Lecture Note 6: Forward Kinematics

Transcription

1 ECE5463: Introduction to Robotics Lecture Note 6: Forward Kinematics Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018 Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 1 / 15

2 Outline Background Illustrating Example Product of Exponential Formula Body Form of the PoE Formula Outline Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 2 / 15

3 Kinematics U (a) (b) S R R R P R R R Product of Exponentials Formula (c) (d) Figure 2.16: Mobile manipulator. S5 re 2.21: A second collection of spatial parallel mechanisms. think of the wheeled base S4 as the same as the rolling coin in Figure 2.11 the it move? wheel (and base) can spin together about an axis perpendicular to the ground, and the wheel rolls without slipping. The base always remains horizontal. (Left 5 The mechanisms of Figures 2.24(a) and 2.24(b). S6 S1 unstated are the means to keep the base horizontal and to spin the wheel and chanism of Figure 2.24(a) consists of six identical squares arranged gle closed loop, connected by revolute joints. The bottom base square about an axis perpendicular S3to the ground.) to ground. Determine the number of degrees of freedom(a) using Ignoring the multi-fingered hand, describe the configuration space of the s formula. mobile manipulator. echanism of Figure 2.24(b) also consists of six identical squares (b) Now suppose that the robot hand rigidly grasps a refrigerator door handle ed by revolute joints, but arranged differently (as shown). Detere number of degrees of freedom using Grübler s formula. Does and, with its wheel and base completely S2 stationary, opens the door using only its arm. With the door open, how many degrees of freedom does the Kinematics is a branch Figure 4.6: of (Left) classical Universal mechanism Robots mechanics formed UR5 6R by robot the that arm. and (Right) describes openshown door have? at itsthe zero motion of of Modern Robotics, Lynch and Park, Cambridge U. Press, position Positive rotations (c) A second about the identical axes indicated mobile manipulator are given by the comes usualalong, right-hand and after parking its points, bodies (objects), rule. W1 is theand distance systems along the ŷ mobile base, s -direction of alsobodies between rigidly grasps (groups the anti-parallel the refrigerator of axes objects) of joints door handle. without How many 1 and 5. W1 = 109 mm, W2 = 82 mm, L1 = 425 mm, L2 = 392 mm, H1 = 89 mm, degrees of freedom does the mechanism formed by the two arms and the considering the mass H2 = 95of mm. each or the forces that caused the motion open refrigerator door have? since e 0 = I. Evaluating, Exercise we 2.7getThree identical SRS open-chain arms are grasping a common object, as shown in Figure (a) 0 Find 1 the number of degrees of freedom 0 1 of 0 this Background Lecture 6 (ECE5463 Sp18) system. Wei Zhang(OSU) 3 / 15

4 Chapter 4. Forward Kinematics 151 Kinematics of Roboticẑb Manipulator L 3 = 60 mm L 2 = 300 mm L 1 = 550 mm ˆxb ˆxs ẑs Wrist J5,J6,J7 Forward Kinematics: calculation of the position W = 45 mm p and 1 orientation R of the end-effector frame Elbow J4 from its joint variables θ = (θ 1,..., θ n ) Shoulder J1,J2,J3 Two commonly adopted approaches Figure 4.8: Barrett Technology s WAM 7R robot arm at its zero configuration (right). At the zero configuration, axes 1, 3, 5, and 7 are along ẑs and axes 2, 4, and 6 are aligned with ŷ s out of the page. Positive rotations are given by the right-hand rule. Axes 1, 2, and 3 intersect at the origin of {s} and axes 5, 6, and 7 intersect at a point 60mm from {b}. The zero configuration is singular, as discussed in Section 5.3. Approach 1: Assign a frame at each link, typically at the joint axis, then calculate the end-effector configuration through intermediate frames. - Denavit-Hartenberg parameters: most widely adopted convention for frame Also, some joints of the WAM are driven by motors placed at the base of the robot, reducing the robot s moving mass. Torques are transferred from the motors to the joints by cables winding around drums at the joints and motors. assignment Because the moving mass is reduced, the motor torque requirements are decreased, allowing low (cable) gear ratios and high speeds. This design is in contrast with that of the UR5, where the motor and harmonic drive gearing for each joint are directly at the joint. Figure 4.8 illustrates the WAM s end-effector frame screw axes B1,..., B7 when the robot is at its zero position. The end-effector frame {b} in the zero Approach 2: Product of Exponential (PoE) formula: directly compute the end-effector frame configuration relative to the fixed frame through screw May 2017 preprint of Modern Robotics, Lynch and Park, Cambridge U. Press, motion along the screw axis associated with each joint. Background Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 4 / 15

5 Forward Kinematics: Basic Setup Suppose that the robot has n joints and n links. Each joint has one degree of freedom 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 T sb (θ 1,..., θ n ) Background Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 5 / 15

6 A Simple Example: Approach 1 (D-H Parameters) {4} φ {3} L 3 θ3 (x, y) {2} L 2 {1} L 1 θ2 {0} θ1.1: Forward kinematics of a 3R planar open chain. For each frame, the ˆxs is shown; the ẑ-axes are parallel and out of the page. cos θ1 sin θ1 0 0 sin θ1 cos θ , T12 = cos θ3 sin θ3 0 L2 sin θ3 cos θ , T34 = cos θ2 sin θ2 0 L1 sin θ2 cos θ L ,. (4.5) that T34 is constant and that each remaining Ti 1,i depends only on variable θi. alternative to this approach, let us define M to be the position and on of frame {4} when all joint angles are set to zero (the home or osition of the robot). Then Illustrating 1 0 Example 0 L1 + L2 + L3 Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 6 / 15

7 A Simple Example: Approach 2 (PoE Formula) {4} φ {3} L 3 θ3 (x, y) {2} L 2 {1} L 1 θ2 {0} θ1.1: Forward kinematics of a 3R planar open chain. For each frame, the ˆxs is shown; the ẑ-axes are parallel and out of the page. cos θ1 sin θ1 0 0 sin θ1 cos θ , T12 = cos θ3 sin θ3 0 L2 sin θ3 cos θ , T34 = cos θ2 sin θ2 0 L1 sin θ2 cos θ L ,. (4.5) that T34 is constant and that each remaining Ti 1,i depends only on variable θi. alternative to this approach, let us define M to be the position and on of frame {4} when all joint angles are set to zero (the home or osition of the robot). Then Illustrating 1 0 Example 0 L1 + L2 + L3 Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 7 / 15

8 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

9 Product of Exponential: Main Idea Chapter 4. Forward Kinematics 141 Goal: Derive T sb (θ 1,..., θ n) θn θn 1 θn 2 θ1 First choose a zero position ( home position): θ 1 = θ 0 1,..., θ n = θ 0 n such that M T sb (θ 0 1,..., θ 0 n) can be easily found. M e [Sn]θn M Without loss of generality, suppose home position is given by θ 0 i = 0, i = 1,..., n. e [Sn 1]θn 1 e [Sn]θn M e [Sn 2]θn 2 e [Sn 1]θn 1 e [Sn]θn M Figure 4.2: Illustration of the PoE formula for an n-link spatial open chain. Let S 1,..., S n be the screw axes expressed in {s}, corresponding to the joint motions when the robot is at its home position Apply screw motion to joint n: T sb (0,..., 0, θ n) = e [Sn]θn M Apply screw motion to joint n 1 to obtain: T sb (0,..., 0, θ n 1, θ n) = e [S n 1]θ n 1 e [Sn]θn M After n screw motions, the overall forward kinematics: T sb (θ 1,..., θ n) = e [S 1]θ 1 e [S 2]θ2 e [Sn]θn M interested in the kinematics, or (2) refer to {s} as frame {0}, use frames {i} fo i = 1,..., n (the frames for links i at joints i), and use one more frame {n + 1} (corresponding to {b}) at the end-effector. The frame {n + 1} (i.e., {b}) i fixed relative to {n}, but it is at a more convenient location to represent the configuration of the end-effector. In some cases we dispense with frame {n + 1} and simply refer to {n} as the end-effector frame {b} First Formulation: Screw Axes in the Base Frame The key concept behind the PoE formula is to regard each joint as applying a screw motion to all the outward links. To illustrate this consider a genera spatial open chain like the one shown in Figure 4.2, consisting of n one-dof joint that are connected serially. To apply the PoE formula, you must choose a fixed base frame {s} and an end-effector frame {b} attached to the last link. Place the robot in its zero position by setting all joint values to zero, with the direction of positive displacement (rotation for revolute joints, translation for prismatic joints) for each joint specified. Let M SE(3) denote the configuration of the end-effector frame relative to the fixed base frame when the robot is in its zero position. Now suppose that joint n is displaced to some joint value θn. The end May 2017 preprint of Modern Robotics, Lynch and Park, Cambridge U. Press, PoE Formula Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 9 / 15

10 Forward Kinematics: Steps for PoE Chapter 4. Forward Kinematics 14 Goal: Derive T sb (θ 1,..., θ n ) M θn θn 1 θn 2 θ1 Step 1: Find the configuration of {b} at the home position M = T sb (0,..., 0) e [Sn]θn M e [Sn 1]θn 1 e [Sn]θn M e [Sn 2]θn 2 e [Sn 1]θn 1 e [Sn]θn M Figure 4.2: Illustration of the PoE formula for an n-link spatial open chain. Step 2: Find screw axes S 1,..., S n expressed in {s}, corresponding to the joint motions when the robot is at its home position Forward kinematics: interested in the kinematics, or (2) refer to {s} as frame {0}, use frames {i} fo i = 1,..., n (the frames for links i at joints i), and use one more frame {n + 1 (corresponding to {b}) at the end-effector. The frame {n + 1} (i.e., {b}) fixed relative to {n}, but it is at a more convenient location to represent th configuration of the end-effector. In some cases we dispense with frame {n + 1 and simply refer to {n} as the end-effector frame {b} First Formulation: Screw Axes in the Base Frame T sb (θ 1,..., θ n ) = e [S1]θ1 e [S2]θ2 e [Sn]θn M The key concept behind the PoE formula is to regard each joint as applyin a screw motion to all the outward links. To illustrate this consider a genera spatial open chain like the one shown in Figure 4.2, consisting of n one-dof joint that are connected serially. To apply the PoE formula, you must choose a fixe base frame {s} and an end-effector frame {b} attached to the last link. Place th robot in its zero position by setting all joint values to zero, with the directio of positive displacement (rotation for revolute joints, translation for prismat joints) for each joint specified. Let M SE(3) denote the configuration of th PoE Formula Lecture 6 (ECE5463 end-effector Sp18) frame relative to the fixed base frame Weiwhen Zhang(OSU) the robot is10 in / its 15zer

11 PoE: Screw Motions in Different Order (1/2) PoE was obtained by applying screw motions along screw axes S n, S n 1,.... What happens if the order is changed? For simplicity, assume that n = 2, and let us apply screw motion along S 1 first: - T sb (θ 1, 0) = e [S 1]θ 1 M - Now screw axis for joint 2 has been changed. The new axis S 2 is given by: PoE Formula Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 11 / 15

12 PoE: Screw Motions in Different Order (2/2) - T sb (θ 1, θ 2) = e [S 2 ]θ 2 T sb (θ 1, 0) PoE Formula Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 12 / 15

13 Chapter 4. Forward Kinematics PoE Example: 3R Spatial Open Chain θ1 ẑ1 ẑ0 ŷ 1 L1 ŷ 0 ˆx0 ˆx1 ẑ2 ˆx2 ŷ 2 θ2 L2 ŷ 3 ẑ3 ˆx3 θ3 Figure 4.3: A 3R spatial open chain Examples We now derive the forward kinematics for some common spatial open ch using the PoE formula. Example 4.1 (3R spatial open chain). Consider the 3R open chain of ure 4.3, shown in its home position (all joint variables set equal to zero). Ch the fixed frame {0} and end-effector frame {3} as indicated in the figure, and press all vectors and homogeneous transformations in terms of the fixed fr The forward kinematics has the form T (θ) = e [S1]θ1 e [S2]θ2 e [S3]θ3 M, where M SE(3) is the end-effector frame configuration when the robot its zero position. By inspection M can be obtained as L1 M = L2. PoE Formula Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 13 / 15

14 Body Form of PoE Formula Fact: e M 1 P M = M 1 e P M Me M 1 P M = e P M Let B i be the screw axis of joint i expressed in end-effector frame when the robot is at zero position Then body-form of the PoE formula is: T sb (θ 1,..., θ n ) = Me [B1]θ1 e [B2]θ2 e [Bn]θn Body Form PoE Lecture 6 (ECE5463 Sp18) Wei Zhang(OSU) 14 / 15

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