Peaucellier-Lipkin linkage

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Peaucellier-Lipkin linkage (bars of identical colour are of equal length)

Geometric diagram of a Peaucellier linkage

The Peaucellier-Lipkin linkage (or Peaucellier-Lipkin cell), invented in 1864, was the first 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 Lippman Lipkin, of Lithuania ^[1] ^[2]. 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.

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' Motion (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.

[edit] 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. 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).

[edit] Mathematical proof of concept

[edit] Collinearity

First, it must be proven that points O, B, D are collinear.

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.

[edit] 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 angle BPA + angle DPA = 180°, then 2 * angle BPA = 180°, angle BPA = 90°, and angle DPA = 90°.

Let:

x = l B P = l P D

y = l O B

h = l A P

Then:

$l_{OB}\cdot l_{OD}=y(y+2x)=y^2+2xy$

${l_{OA}}^2=(y+x)^2+h^2$ (due to the Pythagorean theorem)

${l_{AD}}^2=x^2+h^2$ (Pythagorean theorem)

${l_{OA}}^2-{l_{AD}}^2=y^2+2xy=l_{OB}\cdot l_{OD}$

Since OA and AD are both fixed lengths, then the product of OB and OD is a constant:

$l_{OB}\cdot l_{OD}=k^2$

and since points O, B, D are collinear, then D is the inverse of B with respect to the circle (O,k) with center O and radius k.

[edit] Inversive geometry

Thus, by the properties of inversive geometry, since the figure traced by point D is the inverse of the figure traced by point B, if B traces a circle passing through the center of inversion O, then D is constrained to trace a straight line. But if B traces a straight line not passing through O, then D must trace an arc of a circle passing through O. Q.E.D.