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

Question 5: Dimensional analysis of the lift force 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 $F=f(L,\ V,\ D,\ \alpha,\ \rho,\ \mu,\ c)$ with the dimensions below, expressed in the $M,L,T$ system.

Find. A complete, independent set of dimensionless groups by the Buckingham Pi theorem, and the reduced functional relation, with each group identified by its conventional name wherever one exists.

Approach. Count the variables and the rank of the dimensional matrix to fix the number of groups, choose a dimensionally independent repeating set, and form each group in turn; then recognise the standard aerodynamic parameters among them.

1. Count the variables and find the rank. There are $n=8$ variables including the already dimensionless $\alpha$. The dimensional matrix of the seven dimensional variables spans $M$, $L$ and $T$ — $F$ supplies $M$ and $T$, $L$ supplies pure length, and no row is a combination of the others — so its rank is $j=3$. The Buckingham Pi theorem then gives $$\text{number of groups}=n-j=8-3=\boxed{5}$$ Note that a dimensionless variable such as $\alpha$ is counted in $n$ and emerges as a group in its own right; it is never omitted.
2. Choose the repeating variables. Take $\rho$, $V$ and $D$. They are dimensionally independent (their $3\times3$ dimensional matrix is non-singular), they include one property, one kinematic and one geometric quantity, and none of them is the dependent variable $F$ — all four are the standard requirements. Using $D$ rather than $L$ as the length scale is conventional for a slender body, and only changes which group carries the slenderness ratio.
3. Form the first group from the dependent variable. Writing $\Pi_1=\rho^{a}V^{b}D^{c}F$ and requiring $M^{0}L^{0}T^{0}$ gives, from $M$: $a+1=0$; from $T$: $-b-2=0$; from $L$: $-3a+b+c+1=0$. Hence $a=-1$, $b=-2$, $c=-2$ and $$\Pi_1=\frac{F}{\rho V^{2}D^{2}} .$$ Multiplying by a constant leaves it dimensionless, so this is the lift coefficient, usually written $C_L=F/(\tfrac12\rho V^{2}S)$ with $S$ a reference area proportional to $D^{2}$.
4. Form the remaining groups. The same procedure applied to each of the non-repeating variables in turn gives $$\begin{aligned} \Pi_2&=\frac{L}{D}, &\quad \Pi_3&=\alpha\cr \Pi_4&=\frac{\rho V D}{\mu}, &\quad \Pi_5&=\frac{V}{c} \end{aligned}$$ Each may be checked in one line: $\Pi_2$ and $\Pi_3$ are ratios of like quantities; $\Pi_4$ has dimensions $(ML^{-3})(LT^{-1})(L)/(ML^{-1}T^{-1})=M^{0}L^{0}T^{0}$; and $\Pi_5$ is a ratio of two velocities.
5. Name the groups and write the reduced relation. Four of the five are standard: $\Pi_1$ is the lift coefficient $C_L$, $\Pi_2$ the slenderness (fineness) ratio, $\Pi_4$ the Reynolds number $Re$ and $\Pi_5$ the Mach number $Ma$, while $\Pi_3=\alpha$ is the attitude of the body and is already dimensionless. Therefore $$\boxed{\;\frac{F}{\rho V^{2}D^{2}} =g\!\left(\frac{L}{D},\ \alpha,\ \frac{\rho VD}{\mu},\ \frac{V}{c}\right) \quad\text{or}\quad C_L=g\!\left(\frac{L}{D},\ \alpha,\ Re,\ Ma\right)\;}$$ An experiment that once required mapping eight variables now requires mapping four, and for a given missile shape ($L/D$ fixed) only three.
6. Comment on the practical consequences of the result. Complete similarity between a wind-tunnel model and the full-scale missile demands that $L/D$, $\alpha$, $Re$ and $Ma$ all match. Matching $Re$ and $Ma$ simultaneously in an ordinary tunnel is impossible at reduced scale in the same gas, because $Re\propto\rho VD/\mu$ falls with $D$ while $Ma$ fixes $V$ — hence pressurised or cryogenic tunnels, which raise $\rho$ or lower $\mu$ to recover $Re$. In practice the Reynolds dependence of lift is weak above about $Re=10^{6}$, so supersonic testing matches $Ma$, $\alpha$ and $L/D$ and accepts a Reynolds mismatch, correcting the skin-friction contribution separately. An alternative and equally correct group set replaces $\Pi_5$ by $\Pi_4/\Pi_5$ or uses $L$ as the repeating length; any such set is acceptable provided the groups are independent and five in number.