WEBVTT 00:00.230 --> 00:06.710 The pushover manipulator robot, or rather the position and the orientation of the gripper of a manipulator 00:06.710 --> 00:14.510 robot is defined by six variables, three that define its position and three that define its orientation. 00:15.570 --> 00:21.900 This is a very complex problem to solve, as is the mathematics and the equations that express the pose 00:21.900 --> 00:22.680 of a robot. 00:22.680 --> 00:28.530 And therefore the relationship between the reference frame are attached to the robot's gripper and the 00:28.530 --> 00:32.430 fixed reference frame of the word that we called W. 00:33.080 --> 00:39.290 When engineers and mathematicians are faced with a complex problem to solve, the approach they often 00:39.290 --> 00:44.240 use is to divide it into smaller, more easily solvable sub problems. 00:45.260 --> 00:47.750 That's what we are doing in this case as well. 00:47.990 --> 00:54.170 The problem of determining the pose of a robot in space can be decomposed into two sub problems. 00:54.470 --> 01:01.640 The first one consists of determining only its position and therefore the translation between the reference 01:01.640 --> 01:06.710 frame attached to the robot and the fixed reference frame attached to the award. 01:07.320 --> 01:14.400 The second sub problem consists of determining only the orientation that is the rotation between the 01:14.400 --> 01:18.240 robot's reference frames and the fixed world reference frame. 01:18.830 --> 01:25.610 Once these two simple sub problems are solved, we can combine them and solve the general case of determining 01:25.610 --> 01:32.240 the pose of a robot, which includes the translation and rotation of the reference frame attached to 01:32.240 --> 01:35.930 the robot with the respect to the world reference frame. 01:36.670 --> 01:43.900 Let's start with the study of the translation and see how to mathematically express the X, Y, and 01:43.900 --> 01:46.150 Z coordinates of a robot pose. 01:47.200 --> 01:53.320 The simplified problem we want to solve in this case is to determine only the position of the reference 01:53.320 --> 02:00.640 frame R, which is inertial and is attached to the robot relative to the fixed frame of the word W. 02:01.000 --> 02:08.080 So we are assuming that the irrotational component is zero and thus the reference frame R is always 02:08.080 --> 02:11.590 oriented the same way as the reference frame w. 02:11.980 --> 02:19.780 In fact, we can see that the x, y and z axis of the two reference frame are oriented in the same way 02:19.780 --> 02:21.970 without any rotation among them. 02:22.530 --> 02:29.040 In other words, the problem we want to solve consists of calculating the coordinates of the center 02:29.040 --> 02:33.480 of the reference frame are relative to the world reference frame w. 02:34.230 --> 02:37.320 To describe the three components of the translation. 02:37.320 --> 02:45.120 So X, y, and z, we can draw a vector t that connects the center of the reference frame w with the 02:45.120 --> 02:47.280 center of the reference frame r. 02:48.320 --> 02:55.760 The projection of this vector along the X, y, and z axis of the reference frame W are precisely the 02:55.760 --> 02:56.810 X, y, and z. 02:56.810 --> 03:02.570 Coordinates of the center of the reference frame are relative to the reference frame w. 03:03.880 --> 03:11.620 Mathematically, we can say that the projection of a vector T along the X, Y, and Z axis of the reference 03:11.620 --> 03:18.730 frame w are equal to the coordinates of the center of the reference frame R relative to the world reference 03:18.730 --> 03:19.270 frame. 03:19.480 --> 03:26.560 These three equations can also be rewritten in a matrix form while maintaining the same logical meaning. 03:27.150 --> 03:34.590 Now, if we consider a real case with real values and assign a numerical value to the X, Y, and Z 03:34.590 --> 03:42.360 coordinates, we can easily say that the center of the reference frame are as coordinates three, six 03:42.360 --> 03:44.820 and five within the reference frame. 03:44.820 --> 03:45.450 W. 03:46.020 --> 03:50.280 Let's remember this result and set it aside for a moment. 03:50.310 --> 03:53.340 We will soon see why this vector is useful. 03:54.220 --> 04:01.090 Since the arm of our robot is a certain length, we might also want to know the position in the world 04:01.120 --> 04:04.000 of any point on the robot's arm. 04:04.030 --> 04:10.570 In order, for example, to avoid collision with obstacles, as we know the dimensions of the robot's 04:10.570 --> 04:17.530 arm so we know its length with a bit of trigonometry, we can easily calculate the position of the point 04:17.560 --> 04:28.060 p relative to the reference frame R and therefore we can know its coordinates x, p, y, P and z p. 04:28.420 --> 04:29.620 This information. 04:29.620 --> 04:33.100 So these coordinates are relative coordinates. 04:33.130 --> 04:37.750 They tell us where the point P is located on the robot's arm. 04:37.780 --> 04:44.560 So relative to the our reference frame, which is always attached to the robot's arm and so moves along 04:44.560 --> 04:45.220 with it. 04:45.920 --> 04:49.880 However, we might be interested in another piece of information. 04:50.060 --> 04:56.270 Still, in the case of developing an algorithm that prevents parts of the robot from colliding with 04:56.270 --> 04:57.860 other objects in the world. 04:57.890 --> 05:04.550 We might be interested in knowing the position of the point P and thus the end of the robot's arm relative 05:04.550 --> 05:05.360 to the world. 05:05.390 --> 05:11.840 That is, we might be wondering which are the coordinates of the point P relative to the world reference 05:11.840 --> 05:12.440 frame. 05:13.210 --> 05:20.020 This problem is quite simple and also is intuitive to solve, and it simply involves adding coordinates. 05:20.410 --> 05:27.220 In fact, the position of the point p in the reference frame w is given by the sum of the position of 05:27.220 --> 05:34.720 the point P in the robot reference frame R plus the position of the reference frame R relative to the 05:34.720 --> 05:38.140 reference frame w, which is defined in the vector t. 05:38.800 --> 05:46.780 We can also express this equation in a matrix form by decomposing each vector into its x, y, and z 05:46.780 --> 05:47.770 components. 05:48.160 --> 05:55.000 With this equation we can solve the problem and so determine the position of the point p in the reference 05:55.000 --> 06:00.250 frame W by knowing the position of the point P in the reference frame r. 06:01.010 --> 06:08.180 Once again, if we assign some real values and assume that we know the coordinates of the point P relative 06:08.180 --> 06:17.330 to the reference frame R which are equal to two, four and four, we just need to add these coordinates 06:17.330 --> 06:25.580 to those of the vector t which we calculated previously and which is three, six and five, and we will 06:25.580 --> 06:29.490 have the position of the point P in the reference frame W. 06:30.560 --> 06:38.180 This vector is called translation vector and indicates the distance between the reference frame R and 06:38.180 --> 06:46.820 the reference frame W and thus three components which are its projections along the X, y, and Z axis.