1. Introduction to Robotics Lecture II Alfred Bruckstein Yaniv Altshuler

2. Denavit-Hartenberg Specialized description of articulated figures Each joint has only one degree of freedom rotate around its z-axis translate along its z-axis

3. Denavit-Hartenberg One degree of freedom : very compact notation Only four parameters to describe a relation between two links : link length link twist link offset link rotation

4. Denavit-Hartenberg Link length a i The perpendicular distance between the axes of joint i and joint i+1

5. Link twist α i The angle between the axes of joint i and joint i+1 Angle around x i -axis Denavit-Hartenberg

6. Link offset d i The distance between the origins of the coordinate frames attached to joint i and joint i+1 Measured along the axis of joint i Denavit-Hartenberg

7. Link rotation (joint angle) φ i The angle between the link lenghts α i-1 and α i Angle around z i -axis Denavit-Hartenberg

8. 1. Compute the link vector a i and the link length 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i How to compute the parameters to describe an articulated figure :

9. Denavit-Hartenberg 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1

10. Denavit-Hartenberg 1. Compute the link vector a i and the link length 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1 Let’s do it step by step

11. Denavit-Hartenberg The link length a i is the shortest distance between the joint axes joint i and joint i+1. Let the joint axes be given by the expression : Where p i is a point on axis of joint i and u i is one of its direction vectors (analogous for joint i+1 ).

12. Denavit-Hartenberg

13. There are three methods to compute the link vector a i and the link length

14. Denavit-Hartenberg The shortest distance a i is the length of the vector connecting the two axes, and perpendicular to both of them. Which can be expressed : Method 1 : The Pseudo-naive approach

15. Denavit-Hartenberg Let’s find the points o i and o ai where this distance exists.

16. Denavit-Hartenberg We can go some distance s from p i along axis i, and then the distance a i along the unit vector and finally some distance t along axis i+1 to arrive at point p i+1.

17. Denavit-Hartenberg Multiplying respectively by u i and u i+1, we obtain the two following equations:

18. Denavit-Hartenberg Solution :

19. Denavit-Hartenberg Finally, using and we obtain :

20. Denavit-Hartenberg We can go some distance s from point p i along the axis i, and then go some distance k along u i x u i+1. Finally go some distance t along the axis i+1 to arrive at point p i+1. The vector u i x u i+1 gives the perpendicular vector to both axes. Let’s find out where it is located on the joint axes. Method 2 : The Geometric approach

21. Denavit-Hartenberg We obtain the equation : There are three unknowns.

22. Denavit-Hartenberg Let’s first eliminate the unknown k from the equation : by multiplying by u i :

23. Denavit-Hartenberg by multiplying by u i+1 : Let’s first eliminate the unknown k from the equation :

24. Denavit-Hartenberg by multiplying by u i x u i+1 : Now we shall eliminate the s and t from the equation :

25. Denavit-Hartenberg We have obtained a system of three equations in the unknowns s, t, k :

26. Denavit-Hartenberg From, it can be seen that the shortest distance between joint i and joint i+1 is given by the vector : Where

27. Denavit-Hartenberg From and, we can compute s and t :

28. Denavit-Hartenberg Finally, using and we obtain :

29. Denavit-Hartenberg The distance between two arbitrary points located on the joint axes joint i and joint i+1 is : Method 3 : The Analytic approach

30. Denavit-Hartenberg The link length of link i, a i, is the minimum distance between the joint axes :

31. Denavit-Hartenberg A necessary condition is :

32. Denavit-Hartenberg Which is equivalent to their numerators being equal to 0 :

33. Denavit-Hartenberg Rewriting this system yields :

34. Denavit-Hartenberg Whose solution are :

35. Denavit-Hartenberg Finally, using and we obtain :

36. Denavit-Hartenberg o i and o ai are the closest points on the axes of joint i and joint i+1. We deduce that the link vector a i and the link length a i :

37. Denavit-Hartenberg The link vector a i :

38. Denavit-Hartenberg Calculating the scalar products and, both equal to 0, proves that the vector a i is perpendicular to both axes of joint i and joint i+1

39. Denavit-Hartenberg Three methods How do we actually compute a i and ||ai|| 2 ?

40. Denavit-Hartenberg The link vector a i is perpendicular to both of the axes of joint i and joint i+1. The unit vector : is parallel to the link vector a i.

41. Denavit-Hartenberg Given two points p i and p i+1 on the axes of joint i and joint i+1, the link length can be computed as : And the link vector :

42. Denavit-Hartenberg Special cases : The joint axes intersect The shortest distance a i is equal to zero The link vector is the null vector

43. Denavit-Hartenberg The joint axes are parallel There is no unique shortest distance o i can be chosen arbitrarily, so we should chose values that offset the most of Denavit-Hartenberg parameters

44. Denavit-Hartenberg The first joint There is no link preceding it We use a base link : link 0 Its link frame should coincide with the link frame of link 1 Most of the Denavit-Hartenberg parameters will be equal to zero

45. Denavit-Hartenberg The last joint There is no link succeding it We use arbitrary values so that most of Denavit-Hartenberg parameters are equal to zero

46. Denavit-Hartenberg 1. Compute the link vector a i and the link lenght 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1

47. Denavit-Hartenberg Identify the joint axes Identify the common perpendiculars of successive joint axes Attach coordinate frames to each joint axes

48. Denavit-Hartenberg Identifying the joint axes

49. Denavit-Hartenberg Remember, is the point where the shortest distance to joint i+1 exists Identifying the common perpendiculars

50. Denavit-Hartenberg the origin Attaching the frames

51. Denavit-Hartenberg 1. Compute the link vector a i and the link lenght 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1

52. Denavit-Hartenberg Remember

53. Denavit-Hartenberg Let’s define the function

54. Denavit-Hartenberg We then have :

55. Denavit-Hartenberg 1. Compute the link vector a i and the link lenght 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1

56. Denavit-Hartenberg

57. 1. Compute the link vector a i and the link lenght 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1

58. Denavit-Hartenberg

59. 1. Compute the link vector a i and the link lenght 2. Attach coordinate frames to the joint axes 3. Compute the link twist α i 4. Compute the link offset d i 5. Compute the joint angle φ i 6. Compute the transformation (i-1) T i which transforms entities from link i to link i-1

60. Denavit-Hartenberg This transformation is done in several steps : Rotate the link twist angle α i-1 around the axis x i Translate the link length a i-1 along the axis x i Translate the link offset d i along the axis z i Rotate the joint angle φ i around the axis z i

61. Denavit-Hartenberg

65. Multiplying the matrices :

66. Denavit-Hartenberg By specifying the transformation for all joints we can specify a transformation form link N to the base link 0 : Let’s denote the joint parameters of joint i. We obtain :

67. End effector End effector - the last coordinate system of figure Located in joint N. But usually, we want to specify it in base coordinates.

68. End effector A transformation from the link N to the base :

69. End effector We can also express it as three rotations (around each of the coordinate axes) followed by a translation How can we establish a relation with the other expression ?

70. End effector Location The origin of a coordinate frame relative to some base coordinate frame is specified by the translation :

71. End effector Orientation Any 3D orientation relative to some base coordinate frame can be specified by : three rotations, one around each of the coordinate axes.

72. End effector Orientation These rotations are named roll, pitch, yaw We do them in this order : around x, y, z.

73. End effector Orientation The yaw transformation can be expressed :

74. End effector Orientation The pitch transformation can be expressed :

75. End effector Orientation The roll transformation can be expressed :

76. End effector Orientation The roll, pitch and yaw transformation is then expressed :

77. End effector Finally, the transformation from a coordinate frame to the base frame is expressed :

78. End effector Remember that can be expressed as a 4x4 matrix with elements m ij - functions of joint parameters θ 1,...,θ N. Let’s omit them for readability :

79. End effector As the two transformations and do the same thing :

80. End effector We obtain directly the translation vector :

81. End effector We can obtain the yaw angle : Because : arctan is π-periodic. Let’s use our function arctan 2 to get the right angle.

82. End effector Knowing the yaw angle, we can obtain the pitch angle : Because : Again, let’s use our function arctan 2 :

83. End effector We can obtain the roll angle : Because : Again, let’s use our function arctan 2 :

84. End effector Let’s define the state vector

85. End effector As previously shown, The state vector is composed of elements of this matrix. It’s also a function of joint parameters :