MECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
Assignment 3
Due date: Nov. 14, 2008 in the class
1. To draw the ellipse shown in the figure, you need to derive the expressions for the
...
MECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
Assignment 3
Due date: Nov. 14, 2008 in the class
1. To draw the ellipse shown in the figure, you need to derive the expressions for the
x and y coordinates of the boundary points in parametric equations.
(a) Derive the expression for the ellipse that is located at the origin and oriented
with its major axis in the x direction.
(b) Apply the proper transformations to the expression derived in (a) to get the
parametric equation of the ellipse shown.
Figure 1 Ellipse Location and Orientation
[SOLUTION]
To solve the part (a) of the problem, based on the general form of an ellipse, the
ellipse located at the origin and oriented with its major axis in the X-direction is
defined in the parametric form.
( )
2cos
sin 0 2
0
x y
z
θ θ
θ π
⎧ =
⎪
⎨ = ≤ ≤
⎪
⎩ =
To solve the part (b), the first step is to rotate the ellipse made in part (a) around
the Z-axis by 30 degrees, the transformation matrix is
1
cos30 sin 30 0 0
sin 30 cos30 0 0
0 0 1 0
0 0 0 1
Rz
⎡ ° − ° ⎤
⎢ ⎥
⎡ ⎤ ⎣ ⎦ = ⎢ ⎢ ° ° ⎥ ⎥
⎢ ⎥
⎢ ⎣ ⎥ ⎦
The second step is to translate the ellipse to the location (3, 5, 0), and the
transformation matrix is
2
1 0 0 3
0 1 0 5
0 0 1 0
0 0 0 1
T
⎡ ⎤
⎢ ⎥
⎡ ⎤ ⎣ ⎦ = ⎢ ⎢ ⎥ ⎥
⎢ ⎥
⎢ ⎣ ⎥ ⎦
The third step is to assume a point on the ellipse when it was defined in part (a),
3
5
2
1
300
X
YMECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
the coordinate of the point at the new location is calculated as
3 1
2 2
1 3
2 2
1 0 0 3 cos30 sin 30 0 0 3
0 1 0 5 sin 30 cos30 0 0 5
0 0 1 0 0 0 1 0
1 0 0 0 1 0 0 0 1 1 1
x x x y
y y x y
z z
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ′ ° − ° − + ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ′ = = ° ° ⎢ ⎢ + + ⎥ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ′ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢ ⎣ ⎥ ⎦
Thus the parametric form of the ellipse is
( )
12
3
2
3 cos sin 3
cos sin 5 0 2
0
x
y
z
θ θ
θ θ θ π
⎧ ′ = − +
⎪ ⎪
⎨ ′ = + + ≤ ≤
⎪ ′ =
⎪ ⎩
[END]
2. Consider a Hermite curve on the xy plane defined by the following geometric
coefficients:
(0) 2 3
(1) 4 0
(0) 3 2
3 4
(1)
p p p p
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ′ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎣ ⎦ ′ ⎢ ⎥ ⎣ ⎦ −
JG
JG
JJG
JJG
(a) Find a Bezier curve of degree 3 to represent the given Hermite curve as
exactly as possible. In other words, determine the four control points of the
Bezier curve.
(b) Expand both of the curve equations in polynomial form and compare them.
[SOLUTION]
To find a Bezier curve of degree 3 to approximate the given Hermite curve in part
(a), the four control points of the Bezier curve should be determined. One of the
properties of Bezier curves is the curve passes through the first and last control
points. Thus the two ends of the Hermite curve are the first and last control
points, respectively.
0 3 ( ) 0 , (1) 2 4
3 0
P p P p = = = = ⎡ ⎢ ⎤ ⎡ ⎤ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
JG JG
The first-order derivative of a Bezier curve is provided in the textbook.
1 1 ( ) 1
0
( ) 1
(1 )
n
i n i
i i
i
d p u n
n u u P P
du i
−
− −
+
∑=
⎛ ⎞ −
= ⋅ ⋅ − ⋅ − ⎜ ⎟
⎝ ⎠
JG
When u is equal to zero, 1 0
0
( )
( )
u
d p u
n P P
du
=
= −
JG
; when u is equal to one,MECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
3 2
1
( )
( )
u
d p u
n P P
du
=
= −
JG
. The tangent vectors of the Hermite curve at the ends are
also given. The other two control points ( P P 1 2 , ) can be found with the following
equations.
( )
( )
0 1
( )
' 0
( )
' 1
u u
d p u
p
du
d p u
p
du
= =
= =
JG
JG
JG
JG
Solving the two equations, the control points, P P 1 2 , , can be calculated.
1 2 11 4
3 3
3 3
P P ,
⎡ ⎤ ⎡ ⎤
= = ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
After the four control points are found, the Bezier curve is expressed as
0 ,
0 3 1 2 2 2 1 3 3 0
0 1 2 3
2 3 2 3 2 3 3
0 1 2 3
2 3 2 3
11
( ) ( )
3 3 3 3
(1 ) (1 ) (1 ) (1 )
0 1 2 3
(1 3 3 ) (3 6 3 ) (3 3 )
2 3
(1 3 3 ) (3 6 3 )
3
n
i i n
i
P u P B u
u u P u u P u u P u u P
u u u P u u u P u u P u P
u u u u u u
∑=
= ⋅
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= − + − ⋅ + − + ⋅ + − ⋅ + ⋅
⎡ ⎤
= − + − + − + ⎢ ⎥
⎣ ⎦
( )
2 3 3
4
3 3
2 3
2 3
3 4
(3 3 )
0
2 3 3 2
3 2 9 4 0 u 1
0
u u u
u u u
u u u
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ + − + ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤ + − +
⎢ ⎥
= + − + ≤ ≤ ⎢ ⎥
⎢ ⎥
⎣ ⎦
In part (b) of the question, since the Hermite curve was found before, the equation
of the Hermite curve is
2 3 2 3 2 3 2 3
2 3 2 3 2 3 2 3
3 2
3 2
(0)
(1)
( ) 1 3 2 3 2 2
(0)
(1)
(1 3 2 ) (0) (3 2 ) (1) ( 2 ) '(0) ( ) '(1)
2 3 3 2
(0 1)
4 9 2 3
p p
p u u u u u u u u u u
p p
u u p u u p u u u p u u p
u u u
u
u u u
⎡ ⎤
⎢ ⎥
⎢ ⎥
= − + − − + − + ⎡ ⎤ ⎣ ⎦ ⎢ ⎥
⎢ ⎥ ′
⎢ ⎥ ⎣ ⎦ ′
= − + ⋅ + − ⋅ + − + ⋅ + − + ⋅
⎡ ⎤ − + +
= ≤ ≤ ⎢ ⎥
⎣ ⎦ − + +
JG
JG
JG
JJG
JJG
JG JG JG JG
Comparing the expressions of the Hermite and Bezier curves, the two curves are
the same.
3. A Bezier curve defined by the control points A A A 0 1 2 , , and is to be transformed toMECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
the Bezier curve defined by B B B 0 1 2 , , and shown in the following figure. The
transformation should move point A B 0 0 to and A B 2 2 to . This means that scaling
is also required.
(a) Explain which transformation matrices are applied and in which order.
(b) Calculate the coordinates of control point B1 .
(c) Derive the parametric equation of the resulting curve C2.
[SOLUTION]
In part (a), to move point A B 0 0 to and A B 2 2 to , first translate the A A 0 2 from the
location
A0 to the origin. The transformation matrix is
1
1 0 1
0 1 1
0 0 1
T
⎡ − ⎤
⎡ ⎤ ⎣ ⎦ = ⎢ ⎢ − ⎥ ⎥
⎢ ⎣ ⎥ ⎦
The angle between B B 0 2 and the X direction is 1 tan ( ) 40.89 1 0 2 3 2
7 5
γ −
+ −
= =
−
, and the
angle between A A 0 2 and the X direction is 2 tan ( ) 45 1 0 4 1
4 1
γ −
−
= =
−
. Thus rotate
A A 0 2
about the origin by − − 4.11 40.89 45 D D D ( ). The rotation matrix is
2
cos( 4.11 ) sin( 4.11 ) 0
sin( 4.11 ) cos( 4.11 ) 0
0 0 1
R
⎡ − − − ⎤
⎢ ⎥
⎡ ⎤ ⎣ ⎦ = − − ⎢ ⎥
⎢ ⎥
⎣ ⎦
D D
D D
The module of
A A 0 2 is (4 1) (4 1) 3 2 − + − = 2 2 and the module of B B 0 2 is
(7 5) (2 3 2) 7 − + + − = 2 2 . The ratio of the modules is
7 7
S S S = = = = x y 3 2 18
So the scaling matrix is
7
18
3 7
18
0 0
0 0
0 0 0 0
0 0 1 0 0 1
x
y
S
S S
⎡ ⎤
⎡ ⎤ ⎢ ⎥
⎡ ⎤ ⎣ ⎦ = = ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥
⎢ ⎥ ⎣ ⎦ ⎢ ⎥
⎢ ⎣ ⎥ ⎦
The last step is to translate the point A0 from the origin to the location B0 . The
translation matrix isMECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
4
1 0 5
0 1 2
0 0 1
T
⎡ ⎤
⎡ ⎤ ⎣ ⎦ = ⎢ ⎢ ⎥ ⎥
⎢ ⎣ ⎥ ⎦
The equivalent matrix is calculated as
[ ] 4 3 2 1
7
18
7
18
0 0
1 0 5 cos( 4.11 ) sin( 4.11 ) 0 1 0 1
0 1 2 0 0 sin( 4.11 ) cos( 4.11 ) 0 0 1 1
0 0 1 0 0 1 0 0 1 0 0 1
0.622 0.045 4.333
0.045 0.622 1.423
0 0 1
E T S R T = ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎡ ⎤ − − − −
= − − − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎣ ⎦
⎡ ⎤
⎢ ⎥
= − ⎢ ⎥
⎢ ⎥ ⎣ ⎦
D D
D D
To calculate the coordinate of the location
B1 in the part (b) of the problem,
apply the equivalent matrix [E] on the problem. The coordinate can be
calculated as
1 1 [ ]
0.622 0.045 4.333 2 5.71
0.045 0.622 1.423 3 3.19
0 0 1 1 1
B E A
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
= = − = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎣ ⎥ ⎢ ⎥ ⎢ ⎥ ⎦ ⎣ ⎦ ⎣ ⎦
In part (c) to derive the parametric equation of the resulting curve C2, the three
control points are
0 1 2
5 5.71 7
, , and
2 3.19 2 3
B B B
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= = = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ + ⎦
The form of the Bezier curve isMECH 412 Computer Aided Mechanical Design
Dr. Z. C. Chen, Concordia University
2
0 2 1 2 0
, 0 1 2
0
0 2 1 2 0
0 1 2
2 2 2
0 1 2
2
2 2 2
( ) ( ) (1 ) (1 ) (1 )
0 1 2
2! 2! 2!
(1 ) (1 ) (1 )
0!(2 0)! 1!(2 1)! 2!(2 2)!
(1 ) (2 2 )
5
(1 ) 2
0
i i n
i
p u B B u u u B u u B u u B
u u B u u B u u B
u B u u B u B
u
∑=
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= ⋅ = ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅
− − −
= − ⋅ + − ⋅ + ⋅
⎡⎢
= − ⎢
⎣
JG
2 2
2 2
5.71 7
(2 2 ) 3.19 3.73
0 0
5 1.42 0.58
2 2.38 0.65
0
u u u
u u
u u
⎤ ⎡ ⎤ ⎡ ⎤
⎥ ⎢ ⎥ ⎢ ⎥
⎥ ⎢ ⎥ ⎢ ⎥ + − +
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤ + ⋅ + ⋅
⎢ ⎥
= + ⋅ − ⋅ ⎢ ⎥
⎢ ⎥
⎣ ⎦
4. Determine a Bezier curve of degree 3 that approximates a quarter circle centered
at (0, 0). The end points of the quarter circle are (1, 0) and (0, 1). Calculate the
coordinates of the middle point of this Bezier curve and compare them with those
of the midpoint of the quarter circle.
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