Lecture Note 2: Configuration Space

Transcription

1 ECE5463: Introduction to Robotics Lecture Note 2: Configuration Space Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018 Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 1 / 19

2 Outline Mechanical Structure of a Robot Configuration Space Representation of Configuration Space Outline Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 2 / 19

3 Chapter 4. Forward Kinematics 151 Typical Mechanical 2 Structure ẑb L 3 = 60 mm L 2 = 300 mm ˆxb Wrist J5,J6,J7 W 1 = 45 mm Elbow J4 L 1 = 550 mm ˆxs ẑs Shoulder J1,J2,J3 (a) An open-chain industrial manipulator, (b) Stewart Gough platform. Closed visualized in V-REP [154]. loops are formed from the base platform, through the legs, through the top 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 links, to aligned with ŷ s each other using various types of joints. out of the page. Positive rotations are given by the right-hand rule. platform, and through the legs back to Axes 1, 2, and 3 intersect at the origin of {s} and axes 5, 6, and 7 intersect at a point the base platform. 60mm from {b}. The zero configuration is singular, as discussed in Section 5.3. A robot is mechanically constructed by connecting a set of bodies, called Links are usually modeled as rigid bodies Figure 1.1: Open-chain and closed-chain robot mechanisms. 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 electric winding motors, around drums these at would the joints ideally and be lightweight, operate at relatively low rotational gear ratios speeds and high (e.g., speeds. in the Thisrange designof is hundreds of RPM), and be able to generate motors. Because the moving mass is reduced, the motor torque requirements are decreased, allowing low (cable) thereby causing motion in contrast with that of the UR5, large whereforces of the motor and the andtorques. robot harmonic drive Since gearing most currently available motors operate at low for each joint are directly at the joint. torques and at up to thousands of RPM, speed reduction and torque amplification Theare end-effector required. frame Examples {b} in the zero of such transmissions or transformers include Figure 4.8 illustrates the WAM s end-effector frame screw axes B1,..., B7 when the robot is at its zero position. gears, cable drives, belts and pulleys, and chains and sprockets. These speedreduction devices should have zero or low slippage and backlash (defined as the amount of rotation available at the output of the speed-reduction device Mechanical Structure without motion atlecture the input). 2 (ECE5463 Brakes Sp18) may also be attached to stop Wei Zhang(OSU) the robot 3 / 19 Actuators, such as electric motors, deliver forces and torques to the joints, End-effector, such as gripper or hand, is attached to a specific link

4 Degrees of Freedom of a Robot Typical Joints Revolute Joint (R): Figure 2.3: Typical robot joints. Cylindrical Joint (C): a formula, called Grübler s formula, for determining the number of degrees of freedom of planar and spatial robots. Prismatic Joint (P): Robot Joints Figure 2.3 illustrates the basic joints found in typical robots. Every joint connects exactly two links; joints that simultaneously Spherical connect three Joint or(s): more links are not allowed. The revolute joint (R), also called a hinge joint, allows rotational motion about the joint axis. The prismatic joint (P), also called a sliding or linear joint, allows translational (or rectilinear) motion along the direction of the joint axis. The helical joint (H), also called a screw joint, allows Helical Joint (H): Universal Joint (U): Mechanical Structure Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 4 / 19

5 Outline Mechanical Structure of a Robot Configuration Space Representation of Configuration Space Configuration Space Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 5 / 19

6 Configuration Space Definitions Configuration: a complete specification of the position of every point of the robot. Degree of Freedom (dof): The minimum number of real-valued coordinates needed to represent the configuration Configuration Space (C-space): The space (set) that contains all possible configurations of the robot. Effective representation of the C-space is essential for many aspects of robotics Configuration Space Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 6 / 19

7 (x A, y A ), (x B, y B ), and (x C, y C ). If the points could be placed independently anywhere in the plane, the coin would have six degrees of freedom two for each of the three points. But, according to the definition of a rigid body, the distance Configuration Space Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 7 / 19 How to find the dof? Chapter 2. Configuration Space 13 Example: coin on a table C ŷ d AC C ẑ A A d AB d BC B B ˆx ˆx (a) (b) (c) ŷ Figure 2.2: (a) Choosing three points fixed to the coin. (b) Once the location of A is chosen, B must lie on a circle of radius d AB centered at A. Once the location of B is chosen, C must lie at the intersection of circles centered at A and B. Only one of these two intersections corresponds to the heads up configuration. (c) The configuration of a coin in three-dimensional space is given by the three coordinates of A, two angles to the point B on the sphere of radius d AB centered at A, and one angle to the point C on the circle defined by the intersection of the a sphere centered at A and a sphere centered at B.

8 DoF of Planar and Spatial Rigid Body Chapter 2. Configuration Chapter 2. Configuration Space Space ẑ ŷ ẑ C ŷ d AC C C d AC C A A d BC d A AB A d BC B B d AB ˆx ˆx ŷ B B (a) (b) ˆx (c) ˆx ŷ Figure 2.2: (a) Choosing three points fixed to the coin. (b) Once the location of A is (a) chosen, B must lie on a (b) circle of radius d AB centered at A. Once(c) the location of B is chosen, C must lie at the intersection of circles centered at A and B. Only one of these two intersections corresponds to the heads up configuration. (c) The configuration Figure 2.2: (a) of achoosing coin three-dimensional three pointsspace fixedisto given theby coin. the three (b) coordinates Once the of location A, two angles of A is chosen, B mustolie theon point a circle B on the of sphere radiusofdradius AB centered d AB centered at A. at A, Once and one theangle location to the of point B is chosen, C mustc lie onat the the circle intersection defined by the ofintersection circles centered of the a at sphere A and centered B. Only at A and onea of sphere these centered at B. two intersections corresponds to the heads up configuration. (c) The configuration of a coin in three-dimensional space is given by the three coordinates of A, two angles to the point B (x on A, the y A ), sphere (x B, y B of ), and radius (x C d, y AB C ). centered If the points at A, could andbe one placed angle independently to the point C on the circleanywhere defined by in the theplane, intersection the coin would of thehave a sphere six degrees centered of freedom at A and two for a sphere each of the three points. But, according to the definition of a rigid body, the distance centered at B. between point A and point B, denoted d(a, B), is always constant regardless of where the coin is. Similarly, the distances d(b, C) and d(a, C) must be constant. The following equality constraints on the coordinates (x (x A, y A ), (x B, y B ), and (x C, y C ). If the points could be placed A, y A ), (x independently B, y B ), and (x C, y C ) must therefore always be satisfied: anywhere in the plane, the coin would have six degrees of freedom two for each of the three points. But, according d(a, B) = to (x the A definition x B ) 2 + (y A of ya B ) rigid 2 = d AB body,, the distance Configuration Space Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 8 / 19

9 DoF of Joints Joint can be viewed as providing freedoms to allow one rigid body to move relative to another. Dof of a joint: minimum # of variables needed to represent the configuration of a joint Joint can also be viewed as providing constraints on the possible motions of the two rigid bodies it connects Chapter 2. Configuration Space Degrees of Freedom of a Robot Figure 2.3: Typical robot joints. Constraints c Constraints c between two between two Joint type dof f planar spatial rigid bodies rigid bodies Revolute (R) Prismatic (P) Helical (H) 1 N/A 5 Cylindrical (C) 2 N/A 4 Universal (U) 2 N/A 4 Spherical (S) 3 N/A 3 Table 2.1: The number of degrees of freedom f and constraints c provided by co a formula, called Grübler s formula, for determining the number of degrees of Configuration Space Lecture joints. 2 (ECE5463 Sp18) Wei Zhang(OSU) 9 / 19

10 DoF of Mechanisms (Linkages) dof =(sum of freedoms of the bodies) number of independent constraints (1) Grübler s Formula: dof = m(n 1 J) + J i=1 f i Configuration Space Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 10 / 19

11 2.2. Degrees of Freedom of a Robot DoF Examples (a) Degrees of Freedom of a Robot (a) (a) (b) Figure 2.4: (a) Four-bar linkage. (b) Slider crank mechanism. Figure 2.4: (a) Four-bar linkage. (b) Slider crank mechanism. rübler s formula for the number of degrees of freedom of the robot is N = 4, J = 4, and f i = 1, i = 1,..., 4, into Grü four-bar linkage has one degree of freedom. formula for the number of degrees of freedom J of the robot is The slider crank closed-chain mechanism of F dof = m(n 1) c i two ways: (i) the mechanism consists of three rev }{{} i=1 joint (J = 4 and each f rigid body freedoms }{{ J i = 1) and four links ( } link), or (ii) the mechanism consists of two revol dof = m(n 1) joint constraints c i joint (the RP joint is a concatenation of a revolu }{{} J i=1 f i = 2) and three links (N = 3; remember tha rigid = m(n body freedoms 1) (m f i )}{{} two bodies). In both cases the mechanism has o i=1 joint constraints Example 2.4 (Some classical planar mechanism J J formula to several classical planar mechanisms. = m(n 1 J) + f = m(n 1) (m i. (2.4) f i ) of revolute joints in Figure 2.5(a) (called a kr i=1 has N = k + 1 links (k links plus ground), and J i=1 his formula holds only if all joint constraints are independent. If they are not J (b) (c) Figure 2.5: (a) k-link planar serial chain. (b) Five-b six-bar linkage. (d) Watt six-bar linkage. Configuration Space Lecture 2 (ECE5463 Sp18) May 2017 preprint of Modern Robotics, Wei Zhang(OSU) Lynch and Park, 11 Cambridge / 19

12 Chapter 2. DoF Examples Configuration Space Degrees of Freedom of a Robot (a) (b) Figure 2.6: A planar mechanism with two overlapping joints. joints are revolute, f i = 1 for all i. Therefore, Figure 2.7: (a) A parallelogram linkage. (b) The five-bar linkage in a regular a dof = 3((k + 1) 1 k) + k = k singular configuration. as expected. For the planar five-bar linkage of Figure 2.5(b), N = 5 (four links plus ground), J = 5, and since all joints are revolute, each f i = 1. Therefore, dof = 3(5 1 5) + 5 = 2. (ii) Alternatively, the lower-right revolute prismatic joint pair can be regard For the Stephenson six-bar linkage of Figure 2.5(c), we have N = 6, J = 7, and as a single two-dof joint. In this f i = 1 case for all i, the so thatnumber of links is N = 7, with sev revolute joints, and a single two-dof revolute prismatic dof = 3(6 1 7) + 7 pair. = 1. Substituting in Grübler s formula yields Finally, for the Watt six-bar linkage of Figure 2.5(d), we have N = 6, J = 7, and f i = 1 for all i, so that, like the Stephenson six-bar linkage, dof = 3(6 1 7) + 7 = 1. dof = 3(7 1 8) + 7(1) + 1(2) = 3. Example 2.5 (A planar mechanism with overlapping joints). The planar mechanism illustrated in Figure 2.6 has three links that meet at a single point on the right of the large link. Recalling that a joint by definition connects exactly Example 2.6 (Redundant constraints two links, the joint and at this singularities). point of intersection For shouldthe not beparallelogr Configuration Space Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) regarded 12 as / a 19

13 2.2. Degrees of Freedom of a Robot DoF Examples R R S S S R S Figure 2.8: The Delta robot. lta robot). The Delta robot of Figure 2.8 consists of two er one mobile, the upper one stationary connected by contains a parallelogram closed chain and consists of three spherical joints, and five links. Adding the two platforms, nks and J = 21 joints (nine revolute and 12 spherical). By of = 6( ) + 9(1) + 12(3) = 15. s of freedom, Configuration however, Space only three are visible at the Lecture end- 2 (ECE5463 Sp18) Wei Zhang(OSU) 13 / 19

14 Outline Mechanical Structure of a Robot Configuration Space Representation of Configuration Space Representation Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 14 / 19

15 Issues for Explicit Parameterization Representation of Euclidean space: choose reference frame and represent point as a vector Representation of curved space is more tricky than it appears Explicit parameterization uses the same number of coordinates as the space dimension (suffer from singularity) Sphere S 2 Representation Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 15 / 19

16 Topology Explicit parameterization of sphere (latitude/longitude) suffers from singularity because sphere and plane have different topologies. Roughly, two spaces are topologically equivalent if one can be continuously deformed into the other without cutting or gluing. Topologically distinct 1-d spaces: - circle: - line: - closed interval: Representation Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 16 / 19

17 Topology Configuration Space: Topology and Representation Examples of topologically different 2-d spaces system topology sample representation point on a plane E 2 R 2 ŷ latitude 90 (x, y) ˆx longitude spherical pendulum S 2 [ 180, 180 ) [ 90, 90 ] θ2 2π 0 0 2π θ1 2R robot arm T 2 =S 1 S 1 [0, 2π) [0, 2π) θ 2π... 0 rotating sliding knob E 1 S 1 R 1 [0, 2π)... ˆx Table 2.2: Four topologically different two-dimensional C-spaces and example coordinate Representation representations. In the latitude-longitude representation Lecture 2 (ECE5463 of the sphere, Sp18) the Wei Zhang(OSU) 17 / 19

18 Implicit Representation of C-Space Implicit Representation: View n-dim space as embedded in a higher dimensional Euclidean space subject to constraints. Use more coordinates than the minimum, but can avoid singularities Example: Sphere S 2 In this class, we will primarily use the implicit representation Representation Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 18 / 19

19 Summary Questions What is the configuration space (C-space) of a robot? What is dof of C-space, and how to find dof? What is topological equivalence? Pros and cons for explicit and implicit representation of C-space Further reading: chapter 2 of Lynch and Park. Representation Lecture 2 (ECE5463 Sp18) Wei Zhang(OSU) 19 / 19