A p p ro x im a te L in es

E xa ct L ine M ech a n ism s E xa c t T ra n sla to r M e ch an ism s

In ve rsive M e ch an ism s

In ve rs io n G e om e try

M e cha n ica l S tra igh t L ine G e n e ra to rs

T h e C o m p le te Q ua d ran g le

P ro je c tive G eo m e try

S tra igh t L ine s w ith o u t R u le rs

ATM Conference 2006, OrmskirkMark Dabbs (mark@mfdabbs.com)

Watt’s Parallel Motion• The best known of the

approximations to straight line motion is that of Scottish inventor and engineer James Watt’s 4 Bar Linkage, invented in 1784.

• It took Watt several years to design a straight-line linkage that would change straight-line motion into circular motion. Years later Watt told his son:

"Though I am not over anxious after fame, yet I am more proud of the parallel motion than of any other mechanical invention I have ever made." [Fergusson 1962]

Watt’s Parallel Motion• When in mid-position the

bars AB and CD are parallel, and BC is perpendicular to both.

• Point E is chosen at any point within the bar BC, whose displacement during a small rotation of the bars AB and CD is will be an approximate straight line.

• Since the diagram shows the mechanism at its mid-position we can assume that BC represents the direction of the line generated by point E.

Watt’s Parallel Motion• Ranges of motion of

bars AB and CD are shown.

• The exact position of E’ is found by first assuming a small rotation of the bars so that B moves to B’ and C to C’.

• Let B’C’ cut BC or BC (produced) in E’. Then E’ will be the generating point for the straight line.

• To prove this we therefore need to show that the ratio

is constant for all small rotations of AB and CD about their mid-positions.

Watt’s Parallel Motion

B'E'E'C'

Watt’s Parallel Motion• At these rotation angles of

AB and CD we see that point E’ is just about to diverge from its approximate straight line path. However, the angle has been greatly enlarged for for clarity in this and subsequent diagrams.

Watt’s Parallel MotionProof:

Draw B’H perpendicular to CB produced and C’K to BC.Draw B’F perpendicular to AB and C’G to CD.

Watt’s Parallel Motion

B'E' HB' E'HC'E' KC' E'K

Proof:

Draw B’H perpendicular to CB produced and C’K to BC.Draw B’F perpendicular to AB and C’G to CD.

B'E' BF E'HC'E' CG E'K

Therefore, triangles B’E’H and C’E’K are similar.

Hence

B'E' BF 1

C'E' CG

Watt’s Parallel MotionBy Pythagoras:

2 2 2B'F AB' AF

2 2 2B'F AB' AB BF

2B'F AB AB BF AB AB BF

2B'F BF 2AB BF

2 2 2B'F AB AB BF , Equal radii

but since BF 2AB then

2B'F 2BF AB 2

In a similar manner we have

2C'G 2CG CD 3

Watt’s Parallel Motion

B'E' BF 1

C'E' CG

For small rotations of the bars AB and CD it may be assumed that B’F be equal to C’G, despite the differing lengths of AB and CD.Hence, dividing (2) by (3) gives:

2 2

2 2

B'F B'F2BF AB1

2CG CD C'G B'F

Therefore,

BF CD 4

CG AB

Substituting (4) into (1) gives

B'E' CD A constant

C'E' AB

as required, since AB and CD arethe fixed lengths of the rotating bars.

Watt’s Parallel MotionIn addition to the above work proving the point E generates a straight line approximation for small angles of rotation of the bars AB and CD, we also have a condition for the best possible position of point E within the link BC.

BE CDCE AB

Namely:

That is, E divides the bar BC into the same ratio as the rotating bar lengths

BE: EC CD: AB

Watt’s Beam Engine

Watt’s LinkagePantograph

The Double-Acting, Triple Expansion Steam Engine. The steam travels through the engine from left to right

The engine shown in the picture below is a Double-Acting steam engine because the valve allows high-pressure steam to act alternately on both faces of the piston. The animation below it shows the engine in action:

High-pressure steam in Exhaust steam out

Hyperlinks to Euclidraw Dynamic Linkage Files

Straight Line Approximation EUC Files

Background Theory EUC Files

Exact Line Generator Mechanism EUC Files

Exact Translator Mechanism EUC Files

Link 1 to EUC File Link 10 to EUC File Link 14 to EUC File Link 17 to EUC File

Link 2 to EUC File Link 11 to EUC File Link 15 to EUC File Link 18 to EUC File

Link 3 to EUC File Link 12 to EUC File Link 16 to EUC File Link 19 to EUC File

Link 4 to EUC File Link 13 to EUC File Link 20 to EUC File

Link 5 to EUC File Link 21 to EUC File

Link 6 to EUC File Link 22 to EUC File

Link 7 to EUC File Link 23 to EUC File

Link 8 to EUC File Link 24 to EUC File

Link 9 to EUC File Link 25 to EUC File

The Harmonic RangeIf A, B be two points on a straight line, and C, D two other points on the line, placed such that

AC ADBC BD

then the points A, C, B and D form a Harmonic Range; and C and D are Harmonic Conjugates with respect to A and B. The lengths AC, AB and AD are said to be in Harmonic Progression; and AB is said to be the Harmonic Mean between AC and AD.

A C B D

(Notice: All segments are directed. Thus , etc)

--(1)

AC CB AB ACAD BD AD AB

Eq. (1) can be written in the form:

BC CB

Therefore, in order that the above proportion hold, it can be seen that one, and only one, of the points C, D must lie between A and B. For example:

A C B DO

r r

Extension:

For the Harmonic Range A, C, B and D; set point O as the midpoint of AB. Let |OA| = |OB| = r.

ThenAC AD AC ADBC BD CB BD

can be rewritten in terms of r as:

OC ODOC rD

r rr O

which simplifies to the important result:

2r OC OD

Ceva’s and Menelaus’ TheoremsCeva’s Theorem:

Proof of Ceva’s Theorem

Ceva’s Theorem (The Converse):

Menelaus’ Theorem:

Proof of Menelaus’ Theorem

The Complete Quadrangle

In triangle ABQ Ceva’s Theorem states:

AC BS QR1

CB SQ RA

In triangle ABQ Menelaus’ Theorem states:

AD BS QR1

DB SQ RA

Dividing these results gives:

AC AD AC AD AC ADCB DB BC BD BC BD

Therefore, the “Complete Quadrangle” is a simple geometrical construction that creates the Harmonic

Range A, C, B and D.

Link 26 to EUC File

SoftwareAll of my geometry files were created with Euclidraw

www.euclidraw.com Created by: Prof. Paris Pamfilos, University of Crete, Department of Mathematics

pamfilos@math.uoc.gr