Image Reconstruction for 3D Lung Imaging - Department of Systems ...

with Ci being the following column vectors
⎡
−(x2y3z4 − x2z3y4 − x3y2z4 + x3z2y4 + x4y2z3 − x4z2y3)
⎢
C1 = ⎢ (y3z4 − z3y4 − y2z4 + z2y4 + y2z3 − z2y3)
⎣ −(x3z4 − z3x4 − x2z4 + z2x4 + x2z3 − z2x3)
(x3y4 − y3x4 − x2y4 + y2x4 + x2y3 − y2x3)
C2 =
C3 =
C4 =
⎡
⎢
⎣
⎡
⎢
⎣
⎡
⎢
⎣
(x1y3z4 − x1z3y4 − x3y1z4 + x3z1y4 + x4y1z3 − x4z1y3)
−(y3z4 − z3y4 − y1z4 + z1y4 + y1z3 − z1y3)
(x3z4 − z3x4 − x1z4 + z1x4 + x1z3 − z1x3)
−(x3y4 − y3x4 − x1y4 + y1x4 + x1y3 − y1x3)
−(x1y2z4 − x1z2y4 − x2y1z4 + x2z1y4 + x4y1z2 − x4z1y2)
(y2z4 − z2y4 − y1z4 + z1y4 + y1z2 − z1y2)
−(x2z4 − z2x4 − x1z4 + z1x4 + x1z2 − z1x2)
(x2y4 − y2x4 − x1y4 + y1x4 + x1y2 − y1x2)
(x1y2z3 − x1z2y3 − x2y1z3 + x2z1y3 + x3y1z2 − x3z1y2)
−(y2z3 − z2y3 − y1z3 + z1y3 + y1z2 − z1y2)
(x2z3 − z2x3 − x1z3 + z1x3 + x1z2 − z1x2)
−(x2y3 − y2x3 − x1y3 + y1x3 + x1y2 − y1x2)
In 3D the determinant is equal to six times the tetrahedron’s volume. The determinant of
C is
det(C) = 6A =
x2y3z4 − x2z3y4 − x3y2z4 + x3z2y4 + x4y2z3 − x4z2y3 − x1y3z4
· · · + x1z3y4 + x3y1z4 − x3z1y4 − x4y1z3 + x4z1y3 + x1y2z4 − x1z2y4
· · · − x2y1z4 + x2z1y4 + x4y1z2 − x4z1y2 − x1y2z3 + x1z2y3 + x2y1z3
· · · − x2z1y3 − x3y1z2 + x3z1y2
Substitution of this into 2.17 yields the potential function over the element
U(x,y,z) = � 1 x y z �
⎡ ⎤
a
⎢ b ⎥
⎣ c ⎦
d
= � 1 x y z �
⎡
⎢
⎣
⎤
⎥
⎦
⎤
⎥
⎦
⎤
⎥
⎦
⎤
⎥
⎦
c11 c12 c13 c14
c21 c22 c23 c24
c31 c32 c33 c34
c41 c42 c43 c44
where cij are the elements of C. This can be more easily written as a summation:
U(x,y,z) =
⎤⎡
⎥⎢
⎥⎢
⎦⎣
4�
Uiφi(x,y,z) (2.19)
i=1
where the interpolation functions, φi(x,y,z) i ∈ (1,2,3,4) are given by:
φi = c1i + c2ix + c3iy + c4iz
which are explicitly:
φ1 = 1
6A
⎧
⎪⎨
⎪⎩
−(x2y3z4 − x2z3y4 − x3y2z4 + x3z2y4 + x4y2z3 − x4z2y3)
· · · + (y3z4 − z3y4 − y2z4 + z2y4 + y2z3 − z2y3)x
· · · − (x3z4 − z3x4 − x2z4 + z2x4 + x2z3 − z2x3)y
· · · + (x3y4 − y3x4 − x2y4 + y2x4 + x2y3 − y2x3)z
20
⎫
⎪⎬
⎪⎭
U1
U2
U3
U4
⎤
⎥
⎦