View source for Lower stiffness of smaller machinery

'''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 <br>
the geometry independent stiffness (aka elastic Young's modulus) E [N/m²] as such: <br>
<math>k = E ~ (A/l) \propto L^1</math> <br>
Thus geometry dependent stiffness falls when shrinking the size of machinery (while keeping the same material). <br>
Also covered on page about [[Scaling law]]s.

= Even diamond becomes soft like jelly – Not a problem though =

With scaling down machinery to smaller sizes the stiffness of this machinery falls. <br>
'''One millionth the size => One millionth the stiffness.''' See related page: [[Scaling law]].<br>
'''This makes even diamond jelly soft.''' <br>
Which poses an obvious question: <br> 
'''Q:''' '''Could this maybe be a serious problem?''' <br>
'''A:''' Perhaps surprisingly the answer is: '''No.''' At least for the most part. I.e. only thermal motions are of concern. <br>Math covered on page: [[Same relative deflections across scales]]

= Important are deflection magnitudes rather than spring constants =

For the material astoundingly low spring constants are not a problem because <br>
what is relevant are relative deflections rather the geometry dependent stiffness of the material. <br>
So how do deflections scale?

'''As it turns out the relative deflections / strains ...'''
* from accelerations of machinery scale with L<sup>0</sup> (scale invariant - nice!).
* from gravity scale with L<sup>1</sup>. – <small>(Large machines suffocating under their own weight. A well known macroscale problem.)</small>
* from thermal motions scale with L<sup>-1</sup>. – <small>(Relevant for [[piezomechanosynthesis]] and unguided [[covalent welding]])</small>

For the math deriving these [[scaling laws]] see Page: <br>
'''[[Same relative deflections across scales]]'''

= Example numbers – Jelly indeed =

Example numbers for diamond [[crystolecule]] strut:
* A = 1 nm²
* l = 10 nm
* E = 1000 GPa ≈ 10^12 N/m²
This gives: <br>
<math>k = E ~ (A/l) = (10^{12} N/m^2) · (10^{-18} m^2) / (10^-8 m) = 100N/m = 1daN/dm </math> <br>
Or colloquially: 1kg/dm or 100g/cm. <br>
This is how incredibly soft diamond gets at the nanoscale. <br>

For 10cm long macroscale strut with same aspect ratio (thus 1cm² cross section) that would be a pretty darn low spring constant. <br>
One would need to go to materials like quite soft rubber or jelly to reproduce this low level off a stiffness.<br>
Jelly is probably a better analogy since it tends to rupture somewhere in the low two digit percentual range. <br>
Just like perfect flawless diamond [[crystolecule]]s do. Whereas rubber often can be stretched several 100s of percents.

Related: [[The feel of atoms]]

= Misc =

{{wikitodo|explain the following}}
* The consequences on design constraints based no this falling stiffness

= Related =

* '''[[Same relative deflections across scales]]'''
* '''[[How macroscale style machinery at the nanoscale outperforms its native scale]]'''
* '''[[Applicability of macro 3D printing for nanomachine prototyping]]'''
* [[Macroscale style machinery at the nanoscale]]
* [[Natural scaling of absolute speeds]]
* '''[[Scaling law]]'''
* [[Stiffness]]
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Thermal motion related:
* [[Lattice scaled stiffness]]
* Parallel [[Robotic manipulators]]
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[[Intuitive feel]] related:
* [[The feel of atoms]] – about what "diamond getting jelly soft" intuitively means

[[Category:Pages with math]]
[[Category:Scaling law]]