MIT - Mechanical Engineering Department

2.12 Introduction to Robotics - Fall 2016 - Quiz 1 SolutionsDate: October 19, 2016 Duration: 80 minutes

Notes:· There are three problems totaling 100 points.· You have 80 minutes to solve them.· You are allowed to bring one two-sided handwritten letter-sized paper.· Show the process to get to the answer, and draw a rectangle around the solution.

Problem 1 (35 points)Inverse kinematics of an RPR Manipulator

The planar robot in Figure 1 is comprised of two revolute joints (θ1, θ3) and one prismatic joint (d2). Thefirst link has a 90◦ bend, so that the prismatic axis is always at a constant distance l1 from O. The end effectoris a point at a constant distance l3 from the second revolute joint, with coordinates (xe, ye) and orientationφe with respect the horizontal.

Figure 1: RPR Manipulator.

a. Obtain the kinematic equations that describe the end-effector position (xe, ye) and orientation φe asa function of joint displacements (θ1, d2, θ3).

b. Assume now that the joints are limited to these ranges:

θ1 ∈ [45◦, 135◦] d2 ∈ [l1, 2l1] θ3 ∈ [−90◦, 90◦]

Sketch the workspace of the end-effector in the XY plane (do not consider φe).

c. Find the expression for the inverse kinematics of the manipulator (relating (xe, ye, φe) to (θ1, d2, θ3)).

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Part a: (10 points) This is the problem of the forward kinematics of the manipulator and the expressionsfor the end effector are given by:

xe = l1 cos θ1 + d2 cos(θ1 − π/2) + l3 cos(θ1 + θ3 − π/2)

ye = l1 sin θ1 + d2 sin(θ1 − π/2) + l3 sin(θ1 + θ3 − π/2)

φe = θ1 + θ3 − π/2

or

xe = l1 cos θ1 + d2 sin(θ1) + l3 sin(θ1 + θ3)

ye = l1 sin θ1 + d2 cos(θ1) + l3 cos(θ1 + θ3)

φe = θ1 + θ3 − π/2

Part b: (10 points)

l1

l3l3

l3

l1

l1

x

y

Figure 2: RPR Manipulator workspace.

Part c: (15 points) Labels:

• 90 degree elbow - A

• 3rd joint - B

We will use a geometric approach to find the inverse kinematics of the manipulator. Given the endeffector position and orientation point B is fully defined and the coordinates are given by:

[xbyb

]=

[l3 − cos(φe)le − sin(φe)

]Next carefully inspect the triangle OAB, using Pythagoras theorem:

OB2

= OA2

+ AB2

x2b + y2b = d22 + l21

d22 = x2b + y2b − l21

d2 =√

(l3 − cosφe)2 + (le − sinφe)2 − l21

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Next to find the angle for the first joint we can write:

θ1 = tan−1(ybxb

) + tan−1(

√x2b + y2b − l21

l1)

and finally to find θ3 we can use

θ3 = φe + π/2− θ1

For this problem algebraic approaches are also acceptable using the laws of cosine, i.e. we can write:

θ1 = tan−1(ybxb

) + cos−1(l1√

x2b + y2b

)

Problem 2 (25 points)

Standard reference frames of a robotic manipulator

As a matter of convention, it is helpful to give particular names to frequent reference frames associated witha robot manipulator and its workspace. Figure 3 shows a typical situation in which a robot is carrying out atask on a table, with six common frames:

World frame O Inertial reference frame.Base frame B Located at the base of the robot manipulator.Station frame S Located in the vicinity of the relevant working area. It simplifies the

description of the task. It is usually specified in the world frame OS T .

Wrist frame W Affixed to the last link of the manipulator. It is usually specified in thebase frame, encoding the direct kinematics of the robot arm B

WT .Tool frame T Also known as end-effector frame, is located at a useful point and ori-

entation in the tool, e.g., its tip. It is always specified in the frame WT T .

Goal frame Gi Location where the tool has to go to execute task i. It is usually specifiedin the station frame S

GiT .

Figure 3: (left) Standard reference frames; (right) Calibration socket.

a. Calibration. A normal required step for proper use of a tool is to calibrate its location with respectto the wrist of the robot W

T T . With that intention, we manually guide the robot in order to insert

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the tool into the calibration socket. Once in this calibration configuration (in which frames G1 andT are coincident), we read the position of the robot with respect to its base using direct kinematicsBWT . Assuming that the locations of the station O

S T and calibration goal SG1T are known, and that the

location of the base OBT is given by some localization system, find the expression of the calibration

tool frame WT T as a function of other known transformations.

b. Station coordinates. It is often also useful to figure out the location of the tool with respect to thestation coordinates S

TT . If the robot is in a particular configuration given by the location of its baseOBT and its manipulator B

WT , and if we have already calibrated the tool WT T , and if the location of the

station is known OS T , figure out the transformation S

TT that gives the location of the tool in stationcoordinates.

c. In practice, some of the expressions above might require to invert a matrix. Should we expect prob-lems when inverting them?

Part a: (10 points) We can write:

OG1T =O

S T ·SG1T

OWT =O

B T ·BW TWT T =W

G1T

Consequently:

WT T =W

G1T =O

W T−1 ·OG1T

Part b: (10 points) We can write:

STT = (OS T )−1 ·OB T ·BW T ·WT T

Part c: (5 points) No there is no problem, the homogeneous transformations are always invertible.

Problem 3 (40 points)

Manipulator with long reach

The manipulator in Figure 4 has 2 degrees of freedom (θ1 and θ2). As it stands the manipulator’s end effectorA cannot reach the point B because l3 > l1 + l2. To extend its workspace, we propose to mount it on alinear guide, allowing joint one to move linearly along axis X, where the distance from the origin is denotedby d, as in Figure 5. With this in mind answer the following questions:

a. Derive the forward kinematics and the Jacobian of the manipulator in Figure 4 relating (θ1, θ2) tothe Cartesian position of A = (xA, yA).

b. Construct the Jacobian of the new manipulator in Figure 5 relating now (d, θ1, θ2) to (xA, ˙yA, φ)),where φ is the orientation of the last link with respect the horizontal. Describe the effect on theJacobian from adding the guide as an additional degree of freedom.

c. Compute the singular configurations of both manipulators. Sketch the manipulator in those config-urations.

d. For this problem assume that l1 = l2 = 1 and l3 = 2, and that the robot in Figure 5 is in configurationθ1 = −π/4, θ2 = π/2 with the end effector at B. If we only care about the position of the endeffector and not its orientation, is the robot redundant? If it is, describe the set of joint velocities(d, θ1, θ2) that produce zero velocity in the end effector (xA, yA).

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Figure 4: 2-Link manipulator.

Figure 5: Same 2-Link manipulator on a guide.

Part a: (10 points) Forward kinematics are given by:[xAyA

]=

[l1c1 + l2c12l1s1 + l2s12

]and the Jacobian is found to be:

J =

[−l1s1 − l2s12 −l2s12l1c1 + l2c12 l2c12

]Part b: (10 points) The forward kinematics with the new degree of freedom is:xAyA

φ

=

d+ l1c1 + l2c12l1s1 + l2s12θ1 + θ2

The new Jacobian is:

J =

1 −l1s1 − l2s12 −l2s120 l1c1 + l2c12 l2c120 1 1

The new Jacobian has an additional column due to the new and independent degree of freedom and a row torepresent the variation in the angle of the end effector.

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Part c: (10 points) We already know that the singular configurations of the RR manipulator are givenby θ2 = 0, π. The singular configurations for the new manipulator are given at θ1 = π/2,−π/2 which arefound from deriving the determinant which is given by det(J) = l1c1 = 0. The singular configurations areshown below and are fundamentally different types.

Part d: (10 points) In this part our forward kinematics is simply the first two rows of the forwardkinematics in the previous section: [

xAyA

]=

[d+ l1c1 + l2c12l1s1 + l2s12

]We need to compute the Jacobian and find its null space to evaluate the redundancy of the manipulator:

J =

[1 −l1s1 − l2s12 −l2s120 l1c1 + l2c12 l2c12

]Next we apply the assumptions provided in the problem statement:

J =

[1 0 − 1√

2

0 2√2

1√2

]

Next we find solutions to:

J

dθ1θ2

= 0

Clearly there are 3 unknowns and only 2 equations, we can parametrize the solution using θ2:

N =

1√2

−12

1

θ2

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