Circle with rotating line: Locate section on a tangent with known velocity in the section

Question

I have a hard time phrasing this in the title but let me try to explain. You all probably know the demonstration graphics on the unit circle for trigonomic functions (look here for an example). Now I got an interesting task: Imagine a circle with a constantly spinning pointer that follows the tangent (as if a laser pointer were in the middle of the circle, rotating at a constant RPM and shining onto the tangent). Now I have three points with equal distance between each other somewhere on that tangent and I know the time it takes the beam from one point to the other ($(P_1)$ to $(P_2)$ and $(P_2)$ to $(P_3)$). I have to determine the radius of the circle and the offset of the points from the point where the tangent touches the circle. I tried to get the degrees the pointer turned during the measured times but I always get stuck trying to use the RPM afterwards to get from that information to the actual offset.

Answer 1

Assume for the moment that the radius is 1. Letting the angles be labelled as shown, the heights of $P_1, P_2, P_3$ are given by $\tan \theta, \tan(\theta + \alpha), \tan(\theta + \beta)$, respectively. The angles $\alpha, \beta$ are known from your RPM measurements, and we want to find $h = \tan \theta$. Now, the distance from $P_3$ to $P_2$ is the same as from $P_2$ to $P_1$, so

$$\tan(\theta + \beta) - \tan(\theta + \alpha) = \tan(\theta + \alpha) - \tan \theta$$ or $$\tan \theta + \tan(\theta + \beta) - 2\tan(\theta + \alpha) = 0$$ Applying the tangent sum of angles formula,

$$\tan \theta + \frac{\tan \theta + \tan \beta}{1 - \tan\theta\tan\beta}- 2\frac{\tan \theta + \tan \alpha}{1 - \tan\theta\tan\alpha} = 0$$ Substitute $h = \tan\theta, a = \tan\alpha, b = \tan\beta$ and simplify: $$h + \frac{h + b}{1 - hb}- 2\frac{h + a}{1 - ha} = 0\\h(1-ha)(1-hb) + (h + b)(1 - ha) - 2(h + a)(1-hb) = 0\\ abh^3 +(b-2a)h^2 +abh+(b-2a) = 0\\ (abh + b - 2a)(h^2+1) = 0\\ abh + b-2a = 0\\ h = \frac {2a-b}{ab}$$

The common distance from $P_1$ to $P_2$ and from $P_2$ to $P_3$ is $$\tan(\theta + \alpha) - \tan\theta = \frac{h + a}{1 - ha} - h = \frac{a(h^2 + 1)}{1-ha}$$

Now what happens if the radius $R$ is not $1$? Then the distances all scale accordingly:

• The height of $P_1$ above the tangent point is $Rh = R\frac{2a-b}{ab}$
• The common distance between the points is $R\frac{a(h^2 + 1)}{1-ha}$

If you know the common distance, and the angles $\alpha, \beta$, you can calculate $h, a, b$ as before, then divide the common distance by $\frac{a(h^2 + 1)}{1-ha}$ to find $R$.