In steady level flight lift equals weight and thrust equals drag, and the lift is ½ × air density × speed² × wing area × lift coefficient. For a 400 kg aircraft on 13 m² of wing, holding level flight needs a lift coefficient of 1.23 at 20 m/s but only 0.55 at 30 m/s, so slower flight demands a larger angle of attack. A pitching balance about the centre of gravity must hold as well, or the four forces are not in equilibrium.
Part of Aerodynamics and Flight
The balance of lift, weight, thrust and drag, and the arithmetic that links speed, wing area and lift coefficient.
Lift And Weight Fix The Speed-To-Angle Relationship
Lift L = ½ × ρ × V² × S × CL, where ρ is air density, V speed, S wing area and CL the lift coefficient. The corpus’s Airfoil Profile (Lift vs Drag) article uses ρ = 1.225 kg/m³, the sea-level standard value. Setting L equal to weight W gives the coefficient you need at each speed: CL = 2W ÷ (ρ × V² × S).
Worked Example
Inputs: mass 400 kg, so W = 400 × 9.81 = 3,924 N; S = 13 m² (10 m span, 1.3 m chord); ρ = 1.225 kg/m³. Both the mass and the area sit inside the carrier’s ranges and are illustrative.
| Speed V | ½ρV² | CL needed = W ÷ (½ρV² × S) |
|---|---|---|
| 15 m/s | 137.8 Pa | 3,924 ÷ (137.8 × 13) = 2.19 |
| 20 m/s | 245.0 Pa | 3,924 ÷ (245.0 × 13) = 1.23 |
| 25 m/s | 382.8 Pa | 3,924 ÷ (382.8 × 13) = 0.79 |
| 30 m/s | 551.3 Pa | 3,924 ÷ (551.3 × 13) = 0.55 |
Slowing from 30 to 15 m/s multiplies the coefficient needed by four, because speed enters squared. The maximum coefficient a wing can supply is a measured property of the section and the wing, and it fixes the stall speed (Stall Dynamics); the 2.19 in the first row is not a figure a plain wing should be assumed to have.
Thrust And Drag Fix The Power
Drag is the force opposing motion, thrust the force the propeller makes. In level flight at a steady speed they are equal. The power the propeller must deliver is thrust × speed, and the engine must supply more than that because the propeller is less than fully efficient (Propeller Design). Neither the drag nor the efficiency can be taken from a table in this corpus; both must be measured on the aircraft.
Climbs, Descents And Turns
In a steady climb thrust exceeds drag by the component of weight along the flight path: T = D + W × sin(climb angle). A climb needs excess thrust, not excess lift. In a banked turn the lift must exceed weight, by 1 ÷ cos(bank angle); at 60° bank that is 2 times, so the wing is carrying twice the load (Envelope Expansion).
The Fifth Condition: Moments
The four forces act at different points. Lift acts through the centre of pressure, weight through the centre of gravity, thrust along the propeller axis, drag roughly along the flight path. If their moments about the centre of gravity do not cancel, the nose pitches. The tail’s job is to cancel the remainder (Tail Assembly), and the position of the centre of gravity is checked in Ground Testing.
Common Mistakes
- Equating lift with speed alone — lift depends on angle as well; at lower speeds the same lift needs a larger angle.
- Forgetting that lift exceeds weight in turns — the wing is loaded harder exactly where a pilot is lowest and slowest.
- Using someone else’s drag figure — drag is a property of the whole aircraft, struts and wires included.
Safety
The slow end of the table is where stalls happen
The first row needs a coefficient of 2.19. If the wing cannot make it, the aircraft is already stalled at 15 m/s. Treat any speed that needs more than the section’s published maximum coefficient as unflyable.
What’s Next
- Learn next: Stall Dynamics, the limit on the coefficient
- Combine with: Propulsion Systems, where thrust comes from