background image

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$7<.$

background image

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

background image

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

:=

background image

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

:=