4.1.4 Dimensionless Variables∗
This section shows how solving systems of equations can be used to determine appropriate
dimensionless variables. It is only an introduction to this topic. I got this example from . This
considers a specific example of a simple airplane wing shown below. We assume for simplicity
that it is just a flat plane at an angle to the wind which is blowing against it with speed V as
The angle is called the angle of incidence, B is the span of the wing and A is called the
chord. Denote by l the lift. Then this should depend on various quantities like θ,V,B,A and so
forth. Here is a table which indicates various quantities on which it is reasonable to expect l to
Here m denotes meters, sec refers to seconds and kg refers to kilograms. All of these are likely familiar
except for μ. One can simply decree that these are the dimensions of something called viscosity but it
might be better to consider this a little more.
Viscosity is a measure of how much internal friction is experienced when the fluid moves. It is roughly a
measure of how “sticky” the fluid is. Consider a piece of area parallel to the direction of motion of the fluid.
To say that the viscosity is large is to say that the tangential force applied to this area must be large in
order to achieve a given change in speed of the fluid in a direction normal to the tangential force.
Thus the units on μ are kg sec−1m−1 as claimed above.
Then one would think that you would want
However, this is very cumbersome because it depends on seven variables. Also, it doesn’t make very good
sense. It is likely that without much care, a change in the units such as going from meters to feet would
result in an incorrect value for l. The way to get around this problem is to look for l as a function of
dimensionless variables multiplied by something which has units of force. It is helpful because first
of all, you will likely have fewer independent variables and secondly, you could expect the
formula to hold independent of the way of specifying length, mass and so forth. One looks
where the units on ρV 2AB are
which are the units of force. Each of these gi is of the form
and each gi is independent of the dimensions. That is, this expression must not depend on
meters, kilograms, seconds, etc. Thus, placing in the units for each of these quantities, one
Notice that there are no units on θ because it is just the radian measure of an angle. Hence its dimensions
consist of length divided by length, thus it is dimensionless. Then this leads to the following equations for
Then the augmented matrix for this system of equations is
The row reduced echelon form is then
and so the solutions are of the form
Thus, in terms of vectors, the solution is
Thus the free variables are x2,x3,x5,x7. By assigning values to these, we can obtain dimensionless
variables by placing the values obtained for the xi in the formula 4.11. For example, let x2 = 1 and all the
rest of the free variables are 0. This yields
The dimensionless variable is then A−1B1. This is the ratio between the span and the chord.
It is called the aspect ratio, denoted as AR. Next let x3 = 1 and all others equal zero. This
gives for a dimensionless quantity the angle θ. Next let x5 = 1 and all others equal zero. This
Then the dimensionless variable is V −1V 01. However, it is written as V∕V 0. This is called the Mach
number ℳ. Finally, let x7 = 1 and all the other free variables equal 0. Then
then the dimensionless variable which results from this is A−1V −1ρ−1μ. It is customary to write it as
. This one is called the Reynolds number. It is the one which involves viscosity. Thus we
would look for
This is quite interesting because it is easy to vary Re by simply adusting the velocity or A but it is hard to
vary things like μ or ρ. Note that all the quantities are easy to adjust. Now this could be used,
along with wind tunnel experiments to get a formula for the lift which would be reasonable.
Obviously, you could consider more variables and more complicated situations in the same