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I'm trying to get a curve equation from curve curvature (trying to design a bent air pipe with later boundary layer separation). I should also mention that my math is pretty rusty.

My idea was to have a bend curve that would have a linear decreasing curvature k(s). I checks stackexchange for similar questions (linked bellow) and from my understanding, I would have to define a k(s) function and then integrate it twice.

s - arclength

c(s) - curve equation

k(s) - curvature equation

n(s) - unit normal vector

c''(s) = k(s) * n(s)

How can I get the normal unit vector as a function of arclength s?

Related topics:

Given the curvature, find the equation of any curve in plane

Curve from curvature

user2882635
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    For a curve in the plane you don't need that normal unit vector $n(s)$. The succint answer (copied over here) is: $$\begin{pmatrix} x \ y \end{pmatrix}=\begin{pmatrix} x_0+\int_0^s \cos \left( \int_0^s \kappa(s) , ds \right) , ds \ y_0+\int_0^s \sin \left( \int_0^s \kappa(s) , ds \right) , ds \end{pmatrix}$$ For a curve in $\mathbb R^3$ that $n(s)$ is an extra degree of freedom. In other words, you don't get it, you put it. Take a circle and a helix. They both have the same curvature but different motion of $n(s)$. – Kurt G. Oct 18 '22 at 14:25
  • thanks for the explanation. Do you perhaps know how the answer above is derived ? I'm trying to understand. – user2882635 Oct 26 '22 at 12:36
  • Take derivatives to verify $\ddot c=\kappa\cdot n$. – Kurt G. Oct 26 '22 at 12:44

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