Algebraic solution Consider a 3 link manipulator We

Algebraic solution Consider a 3 -link manipulator We can derive kinematic equations: 1

Algebraic solution • D-H transformation i i-1 di i 1 0 0 0 1 2 0 L 1 0 2 3 0 L 2 0 3 2

Algebraic Solution 3

Algebraic Solution The kinematics of the example seen before are: Assume goal point is the specification of wrist frame, specified by 3 numbers: 4

Algebraic Solution By comparison, we get the four equations: Summing the square of the last 2 equations: From here we get an expression for c 2 And finally: 5

Algebraic Solution The arc cosine function does not behave well as its accuracy in determining the angle is dependant on the angle ( cos(q)=cos(-q)). When sin(q) approaches zero, division by sin(q) give inaccurate solutions. Therefore an arc tangent function which is more consistent is used. 6

Algebraic Solution Using c 12=c 1 c 2 -s 1 s 2 and s 12= c 1 s 2 -c 2 s 1: where k 1=l 1+l 2 c 2 and k 2=l 2 s 2. To solve these eqs, set r=+ k 12+k 22 and =Atan 2(k 2, k 1). 7

Algebraic Solution k 1 l 2 2 k 2 l 1 Then: k 1=r cos , k 2=r sin , and we can write: x/r= cos 1 - sin 1 y/r= cos sin 1 + sin cos 1 or: cos( + 1) = x/r, sin( + 1) =y/r 8

Algebraic Solution Therefore: + 1 = atan 2(y/r, x/r) = atan 2(y, x) And so: 1 = atan 2(y, x) - atan 2(k 2, k 1) Finally, 3 can be solved from: 1+ 2+ 3 = 9

Geometric Solution Two-link Planar Manipulator y a 2 a 1 x 10

Geometric Solution Applying the “law of cosines”: We should try to avoid using the arccos function because of inaccuracy. 11

Geometric Solution The second joint angle 2 is then found accurately as: Because of the square root, two solutions result: 12

Geometric Solution 1 is determined uniquely given 2 1= -y Note the two different solutions for 1 corresponding to the elbow-down versus elbow-up configurations 13

Examples: IK Solution for PUMA 560 Algebraic Solution for UNIMATION PUMA 560 We wish to Solve (4. 54) For i when 0 6 T is given as numeric values 14

Examples: IK Solution for PUMA 560 Paul (1981) suggest pre multiplying the above matrix (4. 54) by its unworn inverse transform successively and determine the unknown angle from the elements of the resultant matrix equation. 15

Examples: IK solution for PUMA 560 By multiplying both sides of (4. 54) Inverting 0 1 T we get: (4. 56) 16

Examples: IK Solution for PUMA 560 1 6 T from (3. 13) is: 17

Examples: IK solution for PUMA 560 By adapting 16 T from (3. 13) and equating elements we have Using trigonometric substitution: We obtain Using difference of angles: 18

Examples: IK Solution for PUMA 560 And so: Solution for 1: Note: two possible solution for 1 19

Examples: IK Solution for PUMA 560 By equating elements (1, 4) and also (3, 4) from Equ. (4. 56): By squaring and addition of resulting equations and Equ. (4. 57): Where 20

Examples: IK Solution for PUMA 560 The above equation is only dependant on 3 so with similar procedure we get: again two different solution for 3. We may rewrite (4. 54) as (4. 70) 21

Examples: IK Solution for PUMA 560 By adapting 36 T from (3. 11) and equating (1, 4) and (2, 4) we have We solve for 23 as Four possible solutions for 2 22

Examples: IK Solution for PUMA 560 By equating elements (1, 3) and also (3, 3) from (4. 70) : 23

Examples: IK Solution for PUMA 560 We continue similar procedure for rest of the Parameters… 24

Examples: IK Solution for PUMA 560 Other possible solutions : After all eight solutions have been computed, some of them may have to be discarded because of joint limitations. Usually the one closest to present manipulator configuration is chosen. 25

Standard Frames {W} {B} {T} {S} {G} 26

Repeatability and accuracy Repeatability: how precisely a manipulator can return to a taught point? Accuracy: the precision with which a computed point can be attained. 27