# Weird assumption in a paper to prove equation [closed]

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: 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 $$\theta$$.

Paper:

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 StevensSep 11 at 13:23

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

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.