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Inversion
O
P
QGiven : OP.OQ==constant
Locus of P is a circle kP.
Let OP be the center of kQ.
S’ S
OP
P’
<S’P’O= <PSO
Since they are angles subtended by the same chord S’P on the circumference of kP.
<S’ OP’ = <POS Since they are vertically opposite angles
Hence S’P’O ~PSO
Let OP intercept kQ at P’.
Let the diameter through OOP intercept kQ at S and S’.
kP
Inversion of a circle is a circle
O
P
QOP.OQ=
S’ S
OP
P’kP
2 2 2 2
. .
.
. .
P P P P
P P P P
OS OP
OP OSOP OP OS OS
OP OP OO O S OO S O [taking direction into consideration]
r OO OP OP r OO OP OP
Hence locus of Q is a scaled locus of P’ and hence P
Therefore if P moves in a circle so does Q.
2 2 .
P P
OQBut OP OQ constant OQ constant×OP
OP r OO
kQ
OQ
Relations for mutually inverse circles
O
P
QOP.OQ=
S’ S
OP
P’kP
kQ
OQ
2 2
2 2 2 2
P P
P P P P
OQ OQ
OP OP r OO
OQ OQ QQ
OP OP P Pr OO r OO
Q
2 2P P P
r λ=
r r -OO
Q’
An alternative proof based on coordinate geometry follows
Further similitude relations :
Relations for mutually inverse circles
O
P
QOP.OQ=
S’ S
OP
P’kP
kQ
OQ
2 2 2 2
2 2
2 2
2 2
2 2 2 2
cos sin cos sin
cos cos
sin sin
sin sin
P P P P
P P P Q Q QP P
P P Q QP P
P QP P
QP Q
PP P P P
OQOP OQ
OP r OO r OO
r OO i r j r OO i r jr OO
r OO r OOr OO
and r rr OO
rr r
rr OO r OO
Q
2P P
r λ=
r r -O 2PO
Q’
Relations for mutually inverse circles
O
P
QOP.OQ=
S’ S
OP
P’kP
kQ
OQ
2 2 2 2
2 2 2 2 2 2
2 2
cos cos
cos cos
P P Q QP P P P
P P P QP P P P P P
P QP P
r OO r OOr OO r OO
r OO r OOr OO r OO r OO
OO OOr OO
Q
2 2P P P
OO λ=
OO r -OO
Q’
When the inverse of a circle is a straight line
And if rP=OOP?Q
2 2P P P
r λ=
r r -OO Q
QP
rr
r Locus of Q is a straight line !
P
Q
S
OPO
kP
Straight line mechanisms with only revolute joints are designed to have O,P,Q as points on the links, with OPP being the crank link.
OP.OQ==constant
2 2
2 cos
cos cos2 cos
2
P
P
P
P
POS PO S
OQOP O P
OT OQO P
OT constantO P
T
General proof
Hencelocus of Qis a straight line
O OP
A
P
B
Q
Peucellier mechanism
PA=PB=AQ=BQ=s, OA=OB=l
C
, ,
int
APO BPO
AQO BQO
O P Q must be collinear
AB PQ
C is the midpo of AB and PQ
2 2
2 2 2 2
2 2 2 2 2 2
2 2
.
.
.
.
.
OP OQ OC PC OC CQ OC PC OC PC
OP OQ OC PC
OP OQ OA AC AP AC
OP OQ OA AC AP AC OA AP
OP OQ l s constant
P moves in a circle, hence Q moves in a straight line
D
C
S
T
O
A
AB=CD
Hart mechanism
B
AC=BD OO2=O2T
BO:OA=BT:TD
Given
O2
.
Draw BM AD, CN AD.
Extend BT to meet AC at S
M
N
, AC=BD
, AMB= CDN=2
AB CD ABC DCA
BAM CDN Also
ABC DCA
BM CN BCNM is a rectangle
BC MN AD
D
C
S
T
O
A
AB=CD
Hart mechanism
B
AC=BD OO2=O2T
BO:OA=BT:TD
Given
O2
M
N
BO BT DA OTBOT BAD
OA TD BA OB
OTDA = BA
OB
2
. . . ..
BAOS OT BC DA k BC DA
OAOB
2 2 2
2 2 22 2 2 2
2 2
cos2 .
cos cos2 2
.2 . 2
.
AD CD ACADC ADC
AD CDND AD MN AD BC
NDC NDC ADCCD CD CD
AD CD AC AD BCAD CD AC AD AD BC
AD CD CD
BC DA AC CD constant
OS.OT = constant
Hence S traces a straight line as T moves in a circle
BO BTAD OT OS OS BC
OA TDBC OS
AOS ABCBA OA
OS
BC = BAOA
