Machine Design

Creating Efficient Graphs With Dimensionless Variables

Although units like kilogram and newton seem indispensable to an equation, plotting a function using dimensionless variables requires much fewer curves

Thomas Szirtes
Thomas Szirtes and Associates Inc.
Toronto, Ont., Canada

Equations and graphs are the building blocks of most product design. Engineers who work with them need to know the relationship between a dependent variable and often several independent variables. Graphs are a helpful way to convey these relationships, especially when dealing with just two or three variables. But as the number of variables increases, plotting the function becomes more and more tedious. Getting useful information from the plots is no easy matter either. Try plugging six values into each variable of a five-variable function and you’ll end up with 216 curves on 36 separate charts. Dimensionless variables can reduce these figures drastically.

A dimensionless variable (DV) is a unitless value produced by (maybe repeatedly) multiplying and dividing combinations of physical variables, parameters, and constants. In design engineering, dimensionless variables are useful for graphing multivariable relations, performing dimensional modeling, and reconstructing explicit relationships among physical variables. An important feature of a dimensionless variable is that its value is independent of the dimensional system in which it is expressed. In other words, the dimensionless variables for a function expressed in metric units will be identical to the dimensionless variables for the function in English units.

A few basics are useful before putting DVs to work. For example, a dimension is a measuring unit acting on the magnitude in a function. In the expression:

total mass = 91 kg

“91” is the magnitude and “kg” is the dimension. According to Buckingham’s Theorem, the number of independent DVs in a physical system is the difference between the number of variables and the number of fundamental dimensions. For instance, if v, q, and L denote speed (m/sec), acceleration (m/sec2), and length (m), then the number of variables is three and the number of dimensions is two (m, sec). Therefore only one (3 – 2 = 1) independent DV can be formed. It should be noted that the dimension of a dimensionless variable is actually one, not zero, as the name falsely implies. A more descriptive name might be unit-dimensional variable. But to follow the existing — albeit inappropriate — practice, the term dimensionless variable is used. It’s conventional to use π, always with a subscript, to denote a DV. Here of course π has nothing to do with the constant 3.1415...

Constructing the Dimensional Set
To construct DVs, first study the problem at hand and determine which variables influence the behavior of the system. In this context, variables also include parameters and constants. It is important to consider all relevant variables because the result will be incorrect if a relevant variable is omitted. If, however, an irrelevant variable is included, calculating the result may be more complex but just as correct. Therefore the prudent advice is: when in doubt, include it.

Consider finding the deflection, U (m), of a cantilever loaded laterally by an external concentrated force, F (N), at its free end. The deflection is influenced by the length, L (m), of the cantilever, the second moment of area, I (m4), of the cross section, and Young’s Modulus, E (N/m2), of the material. Therefore the relevant variables are U, F, I, E, L. Write them in a row from left to right starting with the dependent variable (U in this case). Now write the relevant dimensions, m and N, in a column on the left. Be careful to use only a single dimensional system. In other words, if you express the deflection in meters, you cannot express the beam length in feet. The row of variables and column of dimensions create the boundaries for the “dimensional matrix.” For the above variables and dimensions, the setup is:




Next, fill in the dimensional matrix. Each matrix element is the exponent of a particular variable’s dimension. For example, the variable E has the dimensions N/m2. Therefore, at the intersection of E and N write 1. Similarly, at the intersection of column E and row m write –2. A negative value is used because the fraction, N/m2, is converted to N·m-2 when filling in the matrix. Repeating this process for every variable produces the dimensional matrix:



Now calculate the value of the right-most determinant of order Nd in the dimensional matrix. Nd is the number of dimensions, which in our example is two. Therefore in this example the right-most second-order determinant must be calculated. Verify that this value is not zero. If it is zero then rearrange the columns of the dimensional matrix to obtain a nonzero value. In this example the value in question is –1, so we may continue.

The matrix formed by the right-most determinant is called the A matrix. The B matrix is formed by the remaining elements of the dimensional matrix. In this example they are:



Now use these matrices to determine the C matrix. By the “fundamental formula”:



where A-1 is the inverse of A, the superscript T designates the transpose of the matrix, and the dot indicates matrix multiplication.

The C matrix for this example is:




Next form the “Dimensional Set.” The Dimensional Set consists of four matrices. These are the A, B, and C matrices and the identity matrix, D.

In the Dimensional Set the C matrix is written below the A matrix, and the D matrix is written below the B matrix. This layout produces a Dimensional Set that is also a matrix. In our example the Dimensional Set will be:



Here we have Nv = five variables and Nd = two dimensions. Now, by Buckingham’s Theorem, the number of independent dimensionless variables, Np, is:

Np = NvNd = 5 – 2 = 3

The three independent dimensionless variables constitute a “complete set” as shown above. By this theorem the set unambiguously defines the behavior of the discussed physical system. The three dimensionless variables, π1, π2, and π3, are written in the left-most column immediately beneath the dimensions.

Now form these three dimensionless variables using the elements of the C and D matrices. These elements are used as exponents of the variables under which they fall. They are, accordingly:

For example, to express π3, use the last row of the Dimensional Set. Here the exponent of I is 1, the exponent of L is –4. and the exponents of all other variables are zero. Therefore,

The other dimensionless variables are formed similarly. Thus, each variable can be expressed in terms of its basic units which will cancel out. The result in all cases will be the dimension of one. Hence all of these variables are, indeed, dimensionless.

Note the important distinction that the elements of matrices A and B are the exponents of the respective dimensions, whereas the elements of matrices C and D are the exponents of the respective variables.

As it turns out A and D are always square matrices, whereas B and C may not be. In our example A is a 2 X 2 matrix, B is a 2 X 3 matrix, C is a 3 X 2 matrix, and D is a 3 X 3 matrix.

Economy of graphical presentation
In the graph of a typical relation joining Nv variables, the abscissa is usually the independent variable, the ordinate is the dependent variable, and the parameter is constant along each of the family of curves. If each variable and parameter can assume k distinct values, then the total number of curves needed is:

Ncurv = kNv-2

However, since each chart can have k curves, the number of charts needed is:

Nchart = kNv-3 if Nv > 2

Nchart= 1 if Nv > 2

Although six values for each variable usually generates an adequate plot of a function, it is evident from the above equations that the number of curves and charts increases drastically with the number of variables. For example, four variables of six distinct values each require 36 curves on six charts. Similarly, five variables with six distinct values each can be represented by 216 curves on 36 charts. The table on the following page presents these data.

An illustrative example
Suppose it is necessary to construct a plot for the natural frequency of a transversally vibrating, simply supported prismatic beam. The setup is:

The formula:



contains five variables. Note that π without a subscript is not a dimensionless variable, but the dimensionless and familiar constant, 3.1415... Plotting this relation using six values for each variable would require 216 curves on 36 charts. Not only would this plotting be prohibitively tedious, but extracting information from 36 pages of charts would be another daunting task. However, dimensionless variables provide the same information in a more manageable package. The Dimensional Set, by the preceding protocol, is:




In this formula, there are only two variables. So a single curve, shown at right, can represent the function — a vast improvement over the 216 curves and 36 charts required without using dimensionless variables.

This solitary curve represents the same amount of information as the suggested 36 families of curves. This fact clearly manifests the immense time and effort savings by using dimensionless variables. Let us now see how this plot is used. Say we have an aluminum bar with length, L = 1.126 m and Young’s modulus, E = 6.9 X 1010 kg/(m·sec2). The bar has a rectangular solid cross section 0.06-m wide and 0.08-m thick. The second moment of area of the cross section is I = 2.56 X 10-6 m4. The material’s density, r = 2,770 kg/m3, produces a bar mass, M = 14.9713 kg. With these values:



For this the plot supplies π1 = 0.0105. Hence, using the given composition of π1, we have:



Thus the beam vibrates laterally with a frequency of 142.56 Hz.

© 2010 Penton Media, Inc.

Hide comments


  • Allowed HTML tags: <em> <strong> <blockquote> <br> <p>

Plain text

  • No HTML tags allowed.
  • Web page addresses and e-mail addresses turn into links automatically.
  • Lines and paragraphs break automatically.