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Unit normal to curve

  1. Sep 26, 2012 #1
    Hi,

    I have a curve defined by following parametric equation

    \begin{equation}
    \gamma(\theta) = 1 + 0.5 \times \cos (N \theta) (\cos(\theta),\sin(\theta)), 0 \leq \theta \leq 2 \pi \
    \end{equation}
    where N is an integer. x and y coordinate of any point on the curve are simply

    \begin{align}
    x &= \gamma (1) \nonumber \\
    y &= \gamma (2)
    \end{align}

    Question is how do I compute a unit normal (n) to above curve in eq. (1) and its x (n_x) and y (n_y) components?
    One way to find x and y components of normal n would be to find dx / dn and dy / dn respectively. dx / dn and dy / dn can in turn be found using chain rule as follows -

    \begin{align}
    n_x &= \frac{dx}{dn} \nonumber \\
    & = \frac{dx}{dr} \frac{dr}{dn} \nonumber \\
    & = \frac{dx}{d\theta}\frac{d\theta}{dr}\frac{dr}{dn}
    \end{align}

    and similarly for n_y


    \begin{align}
    n_y &= \frac{dy}{dn} \nonumber \\
    & = \frac{dy}{dr} \frac{dr}{dn} \nonumber \\
    & = \frac{dy}{d\theta}\frac{d\theta}{dr}\frac{dr}{dn}
    \end{align}


    While first two terms on rhs of equations (3) and (4) can be found easily, how to find the dr / dn term? Or if any one has an easy way to compute the normal components of the unit normal using the parametric equation of the curve, it'd be a great help.

    Many thanks for help.
     
  2. jcsd
  3. Sep 26, 2012 #2

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    find the tangent to the curve by differentiation then solve for a unit vector whose inner product with the tangent is zero.
     
    Last edited: Sep 26, 2012
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