Simulation

Terminal Velocity

DynamicsDrag

An object falling through fluid with quadratic drag reaches terminal velocity that depends on mass and cross-section.

Objective

Verify that a falling object subject to quadratic drag (F_d = ½·ρ·C_d·A·v²) asymptotically approaches the analytical terminal velocity v_t = sqrt(2mg / ρ·C_d·A), and that doubling mass raises v_t by only √2 — confirming the square-root dependence. The model assumes a rigid sphere-like body, no buoyancy correction, and constant fluid density.

Setup

  1. Set Mass to 1.0 kg, Drag Area to 0.047 m², and Fluid Density to 1.225 kg/m³ (default air). Record the v_t readout — it should show ≈ 18.46 m/s.
  2. Press Start and observe the v(t) amber curve in the right panel rising and flattening against the forest-green dashed v_t line. Note the time at which the curve visually touches the reference line.
  3. When the simulation stops automatically, check that the v readout ≈ v_t readout and that F_net ≈ 0.00 N — confirming force balance.
  4. Press Reset, then raise Mass to 2.0 kg. The v_t readout should update to ≈ 26.1 m/s (= 18.46 × √2). Run again and verify the curve settles at the new dashed line.
  5. Press Reset, set Fluid Density to 1000 kg/m³ (water). Observe that v_t drops to ≈ 0.646 m/s — the object reaches terminal velocity almost instantly.

Analytical Prediction

Terminal velocity is where drag equals weight: ½·ρ·C_d·A·v_t² = m·g, which gives v_t = sqrt(2·m·g / ρ·C_d·A). With the default settings (m = 1.0 kg, CdA = 0.047 m², ρ = 1.225 kg/m³):

v_t=sqrt(2 · 1.0 · 9.81 / (1.225 · 0.047))
=sqrt(19.62 / 0.057575)
=sqrt(340.8)
18.46 m/s

After doubling mass to 2.0 kg:

v_t=sqrt(2 · 2.0 · 9.81 / 0.057575)
=sqrt(681.6)
26.11 m/s (= 18.46 × √2 ≈ 26.10 m/s)

At ρ = 1000 kg/m³ (water), m = 1.0 kg, CdA = 0.047 m²:

v_t=sqrt(19.62 / (1000 · 0.047))
=sqrt(19.62 / 47)
sqrt(0.4174)
0.646 m/s

The v_t readout must match these values to within 0.05 m/s.

Results Analysis

Once the simulation stops, compare the v readout to the v_t readout — they should agree to within 0.01 m/s (the 0.1% near-terminal threshold). The F_net readout should display ≈ 0.00 N, confirming that weight and drag balance exactly. On the v(t) graph, observe that the amber curve's rate of rise decreases monotonically — early frames show steep acceleration (small drag), late frames show the curve barely moving (drag ≈ weight). The forest-green dashed v_t line is the analytical prediction; the curve must approach it from below without overshooting. For the water-density run (ρ = 1000 kg/m³), the curve flattens within the first second, and the v_t readout should show ≈ 0.646 m/s — a factor of ≈ 28 lower than the air-density case, consistent with the sqrt(1000/1.225) ≈ 28.6 density ratio.

Source of Error

This simulation models the drag force as purely quadratic (F_d = ½·ρ·C_d·A·v²), with a fixed scalar C_d — it omits the Reynolds-number dependence of C_d, which can vary significantly at low speeds (Stokes drag regime) and near the boundary layer transition. Buoyancy (Archimedes force) is not included, which is negligible in air but non-trivial in liquids. The fluid density ρ is held constant throughout the fall — pressure-density stratification of a real atmosphere is ignored. The body is treated as a point mass with no rotation or tumbling. The analytical prediction in the setup section assumes identical idealizations, so both the formula and the sim share the same simplifications — any residual gap between the v_t readout and the formula value is therefore purely numerical, not physical.

Further Exploration