HW4_sol - Texas A& M University Department of Mechanical...

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Unformatted text preview: Texas A & M University Department of Mechanical Engg. MEEN-408 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 Jacobian-Link 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 Jacobian-Link 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 Link-1 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 Link-2 The CM of link-2 has velocity due to rotation about axis fixed to link-1 and velocity due to rotation of link-1 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 link-2 due to rotation of link-1 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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HW4_sol - Texas A& M University Department of Mechanical...

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