Peaucellier–Lipkin Linkage

Consider the device shown in the animation below. It consists entirely of rigid rods joined at pivot points.

It is called the Peaucellier–Lipkin linkage, named after Charles-Nicolas Peaucellier, who invented it in 1864, and Yom Tov Lipman Lipkin, who discovered it independently in 1871.

It has the special property of transforming rotary motion into straight-line motion, and vice versa. This post examines why this works.

Let us first introduce some parameters. Let the two long black rods have length $L$. They are joined at the anchor point $O$. Each rod of the blue rhombus has length $s$. The green rod is anchored at $C$, at distance $r$ from $O$, and has length $r$ as well.

We start by considering the long rods and the rhombus by themselves. We will require that $s < L$, which makes the rhombus fit inside the two long rods.

The long rods join $O$ to the points $A$ and $B$, and the rhombus is $APBP'$. The midpoint of the rhombus is denoted $M$. The diagonals $AB$ and $PP'$ of the rhombus are perpendicular and bisect each other at $M$. Since $O$ is equally far from $A$ and $B$, it lies on the perpendicular bisector of $AB$, which is the line through $P$ and $P'$.

Using the Pythagorean Theorem on the right-angled triangles $OAM$ and $APM$, we get $L^2 = \lvert OM \rvert^2 + \lvert AM \rvert^2$ and $s^2 = \lvert PM \rvert^2 + \lvert AM \rvert^2$. Then we have

$$L^2 - s^2 = \lvert OM \rvert^2 - \lvert PM \rvert^2 = (\lvert OM \rvert - \lvert PM \rvert) (\lvert OM \rvert + \lvert PM \rvert) = \lvert OP \rvert \lvert OP' \rvert.$$

Since $\lvert OM \rvert^2 - \lvert PM \rvert^2 = L^2 - s^2 > 0$, we have $\lvert OM \rvert > \lvert PM \rvert$, so $O$ lies outside the segment $PP'$ and $P$ and $P'$ are on the same ray from $O$.

The two factors $\lvert OM \rvert - \lvert PM \rvert$ and $\lvert OM \rvert + \lvert PM \rvert$ are therefore the distances from $O$ to the closer and the farther of the points $P$ and $P'$. The product does not depend on which is which, so the relation holds whether $P$ is the closer or the farther of the two. This means that the points $P$ and $P'$ are each other's (circle) inverses.

Now for the green rod. It joins $P$ to the anchor point $C$, and since $\lvert OC \rvert = \lvert CP \rvert = r$, the point $P$ is constrained to move on the circle with center $C$ and radius $r$, which passes through $O$. Let $Q$ be the point on this circle opposite $O$, so that $C$ is halfway between $O$ and $Q$, and let $Q'$ be the inverse of $Q$. Now let $P$ be anywhere else on the circle, off the $OQ$ line.

We have $\lvert OQ \rvert \lvert OQ' \rvert = \lvert OP \rvert \lvert OP' \rvert = L^2 - s^2$ and thus

$$\frac{\lvert OP \rvert}{\lvert OQ' \rvert} = \frac{\lvert OQ \rvert}{\lvert OP' \rvert}.$$

Since $Q'$ lies on the ray $OQ$ and $P'$ lies on the ray $OP$, the triangles $OPQ$ and $OQ'P'$ share the angle $\angle QOP = \angle Q'OP'$. This proportionality and the shared angle imply that the triangles $OPQ$ and $OQ'P'$ are similar.

Now for the final part: Since $P$ is a point on the circle where $OQ$ is a diameter, the angle $\angle OPQ$ is a right angle by Thales's Theorem. From the similarity we then have that $\angle OQ'P'$ is a right angle. And since $P$ was chosen arbitrarily (off the $OQ$ line), this will be true for any position of $P$ on the circle.

So $P'$ always lies on the line through $Q'$ perpendicular to $OQ'$. In other words, as $P$ moves on its circle, $P'$ moves on a straight line, at distance $\lvert OQ' \rvert = (L^2 - s^2)/(2r)$ from $O$.

You can also explore the linkage interactively in the lab.