the ground 
te pixel in a 
5 paper the 
of an object 
ope € and 
(€ =0), the 
or 20/2 m. 
adius R 
(1) 
posed that 
the original 
1d generally 
itellite. The 
ave left the 
e may only 
stead of all 
otween the 
f the grid p 
| shift and a 
st from o to 
figure 3(2)). 
oe resolved 
he original 
2 4(1)). The 
en the two 
is distance 
-axis. Then 
and Y =ly,l. 
ect C, can 
-grid, if its 
(2) 
n the 
R should 
T 
p| + 
p 
| 
].«3 
: 
Say (3) 
2 2 
It is known from the above calculations that the 
demanded object radius R has no relationship with a turn 
of the grids around o. Therefore only the grids' shift needs 
to be considered. 
2.2 Shift extent of grids 
The lower bounds of the shift amount X and Y are 
obviously zero (figure 2). What are the upper bounds? 
See the figure 2 and the figure 5. The figure 2 shows the 
original position of the grids. In the figure 5(1) X and Y 
are smaller than a/2. In the figure 5(2) X and Y are 
equal to a/2. Contrast the figure 5(1) and 5(2) with the 
figure 2. It will be known that the demanded object radius 
R is longest in the figure 5(2) and shortest in the figure 2. 
In fact R increases with the increase of the shift's 
distance, if 0O x X «a/2, Ox Y «a/2. How does R 
change, if X 2 a/2, Y 2 a/2? 
ij d Ixpl 0 < ly g 
— xp SQ, O0 «lyol s À 
d Pisa 
The figure 6(1) shows such an example. In this situation 
the neighbour 7, of the grid p is a nearest grid to the 
center of the object C en C can be detected by a 
complete SPOT-detector-grid, if its area is large enough 
to include the grid 7. In fact it may be held that the grids 
are formed by the method that a grid appears repeatedly 
at intervals of 10m or 20m on the x-axis and on the y-axis. 
The grids have the same periodic characteristic on the x- 
axis and on the y-axis in other words. Therefore the 
distance ofa may be treated as the equivalent shift of the 
grid p. But it is tenable: O « Ixigl € a/2 and O « lysel x 8/2. 
a a 
Doce ael sit: 
The figure 6(2) shows an example of this situation. It is 
similar to 1) that the distance 0f¢ may be regarded as 
the equivalent shift of the grid p. It is also tenable: 0 < Ixsl 
€ a/2 and O « lysel x a/2. 
a a 
3) > Ixy € a, 5 « lypl € da x 
See the figure 6(3). Similarly 0f; may be regarded now 
as the equivalent shift of the grid p and it is tenable: 0 < 
Ixzl € a/2 and 0 < lyezl < a/2. 
4) Ixpl > A, Or lypl > A Or Ixpl > A and lypl > A: 
See the figure 6(4). No matter how long the distance of 
the shift is, a grid t' can be always found in the 
neighbourhood of the object's center o, which meets 0 < 
Ixel < a/2 and O< lyrl < a/2. ot’ is the equivalent shift of 
the grid p. 
Therefore the upper bounds of X and Y are equal to 
a/2. The extent of the grids' shift is expressed as Os 
X « a/2, 0 « Y « a/2. In this extent the demanded object 
radius R increases with the increase of X and Y. R 
should reach its maximum, when X - a/2 and Y= a/2 
37 
appear at the same time. The following is its 
mathematical proving. 
2.3 The sufficient area condition 
Suppose Z-R^ (R50). Then the formula (2) is turned 
into the following form: 
Z2 yrs 
= 2 Dur 
Osxedcosrs2,. (4) 
2 2 
The region D, defined by 0x X x a/2, 0x Y « a/2 is 
bounded and closed (figure 7). Therefore Z has a 
minimum and a maximum on D,- The former is already 
known (figure 2). The latter is the sufficient area condition 
for an object on the smooth ground. Let us first consider 
that to which direction the grid p shifts so that Z can get 
a maximal increment. For this purpose the gradient of Z 
at the point (0,0) is calculated: SZ. = (4,4). 
The direction of KZ om is C. m, AS. Suppose 
that the grid p shifts with a very small distance Ó in this 
direction. The new coordinates of the grid p are 
(X,, Y,) Certainly itis tenable: X, 2 Y. VZ at the 
point (X, —Yj is calculated as 
VZ| 
equal to 45^. The direction of VZ does not change, if 
the grid p shifts again by Ó in the direction 45°. If the 
above process goes on, the gradient line originating from 
O can be obtained. It is a straight line from the origin 
O(0,0) to the point K(a/2, a/2) (figure 7). Therefore Z. 
appears at the end of this vector X... — Y,,.- a/2 
(figure 8): 
follows: 
(xx) ^ (2X, t a, 2Y; t a). Its direction is also 
max 
a ZZ ES dE (5) 
553 
is equal to d. The 
sufficient area condition S, is calculated as follows 
(figure 5(2)): 
S, TRUE RU 
Obviously the maximal radius R ax 
(6) 
For a panchromatic SPOT-image there are 
d-1042 (m) and  $,-200z (m. 
For a multispectral SPOT-image there are 
d=202 (m and $,-800z (m. 
An object on the smooth ground can be detected at least 
by a panchromatic (multispectral) SPOT-detector-grid, if 
its area on the ground can include a circle with the radius 
1042 m (2042 m). This radius is equal to the diagonal 
length of a panchromatic (multispectral) SPOT-image 
pixel. 
International Archives of Photogrammetry and Remote Sensing. Vol. XXXI, Part B1. Vienna 1996