Imagine that you have a robot that is of the second link, you have a translational third link that is orthogonal to the plane that the second link is on. That is, if the zero position for the first two links is in the X-Z plane, the rotations of the first two links are around the
Z
and
Y
planes respectively, and the translation of the third link happens in the Y -direction. ( 60 points) (a) What is the benefit of this configuration compared to just having the first two links, aside from being able to reach new coordinates? ( 5 points) (b) Work out the forward and inverse position kinematics for this robot. Assume that you can use known values
L_(1)
and
L_(2)
for the first two links (but do not plug in the actual values yet). Keep in mind that
L_(3)
now is a variable that you need to solve for, just like the first two angles. (Assume that in the zero position, you are at the coordinate [
0L_(3),L_(1)+L_(2)
]. ( 25 points) (c) Work out the forward and inverse velocity kinematics for this robot. It may seem weird at first, but once you do it you will see that it is actually just a variant of the rotational robots we have been studying so far. ( 20 points) (d) Using the information for coordinates of the end effector
(x,y,z)
and lengths for the first two links on the table below, find the angles of the first links and the length of the third link. (5 points) (e) Using the same information, the linear speed information
(V_(x),V_(y),V_(z))
, and the angles and the link length you found in part (d), find the velocities of the three links at corresponding positions. (5 points) Note: If at any part of this question, you find something that looks odd but everything you did seems right, make sure you explain everything to receive the most amount of credit. \table[[
L_(1)
,
L_(2)
,
x
,
y
,
z
,
V_(x)
,
V_(y)
,
V_(z)