Exam preparation

Is any one else besides me finding the amount of material to learn a bit over whelming?

Does any body know how to solve the mathematical question from the Example they gave for the exam? 3a where we have to compute the coordinates of vertices a and b?

you need to write the vertices given in matrix form (4x1 matrix) and then apply the transformations.... Thus you need to know how to setup the matrix and know how the scaling and transformation matrices work.

there's also a trick in the book about the order in which you rotate, scale, transformation as these ops are not commutative.

Assume you working in 2D space, then

to move a point from x,y, to x+xt, y+ty, you can apply the following transformation with matrix math

| 1 0 xt |         | x |     | x + xt |
| 0 1 yt |   *     | y | =   | y + yt |
| 0 0 1  |         | 1 |     | 1      |

The same holds for 3D space, you need 4x4 transformation matrix and a 4x1 vertex matrix.

Note sure if this helps...

I still am stuck with the matrix when performing an anti-clockwise rotation by 90 do you just say for example cos(theta) + 90

For (x,y,z) = (2,-2,2)

Seen from the top

          -z
          | 
          |
          |
-x        |           +x
 ---------y-----2-----
          |
          |     
          |    
          2     * (2,-2,2)
          |
          +z


If rortated along the z axis - is clear that z cannot change, x, y can change

The matrix math looks as follows


¦ cos@  -sin@  0  0  ¦
¦ sin@   cos@  0  0  ¦
¦  0      0    1  0  ¦
¦  0      0    0  1  ¦


¦ x  ¦
¦ y  ¦
¦ z  ¦
¦ 1  ¦

Resultant matrix will be 4 x 1

¦ (cos@ * x)+(-sin@ * y)+(0 * z)+(0 * 1)  ¦
¦ (sin@ * x)+( cos@ * y)+(0 * z)+(0 * 1)  ¦
¦ ( 0   * x)+(  0   * y)+(1 * z)+(0 * 1)  ¦
¦ ( 0   * x)+(  0   * y)+(0 * z)+(1 * 1)  ¦

Simplyfied


¦ (cos@ * x)+(-sin@ * y)  ¦
¦ (sin@ * x)+( cos@ * y)  ¦
¦  z  ¦
¦  1  ¦

The result supports the above observations, z cannot change, but x and y can change


In this example, (2,-2,2) moved to

x = 2*cos(90) + (-2*(-sin(90)) = 2
y = 2*sin(90) + (-2)*cos(90)      = 2
z = 2

(2,2,2)

Hope this helps

The math section is about 5% so I am not going to spend too much time if I have difficulties ... they say bring along a calculator ... transformations ...should I be bringing a scientific calculator for these sin and cos functions?

Yes, any non-programmable calculator. Just memorise the matrices for rotation, translation, scale and shear. I don't think they'll ask anything harder than that because of the time constraints.

On a side note. Why does TL103 state that the paper will count 70 marks? Shouldn't it be 75 since the year mark is 25? In any case, the 5% is still at most 4 marks. So they'll most likely ask only one transformation. And if they give the matrix (as in TL103) it should be an easy 4 marks.

--
"A man is the less likely to become great the more he is dominated by reason: few can achieve greatness - and none in art - if they are not dominated by illusion." Mr. Doctor

In terms of the sample exam paper, the last questions are about the application of texture and lighting.

Is there a guideline/rule as the the absolute minimum that must be in place for

1. Lighting to work
2. Textures to work
3. Materials to work

In the assignments I tried a number of different things and eventually it worked - and broke just as easily..

Any takers ?

Well goodluck to all!
May u all be smiling at the end of the exam and not

Well goodluck to all!
May u all be smiling at the end of the exam and not

They tend to give the matrices that you need - so you don't need to memorize them.