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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.

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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 $$x=\frac {2\theta}{\pi}\sin (\theta)-1$$

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