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 http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


Buy this 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.)

(4)
AC·BD= EF·BD
 = (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.

Further

(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).

References

  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

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2012 Alexander Bogomolny

 40618297

A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help
Games & Puzzles
What Is What
Arithmetic
Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Old and nice bookstore
Other Math sites
Front Page
Movie shortcuts
Personal info
Privacy Policy

Guest book
News sites

Recommend this site

Sites for parents

Education & Parenting

Search:
Keywords:

Google
Web CTK
Supported by
3wVentures