Hart's Inversor

Similar to Peaucellier linkage, Hart's linkage (Hart's inversor or Hart's cell) employs inversion to convert between circular and rectilinear motion. Compared to the Peaucellier linkage, Hart's device uses fewer rods.

The device consists of four rods AB, BC, CD, and AD, such that AB = CD and BC = AD and AD and BC intersect. O, P, Q on AB, AD, and BC satisfy

(1) AO/AB = AP/AD = CQ/BC = m,

for 0 < m < 1. In ΔABD, (1) implies OP||BD. Similarly, in ΔABC, AQ||AC. Because of the symmetry, or since ΔABC = ΔADC), the quadrilateral ABDC is an isosceles trapezoid, so that BD||AC. It then follows from (1) that the three points O, P, Q are collinear and belong to a line parallel to both AC and BD.

The following property of the configuration will be proved later:

(2) OP·OQ = m(1 - m)·(AD2 - AB2).

It indicates that P and Q are mutually inverse under an inversion with center O. This means that, if O is fixed and P traces a curve, Q will trace the inverse image of the curve. If an additional rod SP is so attached to the configuration that

(3) SP = SO

and S is fixed, then P will trace a circle that passes through the center O of inversion. It follows that Q will then describe a segment of a straight line.

The applet below demonstrates this property. The points A, B, D, O, and S are draggable for the purpose of defining (or redefining) the attributes of the configuration. However, when P is dragged both O and S remain fixed.

Note that the dimensions of the rods impose certain limitations on the relative positions of the rods. When these are about to be violated while P is being dragged, the applet stops tracing the points. If this happens, return P into the arc already drawn and reconfigure from here.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

Let's now prove the claim (2), as promised.

Draw AE and CF perpendicular to BD (hence also to AC.)

 = (ED + EB)·(ED - EB)
 = ED2 - EB2.

However, by the Pythagorean theorem,

  ED2 + AE2 = AD2 and
EB2 + AE2 = AB2.

Hence from (4),

(5) AC·BD = AD2 - AB2.


(6) OP/BD = AO/AB = m and
OQ/AC = OB/AB = 1 - m.

Combining (5) and (6) we obtain

OP·OQ = m(1 - m)AC·BD
  = m(1 - m)(AD2 - AB2).


  1. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1996
  2. H. Rademacher and O. Toeplitz, The Enjoyment of Mathematics, Dover, 1990

Inversion - Introduction

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Copyright © 1996-2017 Alexander Bogomolny


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