots of straight coaxial high speed product shootout channels with [[stratified shear bearings]]. It would also need significant active cooling. | ||
+ | |||
+ | === Animation === | ||
+ | |||
+ | Imagine unassembled block-fragments (color coded red) on the bottom of an robotic assembly cell <br> | ||
+ | being assembled by the robotic mechanism inside of the cell <br> | ||
+ | to one product block (color coded green) at the top of the cell. | ||
+ | |||
+ | Taking just one big macroscopic cell this process takes a while since all the parts need to travel the macroscopic distance from the bottom of the big assembly cell to the the top to turn from red to green. | ||
+ | |||
+ | When instead taking many small robotic cells (that have with their robotics inside moving with the same speed) | ||
+ | the distance from red to green is much shorter and thus everything turns from red (unassembled) to green (assembled) much much faster. | ||
+ | |||
+ | '''Advanced Q&A:''' | ||
+ | |||
+ | Why is the transition from red unassembled to green assembled <br> | ||
+ | in a 2D monolayer of robotic assembly cells faster than in one big cell of same 2D cross section? <br> | ||
+ | Isn't it supposed to finish in the same time if it's just a single monolayer? <br> | ||
+ | |||
+ | You're looking a a cross section! <br> | ||
+ | The red to green transition in a single monolayer is indeed way faster. <br> | ||
+ | But that single monolayer needs to repeat that transition many times over in order to <br> | ||
+ | make up for all the other the other layers that would be needed to completely fill up the volume of the big cell. <br> | ||
+ | In the simple model this exactly cancels out. | ||
+ | |||
+ | {{wikitodo|make (that is program) that animation – via OpenSCAD or Blender}} | ||
== From the perspective of diving down into a prospective nanofactory == | == From the perspective of diving down into a prospective nanofactory == |
Revision as of 15:36, 1 April 2021
When production machines are made much smaller
then they can produce much more product per time.
Contents
- 1 Basic math
- 2 Intuition
- 3 From the perspective of diving down into a prospective nanofactory
- 4 Getting silly – questionable and unnecessary productivity levels
- 5 Antagonistic effects/laws – sub microscale
- 6 Lessening the macroscale throughput bottleneck
- 7 Alternate names for this scaling law as a concept
- 8 Related
Basic math
- n = 1 ... one robotic unit – this is just here to not mix it up with other dimensionless numbers
- s ... sidelength of the product blocks that the robotic unit assembles
s^3 ... volume
rho*s^3 ... mass - f ... frequency of operation that is how often per unit of time a block assembly gets finished
- v ... constant speed
Original size
Throughput of one robotic cell:
Q = (n * s^3 * f)
Halve size
Throughput of eight robotic cells that
- are two times smaller each
- fill the same volume
- operate at the same speed:
(same speed means same magnitude of velocity not same frequency)
Q' = (n' * s'^3 * f') where n' = 2^3 n and s' = s/2 and f' = 2f
Q' = (2^3 n) * (s/2)^3 * (2f)
Q' = 2 * (n * s^3 * f)
Q' = 2Q
1/10th size
Throughput of a thousand robotic cells that
- are ten times smaller each
- fill the same volume
- operate at the same speed:
Q'10 = (10^3 n) * (s/10)^3 * (10f)
Q'10 = 10 * (n * s^3 * f)
Q'10 = 10Q
Supplemental: Scaling of frequency
The unexplained scaling of frequency is rather intuitive.
Going back and forth halve the distance with the speed you get double the frequency.
If you really want it formally then here you go:
f = v/(constant*s) ~ v/s
f' ~ v/s' ~ v/(s/2) ~ 2f
Intuition
How the heck can throughput rise when the very first assumption was
that speed of robotic manipulation will be kept constants over all size scales?!
The answer:
- In place assembly is assumed here.
- Removal of the product from the manufacturing structure (the scaffold) is not included.
That is not a problem in the case of basic nanofactories where:
(1) a stack of monolayers (only one layer per assembly level) is the baseline for the approach which
- makes productivity constant over all size scales and
- makes part removal correspond to pusing products up in the next assembly level (where streaming further up continues)
(2) the bottom assembly level where there is a thicker stack of monolayers of the same assembly level has much lower assembly than transport speeds
Warning! you are moving into more speculative areas.
Higher degrees of filling space with nanomachinery than just thin monolayers may not be completely impossible.
It would certainly be way beyond everyday practical.
Transport speeds for product removal would "just" need to be faster than what is robotically possible.
Imaginable would e.g. be a ultra high throughput microcomponent recomposer unit that is a block full of nanomachinery that has lots of straight coaxial high speed product shootout channels with stratified shear bearings. It would also need significant active cooling.
Animation
Imagine unassembled block-fragments (color coded red) on the bottom of an robotic assembly cell
being assembled by the robotic mechanism inside of the cell
to one product block (color coded green) at the top of the cell.
Taking just one big macroscopic cell this process takes a while since all the parts need to travel the macroscopic distance from the bottom of the big assembly cell to the the top to turn from red to green.
When instead taking many small robotic cells (that have with their robotics inside moving with the same speed) the distance from red to green is much shorter and thus everything turns from red (unassembled) to green (assembled) much much faster.
Advanced Q&A:
Why is the transition from red unassembled to green assembled
in a 2D monolayer of robotic assembly cells faster than in one big cell of same 2D cross section?
Isn't it supposed to finish in the same time if it's just a single monolayer?
You're looking a a cross section!
The red to green transition in a single monolayer is indeed way faster.
But that single monolayer needs to repeat that transition many times over in order to
make up for all the other the other layers that would be needed to completely fill up the volume of the big cell.
In the simple model this exactly cancels out.
(wiki-TODO: make (that is program) that animation – via OpenSCAD or Blender)
From the perspective of diving down into a prospective nanofactory
Simplified nanofactory model
Let's for simplicity assume that the convergent assembly architecture in an advanced gem-gum factory is organized
- in simple coplanarly stacked assembly layers.
- that are each only one assembly cell in height, monolayers so to say
- that all operate at the same speed
Here ignored model deviations
There are good reasons to significantly deviate from that simple-most model.
Especially for the lowest assembly levels. E.g.
- high energy turnover in mechanosynthesis and
- fast recycling of pre-produced microcomponents and
- high bearing area
But the focus here is on conveying a baseline understanding.
And for assembly layers above the lowermost one(s) the simple-model above might hold quite well.
Same throughput of successively thinner layers
When going down the convergent assembly level layer stack …
- from a higher layer with bigger robotic assembly cells down
- to a the next lower layer with (much) smaller robotic assembly cells
… then one finds that the throughput capacity of both of these layers needs to be equal.
- If maximal throughput capacity would rise when going down the stack then the upper layers would form a bottleneck.
- If maximal throughput capacity would fall when going down the stack then the upper layers would be underutilized.
See main article: Level throughput balancing
The important thing to recognize here is that
while all the mono-layers have the same maximal product throughput
the thickness of these mono-layers becomes thinner and thinner.
More generally the volume of these layers becomes smaller and smaller.
So the throughput per volume shoots through the roof.
That is a very pleasant surprise!
In a first approximation halving the size of manufacturing robotics doubles throughput capacity per volume.
That means going down from one meter to one micrometer (a factor of a million)
the throughput capacity per volume equally explodes a whopping millionfold.
This is because it's a is a linear scaling law.
As mentioned this can't be extended arbitrarily though.
Below the micrometer level several effects (mentioned above) make
full exploitation of that rise in productivity per volume impossible.
Getting silly – questionable and unnecessary productivity levels
Now what if one would take a super thin microscale (possibly non-flat) assembly mono-"layer" that one finds pretty far down the convergent assembly stack and fills a whole macroscopic volume with many copies of it?
The answer is (in case of general purpose gem-gum factories) that the product couldn't be removed/expulsed fast enough. One hits fundamental acceleration limits (even for the strongest available diamondoid metamaterials) and long before that severe problems with mechanical resonances are likely to occur.
Note that the old and obsolete idea of packing a volume full with diamondoid molecular assemblers wouldn't tap into that potential because these devices are below the microscale level in the nanoscale where the useful behavior of physics of raising throughput density with falling size of assembly machinery is hampered by other effects.
More on silly levels of throughput here:
Macroscale slowness bottleneck
Antagonistic effects/laws – sub microscale
The problem that emerges at the nanoscale is twofold.
- falling size => rising bearing area per volume => rising friction => to compensate: lower operation speed (and frequency) – summary: lower assembly event density in time
- falling size => rising machinery size to part size (atoms in the extreme case) – summary: lower assembly site density in space
Due to the nature of superlubricating friction:
- it scales with the square of speed (halving speed quaters friction losses)
- it scales linear with surface area (doubling area doubles friction)
It makes sense to slow down a bit and compensate by stacking layers for level throughput balancing. A combination of halving speed and doubling the number of stacked equal mono-"layers" halves friction while keeping throughput constant.
Lessening the macroscale throughput bottleneck
There are also effects/laws (located in the macroscale) that can help increase throughput density above the first approximation. Details on that can be found (for now) on the "Level throughput balancing" page.
Alternate names for this scaling law as a concept
- Higher productivity of smaller machinery
- Productivity explosion
The thing is higher throughput does not necessarily means higher productivity in the sense of generation of useful products.
Thus the rename to the current page name "Higher throughput of smaller machinery".