Difference between revisions of "Same relative deflections across scales"

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Scaling speeds linear with L would be way too much <br>
 
Scaling speeds linear with L would be way too much <br>
 
as it would fully counter [[higher throughput of smaller machinery]]
 
as it would fully counter [[higher throughput of smaller machinery]]
 +
 +
'''In all further analysis scale invariance of speeds and deflections is assumed.'''
  
 
= Analyzing scaling behavior =
 
= Analyzing scaling behavior =

Revision as of 09:47, 28 September 2022

Deflections induced by accelerations from mechanical motions are scale invariant
This is one important reason for why macroscale style machinery at the nanoscale works better rather than worse at the nanoscale.
See main article: How macroscale style machinery at the nanoscale outperforms its native scale

Relevance for scale transposed prototyping

Scale invariance of deflections from machine motions is a very important result for applicability of macro 3D printing for nanomachine prototyping.
This is because it means that 3D printed macroscale prototypes will typically vastly underperform nanoscale target systems of equal geometry.
(Slight under-performance is desirable from a exploratory engineering perspective)
From the stiffness aspect, there is no risk to accidentally prototype a macroscale system that then cannot be ported to the nanoscale.
Heck, if anything this may lead to uncircumventable massive over-engineering in macroscale prototypes.
Over-engineering that ...

  • will surely work at the nanoscale, but
  • will also be be very far from optimal.

See: How macroscale style machinery at the nanoscale outperforms its native scale

Same absolute speed across scales

A necessary assumption to get
same relative deflections across scales (aka scale invariance of relative deflections)
is to have same absolute speed across scales (aka scale invariance of absolute speeds).
This is why keeping speeds constant across scales can be seen as a natural choice.

[math]v \propto L^0 \Leftrightarrow \epsilon \propto L^0[/math]

When more or less deviating from that natural choice then relative
deflections scale the same as the speeds are scaled. That is:
When slowing down across scales then one (unsurprisingly) reaps the benefit of smaller relative deflections.

Some such deviation seems well motivated by dynamic friction in gem-gum technology being reducable by
deliberate slowdown at the lowest assembly level‎ & increasing bearing area to decrease friction.
But this happens in a rather nonlinear fashion across scales. Not like a scaling law.
An issue here is the still unclear mesoscale friction gap.

Still, when trying to approximate deliberate slowdown at the lowest assembly level‎ with a scaling law
then it might be roughly a square-root law or weaker:
[math]v \propto L^{1/2} \Leftrightarrow \epsilon \propto L^{1/2}[/math]
Scaling speeds linear with L would be way too much
as it would fully counter higher throughput of smaller machinery

In all further analysis scale invariance of speeds and deflections is assumed.

Analyzing scaling behavior

Without loss of generality only the one dimensional case of tensile strain is covered.

Scaling of deformations from machine motions (spoiler: they are scale invariant)

Accelerations from machine motions go up too worsening the situation.
One millionth the size => One million times the accelerations.

But don't fret just yet. We have a quite literally massive compensating factor. With scaling down machinery to smaller sizes the mass of that machinery is going down.
One millionth the size => (One millionth)³ the mass.


The critical quantity we want to be preserved across scales is relative strain:
[math]\epsilon = \Delta l / l[/math]
Strain epsilon is proportional to stress sigma.
[math]\epsilon = \sigma / E[/math]
Stress is given by applied force F.
[math]\sigma = F / A [/math]
A natural level of force for a scale is zentrifugal force from rotating a mass at that scale.
[math] F = m \omega^2 r = m (2\pi f)^2 r[/math]
Putting it all together we get:
[math]\epsilon = \sigma / E = F / (E A) = m \omega^2 r / (E A)[/math]

With:

  • mass m scales with [math] L^3 [/math]
  • frequency f scales [math] L^{-1} [/math] (assuming absolute speed v is kept scale invariant)
  • radius r scales with [math] L^1 [/math]
  • Tensile modulus being scale invariant [math] L^0 [/math]
  • Area scaling with [math] L^2 [/math]

We get:

  • Force F scales with [math] L^2 [/math]
  • Both stress & strain from machine motions scale with [math] L^0 [/math]

In other words when assuming scale invariant speed then:
Both stress & strain (relative deflections) from machine motions are scale invariant.

Scaling of deformations from gravity

[math] F_{grav} = \rho V g \propto L^3[/math]
[math]\epsilon_{grav} = \sigma_{grav} / E = F_{grav} / (E A) = \rho V g / (E A) \propto L^1 [/math]

Both stress & strain from gravity go down linearly with smaller scales.

Scaling of deformations from thermal motions

(wiki-TODO: Work that out by using equipartitioning theorem on a block the size of the scale L.)