What is Dimensional Analysis?

Date: 11/26/2001 at 21:20:59
From: Danielle
Subject: Dimensional Analysis

What is dimensional analysis and how does it work?

Date: 11/28/2001 at 12:00:37
From: Doctor Greenie
Subject: Re: Dimensional Analysis

Hi, Danielle -

Dimensional analysis is a tool that can be used to determine how to 
manipulate formulas by analyzing the dimensions in the formulas. In 
the physical world, these units are often units of distance, mass (or 
weight) and time; but many other units are possible (see, for example, 
the "fun" problem in the Dr. Math archives for which I have provided a 
link at the end of this response).

One elementary application of dimensional analysis is in the 
memorization of simple geometric formulas. For a particular example, 
many students, when beginning to learn formulas related to circles, 
get confused between the formulas for the circumference and area of a 
circle:

   circumference = pi * diameter

   area = pi * (radius squared)

Using dimensional analysis, it is impossible to confuse these two 
formulas:

(1) "pi times diameter"

In this formula, "pi" is a pure number, and the diameter is a length.  
If the diameter is in inches, then "pi times diameter" is in inches; 
if the diameter is in miles, then "pi times diameter" is in miles.  
The units on "pi times diameter" are always units of length. So "pi 
times diameter" measures a length - so it can't be an area formula; in 
particular, it can't be the formula for the area of a circle,  But it 
can be a formula for a length; and in fact it is the formula for the 
circumference of a circle.

(2) "pi * (radius squared)"

In this formula, "pi" is again a pure number, and the radius is a 
length. If the radius is in inches, then "pi * (radius squared)" is in 
square inches; if the radius is in miles, then "pi * (radius squared)" 
is in square miles. The units on "pi * (radius squared)" are always 
units of (length squared), which are units of area. So "pi * (radius 
squared)" measures an area - so it can't be the formula for any 
length; in particular, it can't be the formula for the circumference 
of the circle. But it can be the formula for an area; and in fact it 
is the formula for the area of a circle.

Another elementary application of dimensional analysis is in 
converting units of measure. For example, young students often get 
confused over whether to divide by 12 or multiply by 12 when 
converting a measurement in inches to a measurement in feet:

    180 inches = ??? feet

You can analyze this problem using dimensional analysis to determine 
whether to multiply or divide by 12. In this application of 
dimensional analysis, you make fractions equivalent to "1" and 
"multiply" your given measurement by those fractions; when you do 
this, you can "cancel" like units in the numerator and denominator, 
just as you can cancel common numerical factors in the numerator and 
denominator of any fraction.

In this example, 1 foot = 12 inches, so the following two fractions 
are equivalent to "1":

       1 foot
(1)   --------- = 1
      12 inches

and

      12 inches
(2)   --------- = 1
       1 foot

Now try multiplying the given measurement, "180 inches," by each of 
these fractions:

(1)

                 1 foot     (180 * 1) (feet * inches)  
   180 inches * --------- = -------------------------
                12 inches          12 inches

Here, the units "inches" cancel in numerator and denominator, leaving 
an answer in feet:

                 1 foot     180
   180 inches * --------- = --- feet = 15 feet
                12 inches    12

or (2)

                12 inches   (180 * 12) (inches * inches)
   180 inches * --------- = ---------------------------- = ????!!!
                 1 foot               feet

Here, there are no common units in numerator and denominator that can 
be cancelled - so we know this is not the right way to convert the 
measurement "180 inches" to feet.

Finally, here is a link to a page in the Dr. Math archives where 
dimensional analysis is used to solve a "fun" problem:

   Dimensional Analysis
   http://mathforum.com/library/drmath/view/56728.html   

I hope this helps.  Write back if you have further questions.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/