Introduction to Robotics Lecture II

Introduction to Robotics

Lecture II

Alfred BrucksteinYaniv Altshuler

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•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

Denavit-Hartenberg• Link length ai

• The perpendicular distance between the axes of jointi and jointi+1

• Link twist αi

• The angle between the axes of jointi and jointi+1

• Angle around xi-axis

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• Link offset di

• The distance between the origins of the coordinate frames attached to jointi and jointi+1

• Measured along the axis of jointi

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• Link rotation (joint angle) φi

• The angle between the link lenghts αi-1 and αi

• Angle around zi-axis

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1.Compute the link vector ai and the link length

2.Attach coordinate frames to the joint axes

3.Compute the link twist αi

• How to compute the parameters to describe an articulated figure :

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4.Compute the link offset di

5.Compute the joint angle φi

6.Compute the transformation (i-1)Ti which transforms entities from linki to linki-1

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1.Compute the link vector ai and the link length






• Let’s do it step by step

Denavit-HartenbergThe link length ai is the shortest

distance between the joint axes jointi and jointi+1.

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 jointi+1).

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•There are three methods to compute the link vector ai and the link length

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The shortest distance ai is the length of the vector connecting the two axes, and perpendicular to both of them.

Which can be expressed :

Method 1 : The Pseudo-naive approach

Denavit-HartenbergLet’s find the points oi and oai where

this distance exists.

Denavit-HartenbergWe can go some distance s from pi along

axisi, and then the distance ai along the unit vector

and finally some distance t along axisi+1 to arrive at point pi+1.

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Multiplying respectively by ui and ui+1, we obtain the two following equations:

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Solution :

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Finally, using and we obtain :

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We can go some distance s from point pi along the axisi, and then go some distance k along ui x ui+1. Finally go some distance t along the axisi+1 to arrive at point pi+1.

The vector ui x ui+1 gives the perpendicular vector to both axes. Let’s find out where it is located on the joint axes.

Method 2 : The Geometric approach

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We obtain the equation :

There are three unknowns.

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Let’s first eliminate the unknown k from the equation :

by multiplying by ui :

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by multiplying by ui+1 :

Let’s first eliminate the unknown k from the equation :

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by multiplying by ui x ui+1 :

Now we shall eliminate the s and t from the equation :

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We have obtained a system of three equations in the unknowns s, t, k :

Denavit-HartenbergFrom

,it can be seen that the shortest distance between jointi and jointi+1 is given by the vector :

Where

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From

and ,

we can compute s and t :

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The distance between two arbitrary points located on the joint axes jointi and jointi+1 is :

Method 3 : The Analytic approach

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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 :

Denavit-HartenbergWhose solution are :

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oi and oai are the closest points on the axes of jointi and jointi+1.

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 jointi+1

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•Three methods

•How do we actually compute ai and ||ai||2 ?

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The link vector ai is perpendicular to both of the axes of jointi and jointi+1. The unit vector :

is parallel to the link vector ai.

Denavit-HartenbergGiven two points pi and pi+1 on the axes

of jointi and jointi+1, the link length can be computed as :

And the link vector :

Denavit-HartenbergSpecial cases :

•The joint axes intersect

•The shortest distance ai is equal to zero

•The link vector is the null vector

Denavit-Hartenberg•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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1.Compute the link vector ai and the link lenght






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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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Remember, is the point where the shortest distance to jointi+1 exists

Identifying the common perpendiculars

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the origin

Attaching the frames

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Remember

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Let’s define the function

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We then have :

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This transformation is done in several steps :

•Rotate the link twist angle αi-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 φi around the axis zi

iiii xxzzi

i RotTransTransRotT 1

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1000

0cossin0

0sincos0

0001

ii

iixi

Rot

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1000

0100

0010

001 i

x

a

Transi

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1000

100

0010

0001

iz d

Transi

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1000

0100

00cossin

00sincos

ii

ii

ziRot

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Multiplying the matrices :

iiii xxzzi

i RotTransTransRotT 1

1000

cossin0

sinsincoscoscossin

cossinsincossincos

1

iii

iiiiiii

iiiiiii

ii

d

a

a

T

Denavit-HartenbergBy specifying the transformation

for all joints we can specify a transformation form linkN to the base link0 :

Let’s denote the joint parameters of jointi.

We obtain :

End effector

•End effector - the last coordinate system of figure

•Located in joint N.

•But usually, we want to specify it in base coordinates.

End effectorA transformation from the link N to the

base :

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 ?

End effector•Location

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

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.


•These rotations are named roll, pitch, yaw

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


•The yaw transformation can be expressed :


•The pitch transformation can be expressed :


•The roll transformation can be expressed :


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

End effector•Finally, the transformation from a

coordinate frame to the base frame is expressed :

End effectorRemember that can be expressed as a 4x4 matrix with elements mij - functions of joint parameters θ1,...,θN.

Let’s omit them for readability :

End effectorAs the two transformations and do the same thing :

End effector

We obtain directly the translation vector :

End effector

We can obtain the yaw angle :

Because :

arctan is π-periodic. Let’s use our function arctan2 to get the right angle.

End effector

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

Because :

Again, let’s use our function arctan2 :

End effector

We can obtain the roll angle :

Because :

Again, let’s use our function arctan2 :

End effectorLet’s define the state vector

End effectorAs previously shown,

The state vector is composed of elements of this matrix. It’s also a function of joint parameters :

Introduction to Robotics Lecture II

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