bars of identical colour are of equal length

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.[1][2]

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.

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

Geometric diagram of a Peaucellier linkage

In the geometric diagram of the apparatus, six bars of fixed length can be seen: OA, OC, AB, BC, CD, DA. The length of OA is equal to the length of OC, and the lengths of AB, BC, CD, and 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 half way 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). On the other hand, 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).

## Mathematical proof of concept

### 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 BD is congruent to itself, side BA is congruent to side BC, and side AD is congruent to side CD. Therefore, angles ABD and CBD are equal.

Next, triangles OBA and OBC are congruent, since sides OA and OC are congruent, side OB is congruent to itself, and sides BA and BC are congruent. Therefore, angles OBA and OBC are equal.

Finally, because they form a complete circle, we have

∠OBA + ∠ABD + ∠DBC + ∠CBO = 360°

but, due to the congruences, angle OBA = angle OBC and angle DBA = angle DBC, thus

2 × ∠OBA + 2 × ∠DBA = 360°
∠OBA + ∠DBA = 180°

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 BD and AC. Therefore, length BP = length PD.

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

Let:

${\displaystyle x=\ell _{BP}=\ell _{PD}}$
${\displaystyle y=\ell _{OB}}$
${\displaystyle h=\ell _{AP}}$

Then:

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

Since OA and AD are both fixed lengths, then the product of OB and OD is a constant:

${\displaystyle \ell _{OB}\cdot \ell _{OD}=k^{2}}$

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

Slider-rocker four-bar acts as the driver of the Peaucellier–Lipkin linkage

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 by 15 by 16 metres (72 ft × 49 ft × 52 ft), weighs 6,600 kilograms (14,600 lb), and can be operated from a control panel accessible to the general public.[3]