Modern Robotics, 课程 3: Robot Dynamics Quizzes & 答案 – Coursera

Join us for a fascinating exploration of robot dynamics 在 课程 3, where the complex interaction of forces and motion shapes 机器人 行为. Immerse yourself in our engaging quizzes and expert answers that shed light on the principles governing the dynamic behaviour of robotic 系统. These quizzes serve as a gateway to understanding the complex 动力学 that govern 机器人 motion from acceleration to torque and beyond.

Whether you are a 机器人 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 机器人 运动. Join us on a journey of discovery as we unravel the 动力学 的 机器人 behaviour and unlock the potential for precise and efficient 机器人 操作.

测验 01: Lecture Comprehension, Lagrangian Formulation of Dynamics (章节 8 通过 8.1.2, 部分 1 的 2)

第一季度. 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} \压裂{\partial \mathcal{大号}}{\partial \dot{\theta}_i} – \frac{\partial \mathcal{大号}}{\partial \theta_i}τ一世​=d吨d​∂θ˙一世​∂L​−∂θ一世​∂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. 因此, 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)2θ1​cos(4θ2​), where \theta_1θ1​ and \theta_2θ2​ are functions of time?

• -8 \点{\theta}_1 \sin(4 \theta_2)−8θ˙1​sin(4θ2​)
• 2 \点{\theta}_1 \cos (4 \theta_2) - 8 \theta_1 \sin(4 \theta_2) \点{\theta}_22θ˙1​cos(4θ2​)−8θ1​sin(4θ2​)θ˙2​
• 2 \点{\theta}_1 \cos (4 \theta_2) - 2 \theta_1 \sin(4 \theta_2) \点{\theta}_22θ˙1​cos(4θ2​)−2θ1​sin(4θ2​)θ˙2​
• 2 \点{\theta}_1 \cos (4 \theta_2) + 2 \theta_1 \cos(4 \theta_2) \点{\theta}_22θ˙1​cos(4θ2​)+2θ1​cos(4θ2​)θ˙2​

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

\tau = M(\theta) \ddot{\theta} + C(\theta,\点{\theta}) + G(\theta)τ=中号(θ)θ¨+C(θ,θ˙)+G(θ).

If the equation for the first joint is

\tau_1 = term1 + term2 + term3 + ...τ1​=吨Ë[R米1+吨Ë[R米2+吨Ë[R米3+...

第四季度. which of the following terms, written in terms of (\theta,\点{\theta},\ddot{\theta})(θ,θ˙,θ¨), 但是开发这些疗法的复杂过程限制了许多领域 不 be one of the terms on the right-hand side of the equation? (The value kķ is not a function of (\theta,\点{\theta},\ddot{\theta})(θ,θ˙,θ¨) and could represent constants like link lengths, masses, or inertias, as needed to get correct units.) 选择所有符合条件的.

• k\ddot{\theta}_2 \cos(\theta_1)kθ¨2​cos(θ1​)
• k\ddot{\theta}_1 \dot{\theta}_1kθ¨1​θ˙1​
• k\sin \theta_3ķsinθ3​
• k\dot{\theta}_1 \dot{\theta}_2 \sin \theta_2kθ˙1​θ˙2​sinθ2​
• k \dot{\theta}_1 \sin \theta_2kθ˙1​sinθ2​

测验 02: Lecture Comprehension, Lagrangian Formulation of Dynamics (章节 8 通过 8.1.2, 部分 2 的 2)

第一季度. Which of the following could be a centripetal term in the dynamics?

• k \dot{\theta}_1^2kθ˙12​
• k \dot{\theta}_1 \dot{\theta}_2kθ˙1​θ˙2​

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

• k \dot{\theta}_1^2kθ˙12​
• k \dot{\theta}_1 \dot{\theta}_2kθ˙1​θ˙2​

Q3. One form of the equations of motion is

\tau = M(\theta)\ddot{\theta} + \点{\theta}^{\rm T} \Gamma(\theta) \点{\theta} + G(\theta).τ=中号(θ)θ¨+θ˙TΓ(θ)θ˙+G(θ).

Which of the following is true about \Gamma(\theta)Γ(θ)? 选择所有符合条件的.

• \Gamma(\theta)Γ(θ) is zero if the mass matrix M中号 has no dependence on \thetaθ.
• \Gamma(\theta)Γ(θ) is an n \times nn×n 矩阵, where nn is the number of joints.
• \Gamma(\theta)Γ(θ) depends on M(\theta)中号(θ) and \dot{\theta}θ˙.

Lecture Comprehension, Understanding the Mass Matrix (章节 8.1.3)测验 03:

第一季度. Which of these is a possible mass matrix M(\theta)中号(θ) for a two-joint robot at a particular configuration \thetaθ? 选择所有符合条件的.

• \剩下[
• 200−1
• \对][20​0−1​]
• \剩下[
• 4123
• \对][41​23​]
• \剩下[
• 3112
• \对][31​12​]
• \剩下[
• 2221
• \对][22​21​]

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

• 真正.
• 假.

Q3. 对或错? 如果你申请 (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.

测验 04: Lecture Comprehension, Dynamics of a Single Rigid Body (章节 8.2, 部分 1 的 2)

第一季度. 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}.
• 一起, \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?

• \压裂{1}{2} \omega_b^{\rm T} \mathcal{一世}_b \omega_b21​ωbT​Ib​ωb​
• \压裂{1}{2} \mathcal{一世}_b \omega_b^221​Ib​ωb2​

第四季度. 对或错? For a given body, there is exactly one orientation of a frame at the center of mass that yields a diagonal rotational inertia matrix.

• 真正.
• 假.

测验 05: Lecture Comprehension, Dynamics of a Single Rigid Body (章节 8.2, 部分 2 的 2)

第一季度. The spatial inertia matrix is the 6×6 matrix

\mathcal{G}_B = \left[

一世b00米一世

\对]G乙​=[一世b​0​0m一世​].

What is the maximum number nn 的 独特 nonzero values the spatial inertia could have? 换一种说法, 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. 表达方式 [\omega_1][\omega_2]-[\omega_2][\omega_1[ω1​][ω2​] - [ω2​][ω1​ is the so(3)所以(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}^3ω1​×ω2​=[ω1​]ω2​∈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? 选择所有符合条件的.

• 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 = (\点{\它们还被证明可以预防抑郁症并提高老年人的认知功能}_b, \点{v}_b) = (0,0)V˙b​=(ω˙b​,v˙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.

第四季度. 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 = (\点{\它们还被证明可以预防抑郁症并提高老年人的认知功能}_b, \点{v}_b) = (0,0)V˙b​=(ω˙b​,v˙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.

测验 06: 章节 8 通过 8.3, Dynamics of Open Chains

第一季度. 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 厘米. Each sphere has a diameter of 20 厘米. Find the approximate rotational inertia matrix \mathcal{一世}_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

\对]⎣⎢⎡​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,\点{\theta}) + G(\theta) = \tau中号(θ)θ¨+C(θ,θ˙)+G(θ)=τ. The Lagrangian \mathcal{大号}(\theta,\点{\theta})大号(θ,θ˙) for the robot can be written in components as

\mathcal{大号}(\theta,\点{\theta}) = \mathcal{大号}^1(\theta,\点{\theta}) + \mathcal{大号}^2 (\theta,\点{\theta}) + \mathcal{大号}^3 (\theta,\点{\theta}) +\ldotsL(θ,θ˙)=L1(θ,θ˙)+L2(θ,θ˙)+L3(θ,θ˙)+...

One of these components is \mathcal{大号}^1 = \mathfrak{米} \点{\theta}_1 \dot{\theta}_2 \sin\theta_2 L1=mθ˙1​θ˙2​sinθ2​.

Find the right expression for the component of the joint torque \tau^1_1τ11​ at joint 1 corresponding to the component \mathcal{大号}^1L1.

• \tau^1_1 = \mathfrak{米} \ddot{\theta_2} \sin \theta_2 – \mathfrak{米} \dot \theta_2^2 \cos \theta_2τ11​=mθ2​¨​sinθ2​−mθ˙22​cosθ2​
• \tau^1_1 = \mathfrak{米} \ddot{\theta_2} \sin \theta_2 + \mathfrak{米} \dot \theta_2^2 \cos \theta_2τ11​=mθ2​¨​sinθ2​+mθ˙22​cosθ2​
• \tau^1_1 = \mathfrak{米} \ddot{\theta_2} \cos \theta_2 + \mathfrak{米} \dot \theta_2^2 \sin \theta_2τ11​=mθ2​¨​cosθ2​+mθ˙22​sinθ2​

Q3. Referring back to Question 2, find the right expression for the component of joint torque \tau^1_2τ21​ at joint 2 corresponding to the component \mathcal{大号}^1L1.

• \tau^1_2 = \mathfrak{米} \ddot{\theta_2} \sin \theta_2 + \mathfrak{米} \dot \theta_2^2 \cos \theta_2τ21​=mθ2​¨​sinθ2​+mθ˙22​cosθ2​
• \tau^1_2 = \mathfrak{米} \ddot{\theta_1} \sin \theta_2τ21​=mθ1​¨​sinθ2​
• \tau^1_2 = \mathfrak{米} \ddot{\theta_1} \sin \theta_2 + \mathfrak{米} \dot \theta_1 \dot \theta_2 \cos \theta_2τ21​=mθ1​¨​sinθ2​+mθ˙1​θ˙2​cosθ2​

第四季度. For a given configuration \thetaθ of a two-joint robot, the mass matrix is

中号(\theta) = \left[

3b一个12

\对],中号(θ)=[3b​一个12​],

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 a一个 and bb satisfy for this to be a valid mass matrix? 选择所有符合条件的.

• 一个 < 6一个<6
• b > 6b>6
• 一个 > b一个>b
• a = b一个=b
• 一个<b一个<b
• 一个 < \sqrt 6一个<6​

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

M_{01} = \left[

100001000010000.0891591

\对], M_{12} = \left[

00−10010010000.280.1358501

\对], M_{23} = \left[

1000010000100−0.11970.3951

\对], 中号01​=⎣⎢⎢⎢⎡​1000​0100​0010​000.0891591​⎦⎥⎥⎥⎤​,中号12​=⎣⎢⎢⎢⎡​00−10​0100​1000​0.280.1358501​⎦⎥⎥⎥⎤​,中号23​=⎣⎢⎢⎢⎡​1000​0100​0010​0−0.11970.3951​⎦⎥⎥⎥⎤​,

M_{34} = \left[

00−1001001000000.142251

\对], M_{45} = \left[

10000100001000.09301

\对], M_{56} = \left[

100001000010000.094651

\对], 中号34​=⎣⎢⎢⎢⎡​00−10​0100​1000​000.142251​⎦⎥⎥⎥⎤​,中号45​=⎣⎢⎢⎢⎡​1000​0100​0010​00.09301​⎦⎥⎥⎥⎤​,中号56​=⎣⎢⎢⎢⎡​1000​0100​0010​000.094651​⎦⎥⎥⎥⎤​,

M_{67} = \left[

100000−10010000.082301

\对],中号67​=⎣⎢⎢⎢⎡​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

\对].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

θ=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢0神父。/6神父。/4神父。/3神父。/22神父。/3⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥,θ˙=⎡⎣⎢⎢⎢⎢⎢⎢⎢0.20.20.20.20.20.2⎤⎦⎥⎥⎥⎥⎥⎥⎥,θ¨=⎡⎣⎢⎢⎢⎢⎢⎢⎢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

\对]⎣⎢⎡​1.112.223.33​⎦⎥⎤​.

1

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

RunReset

周 02: Modern Robotics, 课程 3: Robot Dynamics Coursera Quiz Answers

测验 01: Lecture Comprehension, Forward Dynamics of Open Chains (章节 8.5)

第一季度. To derive the mass matrix M(\theta)中号(θ) of an nn-joint open-chain robot, how many times would we need to invoke the recursive Newton-Euler inverse dynamics algorithm?

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

Q2. When calculating the mass matrix M(\theta)中号(θ) using the Newton-Euler inverse dynamics, which of these quantities must be set to zero? 选择所有符合条件的.

• \thetaθ
• \点{\theta}θ˙
• \ddot{\theta}θ¨
• The gravitational constant.
• The end-effector wrench \mathcal{F}_{{\rm tip}}Ftip​.

测验 02: Lecture Comprehension, Dynamics in the Task Space (章节 8.6)

第一季度. Converting the joint-space dynamics to the task-space dynamics requires an invertible Jacobian, as well as the relationships \mathcal{V} = J(\theta)\点{\theta}V =Ĵ(θ)θ˙ and \dot{\mathcal{V}} = J\ddot{\theta} + \点{Ĵ}\点{\theta}V˙=Ĵθ¨+Ĵ˙θ˙, to find \Lambda(\theta)Λ(θ) and \eta(\theta,\mathcal{V})η(θ,V) in \mathcal{F} = \Lambda(\theta) \点{\mathcal{V}} + \eta(\theta,\mathcal{V})F=Λ(θ)V˙+η(θ,V).

Why do you suppose we left the dependence on \thetaθ, instead of writing it as a dependence on the end-effector configuration T \in SE(3)Ť∈SE(3), which would seem to be more aligned with our task-space view?

• Either TŤ or \thetaθ 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 TŤ.

测验 03: Lecture Comprehension, Constrained Dynamics (章节 8.7)

第一季度. A serial-chain robot has nn links and actuated joints, but it is subject to kķ independent Pfaffian velocity constraints of the form A(\theta)\点{\theta}=0一个(θ)θ˙=0. These constraints partition the nn-dimensional \tauτ space into orthogonal subspaces: a space of forces CC that do not create any forces against the constraints, and a space of forces B乙 that do not cause any motion of the robot. What is the dimension of each of these spaces?

• CC 是 (n-k)(n - ķ)-dimensional and B乙 is kķ-维.
• CC is nn-dimensional and B乙 is kķ-维.
• CC is kķ-dimensional and B乙 是 (n-k)(n - ķ)-维.
• CC is kķ-dimensional and B乙 is nn-维.

Q2. Let the constrained dynamics of a robot be \tau = M(\theta)\ddot{\theta} + H(\theta,\点{\theta}) + A^{\rm T}(\theta)\lambdaτ=中号(θ)θ¨+H(θ,θ˙)+一个Ť(θ)λ, where \lambda \in \mathbb{[R}^kλ∈Rķ. Let P(\theta)P(θ) be the matrix, as discussed in the video, that projects an arbitrary \tau \in \mathbb{[R}^nτ∈Rn to P(\theta)\tau \in CP(θ)τ∈C, where the space CC is the same CC from the previous question. Then what is P(\theta) \tauP(θ)τ? 选择所有符合条件的.

• 中号(\theta)\ddot{\theta} + H(\theta,\点{\theta}) + A^{\rm T}(\theta)\lambda中号(θ)θ¨+H(θ,θ˙)+一个Ť(θ)λ
• P(\theta)(中号(\theta)\ddot{\theta} + H(\theta,\点{\theta}))P(θ)(中号(θ)θ¨+H(θ,θ˙))
• P(\theta)(中号(\theta)\ddot{\theta} + H(\theta,\点{\theta})) +A^{\rm T}(\theta)\lambda)P(θ)(中号(θ)θ¨+H(θ,θ˙))+一个Ť(θ)λ)

测验 04: Lecture Comprehension, Actuation, Gearing, and Friction (章节 8.9)

第一季度. 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? 选择所有符合条件的.

• The mass matrix M(\theta)中号(θ) is increasingly dominated by the apparent inertias of the motors.
• The mass matrix M(\theta)中号(θ) is increasingly dominated by off-diagonal terms.
• The mass matrix M(\theta)中号(θ) is increasingly dominated by constant terms that do not depend on the configuration \thetaθ.
• The robot is capable of higher speeds but lower accelerations.
• The significance of velocity-product (Coriolis and centripetal) terms diminishes.

测验 05: 章节 8.5-8.7 和 8.9, Dynamics of Open Chains

第一季度. A robot system (UR5) 被定义为

M_{01} = \left[

100001000010000.0891591

\对], M_{12} = \left[

00−10010010000.280.1358501

\对], M_{23} = \left[

1000010000100−0.11970.3951

\对], 中号01​=⎣⎢⎢⎢⎡​1000​0100​0010​000.0891591​⎦⎥⎥⎥⎤​,中号12​=⎣⎢⎢⎢⎡​00−10​0100​1000​0.280.1358501​⎦⎥⎥⎥⎤​,中号23​=⎣⎢⎢⎢⎡​1000​0100​0010​0−0.11970.3951​⎦⎥⎥⎥⎤​,

M_{34} = \left[

00−1001001000000.142251

\对], M_{45} = \left[

10000100001000.09301

\对], M_{56} = \left[

100001000010000.094651

\对], 中号34​=⎣⎢⎢⎢⎡​00−10​0100​1000​000.142251​⎦⎥⎥⎥⎤​,中号45​=⎣⎢⎢⎢⎡​1000​0100​0010​00.09301​⎦⎥⎥⎥⎤​,中号56​=⎣⎢⎢⎢⎡​1000​0100​0010​000.094651​⎦⎥⎥⎥⎤​,

M_{67} = \left[

100000−10010000.082301

\对],中号67​=⎣⎢⎢⎢⎡​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

\对].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[

0神父。/6神父。/4神父。/3神父。/22神父。/3

\对]θ=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0神父。/6神父。/4神父。/3神父。/22神父。/3​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

\dot \theta = \left[

0.20.20.20.20.20.2

\对]θ˙=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0.20.20.20.20.20.2​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

\ddot{\theta} = \left[

0.10.10.10.10.10.1

\对]θ¨=⎣⎢⎢⎢⎢⎢⎢⎢⎡​0.10.10.10.10.10.1​⎦⎥⎥⎥⎥⎥⎥⎥⎤​,

\mathfrak{G} = \left[

00−9.81

\对]g=⎣⎢⎡​00−9.81​⎦⎥⎤​,

\mathcal{F}_{\文本{tip}} = \left[

0.10.10.10.10.10.1

\对]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

\对]⎣⎢⎡​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

\对]⎣⎢⎡​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

\对]⎣⎢⎡​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

\对]⎣⎢⎡​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

\对] τ=⎣⎢⎢⎢⎢⎢⎢⎢⎡​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

\对]⎣⎢⎡​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

周 03: Modern Robotics, 课程 3: Robot Dynamics Coursera Quiz Answers

测验 03: Lecture Comprehension, Point-to-Point Trajectories (章节 9 通过 9.2, 部分 1 的 2)

第一季度. A point robot moving in a plane has a configuration represented by (X,和)(X,和). 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)),小号∈[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]小号=2吨,吨∈[0,1/2]. At time t吨, 哪里 0 \leq t \leq 0.50≤吨≤0.5, what is the velocity of the robot (\点{X},\点{和})(X˙,和˙​)?

• (-4 \pi \sin(2 \pi t), 4 \pi \cos(2 \pi t))(−4神父。sin(2πt),4神父。cos(2πt))
• (-2\sin(2 \pi t), 2\cos(2 \pi t))(−2sin(2πt),2cos(2πt))

Q3. 对或错? For a trajectory \theta(小号(吨))θ(小号(吨)), the acceleration \ddot{\theta}θ¨ is \frac{d\theta}{ds}\ddot{小号}dsdθ​小号¨.

• 真正
• 假

第四季度. Let \mathcal{V}_sV小号​ be the spatial twist that takes X_{小号,{\rm start}}Xs,start​ to X_{小号,{\rm end}}Xs,end​ in unit time. Which is an expression for the constant screw path that takes X_{小号,{\rm start}}Xs,start​ (at s=0小号=0) to X_{小号,{\rm end}}Xs,end​ (at s=1小号=1)?

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

测验 02: Lecture Comprehension, Point-to-Point Trajectories (章节 9 通过 9.2, 部分 2 的 2)

第一季度. For a fifth-order polynomial time scaling s(吨)小号(吨), t \in [0,Ť]吨∈[0,Ť], what is the form of \ddot{小号}(吨)小号¨(吨)?

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

测验 03: Lecture Comprehension, Polynomial Via Point Trajectories (章节 9.3)

第一季度. 对或错? Third-order polynomial interpolation between via points ensures that the path remains inside the convex hull of the via points.

• 真正
• 假

Q2. A robot has 3 joints and it follows a motion interpolating 6 点数: a start point, an end point, 和 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 段?

• 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

测验 04: 章节 9 通过 9.3, Trajectory Generation

第一季度. Consider the elliptical path in the (X,和)(X,和)-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]小号∈[0,1] to represent the path.

• x = 3 (1 – \cos 2 \pi s)X=3(1−cos2πs)
• y = \sin 2 \pi s和=sin2πs
• x = 1.5 (1 – \cos 2 \pi s)X=1.5(1−cos2πs)
• y = \sin 2 \pi s和=sin2πs
• x = 1.5 (1 – \cos s)X=1.5(1−cos小号)
• y = \sin s和=sin小号
• x = \cos 2 \pi sX=cos2πs

y = 1.5 (1 – \sin 2 \pi s)和=1.5(1−sin2πs

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

Your answer should be only a mathematical expression, a polynomial in t吨, with coefficients involving TŤ. (Don’t bother to write “s(吨) = 小号(吨)=”, 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, 加速度, and jerk, what would be the minimum order of the polynomial

第四季度. Choose the correct acceleration profile \ddot{小号}(吨)小号¨(吨) for an S-curve time scaling.

• 一个
• 乙
• C
• d

Q5. Given a total travel time T = 5Ť=5 and the current time t = 3吨=3, use the function {\tt QuinticTimeScaling}QuinticTimeScaling in the given software to calculate the current path parameter s小号, 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=10ñ=10 SE(3)小号Ë(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

\对]Xstart​=⎣⎢⎢⎢⎡​1000​0100​0010​0001​⎦⎥⎥⎥⎤​

and the 10th matrix is

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

0100001010001231

\对].Xend​=⎣⎢⎢⎢⎡​0100​0010​1000​1231​⎦⎥⎥⎥⎤​.

The motion is along a constant screw axis and the duration is T_f = 10ŤF​=10. The parameter {\tt method}method equals 3 for a cubic time scaling. Give the 9th matrix (one before X_{{\rm end}}Xend​) 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

\对]⎣⎢⎡​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]]

Q7. Referring back to Question 6, use the function {\tt CartesianTrajectory}CartesianTrajectory in the MR library to calculate another trajectory as a list of N=10ñ=10 SE(3)小号Ë(3) matrices. Besides the same X_{{\rm start}}Xstart​, X_{{\rm end}}Xend​, T_fŤF​ and N = 10ñ=10, we now set {\tt method}method to 5 for a quintic time scaling. Give the 9th matrix (one before X_{{\rm end}}Xend​) 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

\对]⎣⎢⎡​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

周 04: Modern Robotics, 课程 3: Robot Dynamics Coursera Quiz Answers

测验 01: Lecture Comprehension, Time-Optimal Time Scaling (章节 9.4, 部分 1 的 3)

第一季度. When a robot travels along a specified path \theta(小号)θ(小号), torque or force limits at each actuator place bounds on the path acceleration \ddot{小号}小号¨. The constraints due to actuator i一世 can be written

\tau_i^{{\rm min}}(小号,\点{小号}) \leq m_i(小号) \ddot{小号} + c_i(小号)\点{小号}^2 + g_i(小号) \leq \tau_i^{{\rm max}}(小号,\点{小号})τ一世min​(小号,小号˙)≤米一世​(小号)小号¨+C一世​(小号)小号˙2+G一世​(小号)≤τ一世max​(小号,小号˙).

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

• 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, (小号,\点{小号})(小号,小号˙), the constraints on \ddot{小号}小号¨ due to the actuators at the three joints of a robot are: L_1 = -10, U_1 = 10大号1​=−10,该1​=10; L_2 = 3, U_2 = 12大号2​=3,该2​=12; and L_3 = -2, U_3 = 5大号3​=−2,该3​=5. 在 (小号,\点{小号})(小号,小号˙), what is the range of feasible accelerations \ddot{小号}小号¨?

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

Q3. If the robot is at a state (小号,\点{小号})(小号,小号˙) where no feasible acceleration \ddot{小号}小号¨ exists that satisfies the actuator force and torque bounds, 它不能凭空产生脂肪——脂肪必须首先来自某个地方

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

测验 02: Lecture Comprehension, Time-Optimal Time Scaling (章节 9.4, 部分 2 的 3)

第一季度. Consider the figure below, showing 4 motion cones at different states in the (小号,\点{小号})(小号,小号˙) 空间.

Which cone corresponds to U(小号,\点{小号})=4, 大号(小号,\点{小号})=-3该(小号,小号˙)=4,大号(小号,小号˙)=−3?

• 一个
• 乙
• C
• d

Q2. Considering the figure in Question 1, which cone corresponds to U(小号,\点{小号})=4, 大号(小号,\点{小号})=5该(小号,小号˙)=4,大号(小号,小号˙)=5?

• 一个
• 乙
• C
• d

Q3. Which cone corresponds to U(小号,\点{小号})=5, 大号(小号,\点{小号})=2该(小号,小号˙)=5,大号(小号,小号˙)=2?

• 一个
• 乙
• C
• d

第四季度. Which cone corresponds to U(小号,\点{小号})=-2, 大号(小号,\点{小号})=-6该(小号,小号˙)=−2,大号(小号,小号˙)=−6?

• 一个
• 乙
• C
• d

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

• \点{小号} = \sqrt{2小号}小号˙=2小号​.
• \点{小号} = \frac{1}{2}s^2小号˙=21​小号2.

测验 03: 章节 9.4, Trajectory Generation

第一季度. Four candidate trajectories (一个, 乙, C, 和 D) are shown below in the (小号,\点{小号})(小号,小号˙) plane. Select all of the trajectories that cannot be correct, regardless of the robot’s dynamics. 注意: 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!)

• 一个
• 乙
• C
• d

Q2. Four candidate motion cones at \dot{小号} = 0小号˙=0 (一个, b, C, and d) 在里面 (小号,\点{小号})(小号,小号˙) 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.)

• 一个
• b
• C
• d

3.

题 3

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

1 观点

一个

乙

C

d

4.

题 4

Referring back to Question 3, assume the motion starts at (小号, \dot s) = (0, 0)(小号,小号˙)=(0,0) and follows the maximum acceleration U该 for time t吨. Then it follows the maximum deceleration L大号 for time 2t2吨. Then it follows U该 for time t吨. Which of the following best represents the integral curve?

1 观点

一个

乙

C

d

5.

题 5

Below is a time-optimal time scaling \dot{小号}(小号)小号˙(小号) 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.

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

1 观点

• 一个
• 乙
• C
• d
• Ë
• F
• G
• H