The table shows that Class A and C are estimated correctly, 
6/24 represents 25$. The sum of Classes B and D is also correct, 
(9+3)/24 represents 50%. The difference between B and D origi- 
nates from the facts that only half of the photo coverage was 
taken into the sample, and that B bears always towards the nadir, 
and the situation with D is opposite. If the whole photo coverage 
had been taken into illustration all the estimates for class 
proportions would have become correct. Thus, the sampling on 
photo basis is unbias in the conditions of the illustration. 
Vertical photo as a base for sampling - A mathematical 
approach 
For the proof of unbias of a vertical photo, as a base for 
sampling, it is assumed that the inventory area in the field 
is divided into N squares of equal size (Q4-02-Qyw). It is 
further assumed that the true terrain surface of a square is 
level, either horizontal or not. The size of a square is assumed 
very small. In addition, the squares must not be covered by each 
other in the central projection (shielded angles are not accepted 
on the photo area supplied by elements). 
The size of a specific square in the field is noted by Q; - The 
size of a respective square on the photo is noted by gq; (for a 
vertical photo and flat terrain q; = (c/(H-h;))? Qi; where c-a 
camera constant or focal length and H-h;= the altitude of photo- 
graphy over terrain square Q;). The probability, P, that a ran- 
dom dot on an aerial photograph falls on a horizontal square Qi, 
can be written: 
P = PipPif' where 
Pip - probability that a random dot on the photo falls on 
a square Qj, provided that Qj is represented on the 
photo 
z di/Gdp: where gp=the size of the photo and, 
Pig 7» probability that a specific square Qj is represented 
on the specific photo 
z qp/9;/N, where 95/94 = the number which shows how 
many squares of size q; can be placed on a photo. 
By substituting it can be seen that 
P - (qi/dp)(dp/Qj/N) - 1/N