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For the pendulum, the following equations hold: $$\sin\varphi = \frac{x-x_0}{l}\quad\mbox{and}\quad \cos\varphi=\frac{l-(y-y_0)}{l}$$ where $l$ is the pendulum length, $\varphi$ is the angle of the rope, and $(x_0,y_0)$ is the lowest point of the pendulum movement. With the use of $\cos^2\varphi=1-\sin^2\varphi$, it follows that y=y_0+l-\sqrt{l^2-(x-x_0)^2}...

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As your problem is a local regression problem, I would not use a spline fit, but LO(W)ESS. This estimates $f(x)$ at a sample point $x$ by a weighted least squares fit to the points $x_i$ that are the $k$ nearest neighbors of $x$. For details, see Cleveland: "Robust locally weighted regression and smoothing scatterplots." Journal of the American ...

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Thanks to Thor, I figured out the solutions to my questions. As he pointed out, the range of the function can be specified in the fit command. In the above data set, the linear region starts from x=2.63. set term svg set output 'blue.svg' set term svg set output 'blue.svg' f(x) = a*x + b fit[2.63:] f(x) '07-B.txt' via a,b plot[][-1:3] '07-B.txt' pt 7, f(...

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