Corrections for “Introduction to Robotics” by S.K.

Saha

Chapter 1 Introduction
P. 5, Sec. 1.2: Strictly speaking, a robot is a ….

Chapter 2 Serial Robots

P. 23, Item (iii), “… by the ADC …”: ADC Æ DAC

P. 28, Table 2.3, Cylindrical, “φ: height”: φ Æ y
P. 33, Fig. 3.1, “Pc: Input power to motor …”: Pc Æ Pa

Chapter 5 Transformations

P. 104, Table 5.4, Column 3: 0 Æ -π/2

P. 109, Exercises 5.1a-c: Equation nos. (5.55a-c) should be right justified instead in the next lines.
P. 111: Replace Fig. 5.30 with the following figure or correct as instructed below.

Z2
Z1, X3, X4
θ2
X1, X2

O1, O2, O3

θ3
θ1

Z3, Z4

Fig. 5.30 (Line right from O3 towards Z3 is shortened)

Chapter 6 Kinematics
P. 117, eq. (6.9), Expression for T: b1+b2+b3+b4 Æ b1+b2+b3-b4
P. 123, eq. (6.17b): w12 Æ wx2 and w22 Æ wy2
P. 135, after eq. (6.48c): ai, j ≡ ai + L + a j -1 Æ ai, j ≡ ai + ai +1 + L + a j -1
P. 144, Exercise 6.2, 2nd line: transportation Æ transformation

Chapter 7 Statics

P. 153, eq. (7.17b), [n01 ]1 ≡ a1[n12 + a1 × f12 ]2 ... : a1 Æ Q1

Fig. 7.3: It is to be replaced with the following figure or to be corrected as instructed below.

X3

, Z3
f
θ1

Fig. 7.3 (an arrow is added at EE, X3 is moved there)

P. 154, eq. (7.19b), [n12 ]2 ≡ a 2 [a 2 × [f 23 ]3 ... : a 2 Æ Q 2

Chapter 8 Dynamic

h l h l w2
P. 169, eq. (8.15):
=− ∫o ∫o yρdydz
Æ
= − ∫∫ 0 0 2
y ρ dydz
∂L ∂L
P. 172, eq. (8.24): − −
∂q& Æ ∂q
i i
P. 179, after eq. (8.51a): From eq. (8.34), Æ From eq. (8.33),
P. 180, eq. (8.55a): ; Jω,1 Æ ; Jω,2 (The one after ; is to be corrected)
P. 187, eq. (8.85), (3,1)-element: sα i −1cθi −1 Æ sα i −1 sθi −1
P. 188, eq. (8.87a): … × ri-1 …: for a … Æ … × ri-1 …: for a …
eq, (8.87b): L[ω & i −1 ]i-1 × [ri −1 ]i-1 L : Prismatic Æ L[ωi −1 ]i-1 × [ri −1 ]i-1 L : Prismatic
(no ‘dot’ over ‘omega’)
eq. (8.88): [ri ]i Æ [ri ]i +1
P. 189, eq. (8.89a), Prismatic: b&i ωi × ei Æ 2b& ω × e i i i

eq. (8.89b), Prismatic: +b&i ωi × ei ]i Æ +2b&i ωi × ei ]i

eq. (8.90): QTi Æ QTi−1
.
eq. (8.91b): [ωi ]i × [I i ]i [ωi ]i Æ [ωi ]i × [I i ]i [ωi ]i (no ‘dot’ over 2nd ‘omega’)
P. 191, Before (b): [g ]1 ≡ [g 0 0]T Æ [g ]1 ≡ [ g 0 0]T (g is italic, not bold)
⎡ a ⎤ ⎡ a ⎤
⎢ − m 12 ( ⎥ ⎢ −m 2 ( ⎥
⎢ ⎥ ⎢ ⎥
a ⎥ Æ [f ] = ⎢ m a (
After eq. (8.99): [f01 ]1 = ⎢ m ( ⎥
⎢ 12 ⎥ 01 1
⎢ 2 ⎥
⎢ ⎥
0⎥ ⎢ 0 ⎥⎥
⎢ ⎢
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
P. 192, eq. (8.105): I i = Æ [I i ]i +1 =

P. 193, before (b): [g ]2 = [ Æ [g ]2 = QT1 [g ]1 = [

Item (b): [f32 ]2 = [n 32 ]2 Æ [f32 ]3 = [n32 ]3

Chapter 9 Recursive Robot Dynamics

P. 208, eq. (9.2a): M i t& i + Wi M i E i t i = w i Æ M i t& i + Wi M i t i = w i

P. 210, eq. (9.6b): N1 Æ Nl (subscript ‘el’, not ‘one’)
~ ~ % ≡ MΩ % ;M % ≡ NT WMN Æ M % ≡ NT WMN (subscripts ‘el’)
P. 211, eq, (9.9b): M w ≡ M W Æ M w e 1 1 e l l

t′ ≡ ...Nl W... Æ t′ ≡ ...N l Ω...

P. 211, after eq. (9.9b): where N & = WN is used, and the matrices N1,…Æ where
d d
&
N = ΩN ⎯ Ω ≡ diag.[ Ω , L , Ω ] ; Ω ≡ diag.[ω × 1, ω × 1] ⎯is used, and the matrices Nl,…
d d 1 n i i i

~ ~
P. 211, after eq. (9.10a): M ≡ NlTMN1 Æ M ≡ NlTMNl (subscripts are ‘el’ not ‘one’)
~ ~
P. 213, eq. (9.13b): δ i ≡ mi +1c i +1,i + δ i +1 Æ δ% i ≡ m% i +1ci +1,i + δ% i +1
% p i ≡ pT BT M ~
eq. (9.14c): i ≡ pT BT M
ji j ij i j ji p (last ‘j’ will be ’i’)
j ij i i
% ≡ MW
eq. (9.15b): M % % ≡ MΩ
≡ NTl MNl W Æ M % ≡ NT MN Ω
w w l l

P. 214, After eq. (9.15c): Bi+1,1 Æ Bi+1,i

After eq. (9.15d): M% ÆM %
i l
% ≡M
After eq. (9.15e): M % B& Æ H %
j n nj n + 1, n
% %
Expression of Matrix M w : B 21M 2 Æ M % B ;
2 21
%
B M ÆMB ; %
n1 n n n1
% ÆM
B n2 M % B
n n n2

W 1 Æ Ω1 ;
W2 Æ Ω 2 ;
Wn Æ Ω n
Eq. (9.15f), Expression of Matrix M % :M% W ÆM % Ω ;
w 1 1 1 1
T % % Ω ;
B 21M 2 W2 Æ BT21M 2 2
 % W Æ BT M
BTn1M % Ω
n n n1 n n
% %
M 2 B 21 W1 Æ M 2 B 21Ω1
% W ÆM
M % Ω
1 2 2 2
% W Æ BT M
T
B M % Ω
n2n n n2 n n
% %
MB W ÆMB Ω
n n1 1 n n1 1
% B W ÆM
M % B Ω
n n2 2 n n2 2
% %
M W ÆM Ωn n n n

% :B M
P. 215, eq. (9.15g), Expression of Matrix M %' ÆM % 'B ;
e 21 2 2 21
% %
B n1M n Æ M 'n B n1 ;
'

B M %' ÆM % 'B
n2 n n n2

Eq. (9.15h): M % ′ =M % ′+ Æ M % ′ = M' +

i i i i
%
After eq. (9.15h): M i = Æ M i =
' '

Eq. (9.15i): Replace with

ci,j ≡ pTi [BTj +1,i H% % %′
j +1,i + B ji (M j Ω j + M j )]p j
T
if i ≤ j
% +M
ci,j ≡ pTi [H % B Ω +M % ' B ]p otherwise.
ij i ij j i ij j

& = WNd = O. As a result the term M

Before section 9.2.3: … derivative, N % vanishes to
d w
provide simpler expressions. Æ … derivatives vanish. As a result the term WMt vanishes to provide
% does not appear in eq. (9.15a).
simpler expressions, i.e., M e

% ′ = NT1 ( Æ w
Eq. (9.16): w % ′ = NTl ( (subscript ‘one’ is ‘el’)
= (N& + N W ) θ& Æ = ( N
1 1
& + N Ω ) N θ& (subscript ‘one’ is ‘el’)
l l d

P. 217, eq. (9.21): N1 = 1 Æ N l = 1 (subscript ‘one’is ‘el’)

% =
% = Æ I (≡ i ) = pT Mp
Before eq. (9.22a): I (≡ i ) = pT M11 p 11

⎡sθ 2
sθ cθ 0⎤ ⎡cθ 2
sθ cθ 0⎤
⎢ ⎥ ⎢ ⎥
Eq. (9.22d): ⎢ sθ cθ s 2θ 0 ⎥ Æ ⎢ sθ cθ s 2θ 0⎥
⎢ 0 0 1⎥⎦ ⎢ 0 0 1⎥⎦
⎣ ⎣

P. 218, eq. (9.23a): pT (MW + WM )p = θ&eT [I(e × e) + (e × Ie)] Æ

pT (MΩ + WM )p = θ&eT [(Ie × e) + (e × Ie) − md × d]
⎡θ& e × 1 O ⎤ ⎡θ&e × 1 O ⎤ ⎡θ&e × 1 O ⎤
Eq. (9.23b): W = ⎢ ⎥ Æ Ω=⎢ ⎥ and W = ⎢ ⎥
⎣ O O⎦ ⎣ O θ&e × 1⎦ ⎣ O O⎦
After eq. (9.23b): … i.e., e × e = 0 , and e T (e × Ie) = 0 . Æ
… i.e., d × d = 0 , and eT (Ie × e) = 0 or e T (e × Ie) = 0 .
P. 220, eq. (9.29a-d): Replace with
c22 = pT2 (BT32 H% + BT ( M % Ω +M % ′ )p = 0
32 22 2 2 2 2

c21 = pT2 [H% +M % B Ω +M % ′ B ]p = 1 m a a sθ θ&

21 2 21 1 2 21 1 2 1 2 2 1
2
c12 = pT1 [BT31H % + BT (M % Ω +M % ′ )]p = − 1 m a a sθ (θ& + θ& )
32 21 2 2 2 2 2 1 2 2 1 2
2
c11 = pT1 [BT21H % + BT (M % Ω +M % ′ )]p = − 1 m a a sθ θ&
21 11 1 1 12 1 2 1 2 2 2
2
After eq. (9.29d): … term MiWj , for … Æ … term pTi BTji M % Ω p , for …
j j j

P. 231, eq. (9.36d): M% ≡ M + ... Æ M

ˆ ≡ M + ... (‘tilda’ over Mi in the left will be ‘hat’)
i i i i
~ && %
P. 232, before eq. (9.38b): θ ≡ φ Æ θ ≡ φ
&&
n 1 1 1

1 1
P. 238, eq. (9.40d): τ 1g = ... + a2 cθ12 )... + a3 cθ123 )] Æ τ 1g = ... + a2 cθ12 )... + a3 cθ123 )]
2 2
1
τ 2g = ... + a3 cθ123 ) Æ τ 1g = ... + a3 cθ123 )
2
Exercise 9.12: … Example 9.11 … Æ … Example 9.6 …

Chapter 10 Control
P. 243, after eq. (10.5): s1 = 2 and s2 = 3 Æ s1 = -2 and s2 = -3
P. 246: Shaded area after eq. (10.26) until Section 10.3 should not be shaded.
P. 255, before eq. (10.41): constants, i.e., Æ constant τ, i.e.,
Before section 10.5: close to it but less than unity. Æ close to it.
P. 259, eq. (10.50): ( l m s 2 + ... = ... − τ 1 ( s ) Æ ( I m s 2 + ... = ... − τ l ( s ) (‘el’ on left is capital ‘I’ and
η η
subscript ‘one’ on the right is ‘el’)
P. 260, eq. (10.54b): τ 1 Æ τ l (subscript ‘one’ is ‘el’)
η η
Example 10.6: … is 0.02 kg m2 Æ … is 0.01 kg m2
eq. (10.55): = ω n2 Æ = ω n2
+ 2 ςω n + + 2 ςω n s +
P. 261, eqs. (10.56) and (10.57b): ≡ Æ ≡
kpI 2 kpI
uuur 0 s Æ = lim
P. 262, eq. (10.60): = s lim
s→0

P. 269, 1st line: .. perfect (i.e., incomplete … Æ … perfect (i.e., complete …

P. 272, Example 10.9: … the Fig. 10.6 whose Æ … the Fig. 10.7 whose
P. 273, Example 10.11: … in Fig. 10.6 for Æ … in Fig. 10.7 for
P. 275, eq. (10.97): −Bk T Æ −bk T
P. 281, paragraph before Fig. 10.28: … be deined along … Æ … be denied along …
… different crieteria. Æ … different criteria.
P. 282, Exercise 10.8: … Exercise 10.6. Æ … Exercise10.7.
 Chapter 11 Motion Planning

1 1
P. 293, middle of the page: sα Æ
2 2sα
1 1
P. 304, eq. (11.35): θ f + θ&&c (t f − t ) 2 Æ θ f − θ&&c (t f − t ) 2
2 2
P. 314, Exercise 11.6: + a3t Æ + a3t
33 3

Solution Manual for “Introduction to Robotics” by S.K. Saha

Chapter 5 Transformation
Solution to Exercise 5.8, Table S5.1, 3rd row, 2nd column: 0 Æ b2