Robótica moderna, Curso 2: Robot Kinematics Quizzes & Respuestas – Coursera

Bienvenido a Robot Kinematics en Modern Robotics Course 2, where precision meets innovation in robótica. Discover our engaging cuestionarios and expert respuestas 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.

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

Examen 01: Comprensión de conferencias, Product of Exponentials Formula in the Space Frame (Capítulo 4 mediante 4.1.2)

Q1. Verdadero o falso? 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 nnorte to \theta_nθn​, then move joint n-1norte−1 to \theta_{n-1}θn−1​, etcétera, until you move joint 1 to \theta_1i1​.

• Cierto.
• Falso.

Q2. Consider the screw axis \mathcal{S}_iSyo​ used in the PoE formula. Cual de los siguientes es verdadero?

• \matemático{S}_iSyo​ represents the screw axis of joint iyo, expressed in the end-effector frame {si}, when the robot is at its zero configuration.
• \matemático{S}_iSyo​ represents the screw axis of joint iyo, expressed in the end-effector frame {si}, when the robot is at an arbitrary configuration \thetai.
• \matemático{S}_iSyo​ represents the screw axis of joint iyo, expressed in the space frame {s}, when the robot is at its zero configuration.
• \matemático{S}_iSyo​ represents the screw axis of joint iyo, expressed in the space frame {s}, when the robot is at an arbitrary configuration \thetai.

Tercer trimestre. When the robot is at an arbitrary configuration \thetai, does the screw axis corresponding to motion along joint iyo, represented in {s}, depend on \theta_{i-1}θi−1​?

• No.
• Sí.

Examen 02: Comprensión de conferencias, Product of Exponentials Formula in the End-Effector Frame (Capítulo 4.1.3)

Q1. When the robot is at an arbitrary configuration \thetai, does the screw axis corresponding to motion along joint iyo, represented in {si}, depend on \theta_{i-1}θi−1​?

• No.
• Sí.

Q2. When the robot arm is at its home (cero) configuración, the axis of joint 3, a revolute joint, passes through the point (3,0,0)(3,0,0) en el {si} marco. The axis of rotation is aligned with the \hat{{\rm z}}_{{\textrm b}}z^b​-axis of the {si} marco. What is the screw axis \mathcal{segundo}_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)

Examen 03: Comprensión de conferencias, Forward Kinematics Example

Q1. En la imagen de abajo, imagine a frame {do} on the axis of joint 2 and aligned with the {s} marco. What is the screw axis of joint 1 expressed in the frame {do}?

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

Examen 04: Capítulo 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 MMETRO. El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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. Refiriéndose nuevamente a la pregunta 1, determine the screw axes \mathcal{S}_iSyo​ 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[\matemático{S}_1, \matemático{S}_2, \dots, \matemático{S}_6 \right][S1​,S2​,...,S6​], es decir, each column is a screw axis. El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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]]

Tercer trimestre. Refiriéndose nuevamente a la pregunta 1, determine the screw axes \mathcal{segundo}_iByo​ in {si} when the robot is in its zero position. Again L = 1L=1. Give the axes as a matrix with the form \left[\matemático{segundo}_1, \matemático{segundo}_2, \dots, \matemático{segundo}_6 \right][B1​,B2​,...,B6​]. El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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]]

Cuarto trimestre. Refiriéndose nuevamente a la pregunta 1 y 2, given L = 1L=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)i=(-Fr./2,Fr./2,Fr./3,-Fr./4,1,Fr./6), utilizar la función {\tt FKinSpace}FKinSpace in the given software to find the end-effector configuration T \in SE(3)T∈Smi(3). El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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. Refiriéndose nuevamente a la pregunta 1 y 3, given L = 1L=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)i=(-Fr./2,Fr./2,Fr./3,-Fr./4,1,Fr./6), utilizar la función {\tt FKinBody}FKinBody in the given software to find the end-effector configuration T \in SE(3)T∈Smi(3). El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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]]

Semana 02: Robótica moderna, Curso 2: Robot Kinematics Coursera Quiz Answers

Examen 02: Comprensión de conferencias, Velocity Kinematics and Statics (Capítulo 5 Introducción)

Q1. Verdadero o falso? The Jacobian matrix depends on the joint variables.

• Cierto.
• Falso.

Q2. Verdadero o falso? The Jacobian matrix depends on the joint velocities.

• Cierto.
• Falso.

Tercer trimestre. Verdadero o falso? Row iyo of the Jacobian corresponds to the end-effector velocity when joint iyo moves at unit speed and all other joints are stationary.

• Cierto.
• Falso.

Cuarto trimestre. 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?

• Sí.
• No.

Q5. En 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

• cero.
• infinito.

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

Examen 02: Comprensión de conferencias, Statics of Open Chains (Capítulo 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^{\habitación T}(\theta) \matemático{F}t=JT(i)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-dimensional.
• At least 2-dimensional.
• 4-dimensional.
• At least 4-dimensional.

Examen 03: Comprensión de conferencias, Singularities (Capítulo 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. Cuál de las siguientes afirmaciones es verdadera? Seleccione todas las que correspondan.

• 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. Cuál de las siguientes afirmaciones es verdadera? Seleccione todas las que correspondan.

• 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.”

Tercer trimestre. 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}_bVsi​, 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.

Examen 04: Comprensión de conferencias, Manipulability (Capítulo 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 nnorte joints, then the body Jacobian J_bJsi​ is 6 \times n6×norte. We can break J_bJsi​ into two sub-Jacobians, the angular and linear Jacobians:

J_b = \left[

JsiOhJsiv

\Correcto].Jsi=[JsiOh​Jsiv​​].

What is the dimension of J_{bv}J_{bv}^{\habitación 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 nnorte×norte

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? Seleccione todas las que correspondan.

• 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.

Tercer trimestre. 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? Seleccione todas las que correspondan.

• 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.

Examen 05: Capítulo 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} marco. 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​). El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

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

Escriba el vector en el cuadro de respuesta y haga clic en "Ejecutar":

[1.11,2.22,3.33] para izquierda[

1.112.223.33

\Correcto]⎣⎢⎡​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 {si} at its tip.

Considering only the planar twist components (\omega_{bz}, v_{bx}, v_{por})(Ohsiz​,vsiX​,vsiy​) of the body twist \mathcal{V}_bVsi​, the body Jacobian is

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

where s23=sin(i2+i3), etc..

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/2i1=i2​=0,i3=Fr./2,i4​=−Fr./2. The joints generate torques to create the wrench \mathcal{F}_b = (0,0,10, 10,10,0)Fsi=(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​). El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba el vector en el cuadro de respuesta y haga clic en "Ejecutar":

[1.11,2.22,3.33,4.44] para izquierda[

1.112.223.334.44

\Correcto]⎣⎢⎢⎢⎡​1.112.223.334.44​⎦⎥⎥⎥⎤​.

• 1
• [0,0,0,0]

Tercer trimestre. The RRP robot is shown below in its zero position.

Its screw axes in the space frame are

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

Usa la función {\tt JacobianSpace}JacobianSpace in the given software to calculate the 6×3 space Jacobian J_sJs​ when \theta =(90^\circ, 90^\circ, 1)i=(90∘,90∘,1). El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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. Refiriéndose nuevamente a la pregunta 3, the screw axes in the body frame are

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

Usa la función {\tt JacobianBody}JacobianBody in the given software to calculate the 6×3 body Jacobian J_bJsi​ when \theta =(90^\circ, 90^\circ, 1)i=(90∘,90∘,1). El error máximo permitido para cualquier número es 0.01, así que proporcione suficientes decimales cuando sea necesario.

Escriba la matriz en el cuadro de respuestas y haga clic en "Ejecutar":

[[1.11,2.22,3.33],[4.44,5.55,6.66],[7.77,8.88,9.99]] para izquierda[

1.114.447.772.225.558.883.336.669.99

\Correcto]⎣⎢⎡​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_bJsi​ when all joint angles are \pi/2Fr./2 es

J_b = \left[

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

\Correcto]Jsi​=⎣⎢⎢⎢⎢⎢⎢⎢⎡​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, con al menos 2 decimal places for each element in this vector, to represent the direction of the longest principal semi-axis.

Escriba el vector en el cuadro de respuesta y haga clic en "Ejecutar":

[1.11,2.22,3.33] para izquierda[

1.112.223.33

\Correcto]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

• 1
• [0,0,0]

Q7. Refiriéndose nuevamente a la pregunta 5 and its result, give the length, con al menos 2 decimales, of the longest principal semi-axis of that three-dimensional linear velocity manipulability ellipsoid.

Semana 03: Robótica moderna, Curso 2: Robot Kinematics Coursera Quiz Answers

Examen 01: Comprensión de conferencias, Inverse Kinematics of Open Chains (Capítulo 6 Introducción)

Q1. Consider the point (X,y) = (0,2)(X,y)=(0,2). Qué es {\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/2Fr./2
• -\pi/2−Fr./2

Q2. What are advantages of numerical inverse kinematics over analytic inverse kinematics? Seleccione todas las que correspondan.

• 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.

Examen 02: Comprensión de conferencias, Numerical Inverse Kinematics (Capítulo 6.2, Parte 1 de 2)

Q1. Let f(\theta)F(i) be a nonlinear function of \thetai mapping an nnorte-dimensional space (the dimension of \thetai) to an mmetro-dimensional space (the dimension of fF). We want to find a \theta_dire​, which may not be unique, that satisfies x_d = f(\theta_d)Xre=F(ire​), es decir, x_d – f(\theta_d) = 0Xre​-F(ire​)=0. If our initial guess at a solution is \theta^0i0, then a first-order Taylor expansion approximation of f(\theta)F(i) at \theta^0i0 tells us

x_d \approx f(\theta^0) + J(\theta^0)(\theta_d – \theta^0)Xre​≈F(i0)+J(i0)(ire​-i0)

where J(\theta^0)J(i0) is the matrix of partial derivatives \partial f/\partial \theta∂F/∂i evaluated at \theta^0i0. Which of the following is a good next guess \theta^1i1?

• \theta^1 = \theta^0 + J^\dagger(\theta^0) (x_d – f(\theta^0))i1=i0+J†(i0)(xd​-F(i0))
• \theta^1 = \theta^0 – J^\dagger(\theta^0) (x_d – f(\theta^0))i1=i0-J†(i0)(xd​-F(i0))
• \theta^1 = J^{-1}(\theta^0) (x_d – f(\theta^0))i1=J−1(i0)(xd​-F(i0))

Q2. We want to solve the linear equation Ax = bAx=si where AUNA is a 3×2 matrix, XX is a 2-vector, y bsi is a 3-vector. For a randomly chosen AUNA matrix and vector bsi, how many solutions xX can we expect?

• Ninguna.
• Uno.
• More than one.

Tercer trimestre. We want to solve the linear equation Ax = bUNAX=si, dónde

A = \left[

142536

\Correcto]UNA=[14​25​36​]

and b = [7 \;\;8]^{\habitación T}si=[78]T. Since xX is a 3-vector and bsi is a 2-vector, we can expect a one-dimensional set of solutions in the 3-dimensional space of possible xX valores. The following are all solutions of the linear equation. Which is the solution given by x = A^\dagger bX=UNA†si? (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)

Cuarto trimestre. If we would like to find an xX satisfying Ax = bUNAX=si, but AUNA is “tall” (meaning it has more rows than columns, es decir, the dimension of bsi is larger than the dimension of xX), then in general we would see there is no exact solution. En este caso, we might want to find the x^*X∗ that comes closest to satisfying the equation, in the sense that x^*X∗ minimizes\|Ax^* – b\|∥UNAX∗−si∥ (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∗=UNA†si. Which of the two answers below satisfies this condition if

A = \left[

12

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

34

\Correcto]?UNA=[12​],si=[34​]?

• x^* = 2.2X∗=2.2
• x^* = 1X∗=1

Examen 03: Comprensión de conferencias, Numerical Inverse Kinematics (Capítulo 6.2, Parte 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 {si}, we write it as \mathcal{V}_bVsi​. Which of the following is a correct expression?

• \matemático{V}_b = {\rm log} (T_{sb}^{-1}(\theta^i) X_d)Vsi​=log(Tsb−1​(θi)Xd​)
• [\matemático{V}_b] = {\rm log} (T_{sb}^{-1}(\theta^i) X_d)[Vsi​]=log(Tsb−1​(θi)Xd​)
• \matemático{V}_b = {\rm exp} (T_{sb}^{-1}(\theta^i) X_d)Vsi​=exp(Tsb−1​(θi)Distribución de probabilidad

Examen 04: Capítulo 6, Inverse Kinematics

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

F(X,y) = izquierda[

X2−9y2−4

\Correcto]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) con al menos 2 decimal places for each element in the vector. You can do this by hand or write a program.

Escriba el vector en el cuadro de respuesta y haga clic en "Ejecutar":

[1.11,2.22,3.33] para izquierda[

1.112.223.33

\Correcto]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

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• [0,0]

Q2.

Referring to the figure above, find the joint angles \theta_d = (\theta_1,\theta_2,\theta_3)ire=(i1​,i2​,i3​) that put the 3R robot’s end-effector frame {si} a

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

−0.5850.81100−0.811−0.5850000100.0762.60801

\Correcto]T(ire​)=Tsre​=⎣⎢⎢⎢⎡​−0.5850.81100​−0.811−0.58500​0010​0.0762.60801​⎦⎥⎥⎥⎤​

relative to the {s} marco, where linear distances are in meters. (los {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)i0=(Fr./4,Fr./4,Fr./4)=(0.7854,0.7854,0.7854), and tolerances \epsilon_\omega = 0.001ϵOh​=0.001 (0.057 grados) and \epsilon_v = 0.0001ϵv​=0.0001 (0.1 mm). Give \theta_dire​ 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}Tsre​, but we are looking for the solution that is “close” to the initial guess \theta^0 = (\pi/4,\pi/4,\pi/4)i0=(Fr./4,Fr./4,Fr./4), es decir, the solution that will be returned by {\tt IKinBody}IKinBody or {\tt IKinSpace}IKinSpace.)

Escriba el vector en el cuadro de respuesta y haga clic en "Ejecutar":

[1.11,2.22,3.33] para izquierda[

1.112.223.33

\Correcto]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

• 1
• [0,0,0]

semana 04: Robótica moderna, Curso 2: Robot Kinematics Coursera Quiz Answers

Examen 01: Comprensión de conferencias, Kinematics of Closed Chains (Capítulo 7)

Q1. Which of the following statements is true about closed-chain and parallel robots? Seleccione todas las que correspondan.

• 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.

Examen 02: Capítulo 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}i˙, and therefore the inverse Jacobian has nnorte filas (if there are nnorte actuators) y 6 columnas (since a twist is 6-dimensional).

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

• The iyo‘th row of J^{-1}J−1.
• The iyo‘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{R}^2pags∈R2 be the vector p expressed in fixed frame coordinates. Let a_i \in \mathbb{R}^2unayo​∈R2 be the vector a_iyo​ expresed in fixed frame coordinates and b_i \in \mathbb{R}^2siyo​∈R2 be the vector b_iyo​ expressed in the moving body frame coordinates. Define vector \text{re}_i = \text{pags} + R\text{si}_{yo} – \text{una}_{yo}reyo​=p+Rsiyo​−ayo​ for i = 1, 2, 3yo=1,2,3, dónde

R = \left[\begin{formación}{cc}\cos\phi & -\sin\phi \\\sin\phi & \cos\phi \\\end {formación}\Correcto].R=[porqueϕpecadoϕ​−sinϕporqueϕ​].

Derive a set of independent equations relating (\phi, pags)(ϕ,pags) y (\theta_1, \theta_2, \theta_3)(i1​,i2​,i3​). Which of the following is correct?

• ({pags} + R{si}_{yo} - {una}_{yo})^2 = 2L^2(1 + \cos\theta_{yo}), i = 1, 2, 3.(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)2=2L2(1+porqueθi​),yo=1,2,3.
• ({pags} + R{si}_{yo} - {una}_{yo})^\intercal({pags} + R{si}_{yo} - {una}_{yo}) = 2L^2(1 – \sin\theta_{yo}), i = 1, 2, 3.(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)⊺(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)=2L2(1−sinθi​),yo=1,2,3.
• ({pags} + R{si}_{yo} - {una}_{yo})^\intercal({pags} + R{si}_{yo} - {una}_{yo}) = 2L^2(1 – \cos\theta_{yo}), i = 1, 2, 3.(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)⊺(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)=2L2(1−porqueθi​),yo=1,2,3.
• ({pags} + R{si}_{yo} - {una}_{yo})^\intercal({pags} + R{si}_{yo} - {una}_{yo}) = 2L^2(1 + \cos\theta_{yo}), i = 1, 2, 3.(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)⊺(pags+Rbi​-Esta asociación entre una de las principales universidades y desarrolladores de tecnología del mundo y Microsoft es un gran ejemplo de colaboración entre empresas y academia.​)=2L2(1+porqueθi​),yo=1,2,3.