a
−
=
a
l
− ε
⋅
sin
α
( )
⋅
l
ω
( )
⋅
cos
α
( )
⋅
−
l
ε
⋅
sin
α
( )
⋅
−
l
ω
( )
⋅
cos
α
( )
⋅
−
:=
ε
=
ε
l
− ε
⋅
cos
α
( )
⋅
l
ω
( )
⋅
sin
α
( )
⋅
+
l
ω
( )
⋅
sin
α
( )
⋅
+
l
cos
α
( )
⋅
:=
:\]QDF]HQLHSU]\VSLHV]HRJQLZPHchanizmu
V
=
V
l
− ω
⋅
sin
α
( )
⋅
l
ω
⋅
sin
α
( )
⋅
−
:=
ω
=
ω
l
− ω
⋅
cos
α
( )
⋅
l
cos
α
( )
⋅
:=
:\]QDF]HQLHSUGNRFLRJQLZPHFKDQizmu
α
deg
=
α
angle cos
α
2
sin
α
2
,
(
)
:=
cos
α
2
s
l
cos
α
( )
⋅
−
l
:=
sin
α
2
l
−
sin
α
( )
⋅
l
:=
l
sin
α
( )
⋅
l
sin
α
( )
⋅
+
s
=
s
−
=
s
l
⋅
cos
α
( )
⋅
∆
+
:=
s
31
l
⋅
cos
α
( )
⋅
∆
−
:=
∆
−
l
⋅
cos
α
( )
⋅
(
)
l
( )
l
( )
−
⋅
−
:=
ε
−
:=
ω
−
:=
α
deg
⋅
:=
l
:=
l
0.12
:=
.,1(0$7<.$
aS3y
=
aS3x
−
=
aS3y
:=
aS3x
a
:=
3UGNRüLSU]\VSLHV]HQLH URGNDPDV\RJQLZD
aS2y
−
=
aS2x
−
=
aS2y
l
ε
⋅
cos
α
( )
⋅
l
ω
( )
⋅
sin
α
( )
⋅
−
l
ε
cos
α
( )
⋅
ω
( )
sin
α
( )
⋅
−
⋅
+
:=
aS2x
l
− ε
⋅
sin
α
( )
⋅
l
ω
( )
⋅
cos
α
( )
⋅
−
l
ε
sin
α
( )
⋅
ω
( )
cos
α
( )
⋅
+
⋅
−
:=
VS2y
−
=
VS2x
=
VS2y
l
ω
⋅
cos
α
( )
⋅
l
ω
⋅
cos
α
( )
⋅
+
:=
VS2x
l
− ω
⋅
sin
α
( )
⋅
l
ω
⋅
sin
α
( )
⋅
−
:=
3UGNRüLSU]\VSLHV]HQLH URGNDPDV\RJQLZD
aS1y
−
=
aS1x
−
=
aS1y
l
ε
cos
α
( )
⋅
ω
( )
sin
α
( )
⋅
−
⋅
:=
aS1x
l
−
ε
sin
α
( )
⋅
ω
( )
cos
α
( )
⋅
+
⋅
:=
VS1y
−
=
VS1x
=
VS1y
l
ω
⋅
cos
α
( )
⋅
:=
VS1x
l
− ω
⋅
sin
α
( )
⋅
:=
3UGNRüLSU]\VSLHV]HQLH URGNDPDV\RJQLZD
:\]QDF]HQLHSU GNRFLLSU]\VSLHV]H URGNyZPDVRJQLZ
Pbx
2
=
Pby
2
=
Pbx
3
aS3x
−
m
⋅
:=
Pby
3
aS3y
−
m
⋅
:=
Pbx
3
=
Pby
3
=
F
−
:=
rxBS
l
cos
α
( )
⋅
:=
ryBS
l
sin
α
( )
⋅
:=
rxBS
=
ryBS
−
=
rxBC
l
cos
α
( )
⋅
:=
ryBC
l
sin
α
( )
⋅
:=
rxBC
=
ryBC
−
=
rxBS
l
cos
α
+
(
)
⋅
:=
ryBS
l
sin
α
+
(
)
⋅
:=
rxBS
−
=
ryBS
−
=
rxBA
l
cos
α
+
(
)
⋅
:=
ryBA
l
sin
α
+
(
)
⋅
:=
rxBA
−
=
ryBA
−
=
.,1(7267$7<.$
m
:=
m
:=
m
:=
g
:=
J
:=
J
:=
G
g
−
m
⋅
:=
G
g
−
m
⋅
:=
G
g
−
m
⋅
:=
G
−
=
G
−
=
G
−
=
Mb
1
ε
−
J
⋅
:=
Mb
2
ε
−
J
⋅
:=
Mb
1
=
Mb
2
−
=
Pbx
1
aS1x
−
m
⋅
:=
Pby
1
aS1y
−
m
⋅
:=
Pbx
1
=
Pby
1
=
Pbx
2
aS2x
−
m
⋅
:=
Pby
aS2y
−
m
⋅
:=
Mnap
=
Mnap
M1
M2
+
ω
:=
M2
Pbx
(
)
VS1x
⋅
Pby
G
+
(
)
VS1y
⋅
+
Pbx
(
)
VS2x
⋅
Pby
G
+
(
)
VS2y
⋅
+
+
Pbx
(
)
V
⋅
+
−
:=
M1
F V
⋅
Mb
ω
⋅
+
Mb
ω
⋅
+
(
)
−
:=
WYZNACZENIE MOMENTU NAPEDOWGO M
n
W OPARCIU O BILANS
MOCY MECHANIZMU
Find Rx23 Ry23
,
R43
,
Rx12
,
Ry12
,
Rx41
,
Ry41
,
Mn
,
(
)
−
−
−
=
Mb
rxBS
Pby
G
+
(
)
⋅
+
ryBS
Pbx
(
)
⋅
−
rxBA Ry41
(
)
⋅
+
ryBA Rx41
(
)
⋅
−
Mn
+
Ry12
−
Ry41
+
Pby
1
+
G
+
Rx12
−
Rx41
+
Pbx
1
+
Mb
rxBS
Pby
G
+
(
)
⋅
+
ryBS
Pbx
(
)
⋅
−
rxBC
Ry23
−
(
)
⋅
+
ryBC
Rx23
−
(
)
⋅
−
Ry23
−
Ry12
+
Pby
+
G
+
Rx23
−
Rx12
+
Pbx
+
Ry23
G
+
R43
+
Rx23
Pbx
+
F
+
Given
Mn
:=
Ry12
:=
Ry41
:=
Rx41
:=
Rx12
:=
R43
:=
Ry23
:=
Rx23
:=