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HW solution

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HOMEWORK 1SOLUTION Robotics Design
PROBLEM 1
PROBLEM 2 The operation to obtain 𝐴 𝑃 from 𝐵 𝑃 is 𝐴 𝑃 = 𝐵 𝐴 𝑅 𝐵 𝑃 The rotation matrix 𝐵 𝐴 𝑅 can be calculated as: 𝐵 𝐴 𝑅 = 𝑅𝑋(𝜙)𝑅𝑍 𝜃 𝐵 𝐴 𝑅 = 1 0 0 0 cos 𝜙 − sin 𝜙 0 sin 𝜙 cos 𝜙 cos 𝜃 − sin 𝜃 0 sin 𝜃 cos 𝜃 0 0 0 1 𝐵 𝐴 𝑅 = 𝑐𝜃 −𝑠𝜃 0 𝑐𝜙𝑠𝜃 𝑐𝜙𝑐𝜃 −𝑠𝜙 𝑠𝜙𝑠𝜃 𝑠𝜙𝑐𝜃 𝑐𝜙
PROBLEM 3 𝐵 𝐴 𝑇 = 𝑋𝐵. 𝑋𝐴 𝑌𝐵. 𝑋𝐴 𝑍𝐵. 𝑋𝐴 𝐴 𝑃𝐵𝑜𝑟𝑔,𝑥 𝑋𝐵. 𝑌 𝐴 𝑌𝐵. 𝑌 𝐴 𝑍𝐵. 𝑌 𝐴 𝐴 𝑃𝐵𝑜𝑟𝑔,𝑦 𝑋𝐵. 𝑍𝐴 𝑌𝐵. 𝑍𝐴 𝑍𝐵. 𝑍𝐴 𝐴 𝑃𝐵𝑜𝑟𝑔,𝑧 0 0 0 1 𝐵 𝐴 𝑇 = cos 180° cos 90° cos 90° 3 cos 90° cos 180° cos 90° 0 cos 90° cos 90° cos 0° 0 0 0 0 1 𝐵 𝐴 𝑇 = −1 0 0 3 0 −1 0 0 0 0 1 0 0 0 0 1
PROBLEM 4 𝐶 𝐴 𝑇 = 𝑋𝐶. 𝑋𝐴 𝑌𝐶. 𝑋𝐴 𝑍𝐶. 𝑋𝐴 𝐴 𝑃𝐶𝑜𝑟𝑔,𝑥 𝑋𝐶. 𝑌 𝐴 𝑌𝐶. 𝑌 𝐴 𝑍𝐶. 𝑌 𝐴 𝐴 𝑃𝐶𝑜𝑟𝑔,𝑦 𝑋𝐶. 𝑍𝐴 𝑌𝐶. 𝑍𝐴 𝑍𝐶. 𝑍𝐴 𝐴 𝑃𝐶𝑜𝑟𝑔,𝑧 0 0 0 1 𝐶 𝐴 𝑇 = cos 90° cos 120° cos 30° 3 cos 90° cos 30° cos 60° 0 cos 180° cos 90° cos 90° 2 0 0 0 1 𝐶 𝐴 𝑇 = 0 − 1 2 3 2 3 0 3 2 1 2 0 −1 0 0 2 0 0 0 1
PROBLEM 5 𝐶 𝐵 𝑇 = 𝑋𝐶. 𝑋𝐵 𝑌𝐶. 𝑋𝐵 𝑍𝐶. 𝑋𝐵 𝐵 𝑃𝐶𝑜𝑟𝑔,𝑥 𝑋𝐶. 𝑌𝐵 𝑌𝐶. 𝑌𝐵 𝑍𝐶. 𝑌𝐵 𝐵 𝑃𝐶𝑜𝑟𝑔,𝑦 𝑋𝐶. 𝑍𝐵 𝑌𝐶. 𝑍𝐵 𝑍𝐶. 𝑍𝐵 𝐵 𝑃𝐶𝑜𝑟𝑔,𝑧 0 0 0 1 𝐶 𝐵 𝑇 = cos 90° cos 60° cos 150° 0 cos 90° cos 150° cos 120° 0 cos 180° cos 90° cos 90° 2 0 0 0 1 𝐶 𝐵 𝑇 = 0 1 2 − 3 2 0 0 − 3 2 − 1 2 0 −1 0 0 2 0 0 0 1

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Homework 1 Solution.pptx

  • 3. PROBLEM 2 The operation to obtain 𝐴 𝑃 from 𝐵 𝑃 is 𝐴 𝑃 = 𝐵 𝐴 𝑅 𝐵 𝑃 The rotation matrix 𝐵 𝐴 𝑅 can be calculated as: 𝐵 𝐴 𝑅 = 𝑅𝑋(𝜙)𝑅𝑍 𝜃 𝐵 𝐴 𝑅 = 1 0 0 0 cos 𝜙 − sin 𝜙 0 sin 𝜙 cos 𝜙 cos 𝜃 − sin 𝜃 0 sin 𝜃 cos 𝜃 0 0 0 1 𝐵 𝐴 𝑅 = 𝑐𝜃 −𝑠𝜃 0 𝑐𝜙𝑠𝜃 𝑐𝜙𝑐𝜃 −𝑠𝜙 𝑠𝜙𝑠𝜃 𝑠𝜙𝑐𝜃 𝑐𝜙
  • 4. PROBLEM 3 𝐵 𝐴 𝑇 = 𝑋𝐵. 𝑋𝐴 𝑌𝐵. 𝑋𝐴 𝑍𝐵. 𝑋𝐴 𝐴 𝑃𝐵𝑜𝑟𝑔,𝑥 𝑋𝐵. 𝑌 𝐴 𝑌𝐵. 𝑌 𝐴 𝑍𝐵. 𝑌 𝐴 𝐴 𝑃𝐵𝑜𝑟𝑔,𝑦 𝑋𝐵. 𝑍𝐴 𝑌𝐵. 𝑍𝐴 𝑍𝐵. 𝑍𝐴 𝐴 𝑃𝐵𝑜𝑟𝑔,𝑧 0 0 0 1 𝐵 𝐴 𝑇 = cos 180° cos 90° cos 90° 3 cos 90° cos 180° cos 90° 0 cos 90° cos 90° cos 0° 0 0 0 0 1 𝐵 𝐴 𝑇 = −1 0 0 3 0 −1 0 0 0 0 1 0 0 0 0 1
  • 5. PROBLEM 4 𝐶 𝐴 𝑇 = 𝑋𝐶. 𝑋𝐴 𝑌𝐶. 𝑋𝐴 𝑍𝐶. 𝑋𝐴 𝐴 𝑃𝐶𝑜𝑟𝑔,𝑥 𝑋𝐶. 𝑌 𝐴 𝑌𝐶. 𝑌 𝐴 𝑍𝐶. 𝑌 𝐴 𝐴 𝑃𝐶𝑜𝑟𝑔,𝑦 𝑋𝐶. 𝑍𝐴 𝑌𝐶. 𝑍𝐴 𝑍𝐶. 𝑍𝐴 𝐴 𝑃𝐶𝑜𝑟𝑔,𝑧 0 0 0 1 𝐶 𝐴 𝑇 = cos 90° cos 120° cos 30° 3 cos 90° cos 30° cos 60° 0 cos 180° cos 90° cos 90° 2 0 0 0 1 𝐶 𝐴 𝑇 = 0 − 1 2 3 2 3 0 3 2 1 2 0 −1 0 0 2 0 0 0 1
  • 6. PROBLEM 5 𝐶 𝐵 𝑇 = 𝑋𝐶. 𝑋𝐵 𝑌𝐶. 𝑋𝐵 𝑍𝐶. 𝑋𝐵 𝐵 𝑃𝐶𝑜𝑟𝑔,𝑥 𝑋𝐶. 𝑌𝐵 𝑌𝐶. 𝑌𝐵 𝑍𝐶. 𝑌𝐵 𝐵 𝑃𝐶𝑜𝑟𝑔,𝑦 𝑋𝐶. 𝑍𝐵 𝑌𝐶. 𝑍𝐵 𝑍𝐶. 𝑍𝐵 𝐵 𝑃𝐶𝑜𝑟𝑔,𝑧 0 0 0 1 𝐶 𝐵 𝑇 = cos 90° cos 60° cos 150° 0 cos 90° cos 150° cos 120° 0 cos 180° cos 90° cos 90° 2 0 0 0 1 𝐶 𝐵 𝑇 = 0 1 2 − 3 2 0 0 − 3 2 − 1 2 0 −1 0 0 2 0 0 0 1