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I am trying to find out if there is a methodology of finding out the parametric equations of a shape from a picture. To be more specific I am trying to find a way to find x=f(t) and y=f(t) for the picture: Cross section of a tunnel

Is there any way I can do this by hand? I am not interested in doing this by using a software. It is about a coursework and I want to find a way to explore this.

I have only thought of adding this picture to Geogebra on a cartesian plane and then try to figure out if the is a function of time for x coordinate and for y coordinate. But I have not done this before and I don't know how a shape is translated into parametric equations. I didn't manage to find somethinf in textbooks or google.

Have I thought correctly or I am in a totally wrong path? Any help of which steps should I do in order to work out the equations would be very helpful.

Thanks! :)

CiaPan
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kostaus
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1 Answers1

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First of all if you want to do it by hand you need to make a few assumptions regarding the shape - say the top dome is a semicircle of radius $r_1$ and the bottom arc is part of a circle with radius $r_2$

Now, assume at $t=0$ you are at the left most point of the semicircle. Now, you can define traversing the semicircle as

Assume we traverse the path with a constant speed $v$

$$x(t) = r_1(1 - \cos(\omega_1 t))$$

$$y(t) = r_1\sin(\omega_1 t)$$

With limits on $t$ being $t\in \left[0, \frac{\pi}{\omega_1}\right]$

Here, $\omega_1 = \frac{v}{r_1}$

Now for the straight line portions, let their height be $h$. Then we have

$$x(t) = 2r_1$$

$$y(t) = -v\left(t-\frac{\pi}{\omega_1}\right)$$

With limits on $t$ being $t \in \left[\frac{\pi}{\omega_1},\frac{\pi}{\omega_1} + \frac{h}{v} \right]$

You get the drift, right?

Dhanvi Sreenivasan
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