Robot Geometry and Kinematics

Transcription

1 CIS 68/MEAM 50 Robot Geometr and Kinematics CIS 68/MEAM 50 Outline Industrial (conventional) robot arms Basic definitions for understanding -D geometr, kinematics Eamples Classification b geometr Relationship between geometr and functionalit Forward and inverse kinematics Implications for control Other tpes of robots Parallel manipulators End effectors See Notes: Slides available from website

2 Degrees of freedom of a sstem CIS 68/MEAM 50 Definitions The number of independent variables (or coordinates) required to completel specif the configuration of the sstem. Point on a plane Point in -D space Line on a plane Kinematic chain A sstem of rigid bodies connected together b joints. A chain is called closed if it forms a closed loop. A chain that is not closed is called an open chain. Serial chain planar links connected b a pin joint Human shoulder Car If each link of an open chain ecept the first and the last link is connected to two other links it is called a serial chain. Joints CIS 68/MEAM 50 Definitions (continued) Joints are connections between links Revolute, rotar, or pin joint (R) Prismatic, sliding, or telescoping joint (P) Helical or screw joint (H) Spherical or ball joint (S) Planar kinematic chain All the links are constrained to move in or parallel to the same plane. A planar chain can onl allow prismatic and revolute joints. In fact, the aes of the revolute joints must be perpendicular to the plane of the chain while the aes of the prismatic joints must be parallel to or lie in the plane of the chain. Connectivit of a joint The number of degrees of freedom of a rigid bod connected to a fied rigid bod through the joint. 4

3 Planar kinematic chain CIS 68/MEAM 50 The Planar -R manipulator All joints are revolute with connectivit = What is the number of degrees of freedom? END-EFFECTOR R Link R Link Joint Link R Joint Joint ACTUATORS 5 CIS 68/MEAM 50 Connectivit, Mobilit, and Degrees of Freedom Connectivit of a joint The number of degrees of freedom of a rigid bod connected to a fied rigid bod through the joint spherical joint () revolute joint () Mobilit of a chain Number of degrees of freedom of the chain Serial chain Eamples -R chain M= M = n f i i= prismatic joint () helical joint () 6

4 CIS 68/MEAM 50 Connectivit, Mobilit, and Degrees of Freedom Mobilit of a chain Number of degrees of freedom of the chain General epression M = 6 g ( n g) + i= f i n number of moving links g number of joints between the links f i connectivit of joint i Special case of planar geometr M = g ( n g) + i= f i 7 CIS 68/MEAM 50 Eamples: Mobilit (Degrees of Freedom) The Adept 850 Palletizer The G65 Gantr robot manipulator (CRS Robotics) M = 4 M = 8

5 CIS 68/MEAM 50 Eamples: Mobilit (Degrees of Freedom) Planar serial chain number of moving links, n = number of joints, g = connectivit, f i = Planar parallel manipulator number of moving links, n = 7 number of joints, g = 9 connectivit, f i = 4 END-EFFECTOR END-EFFECTOR 7 6 R R P ACTUATORS ACTUATORS 5 M = g ( n g) + i= f i 9 CIS 68/MEAM 50 Eamples: Mobilit (Degrees of Freedom) Ingersoll Rand machine tool (Stewart Platform) number of moving links, n = 9 number of joints, g = 4 connectivit, f i = LEG END EFFECTOR LEG 4 END EFFECTOR S Leg 5 Leg 4 LEG Leg 6 LEG 5 P LEG Leg Leg Leg LEG 6 R R M = 6 g ( n g) + i= f i BASE BASE ( 9 4) M = 6 = LEG NO. i 0

6 CIS 68/MEAM 50 Kinematic modeling Link Actuated joint End effector (EE) Reference point on EE (,) l θ Link (EE) l Joint coordinates Link θ, θ, θ End effector coordinates,, Link lengths (l i ) l θ θ Link CIS 68/MEAM 50 Kinematic modeling Link Actuated joint End effector (EE) Reference point on EE d (, ) Link (EE) Joint coordinates θ, d End effector coordinates,, Link θ RP manipulator

7 CIS 68/MEAM 50 Kinematic transformations Direct kinematics Joint coordinates to end effector coordinates Wh is it useful? Inverse kinematics End effector coordinates Wh is it useful? Direct kinematics CIS 68/MEAM 50 Kinematic transformations Joint coordinates to end effector coordinates Sensors are located at the joints. DK algorithm is used to figure out where the robot is in -D space. Robot thinks in joint coordinates. Programmer/engineer thinks in world coordinates or end effector coordinates. Inverse kinematics End effector coordinates Given a desired position and orientation of the EE, we want to be able to get the robot to move to the desired goal. IK algorithm used to obtain the joint coordinates. Essential for control. 4

8 CIS 68/MEAM 50 Direct kinematics Transform joint coordinates to end effector coordinates (,) l θ l l θ θ = = = l cos θ + l cos( θ + θ) + l cos( θ + θ + θ) l sin θ + l sin( θ + θ ) + l sin( θ + θ + θ) ( θ + θ + θ ) 5 CIS 68/MEAM 50 Direct kinematics Transform joint coordinates to end effector coordinates = d = d = θ cosθ sin θ d (, ) θ RP manipulator 6

9 CIS 68/MEAM 50 Direct kinematics Transform joint coordinates to end effector coordinates = l = = l z = d cosθ sin θ ( θ + θ + θ ) + l + l 4 cos sin ( θ + θ) + l4 cos( θ + θ + θ4) ( θ + θ ) + l sin( θ + θ + θ ) 4 4 Adept palletizer: PRRR manipulator 7 CIS 68/MEAM 50 Direct kinematics Kinematics of trul -D manipulators is difficult PUMA 560 Cincinnati Milacron T- Wh? 8

10 CIS 68/MEAM 50 Inverse kinematics Transform end effector coordinates to joint coordinates (,) l θ l l θ θ = = = l cos θ + l cos( θ + θ) + l cos( θ + θ + θ) l sin θ + l sin( θ + θ ) + l sin( θ + θ + θ) ( θ + θ + θ ) Given,,, solve for θ, θ, θ 9 CIS 68/MEAM 50 Inverse kinematics Transform end effector coordinates to joint coordinates d (, ) = d = d = θ cosθ sin θ θ Given,, solve for θ, d 0

11 CIS 68/MEAM 50 Inverse kinematics = d = d cos θ sinθ (, ) d Given,, solve for θ, θ Solution no. d = + θ = tan + θ Solution no. d = + θ = tan CIS 68/MEAM 50 Inverse kinematics = d = d Given,, solve for d, d d Direct and inverse kinematics are trivial Onl one solution Equations are linear No trigonometric functions Popular geometr CNC machines Gantr robots Plotters, special-purpose transfer devices d

12 CIS 68/MEAM 50 Inverse kinematics is more difficult! Solving nonlinear equations Equations can be ver complicated Equations involve trigonometric functions Trig functions are transcendental Inverse trig functions have multiple solutions tan( ) = tan( + kπ), k = K,,0,,,K ( ) sin = 0,50 cos ( ) = 0, 0 We use the atan function to solve inverse kinematics problems. Eample s = sin, s = Find Define atan function with two arguments, atan c = c = cos (, ) = 0 CIS 68/MEAM 50 Inverse kinematics for the R manipulator (,) l θ θ l θ l = l cos θ + l cos( θ + θ) + l cos( θ + θ + θ) = l sin θ + l sin( θ + θ ) + l sin( θ + θ + θ) = ( θ + θ + θ ) Given,,, solve for θ, θ, θ 4

13 CIS 68/MEAM 50 Inverse kinematics: Step (,) l θ = = = l cos θ + l cos( θ + θ ) + l cos( θ + θ + θ ) l sinθ + l sin( θ + θ ) + l sin( θ + θ + θ ) ( θ + θ + θ ) l Given,,, solve for θ, θ, θ l θ = l cos θ + l = l sinθ + l cos sin ( θ + θ ) + l cos( θ + θ + θ ) ( θ + θ ) + l sin( θ + θ + θ ) θ = ( θ + θ + ) θ l cos = l cos θ + l l sin = l sinθ + l ( θ + θ ) ( θ + θ ) Two equations in two unknowns - θ, θ cos sin 5 CIS 68/MEAM 50 Inverse kinematics: Step (,) l cos = l cosθ + l l sin = l sin θ + l cos sin ( θ + θ ) ( θ + θ ) l θ Rename the left hand sides, = - l cos, = - l sin. l Given,, solve for θ, θ. l θ θ ( l cosθ ) = ( l cos( θ + θ )) + ( l sin θ ) = ( l sin( θ + θ )) ( l ) cosθ + ( l ) sinθ + ( + + l l ) 0 = One equation in one unknown - θ 6

14 CIS 68/MEAM 50 Inverse kinematics: Step Solve for θ ( l ) cos θ + ( l ) sinθ + ( + + l l ) 0 (,) = One equation in one unknown - θ l θ l How do we solve equations of the tpe l θ Pcos α + Qsin α + R = 0 θ 7 CIS 68/MEAM 50 Solutions to the equation Pcos α + Qsin α + R = 0 Define γ so that γ = a tan + Q + Q so that we can rewrite the equation in the form: P Q, P cos γ cos α + sin γsin α + P P R + Q = 0 cos ( α γ) = P R + Q α = γ + σ cos two solutions P R + Q, σ = ± 8

15 CIS 68/MEAM 50 Inverse kinematics: Step ( l ) cos θ + ( l ) sin θ + ( + + l l ) 0 = P Q R (,) l γ = a tan Q P + Q, P P + Q θ l l θ θ θ = γ + σ cos ( + + l l ) l + 9 CIS 68/MEAM 50 Inverse kinematics Step 4: Find θ and θ θ = γ + σcos (,) ( + + l l ) l + l l θ l cos = l cosθ + l cos l sin = l sinθ + l sin ( θ + θ ) ( θ + θ ) l θ l θ = atan sin θ l, cos θ θ l l θ θ = ( θ + θ ) 0

16 CIS 68/MEAM 50 Inverse Kinematics There are two solutions to the inverse kinematics of a R manipulator REFERENC E (,) σ =+ σ =- Ecept at the so-called singular points. CIS 68/MEAM 50 Some unresolved issues Serial chain geometr Direct Kinematics Geometric modeling is difficult Kinematic modeling is relativel simple Inverse Kinematics Difficult Nonlinear problem with multiple solutions Parallel chain geometr Order of magnitude more difficult than inverse kinematics for serial chains Inverse kinematics for each chain Comparable to serial chains (usuall simpler)