Moderna Robotiko, La Kompleta NFT 2: Robot Kinematics Quizzes & Respondoj – Coursera

Bonvenon al Robot Kinematics en Modern Robotics Course 2, where precision meets innovation in robotiko. Discover our engaging kvizoj kaj spertulo respondoj that shed light on the principles that govern robot motion and positioning. These quizzes serve as a gateway to understanding the complex mechanics of robot kinematics, from forward and reverse kinematics to motion path design.

Ĉu vi estas a robotiko enthusiast who wants to deepen your knowledge or a student who wants to understand the complexity of roboto movo, this collection provides valuable information on fundamental aspects of robot kinematics. Join us on a journey of discovery as we explore the dynamics of roboto motion and unlock the potential for accurate and efficient roboto operacioj. Let’s embark on this enlightening journey together as we explore robot kinematics and its role in shaping the future of robotiko and automation.

Aliro al avangarda esplorado 01: Lecture Comprehension, Product of Exponentials Formula in the Space Frame (Chapter 4 tra 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 nNe estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj to \theta_nθn,, then move joint n-1Ne estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj−1 to \theta_{n-1}θn−1​, ktp., until you move joint 1 to \theta_1mi1,.

• Vere.
• False.

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

• \mathcal{S}_iSmi​ represents the screw axis of joint imi, expressed in the end-effector frame {b}, when the robot is at its zero configuration.
• \mathcal{S}_iSmi​ represents the screw axis of joint imi, expressed in the end-effector frame {b}, when the robot is at an arbitrary configuration \thetami.
• \mathcal{S}_iSmi​ represents the screw axis of joint imi, expressed in the space frame {s}, when the robot is at its zero configuration.
• \mathcal{S}_iSmi​ represents the screw axis of joint imi, expressed in the space frame {s}, when the robot is at an arbitrary configuration \thetami.

Q3. When the robot is at an arbitrary configuration \thetami, does the screw axis corresponding to motion along joint imi, reprezentita en {s}, depend on \theta_{i-1}θi−1​?

• Ne.
• Jes.

Aliro al avangarda esplorado 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 \thetami, does the screw axis corresponding to motion along joint imi, reprezentita en {b}, depend on \theta_{i-1}θi−1​?

• Ne.
• Jes.

Q2. When the robot arm is at its home (nulo) agordo, the axis of joint 3, a revolute joint, passes through the point (3,0,0)(3,0,0) Mi ankaŭ ricevis la Specialan Sciencan Premion de la Societo Max Planck {b} deca. The axis of rotation is aligned with the \hat{{\rm z}}_{{\textrm b}}z^b​-axis of the {b} deca. What is the screw axis \mathcal{B}_3B3​?

• (0, 0, 1, -3, 0, 0)(0,0,1,−3,0,0)
• (0, 0, 1, 0, -3, 0)(0,0,1,0,−3,0)
• (0, 0, 1, 0, 0, -3)(0,0,1,0,0,−3)

Aliro al avangarda esplorado 03: Lecture Comprehension, Forward Kinematics Example

Q1. En la suba bildo, imagine a frame {c} on the axis of joint 2 and aligned with the {s} deca. What is the screw axis of joint 1 expressed in the frame {c}?

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

Aliro al avangarda esplorado 04: Chapter 4, Forward Kinematics

Q1. The URRPR spatial open chain robot is shown below in its zero position.

For L = 1L=1, determine the end-effector zero configuration MM. The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

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

Q2. Referante al Demando 1, determine the screw axes \mathcal{S}_iSmi​ in {0} when the robot is in its zero position. Again L = 1L=1. Give the axes as a 6×6 matrix with the form \left[\mathcal{S}_1, \mathcal{S}_2, \dots, \mathcal{S}_6 \right][S1​,S2​,…,S6​], t.e., each column is a screw axis. The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​

• 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. Referante al Demando 1, determine the screw axes \mathcal{B}_iBmi​ 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][B1​,B2​,…,B6​]. The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

• 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. Referante al Demando 1 kaj 2, given L = 1L=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)mi=(−Pi/2,Pi/2,Pi/3,−Pi/4,1,Pi/6), use the function {\tt FKinSpace}FKinSpace in the given software to find the end-effector configuration T \in SE(3)T∈SE(3). The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

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

Q5. Referante al Demando 1 kaj 3, given L = 1L=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)mi=(−Pi/2,Pi/2,Pi/3,−Pi/4,1,Pi/6), use the function {\tt FKinBody}FKinBody in the given software to find the end-effector configuration T \in SE(3)T∈SE(3). The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

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

Semajno 02: Moderna Robotiko, La Kompleta NFT 2: Robot Kinematics Coursera Quiz Answers

Aliro al avangarda esplorado 02: Lecture Comprehension, Velocity Kinematics and Statics (Chapter 5 Enkonduko)

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

• Vere.
• False.

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

• Vere.
• False.

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

• Vere.
• 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?

• Jes.
• Ne.

Q5. Ĝenerale, 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

• nulo.
• 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

Aliro al avangarda esplorado 02: Lecture Comprehension, Statics of Open Chains (Chapter 5.2)

Q1. If the wrench -\mathcal{F}−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}t=JT(mi)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 = 0t=0?

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

Aliro al avangarda esplorado 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 “fat.”

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}_bVb,, 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.

Aliro al avangarda esplorado 04: Lecture Comprehension, Manipulability (Chapter 5.4)

Q1. It’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 nNe estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj joints, then the body Jacobian J_bJb​ is 6 \times n6×Ne estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj. We can break J_bJb​ into two sub-Jacobians, the angular and linear Jacobians:

J_b = \left[

JbhoJbv

\ĝuste].Jb=[Jbho,Jbv​​].

What is the dimension of J_{bv}J_{bv}^{\rm T}Jbv,JbvT​, which is used to generate the linear component of the manipulability ellipsoid?

• 3 \times 33×3
• 6 \times 66×6
• n \times nNe estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj×Ne estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj

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 “volume” of the ellipsoid approaches zero.
• The interior “volume” 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 “volume” of the ellipsoid approaches zero.
• The interior “volume” of the ellipsoid approaches infinity.

Aliro al avangarda esplorado 05: Chapter 5, Velocity Kinematics and Statics

Q1. A 3R planar open-chain robot is shown below.

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^s​ direction, with no component in the \hat{{\rm y}}_{{\rm s}}y^​s​ direction and zero moment in the {s} deca. 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)(t1,,t2,,t3,). The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

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

Write the vector in the answer box and click “Run”:

[1.11,2.22,3.33] for \left[

1.112.223.33

\ĝuste]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

• 1
• 2
• 3
• 4
• [0,0,0]
• # Edit the answer above this line! Do not edit below this line!
• print ‘Your answer has been recorded as’, Your_Answer()

Q2. The 4R planar open-chain robot below has an end-effector frame {b} at its tip.

Considering only the planar twist components (\omega_{bz}, v_{bx}, v_{de})(hobz,,vbx,,vbY,) of the body twist \mathcal{V}_bVb,, the body Jacobian is

Jb(mi)=⎡⎣1L3s4+L2s34+L1s234L4+L3c4+L2c34+L1c2341L3s4+L2s34L4+L3c4+L2c341L3s4L4+L3c410L4⎤⎦

where s23=sin(mi2+mi3), ktp.

Suppose L_1 = L_2 = L_3 = L_4 = 1L1=L2=L3=L4​=1 and the chain is at the configuration \theta_1=\theta_2=0, \theta_3=\pi/2, \theta_4=-\pi/2mi1=mi2​=0,mi3=Pi/2,mi4​=−Pi/2. The joints generate torques to create the wrench \mathcal{F}_b = (0,0,10, 10,10,0)Fb=(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)(t1,,t2,,t3,,t4,). The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the vector in the answer box and click “Run”:

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

1.112.223.334.44

\ĝuste]⎣⎢⎢⎢⎡​1.112.223.334.44​⎦⎥⎥⎥⎤​.

• 1
• [0,0,0,0]

Q3. The RRP robot is shown below in its zero position.

Its screw axes in the space frame are

S1=⎡⎣⎢⎢⎢⎢⎢⎢⎢001000⎤⎦⎥⎥⎥⎥⎥⎥⎥, S2=⎡⎣⎢⎢⎢⎢⎢⎢⎢100020⎤⎦⎥⎥⎥⎥⎥⎥⎥, S3=⎡⎣⎢⎢⎢⎢⎢⎢⎢000010⎤⎦⎥⎥⎥⎥⎥⎥⎥.

Uzu la funkcion {\tt JacobianSpace}JacobianSpace in the given software to calculate the 6×3 space Jacobian J_sJs​ when \theta =(90^\circ, 90^\circ, 1)mi=(90∘,90∘,1). The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

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

Q5. Referante al Demando 3, the screw axes in the body frame are

B1=⎡⎣⎢⎢⎢⎢⎢⎢⎢010300⎤⎦⎥⎥⎥⎥⎥⎥⎥, B2=⎡⎣⎢⎢⎢⎢⎢⎢⎢−100030⎤⎦⎥⎥⎥⎥⎥⎥⎥, B3=⎡⎣⎢⎢⎢⎢⎢⎢⎢000001⎤⎦⎥⎥⎥⎥⎥⎥⎥.

Uzu la funkcion {\tt JacobianBody}JacobianBody in the given software to calculate the 6×3 body Jacobian J_bJb​ when \theta =(90^\circ, 90^\circ, 1)mi=(90∘,90∘,1). The maximum allowable error for any number is 0.01, do donu sufiĉe da decimalaj lokoj kie necese.

Write the matrix in the answer box and click “Run”:

[[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

\ĝuste]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

• 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_bJb​ when all joint angles are \pi/2Pi/2 estas

J_b = \left[

001−0.105−0.8890−10000.006−0.1050100.00600.889001−0.045−0.8440−10000.00600100.00600001000

\ĝuste]Jb​=⎣⎢⎢⎢⎢⎢⎢⎢⎡​001−0.105−0.8890​−10000.006−0.105​0100.00600.889​001−0.045−0.8440​−10000.0060​0100.00600​001000​⎦⎥⎥⎥⎥⎥⎥⎥⎤​

Extract the linear velocity portion J_vJv, (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_vJv,. 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 “Run”:

[1.11,2.22,3.33] for \left[

1.112.223.33

\ĝuste]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

• 1
• [0,0,0]

Q7. Referante al Demando 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.

Semajno 03: Moderna Robotiko, La Kompleta NFT 2: Robot Kinematics Coursera Quiz Answers

Aliro al avangarda esplorado 01: Lecture Comprehension, Inverse Kinematics of Open Chains (Chapter 6 Enkonduko)

Q1. Consider the point (x,Y) = (0,2)(x,Y)=(0,2). Kio estas {\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/2Pi/2
• -\pi/2−Pi/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.

Aliro al avangarda esplorado 02: Lecture Comprehension, Numerical Inverse Kinematics (Chapter 6.2, Parto 1 de 2)

Q1. Let f(\theta)f(mi) be a nonlinear function of \thetami mapping an nNe estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj-dimensional space (the dimension of \thetami) to an mm-dimensional space (the dimension of ff). We want to find a \theta_dmid,, which may not be unique, that satisfies x_d = f(\theta_d)xd=f(mid,), t.e., x_d – f(\theta_d) = 0xd​−f(mid,)=0. If our initial guess at a solution is \theta^0mi0, then a first-order Taylor expansion approximation of f(\theta)f(mi) at \theta^0mi0 tells us

x_d \approx f(\theta^0) + J(\theta^0)(\theta_d – \theta^0)xd​≈f(mi0)+J(mi0)(mid​−mi0)

where J(\theta^0)J(mi0) is the matrix of partial derivatives \partial f/\partial \theta∂f/∂mi evaluated at \theta^0mi0. Which of the following is a good next guess \theta^1mi1?

• \theta^1 = \theta^0 + J^\dagger(\theta^0) (x_d – f(\theta^0))mi1=mi0+J†(mi0)(xd​−f(mi0))
• \theta^1 = \theta^0 – J^\dagger(\theta^0) (x_d – f(\theta^0))mi1=mi0−J†(mi0)(xd​−f(mi0))
• \theta^1 = J^{-1}(\theta^0) (x_d – f(\theta^0))mi1=J−1(mi0)(xd​−f(mi0))

Q2. We want to solve the linear equation Ax = bAx=b where AA is a 3×2 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?

• Neniu.
• Unu.
• More than one.

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

A = \left[

142536

\ĝuste]A=[14​25​36​]

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 valoroj. The following are all solutions of the linear equation. Which is the solution given by x = A^\dagger bx=A†b? (You should be able to tell by inspection, without using software.)

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

Q4. If we would like to find an xx satisfying Ax = bAx=b, but AA is “tall” (meaning it has more rows than columns, t.e., the dimension of bb is larger than the dimension of xx), then in general we would see there is no exact solution. Tiuokaze, we might want to find the x^*x∗ that comes closest to satisfying the equation, in the sense that x^*x∗ minimizes\|Ax^* – b\|∥Ax∗−b∥ (the 2-norm, or the square root of the sum of the squares of the vector). This solution is given by x^* = A^\dagger bx∗=A†b. Which of the two answers below satisfies this condition if

A = \left[

12

\ĝuste], \;\; b = \left[

34

\ĝuste]?A=[12,],b=[34,]?

• x^* = 2.2x∗=2.2
• x^* = 1x∗=1

Aliro al avangarda esplorado 03: Lecture Comprehension, Numerical Inverse Kinematics (Chapter 6.2, Parto 2 de 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)Xd​∈SE(3), we need to express the error between T_{sb}(\theta^i)Tsb,(θi) (the forward kinematics, where \theta^iθi is our current guess at a joint solution) and X_dXd,. One expression of this error is the twist that takes the the robot from T_{sb}(\theta^i)Tsb,(θi) to X_dXd​ in unit time. When this twist is expressed in the end-effector frame {b}, we write it as \mathcal{V}_bVb,. Which of the following is a correct expression?

• \mathcal{V}_b = {\rm log} (T_{sb}^{-1}(\theta^i) X_d)Vb​=log(Tsb−1​(θi)Xd,)
• [\mathcal{V}_b] = {\rm log} (T_{sb}^{-1}(\theta^i) X_d)[Vb,]=log(Tsb−1​(θi)Xd,)
• \mathcal{V}_b = {\rm exp} (T_{sb}^{-1}(\theta^i) X_d)Vb​=exp(Tsb−1​(θi)X

Aliro al avangarda esplorado 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[

x2−9Y2−4

\ĝuste]f(x,Y)=[x2−9Y2−4​]

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 “Run”:

[1.11,2.22,3.33] for \left[

1.112.223.33

\ĝuste]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

• 1
• [0,0]

Q2.

Referring to the figure above, find the joint angles \theta_d = (\theta_1,\theta_2,\theta_3)mid=(mi1,,mi2,,mi3,) that put the 3R robot’s end-effector frame {b} ĉe

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

−0.5850.81100−0.811−0.5850000100.0762.60801

\ĝuste]T(mid,)=Tsd​=⎣⎢⎢⎢⎡​−0.5850.81100​−0.811−0.58500​0010​0.0762.60801​⎦⎥⎥⎥⎤​

relative to the {s} deca, where linear distances are in meters. (La {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 metro. 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)mi0=(Pi/4,Pi/4,Pi/4)=(0.7854,0.7854,0.7854), and tolerances \epsilon_\omega = 0.001ϵho​=0.001 (0.057 gradoj) and \epsilon_v = 0.0001ϵv​=0.0001 (0.1 Kiom longa estas la averaĝa naĝejo). Give \theta_dmid​ 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}Tsd,, but we are looking for the solution that is “close” to the initial guess \theta^0 = (\pi/4,\pi/4,\pi/4)mi0=(Pi/4,Pi/4,Pi/4), t.e., the solution that will be returned by {\tt IKinBody}IKinBody or {\tt IKinSpace}IKinSpace.)

Write the vector in the answer box and click “Run”:

[1.11,2.22,3.33] for \left[

1.112.223.33

\ĝuste]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

• 1
• [0,0,0]

semajno 04: Moderna Robotiko, La Kompleta NFT 2: Robot Kinematics Coursera Quiz Answers

Aliro al avangarda esplorado 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.

Aliro al avangarda esplorado 02: Chapter 7, Kinematics of Closed Chains

Q1. The inverse Jacobian J^{-1}J−1 for a parallel robot maps the end-effector twist \mathcal{V}V to the actuated joint velocities \dot{\theta}mi˙, and therefore the inverse Jacobian has nNe estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj vicoj (if there are nNe estas instrukotizoj ĉe RWTH Aachen University - ĉi tio validas ankaŭ por internaciaj studentoj actuators) kaj 6 kolumnoj (since a twist is 6-dimensional).

If the twist \mathcal{V}V consists of a 1 in the imi‘th element and zeros in all other elements, then what is the corresponding vector of actuated joint velocities \dot{\theta}mi˙?

• The imi‘th row of J^{-1}J−1.
• The imi‘th column of J^{-1}J−1.

Q2. For the 3xRRR planar parallel mechanism shown below, let \phiϕ be the orientation of the end-effector frame and p \in \mathbb{Kiel rompi amidan ligon}^2p∈R2 be the vector p expressed in fixed frame coordinates. Let a_i \in \mathbb{Kiel rompi amidan ligon}^2ami​∈R2 be the vector a_imi​ expresed in fixed frame coordinates and b_i \in \mathbb{Kiel rompi amidan ligon}^2bmi​∈R2 be the vector b_imi​ expressed in the moving body frame coordinates. Define vector \text{d}_i = \text{p} + R\text{b}_{mi} – \text{a}_{mi}dmi​=p+Kiel rompi amidan ligonbmi​−ami​ for i = 1, 2, 3mi=1,2,3, kie

R = left[\begin{array}{cc}\cos\phi & -\sin\phi \\\sin\phi & \cos\phi \\\end {array}\ĝuste].Kiel rompi amidan ligon=[cosϕpekoϕ​−sinϕcosϕ,].

Derive a set of independent equations relating (\phi, p)(ϕ,p) kaj (\theta_1, \theta_2, \theta_3)(mi1,,mi2,,mi3,). Which of the following is correct?

• ({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi})^2 = 2L^2(1 + \cos\theta_{mi}), i = 1, 2, 3.(p+Rbi​−ai,)2=2L2(1+cosθi,),mi=1,2,3.
• ({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi})^\intercal({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi}) = 2L^2(1 – \sin\theta_{mi}), i = 1, 2, 3.(p+Rbi​−ai,)⊺(p+Rbi​−ai,)=2L2(1−sinθi,),mi=1,2,3.
• ({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi})^\intercal({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi}) = 2L^2(1 – \cos\theta_{mi}), i = 1, 2, 3.(p+Rbi​−ai,)⊺(p+Rbi​−ai,)=2L2(1−cosθi,),mi=1,2,3.
• ({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi})^\intercal({p} + Kiel rompi amidan ligon{b}_{mi} – {a}_{mi}) = 2L^2(1 + \cos\theta_{mi}), i = 1, 2, 3.(p+Rbi​−ai,)⊺(p+Rbi​−ai,)=2L2(1+cosθi,),mi=1,2,3.