WEBVTT 00:00.320 --> 00:07.520 Now that we know how to express the translation of our reference frame with the respect to another one 00:07.520 --> 00:15.170 using the translation vector, let's see how to express its orientation using rotation matrices. 00:15.960 --> 00:23.550 In this case, we consider a very simple problem that involves only two reference frames, a global 00:23.550 --> 00:31.290 fixed reference frame called W, and a mobile reference frame that can freely rotate and is oriented 00:31.290 --> 00:34.620 differently from the fixed reference frame of the world. 00:35.290 --> 00:41.710 To simplify the problem, let's assume that these two reference frame have the same origin, the same 00:41.710 --> 00:47.020 center, and therefore the translation vector that connects them is zero. 00:47.620 --> 00:54.190 In this problem, we assume that the translation is zero and therefore the centers of the two reference 00:54.190 --> 00:55.990 frames always overlap. 00:56.350 --> 01:03.100 This way we can focus uniquely on the study of the orientation of the reference frame R that is attached 01:03.100 --> 01:07.450 to the robot with the respect to the global reference frame W. 01:07.480 --> 01:16.540 In other words, we want to find the angular displacement Theta Phi and PSI between the X, y and Z 01:16.570 --> 01:19.300 axis of the two reference frames. 01:19.660 --> 01:26.920 The mathematical tools to solve this problem are complex and are called rotation matrices, but their 01:26.920 --> 01:33.160 derivation is very simple and intuitive, requiring some basic understanding of trigonometry. 01:33.460 --> 01:38.720 It will help you better understand the logic and the manipulation of reference frames. 01:39.910 --> 01:46.870 As always, when faced with a complex problem, let's approach it by breaking it down into simpler and 01:46.870 --> 01:49.040 more easily solvable subproblems. 01:49.060 --> 01:56.740 So let's consider the simple case when we have a fixed reference frame and our mobile reference frame 01:56.740 --> 02:02.500 that is rotated with respect to the Z axis of the fixed reference frame. 02:03.250 --> 02:04.180 In practice. 02:04.180 --> 02:10.930 We are saying that these two reference frame in addition of having the same origin, also have the same 02:10.930 --> 02:11.950 Z axis. 02:11.950 --> 02:15.640 So the Z axis is oriented in the same way. 02:16.250 --> 02:20.260 This is also the rotation axis of the second reference frame. 02:20.270 --> 02:27.460 In other words, we have rotated the second reference frame, the one in red by an angle C with the 02:27.500 --> 02:31.460 respect to the Z axis of the first reference frame. 02:31.460 --> 02:33.560 So the fixed one in black. 02:34.260 --> 02:40.890 If we were to look at these two reference frame from above, we would observe a plane formed by the 02:40.890 --> 02:48.720 X and Y axis of the first reference frame and the Z axis coming out of the plane, so out of the screen. 02:49.360 --> 02:51.790 Consequently the second reference frame. 02:51.790 --> 02:59.140 So the mobile one will be rotated by an angle C with due respect to the first one and therefore the 02:59.140 --> 03:07.540 x first and y first axis of the second reference frame are rotated by an angle C with respect to the 03:07.540 --> 03:09.070 x and Y axis. 03:10.020 --> 03:17.730 Now let's assume that there is a point P on the x y plane of which we project the coordinates along 03:17.730 --> 03:19.410 the x and Y axis. 03:19.890 --> 03:26.940 It will have coordinates, p, x and p y along the axis of the first reference frame. 03:26.940 --> 03:28.380 So the fixed one. 03:29.040 --> 03:36.450 Similarly, we can project the point p onto the second reference frame so the rotated one obtaining 03:36.450 --> 03:40.710 the coordinates p x first and p y first. 03:41.280 --> 03:49.350 Applying some simple trigonometric equations, we know that if we project the vector p x y along the 03:49.350 --> 03:51.810 x axis of the first reference frame. 03:51.810 --> 04:00.930 So the fixed one we obtain that the component of this vector is p x first multiplied by the cosine of 04:00.930 --> 04:01.680 the angle pq. 04:02.580 --> 04:11.670 Similarly, if we project the vector p x first along the y axis of the fixed reference frame, we obtain 04:11.670 --> 04:18.960 that the y component of this vector is p x first multiplied by the sine of the angle PQ. 04:20.160 --> 04:29.490 Similarly, we can project the p first vector along the x and y axis of the fixed reference frame. 04:29.670 --> 04:39.330 So the component along the y axis is given by p y first multiplied by the cosine of the angle C, and 04:39.330 --> 04:47.310 the component along the x axis is given by p y first multiplied by the sine of the angle c. 04:48.400 --> 04:55.600 So far we haven't actually written any kinematic equation, only simple trigonometric derivations. 04:55.960 --> 05:03.760 At this point, knowing these derivations, we ask ourselves what are the values of P, x, P, y, 05:03.760 --> 05:04.810 and Z? 05:04.840 --> 05:07.120 So which are the coordinates of the point? 05:07.150 --> 05:10.270 P with the respect to the fixed reference frame. 05:10.270 --> 05:11.440 So the black one. 05:12.340 --> 05:19.930 Assuming that we know p x first, p, y first and p z first, which are the coordinates of the point 05:19.960 --> 05:23.470 P with the respect to the red reference frame. 05:23.470 --> 05:28.390 So the one that is rotated by an angle C around the Z axis. 05:29.140 --> 05:31.150 Let's start with the easy things. 05:31.360 --> 05:38.860 Let's say that since the Z and Z first axis are equal to each other, so they share the same direction 05:38.860 --> 05:39.790 and origin. 05:39.820 --> 05:46.630 The coordinate of the point P along the Z axis is the same in both reference frames. 05:47.340 --> 05:51.720 Then let's determine the x coordinates of the point. 05:51.750 --> 05:52.080 P. 05:52.840 --> 05:55.480 This is given by the sum of two components. 05:55.600 --> 06:04.030 The first one is p x first, which is the coordinate of the point P along the x first axis multiplied 06:04.030 --> 06:06.140 by the cosine of the angle PC. 06:07.100 --> 06:15.980 The second one is P first, which is the coordinate of the point P along the Y first axis multiplied 06:15.980 --> 06:22.550 by the sine of the angle PC with a minus sign indicating that it has opposite direction. 06:23.410 --> 06:25.630 Finally using the same approach. 06:25.630 --> 06:29.080 We can also determine the y coordinate of the point P. 06:29.560 --> 06:39.280 It is given by p x first, multiplied by the sine of the angle C plus p y, first multiplied by the 06:39.280 --> 06:40.690 cosine of the angle. 06:40.780 --> 06:46.990 C We can express this equation in a matrix form preserving its meaning. 06:47.290 --> 06:54.610 The three coordinates of the point P in the fixed reference frame are given by the product of a certain 06:54.610 --> 06:59.620 matrix by the vector of coordinates of a point P in the second reference frame. 06:59.620 --> 07:00.850 So the mobile one. 07:01.420 --> 07:08.290 In the matrix that connects the coordinate of the point P In the two reference frame, we need to insert 07:08.290 --> 07:12.070 all the components, all the coefficients of the equation. 07:12.910 --> 07:19.900 Since the last row and the last column refer to the components along the Z axis, we can set them to 07:19.900 --> 07:24.010 zero and set the component along the z axis to one. 07:24.280 --> 07:31.120 In practice, we have inserted the first equation into the matrix, which tells us that the coordinate 07:31.120 --> 07:35.500 of P along Z is equal to the coordinate of P along Z. 07:35.530 --> 07:45.640 First, the first row of the matrix consists of the cosine of Z and minus the sign of Z, which corresponds 07:45.640 --> 07:52.480 to the second equation which connects the x coordinate of the point P in the two reference frames. 07:53.400 --> 08:00.330 Finally, the second row consists of the sine of the angle C and the cosine of the angle. 08:00.420 --> 08:07.500 C expressing the relationship between the y coordinate of the point P in the two reference frames. 08:07.920 --> 08:11.250 This matrix is called the rotation matrix. 08:11.280 --> 08:14.810 Specifically an elementary rotation matrix. 08:14.850 --> 08:22.200 It is elementary because we assumed that the two reference frames share their Z axis and therefore they 08:22.200 --> 08:24.630 have a common axis of rotation. 08:25.020 --> 08:34.530 We denote this as R Z of C, which represents an elementary rotation matrix around the Z axis with the 08:34.530 --> 08:37.110 two reference frames rotated by an angle C. 08:38.070 --> 08:44.760 With this tool, we can now freely change the reference frame and express the coordinates of the point 08:44.790 --> 08:46.970 P in the fixed reference frame. 08:46.980 --> 08:53.670 Given the coordinates of the point P in the rotated reference frame and by knowing the rotation angle 08:53.670 --> 08:55.500 between the two reference frames. 08:56.310 --> 09:00.510 This is not the only elementary rotation matrix that we can create. 09:00.600 --> 09:02.100 There are two more. 09:02.740 --> 09:09.460 Let's assume that we have a fixed reference frame still X, Y, and Z in black, and our mobile reference 09:09.460 --> 09:12.850 frame X first, Y first and Z first. 09:12.940 --> 09:17.410 But this time the two reference frame share their y axis. 09:17.410 --> 09:21.600 So the y and y first axis are equal. 09:21.610 --> 09:25.390 They are oriented the same way and they have the same origin. 09:25.900 --> 09:33.130 Then the second reference frame in red is rotated around the y axis by an angle phi. 09:34.350 --> 09:41.250 In this case, if we look at the system from above, imagining ourselves on the perspective of the rotation 09:41.250 --> 09:49.890 axis y, we will see a plane with the X and Z axis and instead the Y axis is coming out of the page 09:49.890 --> 09:51.360 coming out of the screen. 09:51.810 --> 10:01.170 Similarly, we will see the X first and z first axis of the second reference frame rotated by an angle 10:01.200 --> 10:05.730 phi and the Y first axis also coming out of the page. 10:06.470 --> 10:11.960 Again, let's consider describing the coordinates of the point P in both reference frames. 10:11.990 --> 10:22.130 This point will have coordinates, p, z and p x along the z and x axes of the fixed reference frame 10:22.400 --> 10:32.420 and coordinates p, z first and p x first along the z first and x first axes of the rotated reference 10:32.420 --> 10:32.960 frame. 10:33.940 --> 10:40.570 Using some simple trigonometric derivation, we can predict the Z first vector. 10:40.720 --> 10:50.230 Along the Z and x axis of the fixed reference frame and determine the components of the Z first vector 10:50.260 --> 10:52.060 along the Z axis. 10:52.060 --> 10:57.460 And this is given by p z first multiplied by the cosine of the angle phi. 10:58.200 --> 11:05.160 And the component along the x axis is given by p z first multiplied by the sine of the angle phi. 11:06.400 --> 11:14.440 Similarly, we can project the first vector along the x and z axis of the fixed reference frame. 11:15.090 --> 11:23.880 Its component along the Z axis is p x first multiplied by the sine of the angle phi, and the component 11:23.880 --> 11:30.120 along the x axis is p x first multiplied by the cosine of the angle phi. 11:31.040 --> 11:37.880 Once again, we ask ourselves what are the coordinates of the point P in the fixed reference frame? 11:38.030 --> 11:42.500 Specifically, what are P, x, P, y, and Z? 11:43.460 --> 11:48.320 Assuming that we know the coordinates of the point p in the mobile reference frame. 11:48.320 --> 11:56.570 So assuming that we know p x first, p, y first and p z first, we can write a system of three trigonometric 11:56.570 --> 11:58.670 equations to answer this question. 11:59.500 --> 12:06.700 The coordinates of P along the Y axis of the fixed reference frame is equal to the coordinate of P along 12:06.700 --> 12:08.720 the Y first axis. 12:08.740 --> 12:11.800 Since the two y axis are the same. 12:12.760 --> 12:20.200 The coordinate of P along the x axis on the other end is given by the sum of p x first multiplied by 12:20.200 --> 12:25.060 the cosine of P and p, z, first multiplied by the sine of phi. 12:25.720 --> 12:34.270 Similarly, the coordinate of P along the Z axis is given by minus p x first multiplied by the sign 12:34.300 --> 12:39.460 of z plus p, z, first multiplied by the cosine of P. 12:40.480 --> 12:48.130 Once again, we can express this system of equation in a matrix form to connect the coordinates of the 12:48.130 --> 12:54.760 point P in the fixed reference frame to the coordinates of P in the rotated reference frame. 12:55.480 --> 13:00.910 In this matrix since the x axis of the two reference frames are now equal. 13:00.940 --> 13:07.870 The second row and the second column are both zero and one, indicating that the coordinate of P along 13:07.870 --> 13:13.270 the Y axis is equal to the coordinate of P along the Y first axis. 13:14.490 --> 13:22.080 The first row which expressed the relationship between the coordinates of P in x and x first is given 13:22.080 --> 13:26.460 by the cosine of phi zero and the sine of phi. 13:26.880 --> 13:33.060 And finally the last row, which expresses the relationship between the coordinates of P along the Z 13:33.060 --> 13:33.600 and Z. 13:33.600 --> 13:38.970 First axis is given by minus sine of phi and cosine of phi. 13:40.450 --> 13:48.100 This matrix is also called a rotation matrix and in this case an elementary rotation matrix as the two 13:48.100 --> 13:50.080 reference frames share an axis. 13:50.080 --> 13:51.460 So the Y axis. 13:52.280 --> 14:02.190 So we call it our Y of V, which represents an elementary rotation matrix around the Y axis by an angle. 14:03.880 --> 14:10.960 As you may have already realized, there is only one more elementary rotation matrix missing which express 14:10.960 --> 14:15.550 the case when we have two reference frames sharing the x axis. 14:15.970 --> 14:23.590 This means we have a fixed reference frame and a mobile reference frame that share the same x axis and 14:23.590 --> 14:28.480 the mobile frame is rotated around the x axis by an angle theta. 14:29.020 --> 14:36.750 To find the last elementary rotation matrix, we imagine ourselves as an observer located on the x axis. 14:36.760 --> 14:45.070 In this case, we would see a plane with Y first and Z first axis rotated by an angle theta with the 14:45.070 --> 14:47.380 respect to the fixed reference frame. 14:48.300 --> 14:54.000 Once again, let's describe the coordinates of the point P in both reference frame. 14:54.360 --> 15:01.860 This point will have coordinates, P, z and p y along the z and y axis of the fixed reference frame 15:01.860 --> 15:10.740 and coordinates p, z first and p y first along the z first and y first axes of the rotated reference 15:10.740 --> 15:11.160 frame. 15:12.040 --> 15:20.410 Using some simple trigonometric derivations, we can predict the pi first vector along the Y and Z axis 15:20.440 --> 15:28.300 of the fixed reference frame and determine that the component of pi along the y axis is given by pi, 15:28.330 --> 15:35.350 multiplied by the cosine of the angle theta, and the component along the z axis is given by pi, multiplied 15:35.350 --> 15:37.180 with the sine of the angle theta. 15:38.300 --> 15:46.130 Similarly, we can project the first vector along the z and y axis of the fixed reference frame. 15:46.740 --> 15:54.000 Its component along the y axis is p z first multiplied by the cosine of the angle theta. 15:54.720 --> 16:01.800 And the component along the Z axis is p, z, first multiplied by the sine of the angle theta. 16:02.590 --> 16:09.460 Once again, we ask ourselves which are the coordinates of the point p in the fixed reference frame. 16:09.460 --> 16:13.200 So which are P, x, P, y and z. 16:14.290 --> 16:19.030 Assuming that we know the coordinates of the point p in the mobile reference frame. 16:19.030 --> 16:26.980 So p x first, p, y first and p z first, we can write a system of three trigonometric equations to 16:26.980 --> 16:28.240 answer this question. 16:28.600 --> 16:35.020 The coordinate of P along the x axis of the fixed reference frame is equal to the coordinate along the 16:35.020 --> 16:39.370 x first axis since the two x axis are the same. 16:39.850 --> 16:47.170 The coordinate of P along the y axis on the other end is given by the sum of p y first multiplied by 16:47.170 --> 16:53.350 the cosine of theta and minus p z, first multiplied by the sine of the angle theta. 16:53.830 --> 17:01.420 Similarly, the coordinate of P along the z axis is given by p, y first multiplied by the sine of theta 17:01.450 --> 17:05.470 plus p z, first multiplied by the cosine of theta. 17:06.170 --> 17:12.410 Once again, we can express this system of equation in a matrix form to connect the coordinates of the 17:12.410 --> 17:18.260 point P in the fixed reference frame to the coordinate of P in the rotated reference frame. 17:18.850 --> 17:25.480 In this matrix since the x axis of the two reference frames are equal, the first row and the first 17:25.480 --> 17:32.530 column are both zero and one, indicating that the coordinate of P along the x axis is equal to the 17:32.530 --> 17:35.800 coordinate of P along the x first axis. 17:37.410 --> 17:43.350 The second row which expressed the relationship between the coordinates of P in Y and Y. 17:43.380 --> 17:45.630 First is given by zero. 17:45.660 --> 17:49.350 The cosine of theta and minus the sine of theta. 17:49.710 --> 17:56.220 And finally the last row which expresses the relationship between the coordinates of P along Z and Z. 17:56.250 --> 18:01.230 First is given by zero sine of theta and cosine of theta. 18:02.430 --> 18:09.180 Also, this matrix is called a rotation matrix, and this is an elementary rotation matrix as the two 18:09.180 --> 18:12.270 reference frames share their x axis. 18:12.750 --> 18:19.980 So we call it r x of Theta, which represents an elementary rotation matrix around the x axis by an 18:19.980 --> 18:20.640 angle theta. 18:21.820 --> 18:22.690 Together. 18:22.690 --> 18:29.740 These three elementary rotation matrices represents all the possible elementary rotations that a reference 18:29.740 --> 18:32.860 frame can undergo with respect to another. 18:33.250 --> 18:37.210 However, our initial problem was different and more complex. 18:37.420 --> 18:44.530 We wanted to express the arbitrary orientation of one reference frame with respect to another without 18:44.530 --> 18:47.050 necessarily sharing our rotation axis. 18:47.880 --> 18:54.180 It turns out that we can decompose the general problem into three sub problems and determine only the 18:54.180 --> 19:00.360 elementary components of the orientation along the three rotation axis, so X, Y, and Z. 19:00.570 --> 19:07.950 By simply combining, so multiplying the three elementary rotation matrices, we obtain a generic rotation 19:07.950 --> 19:15.450 matrix that allows us to express the orientation of one frame with respect to another in terms of three 19:15.450 --> 19:16.110 angles. 19:16.110 --> 19:23.370 So C around the Z axis, FY, around the Y axis and theta R on the X axis. 19:23.670 --> 19:28.170 These angles are often known as roll pitch and yaw.