In this post I will say something about an interesting mechanic, a linkage.

I will not define what a linkage is. We can consult this page of wikipedia on linkage. From mathematics’ point of view, a linkage is, roughly speaking, a finite set of segments of finite length each as a subset of $\mathbb{R}^3$(most of the time, they are straight segments) such that their union is a connected subspace of $\mathbb{R}^3$. What this mathematical definition does not say is that, these segments can move(rigid motion in Euclidean space), and their intersection points(if they have any) do not change their position on each segment to which the intersection point belongs. These two last points are the essential characteristics of a linkage.

Perhaps the simplest example of a linkage is just one straight segment with one end fixed. If the other end move in a plane, then its locus will be a circle, as the following graph shows(the straight segment is $l$ and the point end fixed is $A$):

A circle

Another famous example is the lever:

A lever

Of course we can again use a lever to draw circles, yet the most important use of levers do not perhaps lie here, we use it to carry heavy things. Of course the mechanism of these two uses are the same, that is, convert a rotary move into another rotary move yet their radii may be different. Using the law of conservation of energy, we can use a little force to move a heavy stone.

So, in some sense, the abstraction of linkages in mechanics is to create a linkage such that it can convert a particular type of movement into another type. Rotary motion into rotary motion, straight-line motion into straight line motion are easy. So the next problem is to convert a rotary motion into a straight-line motion, and vice versa.

The first such linkage was perhaps due to Pierre Frédéric Sarrus, a French mathematician. The linkage is called a Sarrus linkage. This linkage is not well known(on youtube, I find some demonstrations like this and this), mainly because of another much simpler linkage, yet with wider applications. That is Peaucellier linkage. It was also invented by a French(together with a Lithuanian Jews).

Note that Sarrus linkage do not use any guideways to convert rotary motion into straight-line motion. The main reason for this is that there are two hinges not parallel, as the following graph shows:

Note that the two planes $L_1,L_2$(with red segments) can simply be replaced by two segments $b_1,b_2$. The hinge $h_1$ forces the motion of $L_2$ to be in a plane perpendicular to $h_1$, similar for $h_2$, thus the motion of $L_2$ has to be the intersection of these two planes, that is a straight line since $h_1,h_2$ are not parallel. Note that this linkage is inevitably $3$-dimensional, since the mechanism of it is so.

As to Peaucellier linkage, it is a plane linkage, which means that all its segments are in one plane, that is:

Here the segments $OA=OC$(having the same length) and $AB=BC=CD=DA$. The point of this linkage is that, if the point $B$ moves in a circle passing through $O$, then the point $D$ will move on a straight line. If $B$ moves on a straight line not passing through $O$, then $D$ will move on a circle passing through $O$.

This linkage is a beautiful demonstration of the inversion of circles. In some sense, this is just elementary geometry. Look at the following graph:

Here is a simple proof. Suppose that $B$ moves on a circle $l_1$, with center $R$, then by symmetry it is clear to see that $O,B,D$ are co-line. Suppose that $AC\bigcap BD=E$, then since $ABCD$ is a rhombus, thus $AC\perp BD$. So, using Pythagorean theorem, we see that $OB*OD=OE^2-BE^2=OE^2-(AB^2-AE^2)=OA^2-AB^2$ which is a constant. Suppose that the diameter passing through $O$ of $l_1$ is $T$(which is not in the graph) and its intersection with a straight line $l_2$ passing through $D$ is $S$. Now consider the triangles $\triangle OBT\simeq \triangle OSD$. Then we have that $OT*OS=OB*OD$. Since $OT$ is the diameter, a constant, thus $OS$ is also a constant. This means that the projection of $D$ onto the diameter $OT$ is always the same point $S$, thus showing the $D$ always moves on the line $l_2$. For the second part, it is totally similar. In the wikipedia’s article, there is a vivid demonstration of this linkage, I copy it here:

We have said that Peaucellier linkage is a simple application of the inversion of circles. Why? In fact, we have shown that $OB*OD=OA^2-AB^2$, we can set $r^2=OA^2-AB^2$ with $r>0$. Then we can see easily that $B,D$ are inversion points one to the other about the circle, centered at $O$, of radius $r$. Then why the locus of $D$ should be a straight segment when $B$ moves on a circle passing through $O$? This is a very good question, and the answer to this point uses the essential characteristic of circle inversion: it is a conformal mapping. This means that circle inversion conserves angles, sending an angle $\pi/3$ to an angle $\pi/3$, etc. Note that, since the above inversion is about the circle centered at $O$, so this inversion maps $O$ to infinity $\infty$. Since a circle passing through $O$ is sent to a circle passing through $\infty$, so this image must be a straight line(any circle will not pass through a point at infinity), so this explains why the locus of $D$ is a straight segment. And also we require that the locus of $B$ must pass through $O$. This is shown in the following graph:
As for the second case, where $D$ moves on a circle passing through $O$, the mechanism is similar to the previous one, as the following graph shows:
Note that in this case, the straight line $l_1$ of the locus of $B$ must pass through the circle $l_2$ of the locus of $D$.