Dimensional analysis — Buckingham π theorem: Exercises

Before we begin, one reminder of the only pieces of "grammar" we use, in plain words:

The picture below is the mental model behind every exercise: variables go in, cancel their units, and come out as pure numbers.

Figure — Dimensional analysis — Buckingham π theorem

Level 1 — Recognition

Exercise 1.1

State the dimensions of each in $M, L, T$ : force $F$ , pressure $P$ , dynamic viscosity $μ$ , kinematic viscosity $ν$ , power $W$ .

Recall Solution 1.1

WHAT we do: build each from definitions, then read off the exponents.

Force $=$ mass $\times$ acceleration: $[F] = M \cdot L T^{- 2} = M L T^{- 2}$ .
Pressure $=$ force / area: $[P] = \frac{M L T ^{- 2}}{L ^{2}} = M L^{- 1} T^{- 2}$ .
Dynamic viscosity: from $[μ]$ in the drag/stress law $τ = μ \frac{d v}{d y}$ , $[μ] = \frac{M L ^{- 1} T ^{- 2}}{T ^{- 1}} = M L^{- 1} T^{- 1}$ .
Kinematic viscosity $ν = μ / ρ$ : $[ν] = \frac{M L ^{- 1} T ^{- 1}}{M L ^{- 3}} = L^{2} T^{- 1}$ .
Power $=$ energy / time $=$ force $\times$ speed: $[W] = M L T^{- 2} \cdot L T^{- 1} = M L^{2} T^{- 3}$ .

Exercise 1.2

For the sphere-drag problem ${F, d, v, ρ, μ}$ , how many independent dimensionless groups exist? For the pendulum ${T, L, g, m}$ ?

Recall Solution 1.2

WHY the count matters: $p$ tells you how many "knobs" the physics truly has.

Sphere drag: $n = 5$ variables, dimensions present are $M, L, T$ so $k = 3$ , giving $p = 5 - 3 = 2$ .
Pendulum: $n = 4$ , dimensions present $M, L, T$ so $k = 3$ , giving $p = 4 - 3 = 1$ .

Exercise 1.3

Which of these are already dimensionless? (a) $R e = ρ v d / μ$ , (b) $F / (ρ v^{2} d^{2})$ , (c) $v / g L$ , (d) $ρ v d$ .

Recall Solution 1.3

WHAT we check: does every one of $M, L, T$ cancel to power zero?

(a) $\frac{( M L ^{- 3} ) ( L T ^{- 1} ) ( L )}{M L ^{- 1} T ^{- 1}} = \frac{M L ^{- 1} T ^{- 1}}{M L ^{- 1} T ^{- 1}} = 1$ . ✅ dimensionless — this is Reynolds number.
(b) $\frac{M L T ^{- 2}}{( M L ^{- 3} ) ( L ^{2} T ^{- 2} ) ( L ^{2} )} = \frac{M L T ^{- 2}}{M L T ^{- 2}} = 1$ . ✅ dimensionless.
(c) $\frac{L T ^{- 1}}{( L T ^{- 2} ) ( L )} = \frac{L T ^{- 1}}{L T ^{- 1}} = 1$ . ✅ dimensionless (the Froude number).
(d) $[M L^{- 3} \cdot L T^{- 1} \cdot L] = M L^{- 1} T^{- 1}$ . ❌ not dimensionless (it is $μ$ 's dimensions).

Level 2 — Application

Exercise 2.1

Find the single π group for the pendulum ${T, L, g}$ (mass already excluded). Confirm $T \propto L / g$ .

Recall Solution 2.1

WHAT we do: write the group as $π = T g^{a} L^{b}$ and force each dimension to cancel. Dimensions: $[T] = T$ , $[g] = L T^{- 2}$ , $[L] = L$ . $[π] = T \cdot (L T^{- 2})^{a} \cdot L^{b} = L^{a + b} T^{1 - 2 a} .$

$T$ : $1 - 2 a = 0 \Rightarrow a = \frac{1}{2}$ .
$L$ : $a + b = 0 \Rightarrow b = - \frac{1}{2}$ . So $π = T g^{1/2} L^{- 1/2} = T g / L = const$ , hence $T \propto L / g$ . ✅ The constant ( $2 π$ ) is what dimensions cannot give.

Exercise 2.2

A wave on deep water has speed $c$ depending on wavelength $λ$ and gravity $g$ . Find the form of $c$ .

Recall Solution 2.2

WHY dimensional analysis here: solving the full water-wave PDE is hard, but with only ${c, λ, g}$ and dimensions $L, T$ we have $n = 3$ , $k = 2$ , so $p = 1$ — one group means one fixed form. Write $π = c g^{a} λ^{b}$ with $[c] = L T^{- 1}$ , $[g] = L T^{- 2}$ , $[λ] = L$ : $[π] = L T^{- 1} (L T^{- 2})^{a} L^{b} = L^{1 + a + b} T^{- 1 - 2 a} .$

$T$ : $- 1 - 2 a = 0 \Rightarrow a = - \frac{1}{2}$ .
$L$ : $1 + a + b = 0 \Rightarrow b = - \frac{1}{2}$ . $π = \frac{c}{g λ} = const \Rightarrow c \propto g λ .$

Exercise 2.3

Redo the sphere drag but pick repeating variables $μ, v, d$ instead of $ρ, v, d$ . Form the group containing $F$ .

Recall Solution 2.3

WHAT we do: $π = μ^{a} v^{b} d^{c} F$ . With $[μ] = M L^{- 1} T^{- 1}$ , $[v] = L T^{- 1}$ , $[d] = L$ , $[F] = M L T^{- 2}$ : $[π] = (M L^{- 1} T^{- 1})^{a} (L T^{- 1})^{b} L^{c} (M L T^{- 2}) .$

$M$ : $a + 1 = 0 \Rightarrow a = - 1$ .
$T$ : $- a - b - 2 = 0 \Rightarrow 1 - b - 2 = 0 \Rightarrow b = - 1$ .
$L$ : $- a + b + c + 1 = 0 \Rightarrow 1 - 1 + c + 1 = 0 \Rightarrow c = - 1$ . $π = \frac{F}{μv d} .$ WHY this is fine: it is a legal (but less famous) drag group. Note $\frac{F}{μv d} = \frac{F}{ρ v ^{2} d ^{2}} \cdot R e$ — a product of the two standard groups. Both choices of repeaters are valid; they just relabel the same two knobs. See Drag force and drag coefficient.

Level 3 — Analysis

Exercise 3.1

Pressure drop $Δ P$ per unit length $Δ P / ℓ$ in a pipe depends on diameter $D$ , mean speed $v$ , density $ρ$ , viscosity $μ$ . How many groups, and what are they?

Recall Solution 3.1

WHAT we do: variables ${Δ P / ℓ, D, v, ρ, μ}$ , so $n = 5$ . Dimensions $M, L, T$ so $k = 3$ , $p = 2$ . $[Δ P / ℓ] = [P] / L = M L^{- 1} T^{- 2} / L = M L^{- 2} T^{- 2}$ . Repeaters $ρ, v, D$ . Group with $Δ P / ℓ$ : $π_{1} = ρ^{a} v^{b} D^{c} (Δ P / ℓ)$ :

$M$ : $a + 1 = 0 \Rightarrow a = - 1$ .
$T$ : $- b - 2 = 0 \Rightarrow b = - 2$ .
$L$ : $- 3 a + b + c - 2 = 0 \Rightarrow 3 - 2 + c - 2 = 0 \Rightarrow c = 1$ . $π_{1} = \frac{( Δ P / ℓ ) D}{ρ v ^{2}} .$ Group with $μ$ gives the familiar $π_{2} = \frac{μ}{ρ v D} = 1/ R e$ . So $\frac{( Δ P / ℓ ) D}{ρ v ^{2}} = Φ (R e)$ — the basis of the Darcy friction-factor chart.

Exercise 3.2

Show that choosing repeaters ${v, D, ν}$ (with kinematic viscosity $ν = L^{2} T^{- 1}$ ) fails to span all needed dimensions for the drag problem. Which dimension is missing?

Recall Solution 3.2

WHY this is the point: repeaters must (a) together contain all present dimensions and (b) be independent. Dimensions of the trio: $[v] = L T^{- 1}$ , $[D] = L$ , $[ν] = L^{2} T^{- 1}$ . Every one uses only $L$ and $T$ — there is no $M$ anywhere. But the drag problem contains $M$ (in $F$ and $ρ$ ). So these three repeaters cannot cancel mass out of $π_{1} = v^{a} D^{b} ν^{c} F$ — the $M$ -equation is $0 + 1 = 0$ , which is impossible. Missing dimension: $M$ . Fix: include a variable carrying $M$ (like $ρ$ ) among the repeaters.

Exercise 3.3

The dimension matrix $D$ for ${F, d, v, ρ, μ}$ over $(M, L, T)$ is

D = 11 - 2 010 01 - 1 1 - 3 0 1 - 1 - 1 .

Verify $rank (D) = 3$ and hence $p = n - rank (D) = 2$ .

Recall Solution 3.3

WHAT rank means here: the number of dimensionally independent directions among the columns. Take the first three columns (the exponents of $F, d, v$ ):

11 - 2 010 01 - 1 .

Its determinant: expand along the top row $= 1 \cdot (1 \cdot (- 1) - 1 \cdot 0) = - 1 \neq = 0$ . A non-zero $3 \times 3$ minor means $rank (D) = 3$ . Therefore $p = 5 - 3 = 2$ , matching Exercise 1.2. WHY: the null-space dimension of a $k \times n$ matrix is $n - rank$ , and each null-space vector is one independent π. See the parent note's derivation of $p = n - k$ .

Level 4 — Synthesis

Exercise 4.1

A liquid drop of radius $r$ oscillates with frequency $f$ under surface tension $σ$ (dimensions $[σ] = M T^{- 2}$ , force per length) and density $ρ$ . Derive the form of $f$ .

Recall Solution 4.1

WHAT we do: variables ${f, r, σ, ρ}$ , $n = 4$ . Dimensions present: $M, L, T$ ? Check — $[f] = T^{- 1}$ , $[r] = L$ , $[σ] = M T^{- 2}$ , $[ρ] = M L^{- 3}$ . All of $M, L, T$ appear, $k = 3$ , so $p = 1$ : one group, one fixed form. Form $π = f ρ^{a} r^{b} σ^{c}$ : $[π] = T^{- 1} (M L^{- 3})^{a} L^{b} (M T^{- 2})^{c} = M^{a + c} L^{- 3 a + b} T^{- 1 - 2 c} .$

$M$ : $a + c = 0$ .
$T$ : $- 1 - 2 c = 0 \Rightarrow c = - \frac{1}{2} \Rightarrow a = \frac{1}{2}$ .
$L$ : $- 3 a + b = 0 \Rightarrow b = 3 a = \frac{3}{2}$ . $π = f ρ^{1/2} r^{3/2} σ^{- 1/2} = f \frac{ρ r ^{3}}{σ} = const \Rightarrow f \propto \frac{σ}{ρ r ^{3}} .$ Check by feel: more surface tension → stiffer → faster wobble ( $f ↑$ ), bigger/heavier drop → slower ( $f ↓$ ). ✅

Exercise 4.2

Non-dimensionalise: a scale model of a ship is tested. To match the real ship's flow, which π group must be equal on model and prototype if gravity waves dominate? (Hint: use ${v, L, g}$ .)

Recall Solution 4.2

WHY: two flows are dynamically similar when their governing π groups match — the core idea of Model testing and similarity. From Exercise 1.3(c) / 2.2, the gravity group of ${v, L, g}$ is the Froude number $F r = \frac{v}{g L}$ . Matching means $\frac{v _{m}}{g L _{m}} = \frac{v _{p}}{g L _{p}}$ . With $g$ the same, the model speed scales as $v_{m} = v_{p} \frac{L _{m}}{L _{p}} .$ Numeric: a $1 : 100$ model ( $L_{m} / L_{p} = 1/100$ ) of a ship doing $v_{p} = 10 m/s$ must be towed at $v_{m} = 10 1/100 = 1 m/s$ .

Exercise 4.3

Heat transfer: the convective coefficient $h$ (dimensions $M T^{- 3} Θ^{- 1}$ , where $Θ$ = temperature) depends on velocity $v$ , pipe diameter $D$ , fluid density $ρ$ , viscosity $μ$ , specific heat $c_{p}$ ( $L^{2} T^{- 2} Θ^{- 1}$ ), and thermal conductivity $κ$ ( $M L T^{- 3} Θ^{- 1}$ ). How many π groups?

Recall Solution 4.3

WHAT we count: $n = 7$ variables ${h, v, D, ρ, μ, c_{p}, κ}$ . Now four dimensions appear: $M, L, T, Θ$ (temperature enters via $h, c_{p}, κ$ ). So $k = 4$ . $p = n - k = 7 - 4 = 3.$ These three turn out to be the Nusselt, Reynolds and Prandtl numbers — the standard heat-transfer trio. WHY $k = 4$ not 3: temperature $Θ$ is a genuine, independent fundamental dimension here — see the Level 1 trap warning and Fundamental and derived units.

Level 5 — Mastery

Exercise 5.1

The lift force $F_{L}$ on a wing depends on air density $ρ$ , speed $v$ , wing area $S$ (dimensions $L^{2}$ ), viscosity $μ$ , and the speed of sound $a$ ( $L T^{- 1}$ ). Find $p$ and identify each group physically.

Recall Solution 5.1

WHAT we do: $n = 6$ variables ${F_{L}, ρ, v, S, μ, a}$ , dimensions $M, L, T$ so $k = 3$ , giving $p = 6 - 3 = 3$ . Repeaters $ρ, v, S$ (note $[S] = L^{2}$ , still spans $L$ and, with $ρ, v$ , all of $M, L, T$ ).

Group with $F_{L}$ : $π_{1} = ρ^{a} v^{b} S^{c} F_{L}$ . With $[S^{c}] = L^{2 c}$ :
- $M$ : $a + 1 = 0 \Rightarrow a = - 1$ .
- $T$ : $- b - 2 = 0 \Rightarrow b = - 2$ .
- $L$ : $- 3 a + b + 2 c + 1 = 0 \Rightarrow 3 - 2 + 2 c + 1 = 0 \Rightarrow c = - 1$ . $π_{1} = \frac{F _{L}}{ρ v ^{2} S} (lift coefficient C_{L}) .$
Group with $μ$ gives $π_{2} = \frac{μ}{ρ v S}$ (a Reynolds number using $S$ as length scale).
Group with $a$ : $π_{3} = ρ^{a} v^{b} S^{c} a$ → dimensionally $a$ and $v$ are both $L T^{- 1}$ , so the only cancellation gives $π_{3} = \frac{a}{v}$ — i.e. $π_{3} = \frac{1}{M a}$ , the inverse Mach number. So $C_{L} = Φ (R e, M a)$ — three knobs, exactly what wind-tunnel aerodynamics uses.

Exercise 5.2

Degenerate case. A pendulum swinging in vacuum: ${T, L, g}$ but now also list the (irrelevant) air density $ρ$ . Naively $n = 4$ , dimensions $M, L, T$ → $p = 1$ ? Show what actually happens and why.

Recall Solution 5.2

WHY this is subtle: adding $ρ$ introduces the dimension $M$ , but nothing else carries $M$ . So $M$ appears in exactly one variable. Set up any group $π = T^{a} L^{b} g^{c} ρ^{d}$ . The $M$ -equation is $d = 0$ (only $ρ$ has $M$ ). That forces $ρ$ out of every group — just like mass in the plain pendulum. Now the dimension matrix over $(M, L, T)$ has an $M$ -row that is zero except in $ρ$ 's column, and $ρ$ 's column has $L, T$ entries $- 3, 0$ . Effectively $rank (D) = 3$ but one variable ( $ρ$ ) is dead. The real count of usable groups is still $1$ (the same $T g / L$ ). Lesson: including a variable that carries a dimension nothing else shares does not add a knob — it adds a variable that must have exponent zero. Fix: drop variables whose dimension can't be balanced (the parent note's Mistake A).

Exercise 5.3

Full synthesis. Terminal velocity $v_{t}$ of a small sphere (radius $r$ ) settling in a very viscous fluid depends on $r$ , $μ$ , and the effective weight per volume $Δ ρ g$ (buoyancy-corrected, dimensions of $M L^{- 2} T^{- 2}$ ). Predict the form of $v_{t}$ , then compare to Stokes' law.

Recall Solution 5.3

WHAT we do: variables ${v_{t}, r, μ, Δ ρ g}$ , $n = 4$ . Dimensions $M, L, T$ so $k = 3$ , $p = 1$ — one fixed form. Group $π = v_{t} μ^{a} r^{b} (Δ ρ g)^{c}$ with $[v_{t}] = L T^{- 1}$ , $[μ] = M L^{- 1} T^{- 1}$ , $[r] = L$ , $[Δ ρ g] = M L^{- 2} T^{- 2}$ : $[π] = L T^{- 1} (M L^{- 1} T^{- 1})^{a} L^{b} (M L^{- 2} T^{- 2})^{c} = M^{a + c} L^{1 - a + b - 2 c} T^{- 1 - a - 2 c} .$

$M$ : $a + c = 0$ .
$T$ : $- 1 - a - 2 c = 0$ . Substituting $a = - c$ : $- 1 + c - 2 c = 0 \Rightarrow - 1 - c = 0 \Rightarrow c = - 1$ , so $a = 1$ .
$L$ : $1 - a + b - 2 c = 0 \Rightarrow 1 - 1 + b + 2 = 0 \Rightarrow b = - 2$ . $π = \frac{v _{t} μ}{r ^{2} Δ ρ g} = const \Rightarrow v_{t} \propto \frac{Δ ρ g r ^{2}}{μ} .$ Compare: Stokes' law gives $v_{t} = \frac{2}{9} \frac{Δ ρ g r ^{2}}{μ}$ — the exact same shape, with the constant $\frac{2}{9}$ that dimensions alone cannot supply (parent note's Mistake C). ✅

Common mistake Level 5 trap — trusting the naive

$n - k$ over the physics In Exercise 5.2 the naive count says " $p = 1$ , fine" and the student never notices $ρ$ was dead weight; in richer problems (a variable that carries a unique dimension) this inflates $p$ and produces a phantom group. Why it feels right: the formula $p = n - k$ is drilled as gospel. Why it's wrong: the formula is only valid when $rank (D) = k$ and every listed variable is physically relevant. A variable whose dimension nothing else balances is forced to exponent zero — it neither adds a group nor belongs in the list. Fix: always sanity-check by asking "can this dimension actually be cancelled by another variable?" before trusting the count. This is the analysis behind the Navier–Stokes equations non-dimensionalisation, where every retained term earns its place.

Recall Self-test summary (cloze)

The count of independent π groups is == $p = n - rank (D)$ ==, which equals $n - k$ only when ==the dimension matrix has full rank $k$ . Repeaters must together span all present dimensions and be dimensionally independent (form no π). A variable carrying a dimension that no other variable shares== is forced to exponent zero and drops out.

Connections

Reynolds number — the recurring $μ$ -based group in Exercises 2.3, 3.1, 5.1
Drag force and drag coefficient — Exercises 2.3 and 5.1 build $C_{D}$ , $C_{L}$
Dimensional homogeneity — the principle every solution rests on
Model testing and similarity — Exercise 4.2's Froude matching
Navier–Stokes equations — mastery-level non-dimensionalisation
Fundamental and derived units — the $M, L, T, Θ$ basis used in Exercise 4.3