#### 4.1.1. Product of Exponentials Formula in the Space Frame

This video introduces the product-of-exponentials formula, expressed in the space frame, for forward kinematics of an open-chain robot.

Chapter 4 addresses the forward kinematics of open-chain robots, as illustrated in this video of a robot with 6 revolute joints. We define a frame {s} fixed in space, often at the base of the robot, and a frame {b} at the end-effector of the robot arm. If we command the robot to move, the {b}-frame moves. The forward kinematics problem is to find the configuration of the {b}-frame relative to the {s}-frame given the vector of joint angles theta. The transformation matrix representing the {b}-frame in the {s}-frame is T_sb of theta, or just T of theta for short.

To derive a procedure to calculate T of theta, let's use a simple robot arm that moves in a plane. This robot has three joints: a revolute joint, a prismatic joint, and another revolute joint. You can also see the stationary {s}-frame and the end-effector {b}-frame. This is a stick figure version of the robot. The joint variable theta_1 represents the angle of joint 1 relative to the horizontal. Theta_2 represents the extension of the prismatic joint, and theta_3 represents the angle of joint 3. The 3-vector theta is a list of the three joint variables, and T of theta represents the configuration of the {b} frame relative to the {s} frame.

If we set all of the joint variables equal to zero, the robot is in its home position, as shown here. We write the zero configuration of the {b} frame as T of zero, or simply M for short. At the zero configuration, the {b}-frame has the same orientation as the {s}-frame, and the 3 in the top right element of the M matrix means that the {b}-frame is 3 units from the {s}-frame in the xs-direction.

Now say we rotate joint 3 by pi over 4 radians. The theta vector is now zero, zero, pi over 4. This motion of the {b} frame can be represented by a rotation about the screw axis of joint 3. Since it is a revolute joint with no translational motion, the screw axis has zero pitch. Because positive rotation is in the direction indicated in the figure, by the right-hand rule, the screw axis is out of the screen, toward you.

As we learned in chapter 3 videos, a screw axis can be represented in any frame. Let's represent it in the {s}-frame, and let's call joint 3's screw axis S_3, consisting of an angular component omega_3 and a linear component v_3. Since the screw axis involves rotation, omega3 must be a unit vector. Positive rotation is about an axis out of the screen, which is aligned with the {s} frame's z-axis. Therefore, the unit angular component is zero zero one, a unit vector aligned with the zs-axis. To visualize the linear component v_3, imagine the entire space rotating about joint 3, visualized here as a turntable. Then v_3 is the linear velocity of a point at the origin of the {s}-frame when the turntable rotates with unit angular velocity, as shown here. So, v_3 is zero, minus 2, zero, meaning that the origin has a velocity of 2 units in the minus ys-direction. V_3 could also be calculated as minus omega_3 cross q_3, where q_3 is any point on the joint axis represented in the {s}-frame.

Now that we have the screw axis S_3, we can calculate the {b}-frame configuration T of theta. We simply apply the space-frame transformation corresponding to motion along the S_3 screw axis by an angle pi over 4. This transformation is e to the bracket S3 times pi over 4 using the matrix exponential from the chapter 3 videos. Since it is a space-frame transformation, it premultiplies M.

Now suppose we change joint 2, extending it by 0.5 units of distance. The theta vector is now zero, 0.5, pi over 4. The screw axis S_2 corresponding to joint 2 has zero angular component omega_2, so the linear component v_2 must be a unit vector. If we imagine the whole space translating at unit velocity along joint 2, a point at the origin of the {s}-frame would move with a linear velocity v_2 equal to one, zero, zero, expressed in the {s}-frame. Therefore the screw axis S2 is defined as zero, zero, zero, one, zero, zero. The new configuration of the {b}-frame, T of theta, is obtained by left-multiplying the previous configuration by e to the bracket S_2 times theta2.

It's important to notice that the previous motion of joint 3 does not affect the relationship of joint 2's screw axis to the {s}-frame. That's because joint 3 is not between joint 2 and the {s}-frame. Therefore, S_2 is the same as the screw axis of joint 2 when the robot is at its zero configuration.

Finally, let's rotate joint 1 by pi over 6. The theta vector is now pi over 6, 0.5, pi over 4. The screw axis S1 is a pure rotation about an axis out of the screen, so the omega_1 vector is zero, zero, one. Rotation about this axis does not cause any linear motion at the origin of the {s}-frame, so the v_1 vector is zero, zero, zero. The new configuration of the {b}-frame, T of theta, is again given by left-multiplying the previous configuration by the new space-frame transformation.

Again, the previous motions of joints 2 and 3 do not affect the relationship of joint 1's screw axis to the {s}-frame, because they are not in between joint 1 and the {s}-frame. Therefore, S_1 is the same as the screw axis of joint 1 when the robot is at its zero configuration.

For any serial robot, the procedure generalizes directly. First, define the M matrix representing the {b}-frame when the joint variables are zero. Second, define the {s}-frame screw axes S_1 to S_n for each of the n joint axes when the joint variables are zero. Finally, for the given joint values, evaluate the product of exponentials formula in the space frame.

In the next video we will see an alternative version of this formula, in terms of screw axes expressed in the {b}-frame.