Test by dimension is a
well-known, quick and efficient means to check geometrical or physical
formulas. 1.  In order to recall the operation of the test, let us
consider the frustum of a right circular cone. Let

       R be the radius of the lower base,
       r the radius of the upper base,
       h the altitude of the frustum,
       S the area of the lateral surface of the frustum.

If R, r, h, are given, S is visibly determined, we find the
expression 
     S = pi(R+r)sqrt((R-r)^2+h^2)

to which we wish to apply the test by dimension.
      The dimension of a geometric quantity is easily visible.
Thus, R,r,h are lengths, they are measured in centimetres
meters if we use scientific units, their dimension is cm
The area S is measured in square centimetres, is dimension
is cm^2.  Now, pi=3.14159... is mere number if
we wish to ascribe a dimension to a purely numerical
quantity it must be cm^0=1
     Each term of a sum must have the same dimension 
which is also the dimension of the sum. Thus, R, r, and 
R+r have the same dimension, namely cm.  The two
terms (R-r)^2 and h^2 have the same dimension (*as they
must),.cm^2.
     The dimension of a problem is the product of the dimensions 
of its factors, and there is a similar rule about
powers. Replacing the quantities by their dimensions on 
both sides of the formula that we are testing, we obtain

                cm^2 = 1 * cm * sqrt(cm^2)

    This being visibly so, the test could not detect any error
in the formula. The formula passed the test.
     For other examples, see section 14 and  CAN YOU CHECK THE RESULT?,2)
2.  We may apply the test by dimension to the final
result of a problem or to intermediary results, to our own
work or to the work of others (very suitable in tracing
mistakes in examination papers), and also to formulas
that we recollect and to formula that we guess.
     If you recollect the formulas 4pir^2 and 4pir^3/3 for the
area and the volume of the sphere, but are not quite sure
which is which, the test by dimension easily removes the 
doubt. 
     3. The test by dimension is even more important in 
physics than in geometry.
     Let us consider a "simple" pendulum, that is, a small
heavy body suspended by a wire whose length we regard
as inseparable and whose weight we regard as neglible.
Let l stand for the length of the wire, g for the gravitational 
acceleration, and T for the period of the pendulum.
     Mechanical considerations show that T depends on l
and g alone.  But what is the form of the dependence?
We may remember or guess that
  
                   T = cl^mg^n

where c, m, n are certain numerical constants.  That is,
we suppose that T is proportional to certain powers, l^m,
g^n, of l and g.
     We look at the dimensions.  As T is a time, it is measured 
in seconds, its dimension is sec. Thedimensionss of 
the lengths l is cm, the dimension of the acceleration g is 
cm sec^-2 and the dimension of the numerical constant c
is 1.  The test by dimension yield the equation

               sec = 1 * (cm)^m *(cm sec^-2)^n

or 
               sec = (cm)^(m+n) * sec^(-2n)

    Now, we must have the same powers of the fundamental
units cm and sec on both sides, and thus we obtain
 
              o = m + n             1 = -2n

and hence
 
             n = -1/2               m = 1/2

Therefor, the formula for the period T must have the 
form

                T = c l^(1/2) g^(-1/2) = c sqrt(l/g)

The test by dimension yields much in this case but it 
cannot yield everything. First, it gives no information
about the value of the constant c(which is, in fact 2pi). 
Second, it gives no information about the limits of validity;
the formula is valid only for small oscillation of the 
pendulum and only approximately (it is exact or "infinitely
small" oscillations).  In spite of these limitations
there is no doubt that the consideration of the dimensions
has allowed us to foresee quickly and with the most
elementary means an essential part of a result whose
exhaustive treatment demands much more advanced
means.  And this is so in many similar cases.