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I have an object described by the parametric equations $x = x\left(t\right)$ and $y = y\left(t\right)$. I want to divide the arc into $n$ pieces of equal length.

I know that the arc-length is defined by $$l =\int_{0}^{2\pi}\sqrt{\left(x\left(t\right)'\right)^2+\left(y\left(t\right)'\right)^2 }\cdot dt.$$ I know that I need, for the $i$-th sub-arc, to solve the equation $$ l_i=\frac{i}{n}l \overset{!}{=}\int_{0}^{t_i}\sqrt{\left(x\left(t\right)'\right)^2+\left(y\left(t\right)'\right)^2 }\cdot dt $$ for $t_i$.

However, the integral is of a form that, in general, can only be evaluated numerically.

Is there any way to solve for $t_i$ without knowing the analytic result of the integration?

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  • $\begingroup$ It would strongly help your cause to post what your $x(t)$ and $y(t)$ are $\endgroup$
    – Alex R.
    Mar 28, 2016 at 17:53
  • $\begingroup$ They are both continuous and differentiable. $\endgroup$
    – Aaron Wild
    Mar 28, 2016 at 18:27

1 Answer 1

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You can't solve for it analytically, but you are well set up for numerical root finding. Given a $t$ you can compute the arc length up to $t$ and the derivative of the arc length with respect to $t$ is $\sqrt{\left(x\left(t\right)'\right)^2+\left(y\left(t\right)'\right)^2 }$ which you already know how to calculate. This is perfect for Newton's method.

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  • $\begingroup$ But what would be a suited initial guess for the approximation? $\endgroup$
    – Aaron Wild
    Mar 28, 2016 at 19:39
  • $\begingroup$ Absent better info, I would start with $\frac {2\pi}n$ $\endgroup$ Mar 28, 2016 at 21:40

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