How does $s=r\theta$ take into account negative angles or angles greater than $2\pi$

algebra-precalculusangle

Radian is defined as the angle subtended by the arc at the center of the circle.

The formula of radian is $\theta=\dfrac sr$ where $s$ is the arc length and $r$ is the circle's radius.

Since length of arc cannot be more than the circumference, how is a value of radian that is more than $2\pi$ adjusted in the formula?

How about a negative value of radian?

Is a scalar $k$ multiplied in the formula to compensate for negative value and for value more than to $2\pi\;?$ Is this formula correct? $$\theta=k\frac sr$$

Best Answer

You can continue using the $$s=r\theta$$ formula even when the angular displacement $\theta$ rad is negative or exceeds $2\pi$ rad: while the radius $r$ is positive, the signed distance $s$ travelled along the circumference is negative if and only if the travel direction is clockwise.

To elaborate: think of a thread spooling around its reel, where the thread is winding around the reel multiple times; if spooling is considered to be in the positive direction, then unspooling is considered to be in the negative direction.


Addendum (migrating up my comment replies)

  1. $s$ is not displacement.

    Measuring turns

  2. $s$ and $\theta$ are scalars, not vectors in 3D space (though they can be framed as one-dimensional vectors).

    They do not have free direction in 3D space; their direction is conceptual.

    Scalars and Euclidean vectors