I am working with the geometrical situation seen in the attached hand-drawn graphic below.

The given constants are $h, w, \phi, \text{ and }R$. My aim is to find $x$ as a function of those. The general constructive idea, is that the "leg" (hypotenuse) has an angle $\phi$ with vertical. This leg extends to the point where it must intersect with the circle circumference below (the constraint). This circle has radius $R$. The triangle marked in red is isosceles - perhaps worth noting.

Ofcourse we have: $x = R (\sin \frac{\theta_1}{2}-\sin \frac{\theta_0}{2})$, where $\sin \frac{\theta_0}{2} = (w/2 + b)/2R = (w/2 + h\tan \phi)/2R$.

But how to find $\theta_1$ - and from that $x$? The solution should be valid for all values of $\phi$ from $0$ to the limit at which the aforementioned leg becomes tangent to the circle below. By the way, what is that limit?

I have tried to solve this using the laws of sines and cosines. But it always seem that I end up with a triangle-situation with too little information - or it simply becomes too complicated. I have postponed working on it until I receive some feedback. It is likely trivial but nonetheless too time-consuming for me to solve right now :)

Geometrical situation


An elaboration of the accepted answer by André Nicolas:

Using the mechanical coordinate method this suddenly get much easier. And that was all I needed. Great. I will just write the algebraic solutions here.

We now know the following ($R$ is now $r$):
$x = \frac{1}{2} {\sin}^2 \phi \left(2 (h+r) \cot \phi + w {\cot}^2 \phi - \sqrt{(4 r^2 - w^2) {\cot}^2 \phi -4h(h + 2r) - 4(h+r) w \cot \phi} \right)$,
$y = h+r + \frac{1}{2} (w-2 x) \cot \phi$,
$z = x - \frac{1}{2} w - h \tan \phi$.

The angle $\theta_1$ can now be solved directly:
$\sin {\frac{\theta_1}{2}} = \frac{1}{r}\left( \frac{1}{4}w - \frac{1}{2} h \tan \phi - z \right)$


1 Answer 1


The question yields to mechanical coordinate methods. Put the origin at the centre of the circle, and the axes in the usual places. To save some possible confusion, what you call $x$ will be called $z$ in what follows.

Look at the slanty blue (?) line on the right. We find its equation. This line goes through the point $(w/2,R+h)$ and has slope $-\cot \phi$. So we know its equation, say $y=cx+d$. Substitute in $x^2+y^2=R^2$ to find the $x$-coordinates of the points of intersection with the circle. We can easily identify the relevant (smaller) one. So now we know $w/2+b+z$.

To find $w/2+b$, find the $x$-coordinate of the point of intersection of the lines $y=cx+d$ and $y=R$. (A simpler way to find $w/2+b$ is given in your post. Just wanted to stress that we can fully give ourselves over to the method of coordinates.)

Now we know $w/2+b+z$ and $w/2+b$, so we know $z$. If we need to, we can now find an explicit expression for $\theta_1$.

  • $\begingroup$ Thanks. All I needed. It seems to work - but would you be kind to check my added notes in the question above? There ofcourse the problem when $phi$ goes to zero. How could I deal with that? Perhaps by using another more general line description instead of slope intercept? $\endgroup$ May 19, 2012 at 9:01
  • $\begingroup$ Could we compare notes? I used $r=10$, $h=3$, $w=12$, $\phi = \arctan (1/3)$. With this values I get $x=\frac{1}{10} \left(93-\sqrt{39}\right) = 8.6755$, $y = \frac{1}{10} \left(31+3 \sqrt{39}\right) = 4.9735$, and $\sin (\theta_1/2) = 1/100 \left(2 + \sqrt{39}\right) = 0.08245$. According to Mathematica. $\endgroup$ May 19, 2012 at 11:25
  • $\begingroup$ @OleThomsenBuus: The procedure does not work when $\phi=0$, since $\cot 0$ is undefined. We can, in line with your suggestion, use the equation $(y-w/2)\sin\phi+(x-R-h)\cos\phi=0$ for the line. It is simplest, however, if you are writing a general program, to separate out the case $\phi=0$, where we can write down the answer without calculating. You will also need to deal with the fact that for any $h$, after a while you will have tangency, and then no intersection if $\phi$ is large enough. By the way, things are nicer if measure everything in $R$-units. $\endgroup$ May 19, 2012 at 13:40

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