WEBVTT 00:00.260 --> 00:06.860 Now that we have studied and solved separately the translation and the rotation problems, which allowed 00:06.860 --> 00:12.710 us to determine the position and the orientation of one reference frame with respect to another. 00:12.830 --> 00:18.830 Let's see how to approach the general and complex problem of determining the pose of an object in the 00:18.830 --> 00:26.300 space and therefore determining the rotor translation of our reference frame with respect to another. 00:26.330 --> 00:32.900 In this case, we are dealing with a general scenario where we always have a fixed reference frame. 00:32.900 --> 00:39.770 So the word frame and our mobile reference frame that is fixed that is attached to the robot. 00:40.190 --> 00:48.260 This time the mobile frame is both translated and rotated with respect to the fixed world reference 00:48.260 --> 00:48.770 frame. 00:49.670 --> 00:55.790 We can see that the two reference frames are indeed translated from each other, since their origins 00:55.820 --> 00:57.080 do not coincide. 00:57.110 --> 00:59.510 They are not at the same point. 00:59.720 --> 01:06.170 We can also see that they are rotated with respect to each other as the direction of the x, y and z 01:06.200 --> 01:06.780 axis. 01:06.830 --> 01:09.530 Different between the two reference frames. 01:10.170 --> 01:17.120 We can quantitatively indicate the translation and the rotation of the mobile reference frame with the 01:17.160 --> 01:25.020 respect to the fixed reference frame by using the coordinates X, Y, and Z of the mobile frame origin 01:25.020 --> 01:33.240 with the respect to the fixed reference frame as well as the angles theta and phi of rotation around 01:33.240 --> 01:34.260 the three axes. 01:34.830 --> 01:42.570 Once again, the kinematic problem consists of determining the position of a given point p in the global 01:42.570 --> 01:48.030 world reference frame, knowing its position in the inertial reference frame. 01:48.030 --> 01:49.440 So the mobile frame. 01:49.650 --> 01:56.070 And so knowing the translation and the rotation to which the mobile frame is subjected with respect 01:56.070 --> 01:57.150 to the world frame. 01:58.200 --> 02:01.650 We have already solved a simplified version of this problem. 02:01.770 --> 02:07.920 We solved the problem when the two reference frames were oriented the same way, but they were only 02:07.920 --> 02:12.250 translated with respect to each other using the translation vector. 02:12.300 --> 02:18.060 And then we also solved the problem when the two reference frame at the same origin, but they were 02:18.060 --> 02:22.980 only rotated, so they were differently orientated using the rotation matrices. 02:23.790 --> 02:24.990 To solve the problem. 02:24.990 --> 02:32.730 In the general case where the two reference frames are arbitrarily translated and rotated, it is enough 02:32.730 --> 02:40.020 to combine these two equations and thus combine the two matrices to obtain a 4x4 matrix. 02:41.150 --> 02:44.300 The elements of this new matrix are familiar to us. 02:44.330 --> 02:45.530 They are not new. 02:45.560 --> 02:52.550 The first three rows and columns, in fact, represents exactly the rotation matrix that express only 02:52.550 --> 02:57.860 the orientation of the mobile reference frame with respect to the fixed reference frame. 02:58.540 --> 03:05.590 Similarly, the first three rows of the last column represent the translation vector that express the 03:05.590 --> 03:10.750 position of the mobile reference frame with respect to the fixed reference frame. 03:12.790 --> 03:18.820 Finally, the last row of this column is set to zero, except for the last element. 03:18.850 --> 03:25.450 This row actually doesn't add anything to the logic or to the equations for the coordinate transformation 03:25.450 --> 03:27.380 between the two reference frames. 03:27.400 --> 03:34.720 It simply ensures that this matrix is a square matrix, so namely that it has the same number of rows 03:34.720 --> 03:35.800 and columns. 03:36.960 --> 03:43.500 This property will be very useful when we combine and multiply many cumulative transformations. 03:44.840 --> 03:52.310 Overall, these 4x4 matrix is called the Transformation Matrix, and it allows us to obtain the coordinates 03:52.310 --> 03:55.190 of a point in the fixed reference frame. 03:55.190 --> 04:00.890 Given the coordinates of the same point in the mobile reference frame and also vice versa. 04:00.890 --> 04:06.800 So it enables us to obtain the coordinates of a point in any mobile reference frame. 04:06.800 --> 04:13.580 So also if it is rotated or translated given the coordinates of the same point in the fixed reference 04:13.580 --> 04:14.150 frame. 04:14.150 --> 04:20.630 So simply by calculating then the inverse of this matrix, so the inverse of the transformation matrix.