Until this invention, no method existed of producing straight motion without reference guideways, making the linkage especially important as a machine component and for manufacturing. In particular, a piston head needs to keep a good seal with the shaft in order to retain the driving (or driven) medium. The Peaucellier linkage was important in the development of the Steam engine.
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 Sarrus linkage (often misspelled Sarrut), consisting 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.
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 parallelogram. Also, point O is fixed. Then, if point B is constrained to move along a circle (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).
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.
Angle OBA + angle ABD + angle DBC + angle CBO = 360°
but angle OBA = angle OBC and angle DBA = angle DBC, thus
2 * angle OBA + 2 * angle DBA = 360°
angle OBA + angle DBA = 180°
therefore points O, B, and D are collinear.
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°.