So we want to know influence of AR on L/D
L D m a x = 1 K C D 0 {\displaystyle {\frac {L}{D}}_{max}={\sqrt {\frac {1}{KC_{D_{0}}}}}}
K = 1 π A R e ′ {\displaystyle K={\frac {1}{\pi AR\,e'}}}
e ′ = e ′ ( A R , u , s , C D 0 ) {\displaystyle e'=e'(AR,u,s,C_{D_{0}})}
u and s functions of geometry.
V m a x 2 = T A m a x W W S + W S ( T A m a x W ) 2 − 4 C D 0 K ρ ∞ C D 0 {\displaystyle V_{max}^{2}={\frac {{\frac {T_{A_{max}}}{W}}{\frac {W}{S}}+{\frac {W}{S}}{\sqrt {\left({\frac {T_{A_{max}}}{W}}\right)^{2}-4C_{D_{0}}K}}}{\rho _{\infty }C_{D_{0}}}}}
W f u e l W 0 ∗ = W t a k e o f f − f u e l W 0 W c l i m b − f u e l W t a k e o f f − f u e l W c r u i s e − f u e l W c l i m b − f u e l W l o i t e r − f u e l W c r u i s e − f u e l W l a n d − f u e l W l o i t e r − f u e l {\displaystyle {\frac {W_{fuel}}{W_{0}}}^{*}={\frac {W_{takeoff-fuel}}{W_{0}}}{\frac {W_{climb-fuel}}{W_{takeoff-fuel}}}{\frac {W_{cruise-fuel}}{W_{climb-fuel}}}{\frac {W_{loiter-fuel}}{W_{cruise-fuel}}}{\frac {W_{land-fuel}}{W_{loiter-fuel}}}}
