1 00:00:00,090 --> 00:00:05,550 In robotics, generally, we try to describe each of the bodies that we are interested as points and 2 00:00:05,550 --> 00:00:07,380 represent them on different frames. 3 00:00:07,860 --> 00:00:13,920 For example, in this case, the camera is assigned frame eight with frame accesses of eight x a y and 4 00:00:13,950 --> 00:00:20,580 is that the robot is assigned frame B with frame axis of B, XB Y and B's it and so on. 5 00:00:21,030 --> 00:00:25,830 These frames can be inertial or look of moving or stable, depending on our needs. 6 00:00:26,120 --> 00:00:32,160 Generally, the local frames like frames A, B and C are stuck on the bodies of interest, as these 7 00:00:32,160 --> 00:00:38,460 make both kinematics and dynamics calculations easier, which we will see during the future lessons. 8 00:00:39,000 --> 00:00:45,100 The inertial frame generally doesn't move, and it is needed in order to get absolute values of Victorian 9 00:00:45,120 --> 00:00:52,890 quantities like velocity acceleration, angular velocity, angular acceleration and do pictorial calculations 10 00:00:52,890 --> 00:00:53,430 on them. 11 00:00:53,910 --> 00:00:59,910 Because, as you know, in order to do calculations on Victorian quantities, they need to be represented 12 00:00:59,910 --> 00:01:03,600 on the same frames so to have the same unit vectors. 13 00:01:04,050 --> 00:01:10,350 Let's consider a point that's represented with respect to two frames, namely inertial oh zero and look 14 00:01:10,530 --> 00:01:11,100 or one. 15 00:01:11,490 --> 00:01:18,780 Then we can describe that point with respect to two frames like that zero hud j zero had zero had I 16 00:01:18,780 --> 00:01:22,080 want had j one had K one had our unit vectors. 17 00:01:22,080 --> 00:01:28,200 In this case, however, it is not enough to just represent bodies on different frames. 18 00:01:28,200 --> 00:01:34,680 But the key thing that's fundamental concept for robotics is describing the relation and transformation 19 00:01:34,680 --> 00:01:36,000 between these frames. 20 00:01:36,510 --> 00:01:42,600 In order to understand the importance of this issue, let's consider below example as it can be seen 21 00:01:42,600 --> 00:01:43,530 from the picture. 22 00:01:43,530 --> 00:01:49,740 We have inertial frame, which is not moving and local frames assigned to each of the bodies in this 23 00:01:49,740 --> 00:01:54,480 case a robot, some random object robot with camera and fixed camera. 24 00:01:54,780 --> 00:02:00,270 For example, we want to know the distance between the robot and the object on inertial frame. 25 00:02:00,810 --> 00:02:06,330 We can do that firstly, by finding the object's distance from the fixed camera with respect to the 26 00:02:06,330 --> 00:02:12,120 frame of the fixed camera, then find the robot's distance from the fixed camera with respect to frame 27 00:02:12,120 --> 00:02:13,170 of the fixed camera. 28 00:02:14,220 --> 00:02:20,790 Then we can find distance between the robot and the camera from these two vector quantities. 29 00:02:20,790 --> 00:02:27,090 And if we want, we can additional represent this distance in inertial frame in order to do some further 30 00:02:27,090 --> 00:02:27,900 calculations. 31 00:02:28,260 --> 00:02:34,170 As you can see, even for this simple calculation, which is very common in robotics, we use different 32 00:02:34,170 --> 00:02:36,960 transformations and relations between the frames. 33 00:02:38,010 --> 00:02:42,270 In order to do such kind of manipulations, we will need two essential tools. 34 00:02:42,690 --> 00:02:45,780 The first is rotation matrix, which I will explain now. 35 00:02:46,020 --> 00:02:51,090 The second is homogeneous transformation matrix, which will be discussed on the following lessons. 36 00:02:51,630 --> 00:02:53,250 So what is the rotation matrix? 37 00:02:53,730 --> 00:03:00,600 Let's assume that we have a zero coordinate frame, and we rotate this frame by two degrees around zero 38 00:03:00,600 --> 00:03:01,110 axis. 39 00:03:01,350 --> 00:03:04,680 As a result of rotation, we got a new frame of one. 40 00:03:05,040 --> 00:03:10,740 You can ask why not use data itself to describe rotations, which is obvious way of doing that. 41 00:03:11,040 --> 00:03:13,080 However, it has two disadvantages. 42 00:03:13,380 --> 00:03:19,530 Firstly, they can be cases in which small orientation change of epsilon can cause large changes in 43 00:03:19,530 --> 00:03:20,640 the value of data. 44 00:03:21,120 --> 00:03:28,860 Secondly, it is hard to use this representation during 3D cases, so we use less obvious way of rotation 45 00:03:28,860 --> 00:03:32,070 mattresses to describe rotations. 46 00:03:32,370 --> 00:03:33,840 Anyway, let's continue. 47 00:03:34,470 --> 00:03:40,560 So let's try to find the projection of unit vectors of frame or one onto the frame of zero. 48 00:03:41,970 --> 00:03:48,480 As you know, a vector can be represented in a frame by projecting it on the axis of the frame using 49 00:03:48,480 --> 00:03:49,680 simple trigonometry. 50 00:03:50,190 --> 00:03:54,870 We will try to do the same here for each of the axis of the frame or one. 51 00:03:55,350 --> 00:04:00,210 So we have to Project Y one on that zero and then y zero. 52 00:04:00,810 --> 00:04:04,890 Then we project that one onto y zero and then z zero. 53 00:04:05,490 --> 00:04:06,900 In terms of x one. 54 00:04:06,900 --> 00:04:12,870 As the rotation is about x zero axis x zero and x one will be the same. 55 00:04:13,350 --> 00:04:17,100 So we got x one zero one zero and z one zero. 56 00:04:17,530 --> 00:04:22,170 Finally, we can construct our rotation matrix by combining them in a matrix form. 57 00:04:22,380 --> 00:04:25,380 The rotation matrix is depicted by R1 zero. 58 00:04:25,650 --> 00:04:31,500 The meaning of the numbers is that the rotation matrix represents a rotation from frame one to zero. 59 00:04:32,010 --> 00:04:38,970 The same mattresses can be derived for rotation about y zero and z zero axis, as you can do as a Holmberg, 60 00:04:38,970 --> 00:04:41,160 which is very similar to the above derivation. 61 00:04:41,520 --> 00:04:44,730 These mattresses are called element rotation mattresses. 62 00:04:45,300 --> 00:04:51,090 All of the rotation mattresses are not only useful for expressing the orientation of one frame relative 63 00:04:51,090 --> 00:04:51,750 to another. 64 00:04:52,020 --> 00:04:54,720 Let's see there another use of full implementation. 65 00:04:54,930 --> 00:04:59,900 Assume that a point p is represented in two frames, namely o zero and or. 66 00:04:59,970 --> 00:05:07,470 One then we can right, Victor P, with respect to two friends like that, as these two vectors are 67 00:05:07,470 --> 00:05:14,550 the same, we can equalize equations one and two and get equations three, then we can replace unit 68 00:05:14,550 --> 00:05:18,600 vectors of frame one with frame zero and get equation form. 69 00:05:19,020 --> 00:05:25,170 Finally, we can solve four zero one zero zero zero in terms of x one y ones at one. 70 00:05:25,410 --> 00:05:30,960 After doing some algebraic manipulations, we can write the last equations in matrix form. 71 00:05:30,960 --> 00:05:38,220 Like that, you can easily see the rotation matrix of R1 zero, which represents the rotation about 72 00:05:38,220 --> 00:05:39,360 the x zero axis. 73 00:05:39,360 --> 00:05:45,630 This shows another user full implementation of rotation mattresses, which is representing transformation 74 00:05:45,630 --> 00:05:49,740 between the coordinates of a point expressed in two different frames. 75 00:05:50,670 --> 00:05:54,450 Let's see one more implementation case of the rotation mattresses. 76 00:05:54,810 --> 00:05:57,000 Take this scenario into consideration. 77 00:05:57,360 --> 00:06:04,740 We have inertial frame of zero, and moving frame of one frame or one is moving from initial configuration 78 00:06:04,740 --> 00:06:05,300 of all. 79 00:06:05,340 --> 00:06:11,880 One initial equals two or one zero to final configuration of all one final. 80 00:06:12,210 --> 00:06:20,550 Additionally, we have a point p attach to frame or one and move with it from p ie to p f as point p 81 00:06:20,550 --> 00:06:24,690 doesn't move with respect to frame B because it is attached to it. 82 00:06:24,870 --> 00:06:31,290 We can write this equation also from the definition of rotation method as we can write this equation. 83 00:06:32,100 --> 00:06:36,600 Then from the first and second equations, we can get this final equation. 84 00:06:37,050 --> 00:06:43,560 So it is obvious from this relation that rotation mattresses can be used to rotate a vector within the 85 00:06:43,560 --> 00:06:45,000 same coordinate frame. 86 00:06:45,870 --> 00:06:49,860 This is all about the concept and implementation of the rotation mattresses. 87 00:06:49,860 --> 00:06:55,530 In the next lesson, we will discuss some important properties of the new rotation mattresses and their 88 00:06:55,530 --> 00:06:56,340 composition. 89 00:06:56,850 --> 00:06:58,050 See on the next lesson.