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Unformatted text preview: Texas A & M University Department of Mechanical Engg. MEEN408 Intro. to Robotics Solution HW4 Instructor: Dr. Darbha Swaroop 1. Solution to Problem 1 For the given robot,the axes for the links are as shown in fig.1. Also, we have DH parameter table as folows. Figure 1: Coordinate frames Link a d α θ 1 π 2 φ = 90 ˆ φ 2 l 1 π 2 θ Table 1: DH parameters where ˆ φ is the angle z 1 makes with x . Based on DH parameters, the link transformation matrices are as follows T 1 = T z ( θ 1 ) T z ( d 1 ) T x ( a 1 ) T x ( α 1 ) T 1 = cφ sφ sφ cφ 1 1 T 1 2 = T z ( θ 2 ) T z ( d 2 ) T x ( a 2 ) T x ( α 2 ) 1 T 1 2 = cθ sθ sθ cθ 1 l 1 1 T 2 = T 1 T 1 2 T 2 = cφcθ sφ sθcφ l 1 sφ sφcθ cφ sφsθ l 1 cφ sθ cθ 1 (a) Angular Velocity JacobianLink 1 For link 1, the angular velocity is as follows ˜ ω L 1 = ˙ φ ˜ k where ˜ k is the unit vector along z . Therefore, angular velocity of link 1 in terms of principle coordi nates φ,θ is as follows ˜ ω L 1 = J ω 1 ˙ φ ˙ θ ˜ ω L 1 = 0 0 0 0 1 0 ˙ φ ˙ θ (b) Angular Velocity JacobianLink 2 For the second link, the angular velocity is given by ˜ ω L 2 = ˙ φ ˜ k + ˙ θ ˜ k 1 Hence, from the transformation matrix T 1 ,k 1 can be written in the base frame as ˜ k 10 = sφ cφ Therefore, angular velocity jacobian for link 2 is ˜ ω L 2 = J ω 2 ˙ φ ˙ θ ˜ ω L 2 = sφ cφ 1 ˙ φ ˙ θ (c) Velocity Jacobian for Link1 The velocity of the center of mass (CM) for link 1 is given by ˜ V cm 1 = ˜ ω L 1 × ˜ r cm 1 In case of link is ˜ r cm 1 = l 1 2 ˜ k 1 . Hence, we can write coordinates of ˜ k 1 in base coordinate frame as 2 ˜ r cm 1 = l 1 2 sφ ˜ i l 1 2 cφ ˜ j Therefore, assuming ˙ φ = 1 rad/s , we get velocity jacobian as ˜ V cm 1 = ˜ k × ( l 1 2 sφ ) ˜ i ( l 1 2 cφ ) ˜ j ) ˜ V cm 1 = l 1 2 cφ ˜ i + l 1 2 sφ ˜ i ˜ V cm 1 = l 1 2 cφ l 1 2 sφ ˙ φ ˙ θ (d) Velocity Jacobian for Link2 The CM of link2 has velocity due to rotation about axis fixed to link1 and velocity due to rotation of link1 about fixed frame. In order to determine first colummn of the velocity jacobian, we assume ˙ θ = 0 , ˙ φ = 1 rad/s . ˜ V cm 2 = ˜ k × ˜ r cm 2 ˜ r cm 2 = r ˜ k 2 x cm 2 y cm 2 z cm 2 1 = T 2 r 1 = r sθcφ + l 1 cφ r sφsθ l 1 sφ r cθ 1 Therefore velocity of CM of link2 due to rotation of link1 is ˜ V cm 2 = ˜ k × ( rsθcφ + l 1 sφ ) ˜ i ( rsφsθ + l 1 cφ ) ˜ j + rcθ ˜ k ˜ V cm 2 = ( r sφsθ + l 1 cφ ) ˜ i + ( r sθcφ + l 1 sφ ) ˜ j V x cm 2 V y cm 2 V z cm 2 = r sφsθ + l 1 cφ r sθcφ + l 1 sφ ˙ φ Second column of velocity jacobian: In order to determine first colummn of the velocity jacobian, we assume ˙ θ...
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 Spring '09
 SwaroopDarbha
 Energy, Kinetic Energy, Potential Energy, ev, QR, link1

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