Procedure of Creating Dimensionless Groups

1. List all Variables that are included in the problem 2. Express each variable in terms of basic dimension 3. Determine the required number of pi terms 4. Select a number of repeating variables 5. Form a pi term by multiplying one of the non repeating variable by the product of repeating variables each raised to an exponent that will make the combination dimensionless 6. Repeat step 5 7. Check all the resulting pi terms 8. Express the final form as a relationship among the pi terms and think about what it means

Recapitulation

1. A base quantity is a property that is defined in physical terms by two operations: a comparison operation, and an addition operation. The comparison operation is a physical procedure for establishing whether two samples of the quantity are equal or unequal; the addition operation defines what is meant by the sum of two samples of that property.

2. Base quantities are properties for which the following concepts are defined in terms of physical operations: equality, addition, subtraction, multiplication by a pure number, and division by a pure number. Not defined in terms of physical operations are: product, ratio, power, and logarithmic, exponential, trigonometric and other special functions of physical quantities.

3. A base quantity can be measured in terms of an arbitrarily chosen unit of its own kind and a numerical value.

4. A derived quantity of the first kind is a product of various powers of numerical values of base quantities. A derived quantity is defined in terms of numerical value (which depends on base unit size) and does not necessarily have a tangible physical representation.

5. The dimension of any physical quantity, whether base or derived, is a formula that defines how the numerical value of the quantity changes when the base unit sizes are changed. The dimension of a quantity does not by itself provide any information on the quantity's intrinsic nature. The same quantity (e.g. force) may have different dimensions in different systems of units, and quantities that are clearly physically different (e.g. work and torque) may have the same dimension.

6. Relationships between physical quantities may be represented by mathematical relationships between their numerical values. A mathematical equation that correctly describes a physical relationship between quantities is dimensionally homogeneous (see section 2.5). Such equations remain valid when base unit sizes are changed arbitrarily.

7. The categorization of physical quantities as either base or derived is to some extent arbitrary. If a particular base quantity turns out to be uniquely related to some other base quantities via some universal law, then we can, if we so desire, use the law to redefine that quantity as a derived quantity of the second kind whose magnitude depends on the units chosen for the others. All base quantities that are transformed into derived quantities in this way retain their original physical identities (i.e. their comparison and addition operations), but their numerical values are measured in terms of the remaining base quantities, either directly via a defining equation or indirectly by using a unit that is derivable from the remaining base units.

8. A system of units is defined by (a) the base quantities, (b) their units, and (c) the derived quantities, each with either its defining equation or the form of the physical law that has been used to cast the quantity into the derived category. Both the type and the number of base quantities are open to choice.

The comparison and addition operations are physical, but they are required to have certain properties that mimic those of the corresponding mathematical operations for pure numbers:

(1) The comparison operation must obey the identity law (if A=B and B=C, then A=C)

(2) The addition operation must be commutative (A+B=B+A), associative [A+ (B+C) = (A+B) + C], and unique (if A+B=C, there exists no finite D such that A+B+D=C).

The two operations together define, in entirely physical terms:

(1) The concept of larger and smaller for like quantities (if there exists a finite B such that A+B=C, then C>A),

(2) Subtraction of like quantities (if A+B=C, then A C-B),

(3) Multiplication of a physical quantity by a pure number (if B=A+A+A, then B 3A)

(4) Division of a physical quantity by a pure number (if A=B+B+B, then B A/3).

Summary of Dimensional Analysis:

1. Clearly define the problem. 2. Consider the basic laws that governs the phenomenon 3. Start the variables selection process by grouping the variables into 3 broad classes: • Geometrical prop. • Material prop. • External Factors

4. Consider other variables that may not fall into the above categories. 5. Be sure to include all the quantities that enter the problem even though some of them may be held constant. 6. Make sure all variables are independent.

Example:
Deformation of an elastic ball striking a wallSuppose we wish to investigate the deformation that occurs in elastic ballswhen they impact on a wall. We might be interested, for example, infinding out what determines the diameter d of the circular imprint left onthe wall after a freshly dyed ball has rebounded from it (figure 3.1).

Step 1: The independent variables:
The first step is to identify a complete set of independent quantities that determine the imprint radius d. We begin by specifying the problem more clearly. We agree to restrict our attention to (initially) spherical, homogeneous balls made of perfectly elastic material, to impacts at perpendicular to the wall, and to walls that are perfectly smooth and flat and so stiff and heavy that they do not deform or move during the impact process. We also agree to adopt a SI system of unit (L, M, and T).
[pic]

The numerical value of a dependent variable like d will be depend on the values of all quantities that distinguish one impact event from another. Experience suggests that these should include at least the following: the ball's diameter D and velocity V just prior to contact (the initial conditions) and its mass m. The ball’s intrinsic material properties will also play a role. Our theoretical understanding of solid mechanics tells us that the quasi-static response of a perfectly elastic material is characterized by two material properties, the modulus of elasticity E and Poisson's ratio γ, and that the inertial effects which inevitably come into play during collision and rebound will also depend on the material’s density. The properties of the wall are irrelevant if it is indeed perfectly rigid, as we assumed. We know, however, by thinking of how the problem would have to be set up as a theoretical one, that the answer for the numerical value d will also depend on the values of all universal constants that appear in the physical laws that control the ball's impact dynamics. In this case the process is governed by Newton's law of motion and the law of mass conservation. Having chosen SI system of units, we know that Newton's law has the form F=ma and contains neither universal constants nor any physical constants in the law of mass conservation.

We seem to arrive at the conclusion that d depends on six quantities:
D, V, m, E, γ and ρ. This is a complete set, as required, but not an independent set because once the ball's mass and diameter are specified, its density follows.
We must therefore exclude either the density or the mass. (Other quantities like V2, DE1/2, etcetera, all involving quantities that affect the value of d, are excluded for the same reason: they are not independent of the quantities already included.) We conclude that the following relationship expresses the impact diameter in terms of a complete set of independent variables: d = F (D, V, m, E, γ, ρ).

Step 2: Dimensional considerations:
In the type of system of units we have adopted in step 1, the dimensions of the quantities in equation are: Independent: [V] = Lt-1 [ρ] = ML-3 [D] = L [E] = ML-1t-2 [γ] = 1

Dependent: [d] = L

Step 3: π-terms:
Number of π-terms equals to the “k-r”, where “k” is the number of variables (independents and dependents) and “r” is number of basic units used (r=3 in case of using SI system of units (LMT))

Therefore we should have 3 π-terms.

Step 4: Repeating Variables:
Number of repeating variables equals to the number of basic units used.
In this case “r” equals to 3 basic units (LMT)

Inspection of the above shows that the three quantities V, ρ, and D, for example, comprise a complete, dimensionally independent subset of the five independent variables. The dimension of any one of these three cannot be made up of the dimensions of the other two. The dimensions of the remaining independent variables E and γ and the dependent variable d can, however, be made up of those of V, and D as follows: Independent: [E] = ML-1t -2 = (ML-3 )(Lt)2 = [ V2 ] [γ] = 1

Dependent: [d] = L = [D]
In addition it is recommended to use the simplest variables as the repeating variables; this will make the formation of π-terms easier.

Step 5&6&7: Formation of π-terms:

Π0 = d (Va.Db.ρc)
Π1 = E (Va.Db.ρc)
Π2 = γ (Va.Db.ρc)

By solving each equation separately we will get:

Π0 = d.D-1
Π1 = E.V-2.ρ-1
Π2 = γ

Step 8: End of Game

Using the logic that led to Buckingham's p-theorem, we now conclude that

Π0= f (Π1, Π2),

Or

d/D = f(E/ρV2, γ)

The number of independent variables has been reduced from the original n=5 that define the problem to n-(k-1) =2