# Traveling across an arc?

So, I have an arc that's part of a circle that I must "travel across".

The circle has a radius of $15$ - a circumference of $94.248$ (rounding). The arc length in question is equal to $15.708$.

Now, normally, you could sort of walk along the arc, turning with it. But, if I could only travel straight, I need to know how I would get from one part of the arc to $15.708$ units later in the arc of the circle (only being able to travel straight, or make $90$ degree turns).

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Very related. – J. M. Dec 6 '11 at 18:48

For a thousand years, the only extensive trigonometric table was about precisely this question: see Ptolemy's table of chords.

If the central angle is $\theta$ and the diameter is $d$, then the length of the chord is $d\sin(\theta/2)$.

Observe that $\dfrac{15.708}{94.248}=\dfrac 16$, so the central angle is $60^\circ$. For a $60^\circ$ angle, the length of the chord is exactly the radius.

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Pictures should be drawn to match my verbal descriptions.

The situation is particularly simple with your numbers. The circumference $c$ of your circle is $(2\pi)(15)$. If you calculate $\frac{15.708}{c}$, you will get about $0.1666671$. With your slightly rounded circumference, the ratio is $0.1666667$. To the limit of accuracy of our data, the arc is therefore one-sixth of the full circumference.

So if $O$ is the centre of the circle, $A$ the start point of your travels, and $B$ the end point, then $\triangle OAB$ is equilateral. Thus if you can draw the line joining $A$ to the center of the circle, you want $\angle BAO$ to be $60^\circ$. Such an angle is easy to draw with straightedge and compass.

Once $AB$ has been drawn, you can go directly from $A$ to $B$, and your path will be the shortest possible.

A less efficient thing to do is to travel from $A$ along the line $AO$, until you have walked a distance of $7.5$ (half the radius). Then you turn $90^\circ$, and walk steadily towards the circle. The right bisector of $OA$ will meet the circle at the right points.

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