Scale-model ratios re-run: 2026-10-22
A matching shape is only the starting point
Geometric similarity fixes the shape; dynamic similarity also needs matching force ratios, so that pressure, viscous, gravity and inertial forces on corresponding fluid particles form similar polygons (UT Austin similitude notes). With the same fluid and gravity, a model cannot match Re and Fr at once. Match Re for internal flows or bodies far from a free surface, and Fr for free-surface bodies or spillway flow.
Three cases follow: an illustrative 1:10 body, a 1:25 towing-tank ship model and a 1:30 spillway model, with assumed speeds, lengths and discharges. Water properties in Examples 2 and 3 come from the IAEA table (20 °C: ν = 1.0020 × 10⁻⁶ m²/s, ρ = 998.2 kg/m³) and from OpenStax (σ = 0.0728 N/m at 20 °C). No result predicts a real vessel or structure.
Scale factors under each similarity rule
| Quantity | Fr matched | Re matched |
|---|---|---|
| Speed | λ1/2 | λ−1 |
| Time | λ1/2 | λ2 |
| Discharge | λ5/2 | λ |
| Force (ρU²L²) | λ3 | 1 |
| Pressure (ρU²) | λ | λ−2 |
Each row follows from the matched number and equal pressure coefficients. Under Fr similarity the unmatched Re ratio is λ3/2; under Re similarity the unmatched Fr ratio is λ−3/2. The MIT 2.154 notes reach the same result for ships: Fr similitude gives model speed = ship speed/√η (η = full/model length), and matching Re too needs kinematic viscosity reduced by η3/2, which few tank fluids allow.
Three model studies
Example 1. A 1:10 body: which speed matches which ratio?
Given (illustrative). A 10 m surface body at 5 m/s and a geometrically similar 1 m model. Rounded constants for both: ρ = 1000 kg/m³, μ = 0.001 Pa·s, g = 9.81 m/s², so ν = 10⁻⁶ m²/s. L is the body length, not a channel depth.
Full-scale reference
Refull = 1000 × 5 × 10 / 0.001 = 50,000,000
Frfull = 5 / √(9.81 × 10) ≈ 0.5048
A body-length Fr below one does not make the flow subcritical in the open-channel sense.
Match Fr
With equal gravity, vmodel/√Lmodel = vfull/√Lfull. With λ = Lmodel/Lfull = 0.1, the speed ratio is √λ.
Model Fr is then about 0.5048, but model Re is about 1,581,139: Remodel/Refull = λ3/2 ≈ 0.03162, only about 3.16% of full scale.
Match Re
With the same fluid, vmodelLmodel = vfullLfull, so the model needs 5 × 10/1 = 50 m/s. Model Fr becomes 50/√9.81 ≈ 15.9638, about 31.62 times full-scale Fr.
| Condition | Speed (m/s) | Re | Fr |
|---|---|---|---|
| Same speed as full scale | 5 | 5,000,000 | 1.5964 |
| Match Fr | 1.5811 | ≈ 1,581,139 | 0.5048 |
| Match Re | 50 | 50,000,000 | 15.9638 |
The Fr-matched row uses the unrounded speed 5√0.1; entering 1.5811 changes the last digit slightly.
What matching both would require
Keeping the Fr-matched speed and matching Re needs νmodel = λ3/2νfull = 3.1623×10⁻⁸ m²/s. This is an algebraic requirement, not a recommended fluid; if density also changes, use μ = ρν.
Example 2. A 1:25 ship model in a towing tank
Given. Assumed: waterline length 100 m, ship speed 7.5 m/s, model scale 1:25 (Lmodel = 4 m). For simplicity both ship and model are taken in fresh water at 20 °C (IAEA: ν = 1.0020 × 10⁻⁶ m²/s, ρ = 998.2 kg/m³; σ = 0.0728 N/m). Sea water has a slightly different density and viscosity, which this example ignores.
Fr = 7.5/√(9.81 × 100) ≈ 0.2395; vmodel = 7.5 × √(1/25) = 1.5 m/s
Reship = 7.5 × 100 / (1.0020 × 10⁻⁶) ≈ 7.49 × 10⁸; Remodel = 1.5 × 4 / (1.0020 × 10⁻⁶) ≈ 5.99 × 10⁶
Re ratio = (1/25)3/2 = 0.008; ν needed to match both ≈ 8.016 × 10⁻⁹ m²/s
Re-matched model speed in water: 7.5 × 25 = 187.5 m/s
Wemodel = 998.2 × 1.5² × 4 / 0.0728 ≈ 1.23 × 10⁵
Force scale (same density): 25³ = 15,625; time scale √(1/25) = 0.2, so a 10 s full-scale period lasts 2 s in the tank
Interpretation. Fr matching gives a practical 1.5 m/s; Re matching would need an absurd 187.5 m/s, so the model runs at a Reynolds number 125 times smaller than the ship. The MIT notes state that towing tanks routinely use turbulence stimulators so the model boundary layer behaves as it would at ship-scale Re, and then match Fr closely. Frictional resistance measured on the model still cannot be multiplied by 15,625 directly; it has to be handled with a separate friction estimate. With such a large model We, surface tension plays no role, as the MIT notes argue for large lengths.
Example 3. A 1:30 spillway model
Given. Assumed: design discharge 600 m³/s, and at one chute section a mean velocity of 12 m/s with a flow depth of 1.5 m; model scale 1:30. Fresh water at 20 °C in both (IAEA ν, OpenStax σ). For an open channel, Fr uses the flow depth as length.
Qmodel = 600 × (1/30)5/2 = 600/4929.5 ≈ 0.1217 m³/s (121.7 L/s)
vmodel = 12/√30 ≈ 2.191 m/s; depth 1.5/30 = 0.05 m
Fr = 12/√(9.81 × 1.5) ≈ 3.128 at both scales
Re (depth-based): full scale ≈ 1.80 × 10⁷, model ≈ 1.09 × 10⁵, ratio 303/2 ≈ 164.3
Wemodel = 998.2 × 2.191² × 0.05 / 0.0728 ≈ 3291
A 60 min flood hydrograph runs in 60/√30 ≈ 10.95 min in the model
Interpretation. The UT Austin notes list spillway flow as a Fr-similarity case. Fr ≈ 3.1 is well above the critical value of about 1 that USGS associates with the transition between supercritical and subcritical flow, so this section is supercritical at both scales and a hydraulic jump would form downstream at the same scaled position. Model Re is 164 times smaller but still about 10⁵, so the flow stays turbulent. Model We of about 3300 is far below full scale, so surface-tension features such as air entrainment and spray scale worst, and a smaller model makes this worse. Check model We and Re before choosing λ.
Keep a comparison record
- State whether λ means model/full scale or its reciprocal.
- Record the definition of L (body length, waterline length or flow depth), geometric ratios and boundary conditions.
- List both matched ratios and unmatched ratios, with their model and full-scale values.
- Avoid applying circular-pipe Reynolds thresholds or open-channel Froude classifications to body-length calculations.
Matching Fr alone does not validate drag, skin friction or full-scale performance. Surface tension may need a Weber comparison and compressibility a Mach number; choose the phenomenon first.
Questions about matching Re and Fr in model tests
Why not use a different test fluid to match both numbers?
The required kinematic viscosity is extreme: 3.16 × 10⁻⁸ m²/s for the 1:10 model and 8.0 × 10⁻⁹ m²/s for the 1:25 ship, far below water's 1.0 × 10⁻⁶. The MIT notes point out that few fluids with such properties are workable in a large tank.
Which length goes into Fr: body length or depth?
For ships the usual choice is waterline length, as in the MIT notes. For open channels and spillways it is the flow depth (or hydraulic depth). Both are valid, but the critical value Fr ≈ 1 applies only to the open-channel definition.
How do I convert a force measured on a Fr-scaled model?
With the same fluid density, forces scale with λ³ and pressures with λ, because the pressure coefficient is equal at both scales. A 1:25 model force is multiplied by 15,625, but only for the gravity-dominated part; viscous friction needs a separate correction because Re is not matched.
Is a model Re of a few million good enough?
It depends on what is measured. Above some size the model boundary layer is turbulent, and stimulators help fix transition, but skin friction still depends on Re. Report the model Re with every result so the reader can judge the scale effect.
References
- MIT OpenCourseWare, 2.154: Similitude, §5.2 (PDF). Fr on waterline length, Fr-scaled speed, viscosity ratio, turbulence stimulators, surface tension at large length.
- University of Texas at Austin (S. Kinnas): Some Notes on Similitude. Re and Fr cannot both be enforced; when to use each; equal pressure coefficients.
- NASA Glenn Research Center: Reynolds Number. Re as inertial/viscous force ratio and a similarity parameter.
- IAEA (2008): Thermophysical Properties of Materials for Nuclear Engineering. Table 3.3, water at 20 °C (ρ, ν).
- UCF Pressbooks (OpenStax College Physics): Surface Tension. Water at 20 °C, σ = 0.0728 N/m.
- U.S. Geological Survey: Can lava flow like water? Critical flow at Fr ∼ 1 and the supercritical/subcritical transition.