Peaucellier–Lipkin linkage

The Peaucellier–Lipkin linkage (or Peaucellier–Lipkin cell, or Peaucellier–Lipkin inversor), invented in 1864, was the first true planar straight line mechanism – the first planar linkage capable of transforming rotary motion into perfect straight-line motion, and vice versa. It is named after Charles-Nicolas Peaucellier (1832–1913), a French army officer, and Yom Tov Lipman Lipkin (1846–1876), a Lithuanian Jew and son of the famed Rabbi Israel Salanter.

Until this invention, no planar method existed of converting exact straight-line motion to circular motion, without reference guideways. In 1864, all power came from steam engines, which had a piston moving in a straight-line up and down a cylinder. This piston needed to keep a good seal with the cylinder in order to retain the driving medium, and not lose energy efficiency due to leaks. The piston does this by remaining perpendicular to the axis of the cylinder, retaining its straight-line motion. Converting the straight-line motion of the piston into circular motion was of critical importance. Most, if not all, applications of these steam engines, were rotary.

The mathematics of the Peaucellier–Lipkin linkage is directly related to the inversion of a circle.

Earlier Sarrus linkage
There is an earlier straight-line mechanism, whose history is not well known, called the Sarrus linkage. This linkage predates the Peaucellier–Lipkin linkage by 11 years and consists of a series of hinged rectangular plates, two of which remain parallel but can be moved normally to each other. Sarrus' linkage is of a three-dimensional class sometimes known as a space crank, unlike the Peaucellier–Lipkin linkage which is a planar mechanism.

Geometry
In the geometric diagram of the apparatus, six bars of fixed length can be seen: $\overline{OA}$, $\overline{OC}$, $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DA}$. The length of $\overline{OA}$ is equal to the length of $\overline{OC}$, and the lengths of $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ are all equal forming a rhombus. Also, point $O$ is fixed. Then, if point $B$ is constrained to move along a circle (for example, by attaching it to a bar with a length halfway between $O$ and $B$; path shown in red) which passes through $O$, then point $D$ will necessarily have to move along a straight line (shown in blue). In contrast, if point $B$ were constrained to move along a line (not passing through $O$), then point $D$ would necessarily have to move along a circle (passing through $O$).

Collinearity
First, it must be proven that points $O$, $B$, $D$ are collinear. This may be easily seen by observing that the linkage is mirror-symmetric about line $OD$, so point $B$ must fall on that line.

More formally, triangles $△BAD$ and $△BCD$ are congruent because side $\overline{BD}$ is congruent to itself, side $\overline{BA}$ is congruent to side $\overline{BC}$, and side $\overline{AD}$  is congruent to side $\overline{CD}$. Therefore, angles $∠ABD$ and $∠CBD$ are equal.

Next, triangles $△OBA$ and $△OBC$ are congruent, since sides $\overline{OA}$ and $\overline{OC}$ are congruent, side $\overline{OB}$ is congruent to itself, and sides $\overline{BA}$  and $\overline{BC}$  are congruent. Therefore, angles $∠OBA$ and $∠OBC$ are equal.

Finally, because they form a complete circle, we have
 * $$ \angle OBA + \angle ABD + \angle DBC + \angle CBO = 360^\circ$$

but, due to the congruences, $∠OBA = ∠OBC$ and $∠DBA = ∠DBC$, thus
 * $$\begin{align}

& 2 \times \angle OBA + 2 \times \angle DBA = 360^\circ \\ & \angle OBA + \angle DBA = 180^\circ \end{align}$$

therefore points $O$, $B$, and $D$ are collinear.

Inverse points
Let point $P$ be the intersection of lines $AC$ and $BD$. Then, since $ABCD$ is a rhombus, $P$ is the midpoint of both line segments $\overline{BD}$ and $\overline{AC}$. Therefore, length $\overline{BP}$ = length $\overline{PD}$.

Triangle $△BPA$ is congruent to triangle $△DPA$, because side $\overline{BP}$ is congruent to side $\overline{DP}$, side $\overline{AP}$ is congruent to itself, and side $\overline{AB}$ is congruent to side $\overline{AD}$. Therefore, angle $∠BPA$ = angle $∠DPA$. But since $∠BPA + ∠DPA = 180°$, then $2 × ∠BPA = 180°$, $∠BPA = 90°$, and $∠DPA = 90°$.

Let:
 * $$\begin{align}

& x = \ell_{BP} = \ell_{PD} \\ & y = \ell_{OB} \\ & h = \ell_{AP} \end{align}$$

Then:
 * $$\ell_{OB}\cdot \ell_{OD}=y(y+2x)=y^2+2xy $$
 * $${\ell_{OA}}^2 = (y + x)^2 + h^2$$ (due to the Pythagorean theorem)
 * $${\ell_{OA}}^2 = y^2 + 2xy + x^2 + h^2$$ (same expression expanded)
 * $${\ell_{AD}}^2 = x^2 + h^2$$ (Pythagorean theorem)
 * $${\ell_{OA}}^2 - {\ell_{AD}}^2 = y^2 + 2xy = \ell_{OB} \cdot \ell_{OD}$$

Since $\overline{OA}$ and $\overline{AD}$ are both fixed lengths, then the product of $\overline{OB}$  and $\overline{OD}$ is a constant:
 * $$\ell_{OB}\cdot \ell_{OD} = k^2 $$

and since points $O$, $B$, $D$ are collinear, then $D$ is the inverse of $B$ with respect to the circle $(O,k)$ with center $O$ and radius $k$.

Inversive geometry
Thus, by the properties of inversive geometry, since the figure traced by point $D$ is the inverse of the figure traced by point $B$, if $B$ traces a circle passing through the center of inversion $O$, then $D$ is constrained to trace a straight line. But if $B$ traces a straight line not passing through $O$, then $D$ must trace an arc of a circle passing through $O$. Q.E.D.

A typical driver
Peaucellier–Lipkin linkages (PLLs) may have several inversions. A typical example is shown in the opposite figure, in which a rocker-slider four-bar serves as the input driver. To be precise, the slider acts as the input, which in turn drives the right grounded link of the PLL, thus driving the entire PLL.

Historical notes
Sylvester (Collected Works, Vol. 3, Paper 2) writes that when he showed a model to Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”

Cultural references
A monumental-scale sculpture implementing the linkage in illuminated struts is on permanent exhibition in Eindhoven, Netherlands. The artwork measures 22 x, weighs 6600 kg, and can be operated from a control panel accessible to the general public.