Robot Kinematics
"Joint Space"
The Forward
Kinematics
(1,2,3, etc)
 x
 
If we just move linearly, New  y  = Old
 z 
in y and c in z.
"World Space"
(x, y and z for
the end-effector)
 x  a 
 y  b 
  +   where we have moved a in x, b
 z   c 
This approach proves to be more versatile:
 x  1
 y  0
 
 z  0
  
 1  0
1
0

0

0
0 0 a  x 
1 0 b   y 
0 1 c z 
 
0 0 1 1 
0 0 a
1 0 b 
0 1 c

0 0 1
is referred to as the Translation Matrix Trans(a,b,c).
We can also develop matrices which will let us work out what happens to (x,y,z)
when we rotate about an axis.
Rotation about y
Positive rotation - corkscrew or clockwise rule - clockwise as we face in the
positive direction about our axis of rotation.
y
x
z
So positive rotation about y is from z to x.
z
Old z
New z
1

r
x
Old x New x
So new x
= r sin(1+) = r sin(1) cos() + r cos(1) sin()
= old x  cos() + old z  sin()
and new z
= r cos(1+) = r cos(1) cos() - r sin(1) sin()
= old z  cos() - old x  sin()
y is unchanged, as the rotation was about the y axis.
We can produce a matrix to help us calculate the new x and z.
rot(y) =
 cos
 0

 sin 

 0
0 sin 
1
0
0 cos 
0
0
0
0
0

1
We can do similar calculations for rotation about the x and z axes.
Rot(x) =
Rot(z) =
0
1
0 cos

0 sin 

0
0
cos
 sin 

 0

 0
0
 sin 
cos
0
 sin 
cos
0
0
0
0
0

1
0 0
0 0
1 0

0 1
Why bother with the matrices ?
Because:
a) They make the calculations easier if more than one operation is involved.
b) They make it easier to deal with movements involving rotation about more than
one joint of the robot.
To demonstrate, suppose we start at (3,3,4), then rotate 90 about the z axis, then
move linearly such as to increase x by 2, y by 3 and z by 4; then rotate 45 from
the resulting position about the z axis.
The 90 rotation about z requires the following matrix:
cos 90  sin 90
 sin 90 cos 90

 0
0

0
 0
0 0
0 0
1 0

0 1
0  1
1 0

0 0

0 0
=
0 0
0 0
1 0

0 1
The translation requires:
1
0

0

0
0 0 a
1 0 b 
0 1 c

0 0 1
=
1
0

0

0
0 0 2
1 0 3
0 1 4

0 0 1
The 45 rotation about x requires:
0
0
0
1
0
0
1
0 cos  sin  0
0 0.707  0.707



=
0 sin  cos 0
0 0.707 0.707



0
0
1
0
0
0
0
0
0
0

1
It is specified that we rotate about z first, so we end up at:
0  1
1 0

0 0

0 0
0 0  3 
0 0  3 
  =
1 0  4 
 
0 1 1 
 3
3
 
4
 
1
Now we do the translation:
1
0

0

0
0 0 2  3
1 0 3  3 
  =
0 1 4  4 
 
0 0 1  1 
 1
6
 
8
 
1
Now to rotate by 45 about x;
0
0
1
0 0.707  0.707

0 0.707 0.707

0
0
0
0
0
0

1
 1
6
 
8
 
1
=
 1 
 1.414


 9.898 


 1 
(NB - final answer different from the notes - I think I am right !)
We could equally well operate by multiplying the rotation and translation matrices
together and then premultiplying the original position "vector" by the resulting
matrix. In multiplying the matrices, we must be careful to put the one representing
the first move at the right and proceeding leftwards in order of the operations.
0  1
1 0

0 0

0 0
=
0 0  1 0
0 0 0 1

1 0  0 0

0 1  0 0
0 2 1
0
0
0 3 0 0.707  0.707

1 4 0 0.707 0.707

0 1  0
0
0
0
0
0

1
1
0
2 
 0
0.707 0  0.707  0.707


0.707 1
0.707
4.949 


0
0
1 
 0
3
3
and premultiplying   by the above matrix gives us our resulting final position
4
 
1 
vector.
Now for Something More Like a Proper Robot
Member
B
Pivot
Pivot
Member
A
y
Member
C
x
z
Lengths of members
between pivots
A is 0.8 m long
B is 0.6 m long
C is 0.5 m long
We will suppose that the end-effector is at the origin of the (x,y,z) system of coordinates which we are using. Member A ("The trunk") remains vertical and pivots
about its base as shown through an angle 1. Member B rotates about its joint with
member A through an angle 2 (positive clockwise as shown) and Member C
rotates about its joint with Member B through an angle 3 (positive clockwise as
shown).
Example - where will the end-effector be after we have rotated C clockwise 25
with the other joints stiff, then rotated B anticlockwise 40 with the other joints stiff,
then rotated A 55 such as to bring the other members upwards out of the page ?
Answer - first the rotation about the B/C pivot ! Unfortunately the z axis is not
through that point but through the end-effector's original position (known as The
Home Position). We therefore need to move the z axis vertically upwards so it
does go through the pivot. Doing this will leave x and z unchanged but y will now
be -0.5 m, so it is as if we moved the end-effector by -0.5 m.
1
0
Translation matrix 
0

0
0 0
0 
1 0  0.5
,
0 1
0 

0 0
1 
0 
1 0 0
 0 1 0  0 . 5

so new position vector = 
0 0 1
0 


1 
0 0 0
0   0 
 0    0 .5
  = 
.
0   0 
  

1   0 
We now want to rotate through 25 clockwise about the z axis; actually -25 as
anticlockwise is the positive direction !
cos( 25)  sin( 25)
 sin( 25) cos( 25)
Rotation matrix 

0
0

0
0

0 0  0.9063 0.4226
0 0  0.4226 0.9063
=
1 0 
0
0
 
0 1 
0
0
0 0
0 0
1 0

0 1
so the new position vector (NB with the origin at the B/C joint !) will be:
 0.9063 0.4226
 0.4226 0.9063


0
0

0
0

0 0  0    0.2113
0 0   0.5  0.4532 
=

1 0  0  
0

 

0 1  1  
1

We now want to rotate clockwise about the A/B joint, so we need to move the
origin of the co-ordinate system again. We are moving the origin left by 0.6 m, so it
is as if we are moving +0.6 m in x.
1
0

0

0
0 0 0.6   0.2113  0.3887 
1 0 0   0.4532  0.4532 

 = 

0 1 0 0   0 

 

0 0 1 1   1 
Now we are ready for the rotation about the A/B joint.
We are rotating 40 anticlockwise this time about the z axis, so the rotation is
positive.
cos( 40)  sin( 40)
 sin( 40) cos( 40)

 0
0

0
 0
0 0  0.3887   0.5891 
0 0  0.4532   0.0973
=
  0 
1 0 
0
 
 

0 1 
1
  1 
and now the final rotation about y ! Fortunately the y axis goes through our present
origin, so we do not need to move the origin again. Our angle of rotation about y
is negative, as it is anticlockwise as we face in the direction of the y axis.
 cos( 55)

0

 sin( 55)

0

0 sin( 55) 0  0.5891   0.3379 
1
0
0  0.0973  0.0973

 = 


0 cos( 55) 0  0   0.4825 

 

0
0
1  1   1 
So the end-effector is at (0.3379, -0.0973, 0.4825), but that position is relative to
the B/C joint. We would like to know where the end-effector is relative to the
"Home" position ...
We need to move the origin again .. left by 0.6 m and down by 0.5 m.
This reduces x by 0.6 and increases y by 0.5.
1
0

0

0
0 0  0.6  0.3379   0.2621
1 0 0.5   0.0973  0.4027 

 = 

0 1
0   0.4825   0.4825 

 

0 0
1  1   1 
So the final position using the original co-ordinates is (-0.2621, 0.4027, 0.4825).
Your turn ! If we start from the "Home" position again, where will the end-effector
be after we rotate C anticlockwise by 20, then rotate B clockwise by 30, then A
about its base by 35 such as to move B and C out of the paper ?
Would it make any difference if we did the operations in the reverse order ?
The Denavit-Hartenberg or "A" Matrix
This gadget allows us to find what the co-ordinates will be in one system of coordinates (x,y,z) if we know them in another system of co-ordinates (x,y,z) and we
know how the systems relate to each other. It is not necessary for the axes in the
two systems to be parallel !
We can bring the two systems into alignment by at most two rotations and two
translations.
We can demonstrate by using the simple robot example from the previous section.
Member
B
Pivot
Pivot
Member
A
y
Member
C
x
Lengths of members
between pivots
A is 0.8 m long
B is 0.6 m long
C is 0.5 m long
z
For the rotation about the B/C pivot, it is easy to work out where the end-effector
will move to if the origin is at the B/C pivot and the rotation is about one of the
axes (z in this instance). What happens if we do the operations in the reverse
order ? (-55 about the base, +40 about the A/B pivot, -25 about the B/C pivot) ?
Denavit and Hartenberg proposed the following idea. We denote our two sets of
axes by (x1, y1, z1) and (x2, y2, z2) and we want to find (x1, y1, z1) given (x2, y2, z2).
We take it that we changed the system of axes in the following way:
1) We rotated System 1 about its z-axis through an angle  to align the new x axis
correctly.
2) We moved the origin along the z-axis by D.
) To position the new
3) We moved the origin along the new x-axis by L.
) origin correctly.
4) We rotated the system about its new x-axis by an angle  to align the new z
axis correctly.
Their result was:
sin  sin 
L cos    x 
 x
cos   cos  sin 
 y
 sin  cos  cos   sin  cos  L sin    y 
  
 
z
 0
sin 
cos 
D  z 
 

 
0
0
1  1  2
1 1  0
(1 = the original axes, 2 = the changed axes).
In our example, we did not do the first rotation (so  = 0) or shift along z (so D = 0).
We moved the origin 0.6 m left (= -0.6 m) along the new x and we did not rotate
about the new x axis ( = 0).
So the co-ordinates of the end-effector will now be
 x
1
 y

   0
z
0
 

 1  1 0
0 0 0 .6   x 
1 0 0   y 
0 1 0  z 
 
0 0 1  1  2