Similarity — geometric, kinematic, dynamic; Reynolds similarity: Visual walkthrough

We build everything from zero. The only things you need to already know: what a force is (a push or pull), what speed means (distance ÷ time), and that things have size (a length).

Step 1 — Meet the one fluid particle

WHY start here. Similarity is ultimately about forces on this cube. If two different flows push their little cubes in the same balance of ways, the cubes trace the same paths, and the two flows look like scaled copies of each other. So before comparing flows we must name every force acting on the cube.

PICTURE. The cube has a size we call $L$ (a length, in metres), it moves with speed $V$ (metres per second), it sits in fluid of density $ρ$ (mass packed per cubic metre) and stickiness $μ$ (viscosity — how strongly the fluid resists sliding).

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

Step 2 — Estimate the INERTIA force (the "keep going" push)

WHY this tool — why estimate instead of solve? We do not need the exact force, only how it grows with size and speed. A ratio of two such estimates is all similarity needs, and estimates are dead simple: replace every quantity by "roughly this size."

HOW, term by term. Inertia force $=$ mass $\times$ acceleration.

$F_{i} \sim mass of cube ρ L^{3} \times acceleration \frac{V ^{2}}{L} = ρ V^{2} L^{2}$

$ρ L^{3}$ — the cube has volume $\approx L^{3}$ (a box of side $L$ ), and mass = density × volume.
$\frac{V ^{2}}{L}$ — acceleration is change of speed ÷ time. The only natural time here is how long the fluid takes to cross the body: $t \sim L / V$ . So acceleration $\sim V \div (L / V) = V^{2} / L$ .
The $\sim$ symbol means "grows like / same size as," ignoring exact constants like $\frac{1}{2}$ .

PICTURE. The big lavender arrow is the inertia push, driving the cube forward.

Step 3 — Estimate the VISCOUS force (the "stickiness drag")

WHY. Inertia alone is only half the story: a real fluid resists sliding. The competition between "keep going" (inertia) and "stickiness slows me" (viscous) is what decides the whole character of the flow — smooth or turbulent, thin wake or fat wake.

HOW, term by term. Viscous force $=$ shear stress $\times$ area.

$F_{v} \sim shear stress τ μ \frac{V}{L} \times contact area L^{2} = μ V L$

$\frac{V}{L}$ — the velocity gradient: speed changes from $V$ down to $0$ across a thickness $\sim L$ , so the "steepness of the speed profile" is $V / L$ .
$μ \frac{V}{L}$ — viscosity times that steepness gives the shear stress $τ$ (force per unit area of rubbing).
$L^{2}$ — the area of the face where the rubbing happens.

PICTURE. Layers of fluid slide past each other; the coral arrows show the drag on the middle layer, steeper gradient → bigger drag.

Step 4 — Divide the two forces → the Reynolds number is born

WHY divide, not subtract? Similarity cares about ratios, not raw sizes. A ratio is dimensionless — it has no units — so it is the same "3.5" whether you measure in metres or miles. Two flows with the same ratio are in the same balance regime. That is the whole trick.

HOW, term by term.

$Re = \frac{F _{i}}{F _{v}} = \frac{ρ V ^{2} L ^{2}}{μ V L} = \frac{ρ V L}{μ}$

Numerator $ρ V^{2} L^{2}$ — inertia from Step 2.
Denominator $μ V L$ — viscous from Step 3.
One power of $V$ and one power of $L$ cancel top-and-bottom, leaving $ρ V L / μ$ .

PICTURE. The two arrows from Steps 2–3, side by side, feeding into a single dial labelled $Re$ : big Re = inertia wins (turbulent), small Re = stickiness wins (smooth).

Step 5 — Why matching Re copies the WHOLE flow

WHY this proves the magic. If two geometrically similar flows have the same $Re$ , their stretched-unit equations are letter-for-letter identical. Same equation + same shape of boundary ⇒ same dimensionless solution. Nothing else can differ.

HOW. With those four rulers, the non-dimensional momentum equation reads:

$\frac{\partial u ~}{\partial t ~} + (\tilde{u} \cdot \tilde{\nabla}) \tilde{u} = - \tilde{\nabla} \tilde{p} + \frac{1}{Re} \tilde{\nabla}^{2} \tilde{u}$

$\frac{\partial u ~}{\partial t ~}$ and $(\tilde{u} \cdot \tilde{\nabla}) \tilde{u}$ — the inertia terms (speed in units of $V$ , time in units of $L / V$ ). No leftover constant, precisely because we chose those two rulers.
$- \tilde{\nabla} \tilde{p}$ — the pressure push, constant-free because we picked the pressure yardstick $ρ V^{2}$ on purpose so its coefficient becomes $1$ .
$\frac{1}{Re} \tilde{\nabla}^{2} \tilde{u}$ — the viscous term, and its only dial is $1/ Re$ .

Match Re → the two equations coincide → same streamlines, same pressure pattern, same drag coefficient (see Drag Coefficient).

PICTURE. Two boxes of different physical size collapse onto one identical dimensionless picture once Re is matched — the streamlines overlap perfectly.

Step 6 — Reading off the scaling rule (same fluid)

WHY. This is the actual thing an engineer computes: how fast to blow the wind tunnel.

HOW. With the same fluid, $ρ$ and $μ$ are identical on both sides and cancel:

$\frac{ρ V _{m} L _{m}}{μ} = \frac{ρ V _{p} L _{p}}{μ} ⟹ V_{m} L_{m} = V_{p} L_{p} ⟹ V_{m} = V_{p} \frac{L _{p}}{L _{m}}$

$V_{m} L_{m} = V_{p} L_{p}$ — the product of speed and size must be preserved.
So a smaller model ( $L_{m} < L_{p}$ ) needs a faster speed. A $1/5$ model → $5 \times$ the speed.

PICTURE. A small model and a big prototype: the small one's speed arrow is five times longer, keeping $V L$ constant.

Step 7 — The degenerate & edge cases

Case A — different fluids. Then $ρ, μ$ do not cancel. Keep all three factors: $V_{m} = V_{p} \frac{L _{p}}{L _{m}} \frac{ν _{m}}{ν _{p}}$

Case B — there's a free surface (waves). Gravity now matters, so the ruling number becomes the Froude number, not Reynolds.

Re-matching and Froude-matching fight each other (Re wants $V \propto 1/ L$ , Fr wants $V \propto L$ ). With one fluid you cannot satisfy both, so you must pick the dominant force.

Case C — $μ \to 0$ (inviscid limit). Then $Re \to \infty$ and the $1/ Re$ term in Step 5 vanishes. Viscosity survives only in a razor-thin Boundary Layer hugging the surface — the rest of the flow ignores it.

Case D — very slow / tiny flow, $Re \to 0$ . Stickiness dominates; inertia is negligible ("creeping flow"). Streamlines become perfectly reversible.

PICTURE. A regime strip along a $lo g Re$ axis: creeping → laminar → transition → turbulent, with the four cases marked.

The one-picture summary

Everything on one canvas: two forces on a cube → their ratio Re → stretched-unit equation with only $1/ Re$ → matched Re copies the flow → scaling rule $V_{m} = V_{p} (L_{p} / L_{m}) (ν_{m} / ν_{p})$ .

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Picture a tiny cube of water. Two things fight over it: momentum ("I want to keep zooming!") and stickiness ("the water around me drags"). We guessed how big each push is using just size and speed. When we divide the zoom-push by the drag-push, all the units melt away and we're left with one plain number — the Reynolds number. It's like a "flavour" of the flow. Now here's the miracle: if you rewrite the water's rulebook using the body's own size and speed as your ruler, the whole rulebook forgets everything except that one number. So a toy model and a giant real thing, if they share the same Reynolds number and same shape, follow the identical rulebook — their swirls line up perfectly. To make a tiny model share the big thing's number, you blow air faster over the small one (five times smaller ⇒ five times faster). And if the fluids differ, you also multiply by how sticky each fluid is. Watch out at the edges: waves need a different number (Froude), super-fast flow ignores stickiness except in a paper-thin skin, and super-slow flow is all stickiness.