Similarity — geometric, kinematic, dynamic; Reynolds similarity: Worked examples

Before we start, one plain-language refresher of the symbols we will keep using — earned so we never write anything unexplained.

Definition The characters in every equation

$L$ — a characteristic length: one representative size of the body (car length, pipe diameter). Picture a single ruler laid along the object.
$V$ — the flow speed far from the body, in metres per second.
$ρ$ (rho) — density, mass packed into one cubic metre ( $kg / m^{3}$ ).
$μ$ (mu) — dynamic viscosity, how "sticky/gooey" the fluid is.
$ν = μ / ρ$ (nu) — kinematic viscosity, stickiness per unit density ( $m^{2} / s$ ).
$g$ — gravitational acceleration, how hard gravity pulls the fluid down; on Earth $g = 9.81 m/s^{2}$ . Picture a dropped droplet accelerating toward the ground. It appears whenever gravity/waves matter.
$A$ — a reference area: the frontal (or projected) area the flow "sees," in $m^{2}$ . For geometrically similar bodies $A \propto L^{2}$ , so $A_{p} / A_{m} = 1/ L_{r}^{2}$ .
$Q$ — the volume flow rate (discharge): how many cubic metres of fluid pass per second ( $m^{3} / s$ ). It equals speed times cross-section, $Q = V \cdot A$ — we use it in Ex 8.
$σ$ (sigma) — surface tension: the "skin-force per unit length" of a liquid surface ( $N/m$ ). It only matters when the model is tiny and capillary effects appear; it drives the Weber number below.
$c$ — the speed of sound in the fluid ( $m/s$ ); it drives the Mach number, our compressibility warning below.
Subscript $m$ = model (the small test piece), subscript $p$ = prototype (the full-size real thing).
$L_{r} = L_{m} / L_{p}$ — the length-scale ratio. A $1/10$ model has $L_{r} = 0.1$ .

The scenario matrix

Every similarity problem you will ever see falls into one of these cells. Each row is a distinct kind of question; the last column names the example that clears it. The figure below draws the same eight cells as a map so you can see where each example sits.

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

#	Case class	What is tricky about it	Which number	Example
1	Same fluid, find model speed	speed goes up as size goes down	Reynolds	Ex 1
2	Different fluid, find model speed	must keep the $ν$ ratio	Reynolds	Ex 2
3	Force scaling, same fluid	forces come out equal (surprise)	Reynolds + $C_{D}$	Ex 3
4	Force scaling, different fluid	must plug real $ρ, V, A$	Reynolds + $C_{D}$	Ex 4
5	Free-surface / wave flow	Reynolds is the wrong number	Froude (+Weber caveat)	Ex 5
6	Conflict: both Re and Fr wanted	usually impossible with one fluid	choose dominant	Ex 6
7	Degenerate: $L_{r} = 1$ (full scale)	limiting case, everything equals	either	Ex 7
8	Exam twist: hidden regime	you must decide which number	decide first	Ex 8

Two "signs/limits" live inside these: $L_{r} < 1$ (model smaller — the normal case) makes speeds go up under Reynolds; the limit $L_{r} \to 1$ (Ex 7) makes all ratios collapse to $1$ . We hit both.

Case 1 — Same fluid, find model speed

The picture below shows the two hulls to scale with their speeds — look at how the small model needs the big speed arrow.

Pick the number. Submerged body, no free surface → viscous/inertia contest → match Reynolds. Why this step? A submarine deep underwater makes no waves; gravity/free-surface effects are absent, so Re is the controlling ratio.
Write Re-matching and cancel the fluid. $\frac{ρ V _{m} L _{m}}{μ} = \frac{ρ V _{p} L _{p}}{μ} \Rightarrow V_{m} L_{m} = V_{p} L_{p}$ Why this step? Same fluid means identical $ρ$ and $μ$ ; they cancel, leaving the clean product rule $V L = const$ .
Solve. $V_{m} = V_{p} \frac{L _{p}}{L _{m}} = 8 \times \frac{60}{3} = 8 \times 20 = 160 m/s$ Why this step? Smaller body ( $L_{r} = 1/20$ ) must be pushed $20 \times$ faster to keep $ρ V L / μ$ fixed.

Verify: Re should match. $Re_{p} \propto V_{p} L_{p} = 8 \cdot 60 = 480$ ; $Re_{m} \propto V_{m} L_{m} = 160 \cdot 3 = 480$ . ✓ Units: $(m/s) (m)$ on both sides — consistent. Forecast check: yes, bigger — exactly $20 \times$ .

Case 2 — Different fluid, find model speed

Pick the number. Internal pipe flow, no free surface → Reynolds again.
Keep the full Re, do NOT cancel $ν$ . $\frac{V _{m} D _{m}}{ν _{m}} = \frac{V _{p} D _{p}}{ν _{p}}$ Why this step? Fluids differ, so $ν_{m} \neq = ν_{p}$ — the kinematic-viscosity ratio survives and must be carried.
Solve for $V_{m}$ . $V_{m} = V_{p} \frac{D _{p}}{D _{m}} \frac{ν _{m}}{ν _{p}} = 3 \times \frac{0.6}{0.15} \times \frac{1.5 \times 1 0 ^{- 5}}{1.0 \times 1 0 ^{- 6}}$ $= 3 \times 4 \times 15 = 180 m/s$ Why this step? $D_{p} / D_{m} = 4$ (size effect) times $ν_{m} / ν_{p} = 15$ (fluid effect) — both push speed up, and they multiply.

Verify: $Re_{p} = \frac{3 \cdot 0.6}{1 0 ^{- 6}} = 1.8 \times 1 0^{6}$ ; $Re_{m} = \frac{180 \cdot 0.15}{1.5 \times 1 0 ^{- 5}} = 1.8 \times 1 0^{6}$ . ✓ Both dimensionless and equal.

Case 3 — Force scaling, same fluid (the surprise)

Use equal drag coefficients. With Re matched, $C_{D} = \frac{F _{D}}{\frac{1}{2} ρ V ^{2} A}$ (defined above) is identical for model and prototype. Why this step? Matched Re means identical dimensionless solution → identical $C_{D}$ . See Drag Coefficient.
Set the coefficients equal and divide through by the shared factor $\frac{1}{2} ρ$ . $\frac{F _{D, p}}{\frac{1}{2} ρ V _{p}^{2} A _{p}} = \frac{F _{D, m}}{\frac{1}{2} ρ V _{m}^{2} A _{m}}$ The constant $\frac{1}{2}$ and the density $ρ$ (same fluid) are identical on both sides, so dividing every term by $\frac{1}{2} ρ$ removes them cleanly: $\frac{F _{D, p}}{V _{p}^{2} A _{p}} = \frac{F _{D, m}}{V _{m}^{2} A _{m}} \Rightarrow F_{D, p} = F_{D, m} \frac{V _{p}^{2} A _{p}}{V _{m}^{2} A _{m}}$ Why this step? Same fluid → $ρ$ is common; the $\frac{1}{2}$ is a universal constant. Neither survives a term-by-term division, leaving only the speed and area ratios.
Insert same-fluid relations $V_{m} = V_{p} / L_{r}$ and $A = L^{2}$ so $A_{p} / A_{m} = 1/ L_{r}^{2}$ . $\frac{V _{p}^{2}}{V _{m}^{2}} = L_{r}^{2}, \frac{A _{p}}{A _{m}} = \frac{1}{L _{r}^{2}}$ $F_{D, p} = F_{D, m} \cdot L_{r}^{2} \cdot \frac{1}{L _{r}^{2}} = F_{D, m} = 90 N$ Why this step? The higher model speed (which raises force) exactly cancels the smaller model area (which lowers force). The two $L_{r}^{2}$ factors annihilate.

Verify: the cancellation is $L_{r}^{2} \cdot L_{r}^{- 2} = 1$ for any $L_{r}$ — so same-fluid Re-matched drag is always equal. Units: newtons in, newtons out. ✓ Forecast: yes, equal — the counter-intuitive answer.

Case 4 — Force scaling, different fluid

Equal $C_{D}$ again — the $\frac{1}{2}$ cancels as before, but now do NOT cancel $ρ$ . $\frac{F _{D, p}}{\frac{1}{2} ρ _{p} V _{p}^{2} A _{p}} = \frac{F _{D, m}}{\frac{1}{2} ρ _{m} V _{m}^{2} A _{m}} \Rightarrow F_{D, p} = F_{D, m} \frac{ρ _{p} V _{p}^{2} A _{p}}{ρ _{m} V _{m}^{2} A _{m}}$ Why this step? Different fluids means $ρ_{m} \neq = ρ_{p}$ ; every factor is genuine and must be plugged in. Only the shared constant $\frac{1}{2}$ leaves.
Plug numbers. $F_{D, p} = 30 \times \frac{1000 \cdot 4 ^{2} \cdot 5.0}{1.2 \cdot 5 0 ^{2} \cdot 0.20} = 30 \times \frac{1000 \cdot 16 \cdot 5}{1.2 \cdot 2500 \cdot 0.20}$ $= 30 \times \frac{80000}{600} = 30 \times 133. \overline{3} = 4000 N$ Why this step? $ρ$ -ratio and $A$ -ratio push force up; the $V$ -ratio pushes it down — arithmetic decides the winner.

Verify: $\frac{1000 \cdot 16 \cdot 5}{1.2 \cdot 2500 \cdot 0.20} = \frac{80000}{600} = 133.33$ , times $30 = 4000 N$ . ✓ Units cancel to newtons since $C_{D}$ is dimensionless. The lesson: never assume force equality when fluids differ — that shortcut belonged to Ex 3 alone.

Case 5 — Free-surface flow (Reynolds is wrong here)

The figure shows the ship's bow wave: gravity is the force that pulls that raised water back down, which is why $g$ enters.

Pick the number. A ship pushes water up into a wave; the restoring force is gravity (through $g$ ). The inertia/gravity ratio is the Froude Number — this is the controlling number for free surfaces. Why this step? With a free surface, gravity waves dominate the drag; matching Re would ignore the very physics that matters.
Match Froude. $\frac{V _{m}}{g L _{m}} = \frac{V _{p}}{g L _{p}} \Rightarrow V_{m} = V_{p} \frac{L _{m}}{L _{p}} = V_{p} L_{r}$ Why this step? $g$ is the same in tank and sea, so it cancels, leaving $V \propto L$ .
Solve. $V_{m} = 10 \times \frac{1.5}{150} = 10 \times 0.01 = 10 \times 0.1 = 1.0 m/s$ Why this step? $L_{r} = 1/100 = 1/10$ , so the small model goes slower, opposite to the Reynolds case.

Verify: $Fr_{p} = \frac{10}{9.81 \cdot 150} = \frac{10}{38.35} = 0.2608$ ; $Fr_{m} = \frac{1.0}{9.81 \cdot 1.5} = \frac{1.0}{3.836} = 0.2608$ . ✓ Equal. Note the sign of the size effect flipped: Reynolds → faster, Froude → slower.

Case 6 — The conflict: can we match both?

Write both required speeds. $V_{m}^{Re} = V_{p} \frac{1}{L _{r}} = 10 \times 100 = 1000 m/s, V_{m}^{Fr} = V_{p} L_{r} = 1.0 m/s$ Why this step? Reynolds wants $V \propto 1/ L$ ; Froude wants $V \propto L$ — opposite trends in $L$ .
Take the ratio to see the contradiction. $\frac{V _{m}^{Re}}{V _{m}^{Fr}} = \frac{1/ L _{r}}{L _{r}} = L_{r}^{- 3/2} = 10 0^{3/2} = 1000$ Why this step? A single model speed cannot be $1 m/s$ and $1000 m/s$ at once — factor $1000$ apart, so simultaneous matching is impossible with one fluid.
Resolve. Match the dominant force: for a ship's wave drag, that is Froude ( $V_{m} = 1.0 m/s$ ). The viscous (Re) part is corrected separately using Boundary Layer friction-line formulas — this is Froude's classic split of ship resistance.

Verify: $L_{r}^{- 3/2}$ with $L_{r} = 1/100$ gives $(100)^{1.5} = 1000$ . ✓ The mismatch grows fast: even a $1/10$ model gives $1 0^{1.5} \approx 31.6 \times$ .

Case 7 — Degenerate limit $L_{r} \to 1$

Model speed. From Ex 1, $V_{m} = V_{p} / L_{r} \to V_{p} /1 = V_{p}$ . Why this step? If the model is the prototype, it must run at the prototype's speed — a required consistency check.
Force ratio. From Ex 3, $F_{D, p} / F_{D, m} = L_{r}^{2} \cdot L_{r}^{- 2} = 1$ for all $L_{r}$ , so it stays $1$ . Why this step? Same body, same fluid, same speed → identical drag, trivially.
Reynolds ratio. $\frac{Re _{m}}{Re _{p}} = \frac{V _{m} L _{m}}{V _{p} L _{p}} = \frac{( V _{p} / L _{r} ) ( L _{r} L _{p} )}{V _{p} L _{p}} = 1$ . Why this step? Re-matching forces this ratio to $1$ by construction — the limit just confirms nothing breaks.

Verify: every ratio equals $1$ as $L_{r} \to 1$ . ✓ A theory that failed this degenerate limit would be wrong; ours passes.

Case 8 — Exam twist: decide the regime first

Diagnose the physics. "Flowing over a crest, open to the air" = free surface, gravity-driven → Froude, not Reynolds. Why this step? The exam hides the choice; the phrase "free surface / open" is the tell. Submerged/pipe → Re; free surface → Fr.
Match Froude for the velocity. $V_{m} = V_{p} L_{r} = 5 \times 1/25 = 5 \times \frac{1}{5} = 1.0 m/s$ Why this step? Same $L_{r}$ rule as the ship — gravity cancels.
Derive and scale the discharge $Q = V \cdot A$ . Recall $Q$ is volume per second, and it equals speed times cross-section. Take the ratio model-to-prototype and substitute the two scaling laws: $\frac{Q _{m}}{Q _{p}} = \frac{V _{m} A _{m}}{V _{p} A _{p}} = = L_{r} \frac{V _{m}}{V _{p}} \cdot = L_{r}^{2} \frac{A _{m}}{A _{p}} = L_{r} \cdot L_{r}^{2} = L_{r}^{5/2}$ $= (\frac{1}{25})^{5/2} = \frac{1}{2 5 ^{2} 25} = \frac{1}{625 \cdot 5} = \frac{1}{3125}$ Why this step? Velocity scales as $L_{r}$ (Froude) and area as $L_{r}^{2}$ (geometry, since $A \propto L^{2}$ ); their product $L_{r}^{5/2}$ is the standard Froude discharge law.

Verify: $Fr_{p} = \frac{5}{9.81 \cdot 50} = \frac{5}{22.15} = 0.2257$ ; $Fr_{m} = \frac{1.0}{9.81 \cdot 2} = \frac{1.0}{4.429} = 0.2258$ . ✓ And $2 5^{5/2} = 3125$ , so $Q_{m} / Q_{p} = 1/3125$ . ✓

Recall

Recall Quick self-test

Which number for a torpedo running deep? ::: Reynolds (submerged, no free surface). Which number for a breakwater with waves? ::: Froude (free surface, gravity). Same fluid, Re matched, $L_{r} = 1/8$ — model speed factor? ::: $\times 8$ (from $V \propto 1/ L_{r}$ ). Same fluid, Re matched — drag force ratio prototype/model? ::: exactly $1$ . Froude match, $L_{r} = 1/25$ — velocity factor? ::: $\times 1/5$ (from $L_{r}$ ). Froude match, $L_{r} = 1/25$ — discharge ratio $Q_{m} / Q_{p}$ ? ::: $1/3125$ (from $L_{r}^{5/2}$ ). Why can't we match Re and Fr together (same fluid)? ::: they demand $V \propto 1/ L$ vs $V \propto L$ — contradictory (two $Π$ -groups, one knob). When does a fast water-tunnel model stop being faithful? ::: when it cavitates (or in gas, when $Ma ≳ 0.3$ and compressibility appears). When does simple Froude scaling break at tiny scales? ::: when model waves shrink to mm and surface tension (low Weber) distorts them.

See also: Navier-Stokes Equations (why matched Re gives an identical dimensionless equation), and the parent topic note.