A planar elbow manipulator is shown. Write the D H parameters and hence locate the position and orientation of the end E of the manipulator. Given $L_1 = L_2 = 250$ mm and $\theta_1 = \theta_2 = 30^\circ$.[Figure: base frame $X_0, Y_0$ at O. Li…

Question

Q.6 (A) A planar elbow manipulator is shown. Write the D H parameters and hence locate the position and orientation of the end E of the manipulator. Given L₁ = L₂ = 250 mm and θ₁ = θ₂ = 30^° . [Figure: base frame X₀, Y₀ at O. Link 1 (length L₁ ) goes from O to joint A at angle θ₁ measured anticlockwise from X₀ . Link 2 (length L₂ ) goes from A to end point E; θ₂ is measured anticlockwise from the extension of link 1 (dashed line) to link 2.]

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DH parameter table (both joints revolute, all joint axes parallel to Z, so d_i = 0 , α_i = 0 ): Link i θ_i d_i a_i α_i 1 θ₁ = 30^° 0 L₁ = 250 mm0 2 θ₂ = 30^° 0 L₂ = 250 mm0 Link transformation (for α = 0, d = 0 ): ^(i-1)T_i = beginbmatrixcosθ_i & -sinθ_i & 0 & a_icosθ_i\ sinθ_i & cosθ_i & 0 & a_isinθ_i\ 0&0&1&0\0&0&0&1endbmatrix ⁰T₂ = ⁰T₁ ¹T₂ = beginbmatrixcos(θ₁+θ₂) & -sin(θ₁+θ₂) & 0 & L₁cosθ₁ + L₂cos(θ₁+θ₂)\ sin(θ₁+θ₂) & cos(θ₁+θ₂) & 0 & L₁sinθ₁ + L₂sin(θ₁+θ₂)\ 0&0&1&0\0&0&0&1endbmatrix Substituting θ₁ + θ₂ = 60^° : x_E = 250cos30^° + 250cos60^° = 216.51 + 125 = 341.51 mm y_E = 250sin30^° + 250sin60^° = 125 + 216.51 = 341.51 mm ⁰T₂ = beginbmatrix0.5 & -0.866 & 0 & 341.51\ 0.866 & 0.5 & 0 & 341.51\ 0&0&1&0\0&0&0&1endbmatrix Answer: E is at (341.51 mm, 341.51 mm); orientation of the end link = θ₁ + θ₂ = 60^° anticlockwise from X₀ .

Link i	\(\theta_i\)	\(d_i\)	\(a_i\)	\(\alpha_i\)
1	\(\theta_1 = 30^\circ\)	0	\(L_1 = 250\) mm	0
2	\(\theta_2 = 30^\circ\)	0	\(L_2 = 250\) mm	0

A planar elbow manipulator is shown. Write the D H parameters and hence locate the position and orientation of the end E of the manipulator. Given \(L_1 = L_2 = 250\) mm and \(\theta_1 = \theta_2 = 30^\circ\).[Figure: base frame \(X_0, Y_0\) at O. Li…

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