A p p ro xim a te Lines E xa ct L in e M echanism s E xa ct T ra n sla to r M echanism s In ve rsiv e M echanism s In ve rsio n G eom etry M e ch a n ica l S tra ig h t L in e G e ne rato rs T h e C o m p le te Q uadrangle P ro je ctive G eom etry S tra ig h t L in e s w ith o u t R ulers ATM Conference 2006, Ormskirk Mark Dabbs (mark@ mfdabbs .com )
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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
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:
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