WEBVTT 00:00.300 --> 00:06.900 This brief digression into the mathematics that allows us to perform coordinate transformation between 00:06.900 --> 00:11.550 two reference frames has provided us with some very powerful tools. 00:11.580 --> 00:14.910 As we approach solving much more complex problems. 00:15.670 --> 00:22.540 The first one of these problem is how do I find how do I calculate the position of the robot gripper 00:22.540 --> 00:28.690 in the three dimensional space, knowing the angle so the position of each joint of the robot. 00:29.590 --> 00:34.780 The mathematics that is used to answer this question is called forward Kinematics. 00:35.590 --> 00:41.500 The second question we asked ourselves was how do I calculate the angle of each joint of the robot that 00:41.500 --> 00:44.680 has brought the gripper to its current position in space? 00:44.680 --> 00:50.710 Or how do I calculate the angles of the joints if I know the position and the orientation of the gripper 00:50.710 --> 00:51.550 in the space? 00:52.270 --> 00:53.140 The mathematics. 00:53.140 --> 00:53.650 Instead. 00:53.650 --> 00:57.340 That answer this question is called inverse kinematics. 00:57.670 --> 01:01.720 Let's see how to express and solve the first of these two problems. 01:01.720 --> 01:03.550 So the forward kinematics. 01:04.770 --> 01:08.160 Let's start by considering the architecture of our robot. 01:08.640 --> 01:15.270 As we know, it has three degrees of freedom, so it has three movable joints driven by three motors 01:15.270 --> 01:18.510 that allows us to move the robot's gripper in space. 01:19.020 --> 01:24.750 We need to focus on their movement and orientation to solve the forward kinematic problem. 01:25.640 --> 01:32.240 Let's consider a simplified version but still accurate representation of our robot architecture consisting 01:32.240 --> 01:38.810 of three cylinders representing the three rotational joints of the robot, connected by three links 01:38.810 --> 01:40.340 representing its arms. 01:40.940 --> 01:47.480 The problem of forward kinematics is nothing more than a different form of representing the same problem 01:47.480 --> 01:50.690 we have already faced and solved in previous lessons. 01:50.720 --> 01:57.470 That is, how do I express the position and the orientation of our reference frame with respect to one 01:57.470 --> 02:02.090 other after it has arbitrarily rotated and translated? 02:02.980 --> 02:09.940 Therefore, we define two reference frames, one that is fixed and is called word frame, which we call 02:09.970 --> 02:16.540 W in short, and the other one that is mobile and is attached to the robots and the link. 02:16.540 --> 02:24.220 So located at the gripper which we call R and which moves along with the robot gripper by rotating and 02:24.220 --> 02:27.790 translating as the angle of each joint of the robot change. 02:28.700 --> 02:34.670 The goal of the forward kinematics is to know at every moment in time the position of the reference 02:34.670 --> 02:35.120 frame. 02:35.120 --> 02:38.420 R with the respect to the reference frame W. 02:39.410 --> 02:45.500 In fact, since the reference frame R is connected to the robot's gripper and moves along with it. 02:45.530 --> 02:51.650 This implicitly means knowing at every moment in time the position and the orientation of the robot's 02:51.680 --> 02:53.060 gripper in the world. 02:54.160 --> 02:54.880 Correct. 02:55.120 --> 03:01.630 To solve the forward kinematics problem, it is enough to calculate a transformation matrix precisely 03:01.630 --> 03:07.600 the one that expresses how the reference frame, W and R are connected to each other. 03:08.360 --> 03:14.280 However, be aware that this calculation is not trivial, as it depends on three variables. 03:14.300 --> 03:17.540 The three angles of the three joints of the robot. 03:17.960 --> 03:24.110 A change in the position of any of these three joints affect the position and the orientation of the 03:24.110 --> 03:26.760 gripper and therefore of the reference frame. 03:26.780 --> 03:30.470 Ah, and so this change the transformation matrix. 03:31.600 --> 03:37.810 As always when dealing with a very complex problem, Let's try to break it down into many smaller and 03:37.810 --> 03:39.790 more easily solvable problems. 03:40.410 --> 03:44.580 We associate our reference frame to each movable joint. 03:45.030 --> 03:51.000 Although the position and the orientation of this intermediate reference frame are arbitrary, there 03:51.000 --> 03:57.420 are robotics convention, such as the Dynamic Artwork convention that allows us to arrange these frames 03:57.420 --> 04:00.990 in a way that minimize and facilitate calculations. 04:01.290 --> 04:08.370 In general, we position each reference frame so that it is translated and can rotate relative to the 04:08.370 --> 04:11.400 previous reference frame only around one axis. 04:11.640 --> 04:19.110 For example, in this case the frame too can rotate relative to the frame one only around the z axis 04:19.110 --> 04:24.180 and does this when the first rotational joint of the robot said one at the base rotates. 04:24.820 --> 04:31.030 If you recall from theoretical lesson on rotation matrices, this is a very useful result because we 04:31.030 --> 04:38.620 can easily determine the rotation matrix between the frames two and one using just an elementary rotation 04:38.620 --> 04:43.600 matrix around the Z axis that depends only on a single variable. 04:43.600 --> 04:49.420 So on a single angle that is namely the position of the joint at the base of the robot. 04:50.270 --> 04:56.720 At this point, it becomes very easy for us to calculate the intermediate transformation matrices between 04:56.720 --> 04:58.700 each of these reference frames. 04:58.850 --> 05:04.850 They will be composed of a certain translation vector that depends on the physical dimension of the 05:04.850 --> 05:11.900 robots links and ascertained rotation matrix that depends on the rotation angle of each of the movable 05:11.900 --> 05:12.590 joints. 05:13.130 --> 05:20.120 By simplifying the forward kinematics problem, we now only need to calculate four simple and elementary 05:20.120 --> 05:26.660 intermediate transformation matrices instead of dealing with a single complex transformation matrix. 05:28.110 --> 05:31.160 We just need to compose these transformation matrices. 05:31.170 --> 05:37.590 That is, we just need to multiply them to calculate the overall transformation matrix that connects 05:37.590 --> 05:41.870 the word reference frame to the gripper reference frame of the robot. 05:41.880 --> 05:46.410 And this is the matrix that solves the forward kinematic problem. 05:47.440 --> 05:51.220 When composing transformation matrices and calculating their product. 05:51.250 --> 05:55.840 Be aware of one thing The order of the multiplication is crucial. 05:56.530 --> 06:02.410 The transformation matrices must be composed and multiplied in the order that follows the entire chain 06:02.440 --> 06:03.700 of transformation. 06:03.700 --> 06:09.010 So the one that connects the initial frame to the final one in the order they appear. 06:09.850 --> 06:13.360 To better understand why the order of multiplication is crucial. 06:13.390 --> 06:15.430 Let's do a small example. 06:16.450 --> 06:23.770 Let's assume that we have a parallelepiped in a three dimensional reference frame and we apply two successive 06:23.800 --> 06:25.090 transformation to it. 06:25.390 --> 06:33.430 First, we rotate it by 90 degrees around its Z axis, and then we rotate it by another 90 degrees this 06:33.430 --> 06:35.620 time around the Y axis. 06:36.160 --> 06:38.980 Now let's perform the same transformation. 06:38.980 --> 06:45.580 So the same operation with the same parallelepiped in the same initial configuration simply by reversing 06:45.580 --> 06:47.310 the order of rotations. 06:47.320 --> 06:54.760 So first we rotate the parallelepiped 90 degrees around the y axis and then we rotate it around the 06:54.760 --> 06:57.160 Z axis by another 90 degrees. 06:57.640 --> 07:04.870 As you can see, even though we started with the same initial configuration and rotated both parallelepipeds 07:04.870 --> 07:11.410 by the same angle around the same axis, the result and therefore the final configuration of the parallelepiped 07:11.440 --> 07:12.490 is different. 07:13.150 --> 07:16.450 Keep this in mind when composing transformation matrices.