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The science
Lift from the flapping wings
Each wing is driven as a sinusoidal sweep in a horizontal stroke plane, φ(t) = (Φ/2) sin(2πnt), so the peak angular speed is ωm = πΦn. A blade element at radius r moves at rω and carries a mean lift coefficient CL, so the mean vertical force of the two wings is mg = ρ CL (čR³/3) ⟨ω²⟩, with ⟨ω²⟩ = ωm²/2. This gives the lift coefficient the bird needs, CL = 6mg / (ρ č R³ (πΦn)²), or the wingbeat frequency at a chosen CL.
The power budget
Induced power is the cost of throwing air downward: with the actuator-disk area A = ΦR², Pind = k (mg)3/2 / √(2ρA) with k ≈ 1.2. Profile power is the drag on the moving wings: Ppro = 2 · (1/8) ρ CD č R&sup4; ⟨|ω|³⟩ with ⟨|ω|³⟩ = (4/3π) ωm³. Inertial power accelerates the wing mass; muscles do 4nEk of work per wing per second, reduced by the fraction of elastic energy storage. Metabolic power is the mechanical total divided by the muscle efficiency, and nectar energy is the sugar concentration times 16.5 kJ/g.
Honest simplifications: rectangular wings with chord č, quasi-steady aerodynamics with one average lift coefficient (real hummingbirds use unsteady effects such as leading-edge vortices), a horizontal stroke plane, and a rough 1% estimate of wing mass; the muscle efficiency and elastic storage are uncertain in real birds, so use the totals for orders of magnitude, not as measurements.
What to try
- Press Set frequency to hover and then change the body mass: heavier birds need slower, longer wings.
- Shorten the stroke amplitude and watch the wingbeat frequency and the required CL rise.
- Raise the elastic storage to 80% and see the inertial share of the budget shrink.
- Look at the third graph: bigger hummingbirds beat their wings more slowly.