Modern Robotics, Course 2: Robot Kinematics Coursera Quiz Answers 2023 [๐Ÿ’ฏ% Correct Answer]

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Modern Robotics, Course 2: Robot Kinematics Quiz Answers

Week 01: Modern Robotics, Course 2: Robot Kinematics Coursera Quiz Answers

Quiz 01: Lecture Comprehension, Product of Exponentials Formula in the Space Frame (Chapter 4 through 4.1.2)

Q1. True or false? The PoE formula in the space frame only correctly calculates the end-effector configuration if you first put the robot at its zero configuration, then move joint nn to \theta_nฮธnโ€‹, then move joint n-1nโˆ’1 to \theta_{n-1}ฮธnโˆ’1โ€‹, etc., until you move joint 1 to \theta_1ฮธ1โ€‹.

  • True.
  • False.

Q2. Consider the screw axis \mathcal{S}_iSiโ€‹ used in the PoE formula. Which of the following is true?

  • \mathcal{S}_iSiโ€‹ represents the screw axis of joint ii, expressed in the end-effector frame {b}, when the robot is at its zero configuration.
  • \mathcal{S}_iSiโ€‹ represents the screw axis of joint ii, expressed in the end-effector frame {b}, when the robot is at an arbitrary configuration \thetaฮธ.
  • \mathcal{S}_iSiโ€‹ represents the screw axis of joint ii, expressed in the space frame {s}, when the robot is at its zero configuration.
  • \mathcal{S}_iSiโ€‹ represents the screw axis of joint ii, expressed in the space frame {s}, when the robot is at an arbitrary configuration \thetaฮธ.

Q3. When the robot is at an arbitrary configuration \thetaฮธ, does the screw axis corresponding to motion along joint ii, represented in {s}, depend on \theta_{i-1}ฮธiโˆ’1โ€‹?

  • No.
  • Yes.

Quiz 02: Lecture Comprehension, Product of Exponentials Formula in the End-Effector Frame (Chapter 4.1.3)

Q1. When the robot is at an arbitrary configuration \thetaฮธ, does the screw axis corresponding to motion along joint ii, represented in {b}, depend on \theta_{i-1}ฮธiโˆ’1โ€‹?

  • No.
  • Yes.

Q2. When the robot arm is at its home (zero) configuration, the axis of joint 3, a revolute joint, passes through the point (3,0,0)(3,0,0) in the {b} frame. The axis of rotation is aligned with the \hat{{\rm z}}_{{\textrm b}}z^bโ€‹-axis of the {b} frame. What is the screw axis \mathcal{B}_3B3โ€‹?

  • (0, 0, 1, -3, 0, 0)(0,0,1,โˆ’3,0,0)
  • (0, 0, 1, 0, -3, 0)(0,0,1,0,โˆ’3,0)
  • (0, 0, 1, 0, 0, -3)(0,0,1,0,0,โˆ’3)

Quiz 03: Lecture Comprehension, Forward Kinematics Example

Q1. In the image below, imagine a frame {c} on the axis of joint 2 and aligned with the {s} frame. What is the screw axis of joint 1 expressed in the frame {c}?9Tzr5eBSEeeY9RLN7DX 0g 88f5c7d783bc824c90b49530b641c38e MTB still 01

  • (0, 0, 1, 0, 10, 0)(0,0,1,0,10,0)
  • (0, 0, 1, 0, 0, 10)(0,0,1,0,0,10)

Quiz 04: Chapter 4, Forward Kinematics

Q1. The URRPR spatial open chain robot is shown below in its zero position. JA3UTd aEeeK2w4Lcly5FA 98de43fe243493ed9c85de344e61ced5 ex01 01

For L = 1L=1, determine the end-effector zero configuration MM. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]

Q2. Referring back to Question 1, determine the screw axes \mathcal{S}_iSiโ€‹ in {0} when the robot is in its zero position. Again L = 1L=1. Give the axes as a 6ร—6 matrix with the form \left[\mathcal{S}_1, \mathcal{S}_2, \dots, \mathcal{S}_6 \right][S1โ€‹,S2โ€‹,โ€ฆ,S6โ€‹], i.e., each column is a screw axis. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹

  • 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]]

Q3. Referring back to Question 1, determine the screw axes \mathcal{B}_iBiโ€‹ in {b} when the robot is in its zero position. Again L = 1L=1. Give the axes as a matrix with the form \left[\mathcal{B}_1, \mathcal{B}_2, \dots, \mathcal{B}_6 \right][B1โ€‹,B2โ€‹,โ€ฆ,B6โ€‹]. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹.

  • 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]]

Q4. Referring back to Question 1 and 2, given L = 1L=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)ฮธ=(โˆ’ฯ€/2,ฯ€/2,ฯ€/3,โˆ’ฯ€/4,1,ฯ€/6), use the function {\tt FKinSpace}FKinSpace in the given software to find the end-effector configuration T \in SE(3)TโˆˆSE(3). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]

Q5. Referring back to Question 1 and 3, given L = 1L=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)ฮธ=(โˆ’ฯ€/2,ฯ€/2,ฯ€/3,โˆ’ฯ€/4,1,ฯ€/6), use the function {\tt FKinBody}FKinBody in the given software to find the end-effector configuration T \in SE(3)TโˆˆSE(3). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]

Week 02: Modern Robotics, Course 2: Robot Kinematics Coursera Quiz Answers

Quiz 02: Lecture Comprehension, Velocity Kinematics and Statics (Chapter 5 Introduction)

Q1. True or false? The Jacobian matrix depends on the joint variables.

  • True.
  • False.

Q2. True or false? The Jacobian matrix depends on the joint velocities.

  • True.
  • False.

Q3. True or false? Row ii of the Jacobian corresponds to the end-effector velocity when joint ii moves at unit speed and all other joints are stationary.

  • True.
  • False.

Q4. Consider a square Jacobian matrix that is usually full rank. At a configuration where one row of the Jacobian becomes a scalar multiple of another row, is the robot at a singularity?

  • Yes.
  • No.

Q5. In general, a sphere (or hypersphere, meaning a sphere in more than 3 dimensions) of possible joint velocities maps through the Jacobian to

  • a sphere (or hypersphere).
  • a polyehdron.
  • an ellipsoid (or hyperellipsoid).

Q6. Assume a three-dimensional end-effector velocity. At a singularity, the volume of the ellipsoid of feasible end-effector velocities becomes

  • zero.
  • infinite.

Q7. At a singularity,

  • some end-effector forces become impossible to resist by the joint forces and torques.
  • some end-effector forces can be resisted even with zero joint forces or torqu

Quiz 02: Lecture Comprehension, Statics of Open Chains (Chapter 5.2)

Q1. If the wrench -\mathcal{F}โˆ’F is applied to the end-effector, to stay at equilibrium the robot must apply the joint forces and torques \tau = J^{\rm T}(\theta) \mathcal{F}ฯ„=JT(ฮธ)F to resist it. If the robot has 4 one-dof joints, what is the dimension of the subspace of 6-dimensional end-effector wrenches that can be resisted by \tau = 0ฯ„=0?

  • 2-dimensional.
  • At least 2-dimensional.
  • 4-dimensional.
  • At least 4-dimensional.

Quiz 03: Lecture Comprehension, Singularities (Chapter 5.3)

Q1. Consider a robot with 7 joints and a space Jacobian with a maximum rank of 6 over all configurations of the robot. At the current configuration, the rank of the space Jacobian is 5. Which of the following statements is true? Select all that apply.

  • The robot is redundant with respect to the task of generating arbitrary end-effector twists.
  • The robot is kinematically deficient with respect to the task of generating arbitrary end-effector twists.
  • The robot is at a singularity.

Q2. Consider a robot with 7 joints and a space Jacobian with a maximum rank of 3 over all configurations of the robot. At the current configuration, the rank of the space Jacobian is 3. Which of the following statements is true? Select all that apply.

  • The robot is redundant with respect to the task of generating arbitrary end-effector twists.
  • The robot is at a singularity.
  • The space Jacobian is โ€œfat.โ€

Q3. Consider a robot with 8 joints and a body Jacobian with rank 6 at a given configuration. For a given desired end-effector twist \mathcal{V}_bVbโ€‹, what is the dimension of the subspace of joint velocities (in the 8-dimensional joint velocity space) that create the desired twist?

  • 2
  • 0
  • The desired twist cannot be generated.

Quiz 04: Lecture Comprehension, Manipulability (Chapter 5.4)

Q1. Itโ€™s more useful to visualize the manipulability ellipsoid using the body Jacobian than the space Jacobian, since the body Jacobian measures linear velocities at the origin of the end-effector frame, which has a more intuitive meaning than the linear velocity at the origin of the space frame. If the robot has nn joints, then the body Jacobian J_bJbโ€‹ is 6 \times n6ร—n. We can break J_bJbโ€‹ into two sub-Jacobians, the angular and linear Jacobians:

J_b = \left[

Jbฯ‰Jbv

\right].Jbโ€‹=[Jbฯ‰โ€‹Jbvโ€‹โ€‹].

What is the dimension of J_{bv}J_{bv}^{\rm T}Jbvโ€‹JbvTโ€‹, which is used to generate the linear component of the manipulability ellipsoid?

  • 3 \times 33ร—3
  • 6 \times 66ร—6
  • n \times nnร—n

Q2. Consider a robot with a full rank Jacobian as it approaches a singular configuration. As it approaches a singular configuration, what happens to the manipulability ellipsoid? Select all that apply.

  • The length of one principal axis approaches zero.
  • The length of one principal axis approaches infinity.
  • The interior โ€œvolumeโ€ of the ellipsoid approaches zero.
  • The interior โ€œvolumeโ€ of the ellipsoid approaches infinity.

Q3. Consider a robot with a full rank Jacobian as it approaches a singular configuration. As it approaches the singular configuration, what happens to the force ellipsoid? Select all that apply.

  • The length of one principal axis approaches zero.
  • The length of one principal axis approaches infinity.
  • The interior โ€œvolumeโ€ of the ellipsoid approaches zero.
  • The interior โ€œvolumeโ€ of the ellipsoid approaches infinity.

Quiz 05: Chapter 5, Velocity Kinematics and Statics

Q1. A 3R planar open-chain robot is shown below.aGqeL AsEeeY9RLN7DX 0g d5df79628a4c306050173302f5df63b6 ex01 01

Suppose the tip generates a wrench that can be expressed in the space frame {s} as a force of 2 N in the \hat{{\rm x}}_{{\rm s}}x^sโ€‹ direction, with no component in the \hat{{\rm y}}_{{\rm s}}y^โ€‹sโ€‹ direction and zero moment in the {s} frame. What torques must be applied at each of the joints? Positive torque is counterclockwise (the joint axes are out of the screen, so positive rotation about the joints is counterclockwise). Give the torque values in the form (\tau_1, \tau_2, \tau_3)(ฯ„1โ€‹,ฯ„2โ€‹,ฯ„3โ€‹). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Important: Remember that the wrench applied by the robot end-effector has zero moment in the {s} frame. No other frame is defined in the problem. In particular, no frame is defined at the tip of the robot.

Write the vector in the answer box and click โ€œRunโ€:

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]โŽฃโŽขโŽกโ€‹1.112.223.33โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • 2
  • 3
  • 4
  • [0,0,0]
  • # Edit the answer above this line!  Do not edit below this line! 
  • print โ€˜Your answer has been recorded asโ€™, Your_Answer()

Q2. The 4R planar open-chain robot below has an end-effector frame {b} at its tip.m7IE3eB1EeeY9RLN7DX 0g 22e7ceb45566a152517067d8a7d6349a ex02 01

Considering only the planar twist components (\omega_{bz}, v_{bx}, v_{by})(ฯ‰bzโ€‹,vbxโ€‹,vbyโ€‹) of the body twist \mathcal{V}_bVbโ€‹, the body Jacobian is

Jb(ฮธ)=โŽกโŽฃ1L3s4+L2s34+L1s234L4+L3c4+L2c34+L1c2341L3s4+L2s34L4+L3c4+L2c341L3s4L4+L3c410L4โŽคโŽฆ

where s23=sin(ฮธ2+ฮธ3), etc.

Suppose L_1 = L_2 = L_3 = L_4 = 1L1โ€‹=L2โ€‹=L3โ€‹=L4โ€‹=1 and the chain is at the configuration \theta_1=\theta_2=0, \theta_3=\pi/2, \theta_4=-\pi/2ฮธ1โ€‹=ฮธ2โ€‹=0,ฮธ3โ€‹=ฯ€/2,ฮธ4โ€‹=โˆ’ฯ€/2. The joints generate torques to create the wrench \mathcal{F}_b = (0,0,10, 10,10,0)Fbโ€‹=(0,0,10,10,10,0) at the last link. What are the torques at each of the joints? Give the torque values in the form (\tau_1, \tau_2, \tau_3, \tau_4)(ฯ„1โ€‹,ฯ„2โ€‹,ฯ„3โ€‹,ฯ„4โ€‹). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the vector in the answer box and click โ€œRunโ€:

[1.11,2.22,3.33,4.44] for \left[

1.112.223.334.44

\right]โŽฃโŽขโŽขโŽขโŽกโ€‹1.112.223.334.44โ€‹โŽฆโŽฅโŽฅโŽฅโŽคโ€‹.

  • 1
  • [0,0,0,0]

Q3. The RRP robot is shown below in its zero position.T YrjeVTEeeRtwqRjGvJYg 015c1b3cbd893bb6eb8e3b9b932a0bc4 week2 RRP

Its screw axes in the space frame are

S1=โŽกโŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽข001000โŽคโŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅ, S2=โŽกโŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽข100020โŽคโŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅ, S3=โŽกโŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽข000010โŽคโŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅ.

Use the function {\tt JacobianSpace}JacobianSpace in the given software to calculate the 6ร—3 space Jacobian J_sJsโ€‹ when \theta =(90^\circ, 90^\circ, 1)ฮธ=(90โˆ˜,90โˆ˜,1). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [[0,0,0],[0,0,0],[0,0,0],[0,0,0],[0,0,0],[0,0,0]]

Q5. Referring back to Question 3, the screw axes in the body frame are

B1=โŽกโŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽข010300โŽคโŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅ, B2=โŽกโŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽขโˆ’100030โŽคโŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅ, B3=โŽกโŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽข000001โŽคโŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅ.

Use the function {\tt JacobianBody}JacobianBody in the given software to calculate the 6ร—3 body Jacobian J_bJbโ€‹ when \theta =(90^\circ, 90^\circ, 1)ฮธ=(90โˆ˜,90โˆ˜,1). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Write the matrix in the answer box and click โ€œRunโ€:

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] for \left[

1.114.447.772.225.558.883.336.669.99

\right]โŽฃโŽขโŽกโ€‹1.114.447.77โ€‹2.225.558.88โ€‹3.336.669.99โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [[0,0,0],[0,0,0],[0,0,0],[0,0,0],[0,0,0],[0,0,0]]

Q6. The kinematics of the 7R WAM robot are given in Section 4.1.3 in the textbook. The numerical body Jacobian J_bJbโ€‹ when all joint angles are \pi/2ฯ€/2 is

J_b = \left[

001โˆ’0.105โˆ’0.8890โˆ’10000.006โˆ’0.1050100.00600.889001โˆ’0.045โˆ’0.8440โˆ’10000.00600100.00600001000

\right]Jbโ€‹=โŽฃโŽขโŽขโŽขโŽขโŽขโŽขโŽขโŽกโ€‹001โˆ’0.105โˆ’0.8890โ€‹โˆ’10000.006โˆ’0.105โ€‹0100.00600.889โ€‹001โˆ’0.045โˆ’0.8440โ€‹โˆ’10000.0060โ€‹0100.00600โ€‹001000โ€‹โŽฆโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽฅโŽคโ€‹

Extract the linear velocity portion J_vJvโ€‹ (joint rates act on linear velocity). Calculate the directions and lengths of the principal semi-axes of the three-dimensional linear velocity manipulability ellipsoid based on J_vJvโ€‹. Give a unit vector, with at least 2 decimal places for each element in this vector, to represent the direction of the longest principal semi-axis.

Write the vector in the answer box and click โ€œRunโ€:

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]โŽฃโŽขโŽกโ€‹1.112.223.33โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [0,0,0]

Q7. Referring back to Question 5 and its result, give the length, with at least 2 decimal places, of the longest principal semi-axis of that three-dimensional linear velocity manipulability ellipsoid.

Week 03: Modern Robotics, Course 2: Robot Kinematics Coursera Quiz Answers

Quiz 01: Lecture Comprehension, Inverse Kinematics of Open Chains (Chapter 6 Introduction)

Q1. Consider the point (x,y) = (0,2)(x,y)=(0,2). What is {\rm atan2}(y,x)atan2(y,x), measuring the angle from the xx-axis to the vector to the point (x,y)(x,y)?

  • 0
  • \pi/2ฯ€/2
  • -\pi/2โˆ’ฯ€/2

Q2. What are advantages of numerical inverse kinematics over analytic inverse kinematics? Select all that apply.

  • It can be applied to open-chain robots with arbitrary kinematics.
  • It requires an initial guess at the solution.
  • It returns all possible inverse kinematics solutions.

Quiz 02: Lecture Comprehension, Numerical Inverse Kinematics (Chapter 6.2, Part 1 of 2)

Q1. Let f(\theta)f(ฮธ) be a nonlinear function of \thetaฮธ mapping an nn-dimensional space (the dimension of \thetaฮธ) to an mm-dimensional space (the dimension of ff). We want to find a \theta_dฮธdโ€‹, which may not be unique, that satisfies x_d = f(\theta_d)xdโ€‹=f(ฮธdโ€‹), i.e., x_d โ€“ f(\theta_d) = 0xdโ€‹โˆ’f(ฮธdโ€‹)=0. If our initial guess at a solution is \theta^0ฮธ0, then a first-order Taylor expansion approximation of f(\theta)f(ฮธ) at \theta^0ฮธ0 tells us

x_d \approx f(\theta^0) + J(\theta^0)(\theta_d โ€“ \theta^0)xdโ€‹โ‰ˆf(ฮธ0)+J(ฮธ0)(ฮธdโ€‹โˆ’ฮธ0)

where J(\theta^0)J(ฮธ0) is the matrix of partial derivatives \partial f/\partial \thetaโˆ‚f/โˆ‚ฮธ evaluated at \theta^0ฮธ0. Which of the following is a good next guess \theta^1ฮธ1?

  • \theta^1 = \theta^0 + J^\dagger(\theta^0) (x_d โ€“ f(\theta^0))ฮธ1=ฮธ0+Jโ€ (ฮธ0)(xdโ€‹โˆ’f(ฮธ0))
  • \theta^1 = \theta^0 โ€“ J^\dagger(\theta^0) (x_d โ€“ f(\theta^0))ฮธ1=ฮธ0โˆ’Jโ€ (ฮธ0)(xdโ€‹โˆ’f(ฮธ0))
  • \theta^1 = J^{-1}(\theta^0) (x_d โ€“ f(\theta^0))ฮธ1=Jโˆ’1(ฮธ0)(xdโ€‹โˆ’f(ฮธ0))

Q2. We want to solve the linear equation Ax = bAx=b where AA is a 3ร—2 matrix, xx is a 2-vector, and bb is a 3-vector. For a randomly chosen AA matrix and vector bb, how many solutions xx can we expect?

  • None.
  • One.
  • More than one.

Q3. We want to solve the linear equation Ax = bAx=b, where

A = \left[

142536

\right]A=[14โ€‹25โ€‹36โ€‹]

and b = [7 \;\;8]^{\rm T}b=[78]T. Since xx is a 3-vector and bb is a 2-vector, we can expect a one-dimensional set of solutions in the 3-dimensional space of possible xx values. The following are all solutions of the linear equation. Which is the solution given by x = A^\dagger bx=Aโ€ b? (You should be able to tell by inspection, without using software.)

  • (-1.06, -3.89, 5.28)(โˆ’1.06,โˆ’3.89,5.28)
  • (-3.06, 0.11, 3.28)(โˆ’3.06,0.11,3.28)
  • (-5.06, 4.11, 1.28)(โˆ’5.06,4.11,1.28)

Q4. If we would like to find an xx satisfying Ax = bAx=b, but AA is โ€œtallโ€ (meaning it has more rows than columns, i.e., the dimension of bb is larger than the dimension of xx), then in general we would see there is no exact solution. In this case, we might want to find the x^*xโˆ— that comes closest to satisfying the equation, in the sense that x^*xโˆ— minimizes\|Ax^* โ€“ b\|โˆฅAxโˆ—โˆ’bโˆฅ (the 2-norm, or the square root of the sum of the squares of the vector). This solution is given by x^* = A^\dagger bxโˆ—=Aโ€ b. Which of the two answers below satisfies this condition if

A = \left[

12

\right], \;\; b = \left[

34

\right]?A=[12โ€‹],b=[34โ€‹]?

  • x^* = 2.2xโˆ—=2.2
  • x^* = 1xโˆ—=1

Quiz 03: Lecture Comprehension, Numerical Inverse Kinematics (Chapter 6.2, Part 2 of 2)

Q1. To adapt the Newton-Raphson root-finding method to inverse kinematics when the desired end-effector configuration is represented as a transformation matrix X_d \in SE(3)Xdโ€‹โˆˆSE(3), we need to express the error between T_{sb}(\theta^i)Tsbโ€‹(ฮธi) (the forward kinematics, where \theta^iฮธi is our current guess at a joint solution) and X_dXdโ€‹. One expression of this error is the twist that takes the the robot from T_{sb}(\theta^i)Tsbโ€‹(ฮธi) to X_dXdโ€‹ in unit time. When this twist is expressed in the end-effector frame {b}, we write it as \mathcal{V}_bVbโ€‹. Which of the following is a correct expression?

  • \mathcal{V}_b = {\rm log} (T_{sb}^{-1}(\theta^i) X_d)Vbโ€‹=log(Tsbโˆ’1โ€‹(ฮธi)Xdโ€‹)
  • [\mathcal{V}_b] = {\rm log} (T_{sb}^{-1}(\theta^i) X_d)[Vbโ€‹]=log(Tsbโˆ’1โ€‹(ฮธi)Xdโ€‹)
  • \mathcal{V}_b = {\rm exp} (T_{sb}^{-1}(\theta^i) X_d)Vbโ€‹=exp(Tsbโˆ’1โ€‹(ฮธi)X

Quiz 04: Chapter 6, Inverse Kinematics

Q1. Use Newton-Raphson iterative numerical root finding to perform two steps of finding the root of

f(x,y) = \left[

x2โˆ’9y2โˆ’4

\right]f(x,y)=[x2โˆ’9y2โˆ’4โ€‹]

when your initial guess is (x^0,y^0) = (1,1)(x0,y0)=(1,1). Give the result after two iterations (x^2,y^2)(x2,y2) with at least 2 decimal places for each element in the vector. You can do this by hand or write a program.

Write the vector in the answer box and click โ€œRunโ€:

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]โŽฃโŽขโŽกโ€‹1.112.223.33โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [0,0]

Q2.

Referring to the figure above, find the joint angles \theta_d = (\theta_1,\theta_2,\theta_3)ฮธdโ€‹=(ฮธ1โ€‹,ฮธ2โ€‹,ฮธ3โ€‹) that put the 3R robotโ€™s end-effector frame {b} at

T(\theta_d) = T_{sd} = \left[

โˆ’0.5850.81100โˆ’0.811โˆ’0.5850000100.0762.60801

\right]T(ฮธdโ€‹)=Tsdโ€‹=โŽฃโŽขโŽขโŽขโŽกโ€‹โˆ’0.5850.81100โ€‹โˆ’0.811โˆ’0.58500โ€‹0010โ€‹0.0762.60801โ€‹โŽฆโŽฅโŽฅโŽฅโŽคโ€‹

relative to the {s} frame, where linear distances are in meters.  (The {s} frame is located at joint 1, but it is drawn at a different location for clarity.)  The robot is shown at its home configuration, and the screw axis for each joint points toward you (out of the screen). The length of each link is 1 meter. Your solution should use either {\tt IKinBody}IKinBody or {\tt IKinSpace}IKinSpace, the initial guess \theta^0 = (\pi/4,\pi/4,\pi/4) = (0.7854, 0.7854, 0.7854)ฮธ0=(ฯ€/4,ฯ€/4,ฯ€/4)=(0.7854,0.7854,0.7854), and tolerances \epsilon_\omega = 0.001ฯตฯ‰โ€‹=0.001 (0.057 degrees) and \epsilon_v = 0.0001ฯตvโ€‹=0.0001 (0.1 mm). Give \theta_dฮธdโ€‹ as a vector with at least 2 decimal places for each element in the vector. (Note that there is more than one solution to the inverse kinematics for T_{sd}Tsdโ€‹, but we are looking for the solution that is โ€œcloseโ€ to the initial guess \theta^0 = (\pi/4,\pi/4,\pi/4)ฮธ0=(ฯ€/4,ฯ€/4,ฯ€/4), i.e., the solution that will be returned by {\tt IKinBody}IKinBody or {\tt IKinSpace}IKinSpace.)

Write the vector in the answer box and click โ€œRunโ€:

[1.11,2.22,3.33] for \left[

1.112.223.33

\right]โŽฃโŽขโŽกโ€‹1.112.223.33โ€‹โŽฆโŽฅโŽคโ€‹.

  • 1
  • [0,0,0]

week 04: Modern Robotics, Course 2: Robot Kinematics Coursera Quiz Answers

Quiz 01: Lecture Comprehension, Kinematics of Closed Chains (Chapter 7)

Q1. Which of the following statements is true about closed-chain and parallel robots? Select all that apply.

  • For a given set of positions of the actuated joints, there may be more than one configuration of the end-effector.
  • Closed-chain robots are a subclass of parallel robots.
  • Some joints may be unactuated.
  • The inverse kinematics for a parallel robot are generally easier to compute than its forward kinematics.
  • Parallel robots are sometimes chosen instead of open-chain robots for their larger workspace.

Quiz 02: Chapter 7, Kinematics of Closed Chains

Q1. The inverse Jacobian J^{-1}Jโˆ’1 for a parallel robot maps the end-effector twist \mathcal{V}V to the actuated joint velocities \dot{\theta}ฮธห™, and therefore the inverse Jacobian has nn rows (if there are nn actuators) and 6 columns (since a twist is 6-dimensional).

If the twist \mathcal{V}V consists of a 1 in the iiโ€˜th element and zeros in all other elements, then what is the corresponding vector of actuated joint velocities \dot{\theta}ฮธห™?

  • The iiโ€˜th row of J^{-1}Jโˆ’1.
  • The iiโ€˜th column of J^{-1}Jโˆ’1.

Q2. For the 3xRRR planar parallel mechanism shown below, let \phiฯ• be the orientation of the end-effector frame and p \in \mathbb{R}^2pโˆˆR2 be the vector p expressed in fixed frame coordinates. Let a_i \in \mathbb{R}^2aiโ€‹โˆˆR2 be the vector a_iiโ€‹ expresed in fixed frame coordinates and b_i \in \mathbb{R}^2biโ€‹โˆˆR2 be the vector b_iiโ€‹ expressed in the moving body frame coordinates. Define vector \text{d}_i = \text{p} + R\text{b}_{i} โ€“ \text{a}_{i}diโ€‹=p+Rbiโ€‹โˆ’aiโ€‹ for i = 1, 2, 3i=1,2,3, where

R = \left[\begin{array}{cc}\cos\phi & -\sin\phi \\\sin\phi & \cos\phi \\\end {array}\right].R=[cosฯ•sinฯ•โ€‹โˆ’sinฯ•cosฯ•โ€‹].fvdWEOCfEeeK2w4Lcly5FA e60f4d9e78317e908a2f1bfba0587606 3rrr 01

Derive a set of independent equations relating (\phi, p)(ฯ•,p) and (\theta_1, \theta_2, \theta_3)(ฮธ1โ€‹,ฮธ2โ€‹,ฮธ3โ€‹). Which of the following is correct?

  • ({p} + R{b}_{i} โ€“ {a}_{i})^2 = 2L^2(1 + \cos\theta_{i}), i = 1, 2, 3.(p+Rbiโ€‹โˆ’aiโ€‹)2=2L2(1+cosฮธiโ€‹),i=1,2,3.
  • ({p} + R{b}_{i} โ€“ {a}_{i})^\intercal({p} + R{b}_{i} โ€“ {a}_{i}) = 2L^2(1 โ€“ \sin\theta_{i}), i = 1, 2, 3.(p+Rbiโ€‹โˆ’aiโ€‹)โŠบ(p+Rbiโ€‹โˆ’aiโ€‹)=2L2(1โˆ’sinฮธiโ€‹),i=1,2,3.
  • ({p} + R{b}_{i} โ€“ {a}_{i})^\intercal({p} + R{b}_{i} โ€“ {a}_{i}) = 2L^2(1 โ€“ \cos\theta_{i}), i = 1, 2, 3.(p+Rbiโ€‹โˆ’aiโ€‹)โŠบ(p+Rbiโ€‹โˆ’aiโ€‹)=2L2(1โˆ’cosฮธiโ€‹),i=1,2,3.
  • ({p} + R{b}_{i} โ€“ {a}_{i})^\intercal({p} + R{b}_{i} โ€“ {a}_{i}) = 2L^2(1 + \cos\theta_{i}), i = 1, 2, 3.(p+Rbiโ€‹โˆ’aiโ€‹)โŠบ(p+Rbiโ€‹โˆ’aiโ€‹)=2L2(1+cosฮธiโ€‹),i=1,2,3.

We will Update These Answers Soon.

About The Coursera

Coursera, India’s largest online learning platform, started offering students millions of free courses daily. These courses come from a number of well-known universities, where professors and industry experts teach very well and in a way that is easier to understand.


About Modern Robotics, Course 2: Robot Kinematics Course

Want to find out how robots work? Do you want to work in the field of robotics? Are you willing to put in the time and effort to learn basic mathematical modelling techniques that are used in all areas of robotics?

If so, you might be interested in “Modern Robotics: Mechanics, Planning, and Control.” This specialisation, which is made up of six short courses, is a serious way for students who want to work in robotics or go on to more advanced study to prepare. It’s not a tester.

In Course 2 of the specialisation, Robot Kinematics, you will learn how to use the product-of-exponentials formula to solve the forward kinematics. This is the process of figuring out how the robot’s “hand” is set up based on the joint values. Your hard work in Course 1 pays off well, because the tools you’ve learned make forward kinematics easy. This is followed by velocity kinematics and statics, which show how joint velocities, forces, and torques relate to twists and wrenches in the end-effector, inverse kinematics, which shows how to calculate joint values to get a desired “hand” shape, and kinematics of robots with closed chains.

The book “Modern Robotics: Mechanics, Planning, and Control” is used as a guide for this class (Lynch and Park, Cambridge University Press 2017). You can buy the book, or you can use the free PDF preprint. You will build on a library of robotics software in the language of your choice (Python, Mathematica, or MATLAB) and use the free cross-platform robot simulator V-REP, which lets you work with cutting-edge robots at home without spending any money.

Conclusion

Hopefully, this article will help you find all the Week, final assessment, and Peer Graded Assessment Answers for the Coursera Modern Robotics, Course 2: Robot Kinematics Quiz, allowing you to acquire a greater understanding with minimal effort. If this article has helped you, share it on social media and let your friends know about this incredible training.

You may also view our additional course Answers. Follow our Techno-RJ Blog for further updates, and stick with us as we share many more free courses and their exam/quiz solutions.

557 thoughts on “Modern Robotics, Course 2: Robot Kinematics Coursera Quiz Answers 2023 [๐Ÿ’ฏ% Correct Answer]”

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