SinhVienIT.net LXT Bai Tap Robot Cong Nghiep

7/24/2019 SinhVienIT.net LXT Bai Tap Robot Cong Nghiep

1/20

Bi tp robot cng nghip

Chng 1&2

Bi 1:Cho robot Stanford nh hnh 1 gm 2 khpquay v 1 khp tnh tin. Hy xc nh:

Sbc tdo ca robot Cc khnng xoay, tnh tin no trong

hcnh OXYZ ?

Bi 2:Cho robot Elbow nhhnh 2 vi 6 khp xoay.Hy xc nh:

Sbc tdo ca robot

Cc khnng xoay, tnh tin no trong hcnh OXYZ ?

Bi 3:

Vsmt robot (vi cu hnh ti thiu) m khu tc ng cui (End-effector) c khnng tnh tin theo phng Y, tnh tin theo phng Z, v xoay quanh phng X.

Z

Y

X

Hnh 1

XY

Z

Hnh 2


2/20

Chng 3:

Bi 4:

Cho im P biu din bi vect [ ]TA 142=p .

Tnh tin im P theo vect [ ]Th 121= , sau cho im P quanh trc X ca hta {A} mt gc90

0. Xc nh vc-t biu din v tr im P sau 2

bc dch chuyn.

Bi 5:

Cho mt khi lp phng trong h ta OXYZ cnh nhhnh 3. Khi ny c

quay quanh trc OB mt gc 900. Xc nh vc-tbiu din vtr im A (mt nh cakhi lp phng) sau khi thc hin php quay.

Bi 6:

Cho mt khi lp phng trong hta OXYZ cnh nhhnh 3. Tnh tin khi lp

phng theo vc-t [ ]Th 111= sau quay khi lp phng quanh trc OZ mt gc90

0 (lu : hng ca khi lp phng cng sb thay i khi quay). Xc nh vc-t

biu din vtr im A (mt nh ca khi lp phng) sau khi thc hin 2 php bin i.

Bi 7:Cho mt khi lp phng trong h ta OXYZ cnh nhhnh 3. Quay khi lpphng quanh trc OZ mt gc 900sau quay tip quanh trc OX mt gc -900. Xcnh vc-tbiu din vtr im A (mt nh ca khi lp phng) sau khi thc hin 2php bin i.

Bi 8:Cho mt khi lp phng trong h ta OXYZ cnh nhhnh 3. Quay khi lpphng quanh trc OZ mt gc 450sau quay tip quanh vc-tAB (l 1 cnh ca khilp phng) mt gc -900. Xc nh vc-tbiu din vtr im C (mt nh ca khi lpphng) sau khi thc hin 2 php bin i.

Bi 9:

Cho mt khi lp phng trong hta {R: O-XYZ} cnh nhhnh 3. Quay khilp phng quanh trc OX mt gc -450 sau tnh tin khi lp phng theo vc-t

[ ]TR 401=h . Xc nh vc-t biu din v tr im A (mt nh ca khi lpphng) sau khi thc hin 2 php bin i.

Bi 10:

Mt im P= [3 5 7]T trong h ta tham chiu. Sau dch chuyn im P mtkhong cch d= [2 3 4]T. Xc nh vtr mi ca im Ptrong hta tham chiu.


3/20

Bi 11:

Mt hta {A} c m tso vi hta tham chiu {R} bng ma trn bin ithun nht RTA.Xc nh ma trn bin i thun nht

RTAsau khi dch chuyn h{A}

mt khong cch Rd= [5 2 6]T.

=

1000

6100

4001

2010

A

R T

Bi 12:

Cho mt hta {A} c m tso vi hta tham chiu {R} bng ma trn bini thun nht RTA. Hy xc nh cc thnh phn cn thiu.

=

1000

201?

300?

510?

A

R T

Bi 13:

Mt vectApc quay xung quanh trc Z ca h{A} mt gc , v sau c quayxung quanh trc X ca h{A} mt gc . Hy xc nh ma trn quay thhin cc phpquay ny theo thtc cho.

Bi 14:

Mt vect Apc quay xung quanh trc Z ca h{A} mt gc 300, v sau cquay xung quanh trc X ca h{A} mt gc 450. Hy xc nh ma trn quay thhin ccphp quay ny theo thtc cho.

Bi 15:Cho mt hta {B} ban u trng vi hta {R}. Sau quay hta {B} xungquanh trc Z ca n mt gc , v tip theo quay hta {B} xung quanh trc Xca n mt gc . Hy xc nh ma trn quay chuyn i vectthta {B} sanghta {R}.

Bi 16:

Cho mt hta {B} ban u trng vi hta {R}. Sau quay hta {B} xungquanh trc Z ca n mt gc 300, v tip theo quay hta {B} xung quanh trc Xca n mt gc 450. Hy xc nh ma trn quay chuyn i vect th ta {B}sang hta {R}.


4/20

Bi 17:

Cho mi quan hgia cc hta {R}, {A}, {B}, v {C} nhsau:

=

1000

0.8000.1000.0000.0

0.1000.0866.0500.0

0.11000.0500.0866.0

A

R T

=

1000

0.20866.0500.0000.0

0.10500.0866.0000.0

0.0000.0000.0000.1

B

A T

=

1000

0.3866.0433.0250.0

0.3500.0750.0433.0

0.3000.0500.0866.0

R

CT

Xc nh BTC

Chng 4

Bi 18:Cho ccu tay my 2 bc tdo nhhnh 4.Thit lp:

Hta cho tng khu Bng thng sDH H phng trnh ng hc thun cho tay

my.

Bi 19:Cho ccu tay my 3 bc tdo nhhnh 5.Thit lp:

Hta cho tng khu Bng thng sDH

H phng trnh ng hc thun cho taymy.

Bi 20:

Cho ccu tay my 3 bc tdo nhhnh 6.Thit lp:

Hta cho tng khu Bng thng sDH Hphng trnh ng hc thun cho tay my.

Hnh 4

Hnh 5

Hnh 6


5/20

Bi 21:

Cho ccu tay my 3 bc tdo nhhnh 7Thit lp:

Hta cho tng khu Bng thng sDH

Hphng trnh ng hc thun cho tay my.

Bi 22:Cho ccu tay my c cu hnh nhhnh 8. Htocnh l X0Y0Z0. Cc kch thc d2=100mm,d4=100mm v bin khp d3=200mm.

Xc nh vect biu din v tr im E

trong hcnh.

Xc nh ta im E, nu bin khp thnht c gi tr 300, bin khp th hai cgi tr 00, bin khp th ba c gi tr25mm, v ba bin khp tht, thnm, vthsu cn li u bng 0

Bi 23:

Thit lp cc hta v xc nh cc tham sD-Hcho robot 3-DOF trong hnh 9.

Bi 24:Thit lp cc hta v xc nh cc tham sD-H cho robot 3-DOF trong hnh 10.

Hnh 7

Z

Y

X

E

d4

Hnh 8

Hnh 9

Hnh 10


6/20

Bi 25:

Thit lp cc hta v xc nh cc tham sD-H cho robot SCARA trong hnh 11.

Hnh 11


7/20

Bi gii

Bi 1:

Cng thc tnh bc tdoVi n: skhu ng

pi: skhp loi iRobot c 3 khu n = 3

3 khp loi 5 (2 khp quay 1 khp tnh tin) i = 5 ; p5 = 3Vy 33.53.6 ==DOF Robot c 3 bc tdo

Khp 1 quay quanh trc Y, khp 2 tnh tin vy kt hp 2 chuyn ng ny robot

c th tnh tin n v tr bt k trong mt phng XOZ (tnh tin theo X v Z).

Khp 3 quay quanh trc Y do End Effector c thvn n bt kim notrong khng gian 3 chiu. Tng hp li th End Effector ca robot c 3 bc tdol quay tnh tin theo trc X,Y v Z.

Bi 2:

Cng thc tnh bc tdoRobot c 6 khu n = 6

6 khp loi 5 (6 khp quay) i = 5 ; p5 = 6Vy 66.56.6 ==DOF Robot c 6 bc tdo

Vy End Effector ca robot c 6 bc tdo l quay quanh trc X,Y,Z, tnh tin

theo trc X,Y,Z.

Bi 3:

Bi 4:Vtr im P sau php tnh tin

=

=

==

0

2

6

3

1

1

4

2

.

1000

1100

2010

1001

1

.

1000

100

010

001

z

y

x

z

y

x

BB

AA

p

p

p

q

q

q

pTp

Vtr P sau php quay quanh trc X

=5

1

6 iipnDOF

=5

1

6 iipnDOF


8/20

=

=

==

1

6

2

3

1

2

6

3

.

1000

0010

0100

0001

1

.

1000

0cossin0

0sincos0

0001

z

y

x

B

B

AA

p

p

p

pRp

Vy toim P sau 2 php quay lin tip l [ ]TA 623 =p

Bi 5:

Vct biu din im A [ ]TA 202=p . Vct n v ch phng trc quay OB

[ ]TA 1113

1=r . V

y vctbiu din im A sau php quay quanh trc r gc 900l:

++

++

++

==

1

.

1000

0cos)cos1(sin)cos1(sin)cos1(

0sin)cos1(cos)cos1(sin)cos1(

0sin)cos1(sin)cos1(cos)cos1(

2

2

2

z

y

x

zxzyyzx

xzyyzyx

yzxzyxx

BB

AA

p

p

p

rrrrrrr

rrrrrrr

rrrrrrr

pRp

=

=

1

1786.0

333.1

488.2

1

2

0

2

.

1000

03333.09107.02440.0

02440.03333.09107.0

09107.02440.03333.0

pA

Vy

[ ]TA 1786.0333.1488.2=p

Bi 6:

Vtr im A sau php tnh tin

=

=

==

1

3

1

3

1

2

0

2

.

1000

1100

1010

1001

1

.

1000

100

010

001

z

y

x

z

y

x

BB

AA

p

p

p

q

q

q

pTp

Vtr A sau php quay quanh trc Z

=

=

==

1

3

3

1

1

3

1

3

.

1000

0100

0001

0010

1

.

1000

0100

00cossin

00sincos

z

y

x

BB

AA

p

p

p

pRp

Vy toim A sau 2 php bin i l [ ]TA 331=p


9/20

Bi 7:Vtr im P sau php quay quanh trc Z

=

=

==

1

22

0

1

20

2

.

1000

01000001

0010

1

.

1000

010000cossin

00sincos

z

y

x

BB

AA

pp

p

pRp

Vtr P sau php quay quanh trc X

=

=

==

1

2

2

0

1

2

2

0

.

1000

0010

0100

0001

1

.

1000

0cossin0

0sincos0

0001

z

y

x

BB

AA

p

p

p

pRp

Vy toim P sau 2 php quay lin tip l [ ]T

A 220 =p

Bi 8:Sau khi quay quanh trc Z toa im A

=

=

==

1

2

414.1

414.1

1

2

0

2

.

1000

0100

007071.07071.0

007071.07071.0

1

.

1000

0100

00

00

z

y

x

BB

AA

p

p

p

CS

SC

pRp

Sau khi xoay v tnh tin A vO th ta im B l

=

==

1

0

414.1

414.1

1

2

2

2

.

1000

2100

414.107071.07071.0

414.107071.07071.0

pTp BBAA

Do chiu di AB l 2 nn vectn vchphng l:

=

1

0

7071.07071.0

rA

Ma trn quay quanh trc xon r mt gc -900 v tnh tin trvvtr c


10/20

=1000

207071.07071.0

414.17071.05.05.0

414.17071.05.05.0

B

A

R

Ma trn chuyn i cho cc php bin i

=

1000

2100

414.107071.07071.0

414.107071.07071.0

.

1000

207071.07071.0

414.17071.05.05.0

414.17071.05.05.0

BA T

Toim C sau cc php bin i

=

==

1

2

2424.4

414.1

1

0

2

2

.

1000

0001

8282.27071.07071.00

8282.27071.07071.00

pTp BBAA

Bi 9:

Vtr im A sau cc php bin i

=

=

==

1

4142.5

4142.1

3

1

2

0

2

.

1000

47071.07071.00

07071.07071.00

1001

1

.

1000

0

0

001

z

y

x

z

y

x

BB

AA

p

p

p

hCS

hSC

h

pRp

Bi 10:

Vtr im P sau php bin i

1 0 0 1 0 0 2 3 5

0 1 0 0 1 0 3 5 8. .

0 0 1 0 0 1 4 7 11

0 0 0 1 1 0 0 0 1 1 0

x x

y yA A B

B

z z

d p

d p

d p

= = = =

p T p

Bi 11:

Ma trn bin i sau php dch chuyn


11/20

=

=

1000

12100

6001

7010

1000

6100

4001

2010

1000

6100

2010

5001

A

R

T

Bi 12:

=

=

1000

201

300

510

1000

201?

300?

510?

c

b

a

AR T

Ta c[ ]Tcba=u [ ]T100 =v [ ]T001=w

V

0=w.u nn

00.0.0).1( ==++ acba

V

vuw = Nn

kabjcaibccba

kjirrr

).0).1(()).1(.0().0.0(001 ++=

= uw

kjikbjcirrrrrr

1000 +=++= uw

Vy c=0v b=1

Bi 13:

Ma trn quay quanh trc Z

==

100

0cossin0sincos

),(

ZRotBA R

Ma trn quay quanh trc X

==

cossin0

sincos0

001

),(XRotBA R


12/20

Ma trn quay lin tip

==

CCSSS

SCCSC

SC

ZRotXRotBA

0

),().,(R

Bi 14:

Vi 030= v 045=

==

2

2

4

6

4

22

2

4

6

4

2

02

1

2

3

)30,().45,( 00 ZRotXRotBA R

Bi 15:y l php quay EulerPhp quay quanh trc Z ca hB

=

10000100

00cossin

00sincos

),(

ZRot

Php quay quanh trc X ca hB

=

1000

0cossin0

0sincos0

0001

),(

XRot

Tng hp 2 php quay lin tip

==

1000

00

0

0

),().,(

CS

SCCCS

SSCSC

XRotZRotBAT


13/20

Bi 16:

Vi 030= v 045=

Tng hp 2 php quay lin tip

==

1000

07071.07071.00

06124.06124.05.0

03535.03535.0866.0

),().,( XRotZRotTBA

Bi 17:Ta c

BA

AR

RC

BC TTTT =

Vy

=

1000

2.5752-0.6250.64950.433

11.3430.6495-0.1250.75

1.634-0.4330.75-0.5

BCT

V

==

1000

1

M

LLLMLLL

M qRR

TT

CTB

CTB

C

CB

BC

Nn

=

1000

4653.6625.06495.0433.0316.46495.0125.075.0

8053.8433.075.05.0

CB T


14/20

Bi 18:t cc htoln tay my

Bng thng sDH Ma trn chuyn th1 vh0 (khu 1)

=

1000

1010

0001

0100

10

dA

Ma trn chuyn th2 vh1 (khu 2) Ma trn chuyn th2 vh0

=

1000

2100

00100001

21

dA

==

1000

1010

00012100

21

10

20

d

d

AAT

Phng trnh ng hc thun pTp 2200 =

Khu a 0 d 01 0 90 d1 90

2 0 0 d2 0


15/20



=

1000

0010

00

00

11

1 1

10

SC

CS

A

Ma trn chuyn th2 vh1 (khu 2) Ma trn chuyn th3 vh2(khu 3)

+

=

1000

1010

0110

0001

21

dLA

=

1000

0100

0

0

333

33

3

33

32

CaSC

SaCS

A

Ma trn chuyn th3 vh0

32

21

10

30

AAAT =


Khu a 0 d 01 0 90 0 90

0+1

2 0 90 L+d1 1800

3 a3 0 0 90

0+3


16/20



=

1000

010

00

00

1

1

10 11

1

d

CS

SC

A

Ma trn chuyn th2 vh1 (khu 2) Ma trn chuyn th3 vh2 (khu 3)

=

1000

0100

0

0

222

222

2

2

21

SaCS

CaSC

A

=

1000

0100

0

0

333

333

3

3

21

SaCS

CaSC

A


32

21

10

30

AAAT =


Khu a 0 d 01 0 90 d1 12 a2 0 0 23 a3 0 0 3


17/20



=

1000

0010

.0

.0

111

11 1

10

CESC

SECS

A

Ma trn chuyn th2 vh1 (khu 2)

+=

1000

1000010

0001

22

1

dLA


21

10

20

AAT =


Khu a 0 d 01 E 90 0 90

0+1

2 0 0 L+d2 0


18/20

Bi 23: t cc htoln tay my

Bng thng sDH

Khu a 0 d 01 0 90 d1 12 0 0 d2 3


=

1000010

00

00

1

1

10 11

1

d

CS

SC

A

=

1000100

00

00

2

21 33

33

d

CS

SC

A


21

10

20

AAT =



19/20

Bi 24: t cc htoln tay my


=

1000

0010

0000

11

1 1

10

CSSC

A


=

1000100

0

0

1

2

2

21 222

222

d

SaCS

CaSC

A

=

10000100

0

0

333

333

3

3

32

SaCS

CaSC

A


32

21

10

30

AAAT =


Khu a 0

d 0

1 0 90 0 12 a2 0 d1 23 a3 0 0 3


20/20



=

1000

0100

0

0

111

111

1

1

10

SaCS

CaSC

A


=

1000

01000

0

222

222

2

2

21

SaCS

CaSC

A

=

1000

1000010

0001

33

2

dA


32

21

10

30

AAAT =


Khu a 0 d 01 a1 0 0 12 a2 0 0 23 0 0 d3 0

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SinhVienIT.net LXT Bai Tap Robot Cong Nghiep