Found this animation circulating online, and was wondering what shape the rod's end traces out. It seems to be an ellipse, but can that be proved somehow?
$\quad\quad\quad\quad\quad\quad\quad\quad\ $
Asked
Active
Viewed 93 times
5
-
I would start with "The sum of the distances from the foci to any point on the ellipse is a constant." and then go the trig route. – shade4159 Nov 06 '13 at 20:51
-
Out of curiosity, where did you find this? – Keshav Srinivasan Nov 06 '13 at 21:01
-
1@KeshavSrinivasan This is one source. Nice! – Řídící Nov 06 '13 at 21:05
1 Answers
6
Call the two pivots moving on a cross $a$ and $b$. $a$ moves on $\{(0,t)|t\in[0,1]\}$, and $b$ on $\{(s,0)|t\in[0,1]\}$. Their position must furthermore satisfy $$s^2+t^2=1$$ The drawing point is positioned at $$a(t) + (1+l)(b(s)-a(t)) =$$ $$= ((1+l)\sqrt{1-t^2},t+(1+l)t)$$ By parametrizing $t = \cos(\phi)$ we obtain the position $$((1+l)\sin(\phi),(2+l)\cos(\phi))$$ which is indeed an ellipse when we let $\phi$ run from $0$ to $2\pi$.
Sorry if my solution is a bit messy, but I worked it up while writing it.
Daniel Robert-Nicoud
- 29,795