I am trying to make two arcs of the same length such that they will both fit in a right triangle like so:

two arcs in a right triangle

I am given the first radius (r1) and a constant k that is the difference between the leg opposite angle one and the second radius (r2)

I know that the angle of the arcs have to add up to Pi/2. I also know that this becomes impossible at some small value of k - for instance, if k was zero there is no way the two arcs could be of equal length.

If I knew the angle, I could find the second radius, or vice-versa, using trigonometric functions.

In the case where both the angles were Pi/4, I know that k would be r*(sqrt(2) - 1).

In case anyone wants to know why I'm doing this, I'm trying to show a circle that is broken open at a point and the broken ends curve out until they are parallel and a certain distance apart. That distance will be 2*k. I want to make sure that the curved-outward parts are the same length as they would be if they were still completing the circle.


just hint

The same length means that

$$r_1\theta =r_2 (\frac {\pi}{2}-\theta) $$ or $$\theta=\frac {\pi r_2}{2 (r_1+r_2)} $$

in the triangle,

$$\sin (\theta)=\frac {k+r_2}{r_1+r_2} $$

if we put $k=xr_2$ then $$x=\frac {2\theta}{\pi}\sin (\theta)-1$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.