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ECE5463: Introduction to Robotics

Lecture Note 6: Forward Kinematics

Prof. Wei Zhang

Department of Electrical and Computer EngineeringOhio State UniversityColumbus, Ohio, USA

Spring 2018

Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 1 / 15

Outline

• Background

• Illustrating Example

• Product of Exponential Formula

• Body Form of the PoE Formula

Outline Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 2 / 15

Kinematics

R

RR

RR

R

P

S1

S2

S6

S5

S4

S3

Kinematics is a branch of classical mechanics that describes the motion ofpoints, bodies (objects), and systems of bodies (groups of objects) withoutconsidering the mass of each or the forces that caused the motion

Background Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 3 / 15

Kinematics of Robotic Manipulator

• Forward Kinematics: calculation of the positionp and orientation R of the end-effector framefrom its joint variables θ = (θ1, . . . , θn)

• Two commonly adopted approaches

• Approach 1: Assign a frame at each link, typically at the joint axis, thencalculate the end-effector configuration through intermediate frames.

- Denavit-Hartenberg parameters: most widely adopted convention for frameassignment

• Approach 2: Product of Exponential (PoE) formula: directly compute theend-effector frame configuration relative to the fixed frame through screwmotion along the screw axis associated with each joint.


Forward Kinematics: Basic Setup

• Suppose that the robot has n joints and n links. Each joint has one degree offreedom represented by joint variable θ

- Revolute joint: θ represents the joint angle

- Primatic joint: θ represents the joint displacement

• Specify a fixed frame {s}: also referred to as frame {0}

• Attach frame {i} to link i at joint i, for i = 1, . . . , n

• Use one more frame {b} at the end-effector: sometimes referred to as frame{n+1}

• Goal: Find an analytical expression of Tsb(θ1, . . . , θn)


θθ

θθθ

A Simple Example: Approach 1 (D-H Parameters)

{0}

{1}

{2}

{3}

{4}

θ1

θ2

θ3

(x, y)

φ

L 1

L 3

L 2

Illustrating Example Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 6 / 15

A Simple Example: Approach 2 (PoE Formula)

{0}

{1}

{2}

{3}

{4}

θ1

θ2

θ3

(x, y)

φ

L 1

L 3

L 2

Illustrating Example Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 7 / 15

Outline

• Background

• Illustrating Example

• Product of Exponential Formula

• Body Form of the PoE Formula

PoE Formula Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 8 / 15

Product of Exponential: Main Idea

• Goal: Derive Tsb(θ1, . . . , θn)

• First choose a zero position (“home” position):θ1 = θ01, . . . , θn = θ0n such thatM � Tsb(θ

01, . . . , θ

0n) can be easily found.

Without loss of generality, suppose homeposition is given by θ0i = 0, i = 1, . . . , n.

M

θnθn−1

θn−2θ1

e[Sn]θnM

e[Sn−1]θn−1e[Sn]θnM

e[Sn−2]θn−2e[Sn−1]θn−1e[Sn]θn M

• Let S1, . . . ,Sn be the screw axes expressed in {s}, corresponding to the jointmotions when the robot is at its home position

• Apply screw motion to joint n: Tsb(0, . . . , 0, θn) = e[Sn]θnM

• Apply screw motion to joint n− 1 to obtain:

Tsb(0, . . . , 0, θn−1, θn) = e[Sn−1]θn−1e[Sn]θnM

• After n screw motions, the overall forward kinematics:

Tsb(θ1, . . . , θn) = e[S1]θ1e[S2]θ2 · · · e[Sn]θnM


Forward Kinematics: Steps for PoE

• Goal: Derive Tsb(θ1, . . . , θn)

• Step 1: Find the configuration of{b} at the home position

M = Tsb(0, . . . , 0)

M

θnθn−1

θn−2θ1

e[Sn]θnM

e[Sn−1]θn−1e[Sn]θnM

e[Sn−2]θn−2e[Sn−1]θn−1e[Sn]θn M

• Step 2: Find screw axes S1, . . . ,Sn expressed in {s}, corresponding to thejoint motions when the robot is at its home position

• Forward kinematics:

Tsb(θ1, . . . , θn) = e[S1]θ1e[S2]θ2 · · · e[Sn]θnM


PoE: Screw Motions in Different Order (1/2)

• PoE was obtained by applying screw motions along screw axes Sn, Sn−1, . . ..What happens if the order is changed?

• For simplicity, assume that n = 2, and let us apply screw motion along S1

first:

- Tsb(θ1, 0) = e[S1]θ1M

- Now screw axis for joint 2 has been changed. The new axis S ′2 is given by:


PoE: Screw Motions in Different Order (2/2)- Tsb(θ1, θ2) = e[S

′2]θ2Tsb(θ1, 0)


PoE Example: 3R Spatial Open Chain

x0

y0

z0

x1

y1

z1

x2

y2z2

x3

y3

z3

θ1

θ2

θ3

L1

L2


Body Form of PoE Formula

• Fact: eM−1PM = M−1ePM ⇒ MeM

−1PM = ePM

• Let Bi be the screw axis of joint i expressed in end-effector frame when therobot is at zero position

• Then body-form of the PoE formula is:

Tsb(θ1, . . . , θn) = Me[B1]θ1e[B2]θ2 · · · e[Bn]θn

Body Form PoE Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 14 / 15

More Discussions

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Body Form PoE Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 15 / 15

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