 I am Mr. Ninhath Kulkarni working as an assistant professor in Department of Mechanical Engineering at Vultured Institute of Technology, Solaapur. In today's session, I will be taking you through the forward kinematics and I will be demonstrating you how the forward kinematics is done with the help of Roboanalyzer software. At the end of this session, the students will be able to understand and visualize the forward kinematics of various robot configurations and I will be demonstrating you the robot forward kinematics of two DOF configuration. So what is kinematics? As you can see, this is a link, this is link one whose length is denoted by A1, this is link two whose length is denoted by A2. These are the joints which connects link. So this joint connects link one to link two and this is the base joint. So as you can see, there are two joints and two links and we call as an industrial robot as a series of links and joints. So there are series of links connected with series of joints and that creates a kinematic chain. Each link connects two adjacent joints and each joint it connects two adjacent links. Now we have to set up a coordinate frame. What we will call as a coordinate frame is that we have to create a coordinate frame like this one x, y, z axis for this joint as well as this joint. We have to create a coordinate frame for each joint of the robot and we have to find out the relationship between the joints motion that is the joints position and the end effector motion. This is what we call as a end effector denoted by Px and Py. So this end effector is the last part of our industrial robot which performs the certain tasks. The task may be of gripping or any tool. Now we are going to start with the forward kinematics. What is forward kinematics is that you have a robotic arm that is inclined like this. I will just draw this arm. So this is presently inclined in a horizontal axis or aligned with a horizontal x axis. Now you tell the first link that is this link to rotate by an angle of theta 1 and the second link by an angle of theta 2. In forward kinematics what we are required is the joint positions that means the joint angles theta 1 and theta 2 in case of revolute joints. If the joints are revolute we are prescribed with the joint angles. If the joints are prismatic we have to give the or prescribe the displacements. The task is to find out the position Px and Py of the end effector. That means the position and the orientation that is phi of end effector we are required to find out. This is called as forward kinematics. So let it be clear that in forward kinematics the joint positions are given and the end effector positions and configurations are required or we are required to calculate. For this we are going to follow a DH parameter approach that I have told you in the previous lecture. So what is DH parameter approach is first we assign a coordinate frame for each joint that means we denote a coordinate system for each joint like this. And using DH notations the four DH parameters are joint offset, the joint angles then the twist angles and link lengths. These are the four DH parameters that I have told you. Next these are the steps first we attach frames to the links. Next we define the DH parameter or we prepare a DH parameters table. Afterwards we write the homogeneous transformation matrix for each frame i plus 1. So if there is a joint number 2 then we write the homogeneous transformation matrix for joint number 2 with respect to joint number 1 with this data. This is the transformation matrix. After that after we have written all the transformation matrix from starting from joint number 2 to joint number n that is the last joint we multiply all the transformation matrix in order to get the final homogeneous matrix of the end effector. This is the step by step procedure that we are going to follow. Now today we will do forward kinematics of a 2 degree of freedom planar joint that we call as 2R configuration. So let A1 and A2 they represent the link lengths. Theta 1 and theta 2 they represent the joint angles that means these angles and for a planar configuration that means planar configuration is that configuration in which both the links they traverse in a same plane that means there is no joint offset and there is no twist angle. So all for planar configuration the joint offsets and twist angles both are absent. Now we will prepare a DH parameter table. So for joint 1 the DH parameters can be written as the offsets joint offsets are 0 and twist angles are 0. So we write these parameters here. So both the parameters are absent. So both are 0. For jointed arm type the theta 1 angle is given and theta 2 angle is given. So both these angles they are variable. Now I will show you what is mean by variable. Next the link lengths are already prescribed as A1 and A2. So these are the DH parameter table that we have done for both the joints. Next we will prepare an HTM that is homogenous transformation matrix for individual link by using this formula on the right hand side cos theta i minus sin theta i cos alpha i sin theta i sin alpha i ai cos theta i. So this is the homogenous transformation matrix for individual link that can be written. So based on this table that we have written in the previous slide the T1 that means the homogenous transformation matrix for first joint can be calculated as C1 that means cos theta 1 we will write cos theta 1 as C1 sin theta 1 as S1 same for cos theta 2 and sin theta 2. So this T matrix T1 matrix and T2 matrix can be written as follows. Next step is multiplying T1 and T2 because there are only two joints except the base joint there are two joints T1 and T2. So we will multiply these two matrices so T1 this this and T2 is this. After multiplication I will just show you the first row so C1 we will multiply first row of this matrix and first column of this matrix. So C1 into C2 so is written here next minus S1 into minus S2 minus S1 into S2 it can be written here S2 next 0 into 0 here and A1 C1 into 0. So this is the first element that is available similarly the rest of elements can be computed as shown in the matrix. So this C1 C2 minus S1 S2 that means cos theta 1 into cos theta 2 minus sin theta 1 into minus sin theta 2. So this is the formula of cos A plus B cos of bracket A plus B that can be written as in short C12 C12 indicates of cos theta 1 plus theta 2 cos of bracket theta 1 plus theta 2. Similarly minus S1 to 0 A1 C1 plus A2 C12. So this is the matrix that we will get by using the multiplication. From this matrix what we get is the position of the end effector. So the last two parameters that is A1 C1 plus A2 C12 so this is the position of end effector in x direction and this is the position of end effector in y direction. Similarly we will solve a simple problem that is A1 is equal to 300 mm, A2 is equal to 400 mm, theta 1 is equal to 30 degree, theta 2 is equal to 60 degree. By putting A1 C1 plus A2 C12. So we are given C1 that is theta 1 is given as 30 degree, theta 2 is 60 degree, link length 1 is 300 mm that can be converted into meters 0.3 meters and link length 2 is 400 mm that can be converted into meters that is 0.4 mm. So we can easily put in the formula that is Px is equal to A1 C1 plus A2 C12 that can be computed as 0.3 into cos 30 plus 0.4 into cos 30 plus 60. So Px will come as 0.2598 meter similarly Py can come as 0.55 meter. This is the solution of or position of the end effector that can be computed these are the references that I have referred and thank you all.