Dimensional Analysis CUED 1A Lecture notes 4 Dimensionless forms of equations

Chapter 5 Applications of Dimensional Analysis

It is nearly impossible to cover all possible applications of dimensional analysis because of the fundamental nature of its underlying principles. Here we will consider two applications.

5.1 Non-dimensional presentation of experimental data

The scope of the results of an experiment can be greatly expanded with the application of dimensional analysis. Consider the example of thrust generated by an aircraft propeller. Identical analysis also applies to marine propellers.

Propeller thrust

It is intuitive to deduce that thrust, $f$ , generated by a propeller depends on its rotation rate, $\omega$ .

	Parameter:	$\displaystyle f$		depends on	$\displaystyle\left\{\right.\omega$	$\displaystyle\left.\right\}$		(5.1)
	Dimensions:	$\displaystyle\dfrac{ML}{T^{2}}$	$\displaystyle\left\{\right.T^{-1}$	$\displaystyle\left.\right\}$				(5.2)

However, this dependence must be incomplete because no dimensionless number canbe constructed from $f$ and $\omega$ . Nor can the dependence on $\omega$ be eliminated by applying the principle of isolated dimension because it violates our physical intuition. Hence, the thrust must depend on additional parameters. For example, intuitively it is evident that the thrust must depend on the size of the propeller, say its diameter $d$ .

The physics of propulsion with a propeller is as follows (consult a fluid dynamicist to clarify further, if needed). The surface of a propeller blade, as it spins, pushes the fluid back, and the force of reaction pushes the propeller forward. The amount of force needed to push the fluid back depends on its inertia, i.e. how heavy it is. This property is characterized by the fluid’s density. For aviation applications, the density of air is approximately 1.2 kg/m³. For fresh water, the density is 1000 kg/m³, while for sea water in the North Sea it is 1030 kg/m³, so they are quite similar.

The precise angle at which the propeller blade encounters the fluid depends on the speed of the boat and the linear speed of the various sections of the propeller. While the overall kinematics may be complex, it suffices to say that a change in either will influence the thrust. We know from general experience that a fan that spins faster generates a stronger flow. It is reasonable to consider that perhaps the same process operates in a propeller and the thrust depends on the rotation rate. It is also conceivable that the speed of the aircraft or the boat also influences the propeller thrust. This dependence is shown in Figure 5.1(a) using the data generated experimentally for one propeller by Johansson¹¹ 1 Johansson, Marcus (2025). “Experimental Analysis of Rotating Propeller Performance in a Wind Tunnel.” (Degree project), KTH, Stockholm, Sweden.. The data shows that, as the aircraft accelerates through the air, the thrust decreases. Based on this background, the thrust could be said to depend on $d$ , $\omega$ , $v$ and $\rho$ .

The following table shows the variables, their SI units and their dimensions in the MLT system.

Quantity	Symbol	SI Unit	Dimension
Thrust	$f$	N = kg m/s²	MLT^-2
Diameter	$d$	m	L
Angular velocity	$\omega$	rad/s	T^-1
Ferry speed	$v$	m/s	LT^-1
Fluid density	$\rho$	kg/m³	ML^-3

(The reader can try to work this example out in the FLT or the FLV or the FVT system, where F and V are the dimensions of force and velocity, respectively.) Note that radians are defined as the ratio of two lengths and, therefore, are dimensionless.

Figure 5.1: Thrust generated by the APC 11

\times

10E propeller from the work by Marcus Johansson. (a) Thrust versus advance speed for difference rotation rates. (b) The dimensionless thrust coefficient

C_{T}

versus the advance ratio

J

Advanced tip.

There is some flexibility in this choice. For designing a propeller, perhaps the thrust on the model could be specified but the rpm of the model propeller which generates the requisite thrust is to be determined using scale models. This changes whether the thrust is the dependent variable or the rotation rate. The reader is strongly encouraged to consider these possibilities independently.

Question 5.1.

The thrust generated by a propeller depends on the following parameters.

\displaystyle\begin{aligned} \text{Parameter: }&f&&\qquad&&\text{depends on}&&% \left\{\right.d,&&\omega,&&v,&&\rho&&\left.\right\}\\ \text{Dimensions: }&MLT^{-2}&&\qquad&&\text{}&&\left\{\right.L,&&T^{-1},&&LT^{% -1},&&ML^{-3}&&\left.\right\}\end{aligned}

(5.3)

Reduce this dependence to a dimensionless one.

Answer 5.1.

We shall use the recombination-elimination method. Only $f$ and $\rho$ contain the dimension M, so let us introduce $f/\rho$ instead of $f$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{f}{\rho}&&\qquad&&\text% {depends on}&&\left\{\right.d,&&\omega,&&v,&&\rho&&\left.\right\}\\ \text{Dimensions: }&L^{4}T^{-2}&&\qquad&&\text{}&&\left\{\right.L,&&T^{-1},&&% LT^{-1},&&ML^{-3}&&\left.\right\}\end{aligned}

(5.4)

Equation (5.4) shows that the dimension M appears isolated in the parameter $\rho$ and, therefore, the LHS may have no dependence on it. We can thus eliminate $\rho$ as a parameter.

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{f}{\rho}&&\qquad&&\text% {depends on}&&\left\{\right.d,&&\omega,&&v&&\left.\right\}\\ \text{Dimensions: }&L^{4}T^{-2}&&\qquad&&\text{}&&\left\{\right.L,&&T^{-1},&&% LT^{-1}&&\left.\right\}\end{aligned}

(5.5)

Next we seek to eliminate the dimension L and we shall make use of the parameter $d$ for this purpose. Let us construct $\dfrac{f}{\rho d^{4}}$ and $\dfrac{v}{d}$ instead of $\dfrac{f}{\rho}$ and $v$ . Doing so leads to

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{f}{\rho d^{4}}&&\qquad&% &\text{depends on}&&\left\{\right.d,&&\omega,&&\dfrac{v}{d}&&\left.\right\}\\ \text{Dimensions: }&T^{-2}&&\qquad&&\text{}&&\left\{\right.L,&&T^{-1},&&T^{-1}% &&\left.\right\}\end{aligned}

(5.6)

In equation (5.6), the dimension L is isolated in the parameter $d$ (by designing the previous recombination), and thus we can eliminate $d$ from the dependence.

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{f}{\rho d^{4}}&&\qquad&% &\text{depends on}&&\left\{\right.\omega,&&\dfrac{v}{d}&&\left.\right\}\\ \text{Dimensions: }&T^{-2}&&\qquad&&\text{}&&\left\{\right.T^{-1},&&T^{-1}&&% \left.\right\}\end{aligned}

(5.7)

The last iteration is now evident. Recombine the parameters to construct $\dfrac{f}{\rho d^{4}\omega^{2}}$ and $\dfrac{v}{\omega d}$ instead of $\dfrac{f}{\rho d^{4}}$ and $\dfrac{v}{d}$ . This yields

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{f}{\rho d^{4}\omega^{2}% }&&\qquad&&\text{depends on}&&\left\{\right.\omega,&&\dfrac{v}{d\omega}&&\left% .\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.T^{-1},&&1&&\left.\right% \}\end{aligned}

(5.8)

Doing so has isolated the dimension T in the parameter $\omega$ , which may now be eliminated, leading to the final dimensionless result.

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{f}{\rho d^{4}\omega^{2}% }&&\qquad&&\text{depends on}&&\left\{\right.\dfrac{v}{d\omega}&&\left.\right\}% \\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&&\left.\right\}\end{aligned}

(5.9)

The two dimensionless parameters thus identified are modified to match with convention in aerodynamics as follows. Define the thrust coefficient, $C_{T}$ , as

\displaystyle C_{T}=\dfrac{f}{\dfrac{1}{2}\rho\omega^{2}\left(\dfrac{d}{2}% \right)^{2}\left(\dfrac{\pi d^{2}}{4}\right)}=\dfrac{32}{\pi}\dfrac{f}{\rho% \omega^{2}d^{4}}

(5.10)

and the advance ratio, $J$ , as

\displaystyle J=\dfrac{v}{(\omega d/2)}=\dfrac{2v}{\omega d}.

(5.11)

Note that $C_{T}$ and $J$ differ from the dimensionless groups in (5.9) only by factors of constants, so are consistent with dimensional analysis. The thrust coefficient signifies the dimensionless thrust and the advance ratio the dimensionless boat or aircraft speed.

The experimental data is plotted in terms of these dimensionless variables, which leads Figure 5.1(b). Notice how the data for different $\omega$ and $v$ all collapse to a single curve.

This example illustrates how the reduction in number of parameters provided by dimensional analysis is utilized in presenting experimental data. One can readily see the redunduncy in the data, so engineers and scientists usually exploit dimensional analysis to reduce the number of experiments.

The following example on the dip in a hanging cable from Dr John P Longley’s Lecture Notes further illustrates the point. The data and the solution are also from Dr Longley’s notes.

Dip in a hanging cable

An inextensible, but otherwise flexible cable is suspended between two points at the same vertical elevation but a horizontal distance $l$ apart. The ability to predict the dip clearly has many applications, for example in constructing overhead power distribution lines on electrified railway tracks. The dip can be reduced by pulling on the cable with a force $f$ to stretch them taut as shown in Figure 5.2. How much does it dip at the midpoint for a given force $f$ ?

Figure 5.2: A schematic depiction of the setup showing a cable suspended between two fixed points at equal elevation and distance

l

apart, with an force

f

stretching it. The weight of the cable per unit length is

w

. The objective is to characterize the dip in its elevation in the middle.

Dr Longley designed an experiment with a cable that can be suspended between two supports mounted on a desktop, essentially reproducing the setup in Figure 5.2. By varying the tension force $f$ and measure the dip $d$ for a fixed span $l$ , the results he obtained are shown in Table 5.1 and plotted in Figure 5.3 below.

Tension force	Cable dip
$f$ (N)	$d$ (mm)
0.25	59
0.49	29
0.74	20
0.98	15
1.23	12

Table 5.1: Data from the experiment varying

f

and measuring

d

for a fixed

l

= 2.4 m and

w

= 0.02 N/m.

Figure 5.3: Data from Table 5.1 plotted as

f

versus

d

The data can be used to predict the dip for a given tension force, perhaps with a little interpolation if the force did not happen to coincide exactly with the values in the dataset. One may think that the results of this experiments are only useful to one case – the case of the cable used by Dr Longley, with the length spanning the desk. But Dr Longley has been teaching Dimensional Analysis since 1993! He readily anticipated that he could use this data to predict the dip in any inextensible but otherwise flexible cable using Dimensional Analysis.

Let us first determine the dependence in dimensionless form. To enable us to determine the parameters, we start with the hypothesized dependence

\displaystyle\begin{aligned} \text{Parameter: }&d&&\qquad&&\text{depends on}&&% \left\{\right.l,&&f&&\left.\right\}\\ \text{Dimensions: }&L&&\qquad&&\text{}&&\left\{\right.L,&&F&&\left.\right\}% \end{aligned}

(5.12)

where we have chosen the FLT system of base dimensions. Here, we immediately note that $f$ contains the isolated dimension F. Therefore, either $f$ must be eliminated or another parameter containing the dimension of F is missing and must be introduced. The experimental data shows the dependence on $f$ , so it cannot be eliminated, so we must be missing a parameter. Intuitively, we realize that for the same applied force, a heavy cable would dip more than a lighter cable. Afterall, the applied force acts to somehow counteract the force of gravity, so the weight of the cable must enter the dependence. Dr Longley introduces the equivalent parameter, the weight per unit length of the cable, $w$ , thus modifying the hypothesized dependency as

\displaystyle\begin{aligned} \text{Parameter: }&d&&\qquad&&\text{depends on}&&% \left\{\right.l,&&w,&&f&&\left.\right\}\\ \text{Dimensions: }&L&&\qquad&&\text{}&&\left\{\right.L,&&FL^{-1},&&F&&\left.% \right\}\end{aligned}

(5.13)

Afterall, if $w$ is known then the cable weight is $w l$ .

Let us quickly apply Buckingham’s Pi theorem 4.5 to determine the expected number of dimensionless parameters. Here $N=4$ and $M=2$ (because only F and L are independent dimensions representing the parameters), so we expect $K\geq 4-2=2$ independent dimensionless parameters.

We can first recombine $f$ and $w$ into $f/w$ and $f$ , so the modified parameters are

\displaystyle\begin{aligned} \text{Parameter: }&d&&\qquad&&\text{depends on}&&% \left\{\right.l,&&f/w,&&f&&\left.\right\}\\ \text{Dimensions: }&L&&\qquad&&\text{}&&\left\{\right.L,&&L,&&F&&\left.\right% \}\end{aligned}

(5.14)

and isolate the dimension $F$ . (Note that here we made the tacit decision to retain $f$ in the numerator so that the resulting dimensionless number can be interpreted as a dimensionless version of the stretching force.) The parameter with the isolated dimension can now be eliminated.

\displaystyle\begin{aligned} \text{Parameter: }&d&&\qquad&&\text{depends on}&&% \left\{\right.l,&&f/w&&\left.\right\}\\ \text{Dimensions: }&L&&\qquad&&\text{}&&\left\{\right.L,&&L&&\left.\right\}% \end{aligned}

(5.15)

We can then further recombine the remaining parameters into $d/l$ , $l$ and $f/(wl)$ to get

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{d}{l}&&\qquad&&\text{% depends on}&&\left\{\right.l,&&\dfrac{f}{wl}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.L,&&1&&\left.\right\}% \end{aligned}

(5.16)

Doing so isolates the dimension L in the parameter $l$ , which can now be eliminated. Finally, we are left with

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{d}{l}&&\qquad&&\text{% depends on}&&\left\{\right.\dfrac{f}{wl}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&&\left.\right\}\end{aligned}

(5.17)

and the functional dependence is

\displaystyle\dfrac{d}{l}=F\left(\dfrac{f}{wl}\right),

(5.18)

for some unspecified function $F(\cdot)$ . Equation (5.18) confirms the result from Buckingham’s Pi theorem that there are two remaining dimensionless parameters.

Let us now interpret the dimensionless parameters. The first one $d/l$ is simply the dip expressed as a fraction of the span. The second one $f/(wl)$ is nothing more than the stretching force written as a multiple of the cable weight. Intuitively, the amount of force needed to lift the cable must be related to how heavy it is, and that is what this dimensionless relation reflects.

We can now replot the data from the experiment (Table 5.1) in dimensionless terms. It is done in Fig. 5.4 and let us label this dataset as the Desktop experiment.

Figure 5.4: Dimensionless version of the data from the experiment presented in Table 5.1.

To demonstrate the value of presenting experimental data in dimensionless form, Dr Longley conducts another experiment using two different cables and across a different span.

	Weight per length	Dip in the cable	$\dfrac{d}{l}$	$\dfrac{f}{wl}$
	$w$ (N/m)	$d$ (mm)
Blue cable	0.52	210	0.024	5.2
White cable	0.10	40	0.004	27.2

Table 5.2: Data from two further experiments with two different cables.The new span for these two experiments is

l=

8.9 m. The tension force is

f=

24.2 N.

If the premise of this analysis holds, then the new experimental data should agree with the data plotted in Figure 5.4.

Figure 5.5: Comparison of the results of multiple experiments with the hanging cable. This graph includes data from Tables 5.1 and 5.2.

The astute reader can determine the success of Dr Longley’s application of Dimensional Analysis by glancing at Figure 5.5.

Can now do Examples Paper Q7.

5.2 Scale models

Building and testing scale models is a common approach in engineering to learn essential underlying principles and aid in design. The fundamental basis of scale models lies in dimensional analysis. It is dimensional analysis that enables testing of scaled models of Formula One automobiles and aircraft in wind tunnels. We already saw in §5.1 how experiments at different scales are related to each other through dimensional analysis. Here we take that observation one step further and design scale models of full-scale systems to test their performance and aid in design. For the purpose of scale modeling, understanding of two concepts is crucial: (i) geometric similarity, and (ii) dynamic similarity.

The following example illustrates the principles.

Question 5.2.

(from Dr John P Longley’s Lecture Notes)
A beam of length $l$ with a rectangular cross section (breadth $b$ , depth $d$ ), is simply supported on its two ends. A vertical force $f$ is applied to its midpoint to cause it to dip vertically downwards by $\delta$ . Determine the dimensionless dependence of $\delta$ on the independent variables. (The elastic modulus of the material of the beam is $E$ and has dimensions identical to pressure.)

Answer 5.2.

We start with the dimensional dependence as

\displaystyle\begin{aligned} \text{Parameter: }&\delta&&\qquad&&\text{depends % on}&&\left\{\right.f,&&E,&&l,&&d,&&b&&\left.\right\}\\ \text{Dimensions: }&L&&\qquad&&\text{}&&\left\{\right.F,&&FL^{-2},&&L,&&L,&&L&% &\left.\right\}\end{aligned}

(5.19)

Here we work in the FLT base system of dimensions because it reveals most clearly that there are only two independent dimensions (F and L). We can deduce by inspection that $f/El^{2}$ is dimensionless, and so are the ratios $\delta/l$ , $d/l$ and $b/l$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{\delta}{l}&&\qquad&&% \text{depends on}&&\left\{\right.\dfrac{f}{El^{2}},&&\dfrac{d}{l},&&\dfrac{b}{% l}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1,&&1,&&1&&\left.\right% \}\end{aligned}

(5.20)

The diligent student is asked to verify this result before proceeding.

5.2.1 Geometric similarity

A scale model is a replica of the the original in all respects except the overall size. Every dimension of the model is “scaled down” (or up, as the case may be) by a fixed factor. The concept of geometric similarity captures the shape of the object, while letting the overall size vary.

Principle 5.1.

(from Dr John P Longley’s Lecture Notes)
Two objects are geometrically similar when one object is geometrically an exact scale model of the other.

In Question 5.2, if a beam is a scale model of the original, then all their lengths would be reduced by the same factor, thereby leaving the dimensionless numbers $b/l$ and $d/l$ identical to the original. Therefore, for a given set of geometrically similar beams, the two geometric ratios are fixed, i.e. they do not vary. Then, the only variation in equation (5.20) arises from the applied force, leading to

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{\delta}{l}&&\qquad&&% \text{depends on}&&\left\{\right.\dfrac{f}{El^{2}}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&&\left.\right\}\end{aligned}

(5.21)

5.2.2 Dynamic similarity

The matching of geometry is necessary but not sufficient for a successful exercise of scale modeling. It is also necessary to have dynamic similarity for this purpose.

Principle 5.2.

(from Dr John P Longley’s Lecture Notes)
Two situations are dynamically similar when all independent dimensionless parameters have identical values. As a consequence, the value of the dependent dimensionless variable will also be identical.

The following question extends Question 5.2 to illustrate this principle.

Question 5.3.

(from Dr John P Longley’s Lecture Notes)
A one-tenth scale model of a beam is constructed from the same material as the original for the purposes of testing. If the original beam is expected to esperience a force of 10000 N, what force should be applied to the scale model to maintain dynamic similarity?

Answer 5.3.

The question statement says the scale model is one-tenth scale, i.e. each dimension is 1/10 of full-size model. This ensures geometric similarity. The only remaining independent dimensionless parameter is $\dfrac{f}{El^{2}}$ . We must ensure that this parameter has the same value in the scale model that it has for the full-size model.

\displaystyle\dfrac{f_{\text{fs}}}{El_{\text{fs}}^{2}}=\dfrac{f_{\text{sm}}}{% El_{\text{sm}}^{2}}.

(5.22)

Here the subscript fs stands for full-scale and sm stands for scale model. Since we know $l_{\text{sm}}=l_{\text{fs}}/10$ , equation (5.22) implies $f_{\text{sm}}=f_{\text{fs}}/100=100$ N.

5.2.3 A ship’s propeller

Dr John P Longley constructs the following example for the design of a ships propeller to illustrate the use of scale models in engineering design. The same principle also applies to an airplane propeller.

A manufacturer of a ship’s propellers has been commissioned to produce a propeller for a ferry which will operate in the North Sea. It has been determined independently that to cruise at a speed of 12 m/s, the ferry will need a thrust of 25 kN. The manufacturer has just produced a new but untried shape of propeller. For the ferry, the new propeller should have a diameter of 0.6 m and should rotate at 800 rpm.

In order to check the propeller design a 1/10 ${}^{\text{th}}$ scale model is to be built and tested. The model propeller (with diameter 0.06 m) will rotate at 8000 rpm and will be tested in fresh water. The model is to be tested under conditions which simulate the ferry cruising condition.

We need to know:

1.

the speed of the ferry, and
2.

the thrust that the model should produce.

Dr Longley suggests that we use (5.9) and exploit dynamic similarity.

The dimensionless analysis has identified two dimensionless parameters and it is a sound practice to interpret them for developing intuition. The dependent parameter is $\dfrac{f}{\rho d^{4}\omega^{2}}$ , which consists of $f$ , $\rho$ , $d$ and $\omega$ . Dynamic similarity between propellers of two scales implies that

\displaystyle\begin{aligned} \left(\dfrac{f}{\rho\omega^{2}d^{4}}\right)_{% \text{sm}}=\left(\dfrac{f}{\rho\omega^{2}d^{4}}\right)_{\text{fs}}\qquad% \Longrightarrow\qquad f_{\text{sm}}&=f_{\text{fs}}\times\dfrac{\rho_{\text{sm}% }}{\rho_{\text{fs}}}\times\left(\dfrac{d_{\text{sm}}}{d_{\text{fs}}}\right)^{4% }\times\left(\dfrac{\omega_{\text{sm}}}{\omega_{\text{fs}}}\right)^{2}\end{aligned}

(5.23)

It follows from (5.23) and stands to reason that a propeller with a larger diameter (greater $d$ ) generates more thrust, a propeller rotating faster (greater $\omega$ ) generates more thrust, and a propeller pushing a heavier fluid (greater $\rho$ ) generates more thrust.

Similarly, the independent parameter $v/(\omega d)$ is interpreted as the speed of the propeller tip ( $\omega d$ ) relative to the boat speed ( $v$ ). It represents the distance the boat (and with it the propeller) moves forward within one rotation of the propeller, and evokes the image of a screw. The mechanism of the propeller can now be interpreted as that of a screw pushing through a fluid. It is for this reason that a propeller is also colloquially referred to as a screw in nautical and aeronautical circles.

We now insist on dynamically similarity between the two scales.

The independent parameters

There is only one independent parameter, viz. $\dfrac{v}{\omega d}$ . Matching the parameter on the two scales implies

\displaystyle\left(\dfrac{v}{\omega d}\right)_{\text{fs}}=\left(\dfrac{v}{% \omega d}\right)_{\text{sm}}\qquad\Longrightarrow\qquad v_{\text{sm}}=v_{\text% {fs}}\times\dfrac{\omega_{\text{sm}}}{\omega_{\text{fs}}}\times\dfrac{d_{\text% {sm}}}{d_{\text{fs}}}=12\text{ m/s}\times\dfrac{8000}{800}\times\dfrac{0.06% \text{ m}}{0.6\text{ m}}=12\text{ m/s}.

(5.24)

The model ferry should travel at 12 m/s.

The dependent parameter

Matching the dependent parameter between the two scales implies from (5.23)

\displaystyle\begin{aligned} f_{\text{sm}}=25000\text{ N}\times\dfrac{1000}{10% 30}\times\left(\dfrac{0.06}{0.6}\right)^{4}\times\left(\dfrac{8000}{800}\right% )^{2}=242.7\text{ N}.\end{aligned}

(5.25)

5.3 Conclusion

Geometric and dynamic similarity between different scales of the same system allows us to relate their behavior to each other using dimensional analysis. We saw in this chapter how to exploit this principle. Let us conclude with the following maxim by Dr John P Longley:

Principle 5.3.

It is a general principle that the deeper the physical understanding at the start of the problem, the more likely that dimensional analysis will yield a significant result.