Question 5 of 6: Dimensional analysis of the lift on a missile

Question 5: Dimensional analysis of the lift on a missile (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The functional statement $F=f(L,\ V,\ D,\ \alpha,\ \rho,\ \mu,\ c)$, in which the eight variables carry the dimensions $[F]=\mathsf{MLT^{-2}}$, $[L]=[D]=\mathsf{L}$, $[V]=[c]=\mathsf{LT^{-1}}$, $[\rho]=\mathsf{ML^{-3}}$, $[\mu]=\mathsf{ML^{-1}T^{-1}}$, and α is already dimensionless.

Find. A complete set of independent dimensionless groups by the Buckingham Pi theorem, and the functional relation rewritten in terms of them, naming any group that is a recognised similarity parameter.

Approach. Count variables and independent dimensions to fix the number of groups, choose ρ, V and D as the repeating variables, and form each group by solving the exponent equations for M, L and T in turn.

1. Count the variables and the primary dimensions. There are n = 8 variables (F, L, V, D, α, ρ, μ, c) and the dimensions involved are mass, length and time, so k = 3. Buckingham’s theorem then predicts $$n-k=8-3=5\ \text{independent dimensionless groups}.$$
2. Choose repeating variables. Take ρ, V and D: they are three in number, they contain all three primary dimensions between them, they cannot themselves be combined into a dimensionless group, and none of them is the dependent variable. Every remaining variable is then paired with these three in turn.
3. Form the force group. Write $\Pi_1=\rho^{a}V^{b}D^{c}F$ and balance dimensions: $$\mathsf{M}:\ a+1=0,\qquad \mathsf{T}:\ -b-2=0,\qquad \mathsf{L}:\ -3a+b+c+1=0 .$$ These give a = −1, b = −2, c = −2, so $$\boxed{\Pi_1=\frac{F}{\rho V^{2}D^{2}}}$$ a force coefficient. Multiplying it by a constant to replace D² with a reference planform area turns it into the conventional lift coefficient C_L = 2F/(ρV²S).
4. Form the viscous group. With $\Pi_2=\rho^{a}V^{b}D^{c}\mu$ the same procedure gives a = b = c = −1, so $\Pi_2=\mu/(\rho V D)$. Inverting a group leaves it dimensionless, so the conventional form is $$\boxed{\Pi_2=\frac{\rho V D}{\mu}=\mathrm{Re}}$$ the Reynolds number, which measures inertia against viscous diffusion.
5. Form the compressibility group. With $\Pi_3=\rho^{a}V^{b}D^{c}c$ the mass balance gives a = 0, the time balance b = −1 and the length balance c = 0, so $$\boxed{\Pi_3=\frac{V}{c}=\mathrm{Ma}}$$ the Mach number, which measures the flow speed against the speed at which pressure information travels.
6. Form the geometric group and collect the one already dimensionless. Pairing L with the repeating set gives simply $$\Pi_4=\frac{L}{D}\quad(\text{the fineness or slenderness ratio}),$$ and the angle of attack is dimensionless as it stands, so it is its own group, $\Pi_5=\alpha$. That completes the five groups the theorem promised.
7. Rewrite the original statement. The eight-variable relation collapses to a four-argument one: $$\boxed{\frac{F}{\rho V^{2}D^{2}}=\Phi\!\left(\frac{L}{D},\ \alpha,\ \mathrm{Re},\ \mathrm{Ma}\right)}$$ Three of the four arguments are standard named parameters — the fineness ratio, the Reynolds number and the Mach number — and the fourth, the angle of attack, is the primary control variable. Testing therefore requires sweeping four knobs rather than seven, a reduction of an experimental campaign from an impossible size to a routine one.
8. Check the result against a realistic case, and note what it implies for model testing. A missile of D = 0.30 m and L = 3.6 m flying at V = 650 m/s at 10 km altitude (ρ = 0.4135 kg/m³, μ = 1.458 × 10⁻⁵ Pa s, c = 299.5 m/s) has L/D = 12.0, Re = 5.53 × 10⁶ and Ma = 2.17; a 4.2 kN lift then corresponds to Π₁ = 0.267. The practical sting is that matching Re and Ma simultaneously on a scaled model is impossible in an ordinary tunnel, since holding Ma fixed holds V fixed and a smaller D then drops Re in proportion. Because slender-body lift at supersonic speed is dominated by pressure rather than shear, the usual engineering resolution is to match Mach number and accept a Reynolds-number mismatch, correcting the viscous contribution separately — a pressurised or cryogenic tunnel is what is needed if both must be matched.