“Introduction to Robotics ”
Lecture 2
Dr. Greg R. Luecke
Associate Professor
Mechanical Engineering
Iowa State University
©2016
More on rotation: rotation about the x-axis
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More on rotation: rotation about the x-axis
We developed the rotation matrix
from {A} to {B}:
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Unit vectors?
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Where is the point in the frame {A}?
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Rotation about the y-axis
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Rotation about the y-axis
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Rotation about the z-axis
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General motion: rotation +translation
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Homogeneous transformation matrix:
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There are two interpretations:
AT is used to describe P in {A} (transform)
1)
B
A
2)
TB is used to move vector P from some position in
{A} to new position (operator)
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Inverting the transform:
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Compound transformations allow transforms in steps:
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Earth fixed axes-1st X, then Y, finally Z
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More on rotation: Body fixed axes-1st X, then Y, finally Z
“Euler Axes”
Note the order!
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Earth fixed axes-Rotation matrix
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Earth fixed axes-Inversion problem
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Equivalent angle-axis rotations
Rotate by an angle qabout an
axis defined by
K={kx,ky, kz}T
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Well, let’s look at problem 2.6
Huh?
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Fortunately, I have
the answer book:
Huh?
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Let’s try this:
"If we multiply out eq. (1) above, and then
simplify using A2+B2+C2=1, D2+E2+F2=1,
and [A B C]dot[D E F]=0,
[A B C]cross[D E F]=[Kx Ky Kz], we
arrive at eq (2.80) in the book
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"If we multiply out eq. (1) above…
syms A B C D E F kx ky kz UrA Rz th
UrA=[A D kx;B E ky;C F kz]
Rz=[cos(th), sin(th),0;sin(th),cos(th), 0;0 0 1]
ArU=transpose(UrA)
Rot=UrA*Rz*ArU
UrA =
[ A,
[ B,
[ C,
D, kx]
E, ky]
F, kz]
Rz =
[ cos(th), -sin(th), 0]
[ sin(th), cos(th), 0]
[
0,
0, 1]
ArU =
[ A, B, C]
[ D, E, F]
[ kx, ky, kz]
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"If we multiply out eq. (1) above…
Rot =
[ (A*cos(th)+D*sin(th))*A+(-A*sin(th)+D*cos(th))*D+kx^2,
(A*cos(th)+D*sin(th))*B+(-A*sin(th)+D*cos(th))*E+kx*ky,
(A*cos(th)+D*sin(th))*C+(-A*sin(th)+D*cos(th))*F+kx*kz]
[ (B*cos(th)+E*sin(th))*A+(-B*sin(th)+E*cos(th))*D+kx*ky,
(B*cos(th)+E*sin(th))*B+(-B*sin(th)+E*cos(th))*E+ky^2,
(B*cos(th)+E*sin(th))*C+(-B*sin(th)+E*cos(th))*F+ky*kz]
[ (C*cos(th)+F*sin(th))*A+(-C*sin(th)+F*cos(th))*D+kx*kz,
(C*cos(th)+F*sin(th))*B+(-C*sin(th)+F*cos(th))*E+ky*kz,
(C*cos(th)+F*sin(th))*C+(-C*sin(th)+F*cos(th))*F+kz^2]
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Some simplification…
Rot (1,1)=
[ (A*cos(th)+D*sin(th))*A+(-A*sin(th)+D*cos(th))*D+kx^2]
= A^2*cos(th)+AD*sin(th)-AD*sin(th)+D^2*cos(th)+kx^2
= A^2*cos(th)+D^2*cos(th)+kx^2
= cos(th)*(A^2+D^2)+kx^2
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"If we multiply out eq. (1) above…
simple( Rot(:,1) )
[A^2*cos(th)+D^2*cos(th)+kx^2]
[ B*A*cos(th)+E*A*sin(th)-D*B*sin(th)+E*D*cos(th)+kx*ky]
[C*A*cos(th)+F*A*sin(th)-C*D*sin(th)+F*D*cos(th)+kx*kz]
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“…and then simplify using A2+B2+C2=1, D2+E2+F2=1…
Notice that both columns and rows must have a
vector norm of 1: A2+B2+C2=1, A2+D2+kx2=1
A2+D2=1-kx2
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“…and then simplify using A2+B2+C2=1, D2+E2+F2=1…
The other diagonal elements are similar.
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“…and [A B C]dot[D E F]=0, [A B C]cross[D E F]=[Kx Ky Kz]…
Now look at the (1,2) element
The i-element of the cross product is: BF-CE
The j-element is:
(-1)*(AF-CD)
the k-element is
AE-BD
kx= BF-CE
ky= -AF+CD
kz= AE-BD
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Just as the dot product of any two columns of UrA must be
zero, so, too, must the dot product of any two columns of
ArU be zero:
“…and [A B C]dot[D E F]=0,
[A B C]cross[D E F]=[Kx Ky Kz]
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Now look at the (1,2) element
kz= AE-BD
The remaining elements are left as a “trivial” exercise for the student…IOWA STATE UNIVERSITY
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Problem 2.14
"Develop a general
formula to obtain ABT,
where, starting from
initial coincidence, {B}
is rotated by θ about Khat where K-hat passes
through the point AP (not
through the origin of {A}
in general).”
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Problem 2.14
•3) Translate back with the inverse of 1) above.
1) Translate the frame to
any point on the line K
that passes through AP. It
may as well be directly
through the point:
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Problem 2.14
•3) Translate back with the inverse of 1) above.
2) Rotate the frame around
K_hat by an angle θ
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Problem 2.14
•3) Translate back with the inverse of 1) above.
3) Translate back by the
inverse of 1) above.
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Problem 2.14
•3) Translate back with the inverse of 1) above.
Viola!
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