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This is an interesting problem that I thought of myself but I'm racking my brain on it. I recently saw this kinetic energy knick knack in a scene in Iron Man 2: http://www.youtube.com/watch?v=uBxUoxn46A0

And it got me thinking, it looks like either tip of the shorter rod can reach all points within the maximum radius of device.

The axis of either rod is off center, so for simplicity I decided to first simplify the problem by modeling it as having both rods centered. So the radius of the device is $r_1 + r_2$. So I decided to first model the space of points reachable by the tip of the shorter rod as a function of a vector in $\mathbb{R}^2$, consisting of $\theta_1$, the angle of the longer rod, and $\theta_2$, the angle of the shorter rod.

Where I'm getting lost is how to transform this angle vector to its position in coordinate space. How would you describe this mapping? And how would you express the space of all points reachable by the tip of the shorter rod, as the domain encompasses all of $\mathbb{R}^2$?

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You might want to add a better description or a picture. I'm not sure what you mean, and don't really want to watch a video to figure it out... –  copper.hat Jan 9 '13 at 18:47
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@copper.hat: It's essentially a forced double pendulum. If you don't like YouTube, here's a static page with an animation on ThinkGeek. –  Rahul Jan 9 '13 at 18:51
    
@copper.hat Yes, that looks right. But how do you prove that it does or does not encompass all points within the unit circle of radius $r_1 + r_2$? –  Ataraxia Jan 9 '13 at 19:01
    
By the way, the terms "vector space", "isomorph", "vector field" don't mean what you seem to think they mean. What you're looking for is simply a function, or mapping, from the space of angles $(\theta_1,\theta_2) \in \mathbb R^2$ -- or even just $[0,2\pi)\times[0,2\pi)$ -- to the position of the point in $\mathbb R^2$. And in the last sentence you're asking for the range of said function. –  Rahul Jan 9 '13 at 19:10
    
Are you looking for points that are reachable geometrically or physically? Meaning, are you considering a single experiment with specific initial position and velocity, or are you considering some sort of a robotic arm which you can bend in any direction and figure out reachable points? –  Maesumi Jan 9 '13 at 19:24
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2 Answers

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Permit me to quote MathWorld's page on the double pendulum:

enter image description here

The positions of the bobs are given by $$\begin{align} x_1&=l_1\sin\theta_1,\\ y_1&=-l_1\cos\theta_1,\\ x_2&=l_1\sin\theta_1+l_2\sin\theta_2,\\ y_2&=-l_1\cos\theta_1-l_2\cos\theta_2, \end{align}$$

The end of the second rod can't get any closer to the origin than $\lvert l_1-l_2\rvert$, and can't get any farther than $l_1+l_2$. So its range is $$\lvert l_1-l_2\rvert\le x_2^2+y_2^2\le l_1+l_2.$$

To see this, fix $\theta_1$ and observe that as you vary $\theta_2$, the point $(x_2,y_2)$ traces out a circle of radius $l_2$ whose center is at a distance $l_1$ from the origin. This achieves all distances in the range given above, as you can check by drawing a figure (try both cases, $l_1\ge l_2$ and $l_1<l_2$). To reach any other point in that range but not on the circle, just rotate the entire configuration.

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One way to see the reach is to notice that if the configuration $\theta = (\theta_1,\theta_2)$ reaches some spot $x \in \mathbb{R}^2$, and $y$ is a spot obtained by rotating $x$ by $\alpha \in \mathbb{R}$, then the configuration $\theta+(\alpha, \alpha)$ will reach $y$. So, we only need to see what the minimum and maximum radius can be.

For a particular configuration, the radius squared is \begin{eqnarray} (r_1 \cos \theta_1 + r_2 \cos \theta_2)^2 + (r_1 \sin \theta_1 + r_2 \sin \theta_2)^2 &=& r_1^2+r_2^2+ 2r_1r_2 ( \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2)\\ & = & r_1^2+r_2^2+ 2r_1r_2 \cos(\theta_1-\theta_2) \end{eqnarray} Hence the radius lies in $[|r_1-r_2|,|r_1+r_2|]$.

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