Worked examples — Dimensional analysis — Buckingham π theorem

This is the "put it through its paces" companion to the parent topic. The parent showed you the recipe. Here we deliberately hunt down every kind of situation the theorem can hand you — the clean cases, the awkward ones, the traps, and the exam twists — and work each to the last symbol.

We will use only these operations on recipes:

• Multiply two quantities → add their exponents ($L^1{T}^{-1}\cdot L^1=L^2{T}^{-1}$).
• Raise to a power $a$ → multiply every exponent by $a$.
• Dimensionless means every exponent equals $0$.

That is the entire algebra. If you can add and multiply small whole numbers and fractions, you can do all of this.

The scenario matrix

Every problem the theorem throws at you falls into one of these case classes. Each row is a distinct "shape" of situation; the examples below are labelled with the cell they cover so you can see nothing is left out.

Ex 1 — Baseline clean case (Cell C1)

Forecast: guess now — how many independent pure numbers? (Write it down before reading on.)

1. Count everything. $n=5$ variables; base dimensions actually present are $M,L,T$ so $k=3$. Why this step? $p=n-k$ only works once you know both counts. The theorem is a counting statement first.

2. Number of groups. $p=5-3=2$. Why this step? This tells us the physics has exactly two knobs — no more clever variables can add freedom.

3. Pick repeaters $\rho ,v,D$. Why these three? They must (a) together contain $M,L,T$ and (b) form no group among themselves. Look at their recipes: only $\rho$ carries $M$, only $v$ carries $T$, and $D$ supplies a "spare" $L$ to fix lengths. That is exactly the independence we need — no product of them collapses to a pure number.

4. Build ${\pi }_1$ with $G$: write ${\pi }_1={\rho }^a{v}^b{D}^{c}G$ and demand each exponent be zero.

   • $M:a+1=0\Rightarrow a=-1$
   • $T:-b-2=0\Rightarrow b=-2$
   • $L:-3a+b+c-2=0\Rightarrow 3-2+c-2=0\Rightarrow c=1$ $\boxed{{\pi }_1=\frac{GD}{\rho v^2}}$ Why this step? Each base dimension gives one equation; three equations pin down $a,b,c$. This is how a group is forced, not guessed.

5. Build ${\pi }_2$ with $\mu$: ${\pi }_2={\rho }^a{v}^b{D}^c\mu$ gives $a=-1,b=-1,c=-1$, so $\boxed{{\pi }_2=\frac{\mu }{\rho vD}=\frac{1}{Re}}$ the Reynolds number again.

6. The law's shape: $\frac{GD}{\rho v^2}=\Phi (Re)$.

Verify: Check ${\pi }_1$ is truly dimensionless. $[GD]=M{L}^{-2}{T}^{-2}\cdot L=M{L}^{-1}{T}^{-2}$. $[\rho v^2]=M{L}^{-3}\cdot L^2{T}^{-2}=M{L}^{-1}{T}^{-2}$. Ratio $=M^0{L}^0{T}^0$. ✓ And $[\mu ]/[\rho vD]=M{L}^{-1}{T}^{-1}/(M{L}^{-3}\cdot L{T}^{-1}\cdot L)=M{L}^{-1}{T}^{-1}/(M{L}^{-1}{T}^{-1})=1$. ✓

Ex 2 — A variable forced out (Cell C2)

Forecast: does the total mass $m$ appear in the answer? Guess yes or no.

1. Count. $n=4$; dimensions present $M,L,T$, so $k=3$. Then $p=4-3=1$: exactly one group. Why this step? One group means the whole law is $\pi =\text{const}$ — a single locked-in relation.

2. Watch the $M$ dimension. $M$ appears in $F_T,\lambda ,m$ — three carriers, plenty to cancel with. But look what happens when we actually build the group.

3. Repeaters $F_T,\lambda$ (they carry $M,L,T$ between them: $F_T$ has all three, $\lambda$ has $M,L$). Build $\pi =F_T^a{\lambda }^{b}c$:

   • $M:a+b=0$
   • $T:-2a-1=0\Rightarrow a=-\frac{1}{2}$
   • $L:a-b+1=0$ From $a=-\frac{1}{2}$: $b=+\frac{1}{2}$ (from $M$), and check $L$: $-\frac{1}{2}-\frac{1}{2}+1=0$ ✓. $\boxed{\pi =c\sqrt{\frac{\lambda }{F_T}}=\text{const}\Rightarrow c\propto \sqrt{\frac{F_T}{\lambda }}}$

4. Where did $m$ go? Try to include $m$: ${\pi }^{\prime }=F_T^a{\lambda }^{b}m$. Then $T:-2a=0\Rightarrow a=0$; $M:a+b+1=0\Rightarrow b=-1$; $L:a-b=0\Rightarrow b=0$ — contradiction ($b$ cannot be both $-1$ and $0$). No dimensionless group containing $m$ exists. Why this step? $m=\lambda \cdot (\text{length})$, but there is no independent length variable in the list. So $m$ carries a length that nothing can cancel. It is redundant — the physics never uses it.

Verify: $[c\sqrt{\lambda /F_T}]=L{T}^{-1}\cdot \sqrt{M{L}^{-1}/(ML{T}^{-2})}=L{T}^{-1}\cdot \sqrt{L^{-2}{T}^2}=L{T}^{-1}\cdot L^{-1}T=1$. ✓ Answer $c=\sqrt{F_T/\lambda }$ is the known string-wave speed. ✓

Ex 3 — Only length and time survive, $k=2$ (Cell C3)

Forecast: what is $k$ here — 3 or 2? And therefore how many groups?

1. Count dimensions actually present. Scan every recipe: only $L$ and $T$ show up. So $k=2$, not 3. Why this step? The rule is $k=\mathrm{rank}(D)$ = number of independent dimensions present. Blindly writing $k=3$ would give $p=3-3=0$ and wrongly claim zero groups.

2. Groups. $p=n-k=3-2=1$.

3. Build $\pi ={\ell }^a{g}^{b}c$ (repeaters $\ell ,g$; together they carry $L$ and $T$):

   • $T:-2b-1=0\Rightarrow b=-\frac{1}{2}$
   • $L:a+b+1=0\Rightarrow a-\frac{1}{2}+1=0\Rightarrow a=-\frac{1}{2}$ $\boxed{\pi =\frac{c}{\sqrt{g\ell }}=\text{const}\Rightarrow c\propto \sqrt{g\ell }}$

Verify: $[c/\sqrt{g\ell }]=L{T}^{-1}/\sqrt{L{T}^{-2}\cdot L}=L{T}^{-1}/\sqrt{L^2{T}^{-2}}=L{T}^{-1}/(L{T}^{-1})=1$. ✓ (Full theory: $c=\sqrt{g\ell /2\pi }$; the $2\pi$ is exactly what dimensions cannot supply.)

Ex 4 — Bad repeaters kill the method (Cell C4)

Look at the figure: each repeater is drawn as an arrow whose three coordinates are its exponents over $(M,L,T)$. The whole point of "independent repeaters" is that these three arrows must genuinely point in three different directions in this exponent space. The red arrows below refuse to do that.

Forecast: the student expects to build 2 groups. How many can they actually build independently?

1. Write the repeaters' exponent vectors over $(M,L,T)$:

   • $v$: $(0,1,-1)$
   • $d$: $(0,1,0)$
   • $t$: $(0,0,1)$ Why this step? Independence is a statement about these vectors: they must not be flattenable into one another.

2. Test independence. None carries $M$ — the first coordinate is $0$ for all three. So these three live entirely in the $(L,T)$ plane. Three vectors crammed into a 2-dimensional plane cannot be independent (see the red arrows in the figure, all lying in the same grey plane, so one is a combination of the others). Why this step? The repeater sub-matrix now has rank $\le 2<3$. Its determinant is $0$, so it is not invertible — you cannot solve for the exponents uniquely.

3. Show the determinant is zero explicitly. Stack the exponent columns for $v,d,t$ over rows $(M,L,T)$: $\det \left(\begin{array}{ccc}0& 0& 0\\ 1& 1& 0\\ -1& 0& 1\end{array}\right)=0$ because the entire top row is zero. A zero determinant confirms rank $<3$. Why this step? The clean numerical test for "are my repeaters independent?" is: form their exponent matrix and check $\det \ne 0$.

4. Consequence. With rank $2$ repeaters you can only impose 2 constraints, not 3. Worse, they carry no $M$ at all, so any group containing $F_D$ or $\rho$ (both of which carry $M$) is impossible to balance in mass. The method silently loses a whole dimension.

5. Fix. Use the valid choice $\rho ,v,d$ (parent Ex 1). Its exponent matrix over rows $(M,L,T)$, columns $\rho ,v,d$, is $\det \left(\begin{array}{ccc}1& 0& 0\\ -3& 1& 1\\ 0& -1& 0\end{array}\right)=1\cdot (1\cdot 0-1\cdot (-1))=1\ne 0,$ so those repeaters are independent and the method works.

Verify: bad-repeater determinant $=0$ (dependent), good-repeater determinant $=1\ne 0$ (independent). Both are checked numerically at the bottom of this page.

Ex 5 — A fourth base dimension appears (Cell C5)

Forecast: with the "universe has 3 dimensions" reflex you'd say $k=3$. Is that right here?

1. Count base dimensions present. Now $M,L,T,\Theta$ all appear → $k=4$, not 3. Why this step? Ignoring $\Theta$ would over-count groups by one and give the wrong physics.

2. Variables. $n=6$. So $p=n-k=6-4=2$. Why this step? Two groups: exactly the number of knobs convection has.

3. Pick repeaters $\rho ,v,L_0,\kappa$ (four repeaters because $k=4$). Check they span the dimensions: only $\rho$ carries $M$ freely, $v$ supplies $T$, $L_0$ supplies a spare $L$, and $\kappa$ is the only clean source of $\Theta$. They are independent. Why this step? With $k=4$ we need four repeaters, one more than the usual three.

4. Group ${\pi }_1$ with $h$: demand ${\pi }_1={\rho }^a{v}^b{L}_0^c{\kappa }^{e}h$ be dimensionless. Matching $M,L,T,\Theta$ gives $a=0,b=0,c=1,e=-1$, so $\boxed{{\pi }_1=\frac{h{L}_0}{\kappa }=Nu}(\text{the Nusselt number}).$ Why this step? $h$ and $\kappa$ carry the same $M,T,\Theta$ signature; only their length powers differ by one, and $L_0$ closes that gap.

5. Group ${\pi }_2$ with $c_p$: demand ${\pi }_2={\rho }^a{v}^b{L}_0^c{\kappa }^e{c}_p$ dimensionless. Matching gives $a=1,b=1,c=1,e=-1$, so $\boxed{{\pi }_2=\frac{\rho v{L}_0{c}_p}{\kappa }=Re\cdot Pr}$ the product of Reynolds number $Re=\rho v{L}_0/\mu$ and Prandtl number $Pr=\mu c_p/\kappa$ (the $\mu$ cancels in the product, which is why viscosity did not need to be listed separately).

6. The law's shape: $Nu=\Phi (Re\cdot Pr)$ — heat transfer depends on flow and fluid through just these two knobs.

Verify: $[h{L}_0/\kappa ]=M{T}^{-3}{\Theta }^{-1}\cdot L/(ML{T}^{-3}{\Theta }^{-1})=1$ ✓ (dimensionless). And $[\rho v{L}_0{c}_p/\kappa ]=M{L}^{-3}\cdot L{T}^{-1}\cdot L\cdot L^2{T}^{-2}{\Theta }^{-1}/(ML{T}^{-3}{\Theta }^{-1})=1$ ✓. Both checked numerically below.

Ex 6 — One group forces a unique power law (Cell C6)

Forecast: how many free dimensionless groups — and what does having just one group mean for the freedom of the law?

1. Count. Variables in play: $Q,v,\ell$ so $n=3$. Present dimensions: only $L,T$, so $k=2$. Then $p=n-k=3-2=1$. Why this step? One group means there is no free function $\Phi$ of another argument — the law is fixed up to a single constant. (Contrast the drag case, where two groups leave one whole function to determine by experiment.)

2. Build the group $\pi =v^a{\ell }^{b}Q$:

   • $T:-a-1=0\Rightarrow a=-1$
   • $L:a+b+2=0\Rightarrow -1+b+2=0\Rightarrow b=-1$ $\boxed{\pi =\frac{Q}{v\ell }=\text{const}\Rightarrow Q\propto v\ell }$ Why this step? With one group set to a constant, the entire functional form is fixed — dimensions alone give $Q=(\text{number})v\ell$. Nothing is left to guess but that one pure number.

3. The lesson of "few groups". The fewer the groups, the more dimensions dictate the answer. One group → unique power law; and if you ever get $p=0$ (no group at all), it signals your variables can't combine dimensionlessly — a warning that a variable is missing.

Verify: $[Q/(v\ell )]=L^2{T}^{-1}/(L{T}^{-1}\cdot L)=L^2{T}^{-1}/(L^2{T}^{-1})=1$. ✓

Ex 7 — Real-world design: a scale model (Cell C7)

Forecast: must the model go slower or faster than the real sub? Guess before computing.

1. Similarity condition. $C_D$ is a function of $Re$ only. To reproduce the prototype's flow, match the one governing group: $R{e}_m=R{e}_p$. Why this step? Equal π groups ⇒ equal dimensionless physics ⇒ the model faithfully reproduces the prototype. This is Model testing and similarity in action.

2. Solve for $v_m$. Same fluid ⇒ same $\rho ,\mu$, so $Re$ equality reduces to $v_m{D}_m=v_p{D}_p$: $v_m=v_p\frac{D_p}{D_m}=3\times \frac{6}{0.3}=3\times 20=60\text{m/s}.$ Why this step? A $20\times$ smaller model needs $20\times$ the speed to keep $Re$ fixed — the small size is "made up for" by going faster.

3. Scale the drag. With $Re$ matched, $C_D$ is equal for model and prototype, so $F_D=C_D\rho v^2{D}^2$ with the same $C_D$ and same $\rho$. Therefore $\frac{F_{D,p}}{F_{D,m}}=\frac{v_p^2{D}_p^2}{v_m^2{D}_m^2}={\left(\frac{3}{60}\right)}^2{\left(\frac{6}{0.3}\right)}^2={\left(\frac{1}{20}\right)}^2(20{)}^2=1.$ Remarkably, $F_{D,p}=F_{D,m}$: the drag measured on the model equals the real submarine's drag. Why this step? Holding $vD$ fixed also fixes $v^2{D}^2$, so the two forces coincide exactly under this same-fluid matching. The engineer reads the model's drag straight off as the prototype's drag.

Verify: $v_m=60$ m/s (checked). $R{e}_m=R{e}_p$ since $v_m{D}_m=60\times 0.3=18=3\times 6=v_p{D}_p$ (checked). Force ratio $F_{D,p}/F_{D,m}=1$ (checked). All below.

Ex 8 — Exam twist: fractional & negative exponents (Cell C8)

Forecast: guess the power of $t$ in $R\propto t^{?}$ before solving.

1. Count. $n=4$; dimensions $M,L,T$ present, $k=3$. So $p=4-3=1$: one group, one locked law. Why this step? One group again means the whole form is fixed up to a constant (which Taylor later found $\approx 1.03$).

2. Build $\pi =E^a{\rho }^b{t}^{c}R$ and set every exponent to zero, tracking signs carefully:

   • $M:a+b=0$
   • $T:-2a+c=0$
   • $L:2a-3b+1=0$ Why this step? Three base dimensions → three equations → solve the linear system for $a,b,c$.

3. Solve. From $M$: $b=-a$. Put into $L$: $2a-3(-a)+1=0\Rightarrow 5a+1=0\Rightarrow a=-\frac{1}{5}$. Then $b=+\frac{1}{5}$, and from $T$: $c=2a=-\frac{2}{5}$. $\pi =E^{-1/5}{\rho }^{1/5}{t}^{-2/5}R=\text{const}$ Rearranged: $\boxed{R=\text{const}\times {\left(\frac{E{t}^2}{\rho }\right)}^{1/5}\Rightarrow R\propto t^{2/5}}$ Why this step? Isolating $R$ inverts the fractional powers; the negative $-\frac{2}{5}$ on $t$ moves up to $+\frac{2}{5}$ when we solve $\pi =\text{const}$ for $R$.

4. Sanity on signs. Larger energy $E$ (positive power $+\frac{1}{5}$) → bigger blast: correct. Denser air $\rho$ (negative power $-\frac{1}{5}$) → smaller blast: correct, heavier air resists.

Verify: $[(E{t}^2/\rho {)}^{1/5}]=(M{L}^2{T}^{-2}\cdot T^2/(M{L}^{-3}){)}^{1/5}=(L^2\cdot L^3{)}^{1/5}=(L^5{)}^{1/5}=L$ ✓ (matches $[R]=L$). Exponents $a=-1/5,b=1/5,c=-2/5$ checked below.

Recall

Connections

• Reynolds number — reappears in Ex 1, 4, 7 as the master group
• Drag force and drag coefficient — Ex 7 uses $C_D=\Phi (Re)$
• Model testing and similarity — Ex 7 is the worked instance
• Dimensional homogeneity — every "Verify" line is a homogeneity check
• Fundamental and derived units — the $M,L,T,\Theta$ basis behind every recipe
• Navier–Stokes equations — where $Re$ comes from rigorously