Projekt nalotu fotogrametrycznego : BW

I)Dane :

d_x  × d_y = 18×18cm ---> 0,18m×0,18m ; m=8000 ; p =60% ; q=25% ; f=120mm--->0,012m ; V=275km/h ---> $\frac{275000m}{3600s}$ ; t=1/125s

II)Obliczamy :

W/H ; B_x ,  B_y ,  D_x ,  D_y ,  Z ,  P_m ,  P_n ,  N_Z ,  t ,  s 

III)Rozwiązanie :

a)wysokość lotu W/H

$\ \frac{1}{m} = \frac{f}{W}$ ; $\frac{1}{8000} = \frac{0,012m}{W}$ ; W = 8000 × 0, 012m = 96m

b₁)baza podłużna B_x

$B_{x} = \frac{m \times d_{x} \times (100\% - p)}{100\%}\ ;\ B_{x} = \frac{8000 \times 0,18m \times (100\% - 60\%)}{100\%} = \frac{1440m \times 40\%}{100\%} = 576m$

b₂)baza poprzeczna B_y

$$B_{y} = \frac{m \times d_{y} \times (100\% - q)}{100\%}\ ;\ B_{x} = \frac{8000 \times 0,18m \times (100\% - 25\%)}{100\%} = \frac{1440m \times 75\%}{100\%} = 1080m$$

c)zasięg zdjęcia D_x ,  D_y

D_x =  d_x × m ;  D_x = 0, 18m × 8000 = 1440m

D_y = d_y × m ;  D_y = 0, 18m × 8000 = 1440m

d)zakładka Z

Z = 0, 1 × D_y ; Z=0,1×1440m=144m

e)powierzchnia modelu stereoskopowego P_m

P_m = (D_x−B_x) × D_y ;  P_m = (1440m−576m) × 1440m = 1244160m² ≈ 1, 244km²

f)powierzchnia dodana przez każde zdjęcie P_n

P_n = B_x × B_y ;  P_n = 576m × 1080m = 622080m² = 0, 622km²

g)liczba zdjęć N_Z

$N_{Z_{C}} = N_{Z_{S}} \times N_{S}\ ;\ N_{Z_{S}} = \frac{L_{S}}{B_{x}} + 5\ ;\ N_{S} = \frac{L_{y}}{B_{y}}\ ;\left\lbrack L_{s} = D_{x}\ ;\ L_{y} = D_{y} \right\rbrack\ ;\ N_{Z_{C}} = (\frac{D_{x}}{B_{x}} + 5) \times \frac{D_{y}}{B_{y}}$

$N_{Z_{C}} = \left( \frac{1440m}{576m} + 5 \right) \times \frac{1440m}{1080m} = 7,5 \times 1,33 = 9,975 \approx 10\ zdjec$

h)interwał fotografowania t

$t = \frac{B_{x}}{V}\ ;\ t = \frac{576m}{275\frac{\text{km}}{h}} = 576m \times \frac{3600s}{275000m} = 7,54s$

$t = \frac{d_{x} \times m}{V} \times \frac{100\% - p}{100\%}\ ;\ t = \frac{0,18m \times 8000}{275\frac{\text{km}}{h}} \times \frac{100\% - 60\%}{100\%} = 1440m \times \frac{3600s}{275000m} \times \frac{40\%}{100\%} = \frac{10368}{1375}s \approx 7,54s$

i)rozmazanie obrazu s 

$s = V \times t \times \frac{f}{W}\ ;\ s = \frac{275000m}{3600s} \times \frac{1}{125}s \times \frac{0,012m}{96m} = \frac{33}{432000}m = 0,00007638m = 0,076\text{mm}$

$s = \frac{V \times t}{m}\ ;\ s = \frac{\frac{275000m}{3600s} \times \frac{1}{125}s}{8000} = \frac{2750}{4500}m \times \frac{1}{8000} = 0,00007638m = 0,076\text{mm}$