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It is possible to obtain the parameters of a curve in 2d simply by having only its curvature k(s)?

I need to obtain its parametric equations in order to reconstruct the curve but i don't have any idea how or even if its possible?

I try to search on internet but i couldn't find anything related to the problem.I even look into different differential geometry books but i couldn't found anything related to the problem

metamail
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    You'll need to know at least one point in $\mathbb{R}^2$ through which your curve passes. For example, a circle of radius 1 has curvature identically 1 whether its center is the origin $(0,0)$ or some other point. – 211792 Feb 06 '15 at 19:37
  • so it can't be done unless i have a point?but if i try to add it to a Fresnel system? – metamail Feb 06 '15 at 19:42
  • Bear in mind that to reconstruct the curve you will need the signed curvature (often in 2D the curvature is taken unsigned). – user7530 Feb 06 '15 at 19:58

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If you have some initial value like the starting Point $c_0$ of the curve you can reconstruct the curve. If $c(s)$ is the curve parametrized by $s$ (for simplicity, it can be assumed that the curve is parametrized by arclength) then it holds the following equation (if $n(s)$ is the unit normal vector):

$c''(s) = k(s) n(s)$. (' is derivative by $s$)

For arclength-parametrized curves you can assume $c(s) = (cos(\theta(s)),sin(\theta(s)))$ for a function $\theta$. By using above equation you will get the differential equation: $\frac{d}{ds} \theta(s) = k(s)$. By Integration you will obtain $c'$; integrate once again with an initial condition $c(0) = c_0$ and you have the curve.

kryomaxim
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    Note that you need not only $c(0)$ as an initial condition, but also an initial orientation (tangent vector). – user7530 Feb 06 '15 at 19:57
  • I found something but it isn't with a starting point http://mathworld.wolfram.com/NaturalEquation.html so would this work for any curve? – metamail Feb 06 '15 at 19:58
  • It will work for any curve, in theory, but the required integrals will almost always be nasty. To fix the curve in space, you always need a starting point and orientation. These correspond to the two constants of integration that will arise. – bubba Feb 07 '15 at 13:57
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The answer from @kryomaxim is the standard mathematical one. The equation that gives curvature as a function of arclength is known as the "intrinsic equation" of the curve, or sometimes the "Cesàro equation". You can start reading about this here.

But, in practice the mathematical theory by itself is not sufficient. Even with a very simple equation like $\kappa(s) = cs$ (curvature is a linear function of arclength), constructing the curve will require you to numerically compute some nasty integrals. In this particular case, you have to compute Fresnel integrals, and you'll get a curve called a Cornu spiral. These sorts of spirals are used in the design of roads and railway tracks, so they're important.

Various sorts of spirals (with simple Cesàro equations) are also used in font design. See Ralph Levien's thesis for a very nice account of all this.

bubba
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The simple answer to your question is, yes. You can express a two-dimensional curve from the curvature as follows. I choose to use complex variables. The equation for the curve is given by

$$z(s)=\int e^{iP(s)} ds$$

where $\kappa (s)= P'(s)$, i.e., the derivative. For a more detailed description, please see my answer to a post on the Cornu spiral here: Is this Cornu spiral positively oriented or not?. As pointed out above, you need to specify an initial condition, but it will not change the shape of the curve, merely where it resides on the plane.

Community
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Cye Waldman
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