Haptic interaction with rigid bodies by SPIDAR

1. Haptic interaction with rigid bodies by SPIDAR Shoichi Hasegawa Makoto Sato’s group Precision and Intelligence Lab. Tokyo Institute of Technology

2. Contents • Control of SPIDAR • Characteristics of SPIDAR • Position measuring, Force displaying • Update rate of haptic rendering • Rigid body simulation • Contact force modeling • Haptic rendering for 6DOF • Simulation of articulated body • Demo • For demonstration, please visit our booth.

3. Hardware of SPIDAR Motor and encoder Present force to user. Measure length of string. SPIDAR

4. Hardware performance • SPIDAR is the best device in performance • Stiff and light

5. 7DOF 8Strings 3DOF 4Strings 6DOF 8Strings Reconfigurable hardware • Any DOF and arrangements are designable. • Same control algorithm can be used

6. q1 l1 q2 p1 l2 p2 r qm pm lm p3 l3 q3 Position measurement r : posture vector (x,y,z,qx , qy , qz ) pi :tied point of string on the grip qi :position of a motor li:length of string

7. Jp# Jr# Position measurement Solve r by iterative method

8. Displaying force • Simple solution t2 f2 t1 f1 tensions f t3 Directions of strings f3

9. Displaying force • Discontinuous problem Tension Present the same force and move the pointer. Position

10. Displaying force • Simple solution • Use smaller tension • Limitation • Finally Quadratic programming problem

11. λ =0 Displaying force 20 Full presentation area Partial presentation area λ =0.1 Calculated tension [N] 10 λ =0.2 λ =1.0 0 (-0.2,-0.2,-0.2) (0,0,0) (0.2,0.2,0.2) Position of end effecter [m]

12. 6 λ= 0 λ= 0.1 λ= 0.2 4 λ= 0.3 Expanded figure x component of presentation force [N] 2 0 -0.1 0 0.1 0.2 0.3 x coordinate of end effecter [m] Displaying force

13. Haptic rendering and update rate Virtual World Force display 1 2 3 F=kx • Measure finger position • Collision detection and force calculation • Display the force

14. Finger position Out of the object In the object time F=kx penetration Stiff objects (large k) make too much force Haptic rendering and update rate • Problem on slow update rate

15. Finger position Stable contact Out of the object In the object time • Stiff object requires fast update. • It is commonly said that 1kHz or more is needed. Haptic rendering and update rate • Solution by fast update rate

16. 0.2 0.2 0.1 0.1 0 0 -0.2 -0.1 0 0.1 -0.2 -0.1 0 0.1 Effect of fast update • Advantage of stiffness • Display of friction disturbs display of shapes. But, enough stiffness realizes both. Virtual object (gear) Virtual object (gear) 2N/mm, 1kHz 20N/mm, 10kHz Trajectory of the haptic pointer on surfaces with friction (m=0.5)

17. Real-time Rigid Body Simulationfor Haptic Interactions

18. Haptic interaction • Touch the virtual world • User feels contact force from haptic interface Virtual Real

19. Haptic interaction • Touch the virtual world • User feels contact force from haptic interface • The touched object receives force from the user. • The response : Dynamics Virtual Real

20. Video Darumaw

21. Contact force F=mv N=Iw v(t+D t)= v(t) + F/mDt w(t+D t)=w(t)+ I-1NDt Haptic pointer Contact force fordynamics simulation Contact forceto feedback user

22. Contact model Normal force • Normal force • Prevent penetration • Friction force (coulomb’s model) • Static friction • Prevent sliding motion • Dynamic friction • Proportional to normal force Friction force |ff |< |m0fn | |ff |= |mfn |

23. (eq. of motion) (normal) (friction) Solving constraints(1) • Analytical method • David Baraff SIGGRAPH `89 … Drawbacks • Much computation time for one step. O(n3 ) • A virtual coupling is needed to connect a haptic interface. • Coulomb's friction model comes to NP complete problem. Advantages • Object motions are stable.Wide time steps are affordable. • Solves constraints accurately.Completely rigid.

24. Solving constraints(2) • Penalty method . penetration d，penetrating velocityd Spring Damper . slide l，sliding velocityl Spring Damper Drawbacks • Stability and rigidity requires small time steps.(Haptic interfaces also need this.) • Treatment of large contact area makes instability or takes a lot of computation time. Advantages • Very fast for one step. O(n) • Direct connection to haptic interfaces. • Coulomb’s friction model is easily realized. • Integration of other models are easy. (e.g. Featherstone’s method)

25. ? Problem on large contact area • Where should we put spring-damper model? On the most penetrating point

26. Problem on large contact area • Where should we put spring-damper model? ? On vertices Top view

27. Problem on large contact area • Where should we put spring-damper model? Will works well. But, it will takes muchcomputation time and memory. ? Many points

28. ! Distributed model Proposal for the problem • Integrate forces from distributed model for each triangle.

29. Steps • Finding Contact force: • Find contact point and normal. • Find the shape of the contact volume. • Integrate forces over the contact area.

30. Contact detection • Gilbert, Johnson, and Keerthi (GJK) algorithm. • Find closest points of two convex shapes. • A complex shape can be represented by a set of convex shapes. • After the contact, GJK can’t find closest points, So… t=t0 t=t1 New closest points

31. Contact Analysis • Contact part = Intersection of two convexes. • D. E. Muller and F.P.Preparata:“Finding the intersection of two convex” (1978) • For given two convex and a point in the intersection. • Find the intersection.

32. Contact Analysis(2) • Finding the intersection of two convex Half space representation

33. Contact Analysis(3) • Finding the intersection of two convex(2) Half space representation Dual transform Vertex of intersection Dual transform Quick hull

34. Integration of force • Penalty force • Dynamic friction force • Maximum static friction force Integrate forces from distributed model for each triangle.

35. p2 p3 h3 h2 p1 Integration for a triangle p3 Force from spring model: Torque from from spring model:

36. = Distributed model = Static friction force • Spring-damper model for sliding constraint. Two (translation and rotation) models

37. Evaluation • Compare three simulators • Proposed • Penalty method • distributed model. • Point based • Penalty method • A model on the most penetrating point. • Analytic • Analytical method • Open Dynamics Engine (Smith R. 2000)

38. 60 Proposed simulator 50 Point based method Analytical method 40 30 3 Average computation time[ms] 20 10 0 0 5 10 15 5 Number of blocks 13 Computation time

39. angular momentum 2m3 g=9.8m/s 0.1rad Stability on normal force • A cube on a floor. • Measure angular momentum. 1 Proposed method Point based penalty method Analytical method Angular momentum [Nms] 0 -1 0 1 2 time[s]

40. Motion of top

41. Cardboard floor m0=0.265 m =0.160 2.0kg weightwrapped by paper spring 400N/m Tension force Velocity 0.0642m/s Friction force Stick-slip motion • State transition between static and dynamic friction makes stick-slip motion. m0=0.265 m =0.160 2kg 0.0642m/s Spring 400N/m friction force

42. friction force friction force friction force friction force position of object position of object position of object position of object Result Real world Analytical simulator 0.1 0.1 5 5 Force[N] Position[m] Position[m] Force[N] 0 0 -5 -5 0 0 0 1 2 0 1 2 Time[s] Time[s] Proposed simulator Point based simulator 0.1 0.1 5 5 Position[m] Position[m] Force[N] Force[N] 0 0 -5 -5 0 0 0 1 2 0 1 2 Time[s] Time[s]

43. Both hand 6DOF haptic interaction

44. A B The computation time for the Simulation 3 B D A C C 2 Force integration Computation time [ms] 1 Contact volume analysis Contact detection D Simulation iterations Update rate of the simulation • Update rate > 300Hz

45. Articulated body • Constraints of joints should strictly be kept. • Small errors can be magnified by mechanisms. • To solve generic constraints, we need much computation time. But, • Efficient computation method is affordable under some restriction of structure. • Restriction: Tree structure without ring many rings one ring

46. 3 D C 2 B 1 A Generic method A’s eq of motion： Constraint of 1： B’s・・・・・・・・・・・・・・・・ So, we can write: Many 0，sparse matrix: There can be fast calculation method.

47. D D 3 3 C C 2 2 B B 1 1 A A Featherstone’s method • Suppose whole rigid bodies are one rigid body. Then find the acceleration. Find the acceleration of A • Suppose two rigid bodies of A and others. • Find the force applied to the joint 1 • Find the acceleration of B • Suppose three rigid bodies of A and B and others.…

48. 3 D C 2 B 1 A Featherstone’s method • Look the method from a viewpoint of the sparse matrix: 1. From the leaves to the root we merge M 2.Find acceleration .

49. Featherstone’s method 3.Find from the root to the leaves.

50. Demo