assignment book cadcampart2a

8/17/2019 Assignment Book Cadcampart2a

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INTERNATIONAL ISLAMIC UNIVERSITY MALAYSIA

MME 3109 CAD/CAM

MME 3109 CAD/CAM

WORK BOOK Part 2a

Semester 1, 2012/2013

October 2012


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CURVES (HERMITE AND BEZIER)

PROBLEM A:HERMITE CURVE

Based on Figure 26, is a position vector and is a tangent vector of point "0". is a

position vector and is a tangent vector of point "1". At point "0", u = 0 and at point

"1", u = 1.

Figure 26: Vectors definition

Construct a Hermite Curve connecting both points with the vectors value are shown below:

SOLUTION:

Hermite Curve is constructed based on the following Equation:

Create the Table below. Column A contain value of u from 0 to 1 increase by 0.1. Column B

to M are filled with the vectors value (same for each row). Excel can be used to make the

calculation easier.


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Formula of cell N2 and O2 are as follows

N2 = (1-3*A2^2+2*A2^3)*B2+(3*A2^2-2*A2^3)*E2 + (A2-2*A2^2+A2^3)*H2 + (A2^3-

A2^2)*K2

Fill in the other cells in Column N by double click cell N2 (if using MS-Excel)

O2 = (1-3*A2^2+2*A2^3)*C2+(3*A2^2-2*A2^3)*F2 + (A2-2*A2^2+A2^3)*I2 + (A2^3-

A2^2)*L2

Fill in the other cells in Column O by double click cell O2.

Since this is a 2D problem, z values (column P) are zero.

The first and the last X,Y data (highlighted) MUST be the same with the first and second

point respectively.

Draw a Hermite Curve MANUALLY using X, Y data result as shown in the Table above. On

the same graph, draw both of the tangential vectors. Produce EXACTLY THE SAME

GRAPH as example shown below.


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Figure 27: The Hermite curve.

PROBLEM B: BEZIER CURVE

Four control points are defined as shown in Figure 28.

Figure 28: Four control points

where


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Construct a Bezier curve.

SOLUTION

Definition of Bezier Curve:

Create the Table below. Column B contain value of u from 0 to 1 increase by 0.1. Column C

to N are filled with the control points (same for each row). Excel can be used to make the

calculation easier.

Formula of cell O5 and P5 are as follows

O5 = C5*(1-B5)^3+F5*3*B5*(1-B5)^2+I5*3*(B5^2)*(1-B5)+L5*B5^3

Fill in the other cells in Column O by double click cell O5 (if using MS-Excel)

P5 = D5*(1-B5)^3+G5*3*B5*(1-B5)^2+J5*3*(B5^2)*(1-B5)+M5*B5^3

Fill in the other cells in Column P by double click cell P5.

Since this is a 2D problem, z values (column P) are zero.


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The first and the last x(u), y(u) data (highlighted) MUST be the same with the first and lastcontrol point respectively.

Draw a Bezier Curve MANUALLY using x(u), y(u) data result as shown in the Table above.

Produce EXACTLY THE SAME GRAPH as example shown below.

Figure 29: The Bezier curve


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ASSIGNMENT 5: CURVES

1. Based on the previous example, Construct a Hermite Curve connecting both points

with the vectors value are shown below:

Follow step by step procedure as shown in the example.

2. Based on the previous example, Construct a Bezier Curve from 4 control points as

follow:

Follow step by step procedure as shown in the example.


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MEC 3109 CAD/CAM

ASSIGNMENT 5

CURVES

Name:

Matric:

Section:

October 2012


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TRANSFORMATION

PROBLEM: 2D TRANSFORMATION

New Working Coordinate System (WCS) is obtained from Model Coordinate System (MCS)

after translation (2 unit in X direction and 1 unit in Y direction) and rotation (30O

CCW) as

shown in Figure 30.

(1) Define the Transformation Matrix.

(2) A point (3, 1.5) in WCS, equal to what coordinate in MCS?

Figure 30: MCS and WCS

SOLUTION:

Translation Rotation


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Total Transformation

So for a point (x,y,z) in WCS, its coordinates (x*,y*,z*) in the MCS is obtained:

,

The result is shown in the Figure below


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PROBLEM: 3D TRANSFORMATION

In this example, obtaining WCS1 is simpler as it involves translation from point D to point E,

and a rotation CW around X. However obtaining WCS2 involves translation from point D to

point B, rotation CCW around Y, and rotation CW around X.

If AD = AE = 10, AB=15, and ϴ = 60O. If Point I is centre point of the inclined circle and

located in the centre of rectangle BCGF, define its location based on WCS2 and MCS.

Figure 31: 3D Transformation

SOLUTION:

Based on WCS2, Location of point I is half of BC in X2 direction, half of BF in Y2 direction

and 0 in Z2 direction. BC = AD = 10 while BF = EA/sin ϴ = 10/sin(60)=11.55. Therefore, the

coordinate of point I is (5, 5.77, 0)


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Transformation Matrix from MCS to WCS2

Translation from point D to point B, the transformation matrix is:

Rotation 90O

CCW around Y, the transformation matrix is


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Rotation 30O

CW around X, the transformation matrix is

Total Transformation is

[T] = [D][Ry][Rx]CW

Coordinate of Point I according to MCS is


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MEC 3109 CAD/CAM

ASSIGNMENT NO 6

TRANSFORMATION

Name:

Matric:

Section:

October 2012


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ASSIGNMENT 6: TRANSFORMATION

One MCS and three WCS are shown in Figure 32. Plane GHIF has 30O

inclination from

plane CDHG. Length OA=CG=10, AD=AB=20, BE=30. MCS (X-Y-Z) is a Model

Coordinate System, WCS1 (X1-Y1-Z1), WCS2 (X2-Y2-Z2), WCS3 (X3-Y3-Z3) are Working

Coordinate Systems.

a) Define Transformation Matrix of WCS1

b) Define Transformation Matrix of WCS3

c) If Point J (centre of the first circle) is located at 10,10 based on WCS1, what is its location

based on MCS?

d) If Point K (centre of the second circle) is located at 10,10 based on WCS3, what is its

location based on MCS?

Figure 32: MCS and WCS

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