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I have learned the paper,"Shape similarity retrieval under affine transforms". Paper is here

I tried to use the formula (4) in the paper to calculate the curvature of a straight line.But i get the result that the curvature of every point is not zero.It shows that the curvature of the first point and the last point are 5.7 and 7.6.The curvature of others points are the same,1.6. I don't konw why i get this result.Can somebody help me?Thanks very much!

jack fukey
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    Can't access the paper without paying. it would be pretty strange to get curvature non-zero on a straight line ( $0$ curvature is what "straight" means to me! ), but maybe they're measuring the curvature of a line which is straight in one metric ( Euclidean ), but under a different metric. Since it's pay-to-view, could you explain the definitions and show the computation? – Callus - Reinstate Monica Sep 13 '15 at 05:30
  • Thanks for your comment.Try this :http://www.docin.com/p-296411440.html the fomula (2) – jack fukey Sep 13 '15 at 06:44
  • I tried to understand the article and formula, but I can't read chinese, sorry. If you take the time to write out the formula and definitions in English, as well as your computaiton which is leading to the surprising curvature valuations for a straight line, I'd be happy to keep reading. But as it is, I'm guessing about definitions, which is kind of a waste of time. – Callus - Reinstate Monica Sep 13 '15 at 14:43
  • Thank you and very sorry about that.Can you see this website?http://www.tec.u-ryukyu.ac.jp/wp-content/uploads/2010/08/kyou2010-21.pdf The formula is in section 2.I set the σ as 0.6. Maybe this is the reason about my question?But i do not know why the paper using curvature zero crossings to find concavity position.As here what i get the straight line curvature is 1.6 not 0.So why he use curvature zero crossings? – jack fukey Sep 14 '15 at 02:51

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Depends on your defining metric. All certain circles in hyperbolic geometry are deemed to be straight due to a certain defining or modeling metric function.

In a simpler example if you define

$$ \int \dfrac{\sqrt{ r^2 + {(r d \theta)}^2}}{r^2} $$

as your arc length , then any circle through the origin is quite straight.

Narasimham
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