Walking the circle in steps

Question

We're walking on a circular path by making a step forward and turning right a constant angle each time.

If we shorten our step n times, how should the angle change, such that circular path we're walking remains the same?

Answer 1

Let $R$ be the radius of circle, $L$ be the length of step, and $\alpha$ be measure of the correspondent central angle. You can easily see that $\alpha$ is the turning angle.

Note that $L = 2R \sin\dfrac{\alpha}{2}$.

Now, we want to reduce $L$ by $n$ times, $R$ remains unchanged.

Therefore,

$\dfrac{L}{n} = 2R \sin\dfrac{\beta}{2}$, where $\beta$ is a new turning angle.

So, we have $n \cdot \sin\dfrac{\beta}{2} = \sin\dfrac{\alpha}{2}$.

After all, $\beta = 2\sin^{-1}\left(\dfrac{\sin\frac{\alpha}{2}}{n}\right)$.

Answer 2

Well i guess you would be walking in a regular cyclic polygon. So if the number of sides(steps) is n, the angle will be $\frac{n-2}{n}\cdot\theta$ You will be able to calculate the change yourself.

Answer 3

I will assume the start and end points of each step lie on a fixed circle which has radius $R$. Then if each step has length $l$ and the angle turned between one step and the next is $\theta$, then $\theta$ is also the angle of the arc between the start and end points of each step. So

$l = 2 R \sin \left( \frac {\theta}{2} \right)$

If our step length changes to $l'$ then the angle turned between one step with length $l'$ and the next will now be

$\theta' = 2\sin^{-1} \left( \frac{l'}{2R} \right)$

At the point where we change our step length from $l$ to $l'$ then the angle between the last step with length $l$ and the first step with length $l'$ will be

$\frac{\theta + \theta'}{2}$