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Introduction to Robotics (CS223A) Handout
(Winter 2006/2007)
Homework #5 solutions
1. (a) Derive a formula that transforms an inertia tensor given in some frame {C}
into a new frame
{
A
}
. The frame
{
A
}
can differ from frame
{
C
}
by both
translation and rotation. You may assume that frame {C} is located at the
center of mass.
Solving this problem involves using the Parallel Axis Theorem to translate the inertia
tensor to a frame at a different location, and a similarity transformation to rotate it into
the new frame. These operations can be done in either order, as long as we’re careful
that the vectors we use are expressed in the correct frame. However, it is definitely easier
to do the rotation first.
Assume that we have
A
C
T , the transformation from frame {C} coordinates to frame
{A} coordinates, which contains the rotation matrix
A
C
R and the translation vector
A
p
C
which locates the origin of frame {C} with respect to {A}. Let’s first solve the problem
by a rotation followed by a translation. Consider an intermediate frame {C
0
} which has
the same origin as {C}, but whose axes are parallel to frame {A}. Using a similarity
transformation (see p. 134-135 of Lecture Notes), we know that
C
0
I =
C
0
C
R
C
I
C
0
C
R
T
However, since frame {C
0
} has the same orientation as frame {A}, we know that
C
0
C
R =
A
C
R, so
C
0
I =
A
C
R
C
I
A
C
R
T
We now have the inertia tensor expressed in the intermediate frame {C
0
}. Since {C
0
} is
parallel to {A}, we can use the Parallel Axis Theorem to transform
C
0
I to
A
I. To use
this theorem, we just need the vector
A
p
C
0
that locates the center of frame {C
0
} with
respect to {A}, expressed in frame {A}, which yields the formula
A
I =
C
0
I + m
h
(
A
p
T
C
0
A
p
C
0
)I
3
−
A
p
C
0
A
p
T
C
0
i
where m is the total mass of the object and I
3
is the 3 × 3 identity matrix. Since {C
0
}
and {C} have the same origin, the vector
A
p
C
0
is just
A
p
C
. Substituting this value and
our previous expression for
C
0
I yields:
A
I =
A
C
R
C
I
A
C
R
T
+ m
h
(
A
p
T
C
A
p
C
)I
3
−
A
p
C
A
p
T
C
i
Equivalently, we could do this problem with a translation first, and then a rotation. To
do that, we can define an intermediate frame {A
0
}, which has the same origin as {A},
but whose axes are parallel to {C}. We can get the intertia tensor in the intermediate
frame by using the Parallel Axis Theorem. To use it, however, we need the vector
A
0
p
C
which locates the origin of frame {C} with respect to frame {A
0
}, expressed in frame
{A
0
}. Using this formula with the vector expressed in frame {A} is incorrect. We can
get
A
0
p
C
by rotating
A
p
C
with
A
0
A
R =
C
A
R, and then simplify:
A
0
I =
C
I + m
h
(
A
0
p
T
C
A
0
p
C
)I
3
−
A
0
p
C
A
0
p
T
C
i
=
C
I + m
h
(
C
A
R
A
p
C
)
T
(
C
A
R
A
p
C
)I
3
− (
C
A
R
A
p
C
)(
C
A
R
A
p
C
)
T
i
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