Worked calculation notebook

Can a scale model match both Re and Fr?

Calculate two candidate speeds for a 1:10 model and see why the conditions conflict.

Calculations and content checked: 2026-10-09

A matching shape is only the starting point

Imagine testing a geometrically similar, 1 m model of a 10 m surface body moving at 5 m/s. Should the model speed be one tenth of the full-scale speed, or should it stay the same? We will compare Reynolds and Froude numbers to answer a narrower question: what speed matches each ratio? This is an illustrative scaling exercise, not a performance prediction for a particular vessel.

Use ρ = 1000 kg/m³, μ = 0.001 Pa·s and g = 9.81 m/s² for both cases. The kinematic viscosity is ν = 10⁻⁶ m²/s. L denotes a representative body length throughout, not an open-channel depth.

1. Establish the full-scale reference

Refull = 1000 × 5 × 10 / 0.001 = 50,000,000

Frfull = 5 / √(9.81 × 10) ≈ 0.5048

The first ratio concerns viscosity; the second concerns gravity. A body-length Fr below one does not label the surrounding flow as subcritical open-channel flow. Check the chosen length and the conditions attached to a calculator's regime description.

2. Choose a speed that matches Fr

With equal gravity, matching Fr requires vmodel/√Lmodel = vfull/√Lfull. Define λ = Lmodel/Lfull = 0.1. The required speed ratio is √λ.

vmodel = 5 × √0.1 ≈ 1.5811 m/s

At this speed, model Fr is approximately 0.5048, but model Re is approximately 1,581,139. Check the ratio directly: Remodel/Refull = λ3/2 ≈ 0.03162. The viscosity-related scale is only about 3.16% of the full-scale value despite the matching Fr.

3. Choose a speed that matches Re

With the same fluid, matching Re requires vmodelLmodel = vfullLfull. Therefore the model speed must be 5 × 10/1 = 50 m/s. Model Fr then becomes 50/√9.81 ≈ 15.9638, about 31.62 times full-scale Fr.

Three speeds for the 1 m model
ConditionSpeed (m/s)ReFr
Same speed as full scale55,000,0001.5964
Match Fr1.5811≈ 1,581,1390.5048
Match Re5050,000,00015.9638

The Fr-matched row uses the unrounded speed 5√0.1. Entering the displayed 1.5811 produces a small last-digit difference.

4. What would matching both require?

Keeping the Fr-matched speed while matching Re would require νmodel = λ3/2νfull = 3.1623×10⁻⁸ m²/s. This is an algebraic requirement, not a recommendation for an available working fluid. If density also changes, use μ = ρν consistently.

Adjusting speed alone cannot match every important dimensionless ratio. Surface tension may require a Weber-number comparison; compressibility may require Mach number. Choose the phenomenon to reproduce before deciding which additional ratios matter.

Keep a comparison record

A matching Fr alone does not validate total drag, skin friction or full-scale performance. The table quantifies the scaling mismatch under equal-fluid and equal-gravity assumptions; it is a starting point for deciding what further evidence is needed.

Formula and concept references: NASA Glenn — Reynolds Number · MIT OpenCourseWare — Similitude, §5.2 (PDF). The input sets and cross-checks are worked examples constructed for this site, not quoted property tables or experimental results.