Difference between revisions of "Lower stiffness of smaller machinery"
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So how do deflections scale? | So how do deflections scale? | ||
Revision as of 07:21, 28 September 2022
A rod is the stiffer ...
- the bigger its cross-section A (∝L²) is and
- the shorter its length l (∝L¹) is.
The geometry dependent stiffness (aka spring constant) k [N/m] is calculated from
the geometry independent stiffness (aka elastic modulus) E [N/m²] as such:
[math]k = E ~ (A/l) \propto L^1[/math]
Thus geometry dependent stiffness falls when shrinking the size of machinery (while keeping the same material).
Also covered on page about Scaling laws.
Contents
Even diamond becomes soft like jelly - not a problem though
With scaling down machinery to smaller sizes the stiffness of this machinery falls.
One millionth the size => One millionth the stiffness. See related page: Scaling law.
This makes even diamond jelly soft.
Which poses an obvious question:
Q: Could this maybe be a serious problem?
A: Perhaps surprisingly the answer is: No. At least for the most part.
Important are deflection magnitudes rather than spring constants
What is relevant are relative deflections rather the geometry dependent stiffness of the material.
So how do deflections scale?
As it turns out the relative deflections / strains from ...
- accelerations of machinery scale with L0 (scale invariant - nice!).
- gravity scale with L¹. – (Large machines suffocating under their own weight. A known macroscale problem.)
- thermal motions scale with TODO. – (Relevant for piezomechanosynthesis and unguided covalent welding)
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.
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 massive over-engineering in macroscale prototypes that.
Over-engineering that ...
- will surely work at the nanoscale, but
- will also be be very far from optimal.
How macroscale style machinery at the nanoscale outperforms its native scale
Deflections being scale invariant is one important reason for why macroscale style machinery at the nanoscale works better rather than worse at the nanoscale.
As discussed in applicability of macro 3D printing for nanomachine prototyping:
What we absolutely do not want is to accidentally build a prototype macroscale systems with performance so high that
target nanoscale systems of equal geometry will not be able to replicate that performance.
The result (math further below) implies that we do not have to fear that.
In fact for us to build a macroscale system that has the same or higher performance than the target nanoscale systems we would have to use materials ...
- with a tensile modulus as high as diamond
- with a density as low as diamond
- with a maximal strain (bendability) in the double digit percentual range.
Such materials simply do not exist today. Ceramics come closest to stiffness but they're totally not elastic.
(Future gem-gum metamaterials might come close.)
That's the 🤯 degree of how much macroscale style machinery at the nanoscale
works better than our good old macroscale machinery at its native macroscale.
And that does not even factor in that we can easily afford to go a 1000x slower with speeds
by compensating with more nanomachinery (as is possible due to higher throughput of smaller machinery).
This is not an option for the (in comparison extremely voluminous) macromachinery.
In summary:
- steel has less elastic modulus than diamond
- steel has less elasticity than nanoscale flawless diamond
- steel cannot be moves as slow as nanoscale machinery as this slowdown cannot be compensated by mountains of more machinery
Related: Conservative design Exploratory engineering
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 from machine motions is 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.)
Misc
(wiki-TODO: explain the following)
- The consequences on design constraints based no this falling stiffness
