2013-12-09

Practice Final, Problem #3.

Originally Posted By: Henrik
Group members: Arash Zahoory, Henrik Tufte Lien

3) Use De Casteljau's Method to evaluate at t=0.7 the Bezier Curve with control points (0,0), (1,0), (1,1), (0,1).

X:
f = (1 − t) * c0 + t * d0 = 0.3 * 0 + 0.7 * 1 = 0.7
g = (1 − t) * d0 + t * e0 = 0.3 * 1 + 0.7 * 1 = 1
h = (1 − t) * e0 + t * c1 = 0.3 * 1 + 0.7 * 0 = 0.3
m = (1 − t) * f + t * g = 0.3 * 0.7 + 0.7 * 1 = 0.91
n = (1 − t) * g + t * h = 0.3 * 1 + 0.7 * 0.3 = 0.51
c(t) = (1 − t) * m + t * n = 0.3 * 0.91 + 0.7 * 0.51 = 0.630

Y:
f = (1 − t) * c0 + t * d0 = 0.3 * 0 + 0.7 * 0 = 0
g = (1 − t) * d0 + t * e0 = 0.3 * 0 + 0.7 * 1 = 0.7
h = (1 − t) * e0 + t * c1 = 0.3 * 1 + 0.7 * 1 = 1
m = (1 − t) * f + t * g = 0.3 * 0 + 0.7 * 0.7 = 0.49
n = (1 − t) * g + t * h = 0.3 * 0.7 + 0.7 * 1 = 0.91
c(t) = (1 − t) * m + t * n = 0.3 * 0.49 + 0.7 * 0.91 = 0.784

c(0.7) = (0.630, 0.784)
'''Originally Posted By: Henrik''' Group members: Arash Zahoory, Henrik Tufte Lien<br><br>3) Use De Casteljau's Method to evaluate at t=0.7 the Bezier Curve with control points (0,0), (1,0), (1,1), (0,1).<br><br>X:<br>f = (1 &minus; t) * c0 + t * d0 = 0.3 * 0 + 0.7 * 1 = 0.7<br>g = (1 &minus; t) * d0 + t * e0 = 0.3 * 1 + 0.7 * 1 = 1<br>h = (1 &minus; t) * e0 + t * c1 = 0.3 * 1 + 0.7 * 0 = 0.3<br>m = (1 &minus; t) * f + t * g = 0.3 * 0.7 + 0.7 * 1 = 0.91<br>n = (1 &minus; t) * g + t * h = 0.3 * 1 + 0.7 * 0.3 = 0.51<br>c(t) = (1 &minus; t) * m + t * n = 0.3 * 0.91 + 0.7 * 0.51 = 0.630<br><br>Y:<br>f = (1 &minus; t) * c0 + t * d0 = 0.3 * 0 + 0.7 * 0 = 0<br>g = (1 &minus; t) * d0 + t * e0 = 0.3 * 0 + 0.7 * 1 = 0.7<br>h = (1 &minus; t) * e0 + t * c1 = 0.3 * 1 + 0.7 * 1 = 1<br>m = (1 &minus; t) * f + t * g = 0.3 * 0 + 0.7 * 0.7 = 0.49<br>n = (1 &minus; t) * g + t * h = 0.3 * 0.7 + 0.7 * 1 = 0.91<br>c(t) = (1 &minus; t) * m + t * n = 0.3 * 0.49 + 0.7 * 0.91 = 0.784<br><br>c(0.7) = (0.630, 0.784)
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