Let $M_k$ and $M_{k+1}$ be two successive positions. Supposing the road is perfectly planar and horizontal, as the motion is locally circular, we have:

enter image description here

Where $\Delta$ is the length of the circular are followed by $M$, $\omega$ and $\rho$ the radius of the curvature.

My problem is that the paper's author assumed the distance between $M_k$ and $M_{k+1}$ to be equal as $\Delta$. His argument to do so, is the following:

"From basic Euclidean geometry, we know that delta is approximately $|M_kM_{k+1}|$ up to the second order."

Could anybody explain how we can assume they are similar?

Extra explanation:

The symbol $\omega$ is the rotation of the mobile frame and heading angle is denoted by $\theta$.


Data Fusion of Four ABS Sensors and GPS dor an Enhanced Localization of Car-like vehicles.


closed as off-topic by Gert, stafusa, Jon Custer, ZeroTheHero, Aaron Stevens Sep 11 at 13:23

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  • $\begingroup$ Your question needs a lot more context before anyone can give you an answer that would seem reasonable. $\endgroup$ – David White Sep 6 at 2:10
  • $\begingroup$ Updated it. Could you check again? If you have any further questions I will be willing to give additional information. $\endgroup$ – eggrobot78 Sep 6 at 3:30
  • 3
    $\begingroup$ See en.wikipedia.org/wiki/Small-angle_approximation $\endgroup$ – PM 2Ring Sep 6 at 4:48

The authors are making an approximation of the path length by using a straight line between the start and end of the circular arc. They do this to make this math easier and because the difference doesn't matter since the approximation becomes exact when the limit of the path length goes to zero (as in going from a discrete step to a continuous path).

The actual path length along the circular arc is giving by $$\Delta = \rho\omega.$$ The straight-line path is given by the length of the chord: $$|M_kM_{k+1}| = 2\rho\sin\left(\frac{\omega}{2}\right).$$ For angles near zero, we can approximate the $\sin$ function with a Taylor polynomial: $$\sin x = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \cdots$$ When the authors say "up to second order," they mean that the straight-line approximation is the result of using the second-order (maximum degree two) Taylor polynomial of the actual formula. In our case, the second-degree approximation is $$\sin x = x$$ since the quadratic term of the Taylor expansion of $\sin x$ is zero. Plugging this into the approximation results in $$|M_kM_{k+1}| = 2\rho\sin\left(\frac{\omega}{2}\right) \approx 2\rho\frac{\omega}{2} = \rho\omega = \Delta.$$ So, for small distances or time increments (where $\omega \ll 1\,\textrm{rad}$), dividing the circular path into straight-line segments is an accurate approximation.


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