13 Dimensional Analysis, Units, & Conversion Factors


New engineers quickly begin to recognize from even the most casual observations that the solution to engineering problems usually involves the consideration of many parameters as well as parameter dependencies and interdependencies. In other words, many quantities could affect the physical behavior of a given system, and different effects will be obtained when varying one parameter versus another.

In particular, the behavior of fluids is relatively difficult to describe because they deform, primarily as they flow around objects (often referred to as “bodies”). The fluid also produces forces and moments on such bodies, including the familiar lift and drag forces. There is undoubtedly an intuitive expectation that what the flow does (e.g., where the fluid goes) and what effects it has on the body will be determined at least partially by the geometrical shape of the body, as well as its orientation to the flow. The forces will also be expected to depend on the fluid properties already discussed, such as flow velocity (airspeed), density, viscosity, temperature, pressure, speed of sound, etc. However, other parameters may also be involved.

To solve this type of fluid dynamics problem, one approach would be to set down and use the equations of motion governing the behavior of the fluid and then try to solve these equations. These nonlinear partial differential equations are called the Navier-Stokes equations and form the basis of a tool called computational fluid dynamics (CFD). However, this is not a straightforward approach, not just because of the complexity of the equations that must be solved but also because considerable proficiency is needed in the solution methods. Another approach would be to do experiments to find the relationships between the forces on a body and the fluid properties, e.g., in a wind tunnel. At least in principle, either or both of these latter approaches will allow the effects of the fluid on the body to be determined and the reciprocal influences of the body on the fluid. However, even with CFD and the ability to do experiments, sequential variations in the values of the problem parameters to determine cause and effect will take a very long time.

Nevertheless, another method can be used to help determine the general effects of key physical quantities on aerodynamic behavior. This method is called dimensional analysis, which reduces the number of unknowns governing the understanding of the problem into sets of dimensionless or non-dimensional groupings called similarity parameters. These parameters then allow a significant simplification in understanding the effects of the original problem parameters. Dimensional analysis is not restricted to fluid problems and is helpful in many fields of engineering and science.

Learning Objectives

  • Review the system of SI (metric) and USC (British Imperial) units of measurement and their proper use.
  • Understand how to perform dimensional analysis, a method used to bring out the primary dependencies in an engineering problem in terms of dimensionless groupings called similarity parameters.
  • Better understand the significance of similarity parameters and how they can simplify a problem and reduce the number of dependencies, i.e., making a problem easier to understand.
  • Develop and discuss some exemplar problems to become familiar with the method and practice of dimensional analysis and the use of similarity parameters.

Physical Quantities

Science and engineering involve observations of the physical world; engineers aim to express observations in the most general and the simplest type of formulation. The role of mathematics is essential here, and the resulting formulation will stem directly from the properties or quantities considered in the quantitative analysis. The allowed types of properties are called physical quantities.

Physical quantities are of two types: base quantities and derived quantities. Base quantities such as mass, length, time, and temperature can be readily measured. Derived quantities are things like velocity, pressure, energy, and power, all of which are derived from these base quantities. In dimensional analysis, only base quantities are used; those mostly encountered in mechanics and engineering are listed in the table below. Other base quantities that may arise in problem-solving are electrical current (Amp), the intensity of light (Candela), and concentration (Mole).

List of key base quantities used in engineering mechanics.
Base quantity Symbol Base unit SI unit USC unit
Mass M M kilogram (kg) slug (or slugs or sl)
Length L L meter (m) foot (ft)
Time T T second (s) second (s)
Temperature \theta \theta Kelvin (^{\circ}K or K) Rankine (^{\circ}R or R)

Confused about units? Here is a short video lesson on units (SI and USC) from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Uncertain about the differences between mass and weight? Here is a short video lesson on the differences between mass and weight from Dr. Leishman’s “Math & Physics Hints and Tips” series.

All About Units

Before examining the process of dimensional analysis, it is helpful to review the subject of units and measurement systems. A unit is a standardized measurement of a parameter or quantity adopted by consensus and defined by convention, e.g., the international SI system. The expression of quantities measured in standardized units allows their magnitude or numerical value to be evaluated and compared. Adopting a unit system for measuring a broad range of physical parameters allows engineers and scientists worldwide to communicate and work with each other using a common measurement standard.

The value of a particular physical quantity is always given the product of a number and a unit (unless unitless), e.g., 12.4 meters per second (m/s) in SI units is equal to 40.68 feet per second (SI) in USC or British Imperial units, where the conversion factor from meters to feet is 3.281. This means the numerical value of a quantity depends on the system of units in which it is expressed. Engineers in the U.S. routinely use both SI and USC units, and students must learn to use both unit systems. However, there is generally a preference to use SI units because of their widespread adoption and use in scientific research and compliance with international standards. Conversion factors are available so that numerical values in one unit system can be expressed as equivalent values in the other.

SI Units

The Système International or International System of Units (SI), sometimes known as the metric system, has become the international standard for measurement units. The SI system is also called the decimal system because it is based on powers of 10. The SI system comprises seven base units with many other derived units that are given unique names and symbols, with a subset being shown in the table below. The SI system is based on the meter (m) (or the metre if outside the U.S.) as the base unit of length, the kilogram (kg) as the base unit of mass, the Newton (N) as the base unit of force, and the second (s) as the base unit of time. Notice that unit symbols are never italicized.
Quantity Quantity symbol Base units SI unit name Unit symbol
Length L L meter m
Mass M M kilogram kg
Time T T second s
Volume {\cal{V}} L3 meter-cubed m3
Temperature \theta \theta Kelvin K or oK
Velocity V LT-1 meter per second m/s
Acceleration a LT-2 meter per second2 m/s2
Force F MLT-2 Newton N
Energy E ML2T-2 Joule J
Power P ML2T-3 Joule per second = Watt W
Pressure p ML-1T-2 Newtons per m2 = Pascal Pa
Viscosity \mu ML-1T-1 Pascal-second Pa s

USC Units

U.S. customary (USC) units are a system of measurement units most commonly used in the U.S. The USC system was developed from traditional British measurement units. The British measurement system was updated in 1824, which created the so-called Imperial system or British Imperial units of measurement, often known in the U.S. as “English” or “British Engineering” units. Interestingly enough, today all British and European engineers call these “American” units. For engineering use, the USC or “British system” is based on the foot (ft) as the base unit of length, the slug (slug or sl) as the base unit of mass, the pound (lb) as the base unit of force, and the second (s) as the base unit of time, as reviewed in the table below.
Quantity Quantity symbol Base units USC unit name Unit symbol
Length L L foot ft
Mass M M slug slug or sl
Time T T second s
Volume {\cal{V}} L3 foot3 ft3
Temperature \theta \theta Rankine R or oR
Velocity V LT-1 feet per second ft/s
Acceleration a LT-2 feet per second2 ft/s2
Force F MLT-2 pound lb
Energy E ML2T-2 foot-pound ft-lb
Power* P ML2T-3 foot-pounds per second ft-lb/s
Pressure p ML-1T-2 pounds per foot2 lb/ft2
Viscosity \mu ML-1T-1 slug/(ft s) or pound seconds per foot slug/(ft s) or lb s/ft

Notice that the name “slug” as a unit of mass originates from the concept of mass as an inertia, i.e., “sluggish,” and has been referred to as the “engineer’s mass unit.” The use of “pounds-mass” (lbm) as a measure of mass is generally considered to be a source of confusion in the engineering and scientific fields. Therefore, the unit of “pounds-mass” is best avoided in preference of using the “slug” as the base unit of mass. While USC units are largely similar to British Imperial units, there are still differences regarding volumetric measurements.

What is a horsepower?

The mechanical output of an engine is often measured in “horsepower,” which is given the unit symbol “hp.” This unit is attributed to the Scottish engineer James Watt, whose work on steam engines was fundamental to the changes brought about by the Industrial Revolution. James Watt wanted to market his engines by comparing them to the equivalent power produced by a horse. Watt did experiments and determined that a typical farm horse could, on average, steadily lift a 600 lb weight through a simple pulley system an average distance of 63.9 ft in 69 seconds. Therefore, the work done is

    \[ \mbox{Work} = F \, d = 600 \times 63.9 = 38,340~\mbox{ft-lb} \]

Power is the rate of doing work, so the work per unit of time is

    \[ \mbox{Power} = \frac{\mbox{Work}}{\mbox{Time}} = \frac{38,340}{69} = 555.65~\mbox{ft-lb s$^{-1}$} \]

James Watt settled on the result that 1 hp = 550 ft-lb s^{-1}. This measurement unit made sense to farmers and others using horses to move equipment, so for a steam engine of 10 hp, for example, the purchaser knew precisely what they were buying: an engine with the power equivalent to 10 horses. The unit of “horsepower” is still used today almost worldwide.

In 1975, the U.S. government declared the international SI system the “preferred” unit system but did not require it. Indeed, the USC system still prevails in the U.S. as the preferred measurement system, and the U.S. is one of the few countries worldwide that does not exclusively use the SI system. The use of mixed SI and Imperial units in Britain is also common, but less than in the U.S. The SI unit system is preferred in all engineering and scientific work, and in many situations, it may be required by specification, international code, or regulation. However, in today’s global community, engineers and scientists from all countries must be conversant in using both SI and USC units.

NOTE: In aviation, distances are measured in terms of nautical miles, which is neither Imperial nor SI, airspeed is measured in terms of knots (nautical miles per hour), and altitude is measured in feet, the use of such units being in accordance with ICAO aviation and aeronautical standards.

Working with units

Student engineers need to become comfortable in using both SI and USC units. When presented with a problem in SI units, the problem should be solved entirely in SI units. Likewise, when presented with a problem in USC units, then the problem should be solved entirely in USC units. Refrain from converting back and forth to suit yourself because this is when mistakes happen. Sometimes, information (values) will be presented in mixed units, with some values in SI and others in USC. Converting all the information to SI or USC is essential, depending on the required outcome. In either case, it is essential that all information available s converted to base units before evaluations are conducted, in which case the final desired value or values will also be in base units. If unit conversions are required, handy conversion factors can be found by accessing the NIST website from the links below. It is important that when using unit conversions, the full accuracy of the conversion factor should be used. As a general rule, rounding of numerical values should only be done at the end of the calculations.

CGS Units

The centimeter–gram–second or CGS system of units is an increasingly obsolete variant of the metric system. CGS units are based on the centimeter as the base unit of length, the gram as the base unit of mass, and the second as the base unit of time. SI units have largely supplanted the CGS system, but sometimes quantities measured in CGS units are still encountered in engineering and scientific practice. Converting between CGS and SI units is easy because the conversion factors are all powers of 10, as reviewed in the table below. For example, 100 cm = 1 m and 1,000 g = 1 kg. The CGS unit of force is called the dyne, which is 1 g cm/s^2, so a Newton (N) in SI units equals 100,000 dynes in CGS units. Some units in the CGS system are considered depreciated or obsolete, such dyne and erg.

Quantity Quantity symbol Base units CGS name Unit symbol Unit definition SI units
Length L L centimeter cm 1/100 of a meter 10−2 m
Mass M M gram g 1/1000 of a kilogram 10−3 kg
Time T T second s 1 second 1 s
Volume {\cal{V}} L3 liter (litre) l 1/1,000 of a cubic meter 10-3 m3
Temperature \theta \theta Kelvin K or oK K or oK
Velocity V LT-1 centimeter per second cm/s cm/s 10−2 m/s
Acceleration a LT-2 gal (Galileo) Gal cm/s2 10−2 m/s2
Force F MLT-2 dyne dyn g cm/s2 10−5 N
Energy E ML2T-2 erg erg g cm2/s2 10−7 J
Power P ML2T-3 erg per second erg/s g cm2/s3 10−7 W
Pressure p ML-1T-2 barye Ba g/(cm s2) 10−1 Pa
Dynamic viscosity \mu ML-1T-1 poise P g/(cm s) 10−1 Pa s
Kinematic viscosity \nu M2S-1 Stokes St cm2/s 10−4 m2/s

When NASA lost a $100M spacecraft because of a units mistake!

In September of 1999, the NASA Mars Climate Orbiter unexpectedly burned up and crashed into the surface of Mars after almost ten months of space travel. The project cost over $100M. It turned out that engineers had made a massive units mistake, muddling quantities between the SI and USC systems. NASA’s Jet Propulsion Laboratory (JPL) used the SI system, while Lockheed used the USC system. Because of this blunder, the navigation system flew the spacecraft too close to the atmosphere, where it burned up. Read more about this mistake here as well as some other engineering disasters that have occurred because of the muddling of units. Be sure not to make the same mistakes!

Finding Base Dimensions

Many of the parameters routinely used in engineering are derived parameters, in that their units are derived from combinations of base units, e.g., volume, density, force, work, energy, power, etc. From now on, notice that the parameter inside a set of square brackets, i.e., “[parameter],” is used to mean the dimensions of that parameter. Notice also that scientific notation must be used throughout to avoid ambiguity in the final units of the derived parameter, e.g., use L \, T^{-1} and not L / T. Consider the following examples:

  • Volume, {\cal{V}}, is a derived quantity because to calculate the volume of something such as a cuboid, then its width (units of L) must be multiplied by its length (units of L) by its height (units of L). Therefore, the derived units for volume are L^3 or [{\cal{V}}] = L^3.
  • Density, \varrho, is also a derived parameter because the density \varrho of some substance is its mass (M) per unit volume (L^3), so density will have units of M L^{-3}, i..e., [\varrho] = M L^{-3}.
  • Force, F, is a derived parameter, although in some cases, “force” could be carried through the problem as a base unit. A force is the product of a mass and acceleration (Newton’s second law), so in terms of its base units, then [F] = M \,  LT^{-2}.
  • Torque, \tau, is a derived parameter because a torque is a moment, so the product of a force times a distance or “arm.” Therefore, [\tau] = (M LT^{-2}) \, L = M \,  L^2 T^{-2}.

Obtaining the correct base units of any given parameter takes some practice, but it is essential for correctly using dimensional analysis. Again, the scientific notation should be used when expressing the units of a parameter to avoid ambiguity, e.g., always use the form M L T^{-2} and never M L / T^2.

Need to brush up on the concepts of work, energy, and power? Here is a short video lesson on these concepts from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Aerodynamic Force on a Wing

To introduce the subject and methods of dimensional analysis, consider the forces produced on a wing in the wind tunnel, as shown in the photo below. The resultant aerodynamic force on the wing is R. For convenience, it can be assumed that this force acts at a center of pressure, which, by definition, is a point where there are no pitching moments.

A wing set at an angle to the flow will produce some resultant aerodynamic force, the magnitude and direction of which depends on the size of the wing and the free-stream parameters.

On a physical intuitive basis, then it would be expected that the resultant aerodynamic force R on the wing for a given orientation to the flow will depend on:

  1. Free-stream velocity, V_{\infty}.
  2. Free-stream density, \varrho_{\infty}.
  3. The viscosity of the air, i.e., on the coefficient of viscosity \mu_{\infty}.
  4. The size of the wing, as described by its planform area S, which for a rectangular wing is the product of its chord c and span b, i.e., S = b \, c, as shown in the figure below.
  5. The sonic velocity a_{\infty}, i.e., the potential for compressibility effects.

The subscript \infty or “infinity” means undisturbed conditions, i.e., the “free-stream” or reference values of these quantities that are constant well upstream of the wing. This approach makes sense because these values will all likely change as the flow approaches the wing. Notice that the angle of attack of the wing, \alpha, is the angle between the chordline of the wing (i.e., a line running from the nose to the tail of the wing) and the direction of the free-stream velocity V_{\infty}.


The resultant force can also be decomposed into a vertical component that acts perpendicular to the relative wind (called the lift) and a horizontal component parallel to the relative wind (called the drag).

The relationship between R and the air properties may now be written in a general functional form as

(1)   \begin{equation*} R = \phi \left( \varrho_{\infty}, V_{\infty}, S, \mu_{\infty}, a_{\infty} \right) \end{equation*}

where R is referred to as the dependent variable and \varrho_{\infty}, V_{\infty}, \mu_{\infty} and a_{\infty} are called the independent variables. The functional dependence, i.e., the relationship between the variables, denoted by \phi in this case, is still to be determined. Although this latter equation expresses a general relationship, its use in this form is mostly impractical for the direct measurement or calculation of R.

To see why consider how R could be measured in the wind tunnel by a student of aerodynamics. The procedure might start with the student placing a wing at a given angle of attack, \alpha, and then (somehow) systematically varying each of the dependencies V_{\infty}, \varrho_{\infty}, \mu_{\infty} and a_{\infty} (i.e., a total of five parameters) in turn, and in each case measuring the corresponding values of R. Varying the wind speed V_{\infty} can be controlled easily in the wind tunnel, and different wings could be tested to change the wing area S. The other parameters, however, are more challenging to vary directly or systematically. However, if it could be done, the number of possible experimental runs required in the wind tunnel would be 5^5 = 3,125 (!).

Then that process would need to be repeated by the student for different values of the wing origination to the flow, \alpha. If ten angles were selected, over 30,000 runs would be needed! Eventually, the needed relationships between R and all dependent quantities at each angle of attack would be determined. Therefore, it does not take much to figure out that this process of systematically varying each of the dependent parameters, in turn, could take the student a very long time.

This approach would also result in a large quantity of data, from which cross-plotting and perhaps curve-fitting could be used to find the needed functional relationships as to how V_{\infty}, \varrho_{\infty}, etc., would each affect R. If just one dependent variable were used, a single column of numbers would be obtained, and the functional relationship would be easy to establish. However, a book of tables would be needed for four or more dependent variables or many pages of charts and graphs. Hence, it soon becomes apparent to the student that managing an engineering problem with three, four, or more independent variables is challenging.

Basis of Dimensional Analysis

Systematically varying each independent parameter to establish the “cause and effect” on the dependent parameter would be mostly impractical from a measurement perspective, nor would it be attractive from a computational (modeling) approach. Fortunately, the approach to the preceding problem can be simplified by employing dimensional analysis, which can be used to determine sets of dimensionless groups of parameters that affect the problem of interest. Such an approach will be shown also to reduce the number of independent variables, thereby becoming an essential technique for understanding many engineering problems, not just aerodynamics.

Dimensional Homogeneity

Dimensional analysis is based on the fact that if an equation of parameters such as

(2)   \begin{equation*} P_1 + P_2 + P_3 = P_4 \end{equation*}

expresses some physical relation, then the parameters P_1, P_2, P_3, and P_4 must all have the same dimensions and so be dimensionally homogeneous for the equation to be mathematically valid, e.g., P_1, P_2, P_3, and P_4 could all be values of pressures or velocity components. Equation 2 can also be made dimensionless by dividing through by one of the terms, i.e., the same result can be expressed as

(3)   \begin{equation*} \frac{P_1}{P_4} + \frac{P_2}{P_3} + \frac{P_3}{P_4} = \frac{P_4}{P_4} = 1 \end{equation*}

Another advantage of this latter dimensionless form is that all quantities are now expressed as ratios, so the system of units is irrelevant.

Complete and Incomplete Equations

An equation arising from a physical problem is said to be complete if it is stated in such a form that where the physical dimensions of each variable involved can be explicitly recognized. For example, for a falling body accelerating to some velocity, V, under gravity, g, from an initial velocity, U, then

(4)   \begin{equation*} V^2 = U^2 + 2 \, g \, s \end{equation*}

where s is vertical distance, then the form of this equation is said to be complete. However, if the equation is written as

(5)   \begin{equation*} V^2 = U^2 + 64.34 s \end{equation*}

then it is incomplete because the dimensions of the second term in the equation cannot be recognized, i.e., it is impossible to positively conclude that 64.34 is equal to 2g, although that would be a good guess with USC units (g = 32.17 ft/s2). If the equation is complete, then it is possible to conclude that in terms of dimensions then

(6)   \begin{equation*} \left[ V^2  \right] = \left[ U^2 \right] = \left[ 2 \, g \, s \right] \end{equation*}

and so inserting the base units for each parameter gives

(7)   \begin{equation*} \left( \frac{L}{T} \right)^2  = \left( \frac{L}{T} \right)^2 = \left( \frac{L}{T^2} \right) L  = \left( \frac{L}{T} \right)^2 = \left( \frac{L}{T} \right)^2 \end{equation*}

and the equation is, therefore, dimensionally homogeneous. Furthermore, to make sense of any equation, each parameter in the equation must also be expressed numerically in the same units system, i.e., either entirely in SI units or entirely in USC units, in each case with the same base units. However, notice that Eq. 4 can also be written in non-dimensional form as

(8)   \begin{equation*} \left( \frac{V}{U} \right)^2 - \frac{ 2 \, g \, s}{U^2} = 1 \end{equation*}

so that the system units do not matter. When undertaking derivations and developing equations, it is often desirable to work in non-dimensional form.

Satisfying Dimensional Homogeneiety

Consider another problem in which dimensional homogeneity is required. To describe the laminar flow of a fluid between two moving plates, the shear stress, \tau, is some function of the viscosity, \mu, and the velocity difference \Delta V between the two plates, h. The relationship can be expressed as

(9)   \begin{equation*} \tau = \mu^{\alpha} \, \Delta V^{\beta} \, h^{\gamma} \end{equation*}

By satisfying dimensional homogeneity, it is possible to determine the values of \alpha, \beta, and \gamma. In terms of the dimensions

(10)   \begin{equation*} \left[ \tau \right] = \left[ \mu^{\alpha} \right] \left[ \Delta V^{\beta} \right] \left[ \Delta h^{\gamma} \right] \end{equation*}

and so for each parameter, then

(11)   \begin{equation*} \left[ \tau \right] = M L^{-1} T^{-2} \quad\quad \left[ \mu \right] = M L^{-1} T^{-1} \quad\quad \left[ \Delta V \right] = L T^{-1} \quad\quad \left[ h \right] = L \end{equation*}


(12)   \begin{equation*} M L^{-1} T^{-2}   = \left( M L^{-1} T^{-1} \right)^{\alpha} \, \left( L T^{-1} \right)^\beta \, \left( L \right)^\gamma \end{equation*}

To obtain dimensional homogeneity, then

(13)   \begin{eqnarray*} M: \quad 1 & = & \alpha \\ L: \quad -1 & = & -\alpha + \beta + \gamma\\ T: \quad -2 & = & -\alpha -\beta \end{eqnarray*}

so \alpha = 1, \beta = 1, and \gamma = -1. Inserting the values gives the relationship as

(14)   \begin{equation*} \tau = \mu^{\alpha} \, \Delta V^{\beta} \, h^{\gamma} = K \, \mu \left( \frac{\Delta V}{h} \right) \end{equation*}

showing that the shear stress in proportional to the velocity gradient with the constant of proportionality being the value of \mu. These ideas of dimensional analysis and dimensional homogeneity and are intrinsic to the Buckingham \Pi method.

The Buckingham Π Method

Lord Rayleigh and Joseph Bertrand initially set down these foregoing simple but profoundly important ideas in what is called the “Method of Dimensions.” However, they are now formally embodied in what is known as the “Buckingham \Pi Theorem” after Edgar Buckingham, who developed it into a more structured form suitable for general use. The Buckingham \Pi Theorem (usually just called the Buckingham \Pi Method) is a generalized formalization of the method of dimensions, which can now be examined.

Let K equal the number of fundamental dimensions required to describe the physical variables. In engineering, most physical variables can be expressed as a combination of mass, length, time, and temperature, so generally, K=4. In mechanics, then mass, length, and time are involved, so K=3. Let P_{1}, P_{2}, ...P_{N} represent a list of N physical variables in the general physical relationship

(15)   \begin{equation*} P_{1} = \phi \left( P_{2}, .........P_{N} \right) \end{equation*}

where P_{1} is the dependent parameter and P_{2}, .........P_{N} are the independent parameters. The function \phi represents a general functional relationship, as yet unknown. Notice that the previous equation can also be written in an implicit form as

(16)   \begin{equation*} \psi \left( P_{1}, P_{2}, .........P_{N} \right) = 0 \end{equation*}

where \psi represents another functional relationship between the parameters yet to be determined.

The Buckingham \Pi method then states that the relationship in Eq. 16 may be expressed as a relationship of fewer (N-K) dimensionless products called \Pi products, where each \Pi product is a dimensionless product of a selected set of K repeating variables plus one other variable. One key to applying the method, therefore, is how to go about the process of selecting the most suitable set of repeating variables.

To see how the Buckingham \Pi method works, let P_{2}, P_{3}, ....P_{K+1} be the selected K repeating variables. The repeating variables are not unique in most cases, but their selection is not arbitrary and must be selected according to some specific rules, as will be discussed. The primary rule is that the repeating variables cannot include the dependent variable, P_1, which can also only appear in one of the \Pi groups. Therefore,

(17)   \begin{eqnarray*} \Pi_{1} & = & f_{1} \left( \underbrace{ P_{2}, P_{3}......P_{K+1}}_{\mbox{~\scriptsize Repeating}} ,\underbrace{P_{1}}_{\mbox{~\scriptsize Dependent}} \right) \\[6pt] \Pi_{2} & = & f_{2} \left( \underbrace{ P_{2}, P_{3}, .....P_{K+1}}_{\mbox{~\scriptsize Repeating}} ,\underbrace{P_{K+2}}_{\mbox{~\scriptsize Other}} \right) \\[6pt] & & \bigcdot  \\[6pt] & & \bigcdot  \\[6pt] & & \bigcdot  \\[6pt] \Pi_{N-K} & = & f_{N-K} \left( \underbrace{ P_{2}, P_{3}, .......P_{K+1}}_{\mbox{~\scriptsize Repeating}}, \underbrace{P_{N}}_{\mbox{~\scriptsize Remaining}} \right) \end{eqnarray*}

where P_N is the last remaining variable.

Notice that there are N - K new functional relationships, i.e., f_{1}, f_{2}….f_{N-K}, but now they are expressed in terms of the dimensionless \Pi products. Therefore, it can be written that

(18)   \begin{equation*} \phi_1 \left( \Pi_1, \Pi_2, .., \Pi_{N-K} \right) = 0 \end{equation*}

where \phi_1 is the functional relationship (still to be determined), it is now expressed in terms of each of the \Pi products. In explicit form, this latter equation can also be written as

(19)   \begin{equation*} \Pi_1 = \phi_2 \left( \Pi_2, .., \Pi_{N-K} \right) \end{equation*}

where \phi_2 is some other functional relationship.

The upshot of the dimensional analysis process means that the functional relationship has been reduced from the initial N dimensional variables to N-K dimensionless variables. Therefore, reducing the number of dependent parameters significantly simplifies analyzing the original problem. For example, suppose originally N = 6 and K = 3, i.e., the dependent variable is a function of five other parameters as in the aforementioned wing problem. In that case, the dimensional analysis will reduce the problem to three dimensionless groupings. If N = 5 and K = 3, then the dimensional analysis will reduce the problem to two dimensionless groupings in that one grouping becomes a sole function of the other grouping.

Π Method Applied to the Drag on a Sphere

Consider a sphere to be tested in a low-speed wind tunnel to measure its drag, as shown in the photograph below. Perhaps the drag of a sphere sounds boring compared to a wing, but this problem helps introduce the dimensional analysis process without making the process too lengthy.

Furthermore, the aerodynamic characteristics of a sphere are particularly relevant to all sorts of ball games. The motion of balls through the air is a complex fluid dynamic problem that is influenced by many factors, including the density, velocity, viscosity, size of the ball, and its surface finish. For example, the aerodynamics of a golf ball affects the distance, spin, and overall trajectory of the ball. Ever wonder why a golf ball has dimples on its surface? In baseball, the aerodynamics of a baseball can influence the speed and trajectory of a pitch.


A sphere set up in a wind tunnel to measure its drag. The sphere is mounted on a force balance.

Functional Dependency

The objective is to find the dimensionless groupings describing this aerodynamic problem. On a physical intuitive basis, the aerodynamic drag force, D, on the sphere would be expected to depend on:

  1. Free-stream velocity, V_{\infty}.
  2. Free-stream density, \varrho_{\infty}.
  3. Free-stream viscosity, i.e., the coefficient of viscosity \mu_{\infty}.
  4. The size of the sphere, as described by its diameter d.

This list may be incomplete, and other parameters may affect the drag. One the one hand, the omission of a relevant parameter may restrict the validity of the result or may make it totally wrong. On the other hand, the inclusion of unnecessary parameters may make the final result more complicated than it should be.

The relationship between D and the air properties may be written in a general functional form as

    \[ D = \phi\left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d \right) \]

The drag D is the dependent variable. \varrho_{\infty}, V_{\infty}, \mu_{\infty} and d are the independent variables. The functional dependence, in this case, is denoted by \phi

The functional dependence in implicit form is

    \[ \psi \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d,  D \right) = 0 \]

where \psi is some other function. In this case, N = 5, K = 3, N - K = 2, so there are two \Pi products to be determined.

Understanding explicit versus implicit functional forms

What is the difference between writing an equation in the form y = f(x) versus the form g(x,y) = 0? In mathematics, the dependent function, say y, for example, can be written in terms of other parameters, e.g., y = f(x) where f is some function, or more generally as y = f(x_1, x_2, \! ...  x_N), which are called explicit functional dependencies. However, the relationship can also be written in implicit form, i.e., y = f(x) is equivalent to g (x, y) = 0. Notice that the function g is NOT the same as the function f. If y = x^2 +5, for example, then y = f(x), which is an explicit dependency of y on x. Alternatively, it can be written that y - x^2 - 5 = 0, which is in the form g(x,y) = 0, which is the implicit form of the same relationship. Likewise, y = f(x_1, x_2, \! ...  x_N) as g(x_1, x_2, \! ...  x_N, y) = 0. In dimensional analysis, the process starts by writing the functional dependency of the problem parameters in implicit form.

Dimensions of the Variables

Now the dimensions of the variables in this problem must be determined. For each dependency, then

    \begin{eqnarray*} \left[ D \right]  & =  & MLT^{-2} \\ \left[ \varrho_{\infty} \right]  & = & ML^{-3} \\ \left[ V_{\infty} \right] & = & LT^{-1} \\ \left[ \mu_{\infty} \right] & = & ML^{-1}T^{-1} \\ \left[ d \right] & = & L \end{eqnarray*}

Recall that the units of force are obtained from mass times acceleration, i.e., MLT^{-2}.

Dimensional Matrix

The solution to the dimensional analysis problem proceeds by setting up a dimensional matrix with all the parameters in terms of their base units. Setting up the dimensional matrix, in this case, gives

    \[ \begin{array}{l|r|r|r|r|r|r} &   D  & \varrho_{\infty} & V_{\infty} & d &  \mu_{\infty} \\ \hline \mbox{Mass: } M  &   1  & 1 & 0 & 0 & 1 \\ \mbox{Length: } L  &  1  & -3 & 1 & 1 & -1 \\ \mbox{Time: } T  &  -2 & 0 & -1 & 0 & -1 \end{array} \]

One purpose of the dimensional matrix is to lay the problem out clearly and test for linear independence of the repeating variables in terms of the chosen primary dimensions.

Repeating Variables

Although the selection of the repeating variables can be viewed as somewhat arbitrary, in general, they must always be selected based on four essential rules:

  1. The repeating variables must all have a primary dependency on the dependent variable. In this case, the variables \varrho_{\infty}, V_{\infty} and d can all be expected to have strong effects on the dependent variable, D.
  2. The dependent variable, D, cannot be used as a repeating variable, so it must appear only in one group by default. No exceptions.
  3. The repeating variables must include all of the physical base dimensions of the problem, i.e., they must collectively include mass and/or length and/or time, as appropriate, for the particular problem in question. In this case, \varrho_{\infty}, V_{\infty}, and d collectively include all of the base dimensions of mass, length, and time.
  4. The repeating variables must be linearly independent of one another, i.e., repeating variables must not be chosen that will have the same units or products of the same units, e.g., repeating variables with two different length scales or two different reference velocities cannot be used. In this case, \varrho_{\infty}, V_{\infty}, and d are all linearly independent of each other.

The dimensional analysis process will likely succeed if these prior rules are followed. If not, then it will fail, often catastrophically. For complex problems, of which there are many in engineering, the choice of repeating variables may not be obvious. However, they should still be selected as quantities most likely to have substantial effects on the dependent variable. For example, the velocity and density of the flow in an aerodynamic problem are excellent examples of two quantities that will significantly affect the aerodynamic forces.

Formal Check for Linear Independence

To formally check for linear independence (although, in most cases, this will be obvious), the determinant of the sub-matrix formed by the selected repeating variables must be determined, which in this case is

(20)   \begin{equation*} \begin{array}{l|r|r|r|} & \varrho_{\infty} & V_{\infty} & d \\ \hline M & 1 & 0 & 0 \\ L & -3 & 1 & 1 \\ T & 0 & -1 & 0 \end{array} \end{equation*}

If c_1, c_2, and c_3 are non-zero constants in the following equation for linear independence, then

(21)   \begin{equation*} c_1 \left[ \begin{array}{r} 1 \\ -3 \\ 0 \end{array} \right] + c_2 \left[ \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \right] + c_3 \left[ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right] \ne 0 \end{equation*}

Reassembling this in matrix form gives

(22)   \begin{equation*} \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 1 \\ 0 & -1 & 0 \end{array} \right] \left[ \begin{array}{r} c_1 \\ c_2 \\ c_3 \end{array} \right] \ne 0 \end{equation*}

For this matrix equation to have a nontrivial solution then the determinant must be non-zero, i.e., the test is whether

(23)   \begin{equation*} \det \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 1 \\ 0 & -1 & 0 \end{array} \right] \equiv \left| \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 1 \\ 0 & -1 & 0 \end{array} \right| \ne 0 \, ? \end{equation*}

In this case, the determinant is 1.

Therefore, it has now been formally verified that the selected repeating variables \varrho_{\infty}, V_{\infty} and d are indeed linearly independent of each other. In most cases, however, the formal test is not needed because if the variables are chosen such that they do not have the same units or products of the same units, then they will be linearly independent.

Development of the Π Products

Following the Buckingham \Pi method, then the \Pi products in this case will be

    \begin{eqnarray*} \Pi_{1} & = & f_{1} \left( \underbrace{\varrho_{\infty}, V_{\infty}, d}_{\mbox{~\scriptsize Repeating}}  , \underbrace{D}_{\mbox{~\scriptsize Dependent}} \right)  \\[8pt] \Pi_{2} & = & f_{2} \left( \underbrace{\varrho_{\infty}, V_{\infty}, d}_{\mbox{~\scriptsize Repeating}}  , \underbrace{\mu_{\infty}}_{\mbox{~\scriptsize Remaining}} \right) \end{eqnarray*}

First Π Product

For \Pi_{1} then

    \[ \Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} D \]

The values of the coefficients \alpha, \beta, and \gamma must now be obtained to make the equation dimensionally homogeneous and dimensionless.  In terms of the dimensions of the problem, then

    \[ \left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^{\gamma} \left( MLT^{-2} \right) \]

Because the left-hand side of the above equation is dimensionless, it means that the powers or exponents of M, L, and T must add to zero to make the equation dimensionally homogeneous and dimensionless. Therefore, for \Pi_{1} to be dimensionless, then

    \[ M^{0}L^{0}T^{0} = M^{(\alpha+1)} \ L^{(-3\alpha +\beta + \gamma+1)} \ T^{(-\beta-2)} \]

so that

    \begin{eqnarray*} M: 0 & = & \alpha+1  \\ L: 0 & = & -3\alpha +\beta+ \gamma+1  \\ T: 0 & = & -\beta-2 \end{eqnarray*}

By inspection, \alpha = -1, \beta = -2, and \gamma = -2. Therefore, the first \Pi product is

    \[ \Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} D = \varrho_{\infty}^{-1} V_{\infty}^{-2} d^{-2} \, D \]


    \[ \Pi_{1} = \frac{D}{\varrho_{\infty} V_{\infty}^{2} d^2} \]

which is a force coefficient. Usually, aerodynamic force coefficients are defined in terms of the dynamic pressure, i.e., (1/2)\varrho_{\infty} V_{\infty}^2. Also, in this case, a good reference is to use projected area \pi d^2/4. Therefore, the force coefficient would be written as

    \[ \Pi_{1} = C_D = \frac{D}{ \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} (\pi d^2/4)}  = \frac{D}{ \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} A} \]

Dealing with exponents or powers

In dimensional analysis, solving for the dimensionless groupings involves adding the values of the unknown exponents for each base dimension. Exponents are also called the power of the quantity because it represents the number of times a quantity is multiplied by itself. Here are some examples:

    \[ 2^2 \, 2^3 = 2^{2+3} = 2^5 \]

    \[ x^a \, x^b = x^{a+b} \]

    \[ x^a \, y^b \, y^c = x^a \, y^{b+c} \]

    \[ M^{\alpha} \, M^{\beta + 1} = M^{(\alpha + \beta + 1)} \]

    \[ M^{\alpha} \, L^{\gamma} \, M^{\beta -2 } \, L^{-1} = M^{(\alpha + \beta - 2)} \, L^{\gamma -1} \]

When working with exponents, remember that the base parameter has to be the same, and the addition is performed on the exponent.

Second Π Product

For \Pi_{2} then

    \[ \Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} \mu_{\infty} \]

so in terms of dimensions, then

    \[ M^{0} L^{0} T^{0} = \left(ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^\gamma \left( ML^{-1}T^{-1} \right) \]

Equating the powers gives

    \begin{eqnarray*} M: 0 & = & \alpha +1  \\ L: 0 & = & -3\alpha +\beta+ \gamma-1  \\ T: 0 & = & -\beta-1 \end{eqnarray*}

Therefore, in this case \alpha = -1, \beta = -1, and \gamma = -1.

    \[ \Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (S)^{\gamma} \mu_{\infty} = \varrho_{\infty}^{-1} V_{\infty}^{-1} d^{-1} \, \mu_{\infty} \]


    \[ \Pi_{2} = \frac{\mu_{\infty}}{\varrho_{\infty} V_{\infty} d} \]

or by inverting the quantity, a dimensionless grouping is still obtained, i.e.,

    \[ \Pi_{2} = \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}} \]

Recall from previously that this dimensionless parameter is called the Reynolds number.

Final Result for Functional Dependency

As a result of the dimensional analysis of the sphere, then

    \[ \phi_1 \left( \frac {D}{\frac{1}{2} \varrho_{\infty} V_{\infty}^{2} A}, \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}} \right) = 0 \]


    \[ \phi_1 \left( C_{D}, Re \right) = 0 \]

or in explicit form

    \[ C_{D} = \phi_2 \left( Re \right) \]

The functional dependency to be measured is now C_{D} = \phi_2 \left( Re \right). The force can be measured with a balance, and with measurements of the density and viscosity (via measurements of pressure and temperature), the calculations are made of:

1. The drag coefficient, C_D  = \displaystyle{\frac {D}{\frac{1}{2} \varrho_{\infty} V_{\infty}^{2} A}}

2. The Reynolds number, Re = \displaystyle{\frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}}}

The figure below shows the results for the sphere’s drag coefficient as measured in the wind tunnel as a function of Reynolds number. Notice the rapid decrease in the drag coefficient between Re = 3 \times 10^5 and Re = 4 \times 10^5, which suggests that something very profound is happening to the flow about the sphere. Indeed, this behavior is tied to the flow characteristics on the surface of the sphere, called the boundary layer, and the manner in which the flow separates behind the sphere and the creation of a wake.


Drag coefficient of a sphere as a function of Reynolds number based on diameter. Notice the different relationships depending on the surface roughness.

Effects of surface roughness?

Notice also from the results given in the preceding plot that the sphere’s drag coefficient depends on whether the sphere has a smooth or rough surface. Indeed, this is a very interesting outcome as it affects the drag of a golf ball. Therefore, it is helpful to think about what might happen if surface roughness was added as the fifth variable to the parameter list, i.e., in this case, assume that

    \[ D = \phi \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d , \epsilon \right) \]

where \epsilon is a linear measure of the roughness height. The new functional dependence of D in the implicit form is

    \[ \psi\left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d, \epsilon, D \right) = 0 \]

In this case, N = 6, K = 3, N - K = 3, so there are three \Pi products. The first two \Pi products have already been determined, and it is easily shown that now

    \[ C_{D} = \phi_3 \left( Re , \frac{\epsilon}{d} \right) \]

where \epsilon/d is a dimensionless roughness height. Notice that the effects of roughness decrease the drag of a sphere in the range where golf balls fly, which allows them to go further.

Π Method Applied to a Wing

Returning to the wing problem previously discussed, the explicit dependency of the resultant aerodynamic force R (the dependent quantity) in terms of the independent variables is

(24)   \begin{equation*} R = \phi \left( \varrho_{\infty}, V_{\infty}, S, \mu_{\infty}, a_{\infty} \right) \end{equation*}

The functional dependence can also be written in implicit form, i.e.,

(25)   \begin{equation*} \psi \left( \varrho_{\infty}, V_{\infty}, S, \mu_{\infty}, a_{\infty}, R \right) = 0 \end{equation*}

where \psi is some functional dependency to be determined. Following the \Pi method then N = 6, K = 3, N - K = 3, and so three \Pi products must be found in this case.


The resultant force on a wing of a given size and orientation to the flow will be dependent on several parameters, including the velocity, density, and viscosity of the flow.

Dimensions of the Variables

Now, the dimensions of the variables must be determined. Remember that the square brackets [-] around a variable means “the dimensions of” that variable. So, for each dependency in this case, then

    \begin{eqnarray*} \left[ R \right] & = & MLT^{-2} \\ \left[ \varrho_{\infty} \right] & = & ML^{-3} \\ \left[ V_{\infty} \right] & = & LT^{-1} \\ \left[ S \right] & = & L^2 \\ \left[ \mu_{\infty} \right] & = & ML^{-1}T^{-1} \\ \left[ a_{\infty} \right] & = & LT^{-1} \end{eqnarray*}

Notice that the units of force are obtained from the product of mass and acceleration, i.e., [F] = ( M ) ( LT^{-2}). It is critically important that each of the variables is assigned the correct base units. If not, the Buckingham \Pi method will fail.

Forming the Dimensional Matrix

The solution to the dimensional analysis problem proceeds by setting up a dimensional matrix with all of the parameters being given in terms of their base units, i.e.,

(26)   \begin{equation*} \begin{array}{l|r|r|r|r|r|r} & R & \varrho_{\infty} & V_{\infty} & S & \mu_{\infty} & \alpha_{\infty} \\ \hline \mbox{Mass: } M & 1 & 1 & 0 & 0 & 1 & 0 \\ \mbox{Length: } L & 1 & -3 & 1 & 2 & -1 & 1 \\ \mbox{Time: } T & -2 & 0 & -1 & 0 & -1 & -1 \end{array} \end{equation*}

Choice of Repeating Variables

In this problem, a good choice of the repeating variables is \varrho_{\infty}, V_{\infty}, and S, which is a common choice for aerodynamic problems. Remember not to choose the dependent variable as a repeating variable. Remember also that the repeating variables must be linearly independent, i.e., the units of any one repeating variable cannot be linearly obtained from the units of any other. Choosing repeating variables with different dimensions or products of dimensions is necessary. For example, two repeating variables comprising a length and an area or two repeating variables with units, cannot be used, or the method will fail. Furthermore, the repeating variables must collectively include all of the base dimensions of the problem. In this case, the choice of the repeating variables \varrho_{\infty}, V_{\infty}, and S satisfy all of these requirements.

Recap: Rules for choosing the repeating variables

The formal determination that the repeating variables are linearly independent takes some work, although it is just arithmetic. However, in some cases, this outcome can be verified by inspection. Some basic rules can help choose the repeating variables, so the formal test for linear independence can be foregone. These rules are:

  • NEVER choose the dependent variable as a repeating variable! If this rule is ignored, then the Buckingham \Pi method will fail catastrophically.
  • Always choose repeating variables that will, or are likely to, have a primary effect on the dependent variable. For aerodynamic problems, these are usually the density of the flow, the speed of the flow, and some area or length scale.
  • Do NOT choose repeating variables that have the same dimensions, so don’t choose two repeating variables, such as flow velocity and the speed of sound (which both have units of length/unit time), or two length variables, such as a wing chord and a wing span (both have units of length). For example, length and area scales cannot both be used as repeating variables, which are also not linearly independent.
  • Remember that the choice of repeating variables may not, in most cases, be unique! The choice is not arbitrary, but a different choice of repeating variables will inevitably lead to a different set of dimensionless groupings. Even though dimensionless groupings are obtained, the process can be conducted again with a different set of repeating variables.

Development of Π Products

Continuing to follow the Buckingham \Pi procedure, then the \Pi products in this problem are

    \begin{eqnarray*} \Pi_{1} & = & f_{1} \left( \underbrace{\varrho_{\infty}, V_{\infty}, S}_{\mbox{~\scriptsize Repeating}} , \underbrace{R}_{\mbox{\scriptsize ~Dependent}} \right) \\[8pt] \Pi_{2} & = & f_{2} \left( \underbrace{\varrho_{\infty}, V_{\infty}, S}_{\mbox{~\scriptsize Repeating}} , \underbrace{\mu_{\infty}}_{\mbox{~\scriptsize Other}} \right) \\[8pt] \Pi_{3} & = & f_{3} \left( \underbrace{\varrho_{\infty}, V_{\infty}, S}_{\mbox{~Repeating}} , \underbrace{a_{\infty} }_{\mbox{~\scriptsize Remaining}}\right) \end{eqnarray*}

First Π Product

Take the first \Pi product, \Pi_{1}, then

(27)   \begin{equation*} \Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (S)^{\gamma} R \end{equation*}

where the coefficients \alpha, \beta, and \gamma are unknowns, and their values must be obtained appropriately to make the equation dimensionally homogeneous, i.e., finding their values to make the equation dimensionless with units of 1. In terms of the dimensions, then

(28)   \begin{equation*} \left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L^2 \right)^{\gamma} \left( MLT^{-2} \right) \end{equation*}

where \left[x\right] is used to mean “the dimensions of variable x.” Collecting the terms gives

(29)   \begin{equation*} \left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = M^{(\alpha+1)} \, L^{(-3\alpha+\beta+2\gamma+1)} \, T^{(-\beta-2)} \end{equation*}

Because the left-hand side of the above equation is dimensionless, it means that the powers or exponents of M, L, and T must add to zero to make the equation dimensionally homogeneous, i.e.,

    \begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -3\alpha+\beta+2\gamma+1 \\ T: 0 & = & -\beta-2 \end{eqnarray*}

Therefore, in this case by inspection then \alpha = -1, \beta = -2, and \gamma = -1 and so the first \Pi product is

(30)   \begin{equation*} \Pi_{1} = \varrho_{\infty}^{-1} V_{\infty}^{-2} S^{-1} R \end{equation*}


(31)   \begin{equation*} \Pi_{1} = \frac{R}{\varrho V_{\infty}^{2} S} \end{equation*}

Hence, \Pi_{1} is a form of force coefficient. Aerodynamic force coefficients are nearly always defined in terms of the dynamic pressure, i.e., (1/2)\varrho_{\infty} V_{\infty}^2, which is referred to as q_{\infty}, so more usually the force coefficient would be written as

(32)   \begin{equation*} \Pi_{1} = C_R = \frac{R}{ \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} S} = \frac{R}{ q_{\infty} S} \end{equation*}

Notice that multiplying a similarity parameter by a constant value does not change the units of that parameter.Paralleling the above approach used to find \Pi_{1}, the two remaining \Pi products, namely \Pi_{2} and \Pi_{3}, can now be derived.

Second Π Product

For \Pi_2 then

(33)   \begin{equation*} \Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (S)^{\gamma} \mu_{\infty} \end{equation*}


(34)   \begin{equation*} \left[ \Pi_{2} \right] = M^{0} L^{0} T^{0} = \left(ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L^2 \right)^\gamma \left( ML^{-1}T^{-1} \right) \end{equation*}


    \begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -3\alpha+\beta+2\gamma-1 \\ T: 0 & = & -\beta-1 \end{eqnarray*}

Therefore, in this case \alpha = -1, \beta = -1, and \gamma = -1/2 and so this \Pi product becomes

(35)   \begin{equation*} \Pi_{2} = \varrho_{\infty}^{-1} V_{\infty}^{-1} S^{-1/2} \mu_{\infty} \end{equation*}

Notice that this derived \Pi product is a dimensionless parameter that will become infinite as the velocity tends to zero. This behavior is inconvenient in practice, but the \Pi product can be inverted to prevent that. The outcomes are still a dimensionless parameter, but now one that is more convenient, i.e., the \Pi product becomes

(36)   \begin{equation*} \Pi_{2} = \frac{\varrho_{\infty} V_{\infty} S^{1/2}}{\mu_{\infty}} \end{equation*}

Because, in this case, there is a square-root of an area, it can be replaced by a linear dimension, such as the chord of the wing c, i.e., the area of a rectangular wing will be its chord c times the span of the wing b or S = c \, b, so now

(37)   \begin{equation*} \Pi_{2} = \frac{\varrho_{\infty} V_{\infty} \, c}{\mu_{\infty}} \end{equation*}

The choice of the appropriate length scale is somewhat arbitrary (e.g., the wing span could have been used). Still, generally, in an aerodynamic problem, a parameter describing a downstream distance should be used, e.g., wing chord c in this case.

This latter dimensionless parameter is known as the Reynolds number, which is denoted by the symbol Re. The Reynolds number is one of the most significant parameters in aerodynamics because it governs the relative magnitude of viscous effects to inertia effects in the flow.

Third Π Product

Similarly, for the third \Pi product, then

(38)   \begin{equation*} \Pi_{3} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (S)^{\gamma} a_{\infty} \end{equation*}

and so, in terms of dimensions, then

(39)   \begin{equation*} \left[ \Pi_{3} \right] = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right) ^{\beta} \left( L^2 \right)^{\gamma} \left( LT^{-1} \right) \end{equation*}


    \begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha+\beta+2\gamma+1 \\ T: 0 & = & -\beta-1 \end{eqnarray*}

Therefore, in this case \alpha = 0, \beta = -1, and \gamma = 0, i.e.,

(40)   \begin{equation*} \Pi_{3} = \varrho_{\infty}^{0} V_{\infty}^{-1} S^{0} a_{\infty} = \frac{a_{\infty}}{V_{\infty}} \end{equation*}

Again, the \Pi product can be inverted for convenience (to put V_{\infty} on the numerator) and still have adimensionless quantity so

(41)   \begin{equation*} \Pi_{3} = \frac{V_{\infty}}{a_{\infty}} \end{equation*}

This particular dimensionless parameter is called the Mach number (the free-stream Mach number in this case), which is the ratio of the free-stream velocity to the free-stream speed of sound. The Mach number is an important quantity in aerodynamics that affects the magnitude of the forces on a body but also the compressible nature of the flow, in general.

Final Result for Functional Dependency

Therefore, as a result of the dimensional analysis of the wing problem, then

(42)   \begin{equation*} \phi_1 \left( \frac {R}{\frac{1}{2} \varrho_{\infty} V_{\infty}^{2} S}, \frac{\varrho_{\infty} V_{\infty} \, c }{\mu_{\infty}}, \frac{V_{\infty}}{a_{\infty}} \right) = 0 \end{equation*}


(43)   \begin{equation*} \phi _1 \! \left( C_{R}, Re, M_{\infty} \right) = 0 \end{equation*}

or in explicit form

(44)   \begin{equation*} C_{R} = \phi_3 \! \left( Re, M_{\infty} \right) \end{equation*}

In dimensional form, then

(45)   \begin{equation*} R =  \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} S \, \phi_{3} \! \left(Re, M_{\infty} \right) \end{equation*}

which is sometimes called Rayleigh’s equation.

The previous result is a significant outcome because it means the original aerodynamic problem in the wind tunnel has been reduced from one involving five independent variables (for which it is practically impossible to determine the functional relationships to R) to one that now only has two variables! Therefore, with a little effort, it has been shown that to determine the force R on a given wing at a given angle of attack \alpha, then only the dependence of Re and M_{\infty} on R need to be established. The dimensionless parameters Re (the Reynolds number) and M_{\infty} (the Mach number) are known as similarity parameters.

The plot below shows the actual results from the wind tunnel test to measure the resultant force coefficient, C_R, as a function of the angle of attack of the wing, \alpha, to the free stream flow. The measurements were performed at two different values of the Reynolds number, which in this case was obtained by varying the wind speed. Notice the significant effect of a higher Reynolds number, which increases the angle of attack and the maximum force coefficient before the onset of stall. Look carefully at the plot below – why does the resultant force vary depending on whether the angle of attack is increasing versus decreasing? In general, the values of the Reynolds and Mach numbers will substantially affect the aerodynamic forces produced on an airfoil section and on a wing, which is discussed in more detail in the following chapters.


The resultant force coefficient on a wing as a function of its angle of attack at two values of the Reynolds number based on the wing chord.

Adding another dependency to the problem

Think about what might happen if span b was added as a seventh variable to the parameter list in this previous problem, i.e., in this case, assume that

    \[ \psi \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, a_{\infty}, c, b  \right) = 0 \]

Therefore, N = 7, K = 3, N - K = 4, and so four \Pi products will result from the dimensional analysis. The first three have already been determined, so it should be fairly obvious that the fourth dimensionless grouping will be

    \[ \Pi_{4} = \frac{b}{c} = A\!R \]

which is a span-to-chord ratio and is known as the aspect ratio of the wing, A\!R. Notice in the figure below the profound influence of the aspect ratio on the lift coefficient. Wings with a high aspect ratio are more efficient in that for a given angle of attack, and all other parameters held constant, then a wing with a high aspect will produce more lift and less drag.

Diagram of the effects of wing aspect ratio on the lift coefficient of a finite wing.

Practicing Worked Examples

The best way of learning the method of dimensional analysis is by doing examples. Practice checking and balancing equations for dimensional homogeneity to ensure that the terms have the correct dimensions and units. After working through two or three more examples, the student will quickly realize the power of dimensional analysis and how this technique can help in engineering problem-solving. Naturally, the method is not restricted to aerodynamics, and the process of dimensional analysis is helpful in all branches of engineering and the physical sciences.

Worked Example #1 – Balancing equations for dimensional homogeneity

The kinetic energy, KE of a rotating propeller, depends on its moment of inertia, I, and its angular velocity, \Omega. The relationship can be expressed as

    \[ KE = C \, I^{\alpha} \, \Omega^{\beta} \]

where C is a dimensionless constant. Write down the base dimensions of each of the parameters. Then, by satisfying dimensional homogeneity, determine the values of \alpha and \beta. Show all steps in your analysis. Hint: Moment of inertia has dimensions of mass times length squared. Show ALL steps in your analysis.

We are given that

    \[ KE = C \, I^{\alpha} \, \Omega^{\beta} \]

where C is a dimensionless constant. In terms of the dimensions

    \[ \left[ KE\right] = \left[ I^{\alpha} \right] \left[ \Omega^{\beta} \right] \]

In terms of dimensions for each parameter, then

    \[ \left[ KE \right] = M L^2 T^{-2} \quad\quad \left[ I \right] = M L^{2} \quad\quad \left[ \Omega \right] = T^{-1} \]


    \[ M L^2 T^{-2} = \left( M L^{2} \right)^\alpha \, \left( T^{-1}\right)^\beta \]

For dimensional homogeneity

    \begin{eqnarray*} M: \quad 1 & = & \alpha \\ L: \quad 2 & = & 2 \alpha \\ T: \quad -2 & = & -\beta \end{eqnarray*}

so \alpha = 1 and \beta = 2. Inserting the values gives the relationship as

    \[ KE = C \, I^{1} \, \Omega^{2} = C \, I \, \Omega^{2} \]

Therefore, the final answer is

    \[ KE = C \, I \, \Omega^{2} \]

Worked Example #2 – Drag on a tiny aerosol particle

A tiny spherical particle of diameter d (assume in the sub-micron size range) is falling freely vertically at velocity V in air, as shown in the figure below. Find the similarity parameter that governs the drag on the particle.Illustration of a tiny spherical particle falling freely in the vertical direction.

Functional Dependence

The aerodynamic drag on the particle D can be written in functional form as

    \[ D = \phi ( d, V, \mu ) \]

where \mu is the viscosity of the air and \phi is some function to be determined. In implicit form, then

    \[ \psi (d, V, \mu, D ) = 0 \]

where \psi is some other function. In this case, N = 4 and K = 3 (by inspection, all of mass, length, and time are involved in this problem), so there are N - K = 1, i.e., one \Pi product.

Dimensional Matrix

Setting down the dimensional matrix, then

    \[ \begin{array}{l|r|r|r|r|r} & d & V & \mu & D \\ \hline \mbox{Mass} ~M: & 0 & 0 & 1 & 1 \\ \mbox{Length} ~L: & 1 & 1 & -1 & 1 \\ \mbox{Time} ~T: & 0 & -1 & -1 & -2 \end{array} \]

\Pi Product

The dependent variable D in this case) cannot be a repeating variable, and so for the only \Pi product, then

    \[ \Pi = (d)^{\alpha} (V)^{\beta} (\mu)^{\gamma} D \]

and in terms of dimensions, then

    \[ \left[ \Pi \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1})^{\beta} (M L^{-1} T^{-1})^{\gamma} M L T^{-2} \]

For the \Pi group to be dimensionless, then

    \begin{eqnarray*} M: 0 & = & \gamma + 1 \\ L: 0 & = & \alpha + \beta - \gamma + 1 \\ T: 0 & = & -\beta - \gamma - 2 \end{eqnarray*}

Therefore, \gamma = -1, and \beta = -1 and \alpha = -1 so the \Pi product is

    \[ \Pi = (d)^{-1}(V)^{-1} (\mu)^{-1} D = \frac{D}{d V \mu} =  C_D \]

which is a drag coefficient C_d applicable, in this case, to what is known as a Stokes flow.


Notice: It is always good to check the final units of the grouping to be sure it is dimensionless, i.e., it has units of 1. So, in this case

    \[ \left[ C_D \right]  = \frac{M L T^{-2}}{(L) (L T^{-1}) (M L^{-1} T^{-1}} = \frac{M L T^{-2}}{M L T^{-2}}  =  1 \]

which indeed confirms that the grouping is dimensionless.

What happens when things go wrong?

Sometimes when using the Buckingham \Pi method, it is impossible to find a set of repeating variables that are linearly independent or a set that includes all of the fundamental dimensions of the problem.  In some other cases, the solution to the problem of finding the powers of the parameters becomes indeterminate, i.e., we cannot uniquely solve for the dimensionless groupings. The accepted solution to this dilemma is to reduce the number of repeating variables by one and so create an extra \Pi grouping. If this approach is used, then the dimensional analysis process should now work, at least mathematically. However, the groupings may not be unique and some experimentation may be required by eliminating different repeating variables to find acceptable groupings that make physical sense.

Worked Example #3 – Bobbing booster in the ocean

A solid rocket booster parachutes back to Earth and falls into the sea. The booster bobs around in the sea upright at a frequency of \omega. Using dimensional analysis, show that the dimensionless frequency of this motion, k, is given by

    \[ k = \omega \sqrt{\frac{d}{g}} \]

It may be assumed that for this problem, the frequency of motion depends on the diameter of the booster d, the mass of the booster, m, the density of the water \varrho, and acceleration under gravity, g.

Functional Dependency

By using the information provided then, the frequency of motion can be written in an explicit functional form as

    \[ \omega = \phi (D, m, \varrho, g) \]

where \phi is some function to be determined, or in implicit form as

    \[ \psi (d, m, \varrho, g, \omega) = 0 \]

where \psi is some other function. Therefore, N = 5, K = 3 and so there are two \Pi products.

Repeating Variables

The repeating variables must collectively include all the units of mass M, length L, and time T, so the best choice here is d, m, and g. Notice that if \varrho were to be chosen instead of g as a repeating variable, then collectively, the repeating variables would not include time T. So the Buckingham \Pi method will fail in this case. The dependent variable, \omega, cannot be used as a repeating variable.

\Pi Products

Therefore, the two \Pi products are

    \[ \Pi_1 = f_1(d, m, g, \omega) \]


    \[ \Pi_2 = f_2(d, m, g, \varrho) \]

Dimensional Matrix

The dimensional matrix is

    \[ \begin{array}{l|r|r|r|r|r} &   d  & m & g & \omega & \varrho \\ \hline \mbox{Mass} ~M: &    0 & 1 & 0 & 0 & 1 \\ \mbox{Length} ~L: &    1 & 0 & 1 & 0 & -3 \\ \mbox{Time} ~T:  &  0 & 0 & -2 & -1 & 0 \end{array} \]

Derivation of the \Pi Products

The first \Pi product is

    \[ \Pi_{1} = (d)^{\alpha} (m)^{\beta} (g)^{\gamma} \omega \]

where the values of the coefficients \alpha, \beta and \gamma must be obtained to make the equation dimensionally homogeneous and dimensionless.  In terms of the dimensions of the problem, then

    \[ \left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left( M \right)^{\beta} \left( L T^{-2} \right)^{\gamma} \left( T^{-1}\right) \]

For \Pi_{1} to be dimensionless, then the powers or exponents of M, L and T must add to zero, i.e., in this case

    \begin{eqnarray*} M: 0 & = & \beta  \\ L: 0 & = & \alpha + \gamma  \\ T: 0 & = & -2\gamma - 1 \end{eqnarray*}

Therefore, \alpha = 1/2, \beta = 0, and \gamma = -1/2, so the \Pi_1 product is

    \[ \Pi_1 = (d)^{1/2} (m)^{0} (g)^{-1/2} \omega = \omega \sqrt{\frac{d}{g}} \]

For the second \Pi product then

    \[ \Pi_{2} = (d)^{\alpha} (m)^{\beta} (g)^{\gamma} \varrho \]

In terms of the dimensions, then

    \[ \left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left(M \right)^{\beta} \left( L T^{-2} \right)^{\gamma} \left(  ML^{-3}\right) \]

For \Pi_{2} to be dimensionless, then

    \begin{eqnarray*} M: 0 & = & \beta + 1 \\ L: 0 & = & \alpha + \gamma - 3  \\ T: 0 & = & -2\gamma \end{eqnarray*}

Therefore, \beta = -1, \gamma = 0, and \alpha = 3, so the \Pi_2 product is

    \[ \Pi_2 = (d)^{3} (m)^{-1} (g)^{0} \varrho = \frac{\varrho d^3}{m} \]

which is a buoyancy similarity parameter, i.e., the ratio of the mass \varrho d^3 (or weight \varrho d^3  g) of the water displaced to the mass m (or weight m g) of the rocket booster.

Final Dimensionless Functionality

Finally, in dimensionless form, then

(46)   \begin{equation*} \Pi_{1} = \phi_3 (\Pi_2) \end{equation*}

or that

(47)   \begin{equation*} \omega \sqrt{\frac{d}{g}} = \phi_3 \left( \frac{\varrho d^3}{m}\right) \end{equation*}

CAUTION: Pound force versus pound mass? 

In engineering, we generally always use units of “slugs” (or sometimes just “slug”) for mass in the USC system and never “pounds-mass.” Using “slugs” for mass and “pound” for force prevents the muddling of “pounds-force” lb_f and “pounds mass” lb_m, which, if misused, can result in catastrophic errors in engineering calculations. A pound-force is defined as the product of a pound mass and acceleration under gravity, i.e., 1 lb = 1 lb_m x g, i.e., 1 pound force = 1 pound mass times acceleration under gravity or about 32.17 ft s^{-2}. Therefore, a 1 lb force will accelerate a 32.17 lb mass at 1 ft s^{-2}. However, a 1 lb force will also accelerate 1 slug of mass at 1 ft s^{-2}. This means that a “slug” has a mass equivalent to 32.17 “pounds mass.” So, when we are using “pounds mass,” there is always a multiplicative factor of “g” or “32.17 ft s^{-2}” floating around, and most of the time, it causes nothing but problems in calculations for students. However, the multiplicative factor is “1” when using slugs for the above reason. So, we see immediately why the preference is to keep everything simple and in good order by using “slugs” and never “pounds mass,” and so remove the extra “g” or 32.17 from our engineering calculations.

Other Dimensionless Quantities

Some physical variables are naturally dimensionless by their definition, i.e., as ratios of the same dimensional quantities. One example often encountered in thermodynamics and aerodynamics is the ratio of specific heats, \gamma = C_P / C_{\cal{V}}. Another example of a naturally dimensionless parameter used in structural analysis is strain, which is defined as the ratio of the change in length of a structural component under loading, \epsilon, compared to its initial length, L, i.e., the strain is \epsilon / L. Finally, in hydrostatics, the specific gravity, SG, is often used, which is a dimensionless number because it is defined as the ratio of the density of a given liquid relative to the density of water.

Finally, it should be remembered that all angles are dimensionless. In engineering analysis, angles measured in radians are always used, defined as a ratio of the arc length to the radius of a circle. While angles are often measured or reported in degrees, units of degrees must be converted to radians for most engineering calculations. A complete circle subtends an angle of 2\pi radians. which is equivalent to 360 degrees or 360^{\circ}, so the angular conversion is 1 radian = 180/\pi degrees. The same should also be done with angular rates, which should always be converted to radians per second (rad/s) no matter what units the rates are originally given in, e.g., rates are often measured in revolutions per minute (rpm).

Conversion Factors

A ratio that converts one unit of measurement into another unit of measurement is called a conversion factor. Engineers often use conversion factors from common measurement units to engineering units or from USC units to SI units and vice-versa. Examples of conversion factors include:

  • 1 kilogram (kg) = 1,000 grams (g) – units of mass.
  • 1 pound (lb) = 453.592 grams (g) – units of mass.
  • 1 foot (ft) = 12 inches (in) – units of length.
  • 1 inch (in) = 2.54 centimeters (cm) – units of length.
  • 1 hour (hr) = 3,600 seconds (s) – units of time.
  • 1 minute (min) = 60,000 milliseconds (ms) – units of time.
  • 1 atmosphere (atm) = 101.3 kilo-Pascals (kPa) – standard ISA pressure.
  • 1 lb/ft2 (psf) = 144 lb/in2 (psi) – units of pressure.

    Often pressure may be measured in units of “pounds per square inch” or lb/in^2 or psi, but this is not an engineering (base) unit. It usually needs to be converted to “pounds per square foot” or lb/ft^2 or psf, i.e., one psi = 1/144 psf where the conversion factor, in this case, is 1/144 the factor 144 being 12 \times 12 where 12 inches equals one foot. Other units of pressure measurement include a “bar” where 1 bar = 100 kPa = 100,000 Pa with Pa (1 Pascal = 1 Newton per square meter or 1 Pa = 1 N/m^2) being the base engineering unit of pressure measurement. The bar is not an official SI unit but is often used in meteorology and weather forecasting. To obtain pressure units of Pa from measurements of pressure in bars, then 1 Pa = 1/100,000 bars, so the conversion factor, in this case, is 1/100,000 or 10^{-5}. Recall that a Pascal is a small amount of pressure, so the use of hectoPascals (hPa), which is 100 Pa, kilo-Pa (kPa), or even Mega-Pa (MPa), is not uncommon.

    Summary & Closure

    Dimensional analysis can be a powerful tool for engineering analysis because it reduces and simplifies the size and scope of the problem in terms of dimensionless or non-dimensional groupings called similarity parameters. The strength of dimensional analysis as a tool usually becomes apparent when applied to increasingly more complicated engineering problems, such as where the effects of all the dependencies cannot be established explicitly and/or the governing equations for the solution of the problem can become formidably complicated. For example, it is common for engineering problems to have five, six, seven, or even more dependencies. However, sorting out the problem parameters into dimensionless groupings helps understand what parameters are essential as they will affect the physical behavior and what parameters are not.

    The dimensional analysis must be used in a systematic and structured approach, formally embodied in the Buckingham \Pi method. A critical first step is to carefully lay out the problem in terms of the dependent and independent quantities, which must all be assigned the correct base units. While the choice of the repeating quantities is somewhat arbitrary, specific rules must be followed for the method to work. For example, it is critical that the dependent quantity NOT be used as a repeating variable. The repeating variables must also have substantial effects on the dependent quantity and be linearly independent of each other, i.e., two repeating parameters with the same dimensions cannot be used, such as two reference length scales or two reference velocities. If the basic rules of the Buckingham \Pi method are followed, then the outcomes will be groups of dimensionless parameters that should not only help understand the problem but will simplify its scope.

    5-Question Self-Assessment Quickquiz

    For Further Thought or Discussion

    • You are tasked with explaining to a high-school student studying physics the advantages of using dimensional analysis and dimensionless quantities. What two primary arguments would you make?
    • The force on a body in the wind tunnel is known to depend on flow velocity (airspeed) and the static pressure and temperature in the flow. What other parameters may be involved? Hint: Do not suggest parameters such as viscosity or the speed of sound.
    • Use dimensional analysis to show that the drag of a ship’s hull, in part, depends on the Froude number. The Froude number is defined as F\!r = V_{\infty}/\sqrt{L g}, where L is the length of the ship and V_{\infty} is the speed of the ship through the water.
    • You are tasked with determining the hydrodynamic drag force D on a fully submerged, very long, neutrally buoyant cable dragged behind a ship. You propose to do this by doing small-scale experiments in a water tunnel. Determine the dimensionless groupings that will affect this problem.
    • Show that the pressure drop caused by a restrictor in a flow can be expressed in terms of the Euler number Eu, where the pressure drop is expressed in terms of an energy loss in the flow per unit of kinetic energy per volume of the flow.

    Other Useful Online Resources

    To dive deeper into the methods of dimensional analysis, navigate to the following online resources:



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