t L C
T R U IN G D A I HOC SU' PHAM K Y T H U A T
TH AN H PHO HO CHI M IN H
KHOA CO KHI CHE TAO MAY
BO MON CO DIEN TU
BE THI CUOI HOC KY I NAM HOC 2023-2024
Mon: Ky thuat Robot (CLC)
Mambnhoc: R0B0331129
De so/Ma de: 01 De thi co: 02 trang.
Thai gian: 60 phut.
Dugc phep sir dung 1 to A4 tai lieu.
Cau 1: (3.0 diem)
a) Trong toa do cue bo (local frame B), cho toa do H Px thuc hien lan lugt phep xoay
25° quay true x b ta dugc he tga do xiyizi, sau do tiep tuc xoay 30° quanh true yB
thi ta dugc he tga do x:?yiZ2, sau khi thuc hien cac phep xoay thi dugc
B^ = [ - l 2 3]r . Tim tga do H Px trude khi xoay, tren tga do cue bo B. (1.5d)
b) Cho he tga do ban dau la xoyozo, cho tga do diem °NX = [1 0.2 -2 .3 f , thuc
hien phep xoay 0.2 [rad] quay true zn ta dugc he tga do xiyizi, tiep tuc thuc hien
phep xoay 0.4 [rad] quay true xo, ta dugc he tga do xiyizi, tiep tuc thuc hien phep
xoay -0.15 [rad] quay true zo, thi dugc diem 0 N4. Tim tga do diem {) N4 sau khi
thuc hien cac phep xoay. (1.5d)
Cau 2: (3.0 diem)
Trong he tga do toan cue OXYZ (global frame G), cho tga do diem K , thuc hien
phep quay 45° quay Z, ta dugc he tga do X/Y/Zi, sau do tiep tuc quay 45° quay X, ta
dugc he tga do X2Y2Z2. Sau do, he tga do X2Y2Z2 thuc hien phep tinh tien trong tga do
toan ciic (global frame G) {,d = \ l 3 -1.4]r ta dugc he tga do X3Y3Z3 va
3K = [-1.5 -0.6 -1.8]7.
a) Tim ma tran chuyen vi dong nhat rT3 b tga do toan cue sau khi thuc hien cac
phep xoay va tinh tien b tren, tim diem 'K sau khi thuc hien phep chuyen vi tu
m a tran G7’3.
b) Tim ma tran nghich dao cua CT3.
S6 hieu: BM1/QT-PDT-RDTV/02 LAn soat xet: 02 Ngay hieu lire: 15/5/2020 Trang: 1/2
Hinh 1. Cff cau robot
Cau 3: (2.0 diem)
Cho ca cau robot nhu hinh ve ben tren. Ve lai ca cau robot vao bai lam va dat goc toa
do.
a) Xac dinh bang Denavit-Hartenberg (DH) (Standard DH).
b) Xac dinh ma tran chuyen vi °7],1T2 cua tirng khau va cua toan bo robot °T2.
Cau 4: (2.0 diem)
Cho /1 = 1.5 [m], h= 1 [m], d\ = di = 0.15 [m], Va diem cuoi cua robot dang a vi tri khau
so 2 (B2), trong toa do toan ciic co toa do [xoyo zo\1 = [0.8 0.6 0.3]1. Tu "‘Hinh 1. Co"
cau robot”, va ket qua ma tran chuyen vi °T2 cua robot tir cau 3. Tim tat ca cac truong
hop cua goc 6\ va goc G-ft
Ghi chu: Can bo coi thi khong ditac giai thick de thi. Lam trdn chit so thdp phan thie
3 sau dau phay ”
Chuan dau ra cua hoc phan (ve kien thuc) Noi dung kiem tra
[CL02]: Co kha nang ap dung cac kien thuc ve toan
hoc de xay dung mo hinh toan cho robot noi tiep,
robot song song.
Cau 1
[CL02]: Co kha nang ap dung cac kien thuc ve toan
hoc de xay dung mo hinh toan cho robot noi tiep,
robot song»song.
Cau 2
[CL04]: Co kha nang sir dung co so toan hoc cua
phuong phap ma, tran chuyen vi va Denavit -
Hatenberg de tinh bai toan,dong hoc.
Cau 3
[CL05]: Co kha nang mo phong bai toan dong hoc vi
tri
Cau 4
Ngay ^ thang./l2nam 2023
Trirang bo mon
©tL Q _
TRUCJNG DAI HOC SU PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO BIEN TET
DE THI CUOI HOC KY I NAM HOC 2023-2024
Mon: Ky thuat Robot (CLC)
M amon hgc: R0B0331129
De so/M a de: 01 De thi co: 02 trang.
Thdi gian: 60 phut.
Dugc phep su dung 1 to A4 tai lieu.
D A P A N
Cau 1: (3.0 diem)
a) Trong tga do cue bo (local frame B), cho toa do B P] thirc hien lan lugt phep xoay
25° quay true x b ta dirge he toa do x/y/zi, sau do tiep tuc xoay 30° quanh true yB
thi ta dugc he toa do xzy2Z2, sau khi thirc hien cac phep xoay thi dirge
B P} = [-1 2 3]r . Tim tga do H P} trirdc khi xoay, tren tga do cue bo B. (1.5d)
Ma tran xoay quanh true x,y tai tga do cue bo (local frame)
l 0 0 "1 0 0
BK = 0c25 525 =0 0.906 0.423
0-525 c25 0 -0.423 0.906
c30 0 -530“ 0.866 0 -0.5 "
X = 010-0 1 0
530 0 c30 0.5 0 0.866
(0.25d)
(0.25d)
Tga do BPi trirdc khi thirc hien phep xoay
BP3 = BR BP} = ( BRv X ) BPi
’P ^ C R / R , ) ' ' ‘ P,
0.634
0.503
3.653
(Id)
b) Cho he tga do ban dau la xnyozn, cho tga do diem °N ] = [l 0.2 -2.3]r , thirc
hien phep xoay 0.2 [rad] quay true zn ta dugc he tga do xiyizi, tiep tuc thirc hien
phep xoay 0.4 [rad] quay true xo, ta dugc he tga do xzy2Z2, tiep tuc thirc hien phep
xoay -0.15 [rad] quay true zo, thi dugc diem °7V4. Tim tga do diem °A^4 sau khi
So hieu: BM1/QT-PDT-RDTV/02 Lan soat xet: 02 Ngay hieu lire: 15/5/2020 Trang: 1/6
Ma tran xoay quanh true z va x, z tai toa do cue bo (local frame)
0 . 2 n
c
--------
s
180
0 2 * Q
180
'R z =
0 .2 ^
- s
--------
cs
180
° - 2 * 0
180
0 0 1
1 0 0
11
(N
O A t t
0 c
--------
180
0 .4 ^
s
--------
180
O A x
0 - s — —
L 180
0 .4 ;r
C 180 _
—0 .1 5zr —0 .1 5zr
C 180 5 180
% =
—0 .1 5zr - 0 . 1 5 *
180 C 180
0 0
0.98 0.199 0
-0.199 0.98 0
0 0 1
(0.25d)
0
0.921
-0.389
0
0.389
0.921
(0.25d)
0.989
0.194
0
-0.149 0
0.988 0
0 1
(0.25d)
''Neu ra duoc 3 ma tran xoay thi dirac 0.75d, moi ma tran xoay dirffc 0.25d.
Toa do °N, truoc khi thirc hien phep xoay
]N4 = bR X = ( % 2RX 'Rz ) °N] =
1.143
-0.736
-2.117
(0.75d)
Cau 2: (3.0 diem)
Trong he toa do toan cue OXYZ (global frame G), cho toa do diem ( K , thirc hien
phep quay 45° quay Z, ta duoc he toa do X/YiZi, sau do tiep tuc quay 45° quay X, ta
duoc he toa do X2Y2Z2. Sau do, he toa do X2Y2Z2 thirc hien phep tinh tien trong toa do
toan cue (global frame G) °d = [2 3 —1.4] ta dupe he toa do X3Y3Z3 va
2K = [-1.5 -0.6 -1 .8 ]7.
a) Tim ma tran chuyen vj dong nhat °T3 0 toa do toan cue sau khi thirc hien cac
phep xoay va tinh tiln 0 tren, tim diem °K sau khi thirc hien phep chuyen vi tir
matran GT3.
b) Tim ma tran nghich dao cua CT3.
So hieu: BM1/QT-PDT-RDTV/02 Lan soat xet: 02 Ngay hieu luc: 15/5/2020 Trang: 2/6
Ma tran xoay quanh true Z, X tai toa do toan cue
"c45 -545 0" "0.707 -0.707 0"
g r z = 545 c45 0 =0.707 0.707 0
0 0 1 0 0 1
'1 0 0 " 1 0 0
% = 0c45 -545 =0 0.707 -0.707
0 545 c45 0 0.707 0.707
*Neu ra duox moi ma tran xoay thi duox 0.25d, tong la 0.5d
Ma tran xoay tong hop tai toa do toan cue (global frame)
"0.707 -0.707 0
;f l = ( % c « z )= 0.5
0.5
0.5
0.5
-0.707
0.707
(0.5d)
Ma tran chuyen vi T dtra tren ma tran xoay tong hop va vector tinh tien d toa do toan cue
(global frame)
"0.707 -0.707 0 2
~gR Gd 0.5 0.5 -0.707 3
gT3 = =(0.5d)
0 1 0.5 0.5 0.707 -1.4
0 0 0 1
Toa do K, sau khi thirc hien phep chuyen vi
'0.707 -0.707 0 2 —1.5 " 1.364 "
0.5 0.5 -0.707 3 -0.6 3.223
G K = gT33X = —
j0.5 0.5 0.707 -1.4 -1.8 -3.723
0 0 0 1 1 1
Ket luan: CK = [1.364 3.223 -3.723]'
(Id)
*De bai yeu cau lam phmmg phap ma tran chuyen vi T, SV lam khac phmmg phap
chi lay 30% diem.
Ma tran nghjch dao T '1 la
gRtD '
0.707 0.5 0.5 -2.214
-0.707 0.5 0.5 0.614
0-0.707 0.707 3.111
0 0 01
(0.5d)
U T
S6 hieu: BM1/QT-PDT-RDTV/02 Lan soat xet: 02 Ngay hieu lire: 15/5/2020 Trang: 3/6