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surface of the layer and other layers are changed accordingly by 
substracting the depths of the surface So. This operation is 
essential for estimating the volume of a lithology or viewing 
the other layers with respect to a certain layer. 
The following are steps of this operations. 
1) Save the depth offsets of the bottom octants Oy(P, S) 
e Resolve the Peano key P and get the coordinates (x, y, z). 
e Store the z to a working array as Z(x, y). 
e Copy z to Z(xi, y*j) with 1 7 1, 2, ...S-1 and j » 1, 2, 
S1, if S 7 1. 
2) Adjustment operations on all octants O(P, S) 
case 1 : the size of the octant O(P, 1) is 1. 
e Resolve the Peano key P and get the coordinates (x, y, z). 
e Subtract the bottom depth Z^ 7 z - Z(x, y). 
e Interleave the coordinates (x, y, Z^) to calculate the new 
Peano key P". The datum adjusted octant is O’(P’, 1). 
case 2 : the size of the octant O(P, S) is S (S = 2, K > 0) 
e Resolve the Peano key P and get the coordinates (x, y, z). 
e If Z(x, y) » 0 and Z(x, y) « S, subdivide the octant into 8 
smaller octants and go to the beginning of step 2). 
e If Z(x, y) » 0 and the Remainder[Z(x, y), S] « 0, 
subdivide the octant into 8 smaller octants and go to the 
beginning of step 2). 
e Check the values of Z(x+i, y+j) with i = 1, 2, … S-1 and 
j= 1, 2, … S-1. If there is any value different from Z(x, 
y), partition the octant and go to the beginning of the 
steps 2). Otherwise, subtract the bottom depth z' 7 z - 
Z(X, y). Interleave the coordinates (x,y,Z') and get the 
new Peano key P'. The datum adjusted octant becomes 
QE S) 
3) Conformance check and aggregation 
e Set the initial leveli- 1. 
e If the Remainder[P', 8] = 0 and S = 2*!, check the values 
of P^ -- j*8"! (j - 1, 2, ...7). If all octants O(P^4j*8"!, 27!) 
exist, aggregate them and create a new octant O'(P^, 2)). 
e i=i+1 and repeat the above procedure until all the Peano 
keys are processed and no aggregation can be made. 
The following is an example based on a quadtree. 
1) At first, we register the depth offset values of the bottom 
boundary. The octant O(4, 2) has the coordinates (0, 2). We 
have Z(0) = 2 and Z(1) = 2. Similarly, we have Z(2) = 1, Z(3) 
=0,Z(4)= 1, Z(5) = 1, Z(6) = 2, Z(7) = 2. 
2.1) The Peano key 18 can be resolved as its coordinates (1, 4). 
Since the size of the quadrant is 1, y = 4 - Z(1) = 2. The new 
coordinates is (1, 2) corresponding to its Peano key 6. 
Similarly, we process the octants with Peano keys 9, 10, 11, 
33, 35, 44, 45, 46, 54 and 60. Their new Peano keys are 98; 
10210; 11211; 33532; 35234; 44240; 45541; 46212; 
54+51; 6056. 
2.2) As for octants with Peano keys 4, 12, 24, 36, 48 and 56, 
their sizes are S = 2 (K = 1). For example, if P = 12, the 
coordinates are (2, 2). Since Z(2) = 1 <2 = S, the quadrant is 
partitioned to 4 smaller quadrants 12, 13, 14 and 15, which 
511 
have the size of 1. After processing, the smaller quadrants have 
the Peano keys of 9, 12, 14 and 15, respectively. 
Similarly, quadrants with P = 24, 36, 48 produced the 
quadrants with P” = 13, 24, 26, 27, 33, 35, 36, 38, 37, 39, 48, 
50. 
For P = 56, the coordinates are (6, 4). Z(6) 2 2 2 S, Z(6) » 0 
and the remainder [Z(6), 2] = 0. Since Z(7) = 2 = Z(6), y” = 4 - 
2 = 2. The new coordinates are (6, 2). The new octant is O°(44, 
2). 
Finally, octant O(4, 2) becomes O(0, 2). 
3) According to the procedure described above, octants O'(8, 
1), O'(9, 1), O'(10, D, O'(11, 1) can be aggregated to O'(8, 2). 
Similarly we have O'(12, 2), O'(32, 2), O'(36, 2). 
At the level 1 = 2, no aggregation is required. The process is 
thus finished. 
10 
  
Q 1 2 3 4 5 6 7 38 X 
Figure 2. A quadtree before the datum adjustment 
  
  
  
  
  
  
  
Y| 
8 
7 
6 E p 
27 51 
5 
24 | 26 | 48 | SO | 56 
4 
3 
[6 12 36 44 
2 
41 
1 
0 8 32 40 | 42 
  
  
  
  
  
  
  
0 ud uda uIT uuu RX 
Figure 3. The quadtree after the datum adjustment 
International Archives of Photogrammetry and Remote Sensing. Vol. XXXI, Part B4. Vienna 1996