Find the distance through a pipe wall

Question

I am looking to solve y in this question. My problem is that I have a pipe of known wall thickness (t), known internal radius (r) and want to find the distance through the pipe wall from any point on the internal surface of the pipe to a point external to the pipe (point B). Point B is a known distance from the closest internal surface of the pipe (distance a). I want to find out the distance through the pipe wall from a point on the internal surface, at different angles. (i.e. at θ = 0 degrees (x=a, y=t) through to θ = 180 degrees (x=a+2r, y=t+2r). How can I find the distances through this layer for each of the angles between 0 and 180 degrees?I have drawn a picture to show what I am after Does anybody know what formula I can use?

Answer 1

Apply the cosine law(The Law Of Cosines) to the triangle consisting of $B$, the center of the pipe, and the other point whose distance from $B$ is to be found.

This gives that $x^2 = r^2 + (r+a)^2 - 2r(r+a)\cos \theta$. Hence, $x$ can be found using this formula.

To find $y$, let us call the angle between $r$ and $x$ as $\psi$. Then, we have by The Sine Rule, that $\frac{x}{\sin \theta} = \frac{r+a}{\sin \psi}$.

Hence, we know $\sin \psi$, hence $\cos \psi = \sqrt{1- \sin^2 \psi}$, taking positive sign if $\psi< 90^\circ$ and negative otherwise. Now, join the center of the pipe to the point where $y$ ends. This is a line of length $r+t$. To this triangle having side lengths $r,y,r+t$, we will apply the cosine rule again, noting that $\psi$ is opposite the side with length $r+t$: $$ (r+t)^2 = r^2+y^2-2yr \cos \psi \implies 2rt + t^2 = y^2 -2yr \cos \psi $$

This is a quadratic equation in $y$, and can be solved: $$ y = r\cos \psi + \sqrt{r^2 \cos^2 \psi +2 r t+t^2} $$ (The negative is rejected, since the square root is greater than the term outside).

Hence, $y$ can also be solved for.