(or mean value) 
he distribution of 
cni tA 2 din 
variable with a 
pixel (see figure 
AX 
</2 
l position x 
that x is situated 
Ax, if dx € [ x, - 
pectation of the 
d as: 
dx =0 
Ax? 
u/2 303 Jf 
uam : E D 
lard deviation of 
as the maximum 
of it): 
JRM 
in error on the 
n of the image is 
d: when the radii 
(horizontal and 
ntal and vertical 
Is are considered 
ere. 
n by the relative 
> central pixel of 
-xc, being x the 
and xc the exact 
rete longitude of 
z X4 -XC , where 
the X axis. 
The discretization induces also an error r, , namely &r, which 
depends only on the error of x,, since the position of the 
window center (xc) is known without uncertainty: 
Er = r-rm = (x - XxC) - (Xm - XC) = X - Xm = EX 
| EF pax! = \EXmaxl = Ax/2 = 1/2 pixel 
[er ]= E[ex]= 0 
o(er)= Vo (er)=1/VI2 = 0,288 pixels. 
For diagonal radii, we proceed in the same way. The longitude 
of those radii is defined as the distance between the contour 
pixel (x1,y1) and the central pixel of the transformation window 
(xc, yc). 
The real longitude of any diagonal radius is given by 
po (x-xc) *(y- yc). , With (xy) exact position of the 
contour pixel. The measured one is , = (x,— x} +(Yn- y): 
with (x,y), discrete position of the contour. Then, the error 
associated to the longitude of diagonal radii is & = r -r,, and 
using this: 
1 xm+Ax/2 pym+Ay/2 
E[r]- SAT | dx Jax y)-dy = han 
xm—Ax/2 
it can be shown that the expectation of er is null: 
E[er ] E[ r-r4] 2 E[r]- ru 5 4-04 20 
The variance of er is: 
o (&)- 0 (r) * OC (rm) + 2. cov (r, ry) =O (1) 
To calculate 0° (r ) expression (1) is used: 
@(r)= Ep? ]-Efrf = El ]-rn° (2) 
Efr” ] can be calculated as: 
E[? ]  E[(x-xc)? ] * Ely-yo)? ] (3) 
Developing the sum of squares and using that E[ £x ]= 0: 
E[ (x-xc)? ] 2 E[ (x - Xp + Xp - xc)? ] 9 0^ ( € ) * (3 - xc)? 
Similarly, for y: ET (y-yc)? J= ( Ey ) + (Un - yo)? , and using 
these results, expression (3) is reduced to: 
E[? ]- à (x) * o! (£y) «rl 
Going back, expression (2) is: 
o (e)so (Ex) 0 (ey) * rj] rm? = o (ex) * o! (ey) 
Since squared pixels are considered, it is known that G^ (EX) z 
o? (ey). Then: 
O (er ) = 2.1/12 = 0,16 pixels. 
The standard deviation of the error on the longitude of diagonal 
radii is: 
O( er ) = VO ( er ) = 1 /N6 = 0,408 pixels. 
It can be seen that the value of the standard deviation is about 
the half of the maximum expected error: 
1 d = M\EXmad* + \EYmad*)=M(12?+(1/2°)=0,707 pixels 
There results agree with the intuitive perception (given by 
geometry) that the error on diagonal radii due to the 
discretization has a factor V2 with respect to that produced in 
the horizontal (or vertical) radii (which corresponds to the 
relationship between the diagonal and the side of a 1 pixel 
square). 
673 
4. ERROR ON RADII COMPARISON 
Two possibilities have to be considered: on the one hand, 
horizontal and vertical radii; on the other hand, diagonal radii. 
Differences between two horizontal (or vertical, taking again 
advantage of the squared pixels) radii are given by the 
equations: 
f» rl-r2 z (xl-xc) - (x2-xc) 2 x1-x2 
fom ri,-Y259 (Xl,Xc) - (x2,rxc) 9 xl x2, 
where x/ and x2 are the coordinates of the contour pixels inside 
the image, which determine the longitude of r/ and 72 
respectively; x1,, and x2,, are their discrete values and xc is the x 
coordinate of the center of the polar transformation window. 
The error on f,, due to the discretization, namely £f, depends on 
the errors of x/,, and x2,. Remembering that ex = x-x,, . 
& = f-fin =(x1-x2)-(X1 m-X2m) 2(x1- x15)-( x2-x2,) =ex1 - ex2 
Since &f is the difference of two random variables of uniform 
distribution, its probability function follows a triangular 
distribution (figure 3): 
1/AX 
4 
% 
-AX 0 AX 
Figure 3. Triangular distribution for the error on radii 
comparison 
The maximum value is given by the expression: 
|&f pad = 16x] pal + 1EX200l = AX/2 + A/2 = Ax = I pixel 
Again, this is an infrequent value for the error, especially 
considering that this error follows a triangular distribution. The 
expectation may be calculated on the basis of the errors 
associated to x/,, and x2,, as: 
E[ef ] - E[£xl - £x2] 2 0 
B[€x1- Ex2]= 
  
[dxt [7 1 2) 230 i 
x -Ax/2 Y eat = 
Ax.Ax 
and its variance can be evaluated using the independence of x/ 
and x2 [Sanchez, 89]: 
À f )2 o exl )*o'( ex2 ) = 1/12+1/12 = 0,16 pixels? 
So, the standard deviation for the error associated to the 
difference of horizontal (or vertical) radii is: 
0(&f)= VO ( ef) 1/V6 = 0,408 pixels. 
For the case of diagonal radii: 
r= (x- xc) + (y — yc) 
Where (x, y) is the exact position of the pixel contour. 
f-rl-r2. and fn = rl ray 
To calculate &f, previous rl, £&r2 are used: 
& =f fu= (r1-r2)- (rl -12y)= (rl- rly) - (r2-r2,)=¢rl - €r2 
Its maximum value is: |&f ul = ler pul + 1672, =1,41 pixels