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. 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.
Earlier Sarrus linkage 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
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 halfway 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). In contrast, 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). Many different over-all proportions of this linkage are possible. Since points O, B, D must be collinear at all points in the linkage's motion, and countless arm length combinations are viable, then mirror symmetry across OBD isn't necessary. With OBD staying collinear, the only requirement to achieve the intended straight-line motion of D are that AB = AD, that BC = DC, and for B to be constrained to a circular path which crosses O. Otherwise, there is no fixed relationship between the lengths of the sides of the ABCD figure, the radius of the constraining circular path of B, and the lengths of OA or OC.
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
∠ O B A + ∠ A B D + ∠ D B C + ∠ C B O = 360 ∘ {\displaystyle \angle OBA+\angle ABD+\angle DBC+\angle CBO=360^{\circ }}
but, due to the congruences, ∠OBA = ∠OBC and ∠DBA = ∠DBC, thus
2 × ∠ O B A + 2 × ∠ D B A = 360 ∘ ∠ O B A + ∠ D B A = 180 ∘ {\displaystyle {\begin{aligned}&2\times \angle OBA+2\times \angle DBA=360^{\circ }\\&\angle OBA+\angle DBA=180^{\circ }\end{aligned}}}
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 ∠BPA + ∠DPA = 180°, then 2 × ∠BPA = 180°, ∠BPA = 90°, and ∠DPA = 90°. Let:
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