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According to Wolfram MathWorld [0], when talking about principal curvatures, $\kappa_1$ is the maximum and $\kappa_2$ is the minimum normal curvature.

However, I also noticed that Grasshopper documentation [1] defines $K^1$ as the principal direction corresponding to the absolute maximum principal curvature.

This confused me, so I would like to ask whether the principal curvatures $\kappa_1, \kappa_2$ represent the absolute minimum/maximum or just plain minimum/maximum of normal curvature?

What would be the correct value of $\kappa_{1}$ and $\kappa_{2}$ in the example below? Assume $\kappa_{min}$ is the minimum (non-absolute) normal curvature and $\kappa_{max}$ is the maximum (non-absolute) normal curvature? $$ \kappa_{max} = -0.52\\ \kappa_{min} = -4.84\\ \kappa_1 = \text{$\kappa_{min}$ or $\kappa_{max}$?}\\ \kappa_2 = \text{$\kappa_{min}$ or $\kappa_{max}$?} $$

EDIT: By "absolute", I meant the absolute value of a number.

[0] https://mathworld.wolfram.com/PrincipalCurvatures.html

[1] https://grasshopperdocs.com/components/grasshoppersurface/principalCurvature.html

jordi
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  • Describe the difference in the absolute versus "plain" max normal curvature. – Randall Jun 14 '22 at 23:59
  • @Randall If we consider my example, then $\kappa_1 = -4.84, \kappa_2 = -0.52$ for the case with absolute max normal curvature (if we'd adhere to Grasshopper's definition). On the other hand, $\kappa_1 = -0.52, \kappa_2 = -4.84$ when "plain" max normal curvature is used (as per Wolfram MathWorld definition). Which is the correct definition? – jordi Jun 15 '22 at 00:16
  • @Randall What I meant is: let's say we want to find $\kappa_1$ at a point $p$ of a given surface. Out of all possible normal curvatures $\kappa$ at point $p$ (corresponding to different normal sections), we have to select a specific $\kappa$ that will be considered as the principal curvature $\kappa_1$. How do we select such $\kappa$? Is it the $\kappa$ whose absolute value (i.e. regardless of the sign) is the largest (=the absolute max normal curvature)? Or is it the $\kappa$, whose (non-absolute) value is the largest (=the "plain" max normal curvature)? – jordi Jun 15 '22 at 00:34
  • I don’t think that’s what absolute means here. – Randall Jun 15 '22 at 00:35
  • @Randall I apologize for the confusion. I meant "absolute" as in the "absolute value $|\kappa|$". Please let me know if there is something obvious that I'm not understanding (I'm just a student). – jordi Jun 15 '22 at 00:43
  • The curvature, thus its absolute value, can be zero off the principal axes, if you have a saddle. We can say the curvature, and therefore it's absolute value, always has an extremum along the principal axes. – Oscar Lanzi Jun 15 '22 at 00:48
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    Shifrin merely defines $\kappa_1$ and $\kappa_2$ as the two eigenvalues of the shape operator, with no further description of which is $\kappa_1$ and which is $\kappa_2$. So in any case differentiating the two by size is not standard. If I wanted to avoid ambiguity, I would (as you have) simply write $\kappa_{max}$ and $\kappa_{min}$. – Charles Hudgins Jun 15 '22 at 00:51

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