Introduction to Robotics
Analysis, Control, Applications
Solution Manual
Saeed B. Niku
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This solution manual may not be copied, posted, made available to students, placed on
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2
CHAPTER ONE
Problem 1.1
Draw the approximate workspace for the following robot. Assume the dimensions of the
base and other parts of the structure of the robot are as shown.
Estimated student time to complete: 15-25 minutes
Prerequisite knowledge required: Text Section(s) 1.14
Solution:
The workspace shown is approximate.
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3
Problem 1.2
Draw the approximate workspace for the following robot. Assume the dimensions of the
base and other parts of the structure of the robot are as shown.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 1.14
Solution:
The workspace shown is approximate.
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4
Problem 1.3
Draw the approximate workspace for the following robot. Assume the dimensions of the
base and other parts of the structure of the robot are as shown.
Estimated student time to complete: 10-15 minutes
Prerequisite knowledge required: Text Section(s) 1.14
Solution:
The workspace shown is approximate.
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5
CHAPTER TWO
Problem 2.1
Write a unit vector in matrix form that describes the direction of the cross product of
p = 5i + 3k and q = 3i + 4 j + 5k .
Estimated student time to complete: 5-10 minutes
Prerequisite knowledge required: Text Section(s) 2.4
Solution:
⎡i j k⎤
r = p × q = ⎢⎢5 0 3 ⎥⎥ = i ( 0 − 12 ) − j ( 25 − 9 ) + k ( 20 − 0 ) = −12i − 16 j + 20k
⎢⎣3 4 5 ⎥⎦
λ = rx2 + ry2 + rz2 = 144 + 256 + 400 = 28.28
⎡ −12 ⎤
⎢ 28.28 ⎥
⎢
⎥ ⎡ −0.424 ⎤
−16 ⎥ ⎢
⎢
r=
= −0.566 ⎥⎥
⎢ 28.28 ⎥ ⎢
⎢
⎥ ⎢⎣ 0.707 ⎥⎦
⎢ 20 ⎥
⎣⎢ 28.28 ⎦⎥
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6
Problem 2.2
A vector p is 8 units long and is perpendicular to vectors q and r described below.
Express the vector in matrix form.
⎡ 0.3⎤
⎡ rx ⎤
⎢q ⎥
⎢ 0.5⎥
qunit = ⎢ y ⎥
runit = ⎢ ⎥
⎢ 0.4 ⎥
⎢0.4 ⎥
⎢ ⎥
⎢ ⎥
⎣ 0 ⎦
⎣ 0 ⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.4
Solution:
The two vectors given are unit vectors. Therefore, each missing component can be found
as:
q y = 1 − 0.09 − 0.16 = 0.866
rx = 1 − 0.25 − 0.16 = 0.768
Since p is perpendicular to the other two vectors, it is in the direction of the cross product
of the two. Therefore:
j
k ⎤
⎡ i
⎢
λ p = ⎢ 0.3 0.866 0.4 ⎥⎥ = i ( 0.346 − 0.2 ) − j ( 0.12 − 0.307 ) + k ( 0.15 − 0.665 )
⎢⎣0.768 0.5 0.4 ⎥⎦
= i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 )
Since q and r are not perpendicular to each other, the resulting p is not a unit vector.
Vector p can be found as:
λ p = i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 )
λp =
( 0.146 ) + ( 0.187 ) + ( 0.515) = 0.567
2
2
2
8
= 14.1
0.567
p = w ( i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 ) )
w=
p = i ( 2.06 ) + j ( 2.64 ) − k ( 7.27 )
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7
Problem 2.3
Will the three vectors p, q, and r in Problem 2.2 form a traditional frame? If not, find the
necessary unit vector s to form a frame between p, q, and s.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.4
Solution:
As we saw in Problem 2.2, since q × r is not a unit vector, it means that q and r and not
perpendicular to each other, and therefore, they cannot form a frame. However, p and q
are perpendicular to each other, and we can select s to be perpendicular to those two. Of
course, p is not a unit length, therefore we use the unit vector representing it.
λp =
( 0.146 ) + ( 0.187 ) + ( 0.515) = 0.567
2
2
2
1
= 1.764
0.567
p = w ( i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 ) )
w=
p = i ( 0.257 ) + j ( 0.33) − k ( 0.908 )
p = i ( 0.257 ) + j ( 0.33) − k ( 0.908 )
j
k ⎤
⎡ i
⎢
s = ⎢ 0.257 0.33 −0.908⎥⎥ = i ( 0.918 ) − j ( 0.375 ) + k ( 0.124 )
⎢⎣ 0.3 0.866
0.4 ⎥⎦
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8
Problem 2.4
T
Suppose that instead of a frame, a point P = ( 3,5, 7 ) in space was translated a distance
of d = ( 2,3, 4 ) . Find the new location of the point relative to the reference frame.
T
Estimated student time to complete: 5 minutes
Prerequisite knowledge required: Text Section(s) 2.6
Solution:
As for a frame,
⎡1
⎢0
Pnew = ⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
1
0
2⎤ ⎡ 3⎤ ⎡ 5 ⎤
3 ⎥⎥ ⎢⎢ 5 ⎥⎥ ⎢⎢ 8 ⎥⎥
=
4 ⎥ ⎢ 7 ⎥ ⎢11⎥
⎥⎢ ⎥ ⎢ ⎥
1 ⎦ ⎣1 ⎦ ⎣ 1 ⎦
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9
Problem 2.5
T
The following frame B was moved a distance of d = ( 5, 2, 6 ) . Find the new location of
the frame relative to the reference frame.
⎡0
⎢1
B=⎢
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
2⎤
4 ⎥⎥
6⎥
⎥
1⎦
Estimated student time to complete: 5-10 minutes
Prerequisite knowledge required: Text Section(s) 2.6
Solution:
The transformation matrix representing the translation is used to find the new location as:
⎡1
⎢0
Bnew = ⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
1
0
5 ⎤ ⎡0
2 ⎥⎥ ⎢⎢1
6 ⎥ ⎢0
⎥⎢
1 ⎦ ⎣0
1 0
0 0
0 −1
0 0
2⎤ ⎡0
4 ⎥⎥ ⎢⎢1
=
6 ⎥ ⎢0
⎥ ⎢
1 ⎦ ⎣0
1 0 7⎤
0 0 6 ⎥⎥
0 −1 12⎥
⎥
0 0 1⎦
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10
Problem 2.6
For frame F, find the values of the missing elements and complete the matrix
representation of the frame.
⎡ ? 0 −1 5 ⎤
⎢ ? 0 0 3⎥
⎥
F =⎢
⎢ ? −1 0 2 ⎥
⎢
⎥
⎣0 0 0 1 ⎦
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.4
Solution:
⎡ nx
⎢n
F =⎢ y
⎢ nz
⎢
⎣0
0 −1
0 0
−1 0
0 0
5⎤
3 ⎥⎥
2⎥
⎥
1⎦
j k⎤
⎡i
⎢
⎥
From n × o = a
⎢ nx n y nz ⎥ = − i
⎢⎣ 0 0 −1⎥⎦
Or: i ( −n y ) − j ( −nx ) + k ( 0 ) = −i , and therefore: n y = 1, nx = 0, nz = 0
⎡0 0 −1
⎢1 0 0
F =⎢
⎢0 −1 0
⎢
⎣0 0 0
5⎤
3 ⎥⎥
2⎥
⎥
1⎦
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11
Problem 2.7
Find the values of the missing elements of frame B and complete the matrix
representation of the frame.
?
⎡0.707
⎢ ?
0
B=⎢
⎢ ?
−0.707
⎢
0
⎣ 0
0
1
0
0
2⎤
4 ⎥⎥
5⎥
⎥
1⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.4
Solution:
ox
⎡0.707
⎢ n
0
B=⎢ y
⎢ nz
−0.707
⎢
0
⎣ 0
From n × o = a
0
1
0
0
2⎤
4 ⎥⎥
5⎥
⎥
1⎦
j
⎡ i
⎢ 0.707 n
y
⎢
⎢⎣ ox
0
k ⎤
nz ⎥⎥ = j
0.707 ⎥⎦
j
k ⎤
⎡ i
⎢
Therefore: ⎢ 0.707 n y
nz ⎥⎥ = j
⎢⎣ ox
0 0.707 ⎥⎦
i ( 0.707n y ) − j ( 0.5 − nz ox ) + k ( − ny ox ) = j → ny = 0
And
From length equations: n = 1 or
0.707 2 + n y2 + nz2 = 1 → nz = ±0.707
ox2 + 0.5 = 1 → ox = ±0.707
Therefore, there are two possible acceptable solutions:
⎡0.707 0.707
⎢ 0
0
B=⎢
⎢0.707 −0.707
⎢
0
⎣ 0
0
1
0
0
2⎤
4 ⎥⎥
5⎥
⎥
1⎦
and
⎡ 0.707 −0.707
⎢ 0
0
B=⎢
⎢ −0.707 −0.707
⎢
0
⎣ 0
0
1
0
0
2⎤
4 ⎥⎥
5⎥
⎥
1⎦
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12
Problem 2.8
Derive the matrix that represents a pure rotation about the y-axis of the reference frame.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.6.2.
Solution:
From the figure:
px = pn cos θ + pa sin θ
p y = po
pz = − pn sin θ + pa cos θ
and
⎡ px ⎤ ⎡ C
⎢p ⎥ = ⎢ 0
⎢ y⎥ ⎢
⎢⎣ pz ⎥⎦ ⎢⎣ − S
S ⎤ ⎡ pn ⎤
1 0 ⎥⎥ ⎢⎢ po ⎥⎥
0 C ⎥⎦ ⎢⎣ pa ⎥⎦
0
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13
Problem 2.9
Derive the matrix that represents a pure rotation about the z-axis of the reference frame.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.6.2.
Solution:
From the Figure:
px = pn cos θ − po sin θ
p y = pn sin θ + po cos θ
pz = pa
and
⎡ p x ⎤ ⎡C
⎢ p ⎥ = ⎢S
⎢ y⎥ ⎢
⎢⎣ pz ⎥⎦ ⎢⎣ 0
−S
C
0
0 ⎤ ⎡ pn ⎤
0 ⎥⎥ ⎢⎢ po ⎥⎥
1 ⎥⎦ ⎢⎣ pa ⎥⎦
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14
Problem 2.10
Verify that the rotation matrices about the reference frame axes follow the required
constraint equations set by orthogonality and length requirements of directional unit
vectors.
Estimated student time to complete: 10 minutes.
Prerequisite knowledge required: Text Section(s) 2.4.5 and 2.6.2
Solution:
⎡1 0
For Rot ( x, θ ) = ⎢⎢0 C
⎢⎣0 S
0⎤
− S ⎥⎥ we have:
C ⎥⎦
nio = 0 or nx ox + ny oy + nz oz = 0
n ia = 0
ai o = 0
n =1
o = S 2 + C2 = 1
a=
( −S ) + C 2 = 1
2
Rot ( y, θ ) and Rot ( z,θ ) will be the same.
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15
Problem 2.11
Find the coordinates of point P (2,3, 4)T relative to the reference frame after a rotation of
45 D about the x-axis.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.6.3.
Solution:
U
0
0 ⎤ ⎡ 2⎤ ⎡ 2 ⎤
⎡ 2 ⎤ ⎡1
⎢
⎥
⎢
P = Rot ( x, 45) ⎢ 3 ⎥ = ⎢0 0.707 −0.707 ⎥⎥ ⎢⎢ 3 ⎥⎥ = ⎢⎢ −0.707 ⎥⎥
⎢⎣ 4 ⎥⎦ ⎢⎣0 0.707 0.707 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ 4.95 ⎥⎦
Note that the rotation is written in a 3 × 3 , not homogeneous form, because we are only
concerned about the rotation part.
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16
Problem 2.12
Find the coordinates of point P (3,5, 7)T relative to the reference frame after a rotation of
30 D about the z-axis.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.6.3.
Solution:
U
⎡ 3 ⎤ ⎡ 0.866 −0.5 0 ⎤ ⎡ 3 ⎤ ⎡ 0.1 ⎤
P = Rot ( z ,30) ⎢⎢ 5 ⎥⎥ = ⎢⎢ 0.5 0.866 0 ⎥⎥ ⎢⎢ 5 ⎥⎥ = ⎢⎢5.83⎥⎥
⎢⎣7 ⎥⎦ ⎢⎣ 0
0
1 ⎥⎦ ⎢⎣ 7 ⎥⎦ ⎢⎣ 7 ⎥⎦
Note that the rotation is written in a 3 × 3 form not homogeneous form, because we are
only concerned about the rotation part.
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17
Problem 2.13
Find the new location of point P(1, 2,3)T relative to the reference frame after a rotation of
30D about the z-axis followed by a rotation of 60D about the y-axis.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.6.3.
Solution:
U
P = Rot ( y, 60) Rot ( z ,30) P
⎡ 0.5
⎢ 0
U
P=⎢
⎢ −0.866
⎢
⎣ 0
0 0.866 0 ⎤ ⎡0.866 −0.5 0 0 ⎤ ⎡1 ⎤ ⎡ 2.531⎤
1
0
0 ⎥⎥ ⎢⎢ 0.5 0.866 0 0 ⎥⎥ ⎢⎢ 2 ⎥⎥ ⎢⎢ 2.232 ⎥⎥
=
0 0.5 0 ⎥ ⎢ 0
0
1 0 ⎥ ⎢ 3 ⎥ ⎢1.616 ⎥
⎥⎢
⎥⎢ ⎥ ⎢
⎥
0
0
1⎦ ⎣ 0
0
0 1 ⎦ ⎣1 ⎦ ⎣ 1 ⎦
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18
Problem 2.14
A point P in space is defined as B P = (5,3, 4)T relative to frame B which is attached to the
origin of the reference frame A and is parallel to it. Apply the following transformations
to frame B and find AP. Using the 3-D grid, plot the transformations and the result and
verify it. Also verify graphically that you would not get the same results if you apply the
transformations relative to the current frame:
• Rotate 90 D about x-axis, then
• Translate 3 units about y-axis, 6 units about z-axis, and 5 units about x-axis. Then,
• Rotate 90 D about z-axis.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.6.3.
Solution:
A
P = Rot ( z ,90)Trans (5,3, 6) Rot ( x,90) B P
⎡ 0 −1
⎢1 0
A
P=⎢
⎢0 0
⎢
⎣0 0
0 0 ⎤ ⎡1
0 0 ⎥⎥ ⎢⎢0
1 0⎥ ⎢0
⎥⎢
0 1⎦ ⎣0
0 0 5 ⎤ ⎡1
1 0 3 ⎥⎥ ⎢⎢0
0 1 6⎥ ⎢0
⎥⎢
0 0 1 ⎦ ⎣0
0⎤ ⎡5⎤ ⎡ 1 ⎤
0 −1 0 ⎥⎥ ⎢⎢ 3⎥⎥ ⎢⎢10 ⎥⎥
=
1 0 0⎥ ⎢ 4⎥ ⎢ 9 ⎥
⎥⎢ ⎥ ⎢ ⎥
0 0 1 ⎦ ⎣1 ⎦ ⎣ 1 ⎦
0
0
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19
Problem 2.15
A point P in space is defined as B P = (2,3,5)T relative to frame B which is attached to the
origin of the reference frame A and is parallel to it. Apply the following transformations
to frame B and find AP. Using the 3-D grid, plot the transformations and the result and
verify it:
D
• Rotate 90 about x-axis, then
• Rotate 90 D about local a-axis, then
• Translate 3 units about y-, 6 units about z-, and 5 units about x-axes.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.6.4.
Solution:
A
P = Trans (5,3, 6) Rot ( x,90) Rot (a,90) B P
⎡1
⎢0
A
P=⎢
⎢0
⎢
⎣0
0 0 5⎤ ⎡1
1 0 3⎥⎥ ⎢⎢0
0 1 6⎥ ⎢0
⎥⎢
0 0 1 ⎦ ⎣0
0 ⎤ ⎡ 0 −1
0 −1 0 ⎥⎥ ⎢⎢1 0
1 0 0⎥ ⎢0 0
⎥⎢
0 0 1⎦ ⎣0 0
0
0
0 0⎤ ⎡ 2⎤ ⎡ 2 ⎤
0 0 ⎥⎥ ⎢⎢ 3 ⎥⎥ ⎢⎢ −2 ⎥⎥
=
1 0⎥ ⎢5⎥ ⎢ 8 ⎥
⎥⎢ ⎥ ⎢ ⎥
0 1⎦ ⎣1⎦ ⎣ 1 ⎦
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20
Problem 2.16
A frame B is rotated 90D about the z-axis, then translated 3 and 5 units relative to the nand o-axes respectively, then rotated another 90D about the n-axis, and finally, 90D about
the y-axis. Find the new location and orientation of the frame.
⎡0
⎢1
B=⎢
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
1⎤
1⎥⎥
1⎥
⎥
1⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.6.4.
Solution:
Bnew = Rot ( y,90) Rot ( z,90) BTrans(3,5, 0) Rot (n,90)
⎡0
⎢0
Bnew = ⎢
⎢ −1
⎢
⎣0
0 1 0 ⎤ ⎡0 −1
1 0 0 ⎥⎥ ⎢⎢1 0
0 0 0⎥ ⎢0 0
⎥⎢
0 0 1⎦ ⎣0 0
0 0⎤ ⎡0
0 0 ⎥⎥ ⎢⎢1
1 0⎥ ⎢0
⎥⎢
0 1 ⎦ ⎣0
1⎤ ⎡1
0 0 1⎥⎥ ⎢⎢ 0
0 −1 1⎥ ⎢0
⎥⎢
0 0 1⎦ ⎣ 0
1
0
0 0 3⎤ ⎡1
1 0 5⎥⎥ ⎢⎢ 0
0 1 0⎥ ⎢0
⎥⎢
0 0 1⎦ ⎣0
0 ⎤ ⎡ 0 −1 0
0 −1 0⎥⎥ ⎢⎢0 0 −1
=
1 0 0 ⎥ ⎢1 0 0
⎥ ⎢
0 0 1 ⎦ ⎣0 0 0
0
0
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1⎤
6 ⎥⎥
4⎥
⎥
1⎦
21
Problem 2.17
The frame B of problem 2.16 is rotated 90D about the a-axis, 90D about the y-axis, then
translated 2 and 4 units relative to the x- and y-axes respectively, then rotated another
90D about the n-axis. Find the new location and orientation of the frame.
⎡0
⎢1
B=⎢
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
1⎤
1⎥⎥
1⎥
⎥
1⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 2.6.4.
Solution:
Bnew = Trans (2, 4, 0) Rot ( y,90) B Rot ( a,90) Rot ( n,90)
⎡1
⎢0
Bnew = ⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
1
0
2⎤ ⎡ 0
4 ⎥⎥ ⎢⎢ 0
0 ⎥ ⎢ −1
⎥⎢
1⎦ ⎣ 0
0
1
0
0
1
0
0
0
0⎤ ⎡0
0 ⎥⎥ ⎢⎢1
0⎥ ⎢0
⎥⎢
1 ⎦ ⎣0
1 0
0 0
0 −1
0 0
1⎤ ⎡ 0 −1
1⎥⎥ ⎢⎢1 0
1⎥ ⎢ 0 0
⎥⎢
1⎦ ⎣ 0 0
0
0
1
0
0⎤ ⎡1
0 ⎥⎥ ⎢⎢ 0
0⎥ ⎢0
⎥⎢
1⎦ ⎣0
0 0
0 −1
1 0
0 0
0⎤ ⎡ 0 −1
0⎥⎥ ⎢⎢ 0 0
=
0⎥ ⎢ −1 0
⎥ ⎢
1⎦ ⎣ 0 0
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0 3⎤
1 5 ⎥⎥
0 −1⎥
⎥
0 1⎦
22
Problem 2.18
Show that rotation matrices about the y- and the z-axes are unitary.
Estimated student time to complete:
Prerequisite knowledge required: Text Section(s) 2.7.
Solution:
⎡C
A = Rot ( y, θ ) = ⎢⎢ 0
⎢⎣ − S
S⎤
1 0 ⎥⎥
0 C ⎥⎦
0
det A = 1(C 2θ + S 2θ ) + 0 = 1
0
0 ⎤
⎡1
⎢
T
A = ⎢ 0 Cθ Sθ ⎥⎥
⎢⎣ 0 − Sθ Cθ ⎥⎦
adjA = A
T
minor
⎡C 0 − S ⎤
= ⎢⎢ 0 1 0 ⎥⎥ /1 = A−1
⎢⎣ S 0 C ⎥⎦
A−1 = AT
Exact same is true for rotation about the z-axis.
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23
Problem 2.19
Calculate the inverse of the following transformation matrices:
⎡ 0.527 −0.574 0.628 2 ⎤
⎢ 0.369 0.819 0.439 5 ⎥
⎥
T1 = ⎢
⎢ −0.766
0
0.643 3 ⎥
⎢
⎥
0
0
1⎦
⎣ 0
and
⎡ 0.92
⎢ 0
T2 = ⎢
⎢ −0.39
⎢
⎣ 0
0 0.39 5 ⎤
1
0
6 ⎥⎥
0 0.92 2 ⎥
⎥
0
0
1⎦
Estimated student time to complete: 15-20 minutes.
Prerequisite knowledge required: Text Section(s) 2.7.
Solution:
Since transformation matrices are unitary, we calculate the inverses simply by
transposing the rotation part and calculating the position part by:
0.527 × 2 + 0.369 × 5 − 0.766 × 3 = −0.601
−0.574 × 2 + 0.819 × 5 − 0 × 3 = −2.947
0.628 × 2 + 0.439 × 5 + 0.643 × 3 = −5.38
Therefore:
⎡ 0.527 0.369 −0.766 −0.601⎤
⎢ −0.574 0.819
0
−2.947 ⎥⎥
T1−1 = ⎢
⎢ 0.628 0.439 0.643 −5.38 ⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
And similarly,
⎡ 0.92
⎢ 0
−1
T2 = ⎢
⎢ 0.39
⎢
⎣ 0
0 −0.39 −3.82 ⎤
1
0
−6 ⎥⎥
0 0.92 −3.79 ⎥
⎥
0
0
1 ⎦
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24
Problem 2.20
Calculate the inverse of the matrix B of Problem 2.17.
Estimated student time to complete: 5 minutes
Prerequisite knowledge required: Text Section(s) 2.7.
Solution:
⎡0
⎢1
B=⎢
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
1⎤
1⎥⎥
1⎥
⎥
1⎦
⎡0
⎢1
B −1 = ⎢
⎢0
⎢
⎣0
1 0 −1⎤
0 0 −1⎥⎥
0 −1 1 ⎥
⎥
0 0 1⎦
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25
Problem 2.21
Write the correct sequence of movements that must be made in order to restore the
original orientation of the spherical coordinates and make it parallel to the reference
frame. About what axes are these rotations supposed to be?
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 2.9.3.
Solution:
⎡1
⎢0
Tsph (r , β , γ ) Rot (o, − β ) Rot (a, −γ ) = ⎢
⎢0
⎢
⎣0
0
1
0
0
0 rS β Cγ ⎤
0 rS β Sγ ⎥⎥
1 rC β ⎥
⎥
0
1 ⎦
As shown, the rotations are relative to o- and a-axes in the sequence shown. These
rotations must be relative to the current moving frame in order to prevent changing the
position of the frame.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
26
Problem 2.22
A spherical coordinate system is used to position the hand of a robot. In a certain
situation, the hand orientation of the frame is later restored in order to be parallel to the
reference frame, and the matrix representing it is described as:
⎡1 0 0 3.1375⎤
⎢ 0 1 0 2.195 ⎥
⎥
Tsph = ⎢
⎢ 0 0 1 3.214 ⎥
⎢
⎥
1 ⎦
⎣0 0 0
• Find the necessary values of r , β , γ to achieve this location.
• Find the components of the original matrix n, o, a vectors for the hand before the
orientation was restored.
Estimated student time to complete: 20 minutes
Prerequisite knowledge required: Text Section(s) 2.9.3.
Solution:
a) The equations representing position in a spherical coordinates may be set equal to the
given values as:
i. rCγ S β = 3.1375
ii. rSγ S β = 2.195
iii. rC β = 3.214
I. Assuming S β is positive,
γ = 35D
from ii and iii
from iii
II. If S β were negative, then
from i and ii
β = 50D
r=5 units.
γ = 215D
β = −50D
r=5 units.
Since orientation is not specified, no more information is available to check the results.
b) For case I, substitute corresponding values of S β , C β , Sγ , Cγ and r in spherical
coordinates to get:
⎡ 0.5265 −0.5735 0.6275 3.1375⎤
⎢ 0.3687 0.819
0.439 2.195 ⎥⎥
Tsph (r , β , γ ) = Tsph (35,50,5) = ⎢
⎢ −0.766
0
0.6428 3.214 ⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
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electronic media system or the Internet without prior expressed consent of the copyright owner.
27
Problem 2.23
Suppose that a robot is made of a Cartesian and RPY combination of joints. Find the
necessary RPY angles to achieve the following:
⎡ 0.527 −0.574 0.628 4 ⎤
⎢ 0.369 0.819 0.439 6 ⎥
⎥
T =⎢
⎢ −0.766
0
0.643 9 ⎥
⎢
⎥
0
0
1⎦
⎣ 0
Estimated student time to complete: 20 minutes
Prerequisite knowledge required: Text Section(s) 2.10.1
Solution:
The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units
along the x-, y-, and z-axes. The orientation is achieved by RPY rotations, therefore:
φa = ATAN 2(n y , nx ) = ATAN 2(0.369, 0.527) = 35D
or
φa = ATAN 2(−n y , −nx ) = ATAN 2(−0.369, −0.527) = 215D
For φa = 35D
φo = ATAN 2(− nz , (nx Cφa + n y Sφa )) = ATAN 2 ( 0.766, ( 0.527 × 0.819 + 0.369 × 0.574 ) ) = 50D
For φa = 215D
φo = ATAN 2 ( 0.766, −0.643) = 130D
For φa = 35D
φn = ATAN 2(( − a y Cφa + ax Sφa ), (o y Cφa − ox Sφa ))
= ATAN 2 ⎡⎣( −0.439 × 0.819 + 0.628 × 0.574 ) , ( 0.819 × 0.819 + 0.574 × 0.574 ) ⎤⎦
= ATAN 2 ( 0,1) = 0D
For φa = 215D
φn = ATAN 2 ( 0, −1) = 180D
⎧φa = 35D
⎪
Either solution is acceptable: ⎨φo = 50D
⎪ φ = 0D
⎩ n
⎧φa = 215D
⎪
D
⎨φo = 130
⎪φ = 180D
⎩ n
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electronic media system or the Internet without prior expressed consent of the copyright owner.
28
Problem 2.24
Suppose that a robot is made of a Cartesian and Euler combination of joints. Find the
necessary Euler angles to achieve the following:
⎡ 0.527 −0.574 0.628 4 ⎤
⎢ 0.369 0.819 0.439 6 ⎥
⎥
T =⎢
⎢ −0.766
0
0.643 9 ⎥
⎢
⎥
0
0
1⎦
⎣ 0
Estimated student time to complete: 20 minutes
Prerequisite knowledge required: Text Section(s) 2.10.2
Solution:
The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units
along the x-, y-, and z-axes. The orientation is achieved by Euler rotations, therefore:
φ = ATAN 2(a y , ax ) or φ = ATAN 2(−a y , −ax )
⎧⎪φ = ATAN 2(0.439, 0.628) = 35D
or ⎨
D
⎪⎩φ = ATAN 2(−0.439, −0.628) = 215
ψ = ATAN 2[(−nx Sφ + n y Cφ ), (−ox Sφ + o y Cφ )]
⎧⎪ψ = ATAN 2[0,1] = 0
or ⎨
⎪⎩ψ = ATAN 2[0, −1] = 180
θ = ATAN 2[(ax Cφ + a y Sφ ), az )]
D
⎪⎧θ = ATAN 2[(0.628 × 0.819 + 0.439 × 0.573), 0.643)] = 50
or ⎨
D
⎪⎩θ = ATAN 2[(0.628 × ( −0.819 ) + 0.439 × ( −0.573)), 0.643)] = −50
⎧φ = 35D
⎪
Then ⎨ ψ = 0D
D
⎪
⎩θ = 50
⎧φ = 215D
⎪
D
⎨ψ = 180
D
⎪
⎩θ = −50
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29
Problem 2.25
Assume that the three Euler angles used with a robot are 30D , 40D ,50D respectively.
Determine what angles should be used to achieve the same result if RPY is used instead.
Estimated student time to complete: 20-25 minutes
Prerequisite knowledge required: Text Section(s) 2.10
Solution:
For Euler angles:
Euler(φ ,θ ,ψ ) = Rot (a, φ ) Rot (o,θ ), Rot (a,ψ ) = Rot (a,30) Rot (o, 40), Rot (a,50)
⎡ 0.0435 −0.8296 0.5568 0 ⎤
⎢ 0.9096 0.2635 0.3215 0 ⎥
⎥
=⎢
⎢ −0.4134 0.4925 0.766 0 ⎥
⎢
⎥
0
0
1⎦
⎣ 0
φa = ATAN 2(n y , nx ) = ATAN 2(0.9096, 0.0435) = 87.26D
or
φa = ATAN 2(−n y , −nx ) = ATAN 2(−0.9096, −0.0435) = 267.26D
φo = ATAN 2(− nz , (nx Cφa + n y Sφa )) = ATAN 2 ( 0.4134, 0.9106 ) = 24.5D
or
φo = ATAN 2 ( 0.4134, −0.9106 ) = 155.5D
φn = ATAN 2((− a y Cφa + ax Sφa ), (o y Cφa − ox Sφa ))
= ATAN 2 ( 0.5408, 0.8412 ) = 32.7D
or
φn = ATAN 2(−0.5408, −0.8412) = 212.7D
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30
Problem 2.26
A frame UB was moved along its own o-axis a distance of 6 units, then rotated about its
n-axis an angle of 60D , then translated about the z-axis for 3 units, followed by a rotation
of 60D about the z-axis, and finally rotated about x-axis for 45D .
•
Calculate the total transformation performed.
•
What angles and movements would we have to make if we were to create the same
location and orientation using Cartesian and Euler configurations?
Estimated student time to complete: 25-30 minutes
Prerequisite knowledge required: Text Section(s) 2.9- 2.10
Solution:
U
TB = Rot ( x, 45) Rot ( z , 60)Trans(0, 0,3)Trans (0, 6, 0) Rot (n, 60)
0.75
−0.433
−5.196 ⎤
⎡ 0.5
⎢0.6123 −0.4355 −0.6596
0 ⎥⎥
Bnew = ⎢
⎢0.6123 0.789
0.0474 4.242 ⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
The Cartesian translations are: px = −5.196,
px = 0,
px = 4.242 . Euler angles are:
φ = ATAN 2(a y , ax ) or φ = ATAN 2(−a y , −ax )
⎧⎪φ = ATAN 2(−0.6596, 0.75) = −41.33D
or ⎨
D
⎪⎩φ = ATAN 2(0.6596, −0.75) = 138.67
ψ = ATAN 2[(− nx Sφ + n y Cφ ), (−ox Sφ + o y Cφ )]
⎧⎪ψ = ATAN 2[0.79, −0.613] = 127.8
or ⎨
⎪⎩ψ = ATAN 2[−0.79, 0.613] = −52.2
θ = ATAN 2[(ax Cφ + a y Sφ ), az )]
⎧⎪θ = ATAN 2[0.9988, 0.0474] = 87.28D
or ⎨
D
⎪⎩θ = ATAN 2[−0.9988, 0.0474] = −87.28
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electronic media system or the Internet without prior expressed consent of the copyright owner.
31
Problem 2.27
A frame UF was moved along its own n-axis a distance of 5 units, then rotated about its
o-axis an angle of 60D , followed by a rotation of 60D about the z-axis, then translated
about its a-axis for 3 units, and finally rotated 45D about the x-axis.
•
Calculate the total transformation performed.
•
What angles and movements would we have to make if we were to create the same
location and orientation using Cartesian and RPY configurations?
Estimated student time to complete: 20-25 minutes
Prerequisite knowledge required: Text Section(s) 2.11
Solution:
U
TB = Rot ( x, 45) Rot ( z, 60)Trans (5, 0, 0) Rot (o, 60)Trans (0, 0,3)
⎡ 0.25 −0.866 0.433 3.8 ⎤
⎢ 0.918 0.354 0.177 3.59 ⎥
⎥
Bnew = ⎢
⎢ −0.306 0.354 0.884 5.71⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
For Cartesian coordinates: px = 3.8,
p y = 3.59,
pz = 5.71
For RPY angles:
φa = ATAN 2(n y , nx ) = ATAN 2(0.918, 0.25) = 74.8D
or
φa = ATAN 2(−n y , − nx ) = 254.8D
φo = ATAN 2(− nz , (nx Cφa + n y Sφa )) = ATAN 2 ( 0.306, 0.951) = 17.8D
φo = ATAN 2 ( 0.306, −0.951) = 162.2D
φn = ATAN 2((− a y Cφa + ax Sφa ), (o y Cφa − ox Sφa ))
= ATAN 2 [ 0.372, 0.928] = 21.8D
φn = ATAN 2 ( −0.372, −0.928 ) = 201.8D
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32
Problem 2.28
Frames describing the base of a robot and an object are given relative to the Universe
frame.
• Find a transformation RTH of the robot configuration if the hand of the robot is to be
placed on the object.
• By inspection, show whether this robot can be a 3-axis spherical robot, and if so, find
α, β , r .
• Assuming that the robot is a 6-axis robot with Cartesian and Euler coordinates, find
px , p y , pz , φ , θ ,ψ .
⎡1
⎢0
U
Tobj = ⎢
⎢0
⎢
⎣0
0 0
0 −1
1 0
0 0
⎡ 0 −1
⎢1 0
U
TR = ⎢
⎢0 0
⎢
⎣0 0
1⎤
4 ⎥⎥
0⎥
⎥
1⎦
0 2⎤
0 −1⎥⎥
1 0⎥
⎥
0 1⎦
Estimated student time to complete: 20-25 minutes
Prerequisite knowledge required: Text Section(s) 2.11
Solution:
a. We want to place the hand on the part. Therefore, RTH = U TH−1 U Tobj
U
Tobj = U TR RTH = U TR RTobj
⎡0
⎢ −1
R
Substitute: TH = ⎢
⎢0
⎢
⎣0
1
0
0
0
0
0
1
0
TH = U TR−1 U Tobj
→
R
1 ⎤ ⎡1
2 ⎥⎥ ⎢⎢ 0
0 ⎥ ⎢0
⎥⎢
1 ⎦ ⎣0
0 0
0 −1
1 0
0 0
1⎤ ⎡ 0
4 ⎥⎥ ⎢⎢ −1
=
0⎥ ⎢ 0
⎥ ⎢
1⎦ ⎣ 0
0 −1
0 0
1 0
0 0
5⎤
1 ⎥⎥
0⎥
⎥
1⎦
b. No. The 3,2 element of spherical coordinate transformation matrix should be zero. This
is not.
c. px = 5,
p y = 1,
pz = 0
φ = ATAN 2[0, −1] = 180 or φ = ATAN 2[0,1] = 0
ψ = ATAN 2[−1(−1), 0] = 90 or ψ = ATAN 2[−1(1), 0] = 270
θ = ATAN 2[−1(−1), 0] = 90D or θ = ATAN 2[−1(1), 0] = 270D
⎧φ = 180D
⎧ φ = 0D
⎪
⎪
D
Then ⎨ψ = 90D
⎨ψ = 270
⎪ θ = 90D
⎪θ = 270D
⎩
⎩
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electronic media system or the Internet without prior expressed consent of the copyright owner.
33
Problem 2.29
A 3-DOF robot arm has been designed for applying paint on flat walls, as shown.
• Assign coordinate frames as necessary based on the D-H representation.
• Fill out the parameters table.
• Find the UTH matrix.
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified.
#
0-1
1-2
2-H
U
θ
θ1
0
90
d
l4
a
0
0
l6
l5
0
α
0
-90
0
TH = U TR RTH = U TR A1 A2 A3
⎡1
⎢0
=⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
1
0
⎡ 0 −C1
⎢0
S1
=⎢
⎢ −1 0
⎢
0
⎣0
l1 ⎤ ⎡C1
l2 ⎥⎥ ⎢⎢ S1
l3 ⎥ ⎢ 0
⎥⎢
1⎦ ⎣ 0
− S1
C1
0
0
− S1
C1
0
0
0 0 ⎤ ⎡1 0
0 0 ⎥⎥ ⎢⎢ 0 0
1 l4 ⎥ ⎢ 0 −1
⎥⎢
0 1 ⎦ ⎣0 0
− S1l6 + C1l5 + l1 ⎤
C1l6 + S1l5 + l2 ⎥⎥
⎥
l4 + l3
⎥
1
⎦
0
1
0
0
l5 ⎤ ⎡ 0 −1
0 ⎥⎥ ⎢⎢1 0
0 ⎥ ⎢0 0
⎥⎢
1 ⎦ ⎣0 0
0
0
1
0
0⎤
0 ⎥⎥
l6 ⎥
⎥
1⎦
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electronic media system or the Internet without prior expressed consent of the copyright owner.
34
Problem 2.30
In the 2-DOF robot shown, the transformation matrix 0TH is given in symbolic form, as
well as in numerical form for a specific location. The length of each link l1 and l2 is 1 ft.
Calculate the values of θ1 and θ 2 for the given location.
⎡C12
⎢S
0
TH = ⎢ 12
⎢ 0
⎢
⎣ 0
− S12
C12
0
0
0 l2C12 + l1C1 ⎤ ⎡ −0.2924 −0.9563
0 l2 S12 + l1S1 ⎥⎥ ⎢⎢ 0.9563 −0.2924
=
⎥ ⎢ 0
1
0
0
⎥ ⎢
0
1
0
⎦ ⎣ 0
0 0.6978⎤
0 0.8172 ⎥⎥
1
0 ⎥
⎥
0
1 ⎦
y
x
x1
θ2
z
θ1
xH
zH
z1
x0
z0
Figure P.2.30
Estimated student time to complete: 20 minutes
Prerequisite knowledge required: Text Section(s) 2.13
Solution:
This example is very simple, with only two degrees of freedom. Therefore, there is really
no need to use the traditional method of inverse kinematic mentioned in Section 2.13.
The joint angles can be determined as follows:
θ1 + θ 2 = ATAN 2( S12 , C12 ) = ATAN 2(0.9563, −0.2924) = 107D
⎧l2C12 + l1C1 = px
⎨
⎩ l2 S12 + l1S1 = p y
Substitute:
⎧⎪C1 = ( px − l2C12 ) / l1
→ ⎨
→ θ1 = ATAN 2 ( p y − l2 S12 ) / l1 , ( px − l2C12 ) / l1
=
−
S
p
l
S
l
/
(
)
y
2 12
1
⎪⎩ 1
(
θ1 = ATAN 2 ( p y − l2 S12 ) / l1 , ( px − l2C12 ) / l1
(
)
θ1 = ATAN 2 ( ( 0.8172 − 0.9563) , ( 0.6978 + 0.2924 ) ) = ATAN 2 ( −0.1391, 0.9902 ) = −8D
θ 2 = 107 − (−8) = 115D
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electronic media system or the Internet without prior expressed consent of the copyright owner.
)
35
Problem 2.31
For the following SCARA-type robot:
• Assign the coordinate frames based on the D-H representation.
• Fill out the parameters table.
• Write all the A matrices.
• Write the UTH matrix in terms of the A matrices.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified.
#
0-1
1-2
2-H
U
θ
θ1
θ2
θ3
α
d
0
a
d3
0
d4
180
d5
0
0
− S1 0 d3C1 ⎤ ⎡C2
C1 0 d 3 S1 ⎥⎥ ⎢⎢ S 2
0 1
0 ⎥⎢ 0
⎥⎢
0 0
1 ⎦⎣ 0
S2
d 4C2 ⎤ ⎡C3
0 d 4 S 2 ⎥⎥ ⎢⎢ S3
−1
0 ⎥⎢ 0
⎥⎢
0
1 ⎦⎣ 0
0
TH = U T0 0TH = U T0 A1 A2 A3
⎡1
⎢0
⎢
=
⎢0
⎢
⎣0
⎤ ⎡C1
1 0
0 ⎥⎥ ⎢⎢ S1
0 1 d1 + d 2 ⎥ ⎢ 0
⎥⎢
0 0
1 ⎦⎣ 0
0 0
0
−C 2
0
0
0
− S3
C3
0
0
0⎤
0 0 ⎥⎥
1 d5 ⎥
⎥
0 1⎦
0
Multiplying these matrices will yield the total transformation.
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36
Problem 2.32
A special 3-DOF spraying robot has been designed as shown:
• Assign the coordinate frames based on the D-H representation.
• Fill out the parameters table.
• Write all the A matrices.
• Write the UTH matrix in terms of the A matrices.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
Please note that no particular reset position is specified for this robot.
#
0-1
1-2
2-3
3-H
θ
α
d
0
a
0
90
θ3
l2
0
0
0
-90
90
0
l3
0
0
90 + θ1
0
0 0 l1 ⎤
⎡ − S1 0 C1 0 ⎤
⎥
⎢ C 0 S 0⎥
1 0 0⎥
1
⎥
A1 = ⎢ 1
⎢ 0 1 0 0⎥
0 1 0⎥
⎥
⎢
⎥
0 0 1⎦
⎣ 0 0 1 1⎦
⎡1 0 0 0 ⎤
⎡C3 0 S3 0 ⎤
⎡1 0 0 0 ⎤
⎢0 0 1 0 ⎥
⎢ S 0 −C 0 ⎥
⎢0 1 0 0 ⎥
3
⎥
⎥
⎥
A2 = ⎢
A3 = ⎢ 3
A4 = ⎢
⎢ 0 −1 0 l2 ⎥
⎢ 0 1 0 0⎥
⎢ 0 0 1 l3 ⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣0 0 0 1 ⎦
⎣ 0 0 0 1⎦
⎣0 0 0 1 ⎦
U
TH = U T0 0TH = U T0 A1 A2 A3 A4
The total transformation can be summarized by multiplying these matrices.
⎡1
⎢0
U
T0 = ⎢
⎢0
⎢
⎣0
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37
Problem 2.33
For the Unimation Puma 562, 6-axis robot shown,
• Assign the coordinate frames based on the D-H representation.
• Fill out the parameters table.
• Write all the A matrices.
• Find the RTH matrix for the following values:
Base height=27 in, d2=6 in, a2=15 in, a3=1 in, d4=18 in
θ1 = 0D , θ 2 = 45D ,θ3 = 0D ,θ 4 = 0D ,θ5 = −45D ,θ 6 = 0D
Estimated student time to complete: 30-40 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. To express the height from the table, the 27-in base height must be added to
the pz value. Please also note that no particular reset position is specified. Therefore, the
results relate to a position and orientation that would correspond to the given
transformations from a reset position. We assume at reset, there is 90 degrees between x0
and x1 . Please note how frames 2, 3, and 4 relate to each other by referring to the side
view of the arm.
(Reprinted with permission from Staubli Robotics.)
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38
⎡0 0 −1
⎢1 0 0
A1 = ⎢
⎢0 −1 0
⎢
⎣0 0 0
⎡1
⎢0
A4 = ⎢
⎢0
⎢
⎣0
0
0
1
0
⎡0
⎢1
0
T6 = ⎢
⎢0
⎢
⎣0
1
0
0
0
0⎤
⎡0.707 −0.707 0 10.61⎤
⎡1 0 0 0 ⎤
⎢0.707 0.707 0 10.61⎥
⎢0 0 1 0⎥
0 ⎥⎥
⎥
⎥
A2 = ⎢
A3 = ⎢
⎢ 0
⎢ 0 −1 0 0 ⎥
0⎥
0
1
6 ⎥
⎥
⎢
⎥
⎢
⎥
1⎦
0
0
1 ⎦
⎣ 0
⎣0 0 0 1⎦
0 0⎤
⎡1 0 0 0 ⎤
⎡ 0.707 0 0.707 0 ⎤
⎢0 1 0 0 ⎥
⎥
⎢
⎥
−1 0 ⎥
−0.707 0 0.707 0 ⎥
⎢
⎥
⎢
A6 =
A5 =
⎢0 0 1 0⎥
⎢ 0
−1
0 18⎥
0
0⎥
⎥
⎢
⎥
⎥
⎢
0 1⎦
0
0
1⎦
⎣0 0 0 1 ⎦
⎣ 0
−6 ⎤
0
0 −1.41⎥⎥
−1 −24 ⎥
⎥
0
1 ⎦
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39
Problem 2.34
For the given 4-DOF robot:
• Assign appropriate frames for the Denavit-Hartenberg representation.
• Fill out the parameters table.
• Write an equation in terms of A-matrices that shows how U TH can be calculated.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified.
⎡1
⎢0
U
U
U
0
TH = T0 TH = T0 A1 A2 A3 A4 = ⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
1
0
l1 ⎤
l2 ⎥⎥
A1 A2 A3 A4
l3 ⎥
⎥
1⎦
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40
Problem 2.35
For the given 4-DOF robot designed for a specific operation:
•
•
•
Assign appropriate frames for the Denavit-Hartenberg representation.
Fill out the parameters table.
Write an equation in terms of A-matrices that shows how U TH can be calculated.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified.
* L = ( l4 + l5 ) / sin β
The total transformation for the robot can be determined from:
⎡1 0 0 l1 ⎤
⎢0 1 0 0 ⎥
U
U
U
0
⎥AA A A
TH = T0 TH = T0 A1 A2 A3 A4 = ⎢
⎢0 0 1 0 ⎥ 1 2 3 4
⎢
⎥
⎣0 0 0 1 ⎦
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41
Problem 2.36
For the given specialty-designed 4-DOF robot:
• Assign appropriate frames for the Denavit-Hartenberg representation.
• Fill out the parameters table.
• Write an equation in terms of A-matrices that shows how U TH can be calculated.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified.
⎡1
⎢0
U
U
U
0
TH = T0 TH = T0 A1 A2 A3 = ⎢
⎢0
⎢
⎣0
0
1
0
0
0
0
1
0
l1 ⎤
l2 ⎥⎥
A1 A2 A3
0⎥
⎥
1⎦
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electronic media system or the Internet without prior expressed consent of the copyright owner.
42
Problem 2.37
For the given 3-DOF robot:
• Assign appropriate frames for the Denavit-Hartenberg representation.
• Fill out the parameters table.
• Write an equation in terms of A-matrices that shows how U TH can be calculated.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified. Please also note
that there can be an additional frame between frames 1 and H. In that case, the two joint
variables will be represented by two separate matrices.
#
0-1
θ
−90 + θ1
d
−l3 − l4
a
0
α
-90
1-H
θ2
l5 + l6 + l7
0
0
⎡1
⎢0
U
TH = U T0 0TH = U T0 A1 A2 = ⎢
⎢0
⎢
⎣0
0
1
0
0
0 l1 ⎤
0 −l2 ⎥⎥
A1 A2
1 0 ⎥
⎥
0 1 ⎦
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electronic media system or the Internet without prior expressed consent of the copyright owner.
43
Problem 2.38
For the given 4-DOF robot:
• Assign appropriate frames for the Denavit-Hartenberg representation.
• Fill out the parameters table.
• Write an equation in terms of A matrices that shows how U TH can be calculated.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 2.12
Solution:
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore,
intermediate matrices might be different for each case. However, the final answer should
be the same. Please also note that no particular reset position is specified.
#
0-1
θ
θ1
d
l3
1-2
90 + θ 2
2-H
θ3
α
90
−l5
a
l4
0
l6
0
0
0
1
0
0
0 l1 ⎤
0 l2 ⎥⎥
A1 A2 A3
1 0⎥
⎥
0 1⎦
⎡1
⎢0
U
U
U
0
TH = T0 TH = T0 A1 A2 A3 = ⎢
⎢0
⎢
⎣0
90
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44
Problem 2.39
Derive the inverse kinematic equations for the robot of Problem 2.36.
Estimated student time to complete: 20-30 minutes if Problem 2.36 is already solved
Prerequisite knowledge required: Text Section(s) 2.13
Solution:
We assume all positions are made relative to the base of the robot, and therefore, U T0 is
not included in the solution. Using the Denavit-Hartenberg representation and the joint
parameters shown, we get:
⎡C1
⎢S
A1 = ⎢ 1
⎢0
⎢
⎣0
1
0
0
0
0 ⎤
0 ⎥⎥
l3 + l4 ⎥
⎥
1 ⎦
TH = A1 A2 A3
→
A1−1 0TH = A2 A3
0
S1
0 −C1
0
⎡C1
⎢0
⎢
⎢ S1
⎢
⎣0
S1
0
−C1
0
⎤ ⎡ nx
1 −l3 − l4 ⎥⎥ ⎢⎢ n y
0
0 ⎥ ⎢ nz
⎥⎢
0
1 ⎦⎣ 0
0
0
⎡ C2
⎢S
A2 = ⎢ 2
⎢0
⎢
⎣0
ox
ax
oy
ay
oz
az
0
0
0
− S2
0
C2
−1
0
0
0
p x ⎤ ⎡ C2
p y ⎥⎥ ⎢⎢ S 2
=
pz ⎥ ⎢ 0
⎥ ⎢
1⎦ ⎣0
⎡ −C3
⎢ −S
A3 = ⎢ 3
⎢ 0
⎢
⎣ 0
l5C2 ⎤
l5 S 2 ⎥⎥
0 ⎥
⎥
1 ⎦
0
− S2
0
C2
−1
0
0
0
S3
−C3
0
0
l5C2 ⎤ ⎡ −C3
l5 S 2 ⎥⎥ ⎢⎢ − S3
0 ⎥⎢ 0
⎥⎢
1 ⎦⎣ 0
0 ⎤
0
0 ⎥⎥
1 l6 + l7 ⎥
⎥
0
1 ⎦
0
S3
−C3
0
0
0 ⎤
0
0 ⎥⎥
1 l6 + l7 ⎥
⎥
0
1 ⎦
0
Multiplying these matrices we get:
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45
⎡C1nx + S1n y
⎢
nz
LHS = ⎢
⎢ S1nx − C1n y
⎢
0
⎣
⎡ −C2C3
⎢ −S C
RHS = ⎢ 2 3
⎢ S3
⎢
⎣ 0
C1ox + S1o y
oz
S1ox − C1o y
C1ax + S1a y
az
S1ax − C1a y
0
0
− S2
C2
0
0
− S 2 (l6 + l7 ) + l5C2 ⎤
C2 (l6 + l7 ) + l5 S2 ⎥⎥
⎥
0
⎥
1
⎦
C2 S 3
S 2 S3
C3
0
C1 px + S1 p y ⎤
pz − l3 − l4 ⎥⎥
S1 px − C1 p y ⎥
⎥
1
⎦
⎛a ⎞
S1ax − C1a y = 0 → θ1 = tan −1 ⎜ y ⎟
⎝ ax ⎠
From 3,3 or 3,4 elements we get:
⎛p ⎞
S1 px − C1 p y = 0 → θ1 = tan −1 ⎜ y ⎟
⎝ px ⎠
From 1,3 and 2,3 elements we
⎧ S 2 = −(C1ax + S1a y )
get: ⎨
→ θ 2 = ATAN 2 ( −C1ax − S1a y , az )
⎩C2 = az
From 2,1 and 2,2 elements we get:
⎧⎪ S3 = S1nx − C1n y
⎨
⎪⎩C3 = S1ox − C1o y
→ θ3 = ATAN 2 ( S1nx − C1n y , S1ox − C1o y )
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46
Problem 2.40
Derive the inverse kinematic equations for the robot of Problem 2.37.
Estimated student time to complete: 20-30 minutes if Problem 2.37 is already solved
Prerequisite knowledge required: Text Section(s) 2.13
Solution:
We assume all positions are made relative to the base of the robot, and therefore, U T0 is
not included in the solution. Using the Denavit-Hartenberg representation and the joint
parameters shown, we get:
⎡ S1
⎢ −C
A1 = ⎢ 1
⎢ 0
⎢
⎣ 0
0 C1
0 S1
−1 0
0 0
#
0-1
θ
−90 + θ1
d
−l3 − l4
a
0
α
-90
1-H
θ2
l5 + l6 + l7
0
0
0 ⎤
0 ⎥⎥
−l3 − l4 ⎥
⎥
1 ⎦
⎡ C2
⎢S
A2 = ⎢ 2
⎢0
⎢
⎣0
−S2
C2
0
0
0
0
⎤
⎥
0
0
⎥
1 l6 + l6 + l7 ⎥
⎥
0
1
⎦
TH = A1 A2
0
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47
⎡ nx
⎢n
⎢ y
⎢ nz
⎢
⎣0
ox
ax
oy
ay
oz
az
0
0
px ⎤ ⎡ S1
0 C1
0 ⎤ ⎡ C2 − S 2 0
0
⎤
⎥
⎢
⎥
⎢
⎥
p y ⎥ ⎢ −C1 0 S1
0 ⎥ ⎢ S 2 C2 0
0
⎥
=
pz ⎥ ⎢ 0 −1 0 −l3 − l4 ⎥ ⎢ 0
0 1 l5 + l6 + l7 ⎥
⎥ ⎢
⎥⎢
⎥
1⎦ ⎣ 0
0 0
1 ⎦⎣ 0
0 0
1
⎦
⎡ S1C2 − S1S 2 C1 C1 (l5 + l6 + l7 ) ⎤
⎢ −C C
S 2C1 S1 S1 (l5 + l6 + l7 ) ⎥⎥
1 2
⎢
=
⎢ − S2
0
−C2
−l3 − l4 ⎥
⎢
⎥
0
0
1
⎣ 0
⎦
⎧⎪ S1 = a y
From 1,3 and 2,3 elements we get: ⎨
⎪⎩C1 = ax
→ θ1 = ATAN 2 ( a y , ax )
⎧ S = − nz
From 3,1 and 3,2 elements we get: ⎨ 2
⎩C2 = −oz
→ θ 2 = ATAN 2 ( − nz , −oz )
From 1,4 elements we get: C1 (l5 + l6 + l7 ) = px
→ l5 + l6 + l7 =
px
C1
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48
CHAPTER THREE
Problem 3.1
Suppose the location and orientation of a hand frame is expressed by the following
matrix. What is the effect of a differential rotation of 0.15 radians about the z-axis,
followed by a differential translation of [0.1, 0.1, 0.3]? Find the new location of the hand.
⎡0
⎢1
R
TH = ⎢
⎢0
⎢
⎣0
0
0
1
0
1
0
0
0
2⎤
7 ⎥⎥
5⎥
⎥
1⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 3.6, 3.7.
Solution:
For δ z = 0.15, dx = 0.1, dy = 0.1, dz = 0.3, we get:
−0.15
⎡ 0
⎢0.15
0
Δ=⎢
⎢ 0
0
⎢
0
⎣ 0
0 0.1⎤
0 0.1⎥⎥
0 0.3⎥
⎥
0 0⎦
−0.15
⎡ 0
⎢0.15
0
d ⎡⎣ RTH ⎤⎦ = [ Δ ] ⎡⎣ RTH ⎤⎦ = ⎢
⎢ 0
0
⎢
0
⎣ 0
0 0.1⎤ ⎡0
0 0.1⎥⎥ ⎢⎢1
0 0.3⎥ ⎢0
⎥⎢
0 0 ⎦ ⎣0
⎡ −0.15
⎢ 1
⎡⎣ RTH ⎤⎦ = ⎡⎣ RTH ⎤⎦ + d ⎡⎣ RTH ⎤⎦ = ⎢
new
old
⎢ 0
⎢
⎣ 0
0 1 2 ⎤ ⎡ −0.15
0 0 7 ⎥⎥ ⎢⎢ 0
=
1 0 5⎥ ⎢ 0
⎥ ⎢
0 0 1⎦ ⎣ 0
−0.95⎤
0 0.15 0.4 ⎥⎥
0
0
0.3 ⎥
⎥
0
0
0 ⎦
0
0
1.05⎤
0 0.15 7.4 ⎥⎥
1
0
5.3 ⎥
⎥
0
0
1 ⎦
0
1
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49
Problem 3.2
As a result of applying a set of differential motions to frame T shown, it has changed an
amount dT as shown. Find the magnitude of the differential changes made
( dx, dy, dz , δ x, δ y, δ z ) and the differential operator with respect to frame T.
⎡1 0
⎢0 0
T =⎢
⎢ 0 −1
⎢
⎣0 0
0
1
0
0
−0.1 −0.1 0.6 ⎤
⎡ 0
⎢ 0.1
0
0
0.5 ⎥⎥
dT = ⎢
⎢ −0.1 0
0
−0.5⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
5⎤
3⎥⎥
8⎥
⎥
1⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.8.
Solution:
[ dT ] = [ Δ ][T ] → [ Δ ] = [ dT ][T ]
−1
−0.1 −0.1 0.6 ⎤ ⎡1
⎡ 0
⎢ 0.1
0
0
0.5 ⎥⎥ ⎢⎢ 0
⎢
[Δ] = ⎢
0
−0.1 0
−0.5⎥ ⎢ 0
⎢
⎥⎢
0
0
0 ⎦ ⎣0
⎣ 0
0 0 −5⎤ ⎡ 0
−0.1 0.1 0.1⎤
⎥
⎢
0 −1 8 ⎥ ⎢ 0.1
0
0
0 ⎥⎥
=
1 0 −3⎥ ⎢ −0.1 0
0
0⎥
⎥ ⎢
⎥
0 0 1⎦ ⎣ 0
0
0
0⎦
By inspection: d = [0.1,0,0], δ = [0,0.1,0.1].
[ dT ] = [T ] ⎡⎣ T Δ ⎤⎦
⎡1
⎢0
⎡⎣ T Δ ⎤⎦ = ⎢
⎢0
⎢
⎣0
0
0
1
0
→ ⎡⎣ T Δ ⎤⎦ = [T ] [ dT ]
0 −5⎤ ⎡ 0
−0.1 −0.1 0.6 ⎤ ⎡ 0 −0.1 −0.1 0.6 ⎤
⎥
⎢
0
0
0.5 ⎥⎥ ⎢⎢ 0.1 0
0
0.5⎥⎥
−1 8 ⎥ ⎢ 0.1
=
−0.5⎥ ⎢ 0.1 0
0 −3⎥ ⎢ −0.1 0
0
0
0.5⎥
⎥⎢
⎥ ⎢
⎥
0 1 ⎦⎣ 0
0
0
0 ⎦ ⎣0
0
0
0 ⎦
−1
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50
Problem 3.3
Suppose the following frame was subjected to a differential translation of d = [1 0 0.5]
units and a differential rotation of δ = [0 0.1 0] .
a.
What is the differential operator relative to the reference frame?
b. What is the differential operator relative to the frame A?
⎡0
⎢1
A=⎢
⎢0
⎢
⎣0
0
0
1
0
1 10 ⎤
0 5 ⎥⎥
0 0⎥
⎥
0 1⎦
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.8
Solution:
a.
⎡ 0
⎢ 0
Δ=⎢
⎢ −0.1
⎢
⎣ 0
0 0.1 1 ⎤
0 0
0 ⎥⎥
0 0 0.5⎥
⎥
0 0
0⎦
b.
⎡0
⎢0
−1
⎡⎣ A Δ ⎤⎦ = [ A] [ Δ ][ A] = ⎢
⎢1
⎢
⎣0
−5 ⎤ ⎡ 0
0 1 0 ⎥⎥ ⎢⎢ 0
0 0 −10 ⎥ ⎢ −0.1
⎥⎢
0 0 1 ⎦⎣ 0
1 0
0 0.1
0
0
0
0
0
0
1 ⎤ ⎡0
0 ⎥⎥ ⎢⎢1
0.5⎥ ⎢0
⎥⎢
0 ⎦ ⎣0
0 1 10 ⎤
0 0 5 ⎥⎥
1 0 0⎥
⎥
0 0 1⎦
0
0 ⎤
⎡0 0
⎢0 0 −0.1 −0.5⎥
⎥
=⎢
⎢0 0.1 0
1 ⎥
⎢
⎥
0
0 ⎦
⎣0 0
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51
Problem 3.4
The initial location and orientation of a robot’s hand are given by T1, and its new location
and orientation after a change are given by T2.
a.
Find a transformation matrix Q that will accomplish this transform (in the Universe
frame).
b. Assuming the change is small, find a differential operator Δ that will do the same.
c.
By inspection, find a differential translation and a differential rotation that
constitute this operator.
⎡1
⎢0
T1 = ⎢
⎢0
⎢
⎣0
0 0
0 −1
1 0
0 0
5⎤
3⎥⎥
6⎥
⎥
1⎦
⎡1
⎢0.1
T2 = ⎢
⎢0
⎢
⎣0
0 0.1 4.8⎤
0 −1 3.5 ⎥⎥
1 0 6.2 ⎥
⎥
0 0
1 ⎦
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.8
Solution:
a.
T2 = QT1
⎡1
⎢0.1
→ Q = T2T1−1 = ⎢
⎢0
⎢
⎣0
0 0.1 4.8⎤ ⎡1 0
0 −1 3.5 ⎥⎥ ⎢⎢0 0
1 0 6.2 ⎥ ⎢0 −1
⎥⎢
0 0
1 ⎦ ⎣0 0
0 −5⎤ ⎡ 1 −0.1
1 −6 ⎥⎥ ⎢⎢ 0.1 1
=
0 3⎥ ⎢0
0
⎥ ⎢
0 1⎦ ⎣0
0
0 0.1⎤
0 0 ⎥⎥
1 0.2 ⎥
⎥
0 1 ⎦
b.
T2 = T1 + dT = T1 + ΔT1 = ( Δ + I ) T1 = QT1
Therefore: Δ = Q − I
⎡ 0 −0.1 0 0.1⎤
⎢ 0.1 0
0 0 ⎥⎥
⎢
or Δ =
⎢0
0
0 0.2 ⎥
⎢
⎥
0
0 0 ⎦
⎣0
c.
By inspection: d = [0.1, 0, 0.2] and δ = [0, 0, 0.1].
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52
Problem 3.5
The hand frame of a robot and the corresponding Jacobian are given. For the given
differential changes of the joints, compute the change in the hand frame, its new location,
and corresponding Δ .
⎡0
⎢1
T6 = ⎢
⎢0
⎢
⎣0
1 0 10 ⎤
0 0 5 ⎥⎥
0 −1 0 ⎥
⎥
0 0 1⎦
⎡8 0 0 0 0
⎢ −3 0 1 0 0
⎢
⎢
0 10 0 0 0
T6
J =⎢
⎢0 1 0 0 1
⎢0 0 0 1 0
⎢
⎣ −1 0 0 0 0
0⎤
0 ⎥⎥
0⎥
⎥
0⎥
0⎥
⎥
1⎦
⎡ 0 ⎤
⎢ 0.1 ⎥
⎢
⎥
⎢ −0.1⎥
Dθ = ⎢
⎥
⎢ 0.2 ⎥
⎢ 0.2 ⎥
⎢
⎥
⎣ 0 ⎦
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
Solution:
⎡8 0 0 0 0
⎢ −3 0 1 0 0
⎢
⎢ 0 10 0 0 0
⎡⎣ T6 D ⎦⎤ = ⎣⎡ T6 J ⎦⎤ [ Dθ ] = ⎢
⎢0 1 0 0 1
⎢0 0 0 1 0
⎢
⎣ −1 0 0 0 0
0 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎤ ⎡ T6 dx ⎤
⎢
⎥
0 ⎥⎥ ⎢⎢ 0.1 ⎥⎥ ⎢⎢ −0.1⎥⎥ ⎢ T6 dy ⎥
0 ⎥ ⎢ −0.1⎥ ⎢ 1 ⎥ ⎢ T6 dz ⎥
⎥
⎥⎢
⎥=⎢
⎥=⎢
0 ⎥ ⎢ 0.2 ⎥ ⎢ 0.3 ⎥ ⎢ T6δ x ⎥
0 ⎥ ⎢ 0.2 ⎥ ⎢ 0.2 ⎥ ⎢ T6δ y ⎥
⎥
⎥⎢
⎥ ⎢
⎥ ⎢
1 ⎦ ⎣ 0 ⎦ ⎣ 0 ⎦ ⎢⎣ T6δ z ⎥⎦
0
0.2
0 ⎤
⎡ 0
⎢ 0
0 −0.3 −0.1⎥⎥
T6
T6
⎢
Substitute in Δ to get: Δ =
⎢ −0.2 0.3
0
1 ⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
Then
0
0.2
0 ⎤ ⎡ 0
0
−0.3 −0.1⎤
⎡ 0 1 0 10 ⎤ ⎡ 0
⎢1 0 0 5 ⎥ ⎢ 0
0 −0.3 −0.1⎥⎥ ⎢⎢ 0
0
0.2
0 ⎥⎥
⎥⎢
=
dT6 = [T6 ] ⎡⎣ T6 Δ ⎤⎦ = ⎢
⎢ 0 0 −1 0 ⎥ ⎢ −0.2 0.3
−1 ⎥
0
1 ⎥ ⎢0.2 −0.3
0
⎢
⎥⎢
⎥ ⎢
⎥
0
0
0 ⎦ ⎣ 0
0
0
0 ⎦
⎣0 0 0 1 ⎦ ⎣ 0
1
−0.3 9.9 ⎤
⎡0
⎢1
0
0.2
5 ⎥⎥
[T6 ]new = [T6 ]old + [ dT6 ] = ⎢⎢
0.2 −0.3 −1 −1 ⎥
⎢
⎥
0
0
1 ⎦
⎣0
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53
0
0.2
0 ⎤ ⎡0
⎡0 1 0 10 ⎤ ⎡ 0
⎢1 0 0 5 ⎥ ⎢ 0
0 −0.3 −0.1⎥⎥ ⎢⎢1
−1
⎥⎢
Δ = [T6 ] ⎡⎣ T6 Δ ⎤⎦ [T6 ] = ⎢
⎢0 0 −1 0 ⎥ ⎢ −0.2 0.3
0
1 ⎥ ⎢0
⎢
⎥⎢
⎥⎢
0
0
0 ⎦ ⎣0
⎣0 0 0 1 ⎦ ⎣ 0
0
0.3 −0.1⎤
⎡ 0
⎢ 0
0 −0.2
0 ⎥⎥
⎢
=
⎢ −0.3 0.2
0
1 ⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
−5 ⎤
0 0 −10 ⎥⎥
0 −1 0 ⎥
⎥
0 0
1 ⎦
1
0
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54
Problem 3.6
Two consecutive frames describe the old (T1) and new (T2) positions and orientations of
the end of a 3-DOF robot. The corresponding Jacobian relative to T1, relating to
T1
dz , T1δ x, T1δ z is also given. Find values of joint differential motions ds1 , dθ 2 , dθ 3 of the
robot that caused the given frame change.
0.01
1
8.1⎤
⎡0 0 1 8 ⎤
⎡ 0
⎡5 10 0 ⎤
⎢1 0 0 5 ⎥
⎢ 1
⎥
−
0.05
0
5
T
1
⎥
⎥
T1 = ⎢
T2 = ⎢
J = ⎢⎢ 3 0 0 ⎥⎥
⎢0 1 0 2 ⎥
⎢0.05
−0.01 2 ⎥
1
⎢⎣0 1 1 ⎥⎦
⎢
⎥
⎢
⎥
0
0
1⎦
⎣0 0 0 1 ⎦
⎣ 0
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
Solution:
0.01
0
0.1⎤
⎡ 0
⎢ 0
0
0 ⎥⎥
−0.05
⎢
T2 − T1 = dT =
= [ Δ ][T1 ] = [T1 ] ⎡⎣ T1 Δ ⎤⎦
⎢0.05
0
−0.01 0 ⎥
⎢
⎥
0
0
0⎦
⎣ 0
0.01
0
0.1⎤ ⎡ 0
0
0⎤
−0.05
⎡0 1 0 −5⎤ ⎡ 0
⎢ 0 0 1 −2 ⎥ ⎢ 0
⎥
⎢
0
0 ⎥ ⎢ 0.05
0
−0.05
−0.01 0 ⎥⎥
−1
T1
⎢
⎥
⎢
=
⎣⎡ Δ ⎦⎤ = [T1 ] dT = ⎢1 0 0 −8⎥ ⎢ 0.05
0
0.01
0
0.1⎥
−0.01 0 ⎥ ⎢ 0
⎥
⎢
⎥⎢
⎥ ⎢
0
0
0⎦ ⎣ 0
0
0
0⎦
⎣0 0 0 1 ⎦ ⎣ 0
⎧ T1δ z = 0.05
⎪
Therefore, ⎨ T1δ x = 0.01
⎪ T1
⎩ d z = 0.1
→
And: ⎡⎣ T1 D ⎤⎦ = ⎡⎣ T1 J ⎤⎦ [ Dθ ] →
T1
⎡ 0.1 ⎤
D = ⎢⎢ 0.01⎥⎥
⎢⎣ 0.05⎥⎦
−1
[ Dθ ] = ⎡⎣ T J ⎤⎦ ⎡⎣ T D ⎤⎦
1
1
Please see Appendix A for methods to calculate the inverse of a non-unitary matrix.
Substitute to get:
−1
⎡ 5 10 0 ⎤ ⎡ 0.1 ⎤ ⎡ 0.0033 ⎤
−1 T
T1
[ Dθ ] = ⎡⎣ J ⎤⎦ ⎡⎣ 1 D ⎤⎦ = ⎢⎢3 0 0⎥⎥ ⎢⎢ 0.01⎥⎥ = ⎢⎢0.00833⎥⎥
⎢⎣ 0 1 1 ⎥⎦ ⎢⎣ 0.05⎥⎦ ⎢⎣ 0.0417 ⎥⎦
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electronic media system or the Internet without prior expressed consent of the copyright owner.
55
Problem 3.7
Two consecutive frames describe the old (T1) and new (T2) positions and orientations of
the end of a 3-DOF robot. The corresponding Jacobian, relating to dz , δ x, δ z is also
given. Find values of joint differential motions ds1 , dθ 2 , dθ 3 of the robot that caused the
given frame change.
0
1
9.75⎤
⎡0 0 1 10 ⎤
⎡ −0.05
⎢1 0 0 5 ⎥
⎢ 1
−0.1 0.05 5.2 ⎥⎥
⎢
⎥
⎢
T1 =
T2 =
⎢0 1 0 3 ⎥
⎢ 0.1
1
0
3.7 ⎥
⎢
⎥
⎢
⎥
0
0
1 ⎦
⎣0 0 0 1 ⎦
⎣ 0
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
⎡5 10 0 ⎤
J = ⎢⎢ 3 0 0 ⎥⎥
⎢⎣0 1 1 ⎥⎦
Solution:
0
0
−0.25⎤
⎡ −0.05
⎢ 0
−0.1 0.05 0.2 ⎥⎥
⎢
T2 − T1 = dT =
⎢ 0.1
0
0
0.7 ⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
[ dT ] = [ Δ ][T1 ]
0
0
−0.25⎤ ⎡0
⎡ −0.05
⎢ 0
−0.1 0.05 0.2 ⎥⎥ ⎢⎢0
−1
⎢
→ [ Δ ] = [ dT ][T1 ] =
⎢ 0.1
0
0
0.7 ⎥ ⎢1
⎢
⎥⎢
0
0
0 ⎦ ⎣0
⎣ 0
−5 ⎤
0 1 −3 ⎥⎥
0 0 −10 ⎥
⎥
0 0 1 ⎦
1 0
0
0 ⎤
−0.05
⎡ 0
⎢0.05
0
−0.1 0 ⎥⎥
=⎢
⎢ 0
0.1
0
0.2 ⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
⎧d z = 0.2
⎪
Therefore, ⎨δ x = 0.1
⎪δ = 0.05
⎩ z
⎡ 0.2 ⎤
→ D = ⎢⎢ 0.1 ⎥⎥
⎢⎣ 0.05⎥⎦
−1
⎡ 5 10 0 ⎤ ⎡ 0.2 ⎤ ⎡ 0.0334 ⎤
−1
Substitute to get: [ Dθ ] = [ J ] [ D ] = ⎢⎢ 3 0 0 ⎥⎥ ⎢⎢ 0.1 ⎥⎥ = ⎢⎢0.00334 ⎥⎥
⎢⎣ 0 1 1 ⎥⎦ ⎢⎣0.05⎥⎦ ⎢⎣ 0.0467 ⎥⎦
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56
Problem 3.8
A camera is attached to the hand frame T of a robot as given. The corresponding inverse
Jacobian of the robot at this location is also given. The robot makes a differential motion,
as a result of which, the change in the frame dT is recorded as given.
a.
Find the new location of the camera after the differential motion.
b. Find the differential operator.
c.
Find the joint differential motion values associated with this move.
d. Find how much the differential motions of the hand-frame ( T D ) should have been
instead, if measured relative to frame T, to move the robot to the same new location
as in part a.
0
0 0 0 0⎤
⎡1
⎢2
0
−1 0 0 0 ⎥⎥
0
−0.1 0.79 ⎤
⎡ −0.03
⎡0 1 0 3⎤
⎢
⎢
⎢1 0 0 2 ⎥
⎢ 0 −0.2 0 0 0 0 ⎥
0
0.03
0
0.09 ⎥⎥
−1
⎢
⎢
⎥
T=
J =⎢
⎥ dT =
⎢ 0
⎢ 0 0 −1 8 ⎥
−0.1 0
−0.4 ⎥
⎢ 0 −1 0 0 1 0 ⎥
⎢
⎥
⎢
⎥
⎢0
0
0 1 0 0⎥
0
0
0 ⎦
⎣0 0 0 1 ⎦
⎣ 0
⎢
⎥
0
0 0 0 1⎦
⎣1
Estimated student time to complete: 30-40 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
Solution:
−0.1 3.79 ⎤
⎡ −0.03 1
⎢ 1
0.03
0
2.09 ⎥⎥
a. Tnew = Told + dT = ⎢
⎢ 0
−0.1 −1 7.6 ⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
b.
−0.1 3.79 ⎤ ⎡ 0 1 0 −2 ⎤ ⎡ 0
−0.03 0.1 0.05⎤
⎡ −0.03 1
⎢ 1
⎥
⎢
⎥
⎢
0.03
0
2.09 ⎥ ⎢1 0 0 −3⎥ ⎢ 0.03
0
0
0 ⎥⎥
Δ = dT ⋅ T −1 = ⎢
=
⎢ 0
−0.1 −1 7.6 ⎥ ⎢ 0 0 −1 8 ⎥ ⎢ −0.1
0
0 −0.1⎥
⎢
⎥⎢
⎥ ⎢
⎥
0
0
1 ⎦ ⎣0 0 0 1 ⎦ ⎣ 0
0
0
0 ⎦
⎣ 0
c.
D = [ 0.05 0 −0.1 0 0.1 0.03]
0
0
⎡1
⎢2
−1
0
⎢
⎢ 0 −0.2 0
Therefore: Dθ = J −1 ⋅ D = ⎢
⎢ 0 −1 0
⎢0
0
0
⎢
0
0
⎣1
T
0
0
0
0
1
0
0
0
0
1
0
0
0 ⎤ ⎡ 0.05⎤ ⎡ 0.05⎤
0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0.2 ⎥⎥
0 ⎥ ⎢ −0.1⎥ ⎢ 0 ⎥
⎥⎢
⎥=⎢
⎥
0 ⎥ ⎢ 0 ⎥ ⎢ 0.1 ⎥
0 ⎥ ⎢ 0.1 ⎥ ⎢ 0 ⎥
⎥⎢
⎥ ⎢
⎥
1 ⎦ ⎣ 0.03⎦ ⎣ 0.08⎦
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57
d.
⎡0
⎢1
T
T
−1
dT = T ⋅ Δ →
Δ = T ⋅ dT = ⎢
⎢0
⎢
⎣0
0.03
0
0.09 ⎤
⎡ 0
⎢ −0.03
0
−0.1 0.79 ⎥⎥
⎢
=
⎢ 0
0.1
0
0.4 ⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
T
−2 ⎤ ⎡ −0.03
−0.1 0.79 ⎤
0
⎥
⎢
0 0 −3⎥ ⎢ 0
0.03
0
0.09 ⎥⎥
−0.1 0
−0.4 ⎥
0 −1 8 ⎥ ⎢ 0
⎥⎢
⎥
0 0 1 ⎦⎣ 0
0
0
0 ⎦
1
0
D = [ 0.09 0.79 0.4 0.1 0 −0.03]
T
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58
Problem 3.9
A camera is attached to the hand frame T of a robot as given. The corresponding inverse
Jacobian of the robot relative to the frame at this location is also given. The robot makes
a differential motion, as a result of which, the change dT in the frame is recorded as
given.
a.
Find the new location of the camera after the differential motion.
b. Find the differential operator.
c.
Find the joint differential motion values Dθ associated with this move.
⎡0
⎢1
T =⎢
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
3⎤
2 ⎥⎥
8⎥
⎥
1⎦
0
⎡ −0.02
⎢ 0
0.02
dT = ⎢
⎢ 0
−0.1
⎢
0
⎣ 0
0
0
⎡1
⎢2
0
−1
⎢
⎢ 0 −0.1 0
T −1
J =⎢
⎢ 0 −1 0
⎢0
0
0
⎢
0
0
⎣1
−0.1 0.7 ⎤
0
0.08 ⎥⎥
−0.3⎥
0
⎥
0
0 ⎦
0
0
0
0
1
0
0
0
0
1
0
0
0⎤
0 ⎥⎥
0⎥
⎥
0⎥
0⎥
⎥
1⎦
Estimated student time to complete: 30-40 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
Solution:
a.
1
−0.1 3.7 ⎤
⎡ −0.02
⎢ 1
0.02
0
2.08⎥⎥
⎢
Tnew = Told + dT =
⎢ 0
−0.1 −1 7.7 ⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
b.
0
−0.1 0.7 ⎤ ⎡0
⎡ −0.02
⎢ 0
0.02
0
0.08 ⎥⎥ ⎢⎢1
−1
⎢
Δ = dT ⋅ T =
⎢ 0
−0.1 0
−0.3⎥ ⎢0
⎢
⎥⎢
0
0
0 ⎦ ⎣0
⎣ 0
1 0 −2 ⎤ ⎡ 0
−0.02 0.1 −0.06 ⎤
⎥
⎢
0 0 −3⎥ ⎢0.02
0
0
0.02 ⎥⎥
=
0 −1 8 ⎥ ⎢ −0.1
0
0
0 ⎥
⎥ ⎢
⎥
0 0 1⎦ ⎣ 0
0
0
0 ⎦
c.
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59
T
Δ = ⎡⎣T −1 ⎤⎦ [ Δ ][T ] = ⎡⎣T −1 ⎤⎦ [ dT ] =
⎡0
⎢1
=⎢
⎢0
⎢
⎣0
1 0 −2 ⎤ ⎡ −0.02
0
0.02
0
0.08⎤
−0.1 0.7 ⎤ ⎡ 0
0 0 −3⎥⎥ ⎢⎢ 0
0.02
0
0.08 ⎥⎥ ⎢⎢ −0.02
0
−0.1 0.7 ⎥⎥
=
−0.1 0
−0.3⎥ ⎢ 0
0 −1 8 ⎥ ⎢ 0
0.1
0
0.3 ⎥
⎥⎢
⎥ ⎢
⎥
0 0 1 ⎦⎣ 0
0
0
0 ⎦ ⎣ 0
0
0
0 ⎦
Therefore: T D = [ 0.08 0.7 0.3 0.1 0 −0.02]
T
T
D = T J ⋅ Dθ
0
0
⎡1
⎢2
−1
0
⎢
⎢ 0 −0.1 0
→ Dθ = T J −1 ⋅ T D = ⎢
⎢ 0 −1 0
⎢0
0
0
⎢
0
0
⎣1
0
0
0
0
1
0
0
0
0
1
0
0
0 ⎤ ⎡ 0.08 ⎤ ⎡ 0.08 ⎤
0 ⎥⎥ ⎢⎢ 0.7 ⎥⎥ ⎢⎢ −0.14 ⎥⎥
0 ⎥ ⎢ 0.3 ⎥ ⎢ −0.07 ⎥
⎥⎢
⎥=⎢
⎥
0 ⎥ ⎢ 0.1 ⎥ ⎢ −0.7 ⎥
0 ⎥ ⎢ 0 ⎥ ⎢ 0.1 ⎥
⎥⎢
⎥ ⎢
⎥
1 ⎦ ⎣ −0.02 ⎦ ⎣ 0.06 ⎦
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60
Problem 3.10
The hand frame TH of a robot is given. The corresponding inverse Jacobian of the robot at
this location relative to this frame is also shown. The robot makes a differential motion
relative to this frame described as
T
TH
D = [ 0.05 0 −0.1 0 0.1 0.1] .
a.
Find which joints must make a differential motion, and by how much, in order to
create the indicated differential motions.
b. Find the change in the frame.
c.
Find the new location of the frame after the differential motion.
d. Find how much the differential motions (given above) should have been if
measured relative to the Universe, to move the robot to the same new location as in Part
c.
0
0 0 0 0⎤
⎡5
⎢
0
−1 0 0 0 ⎥⎥
⎡ 0 1 0 3⎤
⎢2
⎢ 1 0 0 3⎥
⎢ 0 −0.2 0 0 0 0 ⎥
TH −1
⎥
TH = ⎢
J =⎢
⎥
⎢ 0 0 −1 8 ⎥
⎢ 0 −1 0 0 1 0 ⎥
⎢
⎥
⎢0
0
0 1 0 0⎥
⎣0 0 0 1⎦
⎢
⎥
0
0 0 0 1⎦
⎣1
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
Solution:
a.
0
0 0 0 0 ⎤ ⎡ 0.05⎤ ⎡0.25⎤
⎡5
⎢2
−1 0 0 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0.2 ⎥⎥
0
⎢
⎢ 0 −0.2 0 0 0 0 ⎥ ⎢ −0.1⎥ ⎢ 0 ⎥
−1
[ Dθ ] = ⎡⎣ TH J ⎤⎦ ⎡⎣ TH D ⎤⎦ = ⎢
⎥⎢
⎥=⎢
⎥
−
0
1
0
0
1
0
0
⎢
⎥⎢
⎥ ⎢ 0.1 ⎥
⎢0
0
0 1 0 0 ⎥ ⎢ 0.1 ⎥ ⎢ 0 ⎥
⎢
⎥⎢
⎥ ⎢
⎥
0
0 0 0 1 ⎦ ⎣ 0.1 ⎦ ⎣0.15⎦
⎣1
Therefore, joints 1, 2, 4, and 6 must move as shown.
b. Substituting values from TH D to form TH Δ , we get:
⎡0
⎢1
TH
dT = TH ⋅ Δ = ⎢
⎢0
⎢
⎣0
c.
1 0
0 0
0 −1
0 0
3⎤ ⎡ 0
0
0 ⎤
−0.1 0.1 0.05⎤ ⎡0.1 0
⎥
⎢
⎥
⎢
3⎥ ⎢ 0.1
0
0
0 ⎥ ⎢ 0 −0.1 0.1 0.05⎥⎥
=
8⎥ ⎢ −0.1 0
0 −0.1⎥ ⎢0.1 0
0
0.1 ⎥
⎥⎢
⎥ ⎢
⎥
1⎦ ⎣ 0
0
0
0 ⎦ ⎣0
0
0
0 ⎦
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61
⎡0
⎢1
TH new = TH old + dT = ⎢
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
3⎤ ⎡0.1 0
0
0 ⎤ ⎡0.1 1
0
3 ⎤
⎥
⎢
⎥
⎢
3⎥ ⎢ 0 −0.1 0.1 0.05⎥ ⎢ 1 −0.1 0.1 3.05⎥⎥
+
=
−1 8.1 ⎥
8⎥ ⎢0.1 0
0
0.1 ⎥ ⎢0.1 0
⎥ ⎢
⎥ ⎢
⎥
1⎦ ⎣ 0
0
0
0 ⎦ ⎣ 01
0
0
1 ⎦
d.
dT = TH ⋅ TH Δ = Δ ⋅ TH
0
0 ⎤ ⎡0
⎡0.1 0
⎢ 0 −0.1 0.1 0.05⎥ ⎢1
⎥⎢
Δ = dT ⋅ TH −1 = ⎢
⎢0.1 0
0
0.1 ⎥ ⎢ 0
⎢
⎥⎢
0
0
0 ⎦ ⎣0
⎣0
−3⎤ ⎡ 0
−0.3⎤
0.1 0
⎥
⎢
0 0 −3⎥ ⎢ −0.1 0 −0.1 1.15 ⎥⎥
=
−0.2 ⎥
0 −1 8 ⎥ ⎢ 0
0.1 0
⎥ ⎢
⎥
0 0 1⎦ ⎣ 0
0
0
0 ⎦
1
0
Therefore, D = [ −0.3 1.15 −0.2 0.1 0 −0.1]
T
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62
Problem 3.11
The hand frame T of a robot is given. The corresponding inverse Jacobian of the robot at
this location is also shown. The robot makes a differential motion described as
T
D = [ 0.05 0 −0.1 0 0.1 0.1] .
a.
Find which joints must make a differential motion, and by how much, in order to
create the indicated differential motions.
b. Find the change in the frame.
c.
Find the new location of the frame after the differential motion.
d. Find how much the differential motions (given above) should have been, if
measured relative to Frame T, to move the robot to the same new location as in Part c.
⎡0
⎢1
⎢
T=
⎢0
⎢
⎣0
1 0
0 0
0 −1
0 0
0
0
⎡5
⎢2
0
−1
⎢
⎢ 0 −0.2 0
J −1 = ⎢
⎢ 0 −1 0
⎢0
0
0
⎢
0
0
⎣1
3⎤
3⎥⎥
8⎥
⎥
1⎦
0
0
0
0
1
0
0
0
0
1
0
0
0⎤
0 ⎥⎥
0⎥
⎥
0⎥
0⎥
⎥
1⎦
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 3.6-3.11
Solution:
a.
0
0
⎡5
⎢2
−1
0
⎢
⎢ 0 −0.2 0
Dθ = J −1 ⋅ D = ⎢
⎢ 0 −1 0
⎢0
0
0
⎢
0
0
⎣1
0
0
0
0
1
0
0
0
0
1
0
0
0 ⎤ ⎡ 0.05⎤ ⎡ 0.25⎤
0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0.2 ⎥⎥
0 ⎥ ⎢ −0.1⎥ ⎢ 0 ⎥
⎥⎢
⎥=⎢
⎥
0 ⎥ ⎢ 0 ⎥ ⎢ 0.1 ⎥
0 ⎥ ⎢ 0.1 ⎥ ⎢ 0 ⎥
⎥⎢
⎥ ⎢
⎥
1 ⎦ ⎣ 0.1 ⎦ ⎣ 0.15⎦
Therefore, joints 1, 2, 4, and 6 must move as shown.
b. Using D given above to form Δ ,
−0.1 0.1 0.05⎤ ⎡ 0
⎡ 0
⎢ 0.1
0
0
0 ⎥⎥ ⎢⎢1
⎢
dT = Δ ⋅ T =
⎢ −0.1 0
0 −0.1⎥ ⎢ 0
⎢
⎥⎢
0
0
0 ⎦ ⎣0
⎣ 0
1 0
0 0
0 −1
0 0
3⎤ ⎡ −0.1 0
−0.1 0.55 ⎤
⎥
⎢
3⎥ ⎢ 0
0.1
0
0.3 ⎥⎥
=
−0.1 0
−0.4 ⎥
8⎥ ⎢ 0
⎥ ⎢
⎥
1⎦ ⎣ 0
0
0
0 ⎦
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63
c.
−0.1 3.55⎤
⎡ −0.1 1
⎢ 1
0.1
0
3.3 ⎥⎥
⎢
Tnew = Told + dT =
⎢ 0
−0.1 −1 7.6 ⎥
⎢
⎥
0
0
1 ⎦
⎣ 0
d.
⎡0
⎢1
T
−1
−1
Δ = ⎡⎣T ⎤⎦ [ Δ ][T ] = ⎡⎣T ⎤⎦ [ dT ] = ⎢
⎢0
⎢
⎣0
0.1 0
0.3 ⎤
⎡ 0
⎢ −0.1 0 −0.1 0.55⎥
⎥
=⎢
⎢ 0
0.1 0
0.4 ⎥
⎢
⎥
0
0
0 ⎦
⎣ 0
−3⎤ ⎡ −0.1 0
−0.1 0.55 ⎤
⎥
⎢
0 0 −3 ⎥ ⎢ 0
0.1
0
0.3 ⎥⎥
−0.1 0
−0.4 ⎥
0 −1 8 ⎥ ⎢ 0
⎥⎢
⎥
0 0 1 ⎦⎣ 0
0
0
0 ⎦
1
0
The differential motions should have been:
T
D = [ 0.3 0.55 0.4 0.1 0 −0.1]
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64
Problem 3.12
Calculate the T6 J 21 element of the Jacobian for the revolute robot of Example 2.25.
Estimated student time to complete: 20 minutes
Prerequisite knowledge required: Text Section(s) 3.10
Solution:
The robot has 6 revolute joints. From Equation 3.26:
T6
J 2i = ( −ox p y + o y px ) where for i=1 we use oT6 = A1 A2 A3 A4 A5 A6 given in Equation 2.59.
Substitute to get:
T6
J 21 = − [C1 (−C234C5C6 − S234C6 ) + S1S5 S6 ] × [ S1 (C234 a4 + C23 a3 + C2 a2 ) ]
+ [ S1 (−C234C5C6 − S234C6 ) − C1S5 S6 ] × [C1 (C234 a4 + C23 a3 + C2 a2 ) ]
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65
Problem 3.13
Calculate the T6 J16 element of the Jacobian for the revolute robot of Example 2.25.
Estimated student time to complete: 20 minutes
Prerequisite knowledge required: Text Section(s) 3.10
Solution:
The robot has 6 revolute joints. From Equation 3.26:
T6
J1i = ( −nx p y + n y px ) where for i=6 we use 5T6 = A6 given in Equation 2.57. Substitute
to get:
T6
J 21 = − ⎣⎡( C6 )( 0 ) ⎦⎤ + ⎡⎣( S6 )( 0 ) ⎤⎦ = 0
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66
Problem 3.14
Using Equation (2.34), differentiate proper elements of the matrix to develop a set of
symbolic equations for joint differential motions of a cylindrical robot and write the
corresponding Jacobian.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 3.10-3.11
Solution:
For a cylindrical coordinate (without additional rotation) there can only be 3 variables.
The Jacobian will be a 3 × 3 matrix.
⎡Cα
⎢ Sα
Tcyl (r , α , l ) = ⎢
⎢ 0
⎢
⎣ 0
− Sα
Cα
0
0
0 rCα ⎤
0 rSα ⎥⎥
1
l ⎥
⎥
0
1 ⎦
⎧ px = rCα
⎧dpx = drCα − rSα dα
⎪
⎪
Then ⎨ p y = rSα and ⎨dp y = drSα + rCα dα
⎪
⎪
⎩ pz = l
⎩dpz = dl
⎡ dpx ⎤ ⎡Cα
Therefore, ⎢⎢ dp y ⎥⎥ = ⎢⎢ Sα
⎣⎢ dpz ⎦⎥ ⎣⎢ 0
− rSα
rCα
0
0 ⎤ ⎡ dr ⎤
0 ⎥⎥ ⎢⎢ dα ⎥⎥
1 ⎦⎥ ⎣⎢ dl ⎦⎥
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67
Problem 3.15
Using Equation (2.36), differentiate proper elements of the matrix to develop a set of
symbolic equations for joint differential motions of a spherical robot and write the
corresponding Jacobian.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 3.10-3.11
Solution:
For a spherical coordinate (without additional rotation) there can only be 3 variables. The
Jacobian will be a 3 × 3 matrix.
⎡ C β Cγ
⎢ C β Sγ
Tsph (r , β , γ ) = ⎢
⎢ −S β
⎢
⎣ 0
− Sγ
Cγ
0
0
S β Cγ
S β Sγ
Cβ
0
rS β Cγ ⎤
rS β Sγ ⎥⎥
rC β ⎥
⎥
1 ⎦
⎧ px = rS β .Cγ
⎧dpx = drS β .Cγ + rC β .Cγ d β − rS β .Sγ d γ
⎪
⎪
Then ⎨ p y = rS β .Sγ and ⎨dp y = drS β .S γ + rC β .Sγ d β + rS β .Cγ d γ
⎪
⎪
⎩ pz = rC β
⎩dpz = drC β − rS β d β
⎡ dpx ⎤ ⎡ S β .Cγ
Therefore, ⎢⎢ dp y ⎥⎥ = ⎢⎢ S β .S γ
⎢⎣ dpz ⎥⎦ ⎢⎣ C β
rC β .Cγ
rC β .Sγ
−rS β
− rS β .S γ ⎤ ⎡ dr ⎤
rS β .Cγ ⎥⎥ ⎢⎢ d β ⎥⎥
⎥⎦ ⎢⎣ d γ ⎥⎦
0
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68
Problem 3.16
For a cylindrical robot, the three joint velocities are given for a corresponding location.
Find the three components of the velocity of the hand frame.
r =0.1 in/sec, α =0.05 rad/sec, l =0.2 in/sec, r=15 in, α = 30D , l=10 in.
Estimated student time to complete: 10 or 20 minutes
Prerequisite knowledge required: Text Section(s) 3.10-3.11
Solution:
Referring to the solution of Problem 3.14, and dividing both sides by dt, substitute the
given values into the equation to get:
⎡ p x ⎤ ⎡Cα
⎢ p ⎥ = ⎢ Sα
⎢ y⎥ ⎢
⎢⎣ p z ⎥⎦ ⎢⎣ 0
− rSα
rCα
0
0 ⎤ ⎡ r ⎤ ⎡ 0.866 −7.5 0⎤ ⎡ 0.1 ⎤ ⎡ −0.288⎤
0 ⎥⎥ ⎢⎢α ⎥⎥ = ⎢⎢ 0.5
13 0⎥⎥ ⎢⎢ 0.05⎥⎥ = ⎢⎢ 0.7 ⎥⎥ in/sec
1 ⎥⎦ ⎢⎣ l ⎥⎦ ⎢⎣ 0
0
1 ⎥⎦ ⎢⎣ 0.2 ⎥⎦ ⎢⎣ 0.2 ⎥⎦
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69
Problem 3.17
For a spherical robot, the three joint velocities are given for a corresponding location.
Find the three components of the velocity of the hand frame.
r =2 in/sec, β =0.05 rad/sec, γ =0.1 rad/sec, r=20 in, β = 60D , γ = 30D .
Estimated student time to complete: 10 or 20 minutes
Prerequisite knowledge required: Text Section(s) 3.10-3.11
Solution:
Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the
given values into the equation to get:
⎡ p x ⎤ ⎡ S β Cγ
⎢ p ⎥ = ⎢ S β S γ
⎢ y⎥ ⎢
⎣⎢ p z ⎦⎥ ⎣⎢ C β
rC β Cγ
rC β Sγ
− rS β
8.66 −8.66⎤ ⎡ 2 ⎤ ⎡1.067 ⎤
−rS β Sγ ⎤ ⎡ r ⎤ ⎡ 0.75
⎥
⎢
⎥
⎢
rS β Cγ ⎥ ⎢ β ⎥ = ⎢ 0.433
5
15 ⎥⎥ ⎢⎢ 0.05⎥⎥ = ⎢⎢ 2.616⎥⎥
0 ⎦⎥ ⎣⎢ γ ⎦⎥ ⎣⎢ 0.5 −17.32
0 ⎦⎥ ⎢⎣ 0.1 ⎦⎥ ⎣⎢ 0.134 ⎦⎥
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70
Problem 3.18
For a spherical robot, the three joint velocities are given for a corresponding location.
Find the three components of the velocity of the hand frame.
r =1 unit/sec, β =1 rad/sec, γ =1 rad/sec, r=5 units, β = 45D , γ = 45D .
Estimated student time to complete: 10 or 20 minutes
Prerequisite knowledge required: Text Section(s) 3.10-3.11
Solution:
Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the
given values into the equation to get:
⎡ p x ⎤ ⎡ S β Cγ
⎢ ⎥ ⎢
⎢ p y ⎥ = ⎢ S β Sγ
⎢⎣ p z ⎥⎦ ⎢⎣ C β
rC β Cγ
rC β Sγ
− rS β
2.5
− rS β Sγ ⎤ ⎡ r ⎤ ⎡ 0.5
−3.535⎤ ⎡1⎤ ⎡ −0.535⎤
⎥
⎢
⎥
⎢
rS β Cγ ⎥ ⎢ β ⎥ = ⎢ 0.5
2.5
3.535 ⎥⎥ ⎢⎢1⎥⎥ = ⎢⎢ 6.535 ⎥⎥
0 ⎥⎦ ⎢⎣ γ ⎥⎦ ⎢⎣ 0.707 −3.535
0 ⎥⎦ ⎢⎣1⎥⎦ ⎢⎣ −2.828⎥⎦
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71
Problem 3.19
For a cylindrical robot, the three components of the velocity of the hand frame are given
for a corresponding location. Find the required three joint velocities that will generate the
given hand frame velocity.
x = 1 in/sec, y = 3 in/sec, z = 5 in/sec, α = 45D , r = 20 in, l = 25 in
Estimated student time to complete: 10 or 25 minutes
Prerequisite knowledge required: Text Section(s) 3.12
Solution:
Referring to the solution of Problem 3.16, substitute to get:
⎡ p x ⎤ ⎡Cα
⎢ ⎥ ⎢
⎢ p y ⎥ = ⎢ Sα
⎢⎣ p z ⎥⎦ ⎢⎣ 0
− rSα
rCα
0
0 ⎤ ⎡ r ⎤
⎡ 1 ⎤ ⎡ 0.707 −14.14 0 ⎤ ⎡ r ⎤
⎥
⎢
⎥
0 ⎥ ⎢α ⎥ → ⎢⎢ 2⎥⎥ = ⎢⎢ 0.707 14.14 0 ⎥⎥ ⎢⎢α ⎥⎥
⎢⎣ 3⎥⎦ ⎢⎣ 0
1 ⎥⎦ ⎢⎣ l ⎥⎦
0
1 ⎥⎦ ⎢⎣ l ⎥⎦
⎧0.707 r − 14.14α = 1
⎧r = 2.83 in/sec
⎪
⎪
D
⎨0.707 r + 14.14α = 3 → ⎨α =0.071 /sec
⎪
⎪
in/sec
⎩l = 5
⎩l = 5
Please note that this problem could have been solved by calculating the inverse of the
Jacobian as well.
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72
Problem 3.20
For a spherical robot, the three components of the velocity of the hand frame are given
for a corresponding location. Find the required three joint velocities that will generate the
given hand frame velocity.
x = 5 in/sec, y = 9 in/sec, z = 6 in/sec, β = 60D , r = 20 in, γ =30D
Estimated student time to complete: 15 or 30 minutes
Prerequisite knowledge required: Text Section(s) 3.12
Solution:
Referring to the solution of Problem 3.17, substitute to get:
⎡ p x ⎤ ⎡ S β Cγ
⎢ p ⎥ = ⎢ S β S γ
⎢ y⎥ ⎢
⎢⎣ p z ⎥⎦ ⎢⎣ C β
rC β Cγ
rC β Sγ
− rS β
−rS β Sγ ⎤ ⎡ r ⎤
8.66 −8.66⎤ ⎡ r ⎤
⎡ 5 ⎤ ⎡ 0.75
⎥
⎢
⎥
⎢
⎥
⎢
rS β Cγ ⎥ ⎢ β ⎥ → ⎢9 ⎥ = ⎢ 0.433
5
15 ⎥⎥ ⎢⎢α ⎥⎥
⎢⎣ 6 ⎥⎦ ⎢⎣ 0.5 −17.32
0 ⎥⎦ ⎢⎣ γ ⎥⎦
0 ⎥⎦ ⎢⎣ l ⎥⎦
⎧0.75r + 8.66β − 8.66γ = 5
⎧r = 10.647 in/sec
⎪
⎪
→ ⎨ β = − 0.039 D /sec
⎨0.433r + 5β + 15γ = 9
⎪ ⎪γ = 0.306 D /sec
⎩
⎩0.5r − 17.32 β = 6
Please note that this problem could have been solved by calculating the inverse of the
Jacobian as well.
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73
CHAPTER FOUR
Problem 4.1
Using Lagrangian mechanics derive the equations of motion of a cart with two tires under
the cart as shown:
k
F
m
r, I
r, I
x
Figure P.4.1
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 4.2
Solution:
I. Kinematics:
Since the rollers are in contact with the block at the top, the contact-point velocity is x .
Therefore:
x = 2rω → ω =
x
2r
x
vc
II. Dynamics:
m
I
1 2
1
⎛1
⎞ 1
mx + 2 ⎜ mwvc2 + I cω 2 ⎟ = mx 2 + w x 2 + c2 x 2 = meff x 2
2
2
4
4r
⎝2
⎠ 2
m
I ⎞
⎛1
where ⎜ m + w + c2 ⎟ = meff
4 4r ⎠
⎝2
1
P = kx 2
2
1
Therefore, L = K − P = meff x 2 − kx 2
2
∂L
d ⎛ ∂L ⎞
= 2meff x →
x
⎜ ⎟ = 2meff ∂x
dt ⎝ ∂x ⎠
∂L
= −kx
∂x
F = 2meff x + kx
K=
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74
Problem 4.2
Calculate the total kinetic energy of the link AB, attached to a roller with negligible mass,
as shown.
B
l, I
y
C
x
θ
m
A
Vo
Figure P.4.2
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 4.2
Solution:
First, find the velocity of point C, assuming that the roller has negligible mass too:
l
l
⎛l
⎞
v C = v A + v C / A = v o + θ × = (vo )i + θk × ⎜ cos θ i + sin θ j ⎟
2
2
⎝2
⎠
l sin θ ⎞ l cos θ ⎛l
⎞
⎛l
⎞
⎛
θ ⎟i +
θj
= (vo )i + ⎜ cos θ ⎟ θ j − ⎜ sin θ ⎟ θi = ⎜ vo −
2
2
⎝2
⎠
⎝2
⎠
⎝
⎠
l 2 sin 2 θ 2
l 2 cos 2 θ 2
l2
θ − vol sin θθ +
θ = vo 2 + θ 2 − vo l sin θθ
4
4
4
2
⎞ 1
1
1
1 ⎛
l
K = mvC2 + Iθ 2 = m ⎜ vo 2 + θ 2 − vo l sin θθ ⎟ + Iθ 2
2
2
2 ⎝
4
⎠ 2
vC2 = vo 2 +
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75
Problem 4.3
Derive the equations of motion for the 2-link mechanism with distributed mass, as
shown.
y0
O
I
l 1, 1
x0
θ1 C
y1
m1
A
l2 ,
θ2
D
I2
x1
B
m2
Figure P.4.3
Estimated student time to complete: 1.5-2 hours
Prerequisite knowledge required: Text Section(s) 4.2
Solution:
I. Kinematic: First we write equations describing the kinematic relationships.
l2
⎧
x
l
C
C12
=
+
D
1
1
⎪⎪
2
⎨
⎪ y = l S + l2 S
⎪⎩ D 1 1 2 12
⎧
S − l2 θ + θ S
x
l
θ
=
−
D
1
1 1
1
2
12
⎪⎪
2
→ ⎨
⎪ y = l θ C + l2 θ + θ C
⎪⎩ D 1 1 1 2 1 2 12
(
)
(
)
The velocity of point D will be:
l2 2 2 2
2
2
2
2 2 2
vD = xD + y D = l1 θ1 S1 +
θ1 + θ 2 + 2θ1θ2 S12 2 + l1l2θ1 θ1 + θ2 S1S12
4
2
l
+ l12θ12C12 + 2 θ12 + θ2 2 + 2θ1θ2 C12 2 + l1l2θ1 θ1 + θ2 C1C12
4
2
l
= l12θ12 + 2 θ12 + θ2 2 + 2θ1θ2 + l1l2θ1 θ1 + θ2 C2
4
Rearrange:
⎛
⎞
⎛l 2 ⎞
⎛l 2
⎞
l2
vD2 = θ12 ⎜ l12 + 2 + l1l2C2 ⎟ + θ2 2 ⎜ 2 ⎟ + θ1θ2 ⎜ 2 + l1l2C2 ⎟
4
⎝
⎠
⎝ 4 ⎠
⎝ 2
⎠
(
)
(
(
)
)
(
(
(
)
)
)
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76
II: Kinetics:
Kinetic Energy:
(
)
2
1 2 1
1
I Oθ1 + I D θ1 + θ2 + m2 vD2
2
2
2
2
1⎛1
1⎛ 1
1
⎞
⎞
= ⎜ m1l12 ⎟ θ12 + ⎜ m2l22 ⎟ θ1 + θ2 + m2 vD2
2⎝3
2 ⎝ 12
2
⎠
⎠
K = K1 + K 2 =
(
)
1
1
1
1
⎛1
⎞
⎛1
⎞
⎛1
⎞
= θ12 ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ + θ2 2 ⎜ m2l22 ⎟ + θ1θ2 ⎜ m2l22 + m2l1l2C2 ⎟
6
2
2
2
⎝6
⎠
⎝6
⎠
⎝3
⎠
Potential Energy:
l
l
P = P1 + P2 = m1 g 1 (1 − C1 ) + m2 gl1 (1 − C1 ) + m2 g 2 (1 − C12 )
2
2
Lagrangian:
1
1
1
1
⎛1
⎞
⎛1
⎞
⎛1
⎞
L = K − P = θ12 ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ + θ2 2 ⎜ m2l22 ⎟ + θ1θ2 ⎜ m2l22 + m2l1l2C2 ⎟
6
2
2
2
⎝6
⎠
⎝6
⎠
⎝3
⎠
l
l
l
l
⎛
⎞
+ ⎜ m1 g 1 + m2 gl1 ⎟ C1 + m2 g 2 C12 − m1 g 1 − m2 gl1 − m2 g 2
2
2
2
2
⎝
⎠
Derivatives and equations of motion:
∂L
1
1
1
1
⎛1
⎞
⎛1
⎞
= 2θ1 ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ + θ2 ⎜ m2l22 + m2l1l2C2 ⎟
∂θ
6
2
2
2
⎝6
⎠
⎝3
⎠
1
1
1
1
d ⎛ ∂L ⎞
⎛1 2 1
⎞
⎛1
⎞
2
2
2
⎜ ⎟ = 2θ1 ⎜ m1l1 + m2l2 + m2l1 + m2l1l2C2 ⎟ + θ2 ⎜ m2l2 + m2l1l2C2 ⎟
6
2
2
2
dt ⎝ ∂θ1 ⎠
⎝6
⎠
⎝3
⎠
1
− θ1θ2 m2l1l2 S2 − θ2 2 m2l1l2 S 2
2
∂L
l
l
⎛
⎞
= − ⎜ m1 g 1 + m2 gl1 ⎟ S1 − m2 g 2 S12
2
2
∂θ1
⎝
⎠
Substitute to get:
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77
1
1
⎛1
⎞
⎛1
⎞
T1 = ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ θ1 + ⎜ m2l22 + m2l1l2C2 ⎟ θ2
3
2
⎝3
⎠
⎝3
⎠
1
1
⎛1
⎞
− m2l1l2 S2θ1θ2 − m2l1l2 S 2θ2 2 + ⎜ m1 + m2 ⎟ gl1S1 + m2 gl2 S12
2
2
⎝2
⎠
For the second variable:
∂L 1
1
⎛1
⎞
= m2l22θ2 + θ1 ⎜ m2l22 + m2l1l2C2 ⎟
∂θ 2 3
2
⎝3
⎠
d ⎛ ∂L ⎞ 1
1
⎞ 1
2 2
⎛ 1
⎜ ⎟ = m2l2 θ 2 + θ1 ⎜ m2l2 + m2l1l2C2 ⎟ − m2l1l2 S2θ1θ 2
dt ⎝ ∂θ 2 ⎠ 3
3
2
2
⎝
⎠
1
1
1
∂L
= − m2l1l2 S2θ12 − m2l1l2 S2θ1θ2 − m2 gl2 S12
2
2
2
∂θ 2
Substitute to get:
1
1
1
1
⎛1
⎞
T2 = ⎜ m2l22 + m2l1l2C2 ⎟ θ1 + m2l22θ2 + m2l1l2 S2θ12 + m2 gl2 S12
2
3
2
2
⎝3
⎠
The result can be written in a matrix form too.
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78
Problem 4.4
Write the equations that express U62, U35, U53, U623, and U534 for a 6-axis cylindrical-RPY
robot in terms of the A and Q matrices.
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 4.4.1.
Solution:
For joints 1 and 3:
For joints 2,4,5,6:
U ij =
⎡0
⎢0
Q ( prismatic ) = ⎢
⎢0
⎢
⎣0
0
0
0
0⎥
0
0
1⎥
0
0
0⎦
⎡0 −1
⎢1 0
Q ( revolute ) = ⎢
⎢0 0
⎢
⎣0 0
∂ 0Ti
= A1 A2 ..Q j Aj .. Ai
∂q j
0⎤
0
⎥ = QP
⎥
0
0
0
0
0⎤
0 ⎥⎥
= QR
0⎥
⎥
0⎦
j≤i
U ijk = ∂U ij ∂qk
∂ 0T6 ∂ ( A1 A2 A3 A4 A5 A6 )
=
= A1QR A2 A3 A4 A5 A6
∂α
∂α
∂ 0T3 ∂ ( A1 A2 A3 )
U 35 =
=
= 0 (i ≥ j )
∂φ0
∂φ0
U 62 =
∂ 0T5 ∂ ( A1 A2 A3 A4 A5 )
=
= A1 A2QP A3 A4 A5
∂l
∂l
∂U 62 ∂ ( A1QR A2 A3 A4 A5 A6 )
=
= A1QR A2QP A3 A4 A5 A6
U 623 =
∂l
∂l
∂U 53 ∂ ( A1 A2QP A3 A4 A5 )
=
= A1 A2QP A3QR A4 A5
U 534 =
∂qa
∂qa
U 53 =
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79
Problem 4.5
Using Equations 4.49 to 4.54, write the equations of motion for a 3-DOF revolute robot
and describe each term.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 4.4.4.
Solution:
We have:
T1 = D11θ1 + D12θ2 + D13θ3 + I1( act )θ1 + D111θ12 + D122θ22 + D133θ32
+ 2 D112θ1θ2 + 2 D113θ1θ3 + 2 D123θ2θ3 + D1
T2 = D21θ1 + D22θ2 + D23θ3 + I 2( act )θ2 + D211θ12 + D222θ22 + D233θ32
+ 2 D212θ1θ2 + 2 D213θ1θ3 + 2 D223θ2θ3 + D2
T3 = D31θ1 + D32θ2 + D33θ3 + I 3( act )θ3 + D311θ12 + D322θ22 + D333θ32
+ 2 D312θ1θ2 + 2 D313θ1θ3 + 2 D323θ2θ3 + D3
( D )θi terms are caused by angular accelerations.
( D )θi 2 terms represent centripetal accelerations.
( D )θiθj terms are related to the Coriolis accelerations.
( Di ) terms are gravity terms.
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80
Problem 4.6
Expand the D134 and D15 terms of Equation 4.49 in terms of their constituent matrices.
Estimated student time to complete: 1 hour
Prerequisite knowledge required: Text Section(s) 4.4
Solution:
a. From Equation 4.44: Dijk =
n
∑
p = max( i , j , k )
Trace(U pjk J pU Tpi )
For D134 we have i =1, j =3, k =4, p =4, n =6. Then:
T
D134 = ∑ Trace(U p 34 J pU Tp1 ) = Trace (U 434 J 4U 41
) + Trace (U 534 J 5U 51T ) + Trace (U 634 J 6U 61T )
6
4
∂ 0T4 ∂ ( A1 A2 A3 A4 )
=
= A1 A2Q3 A3 A4
∂q3
∂q3
U 43 =
U 434 =
∂U 43 ∂ ( A1 A2Q3 A3 A4 )
=
= A1 A2Q3 A3Q4 A4
∂q4
∂q4
∂ 0T4 ∂ ( A1 A2 A3 A4 )
=
= Q1 A1 A2 A3 A4
U 41 =
∂q1
∂q1
T
= A4T A3T A2T A1T Q1T
From Equation A.11 of Appendix A: U 41
U 53 =
∂ ( A1 A2 A3 A4 A5 )
= A1 A2Q3 A3 A4 A5
∂q3
U 534 =
U 51 =
U 63 =
∂ ( A1 A2Q3 A3 A4 A5 )
= A1 A2Q3 A3Q4 A4 A5
∂q4
∂ ( A1 A2 A3 A4 A5 )
= Q1 A1 A2 A3 A4 A5
∂q1
T
→ U 51
= A5T A4T A3T A2T A1T Q1T
∂ ( A1 A2 A3 A4 A5 A6 )
= A1 A2Q3 A3 A4 A5 A6
∂q3
U 634 =
∂U 63
= A1 A2Q3 A3Q4 A4 A5 A6
∂q4
T
= A6T A5T A4T A3T A2T A1T Q1T
U 61
Substitute to get:
D134 = Trace ( A1 A2Q3 A3Q4 A4 J 4 A4T A3T A2T A1T Q1T )
+ Trace ( A1 A2Q3 A3Q4 A4 A5 J 5 A5T A4T A3T A2T A1T Q1T )
+ Trace ( A1 A2Q3 A3Q4 A4 A5 A6 J 6 A6T A5T A4T A3T A2T A1T Q1T )
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81
b. From Equation 4.43: Dij =
n
∑ Trace(U J U )
p = max( i , j )
pj
p
T
pi
For D15 we have i =1, j =5, p =5, n =6. Then:
T
D15 = ∑ Trace(U p 5 J pU Tp1 ) = Trace (U 55 J 5U 51
) + Trace (U 65 J 6U 61T )
6
5
U 55 = A1 A2 A3 A4Q5 A5
U 51 = Q1 A1 A2 A3 A4 A5
T
→ U 51
= A5T A4T A3T A2T A1T Q1T
U 65 = A1 A2 A3 A4Q5 A5 A6
U 61 = Q1 A1 A2 A3 A4 A5 A6
T
→ U 61
= A6T A5T A4T A3T A2T A1T Q1T
Substitute to get:
D15 = Trace ( A1 A2 A3 A4Q5 A5 J 5 A5T A4T A3T A2T A1T Q1T )
+ Trace ( A1 A2 A3 A4Q5 A5 A6 J 6 A6T A5T A4T A3T A2T A1T Q1T )
Although a large number of matrices are involved, they are all known.
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82
Problem 4.7
An object is subjected to the following forces and moments relative to the reference
frame. Attached to the object is a frame, which describes the orientation and the location
of the object. Find the equivalent forces and torques acting on the object relative to the
current frame.
⎡0.707 0.707 0 2 ⎤
⎢ 0
0
1 5 ⎥⎥
B=⎢
F T = [10, 0,5,12, 20, 0] N, N.m
⎢0.707 −0.707 0 3 ⎥
⎢
⎥
0
0 1⎦
⎣ 0
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 4.6.
Solution:
Substitute:
f = [10, 0,5]T
m = [12, 20, 0]T
n = [0.707, 0, 0.707]T
i
p = [2,5,3]T
o = [0.707, 0, −0.707]T
a = [0,1, 0]T
j k
f × p = 10 0 5 = i (−25) − j(20) + k (50)
2 5 3
(f × p) + m = −13i + 50k
B
f x = n ⋅ f = 7.07 + 3.54 = 10.61 N
B
f y = o ⋅ f = 7.07 − 3.54 = 3.53 N
B
fz = a ⋅ f = 0
B
mx = n ⋅ [(f × p) + m] = −9.19 + 35.35 = 26.16 Nm
B
my = o ⋅ [(f × p) + m] = −9.19 − 35.35 = −44.54 Nm
B
mz = a ⋅ [(f × p) + m] = 0
and
⎡ 10.61 ⎤
⎢ 3.53 ⎥
⎢
⎥
⎢ 0 ⎥
F =⎢
⎥
⎢ 26.16 ⎥
⎢ −44.54 ⎥
⎢
⎥
⎣ 0 ⎦
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83
Problem 4.8
In order to assemble two parts together, one part must be pushed into the other with a
force of 10 lb in the x-axis direction, 5 lb in the y-direction, and be turned with a torque
of 5 lb ⋅ in along the x-axis direction. The object’s location relative to the base frame of a
robot is described by RT0. Assuming that the two parts must be aligned together for this
purpose, find the necessary forces and moments that the robot must apply to the part
relative to its hand frame.
⎡ 0 −0.707 0.707 4 ⎤
⎢1
0
0
6 ⎥⎥
R
⎢
T0 =
⎢ 0 0.707 0.707 3 ⎥
⎢
⎥
0
0
1⎦
⎣0
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 4.6.
Solution:
Since the robot must be aligned with the object for this operation, then R TH = R TO .
Therefore we have:
f = [10,5, 0]T
m = [5, 0, 0]T
n = [0,1, 0]T
o = [−0.707, 0, 0.707]T
i
p = [4, 6,3]T
a = [0.707, 0, 0.707]T
j k
f × p = 10 5 0 = i (15) − j(30) + k (40)
4
6 3
(f × p) + m = 20i − 30 j + 40k
H
f x = n ⋅ f = 5 units
H
f y = o ⋅ f = −7.07 units
H
f z = a ⋅ f = 7.07 units
H
mx = n ⋅ [(f × p) + m] = −30 units
H
m y = o ⋅ [(f × p) + m] = 14.14 units
H
mz = a ⋅ [(f × p) + m] = 42.42 units
and
⎡ 5 ⎤
⎢ −7.07 ⎥
⎢
⎥
⎢ 7.07 ⎥
F =⎢
⎥
⎢ −30 ⎥
⎢14.14 ⎥
⎢
⎥
⎣ 42.42 ⎦
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84
CHAPTER FIVE
Problem 5.1
It is desired to have the first joint of a 6-axis robot go from an initial angle of 50D to a
final angle of 80D in 3 seconds. Calculate the coefficients for a third-order polynomial
joint-space trajectory. Determine the joint angles, velocities, and accelerations at 1, 2, and
3 seconds. It is assumed that the robot starts from rest, and stops at its destination.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 5.5.1.
Solution:
For a 3rd-order polynomial:
θ (t ) = c0 + c1t + c2t 2 + c3t 3
θ(t ) = c1 + 2c2t + 3c3t 2
Substitute boundary conditions to get:
θi = 50 = c0 + 0
θi = 0 = c1 + 0
θ f = 80 = 50 + 9c2 + 27c3
θ f = 0 = 6c2 + 27c3
Solve to get:
c0 = 5
c1 = 0
c2 = 10
θ (t ) = 50 + 10t − 2.222t
2
c3 = −2.222
3
θ(t ) = 20t − 6.666t 2
θ(t ) = 20 − 13.332t
⎧ θ = 57.78 D
⎪
@t=1 sec ⎨θ = 13.334 D sec
⎪ θ = 6.668 D sec
⎩
⎧ θ = 72.22 D
⎪
@t=2 sec ⎨ θ = 13.34 D sec
⎪θ = −6.664 D sec
⎩
@t=3 sec
⎧ θ = 80 D
⎪ D sec
⎨ θ =0
⎪θ = −20 D sec
⎩
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85
Problem 5.2
It is desired to have the third joint of a 6-axis robot go from an initial angle of 20D to a
final angle of 80D in 4 seconds. Calculate the coefficients for a third-order polynomial
joint-space trajectory and plot the joint angles, velocities, and accelerations. The robot
starts from rest, but should have a final velocity of 5D / sec .
Estimated student time to complete: 20-30 minutes (with plotting)
Prerequisite knowledge required: Text Section(s) 5.5.1.
Solution:
For a 3rd-order polynomial:
θ (t ) = c0 + c1t + c2t 2 + c3t 3
θ(t ) = c1 + 2c2t + 3c3t 2
Substitute boundary conditions to get:
θi = 20 = c0 + 0
θi = 0 = c1 + 0
θ f = 80 = 20 + 16c2 + 64c3
θ f = 5 = 8c2 + 48c3
Solve to get:
c0 = 20
c1 = 0
c2 = 10
θ (t ) = 20 + 10t − 1.5625t
2
c3 = −1.5625
3
θ(t ) = 20t − 4.6875t 2
θ(t ) = 20 − 9.375t
The plot is shown:
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86
Problem 5.3
The second joint of a 6-axis robot is to go from initial angle of 20D to an intermediate
angle of 80D in 5 seconds and continue to its destination of 25D in another 5 seconds.
Calculate the coefficients for third-order polynomials in joint-space. Plot the joint angles,
velocities, and accelerations. Assume the joint stops at intermediate points.
Estimated student time to complete: 30-40 minutes (with plotting)
Prerequisite knowledge required: Text Section(s) 5.5.1
Solution:
There are two segments to this motion. Therefore:
Segment 1
⎧⎪θ (t ) = c0 + c1t + c2t 2 + c3t 3
⎨
2
⎪⎩θ (t ) = c1 + 2c2t + 3c3t
Segment 2
⎧⎪θ (t ) = b0 + b1t + b2t 2 + b3t 3
⎨
2
⎪⎩θ (t ) = b1 + 2b2t + 3b3t
There are eight unknowns; eight equations are needed. Known initial and final conditions
are: θ1i , θ1 f , θ 2i , θ 2 f , θ1i , θ2 f , and θ1 f = θ2i . This results in seven equations. The eighth
equation can be generated by making assumptions such as a maximum allowable
acceleration or an intermediate velocity.
For this problem we will assume that the joint will come to a stop at the intermediate
point. Therefore,
⎧20 = c0 + 0
⎪0 = c + 0
⎪
1
1 ⎨
⎪80 = 20 + 25c2 + 125c3
⎪⎩0 = 10c2 + 75c3
⎧80 = b0 + 0
⎪0 = b + 0
⎪
1
2 ⎨
⎪25 = 80 + 25b2 + 125b3
⎪⎩0 = 10b2 + 75b3
→
→
⎧c0 = 20
⎪
⎪c1 = 0
⎨
⎪c2 = 7.2
⎪⎩c3 = −0.96
⎧b0 = 80
⎪
⎪b1 = 0
⎨
⎪b2 = −6.6
⎪⎩b3 = 0.88
Substitute to get:
1
⎧θ (t ) = 20 + 7.2t 2 − 0.96t 3
⎪
2
⎨θ (t ) = 14.4t − 2.88t
⎪ ⎩θ (t ) = 14.4 − 5.76t
and
2
⎧θ (t ) = 80 − 6.6t 2 + 0.88t 3
⎪
2
⎨θ (t ) = −13.2t + 2.64t
⎪ ⎩θ (t ) = −13.2 + 5.28t
The position, velocities, and accelerations are shown:
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87
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88
Problem 5.4
A fifth-order polynomial is to be used to control the motions of the joints of a robot in
joint-space. Find the coefficients of a fifth-order polynomial that will allow a joint to go
from an initial angle of 0D to a final joint angle of 75D in 3 seconds, while the initial and
final velocities are zero and initial acceleration and final decelerations are 10 D / sec 2 .
Estimated student time to complete: 20-30 minutes (with plotting)
Prerequisite knowledge required: Text Section(s) 5.5.2
Solution:
For a 5th-order polynomial:
⎧θ (t ) = c0 + c1t + c2t 2 + c3t 3 + c4t 4 + c5t 5
⎪
2
3
4
⎨θ (t ) = c1 + 2c2t + 3c3t + 4c4t + 5c5t
⎪ 2
3
⎩θ (t ) = 2c2 + 6c3t + 12c4t + 20c5t
Substitute the given initial and final conditions and get:
⎧0 = c0
⎪75 = 3c + 9c + 27c + 81c + 243c
1
2
3
4
5
⎪
⎪⎪0 = c1
⎨
⎪0 = 6c2 + 27c3 + 108c4 + 405c5
⎪10 = 2c2
⎪
⎪⎩10 = 2c2 + 18c3 + 108c4 + 540c5
Solve to get: c0 = 0, c1 = 0, c2 = 5, c3 = 24.24, c4 = −13.33, c5 = 1.85 . Substitute to
get:
⎧θ (t ) = 5t 2 + 24.44t 3 − 13.33t 4 + 1.85t 5
⎪
2
3
4
⎨θ (t ) = 10t + 73.33t − 53.33t + 9.26t
⎪ 2
3
⎩θ (t ) = 10 + 146.66t − 160t + 37.04t
The position, velocity, and acceleration plots are shown:
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89
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90
Problem 5.5
Joint 1 of a 6-axis robot is to go from an initial angle of θi = 30D to the final angle of
θ f = 120D in 4 seconds with a cruising velocity of ω1 = 30D sec . Find the necessary blending
time for a trajectory with linear segments and parabolic blends and plot the joint
positions, velocities, and accelerations.
Estimated student time to complete: 15-30 minutes (with plotting)
Prerequisite knowledge required: Text Section(s) 5.5.3
Solution:
tb =
θi − θ f + ωt f 30 − 120 + 30 ( 4 )
=
= 1 sec
ω
30
For θ = θi to θ A
For θ = θ A to θ B
For θ = θ B to θ f
1
⎧
2
⎪θ = 30 + 2 ( 30 ) t
⎪⎪
⎨θ = 30t
⎪ ⎪θ = 30
⎪⎩
⎧θ = θ A + 30 ( t − 1)
⎪
⎨θ = 30
⎪ ⎩θ = 0
1
⎧
2
⎪θ = 120 − 2 ( 30 ) (4 − t )
⎪⎪
⎨θ = 30(4 − t )
⎪ ⎪θ = −30
⎪⎩
140
120
100
Position
80
60
Velocity
40
20
0
Acceleration
-20
-40
1
21
2
41
4
Seconds
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91
Problem 5.6
A robot is to be driven from an initial position through two via points before it reaches its
final destination using a 4-3-4 trajectory. The positions, velocities, and time duration for
the three segments for one of the joints are given below. Determine the trajectory
equations and plot the position, velocity, and acceleration curves for the joint.
θ1 = 20D
θ1 = 0
θ1 = 0
θ 2 = 60D
τ 2i = 0
τ2 f = 2
θ 3 = 100D
τ 3i = 0
τ3 f = 1
θ 4 = 40D
θ4 = 0
θ4 = 0
τ 1i = 0
τ1 f = 1
Estimated student time to complete: 50-60 minutes (with plotting)
Prerequisite knowledge required: Text Section(s) 5.5.5
Solution:
Using Equations 5.19 to 5.32 we get the following equations. The same can be found
using the matrix equation of Equation 5.33:
⎧a0 = 20
⎪a = 0
⎪ 1
⎪a2 = 0
⎪
⎪b0 = 60
⎪c = 100
⎪ 0
⎪60 = 20 + a3 + a4
⎪3a + 4a = b
⎪ 3
4
1
⎨
⎪6a3 + 12a4 = 2b2
⎪100 = 60 + 2b1 + 4b2 + 8b3
⎪
⎪b1 + 4b2 + 12b3 = c1
⎪2b + 12b = 2c
3
2
⎪ 2
⎪40 = 100 + c1 + c2 + c3 + c4
⎪
⎪0 = c1 + 2c2 + 3c3 + 4c4
⎪⎩0 = 2c2 + 6c3 + 12c4
→
⎧a3 = 80.95
⎪a = −40.95
⎪ 4
⎪b1 = 79.05
⎪
⎪b2 = −2.85
⎪
⎨b3 = −13.34
⎪c = −92.4
⎪1
⎪c2 = −82.875
⎪
⎪c3 = 202.9
⎪⎩c4 = −87.62
These equations can be solved by elimination or an equation solver. Substituting the
coefficients into each segment we get:
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92
θ1 (t ) = 20 + 80.95t 3 − 40.95t 4
θ 2 (t ) = 60 + 79.05t − 2.85t 2 − 13.34t 3
θ3 (t ) = 100 − 92.4t − 82.875t 2 + 202.9t 3 − 87.62t 4
The results are plotted below.
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93
Problem 5.7
A 2-DOF planar robot is to follow a straight line in Cartesian-space between the start
(2,6) and the end (12,3) points of the motion segment. Find the joint variables for the
robot if the path is divided into 10 sections. Each link is 9 inches long.
Estimated student time to complete: 60-75 minutes (with plotting)
Prerequisite knowledge required: Text Section(s) 5.6
Solution:
The inverse kinematic equations can be found in different ways. We will use a simple
Denavit-Hartenberg representation as follows:
y0
z2
z1
x2
θ2
1-2
⎡C1
⎢S
A1 = ⎢ 1
⎢0
⎢
⎣0
− S1 0 9C1 ⎤
C1 0 9 S1 ⎥⎥
0 1 0 ⎥
⎥
0 0 1 ⎦
θ
θ1
θ2
d
0
a
9
0
9
α
0
0
x1
θ1
z0
#
0-1
x0
⎡ C2
⎢S
A2 = ⎢ 2
⎢0
⎢
⎣0
− S2
C2
0
0
0 9C2 ⎤
0 9 S 2 ⎥⎥
1 0 ⎥
⎥
0 1 ⎦
T2 = A1 A2
0
Multiplying A1 by A2 and setting px and py equal to the components of position for 0T2, we
get:
⎧⎪ px = 9C1C2 − 9 S1S 2 + 9C1 = 9C12 + 9C1
⎨
⎪⎩ p y = 9S1S2 + 9C1C2 + 9S1 = 9S12 + 9S1
(1)
(2)
Squaring these equations and adding them together, and then regrouping, we get:
⎛ px2 + p y2 ⎞
− 1 → θ 2 = cos ⎜
− 1⎟
C2 =
⎜ 162
⎟
162
⎝
⎠
Substitute into (1) to get:
px2 + p y2
−1
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94
px
+ S1S 2
9
(3)
C2 + 1
Substitute (3) into (2) to get:
C1 =
py
S 2 px
+ S1S 22 =
( C2 + 1)
9
9
⎛ p y ( C2 + 1) − S 2 px ⎞
p y ( C2 + 1) − S 2 px
→ θ1 = sin −1 ⎜⎜
S1 =
⎟⎟
18 ( C2 + 1)
18 ( C2 + 1)
⎝
⎠
S1 ( C2 + 1) +
2
Next, calculate the slope of the straight line motion:
y −3
3−6
=
= −0.3
x − 12 12 − 2
The following table shows the coordinates of the intermediate points and the joint angles
from the above equations. The joint angles are shown on the graph as well:
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95
Problem 5.8
The 3-DOF robot of Example 5.7, as shown in Figure 5.18, is to move from point (3, 5,
5) to point (3, -5, 5) along a straight line, divided into 10 sections. Find the angles of the
three joints for each intermediate point and plot the results.
Estimated student time to complete: 30 minutes, depending on programming expertise
Prerequisite knowledge required: Text Section(s) 5.6
Solution:
Using the inverse kinematic equations of the robot from Example 5.7, the joint angles can
be found as shown. The values are also plotted:
#
1
2
3
4
5
6
7
8
9
10
11
Px
3
3
3
3
3
3
3
3
3
3
3
Py
5
4
3
2
1
0
-1
-2
-3
-4
-5
Pz
5
5
5
5
5
5
5
5
5
5
5
Theta 1
30.96
36.87
45
56.31
71.57
90
108.4
123.7
135
143.1
149
Theta 2
48.59
49.87
52.04
54.87
57.51
99.6
122.5
125.1
128
130.1
131.4
Theta 3
137.27
142.2
146.45
149.8
151.98
160.82
151.98
149.8
146.45
142.2
137.27
180
160
Joint 3
140
120
100
80
60
Joint 2
40
Joint 1
20
0
1
2
3
4
5
6
7
8
9
10
11
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96
CHAPTER SIX
Problem 6.1
Derive the inverse Laplace transform of the following equation:
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 6.5 and 6.6
Solution:
F (s) =
a
a
3
3
=
= 1 + 2
( s + 5s + 4) ( s + 1)( s + 4 ) s + 1 s + 4
2
⎡
⎤
3
3
= =1
a1 = ⎢( s + 1)
⎥
( s + 1)( s + 4 ) ⎦ s =−1 3
⎣
⎡
⎤
3
3
=
= −1
a2 = ⎢( s + 4 )
⎥
( s + 1)( s + 4 ) ⎦ s =−4 −3
⎣
1
1
F (s) =
−
s +1 s + 4
−1
1
f (t ) = L−1
+ L−1
= e −t − e−4t
s +1
s+4
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97
Problem 6.2
Derive the inverse Laplace transform of the following equation:
( s + 6)
F (s) =
2
s ( s + 5s + 6)
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 6.5, 6.6
Solution:
F (s) =
=
( s + 6)
( s + 6)
=
s ( s + 5s + 6) s ( s + 5) ( s + 3)
2
a
a1
a
+ 2 + 3
s s+2 s+3
⎡
⎤
6
s+6
a1 = ⎢( s )
⎥ = =1
⎣ s ( s + 2 )( s + 3) ⎦ s =0 6
⎡
⎤
4
s+6
=
= −2
a2 = ⎢ ( s + 2 )
⎥
s ( s + 2 )( s + 3) ⎦ s =−2 −2
⎣
⎡
⎤
3
s+6
=
=1
a3 = ⎢( s + 3)
⎥
s ( s + 2 )( s + 3) ⎦ s =−3 ( −3)( −1)
⎣
1 −2
1
+
F (s) = +
s s+2 s+3
1 ⎞
⎛ 1 −2
−3t
−2 t
+
f (t ) = L−1 ⎜ +
⎟ = u (t ) − e + e
⎝ s s+2 s+3⎠
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98
Problem 6.3
Derive the inverse Laplace transform of the following equation:
1
F (s) =
2
( s + 1) ( s + 2)
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 6.5, 6.6
Solution:
F (s) =
b2
b
a1
1
=
+ 1 +
2
( s + 1) ( s + 2) ( s + 1) ( s + 1) ( s + 2)
2
⎡
⎤
1
2
=1
b2 = ⎢( s + 1)
2
( s + 1) ( s + 2) ⎥⎦ s =−1
⎣
b1 =
−1
d ⎡ 1 ⎤
=
= −1
⎢
⎥
ds ⎣ ( s + 2) ⎦ s =−1 ( s + 2) 2
⎡
⎤
1
=1
a1 = ⎢( s + 2 )
2
( s + 1) ( s + 2) ⎥⎦ s =−2
⎣
1
1
1
−
+
F (s) =
2
( s + 1) ( s + 1) ( s + 2)
⎛ 1
1
1 ⎞
−t
−2 t
−
+
f (t ) = L−1 ⎜
⎟ = ( t − 1) e + e
2
+
+
+
s
s
s
(
1)
(
1)
(
2)
⎝
⎠
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99
Problem 6.4
Derive the inverse Laplace transform of the following equation:
10
F (s) =
( s + 4)( s + 2)3
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.5-6.6
Solution:
F (s) =
b3
b2
b1
a1
10
=
+
+
+
3
3
2
( s + 4)( s + 2)
( s + 2) ( s + 2) ( s + 2) ( s + 4)
⎡
⎤
10
3
=5
b3 = ⎢( s + 2 )
3⎥
( s + 4)( s + 2) ⎦ s =−2
⎣
b2 =
b1 =
⎤
−10
10
d ⎡
d ⎛ 10 ⎞
3
= ⎜
=
( s + 2)
⎟
⎢
3⎥
( s + 4)( s + 2) ⎦ s =−2 ds ⎝ s + 4 ⎠ s =−2 ( s + 4 )2
ds ⎣
1 d 2 ⎛ 10 ⎞
1 d −10
=
⎟
2 ⎜
2 ds ⎝ s + 4 ⎠ s =−2 2 ds ( s + 4 )2
=
s =−2
=
s =−2
−10
= −2.5
4
1 20
5
=
3
2 ( s + 4) s =−2 4
⎡
⎤
10
−5
a1 = ⎢( s + 4 )
=
3⎥
( s + 4)( s + 2) ⎦ s =−4 4
⎣
5
5
5
5
F (s) =
−
+
−
3
2
( s + 2) 2( s + 2) 4( s + 2) 4( s + 4)
⎛ 5
5
5
5 ⎞ ⎛ 5 2 5 5 ⎞ −2t 5 −4t
−
+
−
f (t ) = L−1 ⎜
⎟ = ⎜ t − t + ⎟e − e
3
2
2 4⎠
4
⎝ ( s + 2) 2( s + 2) 4( s + 2) 4( s + 4) ⎠ ⎝ 2
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100
Problem 6.5
Simplify the block diagram of Figure P.6.5.
H2
A
+-
+-
B
+-
G1
G2
H1
H3
Figure P.6.5
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 6.8.
Solution:
The following sequence is used to simplify the block diagram:
H2 /G1
B
A
+-
+-
+-
G2
G1
H1
H3
H2 /G1
A
+-
+-
B
G1
1 + G1 H1
G2
H3
H2 /G1
A
+-
+-
B
G1G2
1 + G1 H1
H3
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101
A
+-
G1G2
1 + G1 H1 + G2 H 2
B
H3
A
G1G2
1 + G1 H1 + G2 H 2 + G1G2 H 3
B
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102
Problem 6.6
Simplify the block diagram of Figure P.6.6.
A
+-
+-
G1
+-
H1
G2
B
H2
H3
Figure P.6.6
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 6.8.
Solution:
The following sequence is used to simplify the block diagram:
A
+-
G2
1 + G2 H 2
G1
1 + G1 H1
B
H3
A
+-
G1G2
(1 + G1H1 )(1 + G2 H 2 )
B
H3
A
+-
K
B
where K is calculated as follows:
G1G2
(1 + G1H1 )(1 + G2 H 2 ) =
G1G2
K=
G1G2 H 3
(1 + G1H1 )(1 + G2 H 2 ) + G1G2 H 3
1+
(1 + G1H1 )(1 + G2 H 2 )
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103
Problem 6.7
Simplify the block diagram of Figure P.6.7.
G4
A
+-
+-
G1
G3
G2
+-
B
H1
H3
Figure P.6.7
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.8.
Solution:
The following sequence is used to simplify the block diagram:
A
+-
G1G2
1 + G1G2 H1
G3 + G4
B
G3 + G4
B
H3
A
+-
G1G2
1 + G1G2 H1
H3
A
+-
K
B
G1G2 ( G3 + G4 )
G1G2 ( G3 + G4 )
1 + G1G2 H1
=
where K =
G G ( G + G4 ) H 3 1 + G1G2 H1 + G1G2 ( G3 + G4 ) H 3
1+ 1 2 3
1 + G1G2 H1
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104
Problem 6.8
Write an equation that describes the output of the system of Figure P.6.8.
U1
U2
+-
G1
++
Y
G2
H1
Figure P.6.8
Estimated student time to complete: 10-15 minutes
Prerequisite knowledge required: Text Section(s) 6.8.
Solution:
( (U − YH ) G + U ) G = Y
1
1
1
2
2
(U1G1 − YG1H1 + U 2 ) G2 = Y
U1G1G2 − YG1G2 H1 + U 2G2 = Y
U1G1G2 + U 2G2 = Y (1 + G1G2 H1 )
Y=
U1G1G2 + U 2G2
1 + G1G2 H1
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105
Problem 6.9
Write the equations that describe the input-output relationships for Figure P.6.9.
U2
U1
++
+
G1
H1
U3
++
Y1
H2
Y2
G2
Figure P.6.9
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 6.8.
Solution:
⎧⎪U1G1 + U 2 + Y2 H 2 = Y1
⎨
⎪⎩(U 3 + U1G1 H1 ) G2 = Y2
⎧U1G1 + U 2 + Y2 H 2 = Y1
⎨
⎩U 3G2 + U1G1G2 H1 = Y2
Substitute Y2 into Y1 to get:
⎧Y1 = U1G1 + U 2 + U 3G2 H 2 + U1G1G2 H1 H 2
⎨
⎩Y2 = U 3G2 + U1G1G2 H1
⎡ Y1 ⎤ ⎡G1 + G1G2 H1 H 2
⎢Y ⎥ = ⎢
G1G2 H1
⎣ 2⎦ ⎣
⎡U ⎤
1 G2 H 2 ⎤ ⎢ 1 ⎥
U
0 G2 ⎥⎦ ⎢ 2 ⎥
⎢⎣U 3 ⎥⎦
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106
Problem 6.10
Sketch the root locus for the following:
K
GH =
s ( s + 1)( s + 3)( s + 4)
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.13.
Solution:
Poles are at p = 0, p = −1, p = −3, p = −4
Number of asymptotes: α = 4 − 0 = 4 , angles of asymptotes: 45, 135, 225, 315.
∑ σ p − ∑ σ Z = ( 0 − 1 − 3 − 4 ) − 0 = −2
Asymptotic center: σ A =
4
α
The root locus, as drawn by MATLAB, is shown:
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107
Problem 6.11
Sketch the root locus for the following:
K ( s + 6)
GH =
s ( s + 4)
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 6.13.
Solution:
Roots are at z = −6, p = 0, p = −4
Number of asymptotes: α = 2 − 1 = 1 , angles of asymptote: 180.
The root locus, as drawn by MATLAB, is shown:
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108
Problem 6.12
Sketch the root locus for the following:
GH =
K ( s + 6)
s ( s + 10 − j10)( s + 10 + j10)( s + 12)
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.13.
Solution:
Roots are at z = −6, p = 0, p = −10 ± j10, p = −12
Number of asymptotes: α = 4 − 1 = 3 , angles of asymptotes: 60, 180, 300.
∑ σ p − ∑ σ Z = ( 0 − 10 − 10 − 12 ) − (−6) = −6.5
Asymptotic center: σ A =
4
α
The root locus, as drawn by MATLAB, is shown:
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109
Problem 6.13
For the system of Problem 6.10, assume that two of the roots are chosen at
s = −5 ± 2.55 j . Find the system’s gain, damping ratio, and natural frequency. Show that
the angle criterion is met. Can you determine from the root locus whether or not the
system is stable?
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 6.13, 6.14.
Solution:
∏ M pi = 52 + 2.552 42 + 2.552 22 + 2.552 12 + 2.552 = 236
K=
∏ M zi
⎛ 2.55 ⎞
⎟ = 180 − 27 = 153
⎝ 5 ⎠
⎛ 2.55 ⎞
θ 2 = 180 − tan −1 ⎜
⎟ = 180 − 32.52 = 147.48
⎝ 4 ⎠
⎛ 2.55 ⎞
θ3 = 180 − tan −1 ⎜
⎟ = 180 − 51.9 = 128.1
⎝ 2 ⎠
⎛ 2.55 ⎞
θ 4 = 180 − tan −1 ⎜
⎟ = 180 − 68.59 = 111.4
⎝ 1 ⎠
∑θtotal = 153 + 147.48 + 128.1 + 111.4 = 540 =3 ×180
It meets the criterion.
θ1 = 180 − tan −1 ⎜
⎛
⎝
ζ = cos θ = cos ⎜ tan −1
2.55 ⎞
⎟ = 0.891
5 ⎠
s = −ζωn ± ωn 1 − ζ 2 j = −0.891ωn ± ωn 1 − 0.8912 j
= −5 ± 2.55 j
ωn = 5.61 rad / sec
The root have to potential to fall on the right side, and therefore, may become unstable..
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110
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111
Problem 6.14
For the system of Problem 6.10, assume that two of the roots are chosen at
s = −4 ± 1.24 j . Find the system’s gain, damping ratio, and natural frequency.
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 6.13,6.14
Solution:
∏ M pi = 42 + 1.242 32 + 1.242 12 + 1.242 1.242 = 26.85
K=
∏ M zi
⎛
⎝
ζ = cos θ = cos ⎜ tan −1
1.24 ⎞
⎟ = 0.955
4 ⎠
s = −ζωn ± ωn 1 − ζ 2 j = −0.891ωn ± ωn 1 − 0.9552 j
= −4 ± 1.24 j
ωn = 4.19 rad / sec
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112
Problem 6.15
For the system of Problem 6.11, assume that the roots are chosen at s = −3 ± 1.73 j . Find
the system’s gain, damping ratio, and natural frequency. Show that the angle criterion is
met. Can you determine whether or not the system may become unstable as the gain
changes?
Estimated student time to complete: 30 minutes
Prerequisite knowledge required: Text Section(s) 6.13, 6.14.
Solution:
K=
∏ M p = 3 + 1.73 1 + 1.73 = 2
3 + 1.73
∏Mz
2
2
2
2
i
2
2
i
⎛
⎝
ζ = cos θ = cos ⎜ tan −1
1.73 ⎞
⎟ = 0.866
3 ⎠
s = −ζωn ± ωn 1 − ζ 2 j = −0.866ωn ± ωn 1 − 0.8662 j
= −3 ± 1.73 j
ωn = 3.46 rad / sec
⎛ 1.73 ⎞
⎟ = 180 − 30 = 150
⎝ 3 ⎠
⎛ 1.73 ⎞
θ p−4 = 180 − tan −1 ⎜
⎟ = 60
⎝ 1 ⎠
⎛ 1.73 ⎞
θ z6 = tan −1 ⎜
⎟ = 30
⎝ 3 ⎠
∑θtotal = 150 + 60 − 30 = 180
θ p = 180 − tan −1 ⎜
0
It meets the criterion.
The roots are always on the left side, and therefore, the system is always stable.
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113
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114
Problem 6.16
For the system of Problem 6.11, find the roots, the gain, and the steady-state error for a
settling time of less that 1 second and overshoot of 4% or less.
Estimated student time to complete: 30-40 minutes
Prerequisite knowledge required: Text Section(s) 6.14
Solution:
GH =
K ( s + 6)
s ( s + 4)
By trial and error,
%OS = e
( −ζπ / 1−ζ ) ×100% = 4% → ζ ≥ 0.715,
2
θ = cos −1 ζ = 44.4D
The chosen roots must be below this value, therefore choose s = −4 ± 2.83 j
4
ζωn (approximate) = = 4
1
System type: 1 → Ess = 0
∏ M p = −4 + 2.83 2.83 = ( 4.9 )( 2.83) = 4
K=
3.465
−2 + 2.83
∏Mz
2
2
2
i
2
2
i
2.83
= 35.28D
4
ζ actual = cos θ = 0.816
θ = tan −1
The root locus, as drawn by MATLAB is shown:
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115
Problem 6.17
For the following system, find the roots, the gain, and steady-state error for the fastest
response and a settling time of less than 2 seconds and an overshoot of less than 4%.
K
GH =
( s + 1)( s + 3)
Estimated student time to complete: 30-40 minutes
Prerequisite knowledge required: Text Section(s) 6.14
Solution:
For root locus: σ A =
−1 − 3
= −2 and θ A = ±90D
2
For overshoot of 4%: %OS = e
( −ζπ / 1−ζ ) ×100% = 4% → ζ ≅ 0.72
2
θ = cos −1 ζ = 44D and Ts =
4
ζωn
=
4
=2
2
The root locus and the limitations are shown:
Choose poles for fastest response: tan 44D =
Therefore: s = −2 ± 1.93 j
K=
l
→ l = 1.93
2
∏ M p = (3 − 2) + 1.93 (1 − 2) + 1.93 = 4.72
1
∏Mz
2
2
2
2
i
i
System type: 0
4.72
= 1.57
s → 0 ( s + 1)( s + 3 )
→ K P = lim G ( s ) = lim
s →0
Ess =
1
= 0.39
1 + 1.57
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116
Problem 6.18
For the system of problem 6.17, select the locations and the proportional and integral
gains to change it to a proportional-plus-integral system with zero steady-state error.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.15
Solution:
We add s = 0 at the origin and a zero near it (location to be found). Therefore,
K (s + z)
K ( s + 0.1)
=
s ( s + 1)( s + 3) s ( s + 1)( s + 3)
Assume ζ = 0.73 for 4% OS even though the system is not second-order (this is close
enough). Therefore,
GH =
θ = 43D approximately at s = −2 ± 1.87 j .
K=
∏ M p = 2 + 1.87 1 + 1.87 1 + 1.87 = 4.6
1.9 + 1.87
∏Mz
2
2
2
2
2
2
i
2
2
i
The gain can be calculated as:
⎛
K ⎞
KP ⎜ s + I ⎟
K P ⎠ 4.6 ( s + 0.1)
K s + KI
⎝
=
=
K= P
s
s
s
K P = 4.6 and K I = 0.46
Since this is Type-1, Ess = 0.
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117
Problem 6.19
For the system of Problem 6.17, we would like to improve the settling time to 1 second
by adding a zero to the system, therefore a proportional-plus-derivative system. Find a
proper location for the zero and the loop gain.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.16
Solution:
Although not accurate, assume that σ =
to be at s = −4 ± 3.85 j .
4
= 4 . Therefore, we arbitrarily choose the poles
1
Angle deficiency:
3.85
= 127.9D
3
3.85
θ −3 = 180 − tan −1
= 104.6D
1
θ z = 127.9 + 104.6 − 180 = 52.5D
θ −1 = 180 − tan −1
3.85
→ z = 6.95
z−4
The loop gain is:
tan 52.5 =
K=
∏ M p = 3 + 3.85 1 + 3.85 = 4
2.95 + 3.85
∏Mz
2
2
2
2
i
2
2
i
GH =
4 ( z + 6.95 )
( s + 1)( s + 3)
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118
Problem 6.20
For the system of problem 6.19, add an integrator to the system to make it into a PID
system in order to achieve a zero steady-state error. Find the location of an additional
zero and proportional, derivative, and integral gains.
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 6.17
Solution:
K ( s + 6.95 )( s + z )
s ( s + 1)( s + 3)
Magnitude criterion:
GH =
Assume:
K=
∏ M p = 4 + 3.85 3 + 3.85 1 + 3.85 = 4.06
2.95 + 3.85 3.9 + 3.85
∏Mz
2
2
2
2
2
2
i
2
2
2
2
i
⎛
K
K ⎞
KD ⎜ s2 + P s + I ⎟
KD
KD ⎠
V
K
⎝
G = a = KP + I + KDs =
E
s
s
2
4.06( s + 6.95)( s + 0.1) 4.06( s + 7.05s + 0.695)
=
=
s
s
⎧
⎪
⎪ K D = 4.06
⎧ K D = 4.06
⎪ KP
⎪
= 7.05 → ⎨ K P = 28.6
⎨
⎪ KD
⎪ K = 2.82
⎩ I
⎪ KI
= 0.695
⎪
⎩ KD
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119
CHAPTER SEVEN
Problem 7.1
A motor with a rotor inertia of 0.030 Kgm2 and maximum torque of 12 Nm is connected
to a uniformly distributed arm with a concentrated mass at its end, as shown in Figure
P.7.1. Ignoring the inertia of a pair of reduction gears and viscous friction in the system,
calculate the total inertia felt by the motor and the maximum angular acceleration it can
develop if the gear ratio is a) 5, b) 50, c) 100. Compare the results.
z
y
0.5 m
Tl
N
x
Im
m=3Kg
Tm
m=3Kg
Figure P.7.1.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 7.2
Solution:
1
1
I load = marml 2 + mmass l 2 = (3)(0.5) 2 + (3)(0.5) 2 = 1 Kgm 2
3
3
a) For N = 5,
b) For N = 50,
c) For N = 100,
1
⎧
⎛ 1 ⎞
2
⎪ I total = I motor + ⎜ N 2 ⎟ I load = 0.03 + 25 =0.07 Kgm
⎪
⎝
⎠
⎨
⎪θ = Tmax = 12 = 171 r / s 2
⎪⎩ max I total 0.07
1
⎧
⎛ 1 ⎞
2
⎪ I total = I motor + ⎜ N 2 ⎟ I load = 0.03 + 2500 =0.0304 Kgm
⎪
⎝
⎠
⎨
⎪θ = Tmax = 12 = 395 r / s 2
⎪⎩ max I total 0.0304
1
⎧
⎛ 1 ⎞
2
⎪ I total = I motor + ⎜ N 2 ⎟ I load = 0.03 + 10000 =0.0301 Kgm
⎪
⎝
⎠
⎨
⎪θ = Tmax = 12 = 399 r / s 2
⎪⎩ max I total 0.0301
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120
Problem 7.2
Repeat Problem 1, but assume that the two gears have 0.002 Kgm2 and 0.005 Kgm2
inertias respectively.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 7.2
Solution:
1
1
I load = marml 2 + mmass l 2 = (3)(0.5) 2 + (3)(0.5) 2 = 1 Kgm 2
3
3
0.005
I gears = 0.002 +
N2
0.005 1
1.005
⎛ 1 ⎞
+ 2 = 0.032 +
I total = I motor + I gears + ⎜ 2 ⎟ I load = 0.03 + 0.002 +
2
N
N
N2
⎝N ⎠
a) For N = 5,
1.005
⎧
2
⎪⎪ I total = 0.032 + 25 =0.0722 Kgm
⎨
⎪θmax = Tmax = 12 = 166 r / s 2
I total 0.0722
⎪⎩
b) For N = 50,
1.005
⎧
2
⎪⎪ I total = 0.032 + 2500 = 0.0324 Kgm
⎨
⎪θmax = Tmax = 12 = 370 r / s 2
I total 0.0324
⎪⎩
c) For N = 100,
1.005
⎧
2
⎪ I total = 0.032 + 10, 000 =0.0321 Kgm
⎪
⎨
⎪θmax = Tmax = 12 = 374 r / s 2
⎪⎩
I total 0.0321
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121
Problem 7.3
The three-axis robot shown in Figure P.7.3. is powered by geared servomotors attached to
the joints by worm gears. Each link is 22 cm long, made of hollow aluminum bars, each
weighing 0.5 Kg. The center of mass of the second motor is 20 cm from the center of
rotation. The gear ratio is 1/3 in the servomotor and 1/5 in the worm gear set. The worst
case scenario for the elbow joint is when the arm is fully extended, as shown. Calculate
the torque needed to accelerate both arms together, fully extended, at a rate of 90 rad/s2.
Assume the inertias of the worm gears are negligible.
Worm gear
N=5
m4=0.5 Kg
m2=0.5 Kg
I1=0.01 Kgm2
N=3
m3=0.5 Kg
Motors
Figure P.7.3.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 7.2
Solution:
Total gear ratio N = 5 × 3 = 15
(
I total = I rotor1 + I arm1+2 + I motor2
) N1
2
2
2⎞
⎛1
= 0.01 + ⎜ ( 2marm )( 2l ) + mmotor ( lmotor ) ⎟
⎝3
⎠
2
2⎤ 1
⎡1
= 0.01 + ⎢ (1)( 0.44 ) + ( 0.5 )( 0.2 ) ⎥ 2 = 0.01038
⎣3
⎦ 15
T = Iθ = 0.01038 × 90 = 0.934 Nm
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122
Problem 7.4
Repeat Problem 3, but suppose the maximum torque this motor can provide is 0.9 Nm.
Therefore, a new motor must be picked. Two other motors are available, one with the
inertia of 0.009 Kgm2 and torque of 0.85 Nm, one with inertia of 0.012 Kgm2 and torque
of 1 Nm. Which one would you use?
Estimated student time to complete:15-20 minutes
Prerequisite knowledge required: Text Section(s) 7.2.
Solution:
For N = 5 × 3 = 15 ,
For case 1:
(
I total = I rotor1 + I arm1+2 + I motor2
) N1
2
2
2⎞
⎛1
= 0.009 + ⎜ ( 2marm )( 2l ) + mmotor ( lmotor ) ⎟
⎝3
⎠
2
2⎤ 1
⎡1
= 0.009 + ⎢ (1)( 0.44 ) + ( 0.5 )( 0.2 ) ⎥ 2 = 0.00938
⎣3
⎦ 15
T = Iθ = 0.00938 × 90 = 0.844 Nm
For case 2:
(
I total = I rotor1 + I arm1+2 + I motor2
) N1
2
2
2⎞
⎛1
= 0.012 + ⎜ ( 2marm )( 2l ) + mmotor ( lmotor ) ⎟
⎝3
⎠
2
2⎤ 1
⎡1
= 0.009 + ⎢ (1)( 0.44 ) + ( 0.5 )( 0.2 ) ⎥ 2 = 0.01238
⎣3
⎦ 15
T = Iθ = 0.01238 × 90 = 1.114 Nm
The smaller motor is better.
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123
Problem 7.5
Estimate how much the torque/inertia ratio of a disk motor might be if it can go from zero
to 2000 rpm in one millisecond, and compare it to the motor of Problem 1.
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 7.2
Solution:
2000 × 2π
= 209.44 rad / sec
60
θ − θ 209.44 − 0
θ = 2 1 =
= 209440 rad / sec 2
0.001
t2 − t1
The torque/inertia ratio is:
θ2 =
T T = Iθ →
= θ = 209440
I
For the motor of Problem 1:
12
T
=
= 400 rad / sec 2
T = Iθ →
I 0.03
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124
Problem 7.6
Using a timer circuit, design a pulse generating circuit that will deliver a range of 5-500
pulses per second to a stepper motor driver.
Estimated student time to complete:15-20 minutes
Prerequisite knowledge required: Text Section(s) 7.6.7.
Solution:
We will pick a 50% duty cycle for each signal. Therefore:
1
tlow = thigh = ttotal
2
ttotal ranges between 0.002 and 0.2 seconds. Therefore tlow = 0.001 sec or tlow = 0.1 sec.
Pick C2 = 0.1 μ F . Therefore:
tlow = 0.693C2 ( R2 )
0.001 = 0.693 ( 0.1× 10−6 ) R2
0.1 = 0.693 ( 0.1× 10−6 ) R2
→ R2 = 14, 430 Ω
→
R2 = 1, 443, 000 Ω
Therefore, a potentiometer with R2 = 1.5 M Ω enables us to achieve the desired values.
For 50% duty cycle, as shown above,
tlow = thigh
0.693C2 ( R2 ) = 0.693C2 ( RH + R2 ) → RH = R1 + R3 + R4 = 0
Pick R1 = R3 = R4 = 1 K Ω . Therefore:
⎧C2 = 0.1 μ F
⎪R = 1 K Ω
⎪⎪ 1
⎨ R2 = 1.5 M Ω
⎪R = 1 K Ω
⎪ 3
⎪⎩ R4 = 1 K Ω
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125
Problem 7.7
Calculate the gear ratio for a Harmonic drive if NL=100, NF=95, N2=90, N3=95.
Estimated student time to complete: 10-15 minutes
Prerequisite knowledge required: Text Section(s) 7.11
Solution:
From Equation 7.25:
e=
ωL N 2 N L − N F N 3 ( 90 )(100 ) − ( 95 )( 95 )
1
=
=
=−
ωA
N2 NL
360
( 90 )(100 )
The gear ratio between the Arm and the Last gear (with Arm as input) is:
ωA
= −360
ωL
The minus sign indicates that the input (Arm) and the output (Last) gears rotate in
opposite directions.
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126
Problem 7.8
Write a program to generate a variable pulse stream to drive a motor with pulse-widthmodulated voltages of 1, 2, 3, 4, and 5 volts for a 5 volt input.
Estimated student time to complete: Variable depending on expertise.
Prerequisite knowledge required: Text Section(s) 7.7.
Solution:
The accuracy with which this program will work depends on the processing speed of the
microprocessor. It also depends on the level of programming language and how
efficiently it is written. We assume that the processor can accept and run a compiled “C”
program, and that it is fast enough such that the execution of commands will not
negatively affect the accuracy of the output timing. Since only five distinct output
voltages of 1, 2, 3, 4, and 5 volts are desired, we only need to generate 5 variations in the
timing loops. We assume the output level will be indicated by an input port. Therefore, if
input port 1 is high, the desired output level is 1 volts, etc.
The following program is written in “C” language for execution by a Mini-Board™
microprocessor. It is the essentials of the program that are important here, not the actual
program and the way it is written. This program must be modified for use for any other
specific microprocessor with its own unique programming syntax and requirements.
In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at
n
VCC . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command
15
“msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port
1 is turned off.
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127
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128
Problem 7.9
Write a program to generate a sinusoidal pulse-width-modulated output for a constant
input voltage.
Estimated student time to complete: Variable depending on expertise.
Prerequisite knowledge required: Text Section(s) 7.7.
Solution:
The accuracy with which this program will work depends on the processing speed of the
microprocessor. It also depends on the level of programming language and how
efficiently it is written. We assume that the processor can accept and run a compiled “C”
program, and that it is fast enough such that the execution of commands will not
negatively affect the accuracy of the output timing. Since a sinusoidal out put is desired
for a 5 volt input voltage, we need to generate a variable control timing for the pulse
width modulation. We calculate the sine value for each degree between zero and 90
degrees. This allows for approximately 10 pulses per second.
The following program is written in “C” language for execution by a Mini-Board™
microprocessor. It is the essentials of the program that are important here, not the actual
program and the way it is written. This program must be modified for use for any other
specific microprocessor with its own unique programming syntax and requirements.
Since Mini-Board can generate voltages in both polarities, the program can generate a
full sine wave. The program must be modified accordingly if the microprocessor cannot
do this.
In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at
n
VCC . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command
15
“msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port
1 is turned off.
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129
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130
Problem 7.10
If you have access to a microprocessor and electronic components such as transistors,
make an H-bridge and write a control program to drive a motor in either direction or to
brake it. Be mindful of the problems associated with an H-bridge’s transistors turning on
and off at inappropriate times.
Estimated student time to complete:
Prerequisite knowledge required: Text Section(s)
Solution:
The following program is written in “C” language for execution by a Mini-Board™
microprocessor. It is the essentials of the program that are important here, not the actual
program and the way it is written. This program must be modified for use for any other
specific microprocessor with its own unique programming syntax and requirements.
In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at
n
VCC . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command
15
“msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port
1 is turned off.
It is assumed that an input to the microprocessor equal to 1 indicates rotation in the CCW
direction, 2 indicates CW rotation, and a 0 means the motor should be braked.
/************************************************************/
/*
H-Bridge Controller
*/
/************************************************************/
.
.
main( )
{
int direction
while(1)
{direction = digital(0);
if (direction = = 1) {msleep (1);
motor (1,15); motor (4,15);}
off (2); off (3);
if (direction = = 2) {msleep (1);
motor (2,15); motor (3,15);}
off (1); off (4);
/* motor rotation direction input */
/* CCW rotation */
/* CW rotation */
if (direction = = 0) {msleep (1);
/* brake the motor */
motor (3,15); motor (4,15);}
off (1); off (2); }}
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131
CHAPTER EIGHT
There are no problems in Chapter 8.
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132
CHAPTER NINE
Problem 9.1
Calculate the necessary memory requirement for a still color image from a camera with
10 megapixels at:
•
8-bits per pixel (256 levels)
•
16-bits per pixel (65536 levels)
Estimated student time to complete: 5 minutes
Prerequisite knowledge required: Text Section(s) 9.5
Solution:
For an 8-bit per pixel system, each pixel requires 1 byte. The total memory needed is 10
megabytes.
For a 16-bit per channel system, each pixel requires 2 bytes. The total memory needed is
20 megabytes.
The assumption here is that the pixel count given is the total number for all three colors.
The actual memory is greatly influenced by the format in which the image is saved. A
bitmap image needs much more memory than JPEG.
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133
Problem 9.2
Consider the pixels of an image, with values as shown in Figure P.9.2, as well as a
convolution mask with the given values. Calculate the new values for the given pixels.
4
3
5
8
7
2
3
6
1
6
2
5
8
6
3
9
1
1
1
0
1
0
1
1
1
Figure P.9.2
Estimated student time to complete: 5-10 minutes
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
4
3
5
8
7
3.3
4.6
6
1
5
4.4
5
8
6
3
9
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134
Problem 9.3
Consider the pixels of an image, with values as shown in Figure P.9.3, as well as a
convolution mask with the given values. Calculate the new values for the given pixels.
Substitute 0 for negative grey levels. What conclusion can you make from the result?
5
5
10
5
5
5
10
5
5
5
10
5
7
8
10
9
8
8
8
9
7
8
10
9
9
8
9
10
8
10
1
0
1
0
-4
0
1
0
1
Figure P.9.3
Estimated student time to complete: 5-10 minutes
Prerequisite knowledge required: Text Section(s) 9.10.
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
5
5
10
5
5
10
0
16
5
12
0
17
7
0
0
1
8
8
8
9
7
3
0
3
9
8
9
10
8
10
This can be an indication of an edge.
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135
Problem 9.4
Repeat Problem 9.3, but substitute the absolute value of negative grey levels. What
conclusion can you make from the result?
Estimated student time to complete: 5 -10 minutes
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
5
5
10
5
5
10
20
16
5
12
13
17
7
0
13
1
8
8
8
9
7
3
6
3
9
8
9
10
8
10
When absolute values of negative pixels are shown, edges are no longer discernable.
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136
Problem 9.5
Repeat Problem 9.3, but apply the mask of Figure P.9.5 and compare your results with
Problem 9.3. Which one is better?
-1
0
-1
0
4
0
-1
0
-1
Figure P.9.5
Estimated student time to complete: 5 -10 minutes
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
5
5
10
5
5
0
20
0
5
0
13
0
7
0
13
0
8
8
8
9
7
0
6
0
9
8
9
10
8
10
This mask also shows the edge, but unlike the one in problem 9.3., the pixels are brighter
than the neighboring pixels.
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137
Problem 9.6
Repeat Problem 9.3, but apply the mask of Figure P.9.6 and compare your results with
Problem 9.3. Which one is better?
0
1
0
1
-4
1
0
1
0
Figure P.9.6
Estimated student time to complete: 5 -10 minutes
Prerequisite knowledge required: Text Section(s)9.10
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
5
5
10
5
5
5
0
8
5
8
0
12
7
0
0
0
8
8
8
9
7
2
0
0
9
8
9
10
8
10
At least for this image, the effect is similar to the mask in problem 9.3.
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138
Problem 9.7
Repeat Problem 9.3, but apply the mask of Figure P.9.7 and compare your results with
Problem 9.3. Which one is better?
1
1
1
1
-8
1
1
1
1
Figure P.9.7
Estimated student time to complete: 5 -10 minutes
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
5
5
10
5
5
15
0
24
5
20
0
29
7
0
0
0
8
8
8
9
7
5
0
3
9
8
9
10
8
10
At least for this image, the effect is similar to Problem 9.3, but darker.
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139
Problem 9.8
An image is represented by values shown below.
a)
Find the value of pixel 2c when mask 1 is applied.
b)
Find the value of pixel 3b when mask 2 is applied.
c)
Find the values of pixels 2b and 3c when a 3 × 3 median filter is applied.
d)
Find the area of the major object that results when a threshold of 4.5 is applied
based on a +4-connectivity (start at the first on-pixel).
1
2
3
4
5
a
b
c
d
e
4
4
6
1
2
9
3
2
6
8
6
7
6
5
4
2
4
5
9
4
6
6
3
2
7
1
1
1
1
1
1
1
1
1
Mask 1
0
1
0
1
-4
1
0
1
0
Mask 2
Figure P.9.8.
Estimated student time to complete: 10 -15 minutes
Prerequisite knowledge required: Text Section(s) 9.10 and 9.17.
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
a) For 2c:
R = (9× 1 + 6× 1 +2× 1 +3× 1 +7× 1 +4× 1 +2× 1 +6× 1 +5× 1) / 9 = 4.9 (or 5)
b) For 3b:
R = (4× 0 + 3× 1 +7× 0 +6× 1 +2× -4 +6× 1 +1× 0 +6× 1 +5× 0) = 13
c) For 2b: 2,3,4,4,6,6,6,7,9. The median is 6.
For 3c: 2,3,4,5,5,6,6,7,9. The median is 5.
d) The resulting image after the application of the threshold is:
a
1
b
c
X
X
X
X
X
X X
X X
X
2
3
4
5
X
X
X
d
e
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140
Applying the +4 connectivity rule starting at the first on-pixel results in: 1b, 1c, 2c, 3c,
3d, 4c, 4d, 4b, 5b, for a total of 9 pixels.
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141
Problem 9.9
An image is represented by values shown below.
a.
Find the value of pixel 3b when mask 1 is applied.
b.
Find the values of pixels 2b, 2c, 2d when mask 2 is applied.
c.
Find the value of pixel 3c when a 5 × 5 median filter is applied.
d.
Find the area of the major object that results when a threshold of 4.5 is applied
based on a × 4-connectivity (start at the first on-pixel).
1
2
3
4
5
a
b
c
d
e
8
4
6
1
2
9
6
7
6
2
5
5
4
2
4
6
9
3
5
6
5
4
2
10
8
0
1
0
1
4
1
0
1
0
Mask 1
1
1
1
-2
-2
-2
1
1
1
Mask 2
Figure P.9.9.
Estimated student time to complete: 10 -15 minutes
Prerequisite knowledge required: Text Section(s) 9.10 and 9.17.
Solution:
The mask will only affect the middle pixels. The remaining pixels on the perimeter will
not be affected. Depending on how a new image is formatted, the pixels on the perimeter
may be set to zero or the current values.
The new values for the image, assuming that the old file is copied into a new file, are:
a) For 3b:
R = (4× 0 + 6× 1 +2× 0 +6× 1 +7× 4 +5× 1 +1× 0 +10× 1 +5× 0)/8 = 6.875 (or 7)
b) For 2b:
R = (8× 1 + 9× 1 +6× 1 +4× -2 +6× -2 +2× -2 +6× 1 +7× 1 +5× 1) = 17
For 2c:
R = (9× 1 + 6× 1 +2× 1 +6× -2 +2× -2 +4× -2 +7× 1 +5× 1 +6× 1) = 11
For 2d:
R = (6× 1 + 2× 1 +5× 1 +2× -2 +4× -2 +6× -2 +5× 1 +6× 1 +5× 1) = 5
c) For 3c: 1,2,2,2,2,3,4,4,4,4,5,5,5,5,6,6,6,6,6,7,8,8,9,9,10. The median is 5.
d) The resulting image after the application of the threshold is:
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142
1
a
b
c
X
X
X
X
X
X
X
2
3
4
5
X
d
e
X
X
X X X
X X
Applying the × 4 connectivity rule starting at the first on-pixel results in: 1a, 2b, 3c, 1c,
3a, 4d, 4b, 3e, for a total of 8 pixels.
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143
Problem 9.10
Write a computer program for the application of a 3× 3 averaging convolution mask unto
a 15× 15 image. Refer to the note on page 415 for more information.
Estimated student time to complete: Variable depending on the expertise
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
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144
Problem 9.11
Write a computer program for the application of a 5× 5 averaging convolution mask unto
a 15× 15 image. Refer to the note on page 415 for more information.
Estimated student time to complete: Variable depending on the expertise
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
145
Problem 9.12
Write a computer program for the application of a 3× 3 high pass convolution mask unto
a 15× 15 image for edge detection. Refer to the note on page 415 for more information.
Estimated student time to complete: Variable depending on expertise
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
146
Problem 9.13
Write a computer program for the application of an n× n convolution mask unto a k× k
image. Refer to the note on page 415 for more information. You should write the routine
such that the user can choose the size of the mask and the values of each mask cell
individually.
Estimated student time to complete: Variable depending on expertise
Prerequisite knowledge required: Text Section(s) 9.10
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
147
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electronic media system or the Internet without prior expressed consent of the copyright owner.
148
Problem 9.14
Write a computer program that will perform the Left-Right search routine for a 15× 15
image. Refer to the note on page 415 for more information.
Estimated student time to complete: Variable depending on expertise
Prerequisite knowledge required: Text Section(s) 9.13
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
149
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electronic media system or the Internet without prior expressed consent of the copyright owner.
150
Problem 9.15
Using the Left-Right search technique, find the outer edge of the object in Figure P.9.15:
a
b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
Figure P.9.15
Estimated student time to complete: 5-10 minutes
Prerequisite knowledge required: Text Section(s) 9.13
Solution:
a
b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
The edge consists of: 2c, 2d, 2e, 2f, 2g, 2h, 2i, 2j, 2k, 3k, 4k, 5j, 6j, 7k, 7j, 8i, 9h, 10h,
11h, 12h, 12g, 12f, 11f, 10g, 9g, 8f, 8e, 7e, 7d, 7c, 6b, 5b, 4b, 4c, 3c.
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151
Problem 9.16
The x and y coordinates of 5 points are given as (2.5, 0), (4,2), (5,4), (7,6), and (8.5,8).
Using the Hough transform, determine which of these points form a line and find its slope
and intercept.
Estimated student time to complete: 20-25 minutes
Prerequisite knowledge required: Text Section(s) 9.15.
Solution:
y
x
x,y
m,c
0
2.5
0=2.5m+c
c= -2.5m+0
2
4
2=4m+c
c= -4m+2
4
5
4=5m+c
c= -5m+4
6
7
6=7m+c
c= -7m+6
8
8.5
8=8.5m+c
c= -8.5m+8
The following graph shows the lines and where they intersect. Line 3 (for point 5,4) does
not intersect the remaining lines.
The approximate slope and intercept of the line on which the other four points lay are 1.4
and -3.5.
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152
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electronic media system or the Internet without prior expressed consent of the copyright owner.
153
Problem 9.17
Write a computer program that will perform a region growing operation based on +4
connectivity. The routine should start at the 1,1 corner pixel, search for a nucleus, grow a
region with a chosen index number, and after finishing that region, must continue
searching for another nuclei until all object pixels have been checked. Refer to the note
on page 415 for more information.
Estimated student time to complete: Variable depending on expertise.
Prerequisite knowledge required: Text Section(s) 9.17
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
In this program, when a nucleus is found, all the pixels connected to it based on a +4connectivity that are “on” (have a value in them and therefore, are part of the image not
the background) are placed in a stack by changing the value of the cell form zero to 1 in
an array of 256x256. The cell’s value is later checked until all are found and the cell’s are
all zero. Each corresponding pixel’s value is changed from 1 to the region number. At the
end of execution, the original image is changed to as many regions as necessary. Since
there will be no cells (pixels) left with a value of 1, execution ends. All segments will be
2 and larger in value.
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154
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electronic media system or the Internet without prior expressed consent of the copyright owner.
155
Problem 9.18
Write a computer program that will perform a region growing operation based on × 4
connectivity. The routine should start at the 1,1 corner pixel, search for a nucleus, grow a
region with a chosen index number, and after finishing that region, must continue
searching for another nuclei until all object pixels have been checked. Refer to the note
on page 415 for more information.
Estimated student time to complete: Variable depending on expertise.
Prerequisite knowledge required: Text Section(s) 9.17
Solution:
Please see the solution for Problem 9.17. The program for growing a region based on × 4
connectivity is very similar to that program, except that the search will follow the × 4
sequence given in Equation 9.7.
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156
Problem 9.19
Using +4 connectivity logic and starting from 1a pixel, write the sequence of pixels in
correct order that will be detected by a region growing routine for the object in Figure
P.9.19:
a b c d e f
1
2
3
4
5
6
7
8
9
g h
4
1
2
3
+4-connectivity
Figure P.9.19
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 9.17
Solution:
The sequence of pixels detected is: 2d, 3d, 3e, 3f, 4e, 3g, 4f, 5e, 5f, 5d, 5g, 6d, 6c, 7c, 7b,
8b.
The remaining pixels are not connected to this region and will not be detected.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
157
Problem 9.20
Using × 4 connectivity logic and starting from 1a pixel, write the sequence of pixels in
correct order that will be detected by a region growing routine for the object in Figure
P.9.20:
a b c d e f
g h
1
2
3
4
5
6
7
8
9
1
4
3
2
x4-connectivity
Figure P.9.20
Estimated student time to complete: 10 minutes
Prerequisite knowledge required: Text Section(s) 9.17
Solution:
The sequence of pixels detected is: 2c, 3d, 4e, 4c, 5f, 5d, 3f, 4b, 6g, 6c, 7b.
The remaining pixels are not connected to this region and will not be detected.
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158
Problem 9.21
Find the union between the two objects in Figure P.9.21.
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Figure P.9.21
Estimated student time to complete: 10-15 minutes
Prerequisite knowledge required: Text Section(s) 9.18
Solution:
The union is shown:
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159
Problem 9.22
Apply a single-pixel erosion based on 8-connectivity on the image of Figure P.9.22.
a b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
Figure P.9.22
Estimated student time to complete: 10-20 minutes
Prerequisite knowledge required: Text Section(s) 9.18
Solution:
The result is shown:
a b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
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160
Problem 9.23
Apply a one-pixel dilation to the result of Problem 9.22 and compare your result to
Figure P.9.22.
Estimated student time to complete: 15-25 minutes
Prerequisite knowledge required: Text Section(s)9.18
Solution:
The result is shown. As you notice, the image is different from the original.
a b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
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161
Problem 9.24
Apply an open operation to Figure P.9.24.
a b c
d e
f
g h
i
j
k
l
m
f
g h
1
2
3
4
5
6
7
8
9
10
11
12
13
Figure P.9.24
Estimated student time to complete: 10-20 minutes
Prerequisite knowledge required: Text Section(s) 9.18
Solution:
The result is shown:
a
1
2
3
4
5
6
7
8
9
10
11
12
13
b c
d e
f
g h
i
j
k
l
m
a
b c
d e
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
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162
Problem 9.25
Apply a close operation to Figure P.9.24.
Estimated student time to complete: 10-20 minutes
Prerequisite knowledge required: Text Section(s) 9.18
Solution:
The result is shown:
a b c
1
2
3
4
5
6
7
8
9
10
11
12
13
d e
f
g h
i
j
k
l
m
a
b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
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163
Problem 9.26
Apply a skeletonization operation to Figure P.9.26.
a b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
Figure P.9.26
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 9.18
Solution:
The result is shown:
a b c
d e
f
g h
i
j
k
l
m
1
2
3
4
5
6
7
8
9
10
11
12
13
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164
Problem 9.27
Write a computer program in which different moments of an object in an image can be
calculated. The program should ask you for moment indices. The results may be reported
to you in a new file, or may be stored in memory. Refer to the note on page 415 for more
information.
Estimated student time to complete: Variable depending on expertise
Prerequisite knowledge required: Text Section(s) 9.21.
Solution:
The following program is written in “C” language. It is the essentials of the programming
that are important here, not the syntax. You may use the proper syntax for the program
you are using. This program may be written in countless other languages, both graphical
and scientific.
In this program, the assumption is that the image input file is read from a data file or it is
entered after initializing the array.
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165
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166
Problem 9.28
Calculate the M 0,2 moment for the result of Problem 9.8d based on +4-connectivity.
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 9.21.
Solution:
The resulting image after the application of the threshold is:
a
1
b
c
X
X
X
X
X
X X
X X
X
2
3
X
X
X
4
5
d
e
Applying the +4 connectivity rule starting at the first on-pixel results in the following
area:
a
1
b
c
X
X
X
X X
X X
2
3
X
X
4
5
d
e
x
y
The moment, calculated relative to the first row and column of pixels with whole-number
distances is:
M 0,2 = ∑ x 0 y 2 = ∑ y 2 = 1(1) 2 + 2(2) 2 + 3(3) 2 + 1(4) 2 = 52
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167
Problem 9.29
For the binary image of a key in Figure P.9.29, calculate the following:
• Perimeter, based on the Left-Right search technique.
2
• Thinness, based on P
.
Area
• Center of gravity
• Moment M 0,1 about the origin (pixel 1,1) and about the lowest pixel of a rectangular
box around the key (2,2).
1
2 3 4 5 6 7 8 9
10
x
1
2
3
4
5
6
7
8
9
10
y
Figure P.9.29
Estimated student time to complete: 15-20 minutes
Prerequisite knowledge required: Text Section(s) 9.13, 9.21.
Solution:
1
2 3 4 5 6 7 8 9
10
x
1
2
3
4
5
6
7
8
9
10
y
The left-right search result is shown. The perimeter based on the pixels on the search line
is P = 23.
The area may be calculated by M 0,0 , by region growing or by counting.
232
P2
=
= 17
Thinness:
Area 31
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168
The 1st moments are:
M 1,0 = ∑ x1 y 0 = ∑ x = 2(1) + 3(2) + 4(3) + 4(4) + 5(5) + 5(6) + 5(7) + 3(8) = 150
M 0,1 = ∑ x 0 y1 = ∑ y = 2(1) + 4(2) + 4(3) + 5(4) + 5(5) + 4(6) + 4(7) + 3(8) =143
Center of gravity is: x =
M 1,0
M 0,0
1
=
150
= 4.84 and
31
2 3 4 5 6 7 8 9
y=
M 0,1
M 0,0
=
143
= 4.61
31
10
x
1
2
3
4
5
6
7
8
9
10
x'
y'
y
M 0,1 @(1,1) = ∑ x 0 y1 =143
M 0,1 @(2, 2) = ∑ x 0 y1 = 4(1) + 4(2) + 5(3) + 5(4) + 4(5) + 4(6) + 3(7) = 112
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169
Problem 9.30
Using moment equations, calculate M 0,2 and M 2,0 about the centroidal axes of the part in
Figure P.9.30.
1
2 3 4 5 6 7 8 9
10
x
1
2
3
4
5
6
7
8
9
10
y
Figure P.9.30
Estimated student time to complete: 20-30 minutes
Prerequisite knowledge required: Text Section(s) 9.21
Solution:
The moment values for the image are:
M 0,0 = ∑ x 0 y 0 = 16(1) =16
M 1,0 = ∑ x1 y 0 = ∑ x = 1(2) + 3(3) + 3(4) + 4(5) + 2(6) + 3(7) =76
M 0,1 = ∑ x 0 y1 = ∑ y = 3(2) + 2(3) + 4(4) + 3(5) + 3(6) + 1(7) =68
x=
M 1,0
M 0,0
=
76
= 4.75
16
and
y=
M 0,1
M 0,0
=
68
= 4.25
16
M 2,0 = ∑ x 2 y 0 = ∑ x = 1(2) 2 + 3(3) 2 + 3(4) 2 + 4(5) 2 + 2(6) 2 + 3(7) 2 = 398
2
M 0,2 = ∑ x 0 y 2 = ∑ y 2 = 3(2) 2 + 2(3) 2 + 4(4) 2 + 3(5) 2 + 3(6) 2 + 1(7) 2 =326
The second moments about the centroids can be calculates as follows:
M 2,0 @ x = M 2,0 − M 0,0 ( x ) = 398 − 16(4.75) 2 = 37
2
M 0,2 @ y = M 0,2 − M 0,0 ( y ) 2 = 326 − 16(4.25) 2 = 37
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170
Problem 9.31
Using moment equations, calculate M 0,2 and M 2,0 about the centroidal axes of the part in
Figure P.9.31.
1
2 3 4 5 6 7 8 9
10
x
1
2
3
4
5
6
7
8
9
10
y
Figure P.9.31
Estimated student time to complete:20 -30 minutes
Prerequisite knowledge required: Text Section(s) 9.21
Solution:
The moment values for the image are:
M 0,0 = ∑ x 0 y 0 = 18(1) =18
M 1,0 = ∑ x1 y 0 = ∑ x = 3(2) + 2(3) + 4(4) + 4(5) + 2(6) + 3(7) =81
M 0,1 = ∑ x 0 y1 = ∑ y = 4(2) + 5(3) + 5(4) + 4(5) =63
x=
M 1,0
M 0,0
=
81
= 4.5
18
and
y=
M 0,1
M 0,0
=
63
= 3.5
18
M 2,0 = ∑ x 2 y 0 = ∑ x = 3(2) 2 + 2(3) 2 + 4(4) 2 + 4(5) 2 + 2(6) 2 + 3(7) 2 = 413
2
M 0,2 = ∑ x 0 y 2 = ∑ y 2 = 4(2) 2 + 5(3) 2 + 5(4) 2 + 4(5) 2 = 241
The second moments about the centroids can be calculates as follows:
M 2,0 @ x = M 2,0 − M 0,0 ( x ) = 413 − 18(4.5) 2 = 48.5
2
M 0,2 @ y = M 0,2 − M 0,0 ( y ) 2 = 241 − 18(3.5) 2 = 20.5
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171
CHAPTER TEN
Problem 10.1
Develop a fuzzy inference system for a robot, where the force exerted at the hand and the
velocity of the hand are the inputs and the % power to the actuators is the output.
Estimated student time to complete:-------Prerequisite knowledge required: Text Section(s) 10.3-10.9
Solution:
As in any other design problem, the solution is not unique. For example, different inputs
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may
be different, and alternate rules may be assigned. Therefore, not only the solution is not
unique, the result will also differ. This suggested solution may only be used as a guide.
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any
other available system.
In this solution, the inputs, outputs, the sets, and the rules are as follows:
Inputs: Force; Small, Average, Large
Velocity; Slow, Fast, Fast, Very-Fast
Outputs: Power; Low, Medium, High, Very-High
The fuzzy sets are (arbitrary numbers):
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172
The rules are:
The resulting output is:
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electronic media system or the Internet without prior expressed consent of the copyright owner.
173
Changing the rules or the membership functions will change this result.
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174
Problem 10.2
Develop a fuzzy inference system for a washing machine. The inputs are how dirty the
fabrics are and how much clothes are being washed, and the output is the wash time.
Estimated student time to complete:-------Prerequisite knowledge required: Text Section(s) 10.3-10.9
Solution:
As in any other design problem, the solution is not unique. For example, different inputs
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may
be different, and alternate rules may be assigned. Therefore, not only the solution is not
unique, the result will also differ. This suggested solution may only be used as a guide.
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any
other available system.
In this solution, the inputs, outputs, the sets, and the rules are as follows:
Inputs: Dirty; Not-Much, Some, Very.
Clothes; Small, Med, Much, A-Lot.
Outputs: Wash-Time; Five, Ten, Twenty, Thirty
The fuzzy sets are:
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175
The rules are:
The resulting output is:
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electronic media system or the Internet without prior expressed consent of the copyright owner.
176
Changing the rules or the membership functions will change this result.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
177
Problem 10.3
Develop a fuzzy inference system for a barbecue. The inputs may be the thickness of the
steak and how cooked or rare it is desired to be. The output may be the temperature of the
flame and/or the time of cooking.
Estimated student time to complete:
Prerequisite knowledge required: Text Section(s) 10.3-10.9
Solution:
As in any other design problem, the solution is not unique. For example, different inputs
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may
be different, and alternate rules may be assigned. Therefore, not only the solution is not
unique, the result will also differ. This suggested solution may only be used as a guide.
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any
other available system.
In this solution, the inputs, outputs, the sets, and the rules are as follows:
Inputs: Thickness; See-Through, Average, Hefty
How Cooked; Rare, Medium-Rare, Medium, Well-Done
Outputs: Temp; Low, Medium, High
Time; Three, Five, Eight, Ten
The fuzzy sets are:
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178
The rules are:
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179
The resulting out puts are:
And
Changing the rules or the membership functions will change this result.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
180
Problem 10.4
Develop a fuzzy inference system for an automatic gearbox. The inputs are the speed of
the car and the load on the engine, and the output is the gear ratio of the transmission.
Estimated student time to complete:
Prerequisite knowledge required: Text Section(s) 10.3-10.9
Solution:
As in any other design problem, the solution is not unique. For example, different inputs
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may
be different, and alternate rules may be assigned. Therefore, not only the solution is not
unique, the result will also differ. This suggested solution may only be used as a guide.
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any
other available system.
In this solution, the inputs, outputs, the sets, and the rules are as follows:
Inputs: Speed; Slow, Medium, Cruise, Fast, Maximum
Load; Low, Average, High
Outputs: Gear-Ratio; Fourth, Third, Second, First
The fuzzy sets are:
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181
The rules are:
The resulting out put is:
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electronic media system or the Internet without prior expressed consent of the copyright owner.
182
Changing the rules or the membership functions will change this result.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
183
Problem 10.5
Develop a fuzzy logic system for a vision system in which the inputs are the intensities of
the three colors of red, green, and blue (RGB) in a color image and the output is the
relationship of the combination to the colors of the rainbow.
Estimated student time to complete:
Prerequisite knowledge required: Text Section(s) 10.3-10.9
Solution:
As in any other design problem, the solution is not unique. For example, different inputs
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may
be different, and alternate rules may be assigned. Therefore, not only the solution is not
unique, the result will also differ. This suggested solution may only be used as a guide.
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any
other available system.
In this solution, the inputs, outputs, the sets, and the rules are as follows:
Inputs: Color-Red; Red-Off, Red-Medium, Red-High
Color-Green; Green-Off, Green-Medium, Green-High
Color-Blue; Blue-Off, Blue-Medium, Blue-High
Outputs: Color; Black, Red, Orange, Yellow, Green, Cyan, Blue, Violet, White
The fuzzy sets are:
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184
The rules are:
The resulting out put is:
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electronic media system or the Internet without prior expressed consent of the copyright owner.
185
Changing the rules or the membership functions will change this result.
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electronic media system or the Internet without prior expressed consent of the copyright owner.
186
Problem 10.6
Develop a fuzzy inference system for grading a robotics course. The inputs are your
effort level in the course and your exam grade, and the output is your letter grade.
Estimated student time to complete:
Prerequisite knowledge required: Text Section(s) 10.3-10.9
Solution:
As in any other design problem, the solution is not unique. For example, different inputs
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may
be different, and alternate rules may be assigned. Therefore, not only the solution is not
unique, the result will also differ. This suggested solution may only be used as a guide.
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any
other available system.
In this solution, the inputs, outputs, the sets, and the rules are as follows:
Inputs: Effort; Low, Medium, High
Exam-Grade; eA, eB, eC, eD, eF
Outputs: Course-Grade; A, B, C, D, F
The fuzzy sets are:
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187
The rules are:
The resulting out put is:
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188
Changing the rules or the membership functions will change this result.
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This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other
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189
APPENDICES
Problem A.1
Show that the determinant of a matrix can be calculated by picking any row or column.
Estimated student time to complete: 10-15 minutes
Prerequisite knowledge required: Text Section(s) A.1
Solution:
For an arbitrary matrix:
⎡a
A = ⎢⎢ d
⎢⎣ g
b
e
h
c⎤
f ⎥⎥
i ⎥⎦
det A = a ( ei − fh ) − b ( di − fg ) + c ( dh − eg )
= aei − afh − bdi + bfg + cdh − ceg
det A = −d ( bi − ch ) + e ( ai − cg ) − f ( ah − bg )
= −dbi + dch + eai − ecg − fah + fbg
det A = −b ( di − fg ) + e ( ai − cg ) − h ( af − cd )
= −bdi + bfg + eai − ecg − haf + hcd
They are all the same.
© Copyrighted 2010.
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other
electronic media system or the Internet without prior expressed consent of the copyright owner.
190
Problem A.2
Calculate the determinant of the following (4 × 4) matrix.
⎡1
⎢0
A=⎢
⎢3
⎢
⎣1
1
1
0
0
0
2
1
0
0⎤
0 ⎥⎥
1⎥
⎥
1⎦
Estimated student time to complete: 5 minutes
Prerequisite knowledge required: Text Section(s) A.1.
Solution:
det A = 1(1(1 − 0 ) − 0 + 0 ) − 1( 0 − 2 ( 3 − 1) − 0 ) + 0 − 0 = 1 + 4 = 5
© Copyrighted 2010.
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other
electronic media system or the Internet without prior expressed consent of the copyright owner.
191
Problem A.3
Calculate the inverse of the following matrix using method 1:
⎡1 1 2⎤
B = ⎢0 1 0⎥
⎢
⎥
⎢⎣2 0 3⎥⎦
Estimated student time to complete: 5-10 minutes
Prerequisite knowledge required: Text Section(s) A.1.
Solution:
det B = 1(3 − 0) − 0 + 2(−2) = 3 − 4 = −1
⎡1 0
B = ⎢⎢1 1
⎢⎣ 2 0
⎡3
adjB = ⎢⎢ 0
⎢⎣ −2
T
2⎤
0 ⎥⎥
3 ⎥⎦
−3 −2 ⎤
−1 0 ⎥⎥
2 1 ⎥⎦
⎡ −3 3 2 ⎤
−1
B = ⎢⎢ 0 1 0 ⎥⎥
⎢⎣ 2 −2 −1⎥⎦
© Copyrighted 2010.
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other
electronic media system or the Internet without prior expressed consent of the copyright owner.
192
Problem A.4
Calculate the inverse of the following matrix using method 2:
⎡1 0 1 ⎤
C = ⎢0 2 1⎥
⎢
⎥
⎢⎣ 3 1 0⎥⎦
Estimated student time to complete: 10-15 minutes
Prerequisite knowledge required: Text Section(s) A.1.
Solution:
⎡1 0 1 ⎤ ⎡ x11
⎢0 2 1 ⎥ ⎢ x
⎢
⎥ ⎢ 21
⎢⎣ 3 1 0 ⎥⎦ ⎢⎣ x31
⎧ x11 + x31 = 1
⎪
⎨2 x21 + x31 = 0
⎪3 x + x = 0
⎩ 11 21
x13 ⎤ ⎡1 0 0 ⎤
x22 x23 ⎥⎥ = ⎢⎢0 1 0 ⎥⎥
x32 x33 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
⎧ x11 = 1/ 7
⎪
→ ⎨ x31 = 6 / 7
⎪ x = −3 / 7
⎩ 21
⎧ x12 + x32 = 0
⎪
⎨2 x22 + x32 = 1
⎪3 x + x = 0
⎩ 12 22
⎧ x31 + x33 = 0
⎪
⎨2 x23 + x33 = 0
⎪3 x + x = 1
⎩ 13 23
1
⎡1
−
⎢7
7
⎢
−3 3
C −1 = ⎢
⎢7
7
⎢
1
⎢6
7
⎣⎢ 7
⎧ x12 = −1/ 7
⎪
→ ⎨ x22 = 3 / 7
⎪ x = 1/ 7
⎩ 32
x12
⎧ x13 = 2 / 7
⎪
→ ⎨ x23 = 1/ 7
⎪ x = −2 / 7
⎩ 33
2⎤
7⎥
⎥
1⎥
7⎥
⎥
−2 ⎥
7 ⎥⎦
© Copyrighted 2010.
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other
electronic media system or the Internet without prior expressed consent of the copyright owner.