Curvature the same in G2 continuity

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In G2-continuous splines(connected curves), people say the curvature is the same at the junction point. In case of C2-continuous splines it is clear because;



$ kappa := fracc' times c'' c' $
, and C2-continuity requires C1-continuity at the same time by construction.
Then $ c'_1(t_0)=c'_2(t_0), ,, and ,, c''_1(t_0) = c''_2(t_0)
,, holds, , thus, $
$ ,, kappa_0(t_0) = kappa_1(t_0) ,, must ,, hold. $



But in case of G2-continuity, the identity of the curvature is not obvious. Rather than just saying that's the definition, is there a clear derivation to show it?



Thanks :)




differential-geometry algebraic-geometry euclidean-geometry analytic-geometry




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    In G2-continuous splines(connected curves), people say the curvature is the same at the junction point. In case of C2-continuous splines it is clear because;



    $ kappa := fracc' times c'' c' $
    , and C2-continuity requires C1-continuity at the same time by construction.
    Then $ c'_1(t_0)=c'_2(t_0), ,, and ,, c''_1(t_0) = c''_2(t_0)
    ,, holds, , thus, $
    $ ,, kappa_0(t_0) = kappa_1(t_0) ,, must ,, hold. $



    But in case of G2-continuity, the identity of the curvature is not obvious. Rather than just saying that's the definition, is there a clear derivation to show it?



    Thanks :)




    differential-geometry algebraic-geometry euclidean-geometry analytic-geometry




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      In G2-continuous splines(connected curves), people say the curvature is the same at the junction point. In case of C2-continuous splines it is clear because;



      $ kappa := fracc' times c'' c' $
      , and C2-continuity requires C1-continuity at the same time by construction.
      Then $ c'_1(t_0)=c'_2(t_0), ,, and ,, c''_1(t_0) = c''_2(t_0)
      ,, holds, , thus, $
      $ ,, kappa_0(t_0) = kappa_1(t_0) ,, must ,, hold. $



      But in case of G2-continuity, the identity of the curvature is not obvious. Rather than just saying that's the definition, is there a clear derivation to show it?



      Thanks :)




      differential-geometry algebraic-geometry euclidean-geometry analytic-geometry




      share|cite|improve this question













      In G2-continuous splines(connected curves), people say the curvature is the same at the junction point. In case of C2-continuous splines it is clear because;



      $ kappa := fracc' times c'' c' $
      , and C2-continuity requires C1-continuity at the same time by construction.
      Then $ c'_1(t_0)=c'_2(t_0), ,, and ,, c''_1(t_0) = c''_2(t_0)
      ,, holds, , thus, $
      $ ,, kappa_0(t_0) = kappa_1(t_0) ,, must ,, hold. $



      But in case of G2-continuity, the identity of the curvature is not obvious. Rather than just saying that's the definition, is there a clear derivation to show it?



      Thanks :)





      differential-geometry algebraic-geometry euclidean-geometry analytic-geometry





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 31 at 10:39









      John Ma

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      asked Jul 31 at 10:17









      Robin

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