1
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
2
, 0). Thus
𝜋
2
= 𝑎11 0 = 𝑎21
(s, t) = (1, 1) (𝜎, 𝜏) = (
𝜋
3
,
𝜋
3
). Thus
𝜋
3
=
𝜋
2
+ 𝑎12 ⇒ 𝑎12 = −
𝜋
6
𝜋
3
= 𝑎22
𝜎 =
𝜋
2
𝑠 −
𝜋
6
𝑡
𝜏 =
𝜋
3
𝑡
𝑋 = 𝑐𝑜𝑠 𝜎 + 5 = 𝑐𝑜𝑠(
𝜋
2
𝑠 −
𝜋
6
𝑡) + 5 ⇒ 𝑠 =
2
𝜋
𝑐𝑜𝑠−1(X − 5) +
𝑡
3
𝑌 = 2𝜏 =
2𝜋
3
𝑡 ⇒ 𝑡 =
3
2𝜋
𝑌
Hence
𝑠 =
2
𝜋
𝑐𝑜𝑠−1(X − 5) +
𝑌
2𝜋
Qn 2
a)
1. F
jk
k
n
1
1
for all j (conservation of energy)
2. A F A Fj jk k kj (uniform light reflection)
2
3. F
jj
0 (for all convex surface patch j)
𝐹12 + 𝐹13 = 1 𝐹13 = 0.6 ⇒ 𝐹12 = 0.4
𝐹13 + 𝐹32 = 1 ⇒ 𝐹32 = 0.4
0.5𝐹12 + 𝐹22 + 0.5𝐹32 = 1 ⇒ 𝐹22 = 0.6
Hence
𝐹 = (
0 0.4 0.6
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 𝛥𝐵𝑗𝐴𝑗
Iteration 1:
B
j due to Bk = 𝜌𝑗𝐵𝑘𝐹𝑘𝑗
𝛥𝐵1 = 0.4 + (𝜌1𝐵2𝐹21) = 0.4 + (0.5)(0.9)(0.2) = 0.49
𝛥𝐵2 = (𝜌2𝐵2𝐹22) = (0.5)(0.9)(0.6) = 0.27
𝛥𝐵3 = (𝜌3𝐵2𝐹23) = (0.5)(0.9)(0.2) = 0.09
𝐵1 = 𝛥𝐵1 = 0.49
𝐵2 = 0.9 + 0.27 = 1.17
𝐵3 = 𝛥𝐵3 = 0.09
0
5.05.0
0
3213
322212
1312
FF
FFF
FF
F
3
𝛥𝐵1𝐴1 = 0.49
𝛥𝐵2𝐴2 = 0.54
𝛥𝐵3𝐴3 = 0.09
Face 2 emits radiosity since it has the largest “stored” radiosity 𝛥𝐵𝑗𝐴𝑗
Iteration 2:
𝛥𝐵3 = 0.09 + (𝜌3𝐵2𝐹23) = 0.09 + (0.5)(1.17)(0.2) = 0.207
𝐵3 = 𝛥𝐵3 = 0.207
c) two of the below:
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.