FORWARD KINEMATICS: THE DENAVIT-HARTENBERG ...

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76CHAPTER 3. FORWARD KINEMATICS: THE DENAVIT-HARTENBERG CONVENTION d θ y0 x0 z0 O0 Figure 3.2: Coordinate frames satisfying assumptions DH1 and DH2. needed from six to four (or even fewer in some cases). In Section 3.2.1 we will show why, and under what conditions, this can be done, and in Section 3.2.2 we will show exactly how to make the coordinate frame assignments. 3.2.1 Existence and uniqueness issues Clearly it is not possible to represent any arbitrary homogeneous transformation using only four parameters. Therefore, we begin by determining just which homogeneous transformations can be expressed in the form (3.10). Suppose we are given two frames, denoted by frames 0 and 1, respectively. Then there exists a unique homogeneous transformation matrix A that takes the coordinates from frame 1 into those of frame 0. Now suppose the two frames have two additional features, namely: (DH1) The axis x1 is perpendicular to the axis z0 (DH2) The axis x1 intersects the axis z0 as shown in Figure 3.2. Under these conditions, we claim that there exist unique numbers a, d, θ, α such that a x1 α z1 O1 A = R z,θTransz,dTransx,aR x,α. (3.11) Of course, since θ and α are angles, we really mean that they are unique to within a multiple of 2π. To show that the matrix A can be written in this y1
3.2. DENAVIT HARTENBERG REPRESENTATION 77 form, write A as A = � R 0 1 O 0 1 0 1 � (3.12) and let ri denote the i th column of the rotation matrix R 0 1. We will now examine the implications of the two DH constraints. If (DH1) is satisfied, then x1 is perpendicular to z0 and we have x1·z0 = 0. Expressing this constraint with respect to o0x0y0z0, using the fact that r1 is the representation of the unit vector x1 with respect to frame 0, we obtain 0 = x 0 1 · z 0 0 = [r11,r21,r31] T · [0,0, 1] T (3.13) (3.14) = r31. (3.15) Since r31 = 0, we now need only show that there exist unique angles θ and α such that ⎡ ⎤ R 0 1 = R x,θR x,α = ⎢ ⎣ cθ −sθcα sθsα sθ cθcα −cθsα 0 sα cα ⎥ ⎦. (3.16) The only information we have is that r31 = 0, but this is enough. First, since each row and column of R 0 1 must have unit length, r31 = 0 implies that r 2 11 + r 2 21 = 1, Hence there exist unique θ, α such that r 2 32 + r 2 33 = 1 (3.17) (r11,r21) = (cθ,sθ), (r33,r32) = (cα,sα). (3.18) Once θ and α are found, it is routine to show that the remaining elements of R 0 1 must have the form shown in (3.16), using the fact that R 0 1 is a rotation matrix. Next, assumption (DH2) means that the displacement between O 0 and O 1 can be expressed as a linear combination of the vectors z0 and x1. This can be written as O 1 = O 0 +dz0 +ax1. Again, we can express this relationship in the coordinates of o0x0y0z0, and we obtain O 0 1 = O 0 0 + dz 0 0 + ax 0 1 (3.19)
Page 1 and 2: Chapter 3 FORWARD KINEMATICS: THE D
Page 3 and 4: 3.1. KINEMATIC CHAINS 73 z1 x0 y1 z
Page 5: 3.2. DENAVIT HARTENBERG REPRESENTAT
Page 9 and 10: 3.2. DENAVIT HARTENBERG REPRESENTAT
Page 11 and 12: 3.2. DENAVIT HARTENBERG REPRESENTAT
Page 13 and 14: 3.3. EXAMPLES 83 Step 9: Form T 0 n
Page 15 and 16: 3.3. EXAMPLES 85 x2 d2 x1 x0 d3 O2
Page 17 and 18: 3.3. EXAMPLES 87 Table 3.3: DH para
Page 19 and 20: 3.3. EXAMPLES 89 where r11 = c1c4c5
Page 21 and 22: 3.3. EXAMPLES 91 T 0 6 is then give
Page 23 and 24: 3.3. EXAMPLES 93 follows. A1 = ⎡
Page 25 and 26: 3.4. PROBLEMS 95 Figure 3.14: Two-l
Page 27 and 28: 3.4. PROBLEMS 97 Figure 3.18: Elbow
Page 29 and 30: 3.4. PROBLEMS 99 Figure 3.21: Rhino
Page 31 and 32: 3.4. PROBLEMS 101 Figure 3.23: GMF

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