1. XAXA A YAYA P YBYB XBXB B Relative Reference Frames (Relative Coordinates Systems) P B = P A P A = P in reference frame A P B = P in reference frame B

2. Two equivalent relative views : 1. B has been translated with transformation T( t x, t y ) away from A. 2. A has been translated with transformation T( -t x, -t y ) away from B. P XAXA A YAYA YBYB XBXB B txtx tyty

3. If P A is known and P B is unknown From A’s perspective, P is fixed in Reference Frame A From B’s perspective, P has been translated in the opposite direction, therefore : P B = T( -t x, -t y ) P A If P B is known and P A is unknown From B’s perspective, P is fixed in Reference Frame B From A’s perspective, P has been translated with Reference Frame B, therefore : P A = T( t x, t y ) P B

4. P B = T( -t x, -t y ) P A P A = T( t x, t y ) P B Therefore, Remember : XAXA A YAYA P YBYB XBXB B txtx tyty P A = T( t x, t y ) P B

5. Two equivalent relative views : 1. B has been rotated with transformation R(  ) away from A. 2. A has been rotated with transformation R(-  ) away from B. XAXA A YAYA P  YBYB XBXB B

6. If P A is known and P B is unknown From A’s perspective, P is fixed in Reference Frame A From B’s perspective, P has been rotated in the opposite direction, therefore : P B = R( -  ) P A If P B is known and P A is unknown From B’s perspective, P is fixed in Reference Frame B From A’s perspective, P has been rotated with Reference Frame B, therefore : P A = R(  ) P B

7. P B = R( -  ) P A P A = R(  ) P B Therefore, Remember : XAXA A YAYA P YBYB XBXB B  P A = R(  ) P B

8. P A = T( t x, t y ) P B P B = R(  ) P C From earlier results : Therefore, P A = T( t x, t y ) P B = T( t x, t y ) R(  ) P C P XAXA A YAYA XBXB B YBYB txtx tyty YCYC XCXC C 

9. Notation : M AB = Transformation Matrix that transforms Reference Frame A to a new Reference Frame B P A = T( t x, t y ) P B P B = T( -t x, -t y ) P A XAXA A YAYA P YBYB XBXB B txtx tyty Example : Therefore : M AB = T( t x, t y ) M BA = T( -t x, -t y ) [M AB ] -1 = M BA P A = M AB P B P B = M BA P A

10. From earlier results : P A = T( 10,5 ) R( 5 o ) P C P C = T( 15,0 ) P D P D = R( 10 o ) P E Therefore, P A = T( 10,5 ) R( 5 o ) T( 15,0 ) R( 10 o ) P E P XAXA A YAYA XBXB B YBYB 10 5 YEYE XEXE E    YCYC 15 YDYD

11. M AB = T( 10,5 ) M BC = R( 5 o ) M CD = T( 15,0 ) M DE = R( 10 o ) P A = M AB P B = M AB M BC P C = M AB M BC M CD P D = M AB M BC M CD M DE P E = T( 10,5 ) R( 5 o ) T( 15,0 ) R( 10 o ) P E Therefore : P A = M AB M BC M CD ……. M HI M IJ M JK P K Also since : P A = M AK P K This mean : M AK = M AB M BC M CD ……. M HI M IJ M JK

12. Answer : P W = T(h,k) R(  ) T(L 1,0) R(  ) T(L 2,0) R(  ) P M P XWXW W YWYW  h k   L1L1 L2L2 If h, k, L 1, L 2, , ,  are parameters for the robotic configuration as shown, what is the relationship between P M and P W ? XMXM YMYM Manipulator

13. P W = T(h,k) R(  ) T(L 1,0) R(  ) T(L 2,0) R(  ) P M If M WM, h, k, L 2, ,  and  are given, what is L 1 ? M WM = T(h,k) R(  ) T(L 1,0) R(  ) T(L 2,0) R(  ) T(h,k) R(  ) T(L 1,0) R(  ) T(L 2,0) R(  ) = M WM T(L 1,0) = [R(  )] -1 [T(h,k)] -1 M WM [R(  )] -1 [T(L 2,0)] -1 [R(  )] -1 T(L 1,0) = R(-  ) T(-h,-k) M WM R(-  ) T(-L 2,0) R(-  )

14. Robot Main Body Transform Arm Transform Hand Transform Leg Transform Head Transform Camera