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T R U IN G D A I HOC SU' PHAM K Y T H U A T
TH A N H PHO HO CHI M IN H
KHOA CO KHI CHE TAO MAY
BO MON CO DIEN TU
DE THI CUOI HOC KY I NAM HOC 2023-2024
Mon: Ky thuat Robot (Dai Tra)
Ma mon hoc: R 0 B 0 3 3 1126
De so/M a de: 01 De thi co: 02 trang.
Thai gian: 60 phut.
Dugc phep sir dung 1 td A4 tai lieu.
Cau 1: (3.0 diem)
a) Trong toa do cue bo (local frame) thuc hien lan luot phep xoayt-25° quay true y,
30° quay true x thi den toa do r, = [1-7 -2 .7 3.7]^. Tim toa do rt trudc khi
thuc hien cac phep xoay. (1.5d)
b) Trong phep quay Euler, cho toa do diem Et =[0.15 -0 .5 -2 .3 ]7 cho xoay
-0.25 [rad] quay true Z, xoay -0.4 [rad] quay true x, xoay -0 .1 5 [rad] quay true
z. Tim toa do diem E2 sau khi thuc hien cac phep xoay tren. (1,5d)
Cau 2: (3.0 diem)
Trong he toa do cue bo (local frame B), cho toa do diem H = [-5 .5 6.6 -8 .8 ]7 ,
thuc hien lien tuc cac phep quay -25 ° quay x, —^45° quay z. Sau do, thuc hien phep
tinh tien trong toa do toan cue (global frame G) d = [3 -3 -1 .2 ] .
a) Tim ma tran chuyen vi dong nhat (’T a toa do toan cue, tim diem H sau khi thuc
hien phep chuyen vi tu m a tran G T .
b) Tim ma tran nghich dao cua (' T . Xac dinh ma tran xoay R (4x4) va ma tran
tinh tien <’ D (4x4) cua ma tran nghich dao (,T . <
Hinh 1. C a cau robot
S6 hieu: BM1/QT-PDT-RDTV/02 Lin soat xet: 02 Ngay hieu luc: 15/5/2020 Trang: 1/2
Cau 3: (2.0 diem)
Cho ca cau robot nhu hinh ve ben tren. Ve lai co cau robot vao bai lam va dat goc toa
do.
a) Xac dinh bang Denavit-Hartenberg (DH).
b) Xac dinh ma tran chuyen vi °7j,' T2 cua tirng khau va cua toan bo robot °T2.
Cau 4: (2.0 diem)
Cho l\= 0.7 [m], /2=1.1 [m], d\ = 0.25 [m], Va diem cuoi P tren khau 2 cua robot dang or
vi tri trong he toa do toan cue co toa do [x0 y 0 z0]y = [0.2 1.2 0.8]7 . T u “Hinh 1.
Ccr cau robot”, va ket qua ma tran chuyen vi °T2 cua robot tir cau 3. Tim tdt ca cac
trudng hop cua goc 6\ va goc ft?
Ghi chit: Can bo coi thi khong dirge giai thick de thi. Lam tron ch it so thap phan thir
3 sau dan phay
Chuan dau ra cua hoc phan (ve kien thirc) o
Noi dung kiem tra
[CL 02]: Co kha nang ap dung cac kien thuc ve toan
hoc de xay dirng mo hinh toan cho robot noi tiep,
robot song song.
Cau 1
[C L02]: Co kha nang ap dung cac kien thuc ve toan
hoc de xay dirng mo hinh toan cho robot noi tiep,
robot song song.
Cau 2
[CL 04]: Co kha nang su dung co so toan hoc cua
phuong phap ma, tran chuyen vi va Denavit -
Hatenberg de tinh bai toan,dong hoc.
Cau 3
[CL 05]: Co kha nang mo phong bai toan dong hoc vi
tri.
Cau 4
Ngay ^ thang^nam 20-£S
Trird’ng bo mon
S6 hieu: BM1/QT-PDT-RDTV/02 Lan soat xet: 02 Ngay hieu luc: 15/5/2020 Trang: 2/2
TRUClNG DA I HOC SU PHAM KY TH U AT
t h An h p h o HO CHI m i n h
DE THI CUOI HOC KY I NAM HOC 2023-2024
KHOA CO KHI CHE TAO MAY Mon: Ky thuat Robot (Hai Tra)
BO MON C ODIEN TIf Ma moil hoc: R 0 B 0 3 3 1 126
De so/Ma de: 01 De thi co: 02 trang.
Thoi gian: 60 phut.
Duoc phep su’ dung 1 to A4 tai lieu.
DAP AN
Cau 1: (3.0 diem)
a) Trong toa do cue bo (local frame B) thuc hien lan luot phep xoay -25° quay true
y, 30° quay true x thi den toa do r2 = [1.7 -2.7 3.7]? . Tim toa do r, truoc khi
thuc hien cac phep xoay. (1.5d)
Ma tran xoay quanh true y,x tai toa do cue bo (local frame)
c(-25) 0 -j(-2 5 )" ' 0.906 0 0.423
010=0 1 0 (0.25d)
J i - 25) 0 c(-25) -0.423 0 0.906
'1 00 ' 'l 0 0
"Rx = 0 c30 x30 =0 0.866 0.5 (0.25d)
0-s30 c30 0 0.5 0.866
Toa do ri truoc khi thuc hien phep xoay
\ = »R\=[BRx°Ry)\
\ =
0.757
-4.188
2.399
(Id)
b) Trong phep quay Euler, cho toa do diem Ex = [0.15 -0.5 -2.3] cho xoay
-0.25 [rad] quay true Z, xoay -0.4 [rad] quay true x, xoay -0.15 [rad] quay true
z. Tim toa do diem E2 sau khi thuc hien cac phep xoay tren. (1,5d)
S6 hieu: BM1 /QT-PDT-RDTV/02 Lan soat xet: 02 Ngay hieu luc: 15/5/2020 Trang: 1/6
Ma tran xoay quanh true Z, x, z cua he toa do Euler
°R 7 =
-0.25 7r
180
-0.25;r
180
0
—s
0
f-0 .4;z^
180
-0.25;r
180
-0.25 n '
180 ,
0
0
-0.47T ''
0.969 0.247
-0.247 0.969
0 0
(0.25d)
-0.4;r
180
*R. =
-0.15;r
180
—s -0.1 5;r
180
0
180
-0.4;r
180
-0.15;r^
v 180 j
^ -0 .15zz-a
10 0
00.921 -0.389
0 0.389 0.921
(0.25d)
v180
0
7
0.989
0.194
0
-0.149 0
0.988 0
0 1
(0.25d)
*Neu ra dutre 3 ma tran xoay thi du’o’c 0.75d, moi ma tran xoay du’o’c 0.25d.
Toa do hE2 sau khi thuc hien cac phep xoay
0.194
BE2 = bRbE, = hR: hRx(gRz Y = 0.518
-2.293
(0.75d)
Cau 2: (3.0 diem)
Trong he toa do cue bo (local frame B), cho toa do diem //, = [-5 .5 6.6 -8 .8]? ,
thuc hien lien tuc cac phep quay -25° quay x, —^45° quay z. Sau do, thuc hien phep
tinh tien trong toa do toan cue (global frame G) d = [3 -3 -1.2]7.
a) Tim ma tran chuyen vi dong nhat °T a toa do toan cue, tim diem H 2 sau khi
thuc hien phep chuyen vi tu ma tran ° T .
b) Tim ma tran nghjch dao cua ‘T . Xac djnh ma tran xoay ° R (4x4) va ma tran
tinh tien '' D (4x4) cua ma tran nghich dao (' T . T f l T
S6 hieu: BM l/QT-PDT-RDTV/02 Lan soat xet: 02 Ngay hieu luc: 15/5/2020 Trang:2/6
Ma tran xoay quanh true x, z tai toa do cue bo (local frame)
’ l 0 0 '1 0 0
“R ,= 0 c(-2 5 ) .,(-2 5) =0 0.906 -0.423
0 -y ( -2 5 ) c (—25) 0 0.423 0.906
1
1
o
x
"0.707 -0.707 O'
11 R: = -x (-4 5 ) c(-4 5 ) 0 =0.707 0.707 0
0 0 1 0 0 1
*Neu ra dirffc 2 ma train xoay thi dirffc 0.25d
Ma tran xoay tong hop tai toa do toan cue (global frame)
~ 0.707 0.707 0
6x = ( x ) " = ( ‘« - x r = -0.641 0.641 0.423 (0.25d)
0.3 -0.3 0.906
Ma tran chuyen vi T dua tren ma tran xoay tong hop va vector tinh tien a toa do toan cue
(global frame)
=> T = C'RU °d
I
0.707
-0.641
0.3
0
0.707
0.641
-0.3
0
0
0.423
0.906
0
3
-3
-1.2 (0.5d)
Toa do H2, sau khi thuc hien phep chuyen vj
G IT
_
Grp H TT
_
ri2 — 1B —
' 0.707 0.707 0 3'-5.5" " 3.778 '
-0.641 0.641 0.423 -3 6.6 1.035
0.3 -0.3 0.906 -1.2 -8.8 -12.791
0 0 0 1 1 1
(id)
*De bai yeu cau lam phyeng phap ma tran chuyen vj T, SV lam khac phyeng phap
chi lay 30% diem.
Ma tran nghich dao T '1 la
% = {(% y - c,RrD - '=
Ma tran xoay: HR =
0.707 -0.641 0.3 -3.685
z r ' - 0.707
0
0.641
0.423
-0.3
0.906
-0.557
2.355
00 0 1
0.707 -0.641 0.3 o'
0.707
0
0.641
0.423
-0.3
0.906
0
0(0.25d)
00 0 1
(0.5d)
i a r
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