Dimensional Analysis

Describes dimensional analysis and explains a method for converting physical units

View version details

Contents

Units
Dimensional Analysis
Physical Equations
Conversion Between Units Of Different Systems
Page Comments

Units

Although the MKS (Meters, Kilograms & Seconds) system is slowly becoming universal, there are actually two different systems and each is split into two variations. Happily nearly ever body uses seconds but decimal minutes are in use for specialist applications.

System	Length	Mass	Time	Force
(metric)
MKS	Metre (m)	Kilo (K)	Second (s)	Newton
CGS	Centi metre (cm)	Gram (s)		Dyne
(imperial)
FPS	Foot (ft)	Pound (lb)	Second (s)	Poundal
F Slug S	Foot (ft)	Slug		Pound

NOTES For those who are not used to the Imperial system.

A Force of one pound is that force which will produce an acceleration of one foot per second squared when acting on a mass of one Slug.
There is considerable variation in how the unit of pressure is shown, so that:

Fourteen pounds per square inch

$= 14\,psi = 14\,lb/sq.in. = 14\,lb\, in^{-2} = 14\,\frac{lb}{in^2}$

Dimensional Analysis

In Applied Mathematics, based on Newton's three laws of motion, there are three dimensions; Mass; Length; and Time and all physical quantities can be expressed in terms of these. It is important to note that all equations must balance dimensionally and each term within the equation must have the same dimensional value.

$Velocity = \frac{L}{T}$	and	$Acceleration = \frac{L}{T^2}$
$Force = \frac{M\:L}{T^2}$	and	$Work = \frac{M\,L^2}{T^2}$
$Power = \frac{M\:L^2}{T^3}$	and	$Stress = \frac{M}{T^2\:L}$
$Density = \frac{M}{L^3}$
$Strain = \frac{\delta\,L}{L} = \frac{L}{L}= 1$ therefore strain is dimensionless

Physical Equations

As all terms in an equation must have the same nature. i.e.When expressed in terms of the fundamental units they must be the same. For example, in the following elementary dynamic equations, the dimensions are shown in brackets.

$v\,\left(\frac{L}{T} \right) = U\:\left(\frac{L}{T} \right)\:+\:a\:\left(\frac{L}{T^2} \right)\,X\:t\:\left(T \right)$

(2)

and

$s\:\left(L \right)\:=\:U\,t\:\left(\frac{L}{t}\,.\:t \right)\:+\:\frac{1}{2}\:a\:t^2\:\left(\frac{L}{t^2}\:.\:t^2 \right)$

(3)

Note that in the first equation each term had a dimensional value of L/t and in the second L

Conversion Between Units Of Different Systems

The system described here is based on the idea of "Unity Brackets". For example 25.4 mm is the same length as 1 inch so we consider the following bracket as having a value of unity.

$\left[\frac{25.4\,mm}{1\,inch} \right]\;has\:a\:value\:of\:unity$

(4)

So if it is required to convert 60 miles per hour to metres per second, the sum looks like this.

$60\left[\frac{miles}{hours} \right]\left[\frac{1 \,hour}{3600\,sec} \right]\left|\frac{1760\,yards}{1\,mile} \right|\left\frac{36\,ins.}{1\,yard} \right|\left[\frac{1\,metre}{39\,ins.} \right]$

(5)

It is now a simple question of canceling out WORDS and leaving the numbers in each bracket. Multiply out and the conversion is complete. The only trick is to place the unity bracket the right way up so that unit names that are not wanted can be canceled. Note this conversion system will work for any quantities including money provided that there is a common zero. It will therefore NOT work to convert Degrees Celsius to degrees Fahrenheit. It will however convert degrees Kelvin to degrees Rankin.