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Roboti za kisasa, Kozi 3: Maswali kuhusu Nguvu za Roboti & Majibu - Coursera

Join us for a fascinating exploration of robot dynamics ndani Kozi 3, where the complex interaction of forces and motion shapes roboti behaviour. Immerse yourself in our engaging quizzes and expert answers that shed light on the principles governing the dynamic behaviour of robotic mifumo. These quizzes serve as a gateway to understanding the complex mienendo that govern roboti motion from acceleration to torque and beyond.

Whether you are a robotiki enthusiast who wants to deepen your understanding or a student who wants to understand the complexity of robot dynamics, this collection provides valuable insight into fundamental aspects of roboti uhifadhi-wa-mitambo-nishati-pendulum. Join us on a journey of discovery as we unravel the mienendo ya roboti behaviour and unlock the potential for precise and efficient roboti shughuli.

Maswali 01: Lecture Comprehension, Lagrangian Formulation of Dynamics (Sura 8 kupitia 8.1.2, Sehemu 1 ya 2)

Q1. The Lagrangian for a mechanical system is

  • the kinetic energy plus the potential energy.
  • the kinetic energy minus the potential energy.

Q2. To evaluate the Lagrangian equations of motion,

\tau_i = \frac{d}{dt} \frac{\partial \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}}{\partial \dot{\theta}_i} – \frac{\partial \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}}{\partial \theta_i}lakini haifanyi yaliyomo katika hisabati kuwa tofautii​=dtd​∂lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙i​∂L​−∂lakini haifanyi yaliyomo katika hisabati kuwa tofautii​∂L​,

you must be able to take derivatives, such as the partial derivative with respect to a joint variable or velocity or a total derivative with respect to time. Kwa hiyo, the product rule and chain rules for derivatives will be useful. (If you have forgotten them, you can refresh your memory with any standard reference, including Wikipedia.) Which of the answers below represents the time derivative of 2 \theta_1 \cos(4 \theta_2)2lakini haifanyi yaliyomo katika hisabati kuwa tofauti1​cos(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.), where \theta_1lakini haifanyi yaliyomo katika hisabati kuwa tofauti1​ and \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti2​ are functions of time?

  • -8 \nukta{\theta}_1 \sin(4 \theta_2)−8lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​sin(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)
  • 2 \nukta{\theta}_1 \cos (4 \theta_2) - 8 \theta_1 \sin(4 \theta_2) \nukta{\theta}_22lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​cos(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)−8lakini haifanyi yaliyomo katika hisabati kuwa tofauti1​sin(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​
  • 2 \nukta{\theta}_1 \cos (4 \theta_2) - 2 \theta_1 \sin(4 \theta_2) \nukta{\theta}_22lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​cos(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)−2lakini haifanyi yaliyomo katika hisabati kuwa tofauti1​sin(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​
  • 2 \nukta{\theta}_1 \cos (4 \theta_2) + 2 \theta_1 \cos(4 \theta_2) \nukta{\theta}_22lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​cos(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)+2lakini haifanyi yaliyomo katika hisabati kuwa tofauti1​cos(4lakini haifanyi yaliyomo katika hisabati kuwa tofauti2.)lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​

Q3. The equations of motion for a robot can be summarized as

\tau = M(\theta) \ddot{\theta} + c(\theta,\nukta{\theta}) + g(\theta)lakini haifanyi yaliyomo katika hisabati kuwa tofauti=kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+c(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+g(lakini haifanyi yaliyomo katika hisabati kuwa tofauti).

If the equation for the first joint is

\tau_1 = term1 + term2 + term3 + …lakini haifanyi yaliyomo katika hisabati kuwa tofauti1​=telakini haifanyi yaliyomo katika hisabati kuwa tofautim1+telakini haifanyi yaliyomo katika hisabati kuwa tofautim2+telakini haifanyi yaliyomo katika hisabati kuwa tofautim3+…

Q4. which of the following terms, written in terms of (\theta,\nukta{\theta},\ddot{\theta})(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙,lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨), inaweza la be one of the terms on the right-hand side of the equation? (The value kk is not a function of (\theta,\nukta{\theta},\ddot{\theta})(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙,lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨) and could represent constants like link lengths, masses, or inertias, as needed to get correct units.) Chagua zote zinazotumika.

  • k\ddot{\theta}_2 \cos(\theta_1)¨2​cos(lakini haifanyi yaliyomo katika hisabati kuwa tofauti1.)
  • k\ddot{\theta}_1 \dot{\theta}_1¨1​lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​
  • k\sin \theta_3ksinlakini haifanyi yaliyomo katika hisabati kuwa tofauti3.
  • k\dot{\theta}_1 \dot{\theta}_2 \sin \theta_2˙1​lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2.
  • k \dot{\theta}_1 \sin \theta_2˙1​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2.

Maswali 02: Lecture Comprehension, Lagrangian Formulation of Dynamics (Sura 8 kupitia 8.1.2, Sehemu 2 ya 2)

Q1. Which of the following could be a centripetal term in the dynamics?

  • k \dot{\theta}_1^2˙12​
  • k \dot{\theta}_1 \dot{\theta}_2˙1​lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​

Q2. Which of the following could be a Coriolis term in the dynamics?

  • k \dot{\theta}_1^2˙12​
  • k \dot{\theta}_1 \dot{\theta}_2˙1​lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​

Q3. One form of the equations of motion is

\tau = M(\theta)\ddot{\theta} + \nukta{\theta}^{\rm T} \Gamma(\theta) \nukta{\theta} + g(\theta).lakini haifanyi yaliyomo katika hisabati kuwa tofauti=kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙TΓ(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙+g(lakini haifanyi yaliyomo katika hisabati kuwa tofauti).

Which of the following is true about \Gamma(\theta)Γ(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)? Chagua zote zinazotumika.

  • \Gamma(\theta)Γ(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) is zero if the mass matrix Mkwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu has no dependence on \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti.
  • \Gamma(\theta)Γ(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) is an n \times nn×n tumbo, where nn is the number of joints.
  • \Gamma(\theta)Γ(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) depends on M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) and \dot{\theta}lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙.

Lecture Comprehension, Understanding the Mass Matrix (Sura 8.1.3)Maswali 03:

Q1. Which of these is a possible mass matrix M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) for a two-joint robot at a particular configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti? Chagua zote zinazotumika.

  • \Ni rahisi sana kutambua miingiliano ya x na y kwenye grafu[
  • 200−1
  • \haki][20​0−1​]
  • \Ni rahisi sana kutambua miingiliano ya x na y kwenye grafu[
  • 4123
  • \haki][41​23​]
  • \Ni rahisi sana kutambua miingiliano ya x na y kwenye grafu[
  • 3112
  • \haki][31​12​]
  • \Ni rahisi sana kutambua miingiliano ya x na y kwenye grafu[
  • 2221
  • \haki][22​21​]

Q2. True or false? If you grab the end-effector of a robot and try to move it around by hand, the apparent mass (kwako) depends on the configuration of the robot.

  • Kweli.
  • Uongo.

Q3. True or false? Ukiomba (by hand) a linear force to the end-effector of a robot, the end-effector will accelerate in the same direction as the applied force.

  • Always true.
  • Always false.
  • Sometimes true, sometimes false.

Maswali 04: Lecture Comprehension, Dynamics of a Single Rigid Body (Sura 8.2, Sehemu 1 ya 2)

Q1. How is the center of mass of a rigid body defined?

  • The point at the geometric centroid of the body.
  • The point at the mass-weighted (or density-weighted) centroid of the body.

Q2. If the body consists of a set of rigidly connected point masses, with a frame {b} at the center of mass, what is the wrench \mathcal{F}_bFb​ needed to generate the acceleration \dot{{\mathcal V}}_bV˙b​ when the body’s current twist is \mathcal{V}_bVb.?

  • The acceleration \dot{{\mathcal V}}_bV˙b​ defines the linear acceleration of each point mass in the inertial frame {b}. The linear component f_bfb​ of \mathcal{F}_bFb​ is the sum of the individual vector forces needed to cause those point-mass accelerations (using f=maf=ma), and the moment m_bmb​ is the sum of the moments the individual linear forces create in {b}.
  • Pamoja, \mathcal{V}_bVb​ and \dot{{\mathcal V}}_bV˙b​ define the linear acceleration of each point mass in the inertial frame {b}. The linear component f_bfb​ of \mathcal{F}_bFb​ is the sum of the individual vector forces needed to cause those point-mass accelerations (using f=maf=ma), and the moment m_bmb​ is the sum of the moments the individual linear forces create in {b}.

Q3. What is the kinetic energy of a rotating rigid body?

  • \frac{1}{2} \omega_b^{\rm T} \mathcal{Mimi}_b \omega_b21​ωbT​Ib.ωb.
  • \frac{1}{2} \mathcal{Mimi}_b \omega_b^221​Ib.ωb2.

Q4. True or false? For a given body, there is exactly one orientation of a frame at the center of mass that yields a diagonal rotational inertia matrix.

  • Kweli.
  • Uongo.

Maswali 05: Lecture Comprehension, Dynamics of a Single Rigid Body (Sura 8.2, Sehemu 2 ya 2)

Q1. The spatial inertia matrix is the 6×6 matrix

\mathcal{G}_B = \left[

Mimib00mMimi

\haki]GB​=[Mimib​0​0mMimi.].

What is the maximum number nn ya kipekee nonzero values the spatial inertia could have? Kwa maneno mengine, even though the 6×6 matrix has 36 entries, we only need to store nn numbers to represent the spatial inertia matrix.

  • 6
  • 7
  • 10
  • 12

Q2. Usemi huo [\omega_1][\omega_2]-[\omega_2][\omega_1[mwangaza1.][mwangaza2.]-[mwangaza2.][mwangaza1​ is the so(3)KOZI(3) 3×3 skew-symmetric matrix representation of the cross product of two angular velocities, \omega_1 \times \omega_2 = [\omega_1]\omega_2 \in \mathbb{R}^3mwangaza1​×mwangaza2​=[mwangaza1.]mwangaza2​∈R3. An analogous expression for twists is [\mathcal{V}_1][\mathcal{V}_2]-[\mathcal{V}_2][\mathcal{V}_1][V1​][V2​]-[V2​][V1​], the 4×4 se(3)se(3) representation of the Lie bracket of \mathcal{V}_1V1​ and \mathcal{V}_2V2​, sometimes written [{\rm ad}_{\mathcal{V}_1}] \mathcal{V}_2 \in \mathbb{R}^6[adV1​​]V2​∈R6. Which of the following statements is true? Chagua zote zinazotumika.

  • The matrix [{\rm ad}_{\mathcal{V}_1}][adV1​​] is an element of se(3)se(3).
  • [{\rm ad}_{\mathcal{V}_1}] \mathcal{V}_2 = -[{\rm ad}_{\mathcal{V}_2}] \mathcal{V}_1[adV1​​]V2​=−[adV2​​]V1​

Q3. The dynamics of a rigid body, in a frame at the center of mass {b}, can be written

\mathcal{F}_b = \mathcal{G}_b \dot{\mathcal{V}}_b – [{\rm ad}_{\mathcal{V}_b}]^{\rm T} \mathcal{G}_b \mathcal{V}_bFb​=Gb​V˙b​−[adVb​​]TGb​Vb..

If \mathcal{V}_b = (\omega_b,v_b) = (0,v_b)Vb​=(ωb.,vb.)=(0,vb.) and \dot{\mathcal{V}}_b = (\nukta{\omega}_b, \nukta{AZ-500}_b) = (0,0)b​=(mwangaza˙b.,AZ-500˙b.)=(0,0), which of the following is true?

  • \mathcal{F}_bFb​ is zero.
  • \mathcal{F}_bFb​ is nonzero.
  • Either of the above could be true.

Q4. The dynamics of a rigid body, in a frame at the center of mass {b}, can be written

\mathcal{F}_b = \mathcal{G}_b \dot{\mathcal{V}}_b – [{\rm ad}_{\mathcal{V}_b}]^{\rm T} \mathcal{G}_b \mathcal{V}_bFb​=Gb​V˙b​−[adVb​​]TGb​Vb..

If \mathcal{V}_b = (\omega_b,v_b) = (\omega_b,0)Vb​=(ωb.,vb.)=(ωb.,0) and \dot{\mathcal{V}}_b = (\nukta{\omega}_b, \nukta{AZ-500}_b) = (0,0)b​=(mwangaza˙b.,AZ-500˙b.)=(0,0), which of the following is true?

  • \mathcal{F}_bFb​ is zero.
  • \mathcal{F}_bFb​ is nonzero.
  • Either of the above could be true.

Maswali 06: Sura 8 kupitia 8.3, Dynamics of Open Chains

Q1. Consider an iron dumbbell consisting of a cylinder connecting two solid spheres at either end of the cylinder. The density of the dumbbell is 5600 kg/m^33. The cylinder has a diameter of 4 cm and a length of 20 sentimita. Each sphere has a diameter of 20 sentimita. Find the approximate rotational inertia matrix \mathcal{Mimi}_bIb​ in a frame {b} at the center of mass with z-axis aligned with the length of the dumbbell. Your entries should be written in units of kg-m^2, and the maximum allowable error for any matrix entry 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]⎣⎢⎡​1.114.447.77​2.225.558.88​3.336.669.99​⎦⎥⎤​.

  • 1
  • [[0,0,0],[0,0,0],[0,0,0]]

Q2. The equations of motion for a particular 2R robot arm can be written M(\theta)\ddot{\theta} + c(\theta,\nukta{\theta}) + g(\theta) = \taukwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+c(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+g(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)=lakini haifanyi yaliyomo katika hisabati kuwa tofauti. The Lagrangian \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}(\theta,\nukta{\theta})na hufika bila nishati inayoweza kutokea na nishati ya kinetic(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙) for the robot can be written in components as

\mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}(\theta,\nukta{\theta}) = \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}^1(\theta,\nukta{\theta}) + \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}^2 (\theta,\nukta{\theta}) + \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}^3 (\theta,\nukta{\theta}) +\ldotsL(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)=L1(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+L2(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+L3(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+…

One of these components is \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}^1 = \mathfrak{m} \nukta{\theta}_1 \dot{\theta}_2 \sin\theta_2 L1=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2..

Find the right expression for the component of the joint torque \tau^1_1lakini haifanyi yaliyomo katika hisabati kuwa tofauti11​ at joint 1 corresponding to the component \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}^1L1.

  • \tau^1_1 = \mathfrak{m} \ddot{\theta_2} \sin \theta_2 – \mathfrak{m} \dot \theta_2^2 \cos \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti11​=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​¨​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​−mlakini haifanyi yaliyomo katika hisabati kuwa tofauti˙22​coslakini haifanyi yaliyomo katika hisabati kuwa tofauti2.
  • \tau^1_1 = \mathfrak{m} \ddot{\theta_2} \sin \theta_2 + \mathfrak{m} \dot \theta_2^2 \cos \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti11​=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​¨​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​+mlakini haifanyi yaliyomo katika hisabati kuwa tofauti˙22​coslakini haifanyi yaliyomo katika hisabati kuwa tofauti2.
  • \tau^1_1 = \mathfrak{m} \ddot{\theta_2} \cos \theta_2 + \mathfrak{m} \dot \theta_2^2 \sin \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti11​=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​¨​coslakini haifanyi yaliyomo katika hisabati kuwa tofauti2​+mlakini haifanyi yaliyomo katika hisabati kuwa tofauti˙22​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2.

Q3. Referring back to Question 2, find the right expression for the component of joint torque \tau^1_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti21​ at joint 2 corresponding to the component \mathcal{na hufika bila nishati inayoweza kutokea na nishati ya kinetic}^1L1.

  • \tau^1_2 = \mathfrak{m} \ddot{\theta_2} \sin \theta_2 + \mathfrak{m} \dot \theta_2^2 \cos \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti21​=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​¨​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​+mlakini haifanyi yaliyomo katika hisabati kuwa tofauti˙22​coslakini haifanyi yaliyomo katika hisabati kuwa tofauti2.
  • \tau^1_2 = \mathfrak{m} \ddot{\theta_1} \sin \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti21​=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti1​¨​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2.
  • \tau^1_2 = \mathfrak{m} \ddot{\theta_1} \sin \theta_2 + \mathfrak{m} \dot \theta_1 \dot \theta_2 \cos \theta_2lakini haifanyi yaliyomo katika hisabati kuwa tofauti21​=mlakini haifanyi yaliyomo katika hisabati kuwa tofauti1​¨​sinlakini haifanyi yaliyomo katika hisabati kuwa tofauti2​+mlakini haifanyi yaliyomo katika hisabati kuwa tofauti˙1​lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙2​coslakini haifanyi yaliyomo katika hisabati kuwa tofauti2.

Q4. For a given configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti of a two-joint robot, the mass matrix is

kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(\theta) = \left[

3ba12

\haki],kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)=[3b.a12.],

which has a determinant of 36-ab36−ab and eigenvalues \frac{1}{2} (15 \pm \sqrt{81 + 4 a b})21.(15±81+4ab.). What constraints must aa and bb satisfy for this to be a valid mass matrix? Chagua zote zinazotumika.

  • a < 6a<6
  • b > 6b>6
  • a > ba>b
  • a = ba=b
  • a<ba<b
  • a < \sqrt 6a<6.

Q5. An inexact model of the UR5 mass and kinematic properties is given below:

M_{01} = \left[

100001000010000.0891591

\haki], M_{12} = \left[

00−10010010000.280.1358501

\haki], M_{23} = \left[

1000010000100−0.11970.3951

\haki], kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu01​=⎣⎢⎢⎢⎡​1000​0100​0010​000.0891591​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu12​=⎣⎢⎢⎢⎡​00−10​0100​1000​0.280.1358501​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu23​=⎣⎢⎢⎢⎡​1000​0100​0010​0−0.11970.3951​⎦⎥⎥⎥⎤​,

M_{34} = \left[

00−1001001000000.142251

\haki], M_{45} = \left[

10000100001000.09301

\haki], M_{56} = \left[

100001000010000.094651

\haki], kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu34​=⎣⎢⎢⎢⎡​00−10​0100​1000​000.142251​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu45​=⎣⎢⎢⎢⎡​1000​0100​0010​00.09301​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu56​=⎣⎢⎢⎢⎡​1000​0100​0010​000.094651​⎦⎥⎥⎥⎤​,

M_{67} = \left[

100000−10010000.082301

\haki],kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu67​=⎣⎢⎢⎢⎡​1000​00−10​0100​00.082301​⎦⎥⎥⎥⎤​,

G_1 = {\tt diag}([0.010267495893,0.010267495893, 0.00666,3.7,3.7,3.7]),G1​=diag([0.010267495893,0.010267495893,0.00666,3.7,3.7,3.7]),

G_2 = {\tt diag}([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),G2​=diag([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),

G_3 = {\tt diag}([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),G3​=diag([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),

G_4 = {\tt diag} ([0.111172755531 ,0.111172755531 ,0.21942, 1.219, 1.219 ,1.219]),G4​=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_5 = {\tt diag} ([0.111172755531 ,0.111172755531, 0.21942 ,1.219 ,1.219 ,1.219]),G5​=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_6 = {\tt diag} ([0.0171364731454 ,0.0171364731454, 0.033822 ,0.1879 ,0.1879, 0.1879]),G6​=diag([0.0171364731454,0.0171364731454,0.033822,0.1879,0.1879,0.1879]),

{\tt Slist} = \left[

001000010−0.08915900010−0.08915900.425010−0.08915900.8172500−1−0.109150.8172500100.00549100.81725

\haki].Slist=⎣⎢⎢⎢⎢⎢⎢⎢⎡​001000​010−0.08915900​010−0.08915900.425​010−0.08915900.81725​00−1−0.109150.817250​0100.00549100.81725​⎦⎥⎥⎥⎥⎥⎥⎥⎤​.

Here are three versions for these UR5 parameters:

UR5_parameter

PY File

UR5_parameters

M File

UR5_parameters

NB File

Given

lakini haifanyi yaliyomo katika hisabati kuwa tofauti=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢0Pi/6Pi/4Pi/3Pi/22Pi/3⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.20.20.20.20.20.2⎤⎦⎥⎥⎥⎥⎥⎥⎥,lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.10.10.10.10.10.1⎤⎦⎥⎥⎥⎥⎥⎥⎥,g=⎡⎣00−9.81⎤⎦,Ftip=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.10.10.10.10.10.1⎤⎦⎥⎥⎥⎥⎥⎥⎥,

use the function {\tt InverseDynamics}InverseDynamics in the given software to calculate the required joint forces/torques of the robot. 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] for \left[

1.112.223.33

\haki]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

1

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

RunReset

Wiki 02: Roboti za kisasa, Kozi 3: Robot Dynamics Coursera Quiz Answers

Maswali 01: Lecture Comprehension, Forward Dynamics of Open Chains (Sura 8.5)

Q1. To derive the mass matrix M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) of an nn-joint open-chain robot, how many times would we need to invoke the recursive Newton-Euler inverse dynamics algorithm?

  • 1 wakati.
  • nn nyakati.
  • It is not possible to derive the mass matrix from the Newton-Euler inverse dynamics.

Q2. When calculating the mass matrix M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) using the Newton-Euler inverse dynamics, which of these quantities must be set to zero? Chagua zote zinazotumika.

  • \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti
  • \nukta{\theta}lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙
  • \ddot{\theta}lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨
  • The gravitational constant.
  • The end-effector wrench \mathcal{F}_{{\rm tip}}Ftip​.

Maswali 02: Lecture Comprehension, Dynamics in the Task Space (Sura 8.6)

Q1. Converting the joint-space dynamics to the task-space dynamics requires an invertible Jacobian, as well as the relationships \mathcal{V} = J(\theta)\nukta{\theta}V=J(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙ and \dot{\mathcal{V}} = J\ddot{\theta} + \nukta{J}\nukta{\theta}V˙=Jlakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+J˙lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙, to find \Lambda(\theta)mwangaza(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) and \eta(\theta,\mathcal{V})η(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,V) in \mathcal{F} = \Lambda(\theta) \nukta{\mathcal{V}} + \eta(\theta,\mathcal{V})F=Λ(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)V˙+η(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,V).

Why do you suppose we left the dependence on \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti, instead of writing it as a dependence on the end-effector configuration T \in SE(3)TSE(3), which would seem to be more aligned with our task-space view?

  • Either TT or \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti could be used; there is no reason to prefer one to the other.
  • The inverse kinematics of an open-chain robot does not necessarily have a unique solution, so we may not know the robot’s full configuration, and therefore the mass properties, given just TT.

Maswali 03: Lecture Comprehension, Constrained Dynamics (Sura 8.7)

Q1. A serial-chain robot has nn links and actuated joints, but it is subject to kk independent Pfaffian velocity constraints of the form A(\theta)\nukta{\theta}=0A(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙=0. These constraints partition the nn-dimensional \taulakini haifanyi yaliyomo katika hisabati kuwa tofauti space into orthogonal subspaces: a space of forces CC that do not create any forces against the constraints, and a space of forces BB that do not cause any motion of the robot. What is the dimension of each of these spaces?

  • CC ni (n-k)(n-k)-dimensional and BB is kk-ya dimensional.
  • CC is nn-dimensional and BB is kk-ya dimensional.
  • CC is kk-dimensional and BB ni (n-k)(n-k)-ya dimensional.
  • CC is kk-dimensional and BB is nn-ya dimensional.

Q2. Let the constrained dynamics of a robot be \tau = M(\theta)\ddot{\theta} + na kuifikisha pale inapohitajika(\theta,\nukta{\theta}) + A^{\rm T}(\theta)\lambdalakini haifanyi yaliyomo katika hisabati kuwa tofauti=kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+na kuifikisha pale inapohitajika(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+AT(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)mwangaza, where \lambda \in \mathbb{R}^kmwangaza∈Rk. Let P(\theta)P(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) be the matrix, as discussed in the video, that projects an arbitrary \tau \in \mathbb{R}^nlakini haifanyi yaliyomo katika hisabati kuwa tofauti∈Rn to P(\theta)\tau \in CP(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofautiC, where the space CC is the same CC from the previous question. Then what is P(\theta) \tauP(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti? Chagua zote zinazotumika.

  • kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(\theta)\ddot{\theta} + na kuifikisha pale inapohitajika(\theta,\nukta{\theta}) + A^{\rm T}(\theta)\lambdakwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+na kuifikisha pale inapohitajika(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙)+AT(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)mwangaza
  • P(\theta)(kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(\theta)\ddot{\theta} + na kuifikisha pale inapohitajika(\theta,\nukta{\theta}))P(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)(kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+na kuifikisha pale inapohitajika(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙))
  • P(\theta)(kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(\theta)\ddot{\theta} + na kuifikisha pale inapohitajika(\theta,\nukta{\theta})) +A^{\rm T}(\theta)\lambda)P(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)(kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨+na kuifikisha pale inapohitajika(lakini haifanyi yaliyomo katika hisabati kuwa tofauti,lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙))+AT(lakini haifanyi yaliyomo katika hisabati kuwa tofauti)mwangaza)

Maswali 04: Lecture Comprehension, Actuation, Gearing, and Friction (Sura 8.9)

Q1. What is the typical reason for putting a gearhead on a motor for use in a robot?

  • To increase torque (simultaneously reducing the maximum speed).
  • To increase speed (simultaneously reducing the maximum torque).

Q2. Compared to a “direct drive” robot that is driven by motors without gearheads (G=1G=1), increasing the gear ratios has what effect on the robot’s dynamics? Chagua zote zinazotumika.

  • The mass matrix M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) is increasingly dominated by the apparent inertias of the motors.
  • The mass matrix M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) is increasingly dominated by off-diagonal terms.
  • The mass matrix M(\theta)kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu(lakini haifanyi yaliyomo katika hisabati kuwa tofauti) is increasingly dominated by constant terms that do not depend on the configuration \thetalakini haifanyi yaliyomo katika hisabati kuwa tofauti.
  • The robot is capable of higher speeds but lower accelerations.
  • The significance of velocity-product (Coriolis and centripetal) terms diminishes.

Maswali 05: Sura 8.5-8.7 na 8.9, Dynamics of Open Chains

Q1. A robot system (UR5) hufafanuliwa kama

M_{01} = \left[

100001000010000.0891591

\haki], M_{12} = \left[

00−10010010000.280.1358501

\haki], M_{23} = \left[

1000010000100−0.11970.3951

\haki], kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu01​=⎣⎢⎢⎢⎡​1000​0100​0010​000.0891591​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu12​=⎣⎢⎢⎢⎡​00−10​0100​1000​0.280.1358501​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu23​=⎣⎢⎢⎢⎡​1000​0100​0010​0−0.11970.3951​⎦⎥⎥⎥⎤​,

M_{34} = \left[

00−1001001000000.142251

\haki], M_{45} = \left[

10000100001000.09301

\haki], M_{56} = \left[

100001000010000.094651

\haki], kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu34​=⎣⎢⎢⎢⎡​00−10​0100​1000​000.142251​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu45​=⎣⎢⎢⎢⎡​1000​0100​0010​00.09301​⎦⎥⎥⎥⎤​,kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu56​=⎣⎢⎢⎢⎡​1000​0100​0010​000.094651​⎦⎥⎥⎥⎤​,

M_{67} = \left[

100000−10010000.082301

\haki],kwa sababu itakuwa na kasi kubwa zaidi na kuwa karibu na Dunia kwa wakati huu67​=⎣⎢⎢⎢⎡​1000​00−10​0100​00.082301​⎦⎥⎥⎥⎤​,

G_1 = {\tt diag}([0.010267495893,0.010267495893, 0.00666,3.7,3.7,3.7]),G1​=diag([0.010267495893,0.010267495893,0.00666,3.7,3.7,3.7]),

G_2 = {\tt diag}([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),G2​=diag([0.22689067591,0.22689067591,0.0151074,8.393,8.393,8.393]),

G_3 = {\tt diag}([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),G3​=diag([0.049443313556,0.049443313556,0.004095,2.275,2.275,2.275]),

G_4 = {\tt diag} ([0.111172755531 ,0.111172755531 ,0.21942, 1.219, 1.219 ,1.219]),G4​=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_5 = {\tt diag} ([0.111172755531 ,0.111172755531, 0.21942 ,1.219 ,1.219 ,1.219]),G5​=diag([0.111172755531,0.111172755531,0.21942,1.219,1.219,1.219]),

G_6 = {\tt diag} ([0.0171364731454 ,0.0171364731454, 0.033822 ,0.1879 ,0.1879, 0.1879]),G6​=diag([0.0171364731454,0.0171364731454,0.033822,0.1879,0.1879,0.1879]),

{\tt Slist} = \left[

001000010−0.08915900010−0.08915900.425010−0.08915900.8172500−1−0.109150.8172500100.00549100.81725

\haki].Slist=⎣⎢⎢⎢⎢⎢⎢⎢⎡​001000​010−0.08915900​010−0.08915900.425​010−0.08915900.81725​00−1−0.109150.817250​0100.00549100.81725​⎦⎥⎥⎥⎥⎥⎥⎥⎤​.

Here are three versions for these UR5 parameters above:

UR5_parameter

PY File

UR5_parameters

M File

UR5_parameters

NB File

Given

\theta = \left[

0Pi/6Pi/4Pi/3Pi/22Pi/3

\haki]lakini haifanyi yaliyomo katika hisabati kuwa tofauti=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0Pi/6Pi/4Pi/3Pi/22Pi/3​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

\dot \theta = \left[

0.20.20.20.20.20.2

\haki]lakini haifanyi yaliyomo katika hisabati kuwa tofauti˙=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0.20.20.20.20.20.2​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

\ddot{\theta} = \left[

0.10.10.10.10.10.1

\haki]lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0.10.10.10.10.10.1​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

\mathfrak{g} = \left[

00−9.81

\haki]g=⎣⎢⎡​00−9.81​⎦⎥⎤​,

\mathcal{F}_{\maandishi{tip}} = \left[

0.10.10.10.10.10.1

\haki]Ftip​=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0.10.10.10.10.10.1​⎦⎥⎥⎥⎥⎥⎥⎥⎤​

use the function {\tt MassMatrix}MassMatrix in the given software to calculate the numerical inertia matrix of the robot. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a matrix in the answer box:

[[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]⎣⎢⎡​1.114.447.77​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]]

Q2. Referring back to Question 1, for the same robot system and condition, use the function {\tt VelQuadraticForces}VelQuadraticForces in the given software to calculate the Coriolis and centripetal terms in the robot’s dynamics. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a vector in the answer box:

[1.11,2.22,3.33] for \left[

1.112.223.33

\haki]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

  • 1
  • [0,0,0,0,0,0]

Q3. Referring back to Question 1, for the same robot system and condition, use the function {\tt GravityForces}GravityForces in the given software to calculate the joint forces/torques required to overcome gravity. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a vector in the answer box:

[1.11,2.22,3.33] for \left[

1.112.223.33

\haki]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

  • 1
  • [0,0,0,0,0,0]

Q4., Referring back to Question 1, for the same robot system and condition, use the function {\tt EndEffectorForces}EndEffectorForces in the given software to calculate the joint forces/torques required to generate the wrench \mathcal{F}_{{\rm tip}}Ftip​. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a vector in the answer box:

[1.11,2.22,3.33] for \left[

1.112.223.33

\haki]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

  • 1
  • [0,0,0,0,0,0]

Q5. Referring back to Question 1, for the same robot system and condition plus the known joint forces/torques

\tau = \left[

0.0128−41.1477−3.78090.03230.03700.1034

\haki] lakini haifanyi yaliyomo katika hisabati kuwa tofauti=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0.0128−41.1477−3.78090.03230.03700.1034​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

use the function {\tt ForwardDynamics}ForwardDynamics in the given software to find the joint acceleration. The maximum allowable error for any number is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a vector in the answer box:

[1.11,2.22,3.33] for \left[

1.112.223.33

\haki]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

  • 1
  • [0,0,0,0,0,0]

Q6. Assume that the inertia of a revolute motor’s rotor about its central axis is 0.005 kg m^22. The motor is attached to a zero-inertia 200:1 gearhead. If you grab the gearhead output and spin it by hand, what is the inertia you feel?

  • 200 kg m^22
  • 1 kg m^22
  • 0.005 kg m^22

Wiki 03: Roboti za kisasa, Kozi 3: Robot Dynamics Coursera Quiz Answers

Maswali 03: Lecture Comprehension, Point-to-Point Trajectories (Sura 9 kupitia 9.2, Sehemu 1 ya 2)

Q1. A point robot moving in a plane has a configuration represented by (x,Y)(x,Y). The path of the robot in the plane is (1+ 2\cos (\pi s), 2\sin(\pi s)), \; s \in [0,1](1+2cos(πs),2sin(πs)),s[0,1]. What does the path look like?

  • An ellipse.
  • A sine wave.
  • A semi-circle.
  • A circle.

Q2. Referring back to Question 1, assume the time-scaling of the motion along the path is s = 2t, \; t \in [0, 1/2]s=2t,t[0,1/2]. At time tt, wapi 0 \leq t \leq 0.50≤t≤0.5, what is the velocity of the robot (\nukta{x},\nukta{Y})(x˙,Y˙​)?

  • (-4 \pi \sin(2 \pi t), 4 \pi \cos(2 \pi t))(−4Pisin(2πt),4Picos(2πt))
  • (-2\sin(2 \pi t), 2\cos(2 \pi t))(−2sin(2πt),2cos(2πt))

Q3. True or false? For a trajectory \theta(s(t))lakini haifanyi yaliyomo katika hisabati kuwa tofauti(s(t)), the acceleration \ddot{\theta}lakini haifanyi yaliyomo katika hisabati kuwa tofauti¨ is \frac{d\theta}{ds}\ddot{s}dsdθ.s¨.

  • Kweli
  • Uongo

Q4. Let \mathcal{V}_sVs​ be the spatial twist that takes X_{s,{\rm start}}Xs,start​ to X_{s,{\rm end}}Xs,end​ in unit time. Which is an expression for the constant screw path that takes X_{s,{\rm start}}Xs,start​ (at s=0s=0) to X_{s,{\rm end}}Xs,end​ (at s=1s=1)?

  • \exp([\mathcal{V}_s s]) X_{s,{\rm start}}, \; s \in [0,1]exp([Vs.s])Xs,start​,s[0,1]
  • X_{s,{\rm start}}\exp([\mathcal{V}_s s]), \; s \in [0,1]Xs,start​exp([Vs.s]),s[0,1]

Maswali 02: Lecture Comprehension, Point-to-Point Trajectories (Sura 9 kupitia 9.2, Sehemu 2 ya 2)

Q1. For a fifth-order polynomial time scaling s(t)s(t), t \in [0,T]t[0,T], what is the form of \ddot{s}(t)s¨(t)?

  • Third-order polynomial
  • Fourth-order polynomial
  • Fifth-order polynomial

Maswali 03: Lecture Comprehension, Polynomial Via Point Trajectories (Sura 9.3)

Q1. True or false? Third-order polynomial interpolation between via points ensures that the path remains inside the convex hull of the via points.

  • Kweli
  • Uongo

Q2. A robot has 3 joints and it follows a motion interpolating 6 pointi: a start point, an end point, na 4 other via points. The interpolation is by cubic polynomials. How many total coefficients are there to describe the motion of the 3-DOF robot over the motion consisting of 5 sehemu?

  • 60
  • 30

Q3. Referring again to Question 2, imagine we constrain the position and velocity of each DOF at the beginning and end of the trajectory, and at each of the 4 intermediate via points, we constrain the position (so the robot passes through the via points) but only constrain the velocity and acceleration to be continuous at each via point. Then how many total constraints are there on the coefficients describing the joint motions for all motion segments?

  • 60
  • 30

Maswali 04: Sura 9 kupitia 9.3, Trajectory Generation

Q1. Consider the elliptical path in the (x,Y)(x,Y)-plane shown below. The path starts at (0,0)(0,0) and proceeds clockwise to (1.5,1)(1.5,1), (3,0)(3,0), (1.5,-1)(1.5,−1), and back to (0,0)(0,0). Choose the appropriate function of s \in [0,1]s[0,1] to represent the path.

KmuEUwnlEeiIOArs7r1YtA 37e2876750ce29ec1ffb0a4889fc703e week3 elliptical path
  • ni sehemu ambazo grafu ya chaguo za kukokotoa au mlinganyo huvuka au "kugusa" mhimili y wa Ndege ya Cartesian. 3 (1 – \cos 2 \pi s)x=3(1−cos2πs)
  • y = \sin 2 \pi sY=sin2πs
  • ni sehemu ambazo grafu ya chaguo za kukokotoa au mlinganyo huvuka au "kugusa" mhimili y wa Ndege ya Cartesian. 1.5 (1 – \cos 2 \pi s)x=1.5(1−cos2πs)
  • y = \sin 2 \pi sY=sin2πs
  • ni sehemu ambazo grafu ya chaguo za kukokotoa au mlinganyo huvuka au "kugusa" mhimili y wa Ndege ya Cartesian. 1.5 (1 – \cos s)x=1.5(1−coss)
  • y = \sin sY=sins
  • x = \cos 2 \pi sx=cos2πs

Ni rahisi sana kutambua miingiliano ya x na y kwenye grafu 1.5 (1 – \sin 2 \pi s)Y=1.5(1−sin2πs

Q2. Find the fifth-order polynomial time scaling that satisfies s(T) = 1s(T)=1 and s(0) = \dot{s}(0) = \ddot{s}(0) = \dot{s}(T) = \ddot{s}(T) = 0s(0)=s˙(0)=s¨(0)=s˙(T)=s¨(T)=0.

Your answer should be only a mathematical expression, a polynomial in tt, with coefficients involving TT. (Don’t bother to write “s(t) = s(t)=”, just give the right-hand side.

Q3. If you want to use a polynomial time scaling for point-to-point motion with zero initial and final velocity, hapa kuna lebo za kitamaduni za tabia mbali mbali za mwili, and jerk, what would be the minimum order of the polynomial

Q4. Choose the correct acceleration profile \ddot{s}(t)s¨(t) for an S-curve time scaling.

  • A98PK t7 EeeK2w4Lcly5FA 36dd58791fdeb9e5078c77589aef5616 ex05 1 01
  • BS eOht8AEeeK2w4Lcly5FA eab5dfe0dfd05a5346dd0e1f3b9a0b30 ex05 2 01
  • CQnjatN8AEeeK2w4Lcly5FA c72cb6462b524c6c55c870e1f930740e ex05 3 01
  • DM388Jd8AEeeVFRLN7DX 0g fdec568a0c601a3f8292087790a225b2 ex05 4 01

Q5. Given a total travel time T = 5T=5 and the current time t = 3t=3, use the function {\tt QuinticTimeScaling}QuinticTimeScaling in the given software to calculate the current path parameter ss, with at least 2 decimal places, corresponding to a motion that begins and ends at zero velocity and acceleration

Q6. Use the function {\tt ScrewTrajectory}ScrewTrajectory in the given software to calculate a trajectory as a list of N=10N=10 SE(3)SE(3) matrices, where each matrix represents the configuration of the end-effector at an instant in time. The first matrix is

X_{{\rm start}} = \left[

1000010000100001

\haki]usambazaji wa uwezekanostart​=⎣⎢⎢⎢⎡​1000​0100​0010​0001​⎦⎥⎥⎥⎤​

and the 10th matrix is

X_{{\rm end}} = \left[

0100001010001231

\haki].usambazaji wa uwezekanoend​=⎣⎢⎢⎢⎡​0100​0010​1000​1231​⎦⎥⎥⎥⎤​.

The motion is along a constant screw axis and the duration is T_f = 10Tf​=10. The parameter {\tt method}method equals 3 for a cubic time scaling. Give the 9th matrix (one before X_{{\rm end}}usambazaji wa uwezekanoend​) in the returned trajectory. The maximum allowable error for any matrix entry is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a matrix in the answer box:

[[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]⎣⎢⎡​1.114.447.77​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]]

Swali 7. Referring back to Question 6, use the function {\tt CartesianTrajectory}CartesianTrajectory in the MR library to calculate another trajectory as a list of N=10N=10 SE(3)SE(3) matrices. Besides the same X_{{\rm start}}usambazaji wa uwezekanostart​, X_{{\rm end}}usambazaji wa uwezekanoend​, T_fTf​ and N = 10N=10, we now set {\tt method}method to 5 for a quintic time scaling. Give the 9th matrix (one before X_{{\rm end}}usambazaji wa uwezekanoend​) in the returned trajectory. The maximum allowable error for any matrix entry is 0.01, so give enough decimal places where necessary.

Use Python syntax to express a matrix in the answer box:

[[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]⎣⎢⎡​1.114.447.77​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]]

RunReset

Wiki 04: Roboti za kisasa, Kozi 3: Robot Dynamics Coursera Quiz Answers

Maswali 01: Lecture Comprehension, Time-Optimal Time Scaling (Sura 9.4, Sehemu 1 ya 3)

Q1. When a robot travels along a specified path \theta(s)lakini haifanyi yaliyomo katika hisabati kuwa tofauti(s), torque or force limits at each actuator place bounds on the path acceleration \ddot{s}s¨. The constraints due to actuator ii can be written

\tau_i^{{\rm min}}(s,\nukta{s}) \leq m_i(s) \ddot{s} + c_i(s)\nukta{s}^2 + g_i(s) \leq \tau_i^{{\rm max}}(s,\nukta{s})lakini haifanyi yaliyomo katika hisabati kuwa tofautiimin​(s,s˙)mi.(s)s¨+ci.(s)s˙2+gi.(s)lakini haifanyi yaliyomo katika hisabati kuwa tofautiimax​(s,s˙).

What is one reason \tau_i^{{\rm min}}τimin​ and \tau_i^{{\rm max}}τimax​ might depend on \dot{s}s˙?

  • The positive torque available from an electric motor typically increases as its positive velocity increases.
  • The positive torque available from an electric motor typically decreases as its positive velocity increases.

Q2. At a particular state along the path, (s,\nukta{s})(s,s˙), the constraints on \ddot{s}s¨ due to the actuators at the three joints of a robot are: L_1 = -10, U_1 = 10na hufika bila nishati inayoweza kutokea na nishati ya kinetic1​=−10,uhifadhi-wa-mitambo-nishati-pendulum1​=10; L_2 = 3, U_2 = 12na hufika bila nishati inayoweza kutokea na nishati ya kinetic2​=3,uhifadhi-wa-mitambo-nishati-pendulum2​=12; and L_3 = -2, U_3 = 5na hufika bila nishati inayoweza kutokea na nishati ya kinetic3​=−2,uhifadhi-wa-mitambo-nishati-pendulum3​=5. Katika (s,\nukta{s})(s,s˙), what is the range of feasible accelerations \ddot{s}s¨?

  • 3 \leq \ddot{s} \leq 53≤s¨≤5
  • -10 \leq \ddot{s} \leq 12−10≤s¨≤12

Q3. If the robot is at a state (s,\nukta{s})(s,s˙) where no feasible acceleration \ddot{s}s¨ exists that satisfies the actuator force and torque bounds, nini kinatokea

  • One or more of the actuators is damaged.
  • The robot leaves the path.
  • The robot must begin to decelerate, \ddot{s}<0s¨<0.

Maswali 02: Lecture Comprehension, Time-Optimal Time Scaling (Sura 9.4, Sehemu 2 ya 3)

Q1. Consider the figure below, showing 4 motion cones at different states in the (s,\nukta{s})(s,s˙) nafasi.

3g3mIuT2Eee78hLOKMVkFA f607c39ba0de00665594846ea21926f8 s plane cones new

Which cone corresponds to U(s,\nukta{s})=4, na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,\nukta{s})=-3uhifadhi-wa-mitambo-nishati-pendulum(s,s˙)=4,na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,s˙)=−3?

  • A
  • B
  • C
  • D

Q2. Considering the figure in Question 1, which cone corresponds to U(s,\nukta{s})=4, na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,\nukta{s})=5uhifadhi-wa-mitambo-nishati-pendulum(s,s˙)=4,na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,s˙)=5?

  • A
  • B
  • C
  • D

Q3. Which cone corresponds to U(s,\nukta{s})=5, na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,\nukta{s})=2uhifadhi-wa-mitambo-nishati-pendulum(s,s˙)=5,na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,s˙)=2?

  • A
  • B
  • C
  • D

Q4. Which cone corresponds to U(s,\nukta{s})=-2, na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,\nukta{s})=-6uhifadhi-wa-mitambo-nishati-pendulum(s,s˙)=−2,na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,s˙)=−6?

  • A
  • B
  • C
  • D

Q5. Assume a time scaling s(t) = \frac{1}{2}t^2s(t)=21​t2. How is this time scaling written as \dot{s}(s)s˙(s)? (Note that this particular time scaling does not satisfy \dot{s}(1) = 0s˙(1)=0.)

  • \nukta{s} = \sqrt{2s}s˙=2s..
  • \nukta{s} = \frac{1}{2}s^2s˙=21​s2.

Maswali 03: Sura 9.4, Trajectory Generation

Q1. Four candidate trajectories (A, B, C, na D) are shown below in the (s,\nukta{s})(s,s˙) ndege. Select all of the trajectories that cannot be correct, regardless of the robot’s dynamics. Kumbuka: It is OK for the trajectory to begin and end with nonzero velocity. (This is consistently one of the most incorrectly answered questions in this course, so think about it carefully!)

gbNGgeTbEeeRtwqRjGvJYg 87571e8df15ba4b3df7727ebbcf85300 s plane errors
  • A
  • B
  • C
  • D

Q2. Four candidate motion cones at \dot{s} = 0s˙=0 (a, b, c, and d) ndani ya (s,\nukta{s})(s,s˙) plane are shown below. Which of these motion cones cannot be correct for any robot dynamics? (Do not assume that the robot can hold itself statically at the configuration.)CDMUTN6jEee5zAog3bNxIA 90ffe47d55d751049fc9a51a521c3f38 s plane errors 02 01

  • a
  • b
  • c
  • d

3.

Swali 3

We have been assuming forward motion on a path, \dot s > 0s˙>0. What if we allowed backward motion on a path, \dot s < 0s˙<0? This question involves motion cones in the (s, \dot s)(s,s˙)-plane when both positive and negative values of \dot ss˙ are available. Assume that the maximum acceleration is U(s, \dot s) = 1uhifadhi-wa-mitambo-nishati-pendulum(s,s˙)=1 (constant over the (s, \dot s)(s,s˙)-ndege) and the maximum deceleration is L(s, \dot s) = -1na hufika bila nishati inayoweza kutokea na nishati ya kinetic(s,s˙)=−1. For any constant ss, which of the following are the correct motion cones at the five points where \dot ss˙ takes the values \{-2, -1, 0, 1, 2\}{−2,−1,0,1,2}?

1 hatua

AlLTGZd6jEeedHw7X0zcEsg 6564b9361b064ed334c7f822ff25f265 ex03 1 01

B00NUK96jEee5zAog3bNxIA 14c9a76822602d2a763f8d6955e38887 ex03 2 01

C579R4N6jEee8ZBJAMiIM0A b7a3d7225eb02ee32507dac4d43f9924 ex03 3 01

DF aW996kEeedHw7X0zcEsg a3b5ea7c6566d38e787c6982d1e7acb8 ex03 4 01

4.

Swali 4

Referring back to Question 3, assume the motion starts at (s, \dot s) = (0, 0)(s,s˙)=(0,0) and follows the maximum acceleration Uuhifadhi-wa-mitambo-nishati-pendulum for time tt. Then it follows the maximum deceleration Lna hufika bila nishati inayoweza kutokea na nishati ya kinetic for time 2t2t. Then it follows Uuhifadhi-wa-mitambo-nishati-pendulum for time tt. Which of the following best represents the integral curve?

1 hatua

AVdJ88t6kEee8ZBJAMiIM0A 6bfbcadd6b7fbf218cf615fab8c1359d ex04 1 01

BaniXUN6kEee5zAog3bNxIA 38b3e2648375f5b76343c4165deb525c ex04 2 01

CfFSUwd6kEee8ZBJAMiIM0A 633b2e9aef47e0d01b2af9033c13ec70 ex04 3 01

D

vW13luYcEeeRtwqRjGvJYg 0735e803e94cb889ab5887ec01fdce1f week4ex04 4 01

5.

Swali 5

Below is a time-optimal time scaling \dot{s}(s)s˙(s) with three switches between the maximum and minimum acceleration allowed by the actuators. Also shown are example motion cones, which may or may not be correct.GzCVSeFjEeeY9RLN7DX 0g ab0d3e400ddb4bcdf2b36a0e27e572ef ex05 01

Without any more information about the dynamics, which motion cones must be incorrect (i.e., the motion cone is inconsistent with the optimal time scaling)? Select all that are incorrect (there may be more than one).

1 hatua

  • A
  • B
  • C
  • D
  • E
  • F
  • G
  • H

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.

    Tazama machapisho yote

Kuhusu 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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