现代机器人作业答案

A B
R(θ
,
φ
)
=
B A
R−1
(θ
,
φ
)
= [ROT (xˆ, − φ)ROT (zˆ, −θ )]−1
= ROT −1(zˆ, −θ )ROT −1(xˆ, − φ)
= ROT (zˆ, θ )ROT (xˆ, φ)
Yet another way of viewing the same operation: 1st rotate by ROT (zˆ, θ )
HITSZ S8301024C Modern Robotics Reference Solutions to Assignment #1
Q-1. A vector AP is rotated about ������! by 30 degrees and is subsequently rotated about ������! by 45 degrees. Give the rotation matrix that accomplishes these rotations in the given order.
" $
c1c23
−c1s23
s1
c1(c2l2 + l1)
% '
03T
$ =$
$
s1c23 s23
−s1s23 c23
−c1 0
s1(c2l2 + l1)
' '.
s2l2
'
$ #
0
00
1
' &
Solution: Case (1):
Case (2):
Q-9. (a) Describe a simple algorithm for choosing the nearest solution from a set of possible solutions.
Solution: (a):
(b):
Q-10. Give the 3 x 3 Jacobian that calculates linear velocity of the tool tip from the three joint rates for the manipulator in Figure Q-10. Give the Jacobian in frame {0}, which coincides with frame {1} at home position.
−0.5
0
0.866
' &
$ #
−0.353
0.707
0.612
' &
Q-2. A frame {B} is located initially coincident with a frame {A}. We rotate {B} about ������! by ������ degrees, and then we rotate the resulting frame about ������! by ������ degrees. Give the rotation matrix that will change the descriptions of vectors from BP to AP.
Solution:
Figure Q-7
Q-8. For the cases of (1) both the desired orientation and position of the tip frame with respect to the base frame are given as:
! #
(b) There exist 6-DOF robots for which the kinematics are NOT closed-form solvable. Does there exist any 3-DOF robot for which the (position) kinematics are NOT closed-form solvable?
Solution:
Figure Q-10
Solution: R = ROT (xˆ, 45)ROT ( yˆ, 30)
" $ =$
1 0
0 0.707
0 −0.707
%" '$ '$
0.866 0
0 1
0.5
% '
" $
0.866
0
0.5
% '
0 ' = $ 0.353 0.707 −0.612 '
$ #
0
0.707
0.707
'$ &#
Figure Q-5
Solution: The frames are as shown and the parameters are on the right-hand side below:
Q-6. For the two-link manipulator shown in Figure Q-6(a), the link-transformation matrices, !!������ and !!������, were constructed. Their product is
Solution: If ������! is an eigenvector of R corresponding to eigenvalue ������, then RVi = λVi
If the eigenvalue with Vi is λ = 1, then RVi = Vi
Hence the vector is not changed by the rotation R. So Vi is the axis of rotation.
2nd rotate by ROT (zˆ, θ )ROT (xˆ, φ)ROT −1(z, θ )
Composing these two rotations = ROT (zˆ,θ )ROT (xˆ, φ)ROT −1(z,θ )ROT (zˆ,θ )
= ROT (zˆ, θ )ROT (xˆ, φ)
=
! # # # # # "
L1c1 + L2c1c2 L1s1 + L2s1c2
L2s2 1
$ & & & & & %
⇒
P0 TIP
=
! # # # "#
L1c1 + L2c1c2 L1s1 + L2s1c2
L2s2
$ & & & %&
Q-7. Sketch the fingertip workspace of the three-link manipulator illustrated in Figure Q-7, for the case of l1 = 15, l2 = 10, and l3 = 3.
Solution: Since rotations are performed about axes of the frame being rotated, these are EulerAngle style rotations:
R = ROT (zˆ, θ )ROT (xˆ, φ)
We might wk.baidu.comlso use the following reasoning:
Q-4. Referring to Figure Q-4, give the value of !!������.
Figure Q-4
" $
0
0 −1
2
% '
Solution:
CAT
=
$ $
−0.5 0.866
0.866 −0.5
0 0
3*0.5 ' −3*0.866 '
$ #
0
00
1
' &
Q-5. Assign link frames to the RPR planar robot shown in Figure Q-5, and give the linkage parameters.
0
0
������!������������! ������!������������! 0
1
The link-frame assignments used are indicated in Figure Q-6(b). Note that frame {0} is
coincident with frame {1} when ������!= 0. The length of the second link is ������!. Find an expression for the vector 0Ptip , which locates the tip of the arm relative to the {0} frame.
" $
cθ
= $ sθ
−sθ cθ
0 0
%" '$ '$
1 0
0 cφ
0
% '
" $
cθ
−sφ ' = $ sθ
−sθ cφ cθ cφ
$ #
0
0
1
'$ &#
0
sφ
cφ
' &
$ #$
0
sφ
sθ sφ
% '
−cθ sφ '
cφ
' &'
Q-3. !!������ is a 3 x 3 matrix with eigenvalues 1, and ������!!", and ������!!", where ������ = −1. What is the physical meaning of the eigenvector of !!������ associated with the eigenvalue 1?
r11
r12
r13
px
$ &
03T
# =#
r21
r22
r23
py & &
# r31 r32 r33 pz &
"# 0 0 0 1 %&
and (2) only the desired position of the tool frame is given as:
3P tool
=[
l3
0
0 ]T ,
derive the inverse kinematics of the 3-link manipulator in Figure Q-1, knowing that
������������!������������! −������������!������������!
!!������ =
������������!������������! ������������!
−������������!������������! ������������!
00
������������! −������������!
Figure Q-6
Solution:
! #
c1c2
! # "#
P0 TIP 1
$!
& %&
=
02T
# "#
P2 TIP 1
$ & %&
=
# # #
s1c2 s2
# "
0
−c1s2 −s1s2
c2 0
s1 −c1 0
0
L1c1 L1s1 0
1
$%&&&&&!"#####
L2 0 0 1
$ & & & & %&