How to find the length of an object in real life from an arc length of a curve?

Question

I am trying to find the length of the side of a guitar from this graph.

Basically I needed to do a project on curve fitting, so I fit a function to the curve of the guitar. I got this 7-degree polynomial $$\small\boxed{ f(x)=-0.005632x^7 + 0.08969x^6-0.5346x^5 + 1.364x^4 -0.8671x^3 -2.005x^2 + 3.038x + 0.4182}$$

I now want to find the length of the side of the guitar from the graph. As in calculate the arc length of the function above and multiply by a factor to get the real life value.

This question solved my problems with the arc length and I got $\boxed{6.72692}$ (I am not $100\%$ certain in this result but the answers from several people have pointed to it).

The guitar is 5 units long on the graph and 48.4 cm long in reality, so the factor is $9.68$.

When I multiply the length by the factor I get 65.11. The actual value measured by a tape measure is 72.3 cm. So the value from the graph is smaller than the value in reality ($9.9 \%$ error). I knew that the values would be different since my curve is a good fit, but not perfect, but I expected a smaller error. Am I doing it correctly?

Is the factor used correct?

Does it have something to do with the factor being in the $x$-axis and the curve being in both $x$ and $y$? (The factor for $y$ is also $9.68$)

Answer 1

The method with the scaling factor is correct - the length also scales by that factor - and Wolfram Alpha agrees with that arc length for the polynomial.

Note however that the polynomial hits the y axis at 0.4182, not at the origin. Also, at x=5 the polynomial evaluates to 0.37695.

If you add 0.37695 and 0.4182 to the polynomial's arc length (effectively making the graph drop vertically from the end points to the x-axis) then after scaling you get a total length of 72.8 cm. This is less than 1% off the measured length of 72.3.

Answer 2

I understand that you needed a project for curve fitting. But in a practical sense, you have achieved a curve fit with four significant figures that probably is greater than the initial $(x,y)$ data. And in the final analysis probably wasn't very good after all. So, to the problem at hand, that of determining the length of the guitar. Given the $(x,y)$ data you could very easily just integrate for the length numerically. This would entail much less effort than doing the curve fit.

This can be done very easily in the complex plane since the arc length is given by

$$ s=\int |\dot z| du $$

With a language such as Matlab, which handles complex variables seamlessly, the code would be

z=x+i*y;    
integrand=abs(gradient(z));    
s=trapz(integrand); % trapezoidal integration

Now, some may argue that numerical integration isn't exactly mathematics, but neither is curve fitting, or starting with numerical data, for that matter. Moreover, those results are clearly incorrect. So what's the point of doing it?