ROBOTICS EXAM STUDY GUIDE
QUESTIONS AND ANSWERS
Joint-space trajectory planning - Answer-When initial and final positions
are specified in Cartesian coordinates these can be converted into joint
coordinates by inverse kinematics. Then, path and trajectory planning
can be carried out in the joint-space between the initial and final joint
positions.
+
Path and trajectory planning between initial and final positions
Smooth motion
-
Difficult to visualise the motion of the end effector
Need to carry out repeated Forward Kinematics repeatedly to visualise
the motion)
Cartesian-space trajectory planning - Answer-Take inital and final
position in Cartesian space, and conduct trajectory planning to
determine Cartesian trajectory. Then use inverse kinematics to
determine joint trajectory.
+
Easy to visualise motion
-
Computationally expensive
Need to convert the calculated trajectory to joint coordinates by applying
inverse kinematics repeatedly for control
May give rise to singularities
Intermediate points may be unreachable (outside workspace of
manipulator)
Challenging motor motion
Cubic polynomial differentials - Answer-θ(t) = a0 + a1t + a2t^2 + a3t^3
θdot(t) = a1 + 2a2t + 3a3t^2
θddot(t) = 2a2 + 6a3t
, Linear segments with parabolic blends - Answer-Split trajectory into
three regions
A ramp up and ramp down velocity at each end, of duration tb
tb = w/a (w is the constant velocity, a is the accleration)
Use SUVAT in each region -> trapezium velocity-time graph
What are via points? - Answer-Intermediate positions along the trajectory
between the initial and final position
Constraints on position, velocity and acceleration may be imposed at
these points
Acceleration threshold for LSPB - Answer-a >= 4(θf -θ0)/tf^2
Gives a triangular velocity-time graph
Cubic trajectory planning coefficients - Answer-For zero velocity at start
and end
a0 = θ0
a1 = 0
a2 = 3(θf -θ0)/tf^2
a3 = -2(θf -θ0)/tf^3
For non-zero velocity
a0 = θ0
a1 = θdot0
a2 = 3(θf -θ0)/tf^2 - 2θdot0/tf - θdotf/tf
a3 = -2(θf -θ0)/tf^3 + 1/tf^2 (θdotf - θdot0)
SUVAT for constant acceleration trajectory - Answer-
Converting from Cartesian to joint space - Answer-Can use a cubic
spline to plan the trajectory in Cartesian-space
Use inverse kinematics calculate initial and final joint angles
For joint accelerations, use forward kinematic equations and Jacobians
Generalised coordinates - Answer-Coordinates at each joint, θ or L
The number of generalised coordinates, N, must be equal to or greater
than the number of degrees of freedom of the system.
Generalised inputs - Answer-Q = [Q1, Q2, ... , QN]
QUESTIONS AND ANSWERS
Joint-space trajectory planning - Answer-When initial and final positions
are specified in Cartesian coordinates these can be converted into joint
coordinates by inverse kinematics. Then, path and trajectory planning
can be carried out in the joint-space between the initial and final joint
positions.
+
Path and trajectory planning between initial and final positions
Smooth motion
-
Difficult to visualise the motion of the end effector
Need to carry out repeated Forward Kinematics repeatedly to visualise
the motion)
Cartesian-space trajectory planning - Answer-Take inital and final
position in Cartesian space, and conduct trajectory planning to
determine Cartesian trajectory. Then use inverse kinematics to
determine joint trajectory.
+
Easy to visualise motion
-
Computationally expensive
Need to convert the calculated trajectory to joint coordinates by applying
inverse kinematics repeatedly for control
May give rise to singularities
Intermediate points may be unreachable (outside workspace of
manipulator)
Challenging motor motion
Cubic polynomial differentials - Answer-θ(t) = a0 + a1t + a2t^2 + a3t^3
θdot(t) = a1 + 2a2t + 3a3t^2
θddot(t) = 2a2 + 6a3t
, Linear segments with parabolic blends - Answer-Split trajectory into
three regions
A ramp up and ramp down velocity at each end, of duration tb
tb = w/a (w is the constant velocity, a is the accleration)
Use SUVAT in each region -> trapezium velocity-time graph
What are via points? - Answer-Intermediate positions along the trajectory
between the initial and final position
Constraints on position, velocity and acceleration may be imposed at
these points
Acceleration threshold for LSPB - Answer-a >= 4(θf -θ0)/tf^2
Gives a triangular velocity-time graph
Cubic trajectory planning coefficients - Answer-For zero velocity at start
and end
a0 = θ0
a1 = 0
a2 = 3(θf -θ0)/tf^2
a3 = -2(θf -θ0)/tf^3
For non-zero velocity
a0 = θ0
a1 = θdot0
a2 = 3(θf -θ0)/tf^2 - 2θdot0/tf - θdotf/tf
a3 = -2(θf -θ0)/tf^3 + 1/tf^2 (θdotf - θdot0)
SUVAT for constant acceleration trajectory - Answer-
Converting from Cartesian to joint space - Answer-Can use a cubic
spline to plan the trajectory in Cartesian-space
Use inverse kinematics calculate initial and final joint angles
For joint accelerations, use forward kinematic equations and Jacobians
Generalised coordinates - Answer-Coordinates at each joint, θ or L
The number of generalised coordinates, N, must be equal to or greater
than the number of degrees of freedom of the system.
Generalised inputs - Answer-Q = [Q1, Q2, ... , QN]
