Concept:
Area covered by one photograph
\({\rm{a}} = \left( {1{\rm{}}-{\rm{}}{{\rm{P}}_{\rm{L}}}} \right){\rm{\;}}\left( {1{\rm{}}-{\rm{}}{{\rm{P}}_{\rm{S}}}} \right){\rm{\;}}\left( {{\rm{L}} \times {\rm{W}}} \right) \times {\rm{}}\frac{1}{{{{\rm{S}}^2}}}\)
L = length of photograph in the direction of flight
W = width of photograph normal to the direction of flight
S = scale of photograph
PL = longitudinal overlap
PS = side overlap
\({\rm{Number\;of\;photographs}} = \frac{{{\rm{total\;area\;to\;be\;covered\;}}\left( {\rm{A}} \right)}}{{{\rm{area\;covered\;by\;one\;photograph}}}}\)
Calculation:
N = 725
\(\frac{{{{\rm{P}}_{\rm{L}}}}}{{{{\rm{P}}_{\rm{S}}}}} = 3\)
PS = 25%
PL = 75%
L = 36 cm
W = 36 cm
\( {\rm{a }} = {\rm{ }}\left( {1{\rm{ }}-\frac{{75}}{{100}}} \right)\left( {1{\rm{ }}-\frac{{25}}{{100}}} \right)\left( {36{\rm{ }} \times {\rm{ }}36} \right) \times {10^{ - 10}} \times {\rm{ }}\frac{1}{{{{\rm{S}}^2}}} \)
\({\rm{a\;}} = \frac{{243 \times {{10}^{ - 10}}}}{{{{\rm{S}}^2}}}\left( {{\rm{k}}{{\rm{m}}^2}} \right)\)
A = 600 Km2
\(725 = {\rm{N}} = \frac{{\rm{A}}}{{\rm{a}}} = \frac{{600 \times {{\rm{S}}^2}}}{{243 \times {{10}^{ - 10}}}}\)
\({\rm{Scale}} = \frac{1}{{5835.84}}\)
i.e. 1 in 5835.84