Editing Dimensional analysis (section)

== Examples ==

=== A simple example: period of a harmonic oscillator ===
What is the period of [[Harmonic oscillator|oscillation]] {{math|''T''}} of a mass {{mvar|m}} attached to an ideal linear spring with spring constant {{math|''k''}} suspended in gravity of strength {{math|''g''}}? That period is the solution for {{math|''T''}} of some dimensionless equation in the variables {{math|''T''}}, {{math|''m''}}, {{math|''k''}}, and {{math|''g''}}.
The four quantities have the following dimensions:  {{mvar|T}}  [T];  {{mvar|m}}  [M]; {{mvar|k}} [M/T<sup>2</sup>]; and  {{math|''g''}} [L/T<sup>2</sup>]. From these we can form only one dimensionless product of powers of our chosen variables, {{math|1=''G''{{sub|1}} = ''T''{{isup|2}}''k''/''m''}} {{nowrap|1=[T<sup>2</sup> · M/T<sup>2</sup> / M = 1]}}, and putting {{math|1=''G''{{sub|1}} = ''C''}} for some dimensionless constant {{math|''C''}} gives the dimensionless equation sought.  The dimensionless product of powers of variables is sometimes referred to as a dimensionless group of variables; here the term "group" means "collection" rather than mathematical [[Group (mathematics)|group]].  They are often called [[dimensionless number]]s as well.

The variable {{mvar|g}} does not occur in the group. It is easy to see that it is impossible to form a dimensionless product of powers that combines {{mvar|g}} with {{mvar|k}}, {{mvar|m}}, and {{mvar|T}}, because {{mvar|g}} is the only quantity that involves the dimension L. This implies that in this problem the {{math|''g''}} is irrelevant. Dimensional analysis can sometimes yield strong statements about the ''irrelevance'' of some quantities in a problem, or the need for additional parameters. If we have chosen enough variables to properly describe the problem, then from this argument we can conclude that the period of the mass on the spring is independent of {{math|''g''}}: it is the same on the earth or the moon. The equation demonstrating the existence of a product of powers for our problem can be written in an entirely equivalent way: {{tmath|1=T = \kappa \sqrt\tfrac{m}{k} }}, for some dimensionless constant {{math|''κ''}} (equal to <math>\sqrt{C}</math> from the original dimensionless equation).

When faced with a case where dimensional analysis rejects a variable ({{math|''g''}}, here) that one intuitively expects to belong in a physical description of the situation, another possibility is that the rejected variable is in fact relevant, but that some other relevant variable has been omitted, which might combine with the rejected variable to form a dimensionless quantity. That is, however, not the case here.

When dimensional analysis yields only one dimensionless group, as here, there are no unknown functions, and the solution is said to be "complete" – although it still may involve unknown dimensionless constants, such as {{math|''κ''}}.

=== A more complex example: energy of a vibrating wire ===
Consider the case of a vibrating wire of [[length]] {{math|''ℓ''}} (L) vibrating with an [[amplitude]] {{math|''A''}} (L).  The wire has a [[linear density]] {{math|''ρ''}} (M/L) and is under [[Tension (physics)|tension]] {{math|''s''}} (LM/T<sup>2</sup>), and we want to know the energy {{math|''E''}} (L<sup>2</sup>M/T<sup>2</sup>) in the wire.  Let {{math|''π''<sub>1</sub>}} and {{math|''π''<sub>2</sub>}} be two dimensionless products of [[Power (mathematics)|powers]] of the variables chosen, given by
: <math>\begin{align}
  \pi_1 &= \frac{E}{As} \\
  \pi_2 &= \frac{\ell}{A}.
\end{align}</math>

The linear density of the wire is not involved. The two groups found can be combined into an equivalent form as an equation
: <math>F\left(\frac{E}{As}, \frac{\ell}{A}\right) = 0 ,</math>
where {{math|''F''}} is some unknown function, or, equivalently as
: <math>E = As f\left(\frac{\ell}{A}\right) ,</math>
where {{math|''f''}} is some other unknown function.  Here the unknown function implies that our solution is now incomplete, but dimensional analysis has given us something that may not have been obvious: the energy is proportional to the first power of the tension.  Barring further analytical analysis, we might proceed to experiments to discover the form for the unknown function&nbsp;{{math|''f''}}.  But our experiments are simpler than in the absence of dimensional analysis.  We'd perform none to verify that the energy is proportional to the tension.  Or perhaps we might guess that the energy is proportional to&nbsp;{{math|''ℓ''}}, and so infer that {{math|1=''E'' = ''ℓs''}}.  The power of dimensional analysis as an aid to experiment and forming hypotheses becomes evident.

The power of dimensional analysis really becomes apparent when it is applied to situations, unlike those given above, that are more complicated, the set of variables involved are not apparent, and the underlying equations hopelessly complex.  Consider, for example, a small pebble sitting on the bed of a river.  If the river flows fast enough, it will actually raise the pebble and cause it to flow along with the water.  At what critical velocity will this occur?  Sorting out the guessed variables is not so easy as before.  But dimensional analysis can be a powerful aid in understanding problems like this, and is usually the very first tool to be applied to complex problems where the underlying equations and constraints are poorly understood. In such cases, the answer may depend on a [[dimensionless number]] such as the [[Reynolds number]], which may be interpreted by dimensional analysis.

=== A third example: demand versus capacity for a rotating disc ===
[[File:Dimensional analysis 01.jpg|thumb|upright=1.5|Dimensional analysis and numerical experiments for a rotating disc]]
Consider the case of a thin, solid, parallel-sided rotating disc of axial thickness {{math|''t''}} (L) and radius {{math|''R''}} (L).  The disc has a density {{math|''ρ''}} (M/L<sup>3</sup>), rotates at an angular velocity {{math|''ω''}} (T<sup>−1</sup>) and this leads to a stress {{math|''S''}} (T<sup>−2</sup>L<sup>−1</sup>M) in the material.  There is a theoretical linear elastic solution, given by Lame, to this problem when the disc is thin relative to its radius, the faces of the disc are free to move axially, and the plane stress constitutive relations can be assumed to be valid.  As the disc becomes thicker relative to the radius then the plane stress solution breaks down.  If the disc is restrained axially on its free faces then a state of plane strain will occur.  However, if this is not the case then the state of stress may only be determined though consideration of three-dimensional elasticity and there is no known theoretical solution for this case.  An engineer might, therefore, be interested in establishing a relationship between the five variables.  Dimensional analysis for this case leads to the following ({{nowrap|1=5 − 3 = 2}}) non-dimensional groups:

: demand/capacity = {{math|''ρR''{{i sup|2}}''ω''{{i sup|2}}/''S''}}
: thickness/radius or aspect ratio = {{math|''t''/''R''}}

Through the use of numerical experiments using, for example, the [[finite element method]], the nature of the relationship between the two non-dimensional groups can be obtained as shown in the figure.  As this problem only involves two non-dimensional groups, the complete picture is provided in a single plot and this can be used as a design/assessment chart for rotating discs.<ref>{{cite web|last1=Ramsay|first1=Angus|title=Dimensional Analysis and Numerical Experiments for a Rotating Disc|url=http://www.ramsay-maunder.co.uk/knowledge-base/technical-notes/dimensional-analysis--numerical-experiments-for-a-rotating-disc/|website=Ramsay Maunder Associates|access-date=15 April 2017}}</ref>