This lesson has several references to inversion geometry. If you are unfamiliar with the subject, you may wish to refer to this guide: Inversion Geometry

This is a geometry challenge that might seem to be simplistic: We want a mechanism that will draw a straight line segment. Your first impulse is to draw it with a straightedge, but that would not be fair. Somebody else had to make the straightedge straight. When you think about it, circles are easy. A compass can be made from any rigid object. Attach a pin to one end of a canoe paddle, and attach a pencil to the other. The paddle will draw a perfect circle. But using a straightedge requires a leap of faith. We cannot be certain that it is straight.

In 1864, the problem was solved by Captain Charles-Nicolas Peucellier, a French artillery officer. Several other linkages followed. Some were simpler, but they
all seem to be variations of the Peucellier linkage. In 1877, they were
summarized in *How to Draw a Straight Line: A Lecture on Linkages*,
by Alfred Bray Kempe. The book was reprinted by the National Council of
Teachers of Mathematics in 1977. All of the linkages presented here are
from that book.

Peucellier approached the problem with inversion geometry. He knew that the inverse of a circle through the center of inversion is a line. It is easy enough to make a point follow a circle, so he only needed to design a linkage that would invert that point. The image below is called the *Peucellier cell*. All of the line segments represent rigid links, like colors are congruent, and the intersection points are hinges. It will be shown that if point *A* is fixed and point *P* moves freely (as far as the links will allow it), then point *Q* is the inverse of point *P*. The center of inversion is *A*. The power of inversion, *r*^{2}, will be derived in the proof.

For the purpose of this proof, let it be given that *AB* > *PB*. That is to say that the blue links are longer than the green links. It must be shown that if *P* moves while *A* is fixed, the
product (*AP*)(*AQ*) is equal to some constant *r ^{2}*. It is clear from the symmetry
that points

The values *AB* and *PB* are measures of rigid links, and
therefore are constant. This means that the product (*AP*)(*AQ*)
is constant. Let *r ^{2}* = (

The only thing left is to force point *P* into the path
of a circle through *A*. The linkage below should do the job. The
small circles represent fixed anchor points. Since they are fixed, the
link between them is unnecessary, but it is there to show the geometric
relationship. Notice that point *P* cannot trace the entire circle
without breaking some of the links. That is why *Q* traces a line
segment rather that an entire line. Click on the image to see a simulation.

Below are some other linkages that draw straight line segments. It will be seen that they are actually variations on the Peucellier linkage.

This version is the simplest mechanism, but it may be more difficult to understand.

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Last update: September 4, 2003 ... Paul Kunkel

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