Roboti za kisasa, Kozi 2: Maswali ya Kinematics ya Roboti & Majibu - Coursera

Karibu Robot Kinematics ndani Modern Robotics Course 2, where precision meets innovation in robotiki. Discover our engaging maswali na mtaalam majibu 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.

Whether you are a robotiki enthusiast who wants to deepen your knowledge or a student who wants to understand the complexity of roboti uhifadhi-wa-mitambo-nishati-pendulum, this collection provides valuable information on fundamental aspects of robot kinematics. Join us on a journey of discovery as we explore the dynamics of roboti motion and unlock the potential for accurate and efficient roboti shughuli. Let’s embark on this enlightening journey together as we explore robot kinematics and its role in shaping the future of robotiki and automation.

Maswali 01: Lecture Comprehension, Product of Exponentials Formula in the Space Frame (Sura 4 kupitia 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_nθn., then move joint n-1n−1 to \theta_{n-1}θn−1, na kadhalika., until you move joint 1 to \theta_1lakini haifanyi yaliyomo katika hisabati kuwa tofauti1..

Kweli.
Uongo.

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

\mathcal{S}_iSi represents the screw axis of joint ii, expressed in the end-effector frame {b}, when the robot is at its zero configuration.
\mathcal{S}_iSi represents the screw axis of joint ii, expressed in the end-effector frame {b}, when the robot is at an arbitrary configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti.
\mathcal{S}_iSi represents the screw axis of joint ii, expressed in the space frame {s}, when the robot is at its zero configuration.
\mathcal{S}_iSi represents the screw axis of joint ii, expressed in the space frame {s}, when the robot is at an arbitrary configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti.

Q3. When the robot is at an arbitrary configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti, does the screw axis corresponding to motion along joint ii, represented in {s}, depend on \theta_{i-1}θi−1?

Hapana.
Ndio.

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

Q1. When the robot is at an arbitrary configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti, does the screw axis corresponding to motion along joint ii, represented in {b}, depend on \theta_{i-1}θi−1?

Hapana.
Ndio.

Q2. When the robot arm is at its home (zero) usanidi, the axis of joint 3, a revolute joint, passes through the point (3,0,0)(3,0,0) ndani ya {b} frame. The axis of rotation is aligned with the \hat{{\rm z}}_{{\textrm b}}z^b-axis of the {b} frame. 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)

Maswali 03: Lecture Comprehension, Forward Kinematics Example

Q1. Katika picha hapa chini, 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)

Maswali 04: Sura 4, Forward Kinematics

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

For L = 1na hufika bila nishati inayoweza kutokea na nishati ya kinetic=1, determine the end-effector zero configuration Mkwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu. 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 “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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

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}_iSi in {0} when the robot is in its zero position. Again L = 1na hufika bila nishati inayoweza kutokea na nishati ya kinetic=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], 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 “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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤

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. Referring back to Question 1, determine the screw axes \mathcal{B}_iBi in {b} when the robot is in its zero position. Again L = 1na hufika bila nishati inayoweza kutokea na nishati ya kinetic=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, so give enough decimal places where necessary.

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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

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. Referring back to Question 1 na 2, given L = 1na hufika bila nishati inayoweza kutokea na nishati ya kinetic=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)lakini haifanyi yaliyomo katika hisabati kuwa tofauti=(-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, so give enough decimal places where necessary.

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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

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

Q5. Referring back to Question 1 na 3, given L = 1na hufika bila nishati inayoweza kutokea na nishati ya kinetic=1 and joint variable values \theta = (-\pi/2, \pi/2, \pi/3, -\pi/4, 1, \pi/6)lakini haifanyi yaliyomo katika hisabati kuwa tofauti=(-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, so give enough decimal places where necessary.

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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

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

Wiki 02: Roboti za kisasa, Kozi 2: Robot Kinematics Coursera Quiz Answers

Maswali 02: Lecture Comprehension, Velocity Kinematics and Statics (Sura 5 Utangulizi)

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

Kweli.
Uongo.

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

Kweli.
Uongo.

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.

Kweli.
Uongo.

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?

Ndio.
Hapana.

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

nyanja (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.

Swali 7. 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

Maswali 02: Lecture Comprehension, Statics of Open Chains (Sura 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}lakini haifanyi yaliyomo katika hisabati kuwa tofauti=JT(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)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 = 0lakini haifanyi yaliyomo katika hisabati kuwa tofauti=0?

2-ya dimensional.
At least 2-dimensional.
4-ya dimensional.
At least 4-dimensional.

Maswali 03: Lecture Comprehension, Singularities (Sura 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? Chagua zote zinazotumika.

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? Chagua zote zinazotumika.

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.

Maswali 04: Lecture Comprehension, Manipulability (Sura 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 nn joints, then the body Jacobian J_bJb is 6 \times n6×n. We can break J_bJb into two sub-Jacobians, the angular and linear Jacobians:

J_b = \left[

JbmwangazaJbAZ-500

\haki].Jb=[Jbmwangaza.JbAZ-500].

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 nn×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? Chagua zote zinazotumika.

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? Chagua zote zinazotumika.

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.

Maswali 05: Sura 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^s direction, with no component in the \hat{{\rm y}}_{{\rm s}}y^s 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)(lakini haifanyi yaliyomo katika hisabati kuwa tofauti1.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti3.). The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Muhimu: 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. Hasa, 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

\haki]⎣⎢⎡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. m7IE3eB1EeeY9RLN7DX 0g 22e7ceb45566a152517067d8a7d6349a ex02 01

Considering only the planar twist components (\omega_{bz}, v_{bx}, v_{na})(mwangazabni mlinganyo halali wa kihisabati ikiwa.,AZ-500bx.,AZ-500bY.) of the body twist \mathcal{V}_bVb., the body Jacobian is

Jb(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)=⎡⎣1na hufika bila nishati inayoweza kutokea na nishati ya kinetic3s4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic2s34+na hufika bila nishati inayoweza kutokea na nishati ya kinetic1s234na hufika bila nishati inayoweza kutokea na nishati ya kinetic4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic3c4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic2c34+na hufika bila nishati inayoweza kutokea na nishati ya kinetic1c2341na hufika bila nishati inayoweza kutokea na nishati ya kinetic3s4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic2s34na hufika bila nishati inayoweza kutokea na nishati ya kinetic4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic3c4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic2c341na hufika bila nishati inayoweza kutokea na nishati ya kinetic3s4na hufika bila nishati inayoweza kutokea na nishati ya kinetic4+na hufika bila nishati inayoweza kutokea na nishati ya kinetic3c410na hufika bila nishati inayoweza kutokea na nishati ya kinetic4⎤⎦

where s23=sin(lakini haifanyi yaliyomo katika hisabati kuwa tofauti2+lakini haifanyi yaliyomo katika hisabati kuwa tofauti3), na kadhalika.

Suppose L_1 = L_2 = L_3 = L_4 = 1na hufika bila nishati inayoweza kutokea na nishati ya kinetic1=na hufika bila nishati inayoweza kutokea na nishati ya kinetic2=na hufika bila nishati inayoweza kutokea na nishati ya kinetic3=na hufika bila nishati inayoweza kutokea na nishati ya kinetic4=1 and the chain is at the configuration \theta_1=\theta_2=0, \theta_3=\pi/2, \theta_4=-\pi/2lakini haifanyi yaliyomo katika hisabati kuwa tofauti1=lakini haifanyi yaliyomo katika hisabati kuwa tofauti2=0,lakini haifanyi yaliyomo katika hisabati kuwa tofauti3=Pi/2,lakini haifanyi yaliyomo katika hisabati kuwa tofauti4=−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)(lakini haifanyi yaliyomo katika hisabati kuwa tofauti1.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti3.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti4.). 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 “Run”:

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

1.112.223.334.44

\haki]⎣⎢⎢⎢⎡1.112.223.334.44⎦⎥⎥⎥⎤.

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

Use the function {\tt JacobianSpace}JacobianSpace in the given software to calculate the 6×3 space Jacobian J_sJs when \theta =(90^\circ, 90^\circ, 1)lakini haifanyi yaliyomo katika hisabati kuwa tofauti=(90∘,90∘,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 “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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

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

Use the function {\tt JacobianBody}JacobianBody in the given software to calculate the 6×3 body Jacobian J_bJb when \theta =(90^\circ, 90^\circ, 1)lakini haifanyi yaliyomo katika hisabati kuwa tofauti=(90∘,90∘,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 “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

\haki]⎣⎢⎡1.114.447.772.225.558.883.336.669.99⎦⎥⎤.

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 ni

J_b = \left[

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

\haki]Jb=⎣⎢⎢⎢⎢⎢⎢⎢⎡001−0.105−0.8890−10000.006−0.1050100.00600.889001−0.045−0.8440−10000.00600100.00600001000⎦⎥⎥⎥⎥⎥⎥⎥⎤

Extract the linear velocity portion J_vJAZ-500. (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_vJAZ-500.. 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

\haki]⎣⎢⎡1.112.223.33⎦⎥⎤.

1
[0,0,0]

Swali 7. 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.

Wiki 03: Roboti za kisasa, Kozi 2: Robot Kinematics Coursera Quiz Answers

Maswali 01: Lecture Comprehension, Inverse Kinematics of Open Chains (Sura 6 Utangulizi)

Q1. Consider the point (x,Y) = (0,2)(x,Y)=(0,2). Nini {\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? Chagua zote zinazotumika.

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.

Maswali 02: Lecture Comprehension, Numerical Inverse Kinematics (Sura 6.2, Sehemu 1 ya 2)

Q1. Let f(\theta)f(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) be a nonlinear function of \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti mapping an nn-dimensional space (the dimension of \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti) to an mm-dimensional space (the dimension of ff). We want to find a \theta_dlakini haifanyi yaliyomo katika hisabati kuwa tofautid., which may not be unique, that satisfies x_d = f(\theta_d)xd=f(lakini haifanyi yaliyomo katika hisabati kuwa tofautid.), i.e., x_d – f(\theta_d) = 0xd−f(lakini haifanyi yaliyomo katika hisabati kuwa tofautid.)=0. If our initial guess at a solution is \theta^0lakini haifanyi yaliyomo katika hisabati kuwa tofauti0, then a first-order Taylor expansion approximation of f(\theta)f(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) at \theta^0lakini haifanyi yaliyomo katika hisabati kuwa tofauti0 tells us

x_d \approx f(\theta^0) + J(\theta^0)(\theta_d – \theta^0)xd≈f(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0)+J(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0)(lakini haifanyi yaliyomo katika hisabati kuwa tofautid−lakini haifanyi yaliyomo katika hisabati kuwa tofauti0)

where J(\theta^0)J(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0) is the matrix of partial derivatives \partial f/\partial \theta∂f/∂lakini haifanyi yaliyomo katika hisabati kuwa tofauti evaluated at \theta^0lakini haifanyi yaliyomo katika hisabati kuwa tofauti0. Which of the following is a good next guess \theta^1lakini haifanyi yaliyomo katika hisabati kuwa tofauti1?

\theta^1 = \theta^0 + J^\dagger(\theta^0) (x_d – f(\theta^0))lakini haifanyi yaliyomo katika hisabati kuwa tofauti1=lakini haifanyi yaliyomo katika hisabati kuwa tofauti0+J†(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0)(xd−f(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0))
\theta^1 = \theta^0 – J^\dagger(\theta^0) (x_d – f(\theta^0))lakini haifanyi yaliyomo katika hisabati kuwa tofauti1=lakini haifanyi yaliyomo katika hisabati kuwa tofauti0-J†(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0)(xd−f(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0))
\theta^1 = J^{-1}(\theta^0) (x_d – f(\theta^0))lakini haifanyi yaliyomo katika hisabati kuwa tofauti1=J−1(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0)(xd−f(lakini haifanyi yaliyomo katika hisabati kuwa tofauti0))

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?

Hakuna.
Moja.
More than one.

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

A = \left[

142536

\haki]A=[142536]

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 maadili. 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, i.e., the dimension of bb is larger than the dimension of xx), then in general we would see there is no exact solution. Kwa kesi hii, 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[

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

\haki]?A=[12.],b=[34.]?

x^* = 2.2x∗=2.2
x^* = 1x∗=1

Maswali 03: Lecture Comprehension, Numerical Inverse Kinematics (Sura 6.2, Sehemu 2 ya 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)usambazaji wa uwezekano

Maswali 04: Sura 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

\haki]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

\haki]⎣⎢⎡1.112.223.33⎦⎥⎤.

1
[0,0]

Q2.

Roboti za kisasa, Kozi 2: Robot Kinematics Coursera Quiz Answers 2023 [💯% Jibu Sahihi] 7

Referring to the figure above, find the joint angles \theta_d = (\theta_1,\theta_2,\theta_3)lakini haifanyi yaliyomo katika hisabati kuwa tofautid=(lakini haifanyi yaliyomo katika hisabati kuwa tofauti1.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti3.) that put the 3R robot’s end-effector frame {b} katika

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

−0.5850.81100−0.811−0.5850000100.0762.60801

\haki]T(lakini haifanyi yaliyomo katika hisabati kuwa tofautid.)=Tsd=⎣⎢⎢⎢⎡−0.5850.81100−0.811−0.5850000100.0762.60801⎦⎥⎥⎥⎤

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 mita. 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)lakini haifanyi yaliyomo katika hisabati kuwa tofauti0=(Pi/4,Pi/4,Pi/4)=(0.7854,0.7854,0.7854), and tolerances \epsilon_\omega = 0.001ϵmwangaza=0.001 (0.057 digrii) and \epsilon_v = 0.0001ϵAZ-500=0.0001 (0.1 mm). Give \theta_dlakini haifanyi yaliyomo katika hisabati kuwa tofautid 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)lakini haifanyi yaliyomo katika hisabati kuwa tofauti0=(Pi/4,Pi/4,Pi/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 “Run”:

[1.11,2.22,3.33] for \left[

1.112.223.33

\haki]⎣⎢⎡1.112.223.33⎦⎥⎤.

1
[0,0,0]

wiki 04: Roboti za kisasa, Kozi 2: Robot Kinematics Coursera Quiz Answers

Maswali 01: Lecture Comprehension, Kinematics of Closed Chains (Sura 7)

Q1. Which of the following statements is true about closed-chain and parallel robots? Chagua zote zinazotumika.

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.

Maswali 02: Sura 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}lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙, and therefore the inverse Jacobian has nn safu (if there are nn actuators) na 6 nguzo (since a twist is 6-dimensional).

If the twist \mathcal{V}V consists of a 1 in the ii‘th element and zeros in all other elements, then what is the corresponding vector of actuated joint velocities \dot{\theta}lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙?

The ii‘th row of J^{-1}J−1.
The ii‘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}^2lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu∈R2 be the vector p expressed in fixed frame coordinates. Let a_i \in \mathbb{R}^2ai∈R2 be the vector a_ii expresed in fixed frame coordinates and b_i \in \mathbb{R}^2bi∈R2 be the vector b_ii expressed in the moving body frame coordinates. Define vector \text{d}_i = \text{lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R\text{b}_{i} – \text{a}_{i}di=p+Rbi−ai for i = 1, 2, 3i=1,2,3, wapi

R = \left[\begin{array}{cc}\cos\phi & -\sin\phi \\\sin\phi & \cos\phi \\\end {array}\haki].R=[cosϕsinϕ−sinϕcosϕ.]. fvdWEOCfEeeK2w4Lcly5FA e60f4d9e78317e908a2f1bfba0587606 3rrr 01

Derive a set of independent equations relating (\phi, lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu)(ϕ,lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu) na (\theta_1, \theta_2, \theta_3)(lakini haifanyi yaliyomo katika hisabati kuwa tofauti1.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.,lakini haifanyi yaliyomo katika hisabati kuwa tofauti3.). Which of the following is correct?

({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i})^2 = 2L^2(1 + \cos\theta_{i}), i = 1, 2, 3.(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)2=2na hufika bila nishati inayoweza kutokea na nishati ya kinetic2(1+cosθi.),i=1,2,3.
({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i})^\intercal({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i}) = 2L^2(1 – \sin\theta_{i}), i = 1, 2, 3.(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)⊺(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)=2na hufika bila nishati inayoweza kutokea na nishati ya kinetic2(1−sinθi.),i=1,2,3.
({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i})^\intercal({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i}) = 2L^2(1 – \cos\theta_{i}), i = 1, 2, 3.(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)⊺(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)=2na hufika bila nishati inayoweza kutokea na nishati ya kinetic2(1−cosθi.),i=1,2,3.
({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i})^\intercal({lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu} + R{b}_{i} - {a}_{i}) = 2L^2(1 + \cos\theta_{i}), i = 1, 2, 3.(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)⊺(lakini inaweza kuwa na faida ya kutosha kwamba helical itashinda juu ya fimbo ya muda mrefu+Rbi−ai.)=2na hufika bila nishati inayoweza kutokea na nishati ya kinetic2(1+cosθi.),i=1,2,3.

Mwandishi

Helen Bassey

Okta, I'm Helena, mwandishi wa blogu ambaye ana shauku ya kuchapisha yaliyomo ndani ya niche ya elimu. Ninaamini kuwa elimu ni ufunguo wa maendeleo binafsi na kijamii, na ninataka kushiriki ujuzi na uzoefu wangu na wanafunzi wa umri na asili zote. Kwenye blogi yangu, utapata makala juu ya mada kama vile mikakati ya kujifunza, elimu mtandaoni, mwongozo wa kazi, na zaidi. Pia ninakaribisha maoni na mapendekezo kutoka kwa wasomaji wangu, kwa hivyo jisikie huru kuacha maoni au kuwasiliana nami wakati wowote. Natumai utafurahiya kusoma blogi yangu na unaona kuwa ni muhimu na ya kutia moyo.
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