Tracking the movement of the watch hand

Question

I love watches, and I had an idea for a weird kind of watch movement (all of the stuff that moves the hands). It is made up of a a central wheel, with one of the hands connected to it (in this case, it will be the hour hand). This hand goes through a pivot, and then displays the time. I attached a video of a 3d mock up here, because it is kinda hard to explain. My question is, is there any functions that would be able to graph the movement of the end of the hand? I don't want to make the real prototype just yet.

Answer 1

I will take the origin to be the place the hand slides through, $y$ vertical positive up, $x$ horizontal positive right. Let the hand have length $L$ and the circle radius $R$. It appears $L$ is a little greater than $2R$, so it sticks out of the pivot even when the left end is at the farthest left point.

The position of the left end is $(R\cos ft-R,R\sin ft)$ where $f=\frac {2 \pi}{ 12 hours}$
The distance from the left end to the pivot is $\sqrt{(R\cos ft-R)^2+(R\sin ft)^2}=\sqrt{2R^2-2R\cos ft}$
The slope of the hand is $\frac {R \sin ft}{R\cos ft-R}=m$
The length of the hand to the right of the pivot is $L-\sqrt{2R^2-2R\cos ft}$
The position of the right end of the hand is $\left(\frac 1{\sqrt{1+m^2}}(L-\sqrt{2R^2-2R\cos ft}),\frac m{\sqrt{1+m^2}}(L-\sqrt{2R^2-2R\cos ft})\right)$

Answer 2

Denote with $l$ the length of the hand and with $R$ the radius of the circle.

Parametric coordinates of point $M$ (as a function of $\alpha\in[0,2\pi)$) are:

$$x_M=(l-2R\cos\frac\alpha2)\cos\frac\alpha2=l\cos\frac\alpha2-R(1+\cos\alpha)$$

$$y_M=-(l-2R\cos\frac\alpha2)\sin\frac\alpha2=-l\sin\frac\alpha2+R\sin\alpha$$

As an exercise you can eliminate angle $\alpha$ and obtain an implicit relation between coordinates of point $M$, but there is not much that you can do with it. It is better to work with parametric equations. Select $l,R$ and calculate coordinates for a range of $\alpha$ angles.