Dimensional Analysis CUED 1A Lecture notes 3 Dimensions 5 Applications of Dimensional Analysis

Chapter 4 Dimensionless forms of equations

Many problems in engineering and physics, which involve equations relating one dependent quantity on several parameters, can be simplified using dimensional analysis, whether or not the precise form of the relationship is known. In most case, the simplification is achieved by expressing the problem in dimensionless terms. This core technique forms the basis of the applications of dimensional analysis in engineering and is explained in this chapter.

4.1 Physical relations between dimensionless parameters

A primitive element of dimensional analysis is the dependence of a quantity on one or more parameters of the problem. The relationship can be specified, such as in the example of an uniformly accelerating car. Consider a car moving at a velocity $u$ , which is then accelerated at a constant rate $a$ . We require the velocity $v$ of the car after a time $t$ , which is given by

\displaystyle v=u+at.

(4.1)

We know this equation from first principles and we can use such examples to glean the principles, which we can apply in situation where the form of the equation is unknown. When it is unknown, we will write such a dependence symbolically as

	Parameter:	$\displaystyle v$		depends on	$\displaystyle\left\{\right.u,$	$\displaystyle a,$	$\displaystyle t$	$\displaystyle\left.\right\}$
	Dimensions:	$\displaystyle LT^{-1}$	$\displaystyle\left\{\right.LT^{-1},$	$\displaystyle LT^{-2},$	$\displaystyle T$	$\displaystyle\left.\right\}$

which stands for $v=f(u,a,t)$ for some unspecified function $f()$ . The second line of the equation lists the dimensions of the quantities above them.

We now state the consequence of dimensional consistency on physical laws.

Principle 4.1.

In a complete statement of a physical law, it is possible to rearrange the terms so that all groups or quantities are dimensionless.

Question 1.5 is an example of this consequence. This result follows directly from the principles 3.2 in §3.5 and principle 3.3 in §3.6. We will further explore this consequence in this section.

4.2 Non-dimensionalizing equations

Here are a few more example.

Question 4.1.

(from Dr John P Longley’s Lecture Notes)
Consider a car moving at a (non-zero) velocity $u$ which then accelerates at a constant rate $a$ . The velocity after time $t$ is $v=u+at$ . Write this in a dimensionless form.

Answer 4.1.

Let us begin by identifying the dimensions of the terms in the equations.

\displaystyle\begin{aligned} &\text{Equation}:\qquad&&v&&=&&u&&+&&at&\\ &\text{Dimensions}:\qquad&&\dfrac{L}{T}&&&&\dfrac{L}{T}&&&&\dfrac{L}{T^{2}}% \times T&\end{aligned}

Let us divide this equation by $u$ to get

\displaystyle\dfrac{v}{u}=1+\dfrac{at}{u}.

Both $v/u$ and $at/u$ are dimensionless (as is the number 1). This equation is equivalent to the original equation and consists only of dimensionless quantities.

Question 4.2.

The range of a projectile shot with speed $u$ at an angle $\theta$ on flat ground is

\displaystyle d=\dfrac{u^{2}\sin 2\theta}{g}.

Express this equation using dimensionless numbers.

Answer 4.2.

Dividing by $d$ yields

\dfrac{u^{2}}{gd}\sin 2\theta=1.

Here $u^{2}/(gd)$ is dimensionless, so is $\sin 2\theta$ . Note that an equally acceptable expression is

\dfrac{gd}{u^{2}}=\sin 2\theta.

Any function of dimensionless numbers is also dimensionless.

Question 4.3.

The trajectory of a projectile is given by the relations

\displaystyle x=ut,\qquad y=vt-\dfrac{1}{2}gt^{2},

(4.2)

where $u$ and $v$ are the initial components of velocity in the $x$ and $y$ directions, respectively, and $t$ is the time elapsed since launch. This equation holds until the projectile hits the ground, say, at $y=0$ . Express it in dimensionless terms.

Answer 4.3.

The first instinct is to divide the $x$ equation by $u t$ and the $y$ equation by $v t$ , and doing so does yield a dimensionless form of the equation

\dfrac{x}{ut}=1,\qquad\dfrac{y}{vt}=1-\dfrac{gt}{2v}.

(4.3)

Here, the dimensionless parameters are $x/(ut)$ , $y/(vt)$ and $gt/(2v)$ . However, sometimes, we wish to retain certain properties of the relation, for example in this case the linear and quadratic dependence on time. So consider dividing the $x$ equation by $uv/g$ (inspired by Question 4.2) and the $y$ equation by $v^{2}/g$ (because that is related to the maximum height reached by the projectile). Both $uv/g$ and $v^{2}/g$ have dimensions of $L$ . This yields

\dfrac{gx}{uv}=\dfrac{gt}{v},\qquad\dfrac{gy}{v^{2}}=\dfrac{gt}{v}-\dfrac{1}{2% }\left(\dfrac{gt}{v}\right)^{2}.

(4.4)

This equation can be expressed in terms of dimensionless variables ${\tilde{{x}}}=gx/(uv)$ , ${\tilde{{y}}}=gy/(v^{2})$ and ${\tilde{{t}}}=gt/v$ in the form

{\tilde{{x}}}=t,\qquad{\tilde{{y}}}={\tilde{{t}}}-\dfrac{1}{2}{\tilde{{t}}}^{2}.

(4.5)

We will discuss an advantage of (4.5) over (4.3) in the §4.3. Here ${\tilde{{x}}}$ , ${\tilde{{y}}}$ and ${\tilde{{t}}}$ are considered to be the dimensionless versions of $x$ , $y$ and $t$ , even though they also contain $g$ , $u$ and $v$ .

Question 4.3 shows that constructing dimensionless forms of equations may not be a matter of simply dividing by one of the terms. And that there is no unique way to non-dimensionalize an equation. While many ways of non-dimensionalizing are technically correct and algebraically equivalent, some are more useful than others, because they reflect the user’s interpretation of the equations. We will see more examples of this in §4.3.

4.3 Non-dimensionalizing charts

(The accelerating car example in this section and the solution upto and including Principle 4.2 is from Dr John P Longley’s Lecture Notes.)

It is possible that a relationship between variables cannot be expressed in terms of nice equations but is available in the form of charts. For our accelerating car example, we can plot $v$ as a function of time $t$ for various $u$ and $a$ . The velocity of a car as it accelerates starting from three different initial velocities and four different accelerations is shown in Figure 4.1.

Figure 4.1: The velocity of a car as a function of time starting with speed

u

and accelerating at a rate

a

. The three panels show three initial speeds (a)

u=10\text{ m/s}

, (b)

u=20\text{ m/s}

, and (c)

u=30\text{ m/s}

. Legend in panel (a) shows the acceleration

a

for each curve (the same legend applies to panels b and c).

For example the graphs in Figure 4.1(a) represents town driving with an initial velocity of 10 m/s and perhaps 4.1(c) represents motorway driving with $u=30\text{ m/s}$ . To present every possible initial velocity $u$ (i.e. driving condition) would require an entire book of graphs!

As a first step, consider plotting the ratio $v/u$ as a function of time, as is presented in Figure 4.2.

Figure 4.2: Same as Figure 4.1 but with

v/u

plotted instead of

u

. The legend, which shows the value of

a/u

, is now different for each panel.

When plotted this way, since initially $v=u$ , all the graphs have an initial condition equal to 1. And the slope of the lines correspond to the ratio $a/u$ , which has dimensions of $T^{-1}$ . Carefully observe the panels of Figure 4.2, and note that the graph corresponding to $a/u=0.5\text{ 1/s}$ is the same in all three panels. It is so because the ratio $v/u$ satisfies

\dfrac{v}{u}=1+\dfrac{a}{u}t,

so that the slope of the lines is the value of $a/u$ . This means that the curves in the three panels in Figure 4.2 are indeed identical for identical values of the parameter $a/u$ , even if $a$ and $u$ may be different. Therefore, they can all be combined in a single graph, where different driving conditions are distinguished by different values of $a/u$ . This is done in Figure 4.3(a).

Figure 4.3: Single graphs that depict the velocity of an accelerating car. (a) The three panels of Figure 4.2 combined into one. The curves correspond to

a/u

= 0, 0.17, 0.25, 0.33, 0.5, 0.75, 1.0, 1.5 1/s. (b) The data from the curves in (a) plotted as a function of

at/u

. (c) The same data as in (a) but with

at/v

plotted against

at/u

Note how rescaling the variables collapsed a family of curves on top of each other making them identical. This raises the question whether there is a way to further collapse the family of curves in Figure 4.3(a) into a single curve. After all, they have a similar fundamental shape. A moments thought will convince you that they can be reduced to a single curve if a different horizontal scale is used for each curve, i.e. a different time axis. One way of achieving this is by plotting $v/u$ against $at/u$ . Figure 4.3(b) shows this graph, which collapses all the curves into a single curve, resulting in a graph that contains all driving conditions. Both axes are now dimensionless and correspond to numbers identified earlier.

This example provides the first real indication of the power of Dimensional Analysis. It suggests that

Principle 4.2.

The number of parameters in a problem is reduced by expressing the relationship in dimensionless form.

However, note that $v/u$ versus $at/u$ is not the only non-dimensionalization that will collapse all the curves. Figure 4.3(c) shows that plotting $at/v$ versus $at/u$ also collapses all the data on a single curve. This curve represents the relationship

\dfrac{at}{v}=\begin{cases}\frac{at}{u}&t<0\\ \dfrac{\frac{at}{u}}{\frac{at}{u}+1}&t>0\end{cases}.

(4.6)

There are in fact an infinite number of combinations that will collapse all the data. Any two independent combinations of the two dimensionless numbers $v/u$ and $at/u$ collapses the data. We will see that this non-uniqueness is a general property of non-dimensionalization of equations.

Purely from algebraic considerations, the collapse of data in Figure 4.3(b) contains the same data as in Figure 4.3(c). There is no algebraic reason to prefer one over the other. The two representations are algebraically equivalent. However, as humans, we prefer the representation in Figure 4.3(b). It is so because we implicitly treat the velocity of the accelerating car as primarily a function of time, and consider $u$ and $a$ as parameters. It was for this reason that we plotted time on the horizontal axis in Figure 4.1 and the different curves correspond to different $u$ and $a$ .

Therefore, it provides us greater insight when the quantity on the vertical axis is proportional to the dependent quantity of interest $v$ , and the one on the horizontal axis is proportional to $t$ . In this representation, it can be readily inferred that $v$ is a constant for $t<0$ and increases linearly with time for $t>0$ , which serves our intuitive understanding of the dynamics. Such insight is not available from the representation in Figure 4.3(c). And for such reasons of human insight, certain non-dimensionalizations are preferred over others. A similar observation is made in Question 4.3.

Principle 4.3.

While the dimensionless parameters representing a problem are not unique, every representation is algebraically equivalent to every other. However, certain representations aid in providing insight into the problem.

Question 4.4.

The height of a projectile launched vertically upwards with initial speed $v$ from ground level ( $y=0$ ) is given by

y=vt-\dfrac{1}{2}gt^{2},

(4.7)

where $t$ is the time since the launch.

Figure 4.4: The height reached by a projectile launched vertically upwards as a function of time. (a) Varying

v

for a fixed

g

, and (b) Varying

g

for a fixed

v

Figure 4.4 shows the height as a function of time for various $v$ and $g$ . Generate a dimensionless version of this graph and, if possible, collapse all these curves to a single master curve.

Answer 4.4.

If equation 4.7 is non-dimensionalized by dividing by $v t$ , we get

\dfrac{y}{vt}=1-\dfrac{gt}{2v},

(4.8)

which suggests plotting $y/(vt)$ against $gt/v$ . This is presented in Figure 4.5(a). All the curves in Figure 4.4 collapse to a single curve, a straight line. It is, however, unclear how to interpret this straight line. Instead consider the non-dimensionalization obtained from dividing equation (4.7) by $v^{2}/g$ , which yields

\dfrac{gy}{v^{2}}=\dfrac{gt}{v}-\dfrac{1}{2}\left(\dfrac{gt}{v}\right)^{2}.

(4.9)

This leads to plotting ${\tilde{{y}}}=2gy/v^{2}$ versus ${\tilde{{t}}}=gt/(2v)$ (the factors of two are included for better interpretation), which also collapses all the curves of Figure 4.4 into a single master curve. This curve is a parabola, which starts at $(0,0)$ and returns to $y=0$ at $gt/(2v)=1$ , after reaching a maximum height of $2gy/v^{2}=1$ halfway between. Thus, this curve can be interpreted as plotting the fraction of the maximum height reached by the projectile as a function of the fraction of its flight duration!

Figure 4.5: The trajectory of a vertically launched projectile plotted in terms of dimensionless variables. (a) Non-dimensionalization based on equation (4.7).

4.4 Dimensionless relationships

(The example of the falling column is from Dr John P Longley’s Lecture Notes, where it was used to illustrate the principle of isolated dimension.)

Dimensional analysis is useful in reducing the complexity of the problem by reducing the number of independent variables whether or not the the relationship between those variables is known before-hand either in the form of charts or equations. The principle of the isolated dimension is instrumental in this process, so let us first learn it.

4.4.1 The principle of the isolated dimension

To illustrate this principle, consider the time taken for a column of height $h$ with mass $m$ to fall over from the vertical to the horizontal under the gravitational acceleration $g$ . We write this relationship as

\displaystyle\begin{aligned} \text{Parameter: }&t&&\qquad&&\text{depends on}&&% \left\{\right.h,&&g,&&m&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.L,&&LT^{-2},&&M&&\left.% \right\}\end{aligned}

(4.10)

The dependent quantity $t$ has dimensions of $T$ , which are independent of $M$ , and of the independent quantities, $m$ includes dimensions of $M$ , which none of the other independent parameters do. No combination of $t$ , $g$ and $h$ can be combined with $m$ to make a dimensionless group. Hence, because the functional relationship must be expressible in terms of dimensionless parameters alone, we must discard $m$ from the relationship.

Thus, we really have

\displaystyle\begin{aligned} \text{Parameter: }&t&&\qquad&&\text{depends on}&&% \left\{\right.h,&&g&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.L,&&LT^{-2}&&\left.% \right\}\end{aligned}

(4.11)

Principle 4.4.

Isolated dimension: When an independent variable contains an isolated dimension, which cannot be combined with any other parameter to render it dimensionless, then the dependence on that variable must be dropped as a matter of dimensional consistency.

Advanced tip.

In this example, we discarded the dependence on $m$ as a redundant variable on the basis of dimensional consistency. While this conclusion is justified in this case, in general this may not be the only logical conclusion. In more complex problems, it may be argued that rather than remove a variable to ensure dimensional consistency, another variable (with appropriate dimensions) may be needed to obtain a complete statement of the physical law. In other words, the variable is perhaps not really isolated and the analysis contains an as of yet unidentified parameter(s) with overlapping dimensions.

This possibility leads to an important point that often arises in the application of dimensional analysis. How does one determine all the independent parameters that the dependent variable depends on? In more complex engineering situations, identifying the independent parameters is not obvious or easy. In fact, that is the most non-trivial step in the application of dimensional analysis, and a user’s physical intuition is perhaps the strongest tool at their disposal for this purpose. The following question demonstrates this idea.

As an example, consider how one would identify the parameters on which the frequency of a stretched string depends. One might approach it purely empirically, and simply vary all the parameters one can think of. In this list should be the mass, diameter, length and elastic properties of the string, the tension in the string. Perhaps one might also suspect that gravitational acceleration and the thermal conductivity of the wire material are also relevant. In such an instance, the thermal conductivity contains the isolated dimension of temperature, and one would need to either eliminate it or identify another parameter with the dimensions of temperature to include in the analysis.

Let us pretend that we do not know the physics behind the vibrations of a stretched string.

Question 4.5.

Identify the parameters on which the natural frequency of oscillation of a stretched string depends. Note that the discussion in this question is quite subjective and is intended to expose the reader to the complexity of the process.

Answer 4.5.

One might approach the process purely empirically, and simply design experiments to vary all the parameters one can think of. In this list should be the mass, diameter, length and elastic properties of the string, the tension in the string. Perhaps one might also suspect that gravitational acceleration and the thermal conductivity of the wire material are also relevant. In such an instance, the thermal conductivity contains the isolated dimension of temperature, and one would need to either eliminate it or identify another parameter with the dimensions of temperature to include in the analysis.

There are a number of difficulties with this approach.

1.

The experiments may be cumbersome to construct and conduct. It might be difficult to isolate only one parameter to vary at a time, while holding all others fixed so that it can be determined whether that parameter enters the dependence. For example, the mass of the string could be varied by changing the material of the string, but that also changes the elastic properties of the string.
2.

Even if one succeeds in varying a single parameter at a time, experimental errors may creep in. Thus, it may not be possible to distinguish the possibility of a genuine dependence from that of experimental errors. This can perhaps be resolved in a statistical manner by conducting a large number of experiments.

Another approach is to develop a conceptual model that connects the independent parameters to the dependent one via a hypothesized physical process. In the case of the stretched string, one can invoke an analogy with a simple harmonic oscillator. The tension in the string is responsible for the spring of the oscillator and provides the restoring force, while the mass of the string represents the mass of the oscillator.

\displaystyle\begin{aligned} \text{Parameter: }&f&&\qquad&&\text{depends on}&&% \left\{\right.\tau,&&m&&\left.\right\}\\ \text{Dimensions: }&T^{-1}&&\qquad&&\text{}&&\left\{\right.MLT^{-2},&&M&&\left% .\right\}\end{aligned}

(4.12)

where, as before, $\tau$ is the tension in the string, $m$ is its mass, and $f$ it’s natural frequency of oscillations. Recognizing that $L$ is an isolated dimension in $\tau$ , the length $l$ is included as an independent parameter as

\displaystyle\begin{aligned} \text{Parameter: }&f&&\qquad&&\text{depends on}&&% \left\{\right.\tau,&&m,&&l&&\left.\right\}\\ \text{Dimensions: }&T^{-1}&&\qquad&&\text{}&&\left\{\right.MLT^{-2},&&M,&&L&&% \left.\right\}\end{aligned}

(4.13)

In this manner, a combination of dimensional analysis and physical intuition is used to construct the parameter list.

Interested students who wishes to sharpen their skills in constructing parameter dependence lists may consider the following systems as challenging examples: (i) The power generated by a wind turbine, (ii) the rotation rate of a Crookes radiometer.

4.4.2 Recombination of parameters

We can now use the principle of isolated dimension to further simplify the dependence in Equation (4.10). Let us undertake the following rearrangement:

\displaystyle\begin{aligned} \text{Parameter: }&t&&\qquad&&\text{depends on}&&% \left\{\right.h,&&g/h&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.L,&&T^{-2}&&\left.\right% \}\end{aligned}

(4.14)

Equation (4.14) represents the same dependence as in (4.11) in terms of $h$ and $g/h$ instead of $h$ and $g$ . Note that since we have retained independent dependence on $h$ , the value of $g$ can be determined from the ratio $g/h$ . Knowledge of $h$ and $g/h$ implies the knowledge of $h$ and $g$ , so no information is lost by this rearrangement.

4.4.3 Iterating to eliminate dimensions

Amongst the many possible recombinations (e.g. $g/h^{2}$ , $h/g$ , $h^{2}/g^{3}$ , etc.), note that Equation (4.14) has chosen a recombination which isolates the dimension $L$ . Applying the principle of isolated dimension, we conclude that $t$ cannot depend on $h$ separately. That is,

\displaystyle\begin{aligned} \text{Parameter: }&t&&\qquad&&\text{depends on}&&% \left\{\right.g/h&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.T^{-2}&&\left.\right\}% \end{aligned}

(4.15)

We now continue with the recombination with the remaining variables by replacing the dependent parameter $t$ with $t\sqrt{\dfrac{g}{h}}$ . The dependent variable is not exempt from recombination. Applying the same logic as in §4.4.2, knowledge of $t\sqrt{\dfrac{g}{h}}$ and $\sqrt{\dfrac{g}{h}}$ implies knowledge of $t$ .

\displaystyle\begin{aligned} \text{Parameter: }&t\sqrt{\dfrac{g}{h}}&&\qquad&&% \text{depends on}&&\left\{\right.g/h&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.T^{-2}&&\left.\right\}% \end{aligned}

(4.16)

The objective of this recombination is to eliminate the dimension of $T$ from the dependent variable. This allows application of isolated dimension to conclude that the recombined dependent variable cannot depend on the remaining independent parameter. That is

\displaystyle\begin{aligned} \text{Parameter: }&t\sqrt{\dfrac{g}{h}}&&\qquad&&% \text{depends on}&&\left\{\right.1&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&&\left.\right\}\end{aligned}

(4.17)

In this way, iterative application of recombination to isolate dimensions and then eliminate the isolated parameter reduces the number of independent parameters by rendering the dependence dimensionless. In this example, we conclude that $t\sqrt{\dfrac{g}{h}}$ is a constant $C$ for all vertical columns

\displaystyle t\sqrt{\dfrac{g}{h}}=C.

(4.18)

4.5 More examples

To reinforce the concepts, let us consider a few more examples. The first we consider is the accelerating car, which appeared in Dr John P Longley’s Lecture Notes.

Question 4.6.

Non-dimensionalize the dependence

\displaystyle\begin{aligned} \text{Parameter: }&v&&\qquad&&\text{depends on}&&% \left\{\right.u,&&a,&&t&&\left.\right\}\\ \text{Dimensions: }&LT^{-1}&&\qquad&&\text{}&&\left\{\right.LT^{-1},&&LT^{-2},% &&T&&\left.\right\}\end{aligned}

(4.19)

Answer 4.6.

We begin by noticing that neither of the two dimensions $L$ and $T$ in the dependence are isolated, thus we must start by recombining the parameters to create isolated dimensions. Let us isolate the dimension of $L$ by replacing $v$ by $v/u$ and $a$ by $a/u$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{v}{u}&&\qquad&&\text{% depends on}&&\left\{\right.u,&&\dfrac{a}{u},&&t&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.LT^{-1},&&T^{-1},&&T&&% \left.\right\}\end{aligned}

(4.20)

Now that the dimension $L$ is isolated, we can eliminate the dependence on $u$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{v}{u}&&\qquad&&\text{% depends on}&&\left\{\right.\dfrac{a}{u},&&t&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.T^{-1},&&T&&\left.\right% \}\end{aligned}

(4.21)

Next we recombine with the intention of isolating the dimension $T$ by replacing $\dfrac{a}{u}$ by $\dfrac{at}{u}$ . Note here that we can also use $\dfrac{u}{at}$ , and the result will be algebraically indistinguishable. However, to aid in intuition, we wish to construct our parameters such that they are proportional to $t$ . Hence we prefer $\dfrac{at}{u}$ over $\dfrac{u}{at}$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{v}{u}&&\qquad&&\text{% depends on}&&\left\{\right.\dfrac{at}{u},&&t&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1,&&T&&\left.\right\}% \end{aligned}

(4.22)

Now that we have isolated the dimension $T$ , we can eliminate the dependence on the variable $t$ to yield

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{v}{u}&&\qquad&&\text{% depends on}&&\left\{\right.\dfrac{at}{u}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&&\left.\right\}\end{aligned}

(4.23)

No further dimensions remain in this dependence to be eliminated. Hence, we conclude that $\dfrac{v}{u}=f\left(\dfrac{at}{u}\right)$ for some unspecified function $f$ . What we have not obtained in the exact form of the relationship between $v/u$ and $at/u$ . However, the dimensionless relationship, $v/u$ depends on only $at/u$ , implies that it must be a single curve on a single graph. Thus we can determine everything about the accelerating car problem after a single experiment that produces the fundamental dimensionless graph.

Of course, we know from Question 4.1 that the functional form is $\dfrac{v}{u}=1+\left(\dfrac{at}{u}\right)$ . The power of dimensional analysis is that we did not have to know this form a priori to simplify the dependence.

The role of dimensional analysis underneath the treatment of Questions 1.1-1.4 should now be transparent. Revisit those questions as a practice. We will treat Question 1.5, which appeared in Dr John P Longley’s Lecture Notes, using methods from this chapter.

Question 4.7.

Reduce the dependence from Question 1.5

\displaystyle\begin{aligned} \text{Parameter: }&v_{3}&&\qquad&&\text{depends % on}&&\left\{\right.m_{1},&&m_{2}&&v_{1}&&v_{2}&&\left.\right\}\\ \text{Dimensions: }&LT^{-1}&&\qquad&&\text{}&&\left\{\right.M,&&M&&LT^{-1}&&LT% ^{-1}&&\left.\right\}\end{aligned}

(4.24)

to its dimensionless form.

Answer 4.7.

Let us construct three recombined parameters $m_{2}/m_{1}$ , $v_{2}/v_{1}$ and $v_{3}/v_{1}$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{v_{3}}{v_{1}}&&\qquad&&% \text{depends on}&&\left\{\right.m_{1},&&\dfrac{m_{2}}{m_{1}}&&v_{1}&&\dfrac{v% _{2}}{v_{1}}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.M,&&1&&LT^{-1}&&1&&\left% .\right\}\end{aligned}

(4.25)

Both $M$ and $L/T$ are now isolated dimensions and they may be eliminated for dimensional consistency. However, note a peculiarity in this dependency. Clearly, $L$ is an isolated dimension because it only occurs in $v_{1}$ and no other parameters in equation (4.25). But once $v_{1}$ is eliminated on account of $L$ being isolated, then $T$ is eliminated along with it. This double-elimination occurred because $L$ and $T$ did not appear independently in the parameters, but only in the fixed combination $L/T$ . This leads to the dependence

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{v_{3}}{v_{1}}&&\qquad&&% \text{depends on}&&\left\{\right.\dfrac{m_{2}}{m_{1}}&\dfrac{v_{2}}{v_{1}}&&% \left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&1&&\left.\right\}\end{aligned}

(4.26)

We arrive at the same conclusion as of Question 1.5, but without knowing the underlying form of the relationship.

Let us look at the case of the vertical projectile from Question 4.4 to see the complexity involved in the various choices in forming dimensionless parameters that are algebraically equivalent.

Question 4.8.

The instantaneous height of a projectile launched vertically upwards depends on the launch velocity, gravitational acceleration and time since launch as

\displaystyle\begin{aligned} \text{Parameter: }&y&&\qquad&&\text{depends on}&&% \left\{\right.v,&&g&&t&&\left.\right\}\\ \text{Dimensions: }&L&&\qquad&&\text{}&&\left\{\right.LT^{-1},&&LT^{-2}&&T&&% \left.\right\}\end{aligned}

(4.27)

Deduce the dimensionless relationship between these parameters, with the intuition that we seek $y$ to be primarily a function of $t$ .

Answer 4.8.

We will explicitly refrain from (i) using $t$ in any recombination with $y$ , and (ii) retain $t$ in the numerator of all recombinations involving it. Therefore, that leaves us with either $v$ or $g$ to combine with $y$ to eliminate the dimension of $L$ . Let us pick $v$ (an we leave it as an exercise to the reader to pick $g$ ). Similarly, let us construct $v/g$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{y}{v}&&\qquad&&\text{% depends on}&&\left\{\right.\dfrac{v}{g},&&g&&t&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.T,&&LT^{-2}&&T&&\left.% \right\}\end{aligned}

(4.28)

Now $L$ is an isolated dimension in the parameter $g$ , so that dependence can be dropped.

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{y}{v}&&\qquad&&\text{% depends on}&&\left\{\right.\dfrac{v}{g},&&t&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.T,&&T&&\left.\right\}% \end{aligned}

(4.29)

Since we cannot use $t$ to remove the dimension of $T$ from $y$ , we will use $v/g$ .

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{gy}{v^{2}}&&\qquad&&% \text{depends on}&&\left\{\right.\dfrac{v}{g},&&\dfrac{gt}{v}&&\left.\right\}% \\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.T,&&1&&\left.\right\}% \end{aligned}

(4.30)

The dimension of $T$ is now isolated in $v/g$ , so that dependence can be dropped. We are then left with the dimensionless relationship

\displaystyle\begin{aligned} \text{Parameter: }&\dfrac{gy}{v^{2}}&&\qquad&&% \text{depends on}&&\left\{\right.\dfrac{gt}{v}&&\left.\right\}\\ \text{Dimensions: }&1&&\qquad&&\text{}&&\left\{\right.1&&\left.\right\}\end{aligned}

(4.31)

leading to $\dfrac{gy}{v^{2}}=f\left(\dfrac{gt}{v}\right)$ , where $f(\cdot)$ is an unspecified function. We thus recover the conclusion of Question 4.4 without knowing the explicit form of the dependence.

4.6 Buckingham’s Pi theorem

Buckingham’s Pi theorem is named so because Buckingham denoted dimensionless groups of parameters by the Greek symbol for capital pi $\Pi$ . The theorem sets an expectation for the number of dependent and independent dimensionless groups that make up a dimensionless form of a relationship based on the postulated dimensional form.

Principle 4.5.

Let us assume we have $N$ variables in circumstances where the statement

\displaystyle\begin{aligned} \text{Parameter: }&p_{1}&&\qquad&&\text{depends % on}&&\left\{\right.p_{2},&&p_{3},&&p_{4},&&\dots&&p_{N}&&\left.\right\},\end{aligned}

(4.32)

expresses a complete relationship between the $p_{1}$ , $p_{2},$ … $p_{N}$ , which require $M$ dimensions. According to principle of dimensional consistency, these can be expressed in dimensionless terms alone. Thus, equation (4.32) is equivalent to

\displaystyle\begin{aligned} \text{Parameter: }&\Pi_{1}&&\qquad&&\text{depends% on}&&\left\{\right.\Pi_{2},&&\Pi_{3},&&\Pi_{4},&&\dots&&\Pi_{K}&&\left.\right% \},\end{aligned}

(4.33)

where there are $K\leq N$ dimensionless groups denoted by $\Pi_{1}$ , $\Pi_{2}$ , $\Pi_{3}$ , … $\Pi_{K}$ . Then

\displaystyle K\geq N-M.

(4.34)

This result is not unexpected once we understand the method of elimination to non-dimensionalize dependencies. Here we present the proof, which appeared in Dr John Longley’s Lecture Notes. In the production of the dimensionless groups, if a variable is discarded each time a dimension is eliminated, we find that the remaining number of dimensionless groups is

K=N-M

If, for any reason, eliminating a variable “costs” more than one dimension (which happened in Question 4.7 – when eliminating $L$ , $T$ is also eliminated), then

K>N-M

So that in general we get equation (4.34).

The following question demonstrate the theorem.

Question 4.9.

Consider the collision between two point masses $m_{1}$ and $m_{2}$ , initially moving at speeds $v_{1}$ and $v_{2}$ , as in Question 1.5 and 4.7. Apply Buckingham’s Pi theorem to the situation with $M$ , $L$ and $T$ as the base dimensions.

Repeat the analysis with $M$ , $L$ and $V$ (velocity) as the base dimensions.

Answer 4.9.

There are five parameters in this dependence $v_{3}$ , $v_{1}$ , $v_{2}$ , $m_{1}$ and $m_{2}$ , so $N=5$ . If we take $M$ , $L$ and $T$ as the base dimensions, then $M=3$ . So Buckingham’s Pi theorem states

K\geq N-M=5-3=2.

We expect at least two dimensionless variables.

Instead, in the $M L V$ system, all the variables may be described using only the dimensions of $M$ and $V$ . The dimension of $L$ is not needed. In this case, $N=5$ as before, but $M$ =2. Hence Buckingham’s Pi theorem concludes that

K\geq N-M=5-2=3.

Thus, we expect at least three dimensionless variables.

At first glance, the two conclusions may appear conflicting. If we did not know the result from Questions 1.5 and 4.7, it may not be clear which of the two ( $K\geq 2$ or $K\geq 3$ ) applies in this case. The answer is that both are correct, and it is so because they are inequalities. But perhaps the more useful result is that $K\geq 3$ , because it subsumes within it the possibility that $K\geq 2$ , but not vice versa. And an additional observation that $M$ and $V$ are the minimum number of independent dimensions required to represent all parameters of the problem, implies that the inequality is, in fact, an equality, i.e. $K=3$ .

4.7 Alternative ways to form dimensionless relationships

The recombination-elimination is perhaps the most systematic method for identifying dimensionless form of parameters, there are other methods, perhaps more opaque but completely equivalent. We will illustrate these methods using the example of the falling column introduced in §4.4.1. The first one is the method of indices.

4.7.1 Indicial method

Reconsider the dependence

\displaystyle\begin{aligned} \text{Parameter: }&t&&\qquad&&\text{depends on}&&% \left\{\right.h,&&g,&&m&&\left.\right\}\\ \text{Dimensions: }&T&&\qquad&&\text{}&&\left\{\right.L,&&LT^{-2},&&M&&\left.% \right\}\end{aligned}

(4.35)

We will then say that there exist exponents $\alpha$ , $\beta$ and $\gamma$ such that $th^{\alpha}g^{\beta}m^{\gamma}$ is dimensionless. The dimensions of $th^{\alpha}g^{\beta}m^{\gamma}$ are

\displaystyle T\times L^{\alpha}\times(LT^{-2})^{\beta}\times M^{\gamma}=M^{% \gamma}L^{\alpha+\beta}T^{1-2\beta}.

(4.36)

Since we expect this combination to be dimensionless, i.e. with dimensions $M^{0}L^{0}T^{0}$ , we obtain for the exponents three equations

\displaystyle\gamma=0,\qquad\alpha+\beta=0,\qquad\text{and }\qquad 1-2\beta=0.

(4.37)

The unique values of the exponents that satisfy these equations simultaneously are $\alpha=-1/2$ , $\beta=1/2$ and $\gamma=0$ . Thus, we find that only one dimensionless combination is $t\sqrt{g/h}$ (and the $m$ is eliminated).

This method is most useful when there is a single dimensionless parameter that can be constructed. When there is more than one, the solution to the equations for the exponents is not unique, so the user must make choices. We will not pursue this method further.

4.7.2 Inspection

Dimensionless parameters may also be constructed by inspection. After some experience, the user may spot that the combination of $h/g$ has dimensions of $T^{2}$ and can use its square-root to form a dimensionless ratio in combination with another quantity with dimensions of $T$ . One thus recovers $t\sqrt{g/h}$ as the dimensionless combination.

4.8 Conclusion

In this chapter, we examined how to construct dimensionless numbers and dimensionless relationships between parameters. The most reliable method for doing so is the method of recombination-elimination, but there also are other methods. The dimensionless combinations are not unique, and in fact any independent combination of dimensionless parameters is yet another set of dimensionless parameters that equally well describes the relationship. However, certain dimensionless combinations may be the best in aiding physical intuition of the users. Indeed, in this chapter it has become more and more clear that physical intuition is the most powerful tool at the user’s disposal and a worthy goal to pursue with the help of dimensional analysis.