13 Dimensional Analysis


New engineers quickly begin to recognize from even the most casual observations that the solution to engineering problems usually involves the consideration of many dependencies. This means that many parameters could affect the physical behavior of a given system, and different effects will be obtained from varying one parameter versus another. In particular, the behavior of fluids is relatively difficult to describe because they deform, especially as they flow around objects (often referred to as “bodies”). The fluid also produces forces and moments on bodies, including the familiar lift and drag forces.

There is certainly an intuitive expectation that what the flow does (e.g., where the fluid goes), 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., although other parameters may be involved.

To solve this type of fluid problem, one approach is to set down and use the equations of motion governing the behavior of the fluid and then try to solve these equations. At least in principle, this approach will allow the effects of the fluid on the body to be determined as well as the effects of the body on the fluid. This approach also requires the specification of appropriate free-stream quantities and other boundary conditions, which may or may not be known in mathematical form. However, this is not a straightforward approach to solving the problem, not just because of the complexity of the equations that must be solved, but also because much practice and proficiency are needed in the solution methods.

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

Objectives of this Lesson

  • Understand how to perform dimensional analysis, a method used to bring out the primary dependencies in an engineering problem in terms of non-dimensional grouping called similarity parameters.
  • Better understand the significance of similarity parameters and how they can be used to 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 that are 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 aerospace 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.
Base quantity Symbol SI unit USC unit
Mass M kilogram (kg) slug (slug or slugs or sl)
Length L meter (m) foot (ft)
Time T second (s) second (s)
Temperature \theta Kevin (^{\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.

Finding Base Dimensions

Consider some examples to determine the primary (base) dimensions of each of the following parameters (often referred to as “quantities” or “variables”) from thermodynamics, that is:

  1. Energy E.
  2. Work W.
  3. Power P.
  4. Heat Q.

Remember that these are all derived quantities in that they are derived from a combination of base quantities. From now on, notice that quantity inside a set of square brackets “[ ]” is used to mean the dimensions of that quantity. Notice also that scientific notation must be used throughout to avoid any ambiguity in the final units of the derived quantity.

1. Energy, E, is the ability to do work and is measured in Joules (J) in the SI system and “foot-pounds” (ft-lb) in the USC system. Energy has the same units of work (force times distance), so \left[ E \right] = \left( MLT^{-2} \right) L = ML^{2} T^{-2} which are the base units of energy. Notice that the units of force are obtained from the product of mass and acceleration, i.e., [F] = ( M) (LT^{-2}) = M L T^{-2}.

2. Work, W, is also measured in Joules (J) in the SI system and “foot-pounds” (ft-lb) in the USC system. Work is equivalent to a force times a distance, so \left[ W \right] = ML T^{-2} L =  M L^{2} T^{-2}, which are the base units of work.

3. Power, P, is the rate of doing work and is measured in Watts (W) in the SI system and “foot-pounds per second” (ft-lb/s or ft-lb s^{-1} in the USC system. Power is equivalent to a force times distance per unit time (or force times velocity), so \left[ P \right] = \left( MLT^{-2} \right) L T^{-1} = M L^{2} T^{-3} which are the base units of power. In practice, power is measured in terms of kiloWatts (kW) in SI and horsepower (hp) in USC, where one horsepower is equivalent to 550 ft-lb/s.

4. Heat Q has units of energy (the ability to do work), which has the same units as work, i.e., units of Joules (J) in the SI system and “foot-pounds” (ft-lb) in the USC system. In base units then\left[ Q \right] = \left( MLT^{-2} \right) L = ML^{2} T^{-2} which are the base units of heat.

Obtaining the correct units for derived parameters: Many of the parameters routinely used in engineering are derived parameters, e.g., volume, density, force, work, energy, power, etc. Use these examples to explain their significance in terms of base units.
  • Volume is 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), and so the derived units for volume are L^3. Density is also a derived parameter because the density of something is its mass (M) per unit volume (L^3), so density will have units of M L^3.
  • In the same sense, force is a derived parameter, although in some cases, “force” can be carried through the problem as a base unit. A force F is the product of a mass and acceleration (Newton’s second law), so in terms of its base units, then [F] = ( M) (LT^{-2}) = M L T^{-2}.
  • Getting the correct base units of any given parameter takes some practice, but it is essential for correctly using dimensional analysis; otherwise, the method will fail. Again, the scientific notation should be used when expressing the units of a parameter to avoid ambiguity, e.g., always use 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, and for convenience, it can be assumed that this force acts at what is known as a center of pressure, which, by definition, a point where there are no pitching moments.

A wing set at an angle to the flow, in this case being tested in the ERAU wind tunnel, will produce some resultant force, R.

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, \rho_{\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 “infinity” or \infty is used to mean 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 as the flow approaches the wing, these values are all likely to change. 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 airfoil) and the direction of the free-stream flow velocity V_{\infty}.

The resultant force R can also be decomposed into a vertical component that acts perpendicular to the wind flow (called the lift, L) and a horizontal component parallel to the direction of the wind (called the drag, D).

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

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

where R would be referred to as the dependent variable and \rho_{\infty}, V_{\infty}, \mu_{\infty} and a_{\infty} are called the independent variables. The functional dependence, i.e., the relationship between the variables, which in this case is denoted by “f,” is still to be determined. Although this latter equation expresses a very 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}, \rho_{\infty}, S, \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} is easy because it 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, then 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. 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 in the wind tunnel, or even by doing parallel calculations using some form of aerodynamic theory.

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}, \rho_{\infty}, S, etc., would each affect R. If just one dependent variable was used, then just a single column of numbers would be obtained, and the functional relationship would be pretty 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. Systematically varying each independent parameter to establish the “cause and effect” on the dependent parameter would be impractical from a measurement perspective, nor would it be attractive from a computational (modeling) approach.

Basis of Dimensional Analysis

Fortunately, the approach to the preceding problem can be simplified by employing dimensional analysis, which can be used to determine sets of non-dimensional (or dimensionless) groups of parameters that affect the problem of interest. Such an approach will also reduce the number of independent variables, so it becomes an essential technique for understanding many engineering problems, not just aerodynamics.

Dimensional analysis is based on the fact that if an equation of parameters like P_1 + P_2 + P_3 = P_4 expresses some physical relation, then the parameters P_1, P_2, P_3, and P_4 must all have the same dimensions (units) for the equation to be mathematically valid, e.g., P_1, P_2, P_3, and P_4 could all be velocity components. This previous equation can be made non-dimensional or dimensionless, or what is called dimensionally homogeneous by dividing through by one of the terms, i.e., the same result can be expressed as

(2)   \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*}

Lord Rayleigh and Joseph Bertrand originally set down these simple but profoundly important ideas as 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 engineering use. The Buckingham \Pi Theorem is a generalized formalization of the method of dimensions, which can now be examined.

The Buckingham Π Theorem

Let K be equal to the number of fundamental dimensions required to describe the physical variables. In mechanics, all physical variables can be expressed in terms of mass, length, and time, so in general K=3. Let P_{1}, P_{2}, ...P_{N} represent N physical variables in the general physical relationship

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

where g represents the functional relationship to be determined. Then the Buckingham \Pi theorem states that the above relation 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 repeating variables plus one other variable.

To see how this works, let P_{1}, P_{2}, ....P_{K} be the selected repeating variables, i.e.,

(4)   \begin{eqnarray*} \Pi_{1} & = & g_{1} \left( P_{1}, P_{2}......P_{K}, P_{K+1} \right) \\[6pt] \Pi_{2} & = & g_{2} \left( P_{1}, P_{2}, .....P_{K}, P_{K+2} \right) \\[6pt] \Pi_{N-K} & = & g_{3} \left( P_{1}, P_{2}, .......P_{K}, P_{N} \right) \end{eqnarray*}

where now there are new functional relationships g_{1}, g_{2} and g_{3} in terms of the \Pi products. Therefore, it can now be written that

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

where \phi is the functional relationship (still to be determined) but now in terms of each of the \Pi products. This means that the functional relationship has been reduced from N variables to N-K variables. The reduction the number of dependent parameters, therefore, is potentially a significant simplification in the analysis of the problem.

The best way of learning the method of dimensional analysis is by example. The more examples you do, the more comfortable you will become with the technique.

The Π Method Applied to a Wing

Returning to the wing problem discussed above, the explicit dependency of the resultant aerodynamic force R in terms of the dependent variables is

(6)   \begin{equation*} R = f \left( \rho_{\infty}, V_{\infty}, S, \mu_{\infty}, a_{\infty} \right) \end{equation*}

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

(7)   \begin{equation*} g \left( \rho_{\infty}, V_{\infty}, S, \mu_{\infty}, a_{\infty}, R \right) = 0 \end{equation*}

where g is some other function. Hence, following the |Pi method then N = 6, K = 3, N - K = 3, and so three \Pi products must be found in this case.

Understanding Explicit Versus Implicit Functional Forms

Explain the differences 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 always starts off the process by writing the functional dependency of the problem parameters in implicit form.

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

    \begin{eqnarray*} \left[ R \right] & = & MLT^{-2} \\ \left[ \rho_{\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}).

Choice of Repeating Variables

In this particular problem, \rho_{\infty}, V_{\infty} and S can be chosen as the repeating variables. Although the selection of the repeating variables can be viewed as somewhat arbitrary, in general, they must always be selected based on three essential rules:

  1. The dependent variable (R in this case) cannot be used as a repeating variable, so it must appear only in one group by default.
  2. The repeating variables must include all of the physical 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.
  3. 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.

If these prior rules are followed, the dimensional analysis process will likely succeed. If not, then it will fail, often catastrophically. For complex problems, of which there are many in engineering, the repeating variables should also be selected to be quantities that are likely to have the most substantial effects on the dependent variable. For example, the velocity and density of the flow in an aerodynamic problem are excellent examples of quantities that will have significant effects, e.g., on the force produced on the wing in this case.

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.,

(8)   \begin{equation*} \begin{array}{l|r|r|r|r|r|r} & R & \rho_{\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*}

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

Linear Independence of Repeating Variables

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. In a practical sense, 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.

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

(9)   \begin{equation*} \begin{array}{l|r|r|r|} & \rho_{\infty} & V_{\infty} & S \\ \hline M & 1 & 0 & 0 \\ L & -3 & 1 & 2 \\ 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

(10)   \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 \\ 2 \\ 0 \end{array} \right] \ne 0 \end{equation*}

Reassembling this in matrix form gives

(11)   \begin{equation*} \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 2 \\ 0 & -1 & 0 \end{array} \right] \left[ \begin{array}{r} c_1 \\ c_2 \\ c_3 \end{array} \right] \ne 0 \end{equation*}

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

(12)   \begin{equation*} \det \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 2 \\ 0 & -1 & 0 \end{array} \right] \equiv \left| \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 2 \\ 0 & -1 & 0 \end{array} \right| \ne 0 \, ? \end{equation*}

In this case, the determinant is 2. Therefore, it has now been formally verified that the selected repeating variables \rho_{\infty}, V_{\infty} and S are linearly independent of each other.

Confirming that Repeating Variables are Linearly Independent

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 be used to 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 non-dimensional groupings. Even though non-dimensional 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 are

    \begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \rho_{\infty}, V_{\infty}, S, R \right) \\ \Pi_{2} & = & g_{2} \left( \rho_{\infty}, V_{\infty}, S, \mu_{\infty} \right) \\ \Pi_{3} & = & g _{3} \left( \rho_{\infty}, V_{\infty}, S, a_{\infty} \right) \end{eqnarray*}

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

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

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

(14)   \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

(15)   \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

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


(17)   \begin{equation*} \Pi_{1} = \frac{R}{\rho 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)\rho_{\infty} V_{\infty}^2 so more usually the force coefficient would be written as

(18)   \begin{equation*} \Pi_{1} = C_R = \frac{R}{ \frac{1}{2} \rho_{\infty} V_{\infty}^{2} 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, the \Pi products \Pi_{2} and \Pi_{3} can be derived. For \Pi_2 then

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


(20)   \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

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

Notice that this derived \Pi product is a non-dimensional parameter that will become infinite as the velocity tends to zero. This behavior is not very convenient in practice, but the \Pi product can simply be inverted to prevent that from happening. The outcomes is still a non-dimensional parameter, but now one that is more convenient, i.e., the \Pi product becomes

(22)   \begin{equation*} \Pi_{2} = \frac{\rho_{\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 s or S = c \, s, so now

(23)   \begin{equation*} \Pi_{2} = \frac{\rho_{\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) but generally in an aerodynamic problem a parameter describing a downstream distance should be used, e.g., wing chord c in this case.

This latter non-dimensional 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 air.

Similarly, for the third \Pi product then

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

and so

(25)   \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.,

(26)   \begin{equation*} \Pi_{3} = \rho_{\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 a non-dimensional quantity so

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

This particular non-dimensional 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 not only 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 on the wing problem then

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


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

or in explicit form

(30)   \begin{equation*} C_{R} = f_{0} \left( Re, M_{\infty} \right) \end{equation*}

The previous result is a significant outcome because it means the original aerodynamic problem in the wind tunnel has been reduced from one that involves five independent variables (for which is practically impossible to determine the functional relationships for R) to one that now only has only 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 non-dimensional parameters Re (the Reynolds number) and M_{\infty} (the Mach number) are known as similarity parameters.

Both the Reynolds number and the Mach number are known to have substantial effects on the aerodynamic forces produced on a wing. In the figure below, the wing lift coefficient (non-dimensional lift) is plotted versus the angle of attack for a series of free-stream Mach numbers from 0.3 (subsonic) up to 0.75 (transonic). Notice that there are two primary effects. The first effect is that increasing the Mach number increases the lift coefficient for a given angle of attack, at least in the lower angle of attack regime. The second effect is that the maximum value of the attainable lift coefficient is reduced with increasing Mach number, which is a consequence of the flow detaching or separating from the wing and is called a stall. In other words, the angle of attack at which stall occurs decreases with increasing Mach number. A discussion of the effects of Reynolds number on the lift will be discussed later.

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

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

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

Worked Examples

The process of dimensional analysis is best learned by means of doing more examples.  After working through two or three more examples, students 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 dimensional analysis is useful in all branches of engineering and the physical sciences.

Worked Example #1 – Viscous Drag on a Small 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.

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

    \[ D = f( d, V, \mu ) \]

where \mu is the viscosity of the air. The drag can be written is explicit functional form as

    \[ D = f( d, V, \mu ) \]

where “f” is some function to be determined. In implicit form then

    \[ f_1 (d, V, \mu, D ) = 0 \]

where “f_1” 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. Setting down the dimensional matrix then

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

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 non-dimensional, 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 non-dimensional

Worked Example #2 – Dynamic Motion of a Floating Object

A solid rocket booster parachutes back to Earth and falls into the sea. The booster bobs around in the sea in an upright position at a frequency of \omega. Using dimensional analysis, show that the non-dimensional 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 \rho, and acceleration under gravity, g.

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

    \[ \omega = f(D, m, \rho, g) \]

where “f” is some function to be determined, or in implicit form as

    \[ f_1 (d, m, \rho, g, \omega) = 0 \]

where “f_1” is some other function. Therefore, N = 5, K = 3 and so there are two \Pi products. 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 \rho were to be chosen instead of g as a repeating variable, then collectively the repeating variables would not include time T and so the Buckingham \Pi method will fail in this case. The dependent variable, \omega, cannot be used as a repeating variable.

The \Pi products are

    \[ \Pi_1 = f_2(d, m, g, \omega) \]


    \[ \Pi_2 = f_3(d, m, g, \rho) \]

The dimensional matrix is

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

So, 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.  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

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

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

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

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

Finally, in non-dimensional form then

(31)   \begin{equation*} \Pi_{1} = f_0 (\Pi_2) \end{equation*}

or that

(32)   \begin{equation*} \omega \sqrt{\frac{d}{g}} = f_0 \left( \frac{\rho d^3}{m}\right) \end{equation*}

Worked Example #3 – Drag on a Banner Towed by an AIrplane

The AIAA Design Build & Fly (DBF) team need to determine the factors that influence the aerodynamic drag on a rectangular banner being towed behind their airplane. The size of the banner is determined by its length, l, and height, h. Use the Buckingham \Pi method to determine the non-dimensional groupings that will govern this problem. You may also assume that the problem is governed by the airspeed of the airplane, as well as the density and viscosity of the air. Assume further that the  banner remains flat and does not flutter in the flow behind the airplane.

The relationship between the drag on the banner D and the air properties can be written in the general functional form as

    \[ D = f \left( \rho_{\infty}, V_{\infty}, \mu_{\infty}, h, l \right) \]

where the size of the banner is represented by its length, l, and height, h. Remember that D is called the dependent variable and \rho_{\infty}, V_{\infty}, \mu_{\infty}, h and l are are the independent variables. The functional dependence of D in implicit form is

    \[ g \left( \rho_{\infty}, V_{\infty}, \mu_{\infty}, h, l, D \right) = 0 \]

Counting up the variables gives N = 6 and K = 3 because there are 3 fundamental dimensions in this problem. Therefore, N - K = 3 and so there will be three \Pi products.

For each variable the dimensions are

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

The dimensional matrix is

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

Choose \rho_{\infty}, V_{\infty} and l as the repeating variables, which will all have primary effects on the drag of the banner. These variables also collectively include all the fundamental dimensions of this problem and are linearly independent of each other.

Following the Buckingham \Pi method, then the three \Pi products are:

    \begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \rho_{\infty}, V_{\infty}, l, D \right)  \\ \Pi_{2} & = & g_{2} \left( \rho_{\infty}, V_{\infty}, l, \mu_{\infty} \right) \\ \Pi_{3} & = & g_{3} \left( \rho_{\infty}, V_{\infty}, l, h \right) \end{eqnarray*}

For \Pi_{1}:

    \[ \Pi_{1} = (\rho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, D \]

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

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

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

    \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

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


    \[ \Pi_{1} = \frac{D}{\rho_{\infty} V_{\infty}^{2}\  l^2} \]

i.e., a form of drag coefficient. The aerodynamic force coefficients are usually defined in terms of the dynamic pressure, i.e., (1/2)\rho_{\infty} V_{\infty}^2 so that more conventionally the force coefficient would be written as

    \[ \Pi_{1} = C_D = \frac{D}{ \frac{1}{2} \rho_{\infty} V_{\infty}^{2} \, l^2} \]

It would also be legitimate to write the drag coefficient as

    \[ C_D = \frac{D}{ \frac{1}{2} \rho_{\infty} V_{\infty}^{2} \, l \, h } = \frac{D}{ \frac{1}{2} \rho_{\infty} V_{\infty}^{2} \, A } \]

where the banner area A = l \, h is used rather than l^2. Ultimately, how  C_D is defined is just matter of convenience and/or consistency with established convention.

For \Pi_{2}:

    \[ \Pi_{2} = (\rho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, \mu_{\infty} \]


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


    \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, so

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


    \[ \Pi_{2} = \frac{\mu_{\infty}}{\rho_{\infty} V_{\infty} l} \]

Inverting the grouping gives

    \[ \Pi_{2} = \frac{\rho_{\infty} V_{\infty} l}{\mu_{\infty}} \]

which in the latter case is a Reynolds number based on the banner length. Notice that the grouping can be inverted if needed, such as to follow established convention or just because it is otherwise convenient to do so.

For \Pi_{3}:

    \[ \Pi_{3} = (\rho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, h \]


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


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

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

    \[ \Pi_{3} = (\rho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, h = \rho_{\infty}^{0} V_{\infty}^{0} l^{-1} \, h = \frac{h}{l} \]


    \[ \Pi_{3} =  \frac{h}{l} \]

or again inverting this grouping (for convenience) giving

    \[ \Pi_{3} =  \frac{l}{h} = A \! R \]

which is a length to height ratio or what would be called an aspect ratio A\!R.

As a result of our dimensional analysis then

    \[ \phi \left( \frac {D}{\frac{1}{2} \rho_{\infty} V_{\infty}^{2} \, A }, \frac{\rho_{\infty} V_{\infty} l }{\mu_{\infty}}, \frac{l}{h} \right) = 0 \]


    \[ \phi \left( C_{D}, Re, A\!R\right) = 0 \]

Finally, in explicit form the drag coefficient can be written as a function of Reynolds number based on banner length and the aspect ratio of the banner, i.e.,

    \[ C_{D} = f_{0} \left( Re, A\!R \right) \]

Note: Try this problem again using \rho_{\infty}, \mu_{\infty} and h as the repeating variables. What happens to the groupings?

Other Dimensionless Quantities

Some physical variables are naturally dimensionless by virtue of 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 that are measured in radians are always used, which is 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).

CAUTION: Pound force versus pound mass? 

In engineering, we 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/s2” 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 calculations.

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, an example being converting from the temperature in Celsius to engineering temperature units of Rankine.

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 needs to be converted to “pounds per square foot” or lb/ft^2 or psf, i.e., 1 psf = 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. A Pascal is a small amount of pressure, so the use of kilo-Pa (kPa) or even Mega-Pa (MPa) is not uncommon. So, 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}.

When NASA lost a 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 blunder, muddling quantities between the SI and USC systems. NASA’s Jet Propulsion Laboratory (JPL) used the SI system, while Lockheed used the USC system. Nobody noticed or bothered to check! Because of this blunder, the navigation system flew the spacecraft too fast and too close to the atmosphere, where it burned up. Read more about this mistake here as well as some other engineering disasters that occurred because of the muddling of units. Be sure not to make the same mistakes!

What is a horsepower?

The output at the shaft 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. Watt wanted to compare the power output of his new steam engines to that of horses to help market them. 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}. Watt was not too concerned about accuracy. All he wanted was a simple but representative quantitative measure of the power delivered by a horse relative to what his steam engines could produce and so better market his engines. The 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 exactly what they were buying. The unit of  “horsepower” is still used today almost worldwide.

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 non-dimensional groupings called similarity parameters. The strength of dimensional analysis as a tool usually becomes apparent when it is 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 seven or more dependencies. However, sorting out the problem parameters into non-dimensional 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, which is 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 adhered to 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, e.g., 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 non-dimensional (unitless) parameters that should not only help in the understanding of 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 non-dimensional 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) as well as 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 the hull of a ship, 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 being dragged behind a ship. You propose to do this by doing small scale experiments in a water tunnel. Determine the non-dimensional 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 in the methods of dimensional analysis, navigate to the following online resources: