1 00:00:00,990 --> 00:00:07,050 With forward kinematics, we have seen how we can get and affect our position and orientation given 2 00:00:07,050 --> 00:00:08,250 the joint variables. 3 00:00:08,500 --> 00:00:13,890 In order to do that, we have utilized homogeneous transformation mattresses under composition. 4 00:00:14,460 --> 00:00:20,730 However, generally we are more interested with finding specific joint variables that will give us desired 5 00:00:20,730 --> 00:00:23,130 and affect their position and orientation. 6 00:00:23,430 --> 00:00:29,040 For example, during point, the point trajectory following we will use inverse kinematics in order 7 00:00:29,040 --> 00:00:30,600 to follow the given trajectory. 8 00:00:31,020 --> 00:00:37,230 So as the name suggests, we need to find inverse kinematics, which is the opposite of forward kinematics 9 00:00:37,710 --> 00:00:44,670 in this formula is the desired and the factor polls and Q is the joint variables. 10 00:00:45,070 --> 00:00:48,720 However, we have to know some realities about inverse kinematics. 11 00:00:48,990 --> 00:00:54,570 First of all, generally, the equations that we have to solve for inverse kinematics are generally 12 00:00:54,570 --> 00:01:01,230 nonlinear and complex, so it's not always possible to find cause form solution for every robot. 13 00:01:01,710 --> 00:01:08,670 Secondly, while the forward kinematics have one solution, the inverse kinematics can have more unique 14 00:01:08,910 --> 00:01:16,110 multiple or infinite solutions, depending on the robot and the desired position, position and orientation, 15 00:01:16,110 --> 00:01:16,530 surely. 16 00:01:17,100 --> 00:01:23,400 Namely, in forward kinematics, you give joint variables and you can get only unique and the vector 17 00:01:23,400 --> 00:01:24,780 position in the orientation. 18 00:01:25,050 --> 00:01:31,680 While in terms of inverse kinematics, in order to reach design and the vector polls, you can have 19 00:01:31,680 --> 00:01:33,510 multiple joint configurations. 20 00:01:33,750 --> 00:01:35,820 Look at the functioning of your arm. 21 00:01:36,030 --> 00:01:41,340 You can take a cop or an object by twisting your arm in multiple ways. 22 00:01:41,670 --> 00:01:48,300 However, in all cases, you reach the desired pose, namely the pause for grabbing the cop or an object. 23 00:01:48,690 --> 00:01:49,830 Anyway, let's continue. 24 00:01:50,160 --> 00:01:56,580 Also, as the number of non-zero the parameters are increasing, the number of admissible solutions 25 00:01:56,580 --> 00:01:57,510 are increasing. 26 00:01:57,810 --> 00:01:59,880 This will be obvious for you in a minute. 27 00:02:01,140 --> 00:02:08,370 And finally, while we can have multiple or infinite solutions, not all of these solutions are achievable 28 00:02:08,370 --> 00:02:14,490 because of physical limits in the robot manipulators such as the joint limits link collisions with each 29 00:02:14,490 --> 00:02:15,780 other and so on. 30 00:02:16,170 --> 00:02:22,470 So after all of these bad news, we have to anyway think about how we can solve this -- problem. 31 00:02:22,890 --> 00:02:25,140 We have two methods in our hands. 32 00:02:25,530 --> 00:02:32,100 Firstly, we can utilize closed form solution to solve our problem, which include geometric and algebraic 33 00:02:32,100 --> 00:02:33,000 manipulations. 34 00:02:33,420 --> 00:02:39,210 Secondly, numerical solution, which includes optimization methods to achieve the solutions. 35 00:02:39,570 --> 00:02:46,050 Both of these solutions have their own advantages and disadvantages, and we will see them during explaining 36 00:02:46,320 --> 00:02:46,650 the. 37 00:02:48,960 --> 00:02:51,300 Let's start with a closed form solution. 38 00:02:51,450 --> 00:02:57,120 First, let's sink what we have in our hands in order to solve India's climatiques problem. 39 00:02:57,420 --> 00:03:02,050 We have the parameters of a robot manipulator that we have seen before. 40 00:03:02,070 --> 00:03:03,030 How to calculate. 41 00:03:03,450 --> 00:03:09,810 And we also have the design and effector position and orientation, which can be represented by rotation 42 00:03:09,810 --> 00:03:16,920 matrix, though let's see how we can use that information to solve our problem. 43 00:03:18,150 --> 00:03:25,110 We can use the parameters in order to find in the vector position and orientation with respect the base 44 00:03:25,110 --> 00:03:27,240 frame in terms of joint variables. 45 00:03:27,600 --> 00:03:31,560 This is just the foreign kinematics problem and we have seen how to do that. 46 00:03:32,070 --> 00:03:37,920 We have to just multiply in the video homogeneous transformation mattresses after we have calculated 47 00:03:37,920 --> 00:03:40,590 forward kinematics of our robot manipulator. 48 00:03:40,590 --> 00:03:44,520 And also we note these are then the vector orientation and position. 49 00:03:44,850 --> 00:03:54,060 We can equalize Eq. 1.0 and Eq. one point one to get joint angles that will give us these are in the 50 00:03:54,060 --> 00:03:55,080 vector pose. 51 00:03:55,530 --> 00:03:59,120 The number of equations has to be 14 normally. 52 00:03:59,130 --> 00:04:00,330 Excuse me, 16. 53 00:04:00,570 --> 00:04:08,340 Normally because homogeneous transformation matrix has 16 entries four by four Matrix. 54 00:04:08,760 --> 00:04:15,720 However, as the last row of the homogeneous matrix is just consist of zeroes and one, we don't need 55 00:04:15,720 --> 00:04:17,700 to take them into consideration. 56 00:04:18,000 --> 00:04:24,000 If you couldn't visualize the solution, don't worry, we will do it in MATLAB in a minute and it will 57 00:04:24,000 --> 00:04:25,470 be clearer for you. 58 00:04:26,880 --> 00:04:29,280 However, we have a big problem with that solution. 59 00:04:29,520 --> 00:04:35,340 As you can notice, as the number of joint variables will increase or degree of freedom will increase. 60 00:04:35,580 --> 00:04:38,040 We will have extremely complex problem to solve. 61 00:04:38,370 --> 00:04:44,880 So this is the downside of this solution or as an advantage, we can get all of our admissible solutions 62 00:04:44,880 --> 00:04:45,690 by this method. 63 00:04:46,110 --> 00:04:51,420 While this method become extremely difficult to use for robot manipulator with more than six degrees 64 00:04:51,420 --> 00:04:52,020 of freedom. 65 00:04:52,290 --> 00:04:58,440 We can find easier application of these for the robots with six degrees of freedom and spherical wrist. 66 00:04:58,800 --> 00:05:01,260 This method is called cinematic decoupling. 67 00:05:01,560 --> 00:05:05,940 I will not give its deep explanation to you because it is not important to know. 68 00:05:06,120 --> 00:05:10,050 But if you are interested, you can find it from standard robotics textbooks. 69 00:05:10,440 --> 00:05:14,490 And based on this method, we divide our robot into two parts. 70 00:05:14,940 --> 00:05:22,650 Namely, the first three joints are responsible for the position of the robot in the vector only so 71 00:05:22,650 --> 00:05:25,080 we can use position inverse kinematics here. 72 00:05:25,410 --> 00:05:31,530 The last three joints are responsible for the orientation of the end effector, and we can utilize orientation 73 00:05:31,530 --> 00:05:33,900 inverse kinematics in this case. 74 00:05:34,470 --> 00:05:37,800 This will simplify the calculations. 75 00:05:38,350 --> 00:05:43,830 However, let me repeat this is valid for robots with six degrees of freedom and spherical wrist. 76 00:05:44,370 --> 00:05:46,890 So after all, these are your ethical issues. 77 00:05:47,070 --> 00:05:50,640 Let's switch to MATLAB and try to apply all of these. 78 00:05:51,750 --> 00:05:56,820 OK, let's not try to apply in mock, but what we have learned. 79 00:05:57,180 --> 00:05:59,580 Let's first clear variables. 80 00:05:59,880 --> 00:06:00,960 Clear the screen. 81 00:06:01,320 --> 00:06:03,960 OK, let's first start. 82 00:06:04,320 --> 00:06:12,510 Um, see the application of inverse kinematics in 2D plugin or manipulator because it will make our 83 00:06:12,510 --> 00:06:19,060 understanding, you know, it will make us understand the problem easier. 84 00:06:19,080 --> 00:06:19,460 OK. 85 00:06:22,020 --> 00:06:27,000 And let's first import e t S2. 86 00:06:27,450 --> 00:06:31,050 And what the star means to mean import all the things. 87 00:06:31,080 --> 00:06:31,530 OK. 88 00:06:32,310 --> 00:06:34,740 This is just standard library from the petrol corpus. 89 00:06:34,890 --> 00:06:35,400 OK. 90 00:06:35,670 --> 00:06:37,980 As you know, we are working with Politico.com's library. 91 00:06:38,400 --> 00:06:40,830 We here define the link lengths. 92 00:06:41,070 --> 00:06:41,580 OK. 93 00:06:42,000 --> 00:06:46,330 And here we define our robot. 94 00:06:46,350 --> 00:06:46,740 OK. 95 00:06:47,550 --> 00:06:49,350 This is our robot manipulator model. 96 00:06:49,740 --> 00:06:54,660 And you can visualize it by doing E dot plot. 97 00:06:54,750 --> 00:06:55,620 Zero zero. 98 00:06:55,650 --> 00:06:55,990 OK. 99 00:06:56,020 --> 00:06:56,880 No, no positions. 100 00:06:56,880 --> 00:06:58,920 Because our robot is. 101 00:06:59,370 --> 00:07:05,130 Let's just write it here to two degree of freedom. 102 00:07:05,530 --> 00:07:05,930 Robot. 103 00:07:05,940 --> 00:07:06,270 OK? 104 00:07:07,260 --> 00:07:07,980 And clutter. 105 00:07:08,040 --> 00:07:08,490 OK. 106 00:07:09,660 --> 00:07:11,370 Planner two degree of freedom robot. 107 00:07:11,820 --> 00:07:15,330 Let's visualize it in our joint, but I are zero zero. 108 00:07:15,900 --> 00:07:17,430 OK, let's press f nine. 109 00:07:17,430 --> 00:07:20,640 And you can see that it's nothing but a planet, a robot. 110 00:07:21,240 --> 00:07:27,750 As you can see, the red ones are red circles are our joints, and they are, as you can see, a rotating 111 00:07:27,750 --> 00:07:28,970 about z axis. 112 00:07:28,980 --> 00:07:36,660 OK, because we have defined it like that z Q1 and Q2 or Q1 and Q2 are our joint variables. 113 00:07:36,960 --> 00:07:43,440 And this is rotating joined by robots and our translation about A1 and translation about A2. 114 00:07:43,710 --> 00:07:44,160 OK. 115 00:07:44,190 --> 00:07:46,200 These are translation about A1. 116 00:07:46,500 --> 00:07:48,060 Excuse me, a translation of A1. 117 00:07:48,120 --> 00:07:51,570 About X and translation of it, about X, OK? 118 00:07:51,900 --> 00:07:58,530 As you can see, one and one VE created a simple plan or two degrees of freedom robot manipulator anyway. 119 00:07:59,400 --> 00:08:02,640 Let's not try to create some symbolic variables. 120 00:08:02,640 --> 00:08:06,510 Q1 and Q2 are our joint variables X and Y. 121 00:08:06,840 --> 00:08:11,520 What is X and Y, X and Y as our robots manipulator will move? 122 00:08:11,580 --> 00:08:15,690 Let me show you again, OK, our robot manipulator will move. 123 00:08:15,700 --> 00:08:18,420 As you can see, its X and Y coordinates will change. 124 00:08:18,420 --> 00:08:24,720 OK, X and Y represents this one because we want our what we are doing is inverse kinematic. 125 00:08:24,730 --> 00:08:32,430 So we are given some end effector position and orientation, and we try to find joint variables that 126 00:08:32,430 --> 00:08:35,490 will give us that end effector post, OK. 127 00:08:36,090 --> 00:08:45,480 For example, if our robot manipulators pose is at this position, OK, for example, x of one and y 128 00:08:45,480 --> 00:08:53,700 of one, then we will try to find what will be our Q1 and Q2 that will give us that end effector position, 129 00:08:53,700 --> 00:08:54,120 OK. 130 00:08:55,500 --> 00:08:58,470 So X and Y represent this one, and they are real numbers. 131 00:08:58,560 --> 00:09:04,750 So what we are doing here, we are finding first forward kinematics, OK in symbolic weight. 132 00:09:04,770 --> 00:09:05,640 OK, let's see. 133 00:09:05,640 --> 00:09:07,830 What do I mean by symbolic weight? 134 00:09:07,890 --> 00:09:08,970 Let's run this one. 135 00:09:09,300 --> 00:09:12,910 And as you can see it, this is the forward kinematics of our robot. 136 00:09:12,930 --> 00:09:18,360 OK, so this is the end effectors position with this position and orientation. 137 00:09:18,660 --> 00:09:20,910 OK, with respect to what? 138 00:09:21,180 --> 00:09:22,990 With respect to base frame? 139 00:09:23,010 --> 00:09:27,360 OK, so what's our translation part? 140 00:09:27,360 --> 00:09:29,520 Because we need translation Part X and Y. 141 00:09:29,730 --> 00:09:39,020 The translation part, as you know, is the last OK column of our homogeneous transformation matrix. 142 00:09:39,030 --> 00:09:39,540 OK? 143 00:09:39,720 --> 00:09:44,840 So this is our X, OK and this is our Y. 144 00:09:45,390 --> 00:09:52,700 So what we are doing here, we are defining here that our X here, we just define Eq.. 145 00:09:52,710 --> 00:09:53,040 OK? 146 00:09:53,250 --> 00:09:55,890 This double equality means that this is an equation. 147 00:09:55,890 --> 00:10:02,820 So this is an equation of X equals two blah blah blah blah or this one and y equals two blah blah blah. 148 00:10:02,820 --> 00:10:04,160 Or this one, OK? 149 00:10:04,830 --> 00:10:09,030 This is our X, this is our Y, and we are making e one. 150 00:10:09,090 --> 00:10:12,150 This is our first equation, and this is our second equation. 151 00:10:12,450 --> 00:10:16,490 We will use this E one and E two in order to solve our equations. 152 00:10:16,500 --> 00:10:19,230 So our equations are e one and E two. 153 00:10:19,470 --> 00:10:28,590 We will solve them with respect to what Q1 and Q2 because we want Q1 and Q2, variety solutions are 154 00:10:28,820 --> 00:10:30,780 to have to be Q1 and Q2. 155 00:10:30,790 --> 00:10:33,870 That will give us the desired and the vector post, OK. 156 00:10:34,080 --> 00:10:41,880 And this is our solution as one and as to let's try to run all of these press F9 again. 157 00:10:42,540 --> 00:10:44,790 OK, no need to see this one. 158 00:10:45,180 --> 00:10:48,220 And then now let's check out OK. 159 00:10:48,290 --> 00:10:58,230 As you can see now, as calculated S1 and as two, OK, let's look at the lengths of as one. 160 00:10:58,440 --> 00:10:59,010 OK. 161 00:10:59,040 --> 00:11:02,520 And let look the lengths of S2. 162 00:11:02,880 --> 00:11:06,600 As you can see, we have two solutions for S1. 163 00:11:06,600 --> 00:11:11,490 And as to the first one S1, one corresponds to S2 one. 164 00:11:11,790 --> 00:11:14,430 And as one two corresponds to S2 two. 165 00:11:14,460 --> 00:11:14,880 OK. 166 00:11:14,910 --> 00:11:15,420 Like that? 167 00:11:15,780 --> 00:11:21,270 So we have two solutions, as you can see if we have multiple solutions here, not a unique solution 168 00:11:21,270 --> 00:11:22,950 that we have said before. 169 00:11:23,190 --> 00:11:32,790 Both of these solutions, OK meets our criterion, so give us the desired the and the vector position. 170 00:11:33,750 --> 00:11:34,170 OK. 171 00:11:35,070 --> 00:11:39,900 As you can see, this is the inverse kinematics, but in in a closed form. 172 00:11:40,350 --> 00:11:48,270 But it will be very, very difficult if we applied these robots with more than six degrees of freedom 173 00:11:48,270 --> 00:11:50,370 or also in six degrees of freedom. 174 00:11:50,970 --> 00:11:59,250 If we don't separated into, you know, if it doesn't have spherical Reese and B doesn't do kinematic 175 00:11:59,250 --> 00:12:08,250 decoupling as a homework, I will give you in order to apply this the same, you can do that and you 176 00:12:08,250 --> 00:12:11,030 can see that it's really very difficult to do that. 177 00:12:11,040 --> 00:12:12,990 You can do this method. 178 00:12:12,990 --> 00:12:16,120 Apply this method to promote 560. 179 00:12:17,730 --> 00:12:26,040 So as you know, in human five six, this is really at, um, pivotal caucus library. 180 00:12:26,040 --> 00:12:27,270 So you can take this. 181 00:12:27,270 --> 00:12:33,210 You can calculate forward kinematics, find the end effector pause, OK? 182 00:12:33,480 --> 00:12:38,490 And after you can find for re kinematics, you can take desired pause. 183 00:12:38,490 --> 00:12:46,080 For example, let's take you can find the desired in the vector position that is um p five six the dot 184 00:12:46,560 --> 00:12:47,970 f kinematics. 185 00:12:48,150 --> 00:12:48,930 Okay. 186 00:12:49,320 --> 00:12:51,240 Just for nominal polls, okay. 187 00:12:51,600 --> 00:12:56,640 This is your desired in the factor position and orientation. 188 00:12:56,910 --> 00:12:58,090 You can equalize. 189 00:12:58,110 --> 00:13:03,060 You will have 12 equations and you have to solve equations and you will see that. 190 00:13:03,780 --> 00:13:06,150 I don't know whether you can do that or not. 191 00:13:06,420 --> 00:13:13,400 I mean, not as surely you can do that, but in terms of MATLAB, I mean, you have to try this. 192 00:13:13,410 --> 00:13:17,700 You can try and see that it will be very difficult to do and think about that. 193 00:13:17,700 --> 00:13:21,040 What will happen when we do that with the redundant robots? 194 00:13:21,060 --> 00:13:25,320 OK, more than six degrees of freedom robots, it will be impossible to solve. 195 00:13:25,350 --> 00:13:27,000 Almost impossible to solve. 196 00:13:27,300 --> 00:13:27,690 OK. 197 00:13:27,960 --> 00:13:30,600 As you can see, here we are using now. 198 00:13:30,600 --> 00:13:32,490 We came to three degrees of freedom. 199 00:13:32,700 --> 00:13:34,230 Excuse me, six degrees of freedom. 200 00:13:34,230 --> 00:13:36,660 Robot, OK, in 3-D workspace. 201 00:13:36,900 --> 00:13:37,250 Hmm. 202 00:13:38,450 --> 00:13:40,830 OK, let's clear everything. 203 00:13:41,220 --> 00:13:43,740 Clear and cool everything. 204 00:13:43,740 --> 00:13:50,240 If there is anything, OK, now what we are doing, we will first import our in the output of our 60 205 00:13:50,250 --> 00:13:51,840 and we will use. 206 00:13:52,230 --> 00:14:00,270 OK, now we will first find the desert and the vector orientation, and we want this out in the vector 207 00:14:00,270 --> 00:14:00,900 position. 208 00:14:01,270 --> 00:14:08,130 And when our joints are at nominal or position, OK or configuration. 209 00:14:08,430 --> 00:14:09,720 Q And is this one? 210 00:14:09,870 --> 00:14:13,500 So our first joint, this is our second, third, fourth, fifth and sixth. 211 00:14:13,830 --> 00:14:21,330 When our joints in this position are in the vector will be in that or postdoc position and orientation. 212 00:14:22,290 --> 00:14:25,960 So we want to find inverse kinematic of this one. 213 00:14:25,980 --> 00:14:32,210 Let's find inverse kinematic of this by I kinematics six as power. 214 00:14:32,280 --> 00:14:35,020 Be careful this is for spherical robots. 215 00:14:35,040 --> 00:14:35,390 OK. 216 00:14:36,030 --> 00:14:40,390 The robot with six degrees of freedom and spherical wrist, as we have see. 217 00:14:40,440 --> 00:14:45,810 Then we can apply what kinematic decoupling and solve our English climatiques easily. 218 00:14:46,050 --> 00:14:51,180 What this function does is the same as kinematic decoupling. 219 00:14:51,180 --> 00:14:55,020 Forget this is for all this vertical wrists and six degrees of freedom robots. 220 00:14:55,020 --> 00:15:01,350 You cannot apply this to redundant robots or robots without, um, spherical wrist. 221 00:15:01,360 --> 00:15:01,800 OK? 222 00:15:02,130 --> 00:15:07,860 As you can see now, let's be calculated the inverse kinematics and get our desired joints. 223 00:15:08,160 --> 00:15:16,450 As you can see, our design joints are very, very different from Dan from that our original one. 224 00:15:16,470 --> 00:15:16,810 OK. 225 00:15:16,830 --> 00:15:17,190 Why? 226 00:15:17,460 --> 00:15:24,480 Because we have seen that our robot with during in the market, we can have multiple or even infinite 227 00:15:24,480 --> 00:15:25,470 solutions, OK? 228 00:15:25,650 --> 00:15:30,390 As you can see, this is one of the multiple solutions which is different from CU end. 229 00:15:30,600 --> 00:15:39,780 So when you give that joined the values to your robot manipulator, you will get this in and the picture 230 00:15:39,780 --> 00:15:40,530 orientation. 231 00:15:40,530 --> 00:15:47,610 And in the same way, if you applied these joint variables, you will get again the same joint, excuse 232 00:15:47,610 --> 00:15:48,750 me, and the vector pose. 233 00:15:48,990 --> 00:15:51,360 Let's check whether it's correct or not. 234 00:15:51,540 --> 00:15:57,350 Let's find again forward kinematics of our robot manipulator AQI. 235 00:15:57,630 --> 00:16:06,450 If Q and Q and do give the same and the vector pose, then the answer of this will have be have to be 236 00:16:06,450 --> 00:16:07,740 equals to T. 237 00:16:07,770 --> 00:16:10,500 OK, let's calculate press f nine. 238 00:16:11,010 --> 00:16:14,790 Oh, excuse me, but do it in this way and press f nine. 239 00:16:15,780 --> 00:16:18,000 OK, now let's see our t. 240 00:16:18,030 --> 00:16:21,500 As you can see, they are almost the same. 241 00:16:21,510 --> 00:16:22,020 OK? 242 00:16:22,050 --> 00:16:23,400 They are almost the same. 243 00:16:23,610 --> 00:16:24,660 What does it mean? 244 00:16:24,900 --> 00:16:32,090 This means that two AI and Q n while they are very different joint variables. 245 00:16:32,100 --> 00:16:35,160 OK, as you can see, Q and I are very different. 246 00:16:35,190 --> 00:16:42,810 However, when they are applied to our robot manipulator, they give the same end effector pause. 247 00:16:42,870 --> 00:16:47,970 OK, now let's try to visualize Visualize. 248 00:16:49,980 --> 00:16:56,850 OK, now you can see that here our robot manipulator have different configurations and they're, you 249 00:16:56,850 --> 00:17:03,240 know, different solutions will give different configurations to our robot, OK, in order to reach 250 00:17:03,240 --> 00:17:03,660 the same. 251 00:17:03,660 --> 00:17:06,290 And the factor, for example, what does it mean? 252 00:17:06,300 --> 00:17:14,490 Are you, for example, our robot manipulator can have, um, right side and arm up, OK, elbow up 253 00:17:14,700 --> 00:17:21,180 or lift or lift and elbow up or right elbow down or lift. 254 00:17:21,180 --> 00:17:22,130 Elbow down. 255 00:17:22,480 --> 00:17:24,870 OK, let's try to visualize this one. 256 00:17:24,870 --> 00:17:36,810 For example, you will see what I mean, and let's try to p 562 dot plot three d July one. 257 00:17:37,260 --> 00:17:41,010 As you can see now, you can see that it is right, OK? 258 00:17:41,040 --> 00:17:44,160 As you can see, this is right and the elbow up. 259 00:17:44,160 --> 00:17:45,690 As you can see, the elbow is up. 260 00:17:45,840 --> 00:17:46,500 Let's see. 261 00:17:46,500 --> 00:17:47,820 Now let's. 262 00:17:48,030 --> 00:17:50,490 And elbow up, what does it mean? 263 00:17:51,550 --> 00:17:53,400 Okay, try to visualize this one. 264 00:17:53,730 --> 00:17:59,270 Oh, excuse me, this is one, but we want Q I too cute. 265 00:17:59,310 --> 00:18:00,650 I too. 266 00:18:00,690 --> 00:18:04,440 As you can see, now it is left it. 267 00:18:04,440 --> 00:18:05,580 Can you see the difference? 268 00:18:07,200 --> 00:18:13,740 Let me just do this is the same the problematic side with the competitive cookies library that if I 269 00:18:14,070 --> 00:18:17,610 do again, plot 3-D, it will update us on this one. 270 00:18:17,610 --> 00:18:20,580 So you have to watch it separately, but be careful. 271 00:18:20,640 --> 00:18:24,660 You can see that it is the left side, OK and elbow up. 272 00:18:25,020 --> 00:18:28,470 As you can see, there are two completely different configuration. 273 00:18:28,470 --> 00:18:32,370 But and for Q1, as you can see, this is right side of the robot. 274 00:18:33,180 --> 00:18:38,810 There are two different configurations, but end effector is the same pose. 275 00:18:38,820 --> 00:18:40,620 OK, we achieve the same pose. 276 00:18:40,980 --> 00:18:43,860 This me, this is what does it mean? 277 00:18:44,130 --> 00:18:46,320 Inverse kinetics have multiple solutions. 278 00:18:46,560 --> 00:18:50,010 OK, you can also have elbow down configuration. 279 00:18:50,040 --> 00:18:52,860 Let's check what will be elbow tone. 280 00:18:53,520 --> 00:18:54,040 OK. 281 00:18:54,060 --> 00:18:56,310 As you can see now, our elbow down. 282 00:18:56,610 --> 00:18:59,340 Yeah, you can see right and elbow down. 283 00:18:59,880 --> 00:19:03,900 Now let's check for the left and elbow down. 284 00:19:07,360 --> 00:19:14,710 You can see that all of these gave the scene and the orientation now, as you can see, this is don 285 00:19:15,040 --> 00:19:17,350 elbow down, OK and left side. 286 00:19:17,920 --> 00:19:24,730 You can also check with wrist rotated or wrist not rotated like that. 287 00:19:24,820 --> 00:19:25,300 OK. 288 00:19:25,330 --> 00:19:33,220 As you can see, for one and the victor pause in order to get one in the fact we can have multiple joint 289 00:19:33,280 --> 00:19:34,270 configurations. 290 00:19:34,690 --> 00:19:40,720 So this is what inverse that means that in the kinetics have multiple or infinite solutions. 291 00:19:42,100 --> 00:19:44,200 What if we do this one? 292 00:19:44,710 --> 00:19:51,490 Because there are some cases that are we can want our in the vector to achieve, but our robot's physical 293 00:19:51,490 --> 00:19:52,900 limits don't allow that. 294 00:19:53,170 --> 00:19:53,680 OK? 295 00:19:53,920 --> 00:20:01,490 For example, if you try to reach something that's very far from you, your arm cannot reach it. 296 00:20:01,510 --> 00:20:05,620 OK, so your inverse kinematics will give an error. 297 00:20:05,830 --> 00:20:13,600 OK, your hands in the schematics will give an error or your brain because your brain calculates inverse 298 00:20:13,600 --> 00:20:14,080 kinematics. 299 00:20:14,080 --> 00:20:20,590 For example, if some object is very far from you and you want to grab it, hold your arm in length 300 00:20:20,590 --> 00:20:24,630 is not sufficient, so your brain will see to you that you cannot reach it. 301 00:20:24,640 --> 00:20:28,770 OK, because there is a problem, there is no solution for your innermost kinematics. 302 00:20:28,780 --> 00:20:34,990 The same way if we give, for example, translation on X X is three units. 303 00:20:35,290 --> 00:20:40,210 Our robot manipulator will not reach three metres. 304 00:20:40,240 --> 00:20:43,900 OK, our robot, the manipulator cannot reach at the end. 305 00:20:43,900 --> 00:20:49,330 The fact that we have not, we will not be able to reach to that end effector position. 306 00:20:49,570 --> 00:20:52,990 OK, so our English schematics will return. 307 00:20:53,290 --> 00:20:53,890 None. 308 00:20:54,040 --> 00:20:54,700 OK. 309 00:20:56,530 --> 00:20:59,140 As you can see, turning point not reachable. 310 00:20:59,350 --> 00:21:07,660 This can happen when during the robot reaches two singularities or in the singularities or even our 311 00:21:07,660 --> 00:21:13,450 links are, you know, colliding with each other. 312 00:21:13,660 --> 00:21:19,050 So that's why sometimes the inverse kinematics have no solution. 313 00:21:19,060 --> 00:21:25,780 Or it can also happen that the solutions that are formed by inverse schematic, some of them are not 314 00:21:25,780 --> 00:21:32,530 applicable because of, as you know, the physical limits of our robot manipulator, because during 315 00:21:32,530 --> 00:21:38,950 inward schematics, we don't take into account account the internal collisions of links or, uh, you 316 00:21:38,950 --> 00:21:43,000 know, the length of the robots manipulators. 317 00:21:44,110 --> 00:21:48,810 The physical limits of our robotic manipulator in terms of length. 318 00:21:49,210 --> 00:21:59,440 So that's why our inverse clematis can also don't find any solution to our end effector posent orientation.