Length of a curve without function?

Question

I need to find the length of this curve:

I don't have a function but I do have 3 sets of coordinates:

$(0, 51)$, $(337, 674)$, and $(1022, 1022)$

The $(337, 674)$ set refers to the tangent and it makes the curve be equidistant.

I know nothing beyond basic math, however I'm working on this tablet pressure sensitivity thing and there's no other way out. I have to find the length of the curve or figure out the function which creates that curve.

Thank you fellas!

Answer 1

I'm not sure exactly what you mean when you claim not to know anything beyond "basic math," but staying on the safe side, I think your best bet is to follow copper.hat's suggestion and just use a string. It looks like your window is already square, so just measure across the square, which will give you a conversion factor between string length (in cm or whatever other physical units you like) and the abstract measure of distance employed in the diagram (you already know that the square measures 1022 across). Then, overlay a string onto the curve, cut it or mark it up, and determine its length in physical units. Then, convert that with the factor you determined earlier.

Answer 2

Fits a quadratic function $f(x) = ax^2 + bx + c$ as follow: $$ \begin{cases} y_1 = f(x_1) = ax_{1}^2 + bx_1 + c\\ y_2 = f(x_2) = ax_{2}^2 + bx_2 + c\\ y_3 = f(x_3) = ax_{3}^2 + bx_3 + c\\ \end{cases} $$ Solve the linear system to find $a$, $b$ e $c$. Then use the formula $$ L = \int_{x_1}^{x_3}\sqrt{1 + f^{\prime}(x)^2} \ dx $$ Note that, $f^{\prime}(x)^2 = (2ax + b)^2$. Thus, $$ L = \int_{x_1}^{x_3}\sqrt{(2ax + b)^2 + 1} \ dx $$ Write $2ax + b = \tan \theta$ and make the substitution of variables.

Answer 3

Use a Riemann sum for the arc length integral, making estimates of the slope at test points.

Sorry if you are unfamiliar with that vocabulary, but you say "I can't believe Math hasn't come up with a way to solve my issue", and in fact it has, and it uses this vocabulary.

Answer 4

If you know (or can make good estimates for) other points on the curve, you can make decent approximations using triangles.

E.g. if you divide it up into 1, 2, or 4 triangles (you can use however many you want; more triangles is more accurate), it might look like this:

!!

To find the length of the curve, just add the lengths of the hypotenuses of the triangles. For each triangle, just take the $x$ length and $y$ length and plug them into $L = \sqrt{x^2 + y^2}$.