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In The Physiology of the Joints by Kapandji, he provides a diagram on page 21

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He does not formally define the concept of "resistance" but just creates this variable $R$ and claims that, if $n$ is the number of curves,

$$R = n^2 +1$$

enter image description here

He also references an equation called "Delmas Index"

$$DI = \frac{\text{length of vertebral column from tip to tip with curves}}{\text{length of vertebral column from tip to tip fully extended}} \times 100$$

Unfortunately, he provides no citation for any of this. Upon Googling both of these a fair amount, I have sadly found other physiology textbooks reference only this book. There doesn't seem to be any actual scientific analysis given.

There are no references given anywhere in the book.

My question is,

Does an actual quadratic relationship exist between the curves and the resistance of the rod to compression? If so, what is the real equation used in fields like continuum mechanics?

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    $\begingroup$ Google for references to "axial stiffness of curved beams" or "end loaded curved beams". For example you could probably convert the results in pages.hmc.edu/dym/20-JSE2011EndLoadedBeams.pdf to use the Delmas Index etc. But the quote in your OP is getting close to "engineering illiteracy" IMHO - it seems to be making up terminology on the fly without bothering to define it properly (Cue "Trust me, I'm a doctor" joke?) $\endgroup$
    – alephzero
    Aug 16 '17 at 13:41
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When a curved solid is subjected to compressive stress, the displacement depends on the curvature (all other things being equal). When you have a section of material with a certain bending stiffness, and the entire section is curved with a radius $R$ and a subtended angle $2\phi$:

enter image description here

then an axial load F will give a bending moment $M=Fh$ and change $\phi$ by $\Delta \phi \propto \frac{M}{EI}$. Displacement of the ends of the beam (towards each other) will depend on $\phi$ and $R$,

$$\Delta x = 2R\left(\sin(\phi+\Delta \phi) - \sin\phi)\right)\\ \approx 2R\cos\phi \Delta \phi$$

To relate this to the force, we need to know $h=R(1-\cos\phi)$, where I am assuming the initial curvature (at $\phi = \phi_0$, if you like) occurs without needing to apply any stress.

Then

$$\Delta x = 2R\cos\phi \frac{F R(1-\cos\phi)}{EI}$$

How you can translate this into the equations you quote really depends on the assumptions you make. If you assume the radius of curvature to be constant, then two shorter arcs with the same curvature have much lower value of $h$ and will be stiffer; but if you assume that the displacement $h$ is the same, then the radius of curvature must be much smaller, and the object will be more compliant. If you assume constant total length $L$, you can convert $\frac{L}{R}=\phi$.

I hope you can play around with these equations to convince yourself that the "standard rule" requires a lot of assumptions. I am not sure it's worth going much deeper into this - but ask questions in the comments and I can clarify if needed.

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