Answers to Assignment 4
Qn 1
𝜎 = 𝑎11𝑠 + 𝑎12𝑡 + 𝑎13 𝜏 = 𝑎21𝑠 + 𝑎22𝑡 + 𝑎23
(s, t) = (0, 0) (𝜎,𝜏) = (0,0). Thus 0 = 𝑎13 0 = 𝑎23
(s, t) = (1, 0) (𝜎, 𝜏) = (𝜋 , 0). Thus 𝜋2
2=𝑎11 0=𝑎21
(s, t) = (1, 1) (𝜎, 𝜏) = (𝜋 , 𝜋). Thus 33
𝜋 = 𝜋 + 𝑎12 ⇒ 𝑎12 = − 𝜋 326
𝜋 = 𝑎22 3
𝜎=𝜋𝑠−𝜋𝑡 26
𝜏=𝜋𝑡 3
𝑋 = 𝑐𝑜𝑠 𝜎 + 5 = 𝑐𝑜𝑠( 𝜋 𝑠 − 𝜋 𝑡) + 5 ⇒ 𝑠 = 2 𝑐𝑜𝑠−1(X − 5) + 𝑡 26𝜋3
𝑌 = 2𝜏 = 2𝜋 𝑡 ⇒ 𝑡 = 3 𝑌 3 2𝜋
Hence
𝑠 = 2 𝑐𝑜𝑠−1(X − 5) + 𝑌 𝜋 2𝜋
Qn 2
a)
n
1.
F jk
1
for all j
(conservation of energy)
(uniform light reflection) 1
k1
jjk kkj
2. A F
A F
3. Fjj 0 0FF
(for all convex surface patch j)
𝐹 +𝐹 =1𝐹 =0.6⇒𝐹 =0.4 121313 12
𝐹 +𝐹 =1⇒𝐹 =0.4 1332 32
0.5𝐹 +𝐹 +0.5𝐹 =1⇒𝐹 =0.6 12 22 32 22
Hence
0 0.4 0.6 𝐹 = (0.2 0.6 0.2)
0.6 0.4 0
b)
Initial step
𝐵1 = 0.4 𝐵2 = 0.9 𝐵3 = 0
𝛥𝐵1𝐴1 = 0.4 𝛥𝐵2𝐴2 = 1.8 𝛥𝐵3𝐴3 = 0
Face 2 emits radiosity it has the largest “stored” radiosity 𝛥𝐵 𝐴 𝑗𝑗
12 13 F 0.5F F 0.5F
12 22 FF0
32 1332
Iteration 1:
B due to B = 𝜌 𝐵 𝐹
j k𝑗𝑘𝑘𝑗
𝛥𝐵 =0.4+(𝜌 𝐵 𝐹 )=0.4+(0.5)(0.9)(0.2)=0.49
1 1 2 21
𝛥𝐵 = (𝜌 𝐵 𝐹 ) = (0.5)(0.9)(0.6) = 0.27 2 2222
𝛥𝐵 = (𝜌 𝐵 𝐹 ) = (0.5)(0.9)(0.2) = 0.09 3 3223
𝐵1 = 𝛥𝐵1 = 0.49
𝐵2 =0.9+0.27=1.17 𝐵3 = 𝛥𝐵3 = 0.09
2
𝛥𝐵1𝐴1 = 0.49 𝛥𝐵2𝐴2 = 0.54 𝛥𝐵3𝐴3 = 0.09
Face 2 emits radiosity since it has the largest “stored” radiosity 𝛥𝐵 𝐴 𝑗𝑗
Iteration 2:
𝛥𝐵 = 0.09 + (𝜌 𝐵 𝐹
3 3 2 23
𝐵3 = 𝛥𝐵3 = 0.207
c) two of the below:
) = 0.09 + (0.5)(1.17)(0.2) = 0.207
1) It does not have to solve the large system of linear equations. Hence it is much faster. 2) It does not need to calculate all the form factors. It calculates form factors when needed 3) It does not need to store all the form factors.
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