Talk:Curvature

(Layout problem)
The illustration and the text are interfering with each other, as viewed from Netscape. I've tried putting a colon before the "div", and I've tried putting "br" before and after it, to no avail. Michael Hardy 20:12 Mar 14, 2003 (UTC)


 * This seems to be affecting a number of images that used to work correctly in Netscape (they still work as expected in IE). Was something changed in the Wiki software that is affecting this? I'll change it to using a table. Chas zzz brown 22:50 Mar 14, 2003 (UTC)

Earth's curvature
Seeking information on the Earth's curvature, but no linkage from this page. I've read that "The earth's curvature is not visible from altitudes lower than about 20 miles.", but I'd really like a cite. ~ender 2007-08-21 12:06:PM MST —The preceding unsigned comment was added by Special:Contributions/ (talk)

Signed curvature in three dimensions
It seems noteworthy to me that the local curvature can easily be obtained by adding an obvious term. If one extends the given equation $$\kappa = \frac{|\dot{r} \times \ddot{r}|}{|\dot{r}|^3}$$ by the directional vector normalized to unit length the curvature vector becomes as signed quantity: $$k_{3D} = \frac{\dot{r} \times \ddot{r}}{|\dot{r} \times \ddot{r}|^3} \frac{|\dot{r} \times \ddot{r}|}{|\dot{r}|^3}$$

Where the added term makes $$k_{3D}$$ consistent to the sign in the signed curvature k for the two dimensional case:


 * $$k = \frac{x'y-y'x}{(x'^2+y'^2)^{3/2}}.$$

Thus it is possible to give also a signed curvature for a three dimensional curve. Then one can integrate this and obtain, for example the 'net' curvature for a Lissajous (1:2) figure to be (0.0,0.0,0.0) instead of the unsigned case, where the curvature adds up.

I verified this 'experimentally' in Mathematica. However, can this be found in literature? User:Aritglanor Friday, June 19, 2009 at 3:44:14 PM (UTC)

the next derivative
If (signed) curvature of a plane curve is the first derivative of tangent angle with respect to arc length, is there a common word for the next derivative, i.e. the first derivative of curvature? —Tamfang (talk) 19:19, 24 January 2023 (UTC)


 * Torsion of a curve? –jacobolus (t) 20:59, 24 January 2023 (UTC)


 * I now bolded the thing you may have missed. —Tamfang (talk) 21:10, 24 January 2023 (UTC)


 * The next-derivative analog of curvature (a naturally bivector-valued quantity) is torsion (a naturally trivector-valued quantity). In the plane of course torsion vanishes (any wedge product of 3 coplanar vectors is 0). You can come up with various other planar concepts involving higher derivatives, but IMO they aren’t really natural analogs of curvature. –jacobolus (t) 03:00, 25 January 2023 (UTC)
 * Raph Levien's thesis has a lot of analysis about changes in curvature with respect to arclength, but I am not sure if there are any specific names like what you are looking for. –jacobolus (t) 03:08, 25 January 2023 (UTC)
 * Heh, reading Levien's work prompted the question. —Tamfang (talk) 00:19, 30 January 2023 (UTC)
 * Next time I see him, I’ll try to remember to ask if there’s a name for this. No promises though. –jacobolus (t) 02:34, 30 January 2023 (UTC)

Problematic description
The section Gaussian curvature begins as follows:

"In contrast to curves, which do not have intrinsic curvature, but do have extrinsic curvature (they only have a curvature given an embedding), surfaces can have intrinsic curvature, independent of an embedding. The Gaussian curvature, named after Carl Friedrich Gauss, is equal to the product of the principal curvatures, k1k2."

So: Immediately after the reader is told that Gaussian curvature is intrinsic to a surface, it is defined extrinsically.

This strikes me as rather confusing to the reader.

I hope someone knowledgeable about this subject can fix this.