
U(p; q)
V (p; q)
W (p; q)
3
7775 ;
r(p; 0) = r(p; Q)
r(0; 0) = r(0; q)
r(P ? 1; 0) = r(P ? 1; q)
; p = ; 1; : : : ; P ? 1
q = ; 1; : : : ; Q (1)
2
Figure 1: Recovery of 3D vehicle shapes from images. Each target object is shown in a frame from the corresponding sequence. Below this frame is shown the obtained shape model.
The model is mirror symmetrical about a vertical plane through the sample points
r(p; 0) and r(p;
hQ2
i
), p = ; 1; : : : ; P ? 1.
2.2 Extracting a shape vector
To concisely represent the shapes in chosen training set, a shape vector is extracted
from each shape instance. The shape vector consists of a set of control points, X ij ,
<= i < M and <= j < N , which defines a bicubic Bspline surface S(u; v) =
fSx(u; v); Sy(u; v); Sz(u; v)gT approximating the model r(p; q), <= p < P and <=
q <= Q. The spline S(u; v) is expressed as [13]
S(u; v) =
M?1
X
i=0
N?1
X
j=0
X ij Bi(u)Bj(v) (2)
where, Bi(u) and Bj(v) are the 4th order basis functions defined on the following knot vectors respectively:
u = [0; ; ; ; 1; 2; ? ? ? ; M ? 3; M ? 3; M ? 3; M ? 3]
v = [0; 1; 2; ? ? ? ; N ? 1; N;N + 1; N + 2]