Introduction to Robotics Lecture II

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Introduction to RoboticsLecture II

• Alfred Bruckstein
• Yaniv Altshuler

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Denavit-Hartenberg

• Specialized description of articulated figures
• Each joint has only one degree of freedom
• rotate around its z-axis
• translate along its z-axis

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• One degree of freedom very compact notation
• Only four parameters to describe a relation
  between two links
• link length
• link twist
• link offset
• link rotation

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• Link length ai
• The perpendicular distance between the axes of
  jointi and jointi1

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• Link twist ai
• The angle between the axes of jointi and jointi1
• Angle around xi-axis

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• Link offset di
• The distance between the origins of the
  coordinate frames attached to jointi and jointi1
• Measured along the axis of jointi

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• Link rotation (joint angle) fi
• The angle between the link lenghts ai-1 and ai
• Angle around zi-axis

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• How to compute the parameters to describe an
  articulated figure

1. Compute the link vector ai and the link length
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai

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1. Compute the link offset di
2. Compute the joint angle fi
3. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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• Lets do it step by step

1. Compute the link vector ai and the link length
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai
4. Compute the link offset di
5. Compute the joint angle fi
6. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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The link length ai is the shortest distance
between the joint axes jointi and jointi1.
Let the joint axes be given by the expression
Where pi is a point on axis of jointi and ui is
one of its direction vectors (analogous for
jointi1).
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• There are three methods to compute the link
  vector ai and the link length

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Method 1 The Pseudo-naive approach
The shortest distance ai is the length of the
vector connecting the two axes, and perpendicular
to both of them.
Which can be expressed
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Lets find the points oi and oai where this
distance exists.
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Multiplying respectively by ui and ui1, we
obtain the two following equations
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Solution
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Finally, using and
we obtain
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Method 2 The Geometric approach
The vector ui x ui1 gives the perpendicular
vector to both axes. Lets find out where it is
located on the joint axes.
We can go some distance s from point pi along the
axisi, and then go some distance k along ui x
ui1. Finally go some distance t along the
axisi1 to arrive at point pi1.
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We obtain the equation
There are three unknowns.
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Lets first eliminate the unknown k from the
equation
by multiplying by ui
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Lets first eliminate the unknown k from the
equation
by multiplying by ui1
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Now we shall eliminate the s and t from the
equation
by multiplying by ui x ui1
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We have obtained a system of three equations in
the unknowns s, t, k
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From
,
it can be seen that the shortest distance between
jointi and jointi1 is given by the vector
Where
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From
and
, we can compute s and t
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Finally, using and
we obtain
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Method 3 The Analytic approach
The distance between two arbitrary points located
on the joint axes jointi and jointi1 is
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The link length of linki, ai, is the minimum
distance between the joint axes
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A necessary condition is
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Which is equivalent to their numerators being
equal to 0
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Rewriting this system yields
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Whose solution are
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Finally, using and
we obtain
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oi and oai are the closest points on the axes of
jointi and jointi1. We deduce that the link
vector ai and the link length ai
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The link vector ai
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Calculating the scalar products and,
both equal to 0, proves that the vector ai is
perpendicular to both axes of jointi and jointi1
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• Three methods
• How do we actually compute ai and ai2 ?

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The link vector ai is perpendicular to both of
the axes of jointi and jointi1. The unit vector
is parallel to the link vector ai.
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Given two points pi and pi1 on the axes of
jointi and jointi1, the link length can be
computed as
And the link vector
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• Special cases
• The joint axes intersect
• The shortest distance ai is equal to zero
• The link vector is the null vector

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• The joint axes are parallel
• There is no unique shortest distance

oi can be chosen arbitrarily, so we should chose
values that offset the most of Denavit-Hartenberg
parameters
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• The first joint
• There is no link preceding it
• We use a base link link0
• Its link frame should coincide with the link
  frame of link1
• Most of the Denavit-Hartenberg parameters will be
  equal to zero

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• The last joint
• There is no link succeding it
• We use arbitrary values so that most of
  Denavit-Hartenberg parameters are equal to zero

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Denavit-Hartenberg

1. Compute the link vector ai and the link lenght
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai
4. Compute the link offset di
5. Compute the joint angle fi
6. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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• Identify the joint axes
• Identify the common perpendiculars of successive
  joint axes
• Attach coordinate frames to each joint axes

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Identifying the joint axes
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Identifying the common perpendiculars
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Attaching the frames
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Denavit-Hartenberg

1. Compute the link vector ai and the link lenght
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai
4. Compute the link offset di
5. Compute the joint angle fi
6. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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Remember
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Lets define the function
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We then have
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1. Compute the link vector ai and the link lenght
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai
4. Compute the link offset di
5. Compute the joint angle fi
6. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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Denavit-Hartenberg
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1. Compute the link vector ai and the link lenght
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai
4. Compute the link offset di
5. Compute the joint angle fi
6. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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Denavit-Hartenberg
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Denavit-Hartenberg

1. Compute the link vector ai and the link lenght
2. Attach coordinate frames to the joint axes
3. Compute the link twist ai
4. Compute the link offset di
5. Compute the joint angle fi
6. Compute the transformation (i-1)Ti which
   transforms entities from linki to linki-1

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This transformation is done in several steps

• Rotate the link twist angle ai-1 around the axis
  xi
• Translate the link length ai-1 along the axis xi
• Translate the link offset di along the axis zi
• Rotate the joint angle fi around the axis zi

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Multiplying the matrices
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By specifying the transformation
for all joints we can specify a
transformation form linkN to the base link0

Lets denote the joint
parameters of jointi.
We obtain
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End effector

• End effector - the last coordinate system of
  figure
• Located in joint N.
• But usually, we want to specify it in base
  coordinates.

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End effector

• A transformation from the link N to the base

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End effector

• We can also express it as
• three rotations (around each of the coordinate
  axes)
• followed by a translation

• How can we establish a relation with the other
  expression ?

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End effector

• Location
• The origin of a coordinate frame relative to some
  base coordinate frame is specified by the
  translation

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End effector

• Orientation
• Any 3D orientation relative to some base
  coordinate frame can be specified by
• three rotations, one around each of the
  coordinate axes.

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End effector

• Orientation
• These rotations are named roll, pitch, yaw

• We do them in this order around x, y, z.

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End effector

• Orientation
• The yaw transformation can be expressed

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End effector

• Orientation
• The pitch transformation can be expressed

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End effector

• Orientation
• The roll transformation can be expressed

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End effector

• Orientation
• The roll, pitch and yaw transformation is then
  expressed

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End effector

• Finally, the transformation from a coordinate
  frame to the base frame is expressed

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End effector

• Remember that can be
  expressed as a 4x4 matrix with elements mij -
  functions of joint parameters ?1,...,?N.
• Lets omit them for readability

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End effector

• As the two transformations
  and do the same thing

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End effector

• We obtain directly the translation vector

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End effector

• We can obtain the yaw angle

Because
arctan is p-periodic. Lets use our function
arctan2 to get the right angle.
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End effector

• Knowing the yaw angle, we can obtain the pitch
  angle

Because
Again, lets use our function arctan2
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End effector

• We can obtain the roll angle

Because
Again, lets use our function arctan2
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End effector

• Lets define the state vector

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End effector

• As previously shown,

The state vector is composed of elements of this
matrix. Its also a function of joint parameters