1 00:00:00,250 --> 00:00:04,500 Less time we have seen what is inverse kinematics and how to solve it. 2 00:00:04,740 --> 00:00:11,520 And we have also seen two methods of solving inverse kinematics, namely analytical way and numerical 3 00:00:11,520 --> 00:00:13,690 way in terms of analytical way. 4 00:00:13,710 --> 00:00:20,220 We have said that while it is faster and more initiative, it becomes extremely complex in terms of 5 00:00:20,220 --> 00:00:24,690 redundant robot manipulators because they have higher degree of freedoms. 6 00:00:24,990 --> 00:00:30,390 While we have say that we can utilize kinematic decoupling to make the process easier. 7 00:00:30,630 --> 00:00:36,390 This method belongs to specific robots, namely six degree of freedom and spherical wrist. 8 00:00:37,020 --> 00:00:43,080 So what we will do if we have redundant robots, which may not have closed loop solution or if there 9 00:00:43,080 --> 00:00:49,080 is but very complex, what will we do if our robot doesn't have spherical wrist? 10 00:00:49,380 --> 00:00:53,740 How we can solve inverse kinematics problem in such kind of situations? 11 00:00:53,760 --> 00:00:55,530 We will use a numerical approach. 12 00:00:55,770 --> 00:00:58,200 Let's explain this method more clearly. 13 00:00:58,590 --> 00:01:06,870 Assume that we have a function of GQ, which is equal to X the minus f q here X is desirable and the 14 00:01:06,870 --> 00:01:09,300 vector pose that we want to obtain. 15 00:01:09,300 --> 00:01:12,930 While F Q is nothing but forward, kinematics function. 16 00:01:13,290 --> 00:01:22,740 So we give Q joint configurations to function f it returns US end effector pose at the joint configuration. 17 00:01:23,100 --> 00:01:27,180 Our goal is to make that function, namely GQ zero. 18 00:01:27,390 --> 00:01:34,290 Because then we will have our design joint configuration of Q D, which will make our robot reach the 19 00:01:34,290 --> 00:01:37,200 desired and the vector pose of XD. 20 00:01:37,440 --> 00:01:41,190 So we will eventually lasso our inverse kinematics problem. 21 00:01:41,550 --> 00:01:45,210 No, the problem is how we can make G.Q becomes zero. 22 00:01:45,630 --> 00:01:47,670 We have two methods for doing that. 23 00:01:48,150 --> 00:01:51,330 The first is Jacobean inverse method. 24 00:01:51,660 --> 00:01:54,150 The second is Jacobean transpose method. 25 00:01:54,420 --> 00:02:00,300 Both of these solutions are iterative because numerical inverse climatiques iterative solution. 26 00:02:02,310 --> 00:02:04,470 Let's start with Jacob in inverse method. 27 00:02:04,800 --> 00:02:11,810 First, we expand forward kinematics function of F with Taylor series about a configuration of Kudi, 28 00:02:11,820 --> 00:02:13,740 which is desired configuration. 29 00:02:14,340 --> 00:02:21,870 Higher order terms, which is denoted by h o t because, as you know, we can go without stopping the 30 00:02:21,870 --> 00:02:23,220 Taylor expansion. 31 00:02:24,540 --> 00:02:28,740 So if we hold, we don't need higher order terms. 32 00:02:29,430 --> 00:02:36,750 So if we neglect higher order terms, then we will have this equation, which contains inverse of Jacobean. 33 00:02:37,260 --> 00:02:38,340 Be careful here. 34 00:02:38,580 --> 00:02:42,210 We assume that all the Jacobean is square and invert. 35 00:02:42,810 --> 00:02:51,450 So in order to find true, we start with initial guess of Q zero and then find Q1, Q2 and so on up 36 00:02:51,450 --> 00:02:57,380 to when we converge q d through an iterative way in order to that code. 37 00:02:57,450 --> 00:03:01,800 In order to do that calculation, this equation is needed, which is iterative. 38 00:03:01,800 --> 00:03:10,980 Way of converging to the here over is nothing but step size, and it's greater than zero. 39 00:03:11,580 --> 00:03:17,790 It can be chosen as constant or K dependent, so it changes at each iteration. 40 00:03:18,540 --> 00:03:24,860 When we neglect the higher order terms, we can expect local convergence, namely Q, has to be near 41 00:03:24,870 --> 00:03:26,510 solution or kudi. 42 00:03:27,360 --> 00:03:33,780 Additionally, as inverse kinematics have multiple solutions, we can have different kudi depending 43 00:03:33,780 --> 00:03:35,820 on initial guess of Q zero. 44 00:03:36,660 --> 00:03:39,510 What if Jacobean is singular and non-convertible? 45 00:03:39,840 --> 00:03:41,310 Then we will use less. 46 00:03:41,490 --> 00:03:43,830 So the inverse Jacobean instead? 47 00:03:45,570 --> 00:03:48,360 Now let's switch to the Giacobbe and transpose method. 48 00:03:48,420 --> 00:03:53,490 So we will first define quadratic optimization problem, which is given like this. 49 00:03:54,120 --> 00:04:02,370 So if you have read about optimization, one of the widely used ways of solving them is through gradient 50 00:04:02,370 --> 00:04:09,930 descent, which requires us to know the direction of minimization and steps that we take at each iteration 51 00:04:09,930 --> 00:04:12,300 through that direction. 52 00:04:12,690 --> 00:04:18,720 We can find direction of minimization through finding gradient of the cost function, and it is given 53 00:04:18,720 --> 00:04:22,350 with that equation, which contains transpose of Jacobean. 54 00:04:23,340 --> 00:04:30,420 Then we can find optimal solution, which is kudi using gradient descent algorithm, which is also iterative. 55 00:04:32,130 --> 00:04:38,400 As you can see, this method includes transpose of Jacobean, which have advantages of being much more 56 00:04:38,400 --> 00:04:39,570 easier to calculate. 57 00:04:39,870 --> 00:04:43,080 Also, we don't have issues of Jacobean singularity. 58 00:04:43,710 --> 00:04:50,340 However, while we have these advantages, we have this advantage of slower convergence. 59 00:04:53,350 --> 00:04:54,990 See you on the next lesson.