L2 Exercise Answers - OXBR Website

License: CC-BY-NC-SA 4.0

I found an issue¶

Thank you! Please report it at https://github.com/MarinhoLab/OpenExecutableBooksRobotics/issues

Latex Macros¶

$\providecommand{\myvec}[1]{{\mathbf{\boldsymbol{{#1}}}}}$ $\providecommand{\mymatrix}[1]{{\mathbf{\boldsymbol{{#1}}}}}$

\providecommand{\myvec}[1]{{\mathbf{\boldsymbol{{#1}}}}}
\providecommand{\mymatrix}[1]{{\mathbf{\boldsymbol{{#1}}}}}

Valid imports¶

from math import pi, sin, cos
import numpy as np

Exercises¶

Exercise a¶

θ_a = pi/4.0

R_a =  np.array([[cos(θ_a),-sin(θ_a)],
                 [sin(θ_a), cos(θ_a)]])

# Printing the result is NOT a mandatory part of the answer.
print(f'R_a = {R_a}')

R_a = [[ 0.70710678 -0.70710678]
 [ 0.70710678  0.70710678]]

Exercise b¶

θ_b1 = pi/12.0
θ_b2 = -pi/2.0

R_b1 =  np.array([[cos(θ_b1),-sin(θ_b1)],
                  [sin(θ_b1), cos(θ_b1)]])

R_b2 =  np.array([[cos(θ_b2),-sin(θ_b2)],
                  [sin(θ_b2), cos(θ_b2)]])

R_b = R_b1 @ R_b2

print(f'R_b = {R_b}')

R_b = [[ 0.25881905  0.96592583]
 [-0.96592583  0.25881905]]

Exercise c¶

θ_c = pi/3.0
x_c = 2.0
y_c = 5.0

H_c1 = np.array([[cos(θ_c),-sin(θ_c), 0],
                 [sin(θ_c), cos(θ_c), 0],
                 [0,        0,        1]])

H_c2 = np.array([[1,0,x_c],
                 [0,1,y_c],
                 [0,0,1]])

H_c = H_c1 @ H_c2

print(f'H_c = {H_c}')

H_c = [[ 0.5        -0.8660254  -3.33012702]
 [ 0.8660254   0.5         4.23205081]
 [ 0.          0.          1.        ]]

Exercise d¶

θ_d = pi/3.0
x_d = 2.0
y_d = 5.0


H_d1 = np.array([[1,0,x_d],
                 [0,1,y_d],
                 [0,0,1]])

H_d2 = np.array([[cos(θ_d),-sin(θ_d), 0],
                 [sin(θ_d), cos(θ_d), 0],
                 [0,        0,        1]])


H_d = H_d1 @ H_d2

print(f'H_d = {H_d}')

H_d = [[ 0.5       -0.8660254  2.       ]
 [ 0.8660254  0.5        5.       ]
 [ 0.         0.         1.       ]]

H_c is not the same as H_d. This indicates that the order of operations matter. That is, sequential pose transformations are not commutative.

Extra challenge 1¶

R = \left[\begin{array}{cc} \cos\left(\sin(t) + 2\cos(t)\right) & -\sin\left(\sin(t) + 2\cos(t)\right) \\ \sin\left(\sin(t) + 2\cos(t)\right) & \cos\left(\sin(t) + 2\cos(t)\right) \end{array}\right].

(1)

t = 10.0

θ = sin(t) + 2 * cos(t)

R = np.array([[cos(θ),-sin(θ)],
              [sin(θ), cos(θ)]])

Extra challenge 2¶

See DH parameters in lesson 3.

θ = pi/10.0
d = 0.3
a = 0.5
α = -pi/2.0

H1 = np.array(
    [[cos(θ), -sin(θ), 0, 0],
    [ sin(θ),  cos(θ), 0, 0],
    [ 0,       0,      1, 0],
    [ 0,       0,      0, 1]]
)

H2 = np.array(
    [[1, 0, 0, 0],
     [0, 1, 0, 0],
     [0, 0, 1, d],
     [0, 0, 0, 1]]
)

H3 = np.array(
    [[1, 0, 0, a],
     [0, 1, 0, 0],
     [0, 0, 1, 0],
     [0, 0, 0, 1]]
)

H4 = np.array(
    [[1, 0,       0,      0],
     [0, cos(α), -sin(α), 0],
     [0, sin(α),  cos(α), 0],
     [0, 0,       0,      1]]
)

H = H1 @ H2 @ H3 @ H4