Similarity — geometric, kinematic, dynamic; Reynolds similarity: Question bank

Before you start, one anchor to keep in mind: similarity is about dimensionless ratios, never about matching raw speeds, sizes, or forces. Almost every trap below is a disguised violation of that single idea.

Definition Symbols and terms used on this page — read once first

Length-scale ratio $L_{r} = \frac{L _{m}}{L _{p}}$ — one constant that relates every model length $L_{m}$ to the corresponding prototype length $L_{p}$ . A " $1/10$ model" means $L_{r} = 1/10$ .
Geometric similarity — model and prototype have the same shape (all lengths related by the single $L_{r}$ ).
Kinematic similarity — the motion is similar: velocity fields share the same streamline pattern, differing only by a constant magnitude scale. This requires geometric similarity.
Dynamic similarity — the forces are similar: every pair of corresponding forces has the same ratio in model and prototype. This implies kinematic (and hence geometric) similarity.
Mach number $Ma = \frac{V}{c}$ (flow speed $V$ over sound speed $c$ ) — the inertia/elastic force ratio; matters when compressibility matters. See Re for the dimensionless-number recipe.
Weber number $We = \frac{ρ V ^{2} L}{σ}$ (with surface tension $σ$ ) — the inertia/surface-tension force ratio; matters for droplets, bubbles, thin films.

Below, the picture pins these three levels of similarity to one image so the True/False items have a visual to lean on.

Figure — Similarity — geometric, kinematic, dynamic; Reynolds similarity

True or false — justify

Two geometrically similar bodies automatically have similar streamline patterns.

False — geometric (shape) similarity is necessary but not sufficient for kinematic similarity; you also need matched motion, which for a submerged body means matched Re.

If two flows have the same Reynolds number, they must be geometrically similar.

False — equal Re is required for dynamic similarity but only given geometric similarity; two differently shaped bodies can share a Re number and still behave completely differently.

Dynamic similarity (equal force ratios) implies kinematic similarity (matched streamline patterns).

True — matched force ratios (Newton's law non-dimensionalised) force the velocity fields to have the same pattern, so motion is automatically similar.

Kinematic similarity (matched motion) implies dynamic similarity (matched force ratios).

False — the ladder only climbs downward: shape → motion → forces. Matched motion does not guarantee matched force ratios (e.g. gravity/inertia could differ).

Given geometric similarity, matching Reynolds number makes the drag coefficient

C_{D}

equal in model and prototype.

True — with geometric similarity and equal Re the Navier–Stokes equation becomes the identical dimensionless equation, so every dimensionless coefficient including $C_D$ matches.

Matching Reynolds number makes the drag force equal in model and prototype.

False in general —

C_{D}

matches, but

F_{D} = \frac{1}{2} ρ V^{2} A C_{D}

still depends on the actual

ρ, V, A

; the forces are equal only in the special same-fluid case where the higher model speed exactly cancels the smaller area.

You can always match Reynolds number and Froude number simultaneously with the same fluid.

False — Re-matching demands

V \propto 1/ L

while Fr-matching demands

V \propto L

; these two speed laws are incompatible for a scaled model in the same fluid.

The Reynolds number is dimensionless.

True —

\frac{ρ V L}{μ}

has units

\frac{( kg/m ^{3} ) ( m/s ) ( m )}{kg/(m\cdots)} = 1

, which is exactly why it can be equated across differently sized flows.

Under geometric similarity, areas scale as

L_{r}

and volumes as

L_{r}^{2}

False — every length scales by

L_{r}

, so areas (length²) scale as

L_{r}^{2}

and volumes (length³) as

L_{r}^{3}

; mixing up the exponents is a classic slip.

Spot the error

"The model runs at the same 30 m/s as the car, so the flows are similar."

The error is matching raw velocity instead of Re; a smaller model at the same speed has a smaller

ρ V L / μ

, so it sits in a different flow regime — not similar at all.

"Since the shapes are identical, the measured model drag equals the prototype drag."

Geometric similarity alone fixes shape but not the force ratios; without dynamic (Re) similarity the

C_{D}

values differ, so the drags cannot simply be equated.

"To be safest, we should match Re, Fr, Ma and We all at once."

More matching is not automatically better — these criteria (Reynolds, Froude, Mach

Ma = V / c

, Weber

We = ρ V^{2} L / σ

) impose contradictory speed/fluid requirements; you must instead identify the dominant force and match only its number.

"Inertia force scales as

ρ V L^{2}

, so Re

= ρ V L^{2} / μ

The inertia force is

ρ V^{2} L^{2}

(an extra factor of

V

from acceleration

\sim V^{2} / L

); dividing by viscous

μ V L

then gives the correct

ρ V L / μ

"For a ship with waves, use Reynolds similarity because water is viscous."

A free surface means gravity/wave forces dominate, so Froude similarity governs; viscosity is present but is not the controlling force ratio for the wave pattern.

"Kinematic viscosity

ν

always cancels in Re-matching, so ignore it."

ν = μ / ρ

cancels only when model and prototype use the same fluid; with different fluids the full

ν_{m} / ν_{p}

ratio must be kept in the Re equation.

Why questions

Why does dynamic similarity sit at the top of the hierarchy?

Because matching force ratios non-dimensionalises Newton's second law identically, which automatically produces matched motion (kinematic) — achieving it delivers everything below for free.

Why is matching Re enough to guarantee identical dimensionless solutions?

When you non-dimensionalise Navier–Stokes with

L

V

ρ V^{2}

, every coefficient vanishes except

1/ Re

before the viscous term, so equal Re means the same equation and hence the same solution.

Why must a 1/10 scale model in the same fluid be tested at 10× the prototype speed?

Re-matching with equal

ρ, μ

gives

V_{m} L_{m} = V_{p} L_{p}

, so shrinking

L

by 10 (i.e.

L_{r} = 1/10

) forces

V

up by 10 to keep the ratio fixed.

Why do engineers often prefer water tunnels or pressurised air tunnels over ordinary wind tunnels?

Same-fluid Re-matching drives model speeds impractically high; a denser or more suitable fluid (higher

ρ

, lower

ν

) lets you hit the target Re at a reasonable, safer speed.

Why does the Boundary Layer concept matter for whether Reynolds similarity is trustworthy?

The boundary layer's behaviour (laminar vs turbulent transition) depends on Re, so if the model's Re is far from the prototype's, the layers differ and the extrapolated

C_{D}

becomes unreliable.

Why do we build dimensionless numbers by taking ratios of forces rather than the forces themselves?

Raw forces carry units and change with scale, but their ratios are pure numbers that can be equated across sizes — this is the Π-theorem idea that physics depends on dimensionless groups.

Edge cases

If the model velocity from Re-matching comes out faster than the speed of sound, is Reynolds similarity still valid?

No — once flow approaches the sound speed, compressibility enters (Mach number

Ma = V / c

becomes significant) and Re-matching alone no longer captures the physics; you have left the low-speed regime the criterion assumes.

What happens to "similarity" if the prototype flow is turbulent but the model's Re lands it in laminar flow?

They are in different regimes, so despite matched geometry the dimensionless solutions diverge — this is why you must reach the correct Re, not merely a proportional one.

For creeping (very low Re) flow where inertia is negligible, does matching Re still matter?

In the limit Re → 0 the inertia term drops out of the equation entirely, so the flow is governed by the viscous balance alone and results become nearly Re-independent — matching becomes trivial rather than the key constraint.

If two flows use the same fluid and the same size body, what velocity keeps Re matched?

With

L

and

ν

identical, Re-matching forces

V_{m} = V_{p}

— there is no scale factor to compensate, so identical size in the same fluid simply means identical speed.

Can you have geometric similarity with a scale factor

L_{r} = 1

but still lack dynamic similarity?

Yes — same size (a full replica,

L_{r} = 1

) can still run at a different speed or fluid, giving a different Re, so the shapes match while the force ratios do not.

Is a perfectly symmetric body needed for similarity, or just the same shape at all scales?

Only the same shape scaled by one constant $L_{r}$ is required; symmetry is irrelevant — an asymmetric body and its scaled copy are geometrically similar as long as every length uses the same factor.