The frame assignment of the robot joints is as shown in figure(1). The assignments of the frame are strictly based on the Denavit - Hartenburg (DH) convention. In this section, the homogeneous transformation of the joint frames is discussed. The table below gives the DH parameters of the arm.
The homogeneous transformation of the frame is calculated by using DH matrix as given in the equation no.
figure(4). The assignment of the frame is according to the Denavit Hartenburg convention. Note that there is an intermediate frame at the wrist centre. It is provided to map the 2nd frame to 3rd frame by multiplying the DH matrix.
The Denavit Hartenburg Matrices are computed as follows
Thus for each joint transformation matrices are given by
The final Transformation matrix for the end-effector frame is calculated with respect to the base frame.
Using the six joint angle parameters (q1, q2,....,q6) the configuration of the robot is controlled.
The model in the video is controlled by joint angles using the forward kinematics only.
The approach to the solution of inverse kinematics is using kinematic decoupling. That is breaking the solution into two parts. They are
1. Geometric approach
2. Inverse Orientation
The first two joints from the base of the arm constitute the shoulder, the third joint is of the elbow and the fourth, fifth and sixth joints together constitute the wrist. Here kinematic decoupling will be used to estimate the joint angle parameters of the shoulder, elbow and wrist.
Arm consisting of shoulder and elbow is shown in figure(2)
figure(5) elbow manipulator. Note that the origin is at the joint2 which is in the middle of the three joint. But for this illustrative purpose, it was shown as in the figure
The possible solution of the q1 will immediately obtained from the figure(2)
figure(6). Projecting the onto the plane formed by links 2 and 3
The possible solution of the q3 from the figure(6) , using law of cosines.
From the figure(6), angle q2 is obtained from using the value q3
The orientation of the wrist center in terms of joint angle parameters is obtained from forward kinematics. Also from above results one can easily obtain the orientation of frame 3 with respect to the base frame as follows.
But we can only determine the pose of the end-effector. So, it can be represented as given below. Thus one can obtain the three remaining angles.
Note:
We found the total of four set of solutions
s_θ is sin(θ)
c_θ is cos(θ)
r = square root(Xc^2+Yc^2)