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2014-03-14

- Link:
- http://jbuckland.com/pantograph/delta.html
- GitHub:
- http://github.com/ambuc/pantograph

I was inspired by the above video (depicting a set of record players in control of a pantograph) to derive a closed-form solution for the vector position of the pen in terms of the angles of rotation for each of the arms

For our pantograph assembly, we define the centers of two circles $A$ and $B$, each with a radius $r_a$ and $r_b$, respectively. The current positions of nodes $P$ and $Q$ are determined by angles $\theta_a$ and $\theta_b$.

Taking the position of $A$ as the origin, we can find

\[\vec{A} = \Big(0,\ 0\Big)\] \[\vec{B} = \Big(0,\ (r_a + d + r_b)\Big)\] \[\vec{P} = \vec{A} + \Big( r_a\ \angle\ \theta_a \Big)\] \[\vec{Q} = \vec{B} + \Big( r_b\ \angle\ \theta_b \Big)\] \[\vec{M} = \frac{1}{2}\left( \vec{P} + \vec{Q}\right) = \Big( \tfrac{1}{2}(P_x + Q_x), \tfrac{1}{2}(P_y + Q_y) \Big)\]Attached to the points $P$, $Q$ are two rigid bodies of length $\ell_1$, which meet at variable point $R$. By finding the midpoint $M$ between $P$ and $Q$, we can find the right triangle $\triangle PMR$. From the definition of slope, we find $\overline{PQ}$ to be \(m_{\overline{PQ}} = \dfrac{Q_y - P_y}{Q_x - P_x}\)

Because $\overline{MR} \perp \overline{PQ}$, we can find the slope of the line $\overline{MR}$ to be

\[m_{\overline{MR}} = \left(m_{\overline{PQ}}\right)^{-1}\]In addition, we can construct the right triangle $\triangle PMR$ with known side lengths $a$ and $g$, where

\[g = \tfrac{1}{2}|\overline{PQ}| = \tfrac{1}{2}{\sqrt{(Q_x - P_x)^2 + (Q_y - P_y)^2}}\]Thus, $h = \sqrt{(\ell_1)^2 - g^2}$. This allows us to find

\[\vec{R} = \vec{M} + \left( h\ \angle\ \tan^{-1}\left(m_{\overline{MR}}\right) \right)\] \[\vec{R} = \vec{M} + \left( \sqrt{a^2 - g^2}\ \angle\ \tan^{-1}\left( \dfrac{-1}{m_{\overline{PQ}}} \right) \right)\]We can easily find $\varphi_a$, $\varphi_b$ from the slopes of lines $\overline{PR}$ and $\overline{QR}$, respectively.

\[\varphi_a = \tan^{-1}\left( \dfrac{R_y - P_y}{R_x - P_x} \right)\] \[\varphi_b = \tan^{-1}\left( \dfrac{R_y - Q_y}{R_x - Q_x} \right )\]We can then write nodes $S$ and $T$ as \(\vec{S} = \vec{R} + \left( \ell_2\ \angle\ \varphi_b\right)\) \(\vec{T} = \vec{R} + \left( \ell_2\ \angle\ \varphi_a\right)\)

It is trivial to present a construction of triangle $\triangle TUS$ that directly mirrors that of triangle $\triangle QRP$. We find

\[\vec{N} = \frac{1}{2}\left(\vec{S} + \vec{T}\right)\] \[d = \sqrt{(\ell_3)^2 - c^2}\] \[m_{\overline{NU}} = \dfrac{-1}{m_{\overline{ST}}} = -\dfrac{T_x - S_x}{T_y - S_y}\]Thus, we can write node $U$ as \(\vec{U} = \vec{N} + \left(d\ \angle\ \tan^{-1}\left( m_{\overline{NU}} \right)\right)\)

Distressingly, this appears to be a function that cannot be resolved into a one-to-one mapping from the vector space $(\phi_a, \phi_b)$ onto the $(x,y)$ coordinate system of the pen by any conventional means: that is, for each $(x,y)$ on the pad, there are two or three sets of $(\phi_a, \phi_b)$ positions that could achieve that position.

At a loss with the tools of continuous mathematics, I turned to discrete simulations: a set of three processing.js simulations illustrating what turned out to be a very complex space.

- Demo $\alpha$: Mouse-driven geometry demo of pantograph behavior
- Demo $\beta$: Mouse-driven geometry demo of point names
- Demo $\delta$: Sample space of points available from an array of rotational values

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