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Homework Help: Radius of curvature

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.


    2. Relevant equations
    Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:

    [tex]\textbf{u}=\frac{d\textbf{r}}{ds}[/tex]

    and

    [tex]\frac{d\textbf{u}}{ds}=k\textbf{n}[/tex]

    where k is called the curvature.

    We can expand r(s) around a certain point:

    [tex]\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...[/tex]

    using the equations for u and n:

    [tex]\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...[/tex]

    which is the same as:

    [tex]\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...[/tex]

    And from here they conclude that this is the equation of a circle with radius [tex]\frac{1}{k}[/tex], which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.

    Thanks to advance to whoever is able to help.
     
  2. jcsd
  3. Mar 29, 2010 #2
    The parametric equation of a circle is ( a+ Rcosu, b+Rsinu).
     
  4. Mar 29, 2010 #3
    I'm wondering if your latter equation is well derived. Did you take it from your book?
     
  5. Mar 29, 2010 #4
    Yes, it is from the book and I was able to derive it by myself.

    @Eynstone, that accounts only for the sin term, not for the -cos term
     
  6. Mar 29, 2010 #5
  7. Mar 29, 2010 #6
    Oh wait! it's not expressed differently, it's just rearranged so that you can manage to get the radius of curvature, check that first website I handled, and enjoy up to point (31). ;)
     
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