One important thing to think about is what physical quantities determine variation in forces and moments? When studying something such as an airfoil on an airplane, this would be a valid thing to ask for structural and safety analysis, and for controls systems as well. One way we can do this is through dimensional analysis, or the analysis of fundamental quantities such as mass, length, time, and temperature. Let's take a look at an example to start our journey.
Let's say there is an airfoil in a wind tunnel and we want to determine the resultant aerodynamic force, R, on the airfoil. We would say that R would depend on
Since R is dependent on these values, we can write it as a function of the 5 parameters, as so:
[ 1 ]
$$ R = f(\rho_\infty, v_\infty, c, \mu_\infty, a_\infty) $$
Now you may be wondering where the dimensional analysis comes in here. Well you see, that is how the Buckingham Pi Theorem comes into play. We can use the fundamental quantities of each of the 5 parameters and use them to derive important, fundamental unitless variables such as Mach number or Reynolds' number.
The Buckingham Pi Theorem starts off with the equation
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$$ N - k = \Pi 's $$
where N is the number of related parameters, k is the number of fundamental dimensions, and $ \Pi $ is the number of pi equations that we will need to derive. Now, before we start, since we are trying to parameterize the resultant force, R, we will need to add R to our list of parameters from before and make a function g as so:
[ 3 ]
$$ g(\rho_\infty, v_\infty, c, \mu_\infty, a_\infty, R) = 0 $$
Now that we have all of our parameters written out, we can write that we have 6 related parameters and we have 3 fundamental dimensions in this case: mass, length, and time. So, we can solve Eq. (2) by doing $ 6 - 3 = 3 $. Hence, we have 3 $ \Pi $ equations that we will need to solve for the example.
What we need to now do is write out the fundamental dimensions for each parameter for the analysis. We will denote mass by $ m $, length by $ l $, and time by $ t $. We do not have temperature in this example, but if we did, we would denote it by $ k $. We can write them as so
[ 4 ]
$$ [R ] = m l t^{-2} $$ $$ [\rho_\infty] = m l^{-3} $$ $$ [v_\infty] = l t^{-1} $$ $$ [c] = l $$ $$ [\mu_\infty] = m l^{-1} t^{-1} $$ $$ [a_\infty] = l t^{-1} $$
Now we need to define our $ \Pi $ functions. Since we have 3 fundamental dimensions in this example, mass, length, and time, we will have to select 3 of our parameters above to include in each equation and then the other 3 parameters will be split up between the $ \Pi $ equations.
What we want to make sure is that each fundamental dimension is included in the 3 parameters that we choose to include in all 3 $ \Pi $ equations. For example, let's choose $ \rho_\infty $, $ v_\infty $, and $ c $ to be in all 3 equations. Notice we have a mass, length, and time unit present between the 3. We can then write our $ \Pi $ functions as so
[ 5 ]
$$ \Pi_1 = f_1(\rho_\infty, v_\infty, c, R) $$ $$ \Pi_2 = f_2(\rho_\infty, v_\infty, c, \mu_\infty) $$ $$ \Pi_3 = f_3(\rho_\infty, v_\infty, c, a_\infty) $$
You may be wondering why those 3 parameters were selected to be in each $ \Pi $ equation, and that is a valid question. The truth is, we could select any of the parameters to take the place of each other and we would get a result, but some are more useful than others. It's just a little bit of experimenting to see what derives out!
Now that we have our $ \Pi $ functions set up, we can start to derive the associated unitless variable. We can do this by assigning each parameter in the function a power of something and then insert their fundamental dimensions. Let's see what I mean:
[ 6 ]
$$ \Pi_1 = \rho_\infty^a v_\infty^b c^d R $$ $$ \Rightarrow m^0 l^0 t^0 = [m l^{-3}]^a [l t^{-1}]^b [l]^d [m l t^{-2}] $$ $$ = m^{a+1} l^{-3a+b+d+1} t^{-b-2} $$
What we did above was we set one side of our equation equal to $ m^0 l^0 t^0 $. The reason we did this is because we are solving for a unitless variable, hence, it will have no fundamental dimensions, so that is why they are all to the power of 0. Now on the right-hand side, we started by assigning a variable exponent to each parameter in the $ \Pi_1 $ equation and then substituted the fundamental dimensions of each parameter.
When substituted, we can then sort the exponents for each respective fundamental dimension. Notice that $ R $ has an exponent of 1 instead of a variable. This is because we are parametrizing $ R $, so we are fixing it to only have a possible power of 1. We will do this with each parameter that is not fixed. Now, you may be wondering, why do this? Because then we can make a system of equations to solve! Let's do it.
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$$ m: 0 = a+1 $$ $$ l: 0 = -3a+b+d+1 $$ $$ t: 0 = -b-2 $$
From this system, we can then solve it to find that
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$$ a = -1 $$ $$ b = -2 $$ $$ d = -2 $$
Now we can input our exponent value back into the first part of Eq. (6).
[ 9 ]
$$ \Pi_1 = \rho_\infty^{-1} v_\infty^{-2} c^{-2} R $$ $$ = \frac{R}{\rho_\infty v_\infty^2 c^2} $$
Now this equation looks familiar, right? It looks like the equation for $ C_R $, or the coefficient for the resultant force. Although it is missing one thing, the factor of $ \frac{1}{2} $ in the denominator. Now, since we are working with a unitless variable, we can actually multiply this factor of $ \frac{1}{2} $ without affecting the coefficient. I know it feels odd, but it will start to feel natural as we practice this more. Let's take the second part of Eq. (9) and make this change as well as replacing our $ c^2 $ with our surface area $ S $.
[ 10 ]
$$ \Pi_1 = \frac{R}{\frac{1}{2} \rho_\infty v_\infty^2 S} $$ $$ \Rightarrow \Pi_1 = \frac{R}{q_\infty S} $$
Hence, we can say
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$$ \Pi_1 = C_R = \frac{R}{q_\infty S} $$
It took some time, but we just derived the equation for the coefficient of the resultant force! Now, there are two more equations that we are going to derive.
We will now derive the $ \Pi_2 $ equation, but we will be a little bit speedier with the derivation. It is good practice to try the derivation for yourself before seeing the solution!
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$$ \Pi_2 = \rho_\infty^a v_\infty^b c^d \mu_\infty $$ $$ \Rightarrow m^0 l^0 t^0 = [m l^{-3}]^a [l t^{-1}]^b [l]^d [m l^{-1} t^{-1}] $$ $$ = m^{a+1} l^{-3a+b+d-1} t^{-b-1} $$
Now that we have our system of equations, let's solve for our exponents.
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$$ a = -1 $$ $$ b = -1 $$ $$ d = -1 $$
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$$ \Pi_2 = \rho_\infty^{-1} v_\infty^{-1} c^{-1} \mu_\infty $$ $$ = \frac{\mu_\infty}{\rho_\infty v_\infty c } $$
Now that we have an expression for $ \Pi_2 $, it looks familiar, right? This is because it is the inverse of the Reynolds number equation, and since our Reynolds number is unitless, we can take the inverse to write
[ 15 ]
$$ \Pi_2 = R_e = \frac{\rho_\infty v_\infty c }{\mu_\infty} $$
Now, one more equation left to derive!
For the third equation, this is the same process, except now we are parameterizing around $ f_3 $ from Eq. (5). Which unitless variable do you think will result? Can you try deriving it before looking on? I think you can!
After you've given it a try, let's groove on.
[ 16 ]
$$ \Pi_3 = \rho_\infty^a v_\infty^b c^d a_\infty $$ $$ \Rightarrow m^0 l^0 t^0 = [m l^{-3}]^a [l t^{-1}]^b [l]^d [l t^{-1}] $$ $$ = m^{a} l^{-3a+b+d+1} t^{-b-1} $$
Now, solving our system of equations we get
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$$ a = 0 $$ $$ b = -1 $$ $$ d = 0 $$
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$$ \Pi_3 = \rho_\infty^{0} v_\infty^{-1} c^{0} a_\infty $$ $$ = \frac{a_\infty}{v_\infty} $$
Now this one looks awfully like the Mach equation, but it is inversed. Again, we can flip inverse the equation because $ \Pi_3 $ is a unitless variable to get
[ 19 ]
$$ \Pi_3 = M_\infty = \frac{v_\infty}{a_\infty} $$
Congrats, you have now learned how to use the Buckingham Pi Theorem! What is interesting with this theorem is you do not always know what variable will be derived. Most are used today in all sorts of equations and theories, but there are many combinations that can be made for derivation.