EE 4315 / EE 5325 Robotics
Homework 1 Solution
February 18, 2015
Q3:
Kinematics is the science of motion based on geometric description, regardless of the forces
which cause it. Kinematics deals with positions and its derivatives (velocity/acceleration).
Workspace is the total volume swept out by the end-effector of the manipulator. The traditional
concept is that of the “reachable” workspace, composed of 3D points. Alternatively, we can also
add the set of reachable points and orientations to define the “dextrous” workspace.
DOFs is the number of the manipulator equals the number of independent position variables that
would have to be specified in order to locate all parts of the mechanism. It equals the number of
joints in an open kinematic chain.
The number of DOF is
nn  1
. Because the number of DOF’s is given by the number of
2
independent parameters that define positions and orientations in Rn, that number is given by:


n independent parameters defining translations. DOF_T=n
nn  1
independent parameters defining rotations. We arrive at this number by
2
observing that rotations are matrices in SO(n), and therefore are composed of orthonormal row (or
column) vectors.
 r11
r
R   21


rn1
r12  r1n 
r22  r2 n 
, RR T  I n


 

rn 2  rnn 
Chose a first row vector by picking its n entries. Due to the fact that this vector has length 1, one
of the entries is dependent on the others, that is:
r11   (1   r12i ) . This constraint leaves a remaining n-1 independent entries in the first row.
i 1
Now chose a second row. One of the second row entries will also be dependent on the rest through
a length 1 vector constraint. In addition, we also impose the orthogonality condition between the
rows, therefore
r11r21  r12 r22  ...  r1n r2n  0. Since not all of ri1 are zero (otherwise the length of the row
vector will be 0!), at least one entry is nonzero. Suppose that non-zero entry is r11 . Therefore
r21  
r r
i 1
1i 2 i
r11
, and , of course, r22   (1 
independent entries in the second row.
r
i2
2
2i
) . These constraints leave n-2
Now keep adding row vectors to the matrix R. Because of the orthogonality/orthonormality
constraints, additional rows will have n-3, n-4…1 independent entries. Therefore the total number
of independent parameters necessary to reconstruct the rotation matrix R is
DOF_R= (n  1)  ( n  2)  ...  1 
Finally, DOF=DOF_R+DOF_P=
n(n  1)
.
2
n(n  1)
n(n  1)
.
n
2
2
When n  2 , DOF=3, n  3 , DOF=6.
Q4:
To prove that SO(3) is group, we must show that the matrix multiplication has the following
properties when applied to members of SO(3):
Closure:
x, y  G  x  y  G
Associativity:
x  y   z  x   y  z 
x  G , x 1 , x  x 1  I
Existence of identity and inverse:
For SO(3), the 3x3 identity matrix is obviously a member of SO(3). Also, matrix multiplication
is associative by default. Remaining to show are the closure and inverse properties.
Closure:  R1 , R2  SO(3) , we must show that R1 R2  SO(3) . Members of SO(3) have the
T
orthonormal property, therefore we must show that ( R1 R2 )( R1 R2 )  I 3 . However,
( R1 R2 )( R1 R2 ) T  ( R1 R2 )( R2T R1T )  R1 ( R2 R2T ) R1T  R1 ( I 3 ) R1T  R1 R1T  I 3
Inverse: R  SO(3), the matrix inverse R
1
1
exists since det( R )  1 and also R = R !
1
T
Since, ( R 1 ) T R 1  ( R T ) T R 1  RR 1  I , and det( R ) 1 / det( R)  1 , the inverse of R is
also in SO(3).
Q5:
Let the columns of R be denoted by 𝑟𝑖 where 𝑖 ∈ {1,2,3, … , 𝑛}. Then
𝑅 = [𝑟1 𝑟2 ⋯ 𝑟𝑛 ]
Therefore
𝑇
𝑅 𝑅 = [𝑟1 𝑟2 ⋯
𝑟𝑛 ]𝑇 [𝑟1 𝑟2
𝑟1 𝑇 𝑟1
⋯ 𝑟𝑛 ] = [ ⋮
𝑟𝑛 𝑇 𝑟1
⋯ 𝑟1 𝑇 𝑟𝑛
1 ⋯ 0
⋱
⋮ ] = [⋮ ⋱ ⋮] = 𝐼
⋯ 𝑟𝑛 𝑇 𝑟𝑛
0 ⋯ 1
By equating the elements we can see that
1
𝑟𝑖 𝑇 𝑟𝑗 = {
0
𝑖𝑓 𝑖 = 𝑗
𝑖𝑓 𝑖 ≠ 𝑗
Therefore, ‖𝑟𝑖 ‖2 = √𝑟𝑖 𝑇 𝑟𝑖 = 1. Therefore the columns are of length 1. Also 𝑟𝑖 ∙ 𝑟𝑗 = 𝑟𝑖 𝑇 𝑟𝑗 = 0
for 𝑖 ≠ 𝑗 are that the columns are mutually perpendicular.
Q6:
Applying the law of cosines to 𝐶̂ and using symmetry,
𝑑 2 = 𝑙 2 + 𝑙 2 − 2𝑙 2 𝐶𝑜𝑠(𝐶)
2
2
𝑑 = 2𝑙 − 2𝑙 2𝐶𝑜𝑠(𝜃) = 2𝑙 2(1 − 𝐶𝑜𝑠(𝜃))
𝑑 = 𝑙√2(1 − 𝐶𝑜𝑠(𝜃))
a) With 𝜃 = 90° and 𝑙 = 1𝑚,
𝑑 = 1√2(1 − 𝐶𝑜𝑠(90°)) = √2 = 1.414 𝑚
Assuming uniformity in resolution for the linear link,
Resolution =
𝑑
210
=
1.414
1024
= 0.00138𝑚 = 1.38𝑚𝑚
b) Angular resolution for 𝜃 = 90° and 8-bit encoder connected to the shaft:
Control resolution for the shaft =
𝜃
28
=
90
256
= 0.352°
(Although the question asks for the control resolution for the linear link, we will stop at the
control resolution for the shaft since the linear resolution requires more knowledge about the joint.
A proper calculation for this linear resolution with the encoder connected to the shaft is uploaded
separately).
c) The encoder can provide a feedback as to how close the actual position is to the desire position.
This information can be used drive the link in the direction that minimizes the error which in
turn increases the accuracy. This is the basic idea of feedback control.