1 00:00:00,150 --> 00:00:06,420 OK, after we have seen some preliminary knowledge related dynamics or a little organized dynamics, 2 00:00:06,420 --> 00:00:10,950 let's jump into the derivation of Eyler Lagrange modeling of robotic manipulators. 3 00:00:11,520 --> 00:00:17,190 As we have said before, choice of generalized coordinates during dynamic modeling is very important. 4 00:00:17,580 --> 00:00:23,760 If we don't see a suitable set of generalized coordinates or dynamic equations can become unnecessarily 5 00:00:23,760 --> 00:00:24,900 difficult to obtain. 6 00:00:25,260 --> 00:00:31,380 And we don't want that because dynamic modeling of robotic manipulators is already a complex problem. 7 00:00:31,860 --> 00:00:38,940 We have two choices for generalized coordinates either choose work space coordinates, namely X, Y 8 00:00:38,940 --> 00:00:43,590 and Z, or choose joint angles Q as generalized coordinates. 9 00:00:44,040 --> 00:00:49,950 Indeed, doing calculations with joint angles are easier because we will not use inverse or Jacobean 10 00:00:49,950 --> 00:00:53,110 mattresses, which can cause problems during singularities. 11 00:00:53,160 --> 00:00:57,480 Let's start with the formula of kinetic energy, which we have seen before. 12 00:00:57,720 --> 00:01:04,410 However, as we have two joint angles as generalized coordinates, we need to express linear velocity 13 00:01:04,410 --> 00:01:09,160 and angular velocity terms in terms of joint angles or joint velocities. 14 00:01:09,180 --> 00:01:09,720 Excuse me. 15 00:01:10,590 --> 00:01:17,640 And here we will use Jacoby mattresses in order to convert from joint space velocities to the linear 16 00:01:17,640 --> 00:01:20,550 and angular velocities at work space. 17 00:01:20,760 --> 00:01:25,530 Mainly, we will use velocity kinematics, as you can see from these formulas. 18 00:01:26,010 --> 00:01:31,620 I hope you can remember these formulas because we have seen them during velocity kinematics tutorial 19 00:01:32,220 --> 00:01:32,540 here. 20 00:01:32,550 --> 00:01:38,850 PCI is nothing but the velocity of the center of mass of the link, i.e. here is the velocity and angular 21 00:01:38,850 --> 00:01:39,870 velocity Jacobean. 22 00:01:40,590 --> 00:01:47,190 If you analyze these mattresses, you will see they have interesting structure in this structure. 23 00:01:47,200 --> 00:01:53,010 Take into account the coupling between velocities of the joints, and the structure is very logical 24 00:01:53,010 --> 00:01:54,260 and easy to understand. 25 00:01:54,270 --> 00:01:58,920 Indeed, as you can see, the Jacobins after the link is zero. 26 00:01:58,950 --> 00:02:03,540 However, before the link, i.e. they are non-zero, let's look this example. 27 00:02:04,080 --> 00:02:08,690 Let's assume that we want to find the linear and angular velocity Jacobins falling. 28 00:02:08,700 --> 00:02:15,480 I then I +1 and I +2 are the next links and I am minus one is the previously. 29 00:02:15,780 --> 00:02:20,740 The motion of next means will not affect the velocity of Link I. 30 00:02:20,850 --> 00:02:23,190 You can observe this from your hand. 31 00:02:23,310 --> 00:02:27,390 Do your fingers motion contribute to the velocity of your wrist? 32 00:02:27,690 --> 00:02:31,530 No, because fingers are next things to your wrist. 33 00:02:32,070 --> 00:02:33,660 How are your shoulder? 34 00:02:33,660 --> 00:02:39,450 Is previous linked with respect to your wrist and its motion will contribute to the velocity of the 35 00:02:39,450 --> 00:02:39,840 wrist. 36 00:02:40,470 --> 00:02:42,420 The same logic is valid here. 37 00:02:42,630 --> 00:02:44,940 OK, let's write the equations here again. 38 00:02:45,330 --> 00:02:52,800 If we consider the four equations of VCI and Omega IV, you can plug them into the form of kinetic energy 39 00:02:52,800 --> 00:02:53,650 and write them. 40 00:02:53,650 --> 00:03:02,700 In this way, you can see that both of the terms contain pure dot and its transpose, so we can take 41 00:03:02,700 --> 00:03:06,720 them out as a common multiplier and write the equation in this way. 42 00:03:07,020 --> 00:03:12,720 And finally, let's call the middle matrix as M and write equation in much simpler way. 43 00:03:13,500 --> 00:03:15,600 Does this equation remind you something? 44 00:03:16,080 --> 00:03:21,810 I think so, because this is very similar to the formula of kinetic equation from basic physics, which 45 00:03:21,810 --> 00:03:29,850 was one over two m v squared instead of linear velocity V if you have Q Dot here, which is also velocity 46 00:03:29,850 --> 00:03:30,210 term. 47 00:03:30,930 --> 00:03:34,380 So the middle term has to be must related symptom. 48 00:03:34,800 --> 00:03:36,730 Indeed, your assumption is correct. 49 00:03:37,270 --> 00:03:45,360 M fix is called inertia matrix and it is square symmetric and the positive definite it has to be positive 50 00:03:45,360 --> 00:03:48,480 definite because must always be positive definite. 51 00:03:49,020 --> 00:03:55,440 Additionally, its configuration dependent because, as you can see from the formula, it is function 52 00:03:55,440 --> 00:03:57,630 of Q namely configuration. 53 00:03:57,900 --> 00:04:01,560 Indeed, we can decompose it and write in this way. 54 00:04:02,190 --> 00:04:09,810 This indeed shows us an interesting property of kinetic energy, which is unusual for us because from 55 00:04:09,810 --> 00:04:14,370 the basic physics, we know that the kinetic energy is velocity dependent on it. 56 00:04:14,580 --> 00:04:20,910 However, from this formulation, we can see that it is also configuration dependent as we have expressed 57 00:04:20,910 --> 00:04:24,580 kinetic energy with joint velocities Q Dot. 58 00:04:24,840 --> 00:04:31,410 Then we can find Lagrange Equation because for potential energy, we don't need to express it in terms 59 00:04:31,410 --> 00:04:39,210 of joint angles, I mean, generalized Corness as it is all expressed in joint angles PC II, which 60 00:04:39,210 --> 00:04:45,420 is this with the centre of mass of being IE, is function of just joint angles. 61 00:04:46,140 --> 00:04:49,050 So here is the expression for Lagrangian function. 62 00:04:49,680 --> 00:04:53,640 Here's the formulation for a little Lagrange form that we have seen before. 63 00:04:53,940 --> 00:04:55,560 Let's calculate the first term. 64 00:04:56,190 --> 00:04:59,820 As you can see, it is nothing but the partial derivative of like. 65 00:04:59,980 --> 00:05:06,160 Engine function with respect, joint velocities, and you know that only the kinetic energy is a function 66 00:05:06,160 --> 00:05:08,650 of joint velocities, not potential energy. 67 00:05:09,430 --> 00:05:10,600 So potential. 68 00:05:10,810 --> 00:05:16,030 So partial derivative of potential energy with respect to joint velocities will be zero. 69 00:05:16,270 --> 00:05:22,540 And we will have only the some of the products of mass mattresses with joint velocities at the end. 70 00:05:22,780 --> 00:05:25,390 Do you interested how this last term obtained? 71 00:05:25,900 --> 00:05:32,140 Indeed, it is very easy to see that, but let's take the example of two degrees of freedom manipulator 72 00:05:32,410 --> 00:05:40,540 and calculate partial derivative of kinetic energy with respect to joint velocities when K is equal 73 00:05:40,540 --> 00:05:41,110 to one. 74 00:05:41,470 --> 00:05:46,480 You can see here we have two variables, namely and one two and and two one. 75 00:05:47,410 --> 00:05:54,040 Or as initial matrix is symmetric, they will be the same so we can sum them together and we will get 76 00:05:54,040 --> 00:05:55,600 this final expression. 77 00:05:56,260 --> 00:06:02,290 Now, after we have finished with partial derivative part, let's find its time derivative, which is 78 00:06:02,290 --> 00:06:03,070 very easy. 79 00:06:03,580 --> 00:06:06,460 Don't forget the rule of derivative of summation. 80 00:06:06,940 --> 00:06:13,660 Here we have to find time derivative of inertia matrix also, because as we have said before, it is 81 00:06:13,660 --> 00:06:17,200 configuration dependent, so changes with respect to time. 82 00:06:17,830 --> 00:06:20,650 So it's time derivative will be found in this way. 83 00:06:20,890 --> 00:06:27,640 Here we have used chain rule as inertia matrix, which is a function of configuration, which is also 84 00:06:27,640 --> 00:06:29,260 it serves a function of time. 85 00:06:29,740 --> 00:06:31,270 Here is the chain formula. 86 00:06:31,880 --> 00:06:35,560 I'm sorry for putting it up because there was no free space. 87 00:06:36,010 --> 00:06:41,950 Anyway, at the end of the at the end, the resulting formulation will be like that. 88 00:06:42,070 --> 00:06:46,120 As we have finished with first term, we can continue with the second one. 89 00:06:46,160 --> 00:06:50,980 Namely, partial derivative of Lagrangian function with respect to joint variables. 90 00:06:51,430 --> 00:06:57,430 In order to calculate this term, we have to again write Lagrangian function in terms of kinetic and 91 00:06:57,430 --> 00:06:58,690 potential energies. 92 00:06:59,170 --> 00:07:04,570 Then we can individually find their partial derivatives with respect to joint variables. 93 00:07:05,200 --> 00:07:11,980 The partial derivative of potential energy will be very easy, as it is only a function of configuration, 94 00:07:12,400 --> 00:07:14,830 so we will get only gravity vector. 95 00:07:15,610 --> 00:07:17,800 Now we can combine two terms. 96 00:07:17,830 --> 00:07:23,500 We have calculated and obtained earlier Legrand's dynamic model of the remote manipulator. 97 00:07:24,220 --> 00:07:31,420 If we called the middle term as H K.J IQ, we can ride the formula in simpler way. 98 00:07:32,020 --> 00:07:38,560 As we have said before, C K is nothing but the conservative forces or some of the conservative forces, 99 00:07:38,770 --> 00:07:45,160 meaning the joint actuator towards external or contact forces and joint friction forces. 100 00:07:45,400 --> 00:07:51,940 Don't forget that there there are other forces also, but we cannot model each and every of them. 101 00:07:52,180 --> 00:07:59,350 This is because, firstly, it is impossible to model every one of them in the exact manner. 102 00:07:59,560 --> 00:08:07,000 Secondly, if we even can do that, modelling them will make the dynamic equation complex. 103 00:08:07,390 --> 00:08:15,010 So it is enough for us for now to model just these key components of non-conservative forces. 104 00:08:15,670 --> 00:08:21,520 So finally, we can write a little like large dynamic model of the robot manipulator in this way. 105 00:08:22,210 --> 00:08:27,250 Try to compare this formula because we will need it almost in every future topics. 106 00:08:27,490 --> 00:08:32,380 Before finishing the video, I want to just give short note about Aristotle's symbols. 107 00:08:33,010 --> 00:08:39,580 Let us know the dynamic model here again now focused on this Si matrix, which is called carryall this 108 00:08:39,580 --> 00:08:42,730 matrix and it can be decomposed in this way. 109 00:08:42,850 --> 00:08:49,930 Here, C I J K is nothing but components of this matrix that can be calculated in this way. 110 00:08:50,860 --> 00:08:54,580 This C IG K components are called crystal symbols. 111 00:08:54,850 --> 00:08:56,560 You don't have to memorize them. 112 00:08:57,070 --> 00:09:03,460 Even it would be possible to miss this detail, but I have mentioned it for one important thing that 113 00:09:03,460 --> 00:09:05,050 we will use in future lessons. 114 00:09:05,770 --> 00:09:11,710 We can just see Matrix not only by a combination of C IG K elements, but in other ways. 115 00:09:11,740 --> 00:09:21,040 Also, as long as this equality is valid, then you can ask Why won't we choose see Matrix as combination 116 00:09:21,040 --> 00:09:22,240 of Christoffer symbols? 117 00:09:22,660 --> 00:09:28,630 In order to know this, let's assume that we have calculated the difference between derivative of inertia 118 00:09:28,630 --> 00:09:31,180 matrix and two times Coriolis matrix. 119 00:09:31,510 --> 00:09:37,300 The resultant matrix will be called KN know the important detail is here. 120 00:09:37,570 --> 00:09:43,690 If we are to see as combination of Christoph Waltz symbols, then it will be skewed symmetric. 121 00:09:44,230 --> 00:09:49,870 Namely, its symmetric elements will have different size and diagonal elements will be zero. 122 00:09:50,380 --> 00:09:53,770 Then this equality will hold for any vector of X. 123 00:09:54,250 --> 00:09:59,770 However, if we didn't choose Christoph, all symbols to construct the Coriolis matrix, then end matrix. 124 00:09:59,860 --> 00:10:06,380 Will not be symmetric and previous equality will hold when X is vector of joint velocities only. 125 00:10:07,040 --> 00:10:09,170 Now you can say, OK, so what? 126 00:10:09,830 --> 00:10:16,850 Yes, it can be seen as useless detail for now, but just keep this property in mind and you will see 127 00:10:16,850 --> 00:10:19,490 how useful is it in future lessons? 128 00:10:19,700 --> 00:10:20,750 Well, one more thing. 129 00:10:21,320 --> 00:10:28,070 Don't forget that during this course, we will always assume that sea matrix is constructed by Christopher 130 00:10:28,070 --> 00:10:32,930 symbols so skewed symmetric property of any matrix will always hold.