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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.

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    $\begingroup$ That watch looks awesome but do you think it will be physically implementable? $\endgroup$ – Mohammad Zuhair Khan Jan 22 at 5:34
  • $\begingroup$ I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird. $\endgroup$ – Aubrey Champagne Jan 22 at 5:37
  • $\begingroup$ Which end of the hand are you talking about? The one that actually shows the hour, or the other end? $\endgroup$ – Arthur Jan 22 at 6:00
  • $\begingroup$ The side of the hand that isn’t attached to the whee $\endgroup$ – Aubrey Champagne Jan 22 at 6:07
  • $\begingroup$ I am also worried about the friction as it may slow down the transition. Use plenty of graphite. $\endgroup$ – Mohammad Zuhair Khan Jan 22 at 6:16
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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 $\left(\frac 1{\sqrt{1+m^2}}(L-\sqrt{2R^2-2R\cos ft}),\frac m{\sqrt{1+m^2}}(L-\sqrt{2R^2-2R\cos ft})\right)$

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  • $\begingroup$ Yep. graphed it out, and that's the path. Thanks! $\endgroup$ – Aubrey Champagne Jan 22 at 19:24
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enter image description here

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.

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