CD_cruise = 0.025 + 0.0486 × 0.0966² ≈ 0.025 + 0.00045 ≈ 0.02545 Drag D = q S CD = 5,300 ×14 ×0.02545 ≈ 1,887 N
Trade: reduce cruise speed to 140 kt (72 m/s) to get realistic engine size: Recompute q = 0.5×1.225×72² ≈ 3,178 CL = 7,166 / (3,178 ×14) ≈ 0.161 CD = 0.025 + 0.0486×0.161² ≈ 0.025 + 0.00126 ≈ 0.02626 D = 3,178×14×0.02626 ≈ 1,169 N Power = 1,169×72 ≈ 84,200 W = 113 hp shaft → engine ~130–160 hp (accounting prop eff and climb reserve). This is a practical choice (e.g., Lycoming IO-320/IO-360 class). general aviation aircraft design snorri gudmundsson pdf
At cruise speed V = 180 kt ≈ 93 m/s. Dynamic pressure q = 0.5ρV² ≈ 0.5×1.225×93² ≈ 5,300 N/m² Lift = W = q S CL → CL_cruise = W / (q S) = 7,166 / (5,300 ×14) ≈ 0.0966 CD_cruise = 0
Choose S ≈ 14 m².
Power required (shaft) = D × V = 1,887 × 93 ≈ 175,500 W ≈ 235 hp (assuming propulsive eff 0.8, required engine power ≈ 235 / 0.8 ≈ 294 hp). That’s high for a light two-seater. Dynamic pressure q = 0
Aspect ratio: for GA, AR = 7–9 is typical. Pick AR = 8. Wing span b = sqrt(AR × S) = sqrt(8 × 14) ≈ sqrt(112) ≈ 10.6 m.