Now for another post that nobody will want to read. But unlike the previous post about programming languages, this one you should read, because you might learn something very useful!

I graduated from high school 40 years ago, and looking back, it’s easy to recognize that some of the things that I was being taught were far more valuable than I recognized at the time whereas others I’ve never had a use for. I want to talk about the one thing that I learned in high school which has proved more valuable than any other thing I was taught: dimensional analysis. Dimensional analysis is a mathematical technique that is simple enough that anyone can learn it. The name makes it sound complex… but in fact, it can be learned very simply. More importantly, dimensional analysis can be used to solve many common everyday problems; and it can even help you solve problems that you don’t fully understand.

So, in 11th grade, I took AP Chemistry, and my teacher (I wish I could remember his name, but I cannot) taught us a method of solving calculations called dimensional analysis. Dimensional analysis is a method of analyzing and solving problems that contain dimensions.

In dimensional analysis, a dimension refers to what the number is counting. So, for example, the value ‘5 miles’ is a dimensional quantity. The number 5 is counting something, and in this case, it is ‘miles’. The dimension here is ‘miles’ (or perhaps more accurately, ‘distance’). An even more specific dimensional quantity would be ‘3 gallons milk’ where the dimension is the volume of milk.

My teacher introduced us to dimensional analysis on practically the first day of class, and he kept teaching it, and teaching it, and teaching it. We studied dimensional analysis for literally the first 3 months of the AP chemistry class, and I have to say, at the time I was pretty annoyed. I wanted to prepare for the AP chemistry exam at the end of the year, and spending basically a third of the class on a math technique (and not even one that seemed that important to me at the time) felt like he was failing to teach us what we needed to pass the test.

Of course, looking back, I see that he was using it to help us solve harder and harder chemistry problems, while at the same time, drilling it into us until it became second nature. At the end of the year, I passed my AP exam, and had decided that I would become a chemist.

Pure math is largely about numbers. When you do an arithmetic problem, an algebra problem, or a calculus problem, the problems typically are about numbers without dimensions.

Dimensional analysis helps you look at any problem with dimensional quantities and helps you understand the problem, and usually can help guide you in solving it. What types of problems are these? Real life problems. In real life, you’re always talking about dimensions. We drive 60 miles per hour. There are 800 calories in a hamburger. Understanding dimensional analysis helps us to make sense of the numbers we run into every day. They help us to understand and solve those most hated of all math problems: story problems.

How helpful is dimensional analysis? Let me tell you two stories.

Immediately after graduating from high school, I knew I wanted to study chemistry. I needed a job for the summer, so I went up to the chemistry department at the university (I lived in a university town) and walked into the department chair’s office to see if I could get a job in a chemistry lab. As a clueless high school student, I had no idea then that department chair’s tend to be pretty busy, so I’m amazed in retrospect that she would take time to meet with a high school student who showed up unannounced. But she did. She spent 15 minutes listening to what I wanted, and she thought for a moment and then gave me a list or 5 or 6 professors who might have positions available. Then, I went and talked to each of those professors and with one exception, every single one of them was willing to meet with me that day. The one exception was the last one on the list who wanted me to meet with one of her post docs who largely ran the lab. The professors I met with were all nice, but none actually had a position for me.

However, I had a meeting with the post doc (Carl Wittwer) the following day, and when I went to meet with him, he immediately launched into what they were doing and what my job would entail as though I were already hired. I’m still not certain if the professor told him to hire me, or if there was a communication problem and the professor asked Carl to evaluate whether or not to hire me, but he understood I was already hired. In any case, I got the job and ended up working there for 3 years as an undergrad.

Now, going in to that job with only high school level chemistry and no lab experience at all, I was expecting to be a dishwasher/gofer. Instead, Carl (who became a close friend) put me to work doing actual chemistry research.

It was an amazing experience. The lab was studying enzyme kinetics. The problem was… I did not know what an enzyme was. I didn’t know what kinetics were. And I didn’t know any of the equations that we used to determine what results were. But, because dimensional analysis was 2nd nature to me (and all experimental data have dimensions associated with them corresponding to whatever we were measuring), I was able to understand far more of what was going on than I would have expected. True, I didn’t understand all of the theory behind what was going on… but I understood what we were measuring and how much of the math worked, because I understood the dimensions.

The year after I started working there, Carl interviewed another student who had graduated a year after me who had had the same high school chemistry teacher, and the same grasp of dimensional analysis. I was there for some of the interview and the he was able to answer questions because of dimensional analysis. Both of us were prepared to work at a level far beyond what a typical high school graduate could do.

Another similar illustration of the power of dimensional analysis happened a few years later when I was a graduate student. I was taking graduate level classes but also working as a teaching assistant. I was assigned to teach a lab for a freshman level course, including reviewing material they had learned in the class and preparing the lab final exam. One of the things they had been taught in class was dimensional analysis.

On one of my graduate level exams that I took shortly before finals, one of the problems could be solved entirely by dimensional analysis (though a full understanding of the problem used information from the graduate level class). When I prepared the final exam for my freshman students, I took that question and put it on the final exam for the freshmen. The question was unmodified except that I added a small amount of explanation because they obviously didn’t have the graduate level class under them. The majority of the students were able to solve the problem from a chemistry class 4 years above them!

And just so you know, dimensional analysis is not useful only in chemistry. Over the years, I have used dimensional analysis to understand problems from people who were in significantly different fields than anything I have studied, and it is always an interesting experience when I can explain how to solve a problem to them in their field.

So, hopefully I have made you interested in dimensional analysis enough that you’re asking “so what exactly is it?”. Well, I’m glad you asked.

MATH WITH DIMENSIONAL QUANTITIES

The first thing to know is that you can do math with dimensional quantities in almost the same way that you do with numbers. Dimensional quantities can be multiplied and divided in very much the same way that you multiply and divide numbers. So, for example, just as you can multiply two numbers together:

     5 x 6

you can also multiply two dimensional quantities together:

     5 inch x 6 gallon

It may not seem to make sense… but it’s still valid (and don’t worry, the dimensions will make more sense shortly). Division works the same. You can also add and subtract, but it turns out that you don’t do that very often in dimensional analysis.

Second, you can cancel out dimensions just like you cancel out variables in algebra. In algebra, you might have an equation that looks like:

     7 A
     ---
     5 A

Since there is an A on top and bottom, they cancel each other out (because A/A = 1). So it simplifies to:

     7 A     7
     ---  =  -
     5 A     5

Likewise, you can simplify dimensional quantites, so:

     12 inch
     -------   =  4
      3 inch

     12 inch
     -------   =  4 inch
      3

Intuitively, you can understand these two simplifications pretty easily. In the first, you have a 12 inch string and you are dividing it by a 3 inch string. In plain English, you are asking how many 3 inch strings there are in the 12 inch string, and the answer is that there are 4 of them. Notice that the 4 is dimensionless because the dimensions cancelled out.

The second one is saying if you divided the 12 inch string by three, what do you have? The answer is that you have 4 inch pieces of string.

So, if you have a long string of dimensional quantities multiplied together, you just cancel out all the units possible, multiply and divide the numbers, and what you’re left with in the end is the answer. For example if you have the following:

   11 cup     16 tblsp     3 tsp
           x  --------  x  -------
               1 cup       1 tblsp

You can simplify it by removing the cups and tblsp units (since they exist on the top and bottom, they cancel out), and this simplifies to:

     11 x 16 x 3 tsp   =   528 tsp

CONVERSION FACTORS

In addition to multiplying and dividing dimensional quantities, you can set up complete equations using them. For example:

     12 inch = 1 foot

We recognize that this is in fact true, but it helps us to understand that dimensions have an interesting affect on the equation. We understand that the counting numbers 12 and 1 are not equal, but the dimensional quantities 12 inches and 1 foot ARE equivalent, so we can set them equal to each other in this equation.

This equation, where both sides are known, will produce something called a conversion factor.

We know that it is a basic mathematical property that if you have two equal quantities, you can divide both by the same value and still be equal. So let’s divide both sides of this equation by ’12 inch’. You end up with:

     12 inch     1 foot
     -------  =  ------
     12 inch     12 inch

which simplifies to:

           1 foot
     1  =  -------
           12 inch

Alternately, if you divide both sides by '1 foot', you get:

           12 inch
     1  =  -------
           1  foot

These two values:

     1 foot          12 inch
     ------   and    -------
     12 inch         1 foot

are known as conversion factors and are very useful. These allow you to convert something measure in one unit (like inches) to another unit (like feet).

Any time you have two different measures which are equal, then you can create a conversion factor with one on top and the other on the bottom (and it doesn’t matter which is which).

You can always come up with a conversion factor that will convert one measure to another, provided they are the same type of unit. For example, if you are working with units of length or distance, there is a conversion factor. You can convert miles to km, light years to millimeters, etc. Coming up with the actual conversion factor isn’t interesting. You basically just look it up in a table (or ask your favorite computer assistant).

Alternately, you can use facts you may already know. For example, if you want to know how many teaspoons are in a gallon, you can just ask Google (or Alexa, Siri, etc.). Or you can construct it from facts that you know. For example, because I cook a lot, I know:

     1 gallon = 4 quart
     1 quart  = 2 pint
     1 pint   = 2 cup
     1 cup    = 16 tblsp
     1 tblsp  = 3 tsp

So, I can create a conversion factor from each one of these and multiply them together as such:

     1 gallon     1 quart     1 pint     1 cup        1 tbls
     --------  x  -------  x  ------  x  --------  x  -------
     4 quart      2 pint      2 cup      16 tblsp     3 tsp

This simplifies to:

     1 gallon
     --------
     768 tsp

If you ask Google how many teaspoons are in a gallon, you’ll get 768.

Conversion factors will be very useful in the next section.

CONVERSION PROBLEMS

The most common use of dimensional analysis is conversions. This is actually a pretty simple problem, but it seems quite a bit more complicated because it can be worded in so many different ways.

Let’s do a simple conversion. Using the facts above (don’t ask Google), how many teaspoons are in 3 quarts? The first step is to write an equation, but this time we’ll have to use a variable because we don’t know the answer. We have a certain number of teaspoons (call it A) and it is equal to 3 quarts. So, the equation is:

     A tsp = 3 quart

Now, we can apply conversion factors from the facts above:

     A tsp = 3 quart   2 pint    2 cup   16 tblsp   3 tsp
             ------- x ------  x ----- x -------- x -------
                       1 quart   1 pint  1 cup      1 tblsp

and now we see that all of the units cancel out on the right side except for tsp. Multiply all of the numbers together and we get:

     A tsp = 576 tsp

and that is how many teaspoons are in 3 quarts.

Sometimes, there are multiple units that need to be converted. Let’s start with a very common conversion. Imagine that you are driving a car in Europe. Since they do everything in metrics (because they have far more sense than us Americans do), they mark everything in kilometers. You wonder how fast you are driving in miles per hour. Well, you see a road sign that says that you are 5 km from the nearest town. You note that it is exactly 5:04 PM when you see that sign. A few minutes later, you arrive at the town and see that it is now 5:11 PM. So how fast are you travelling?

The way you solve this problem is you start by determining what you want to find. In this case, you want to find your speed as us non-metric Americans would express it. In other words:

     X mile
     ------
       hr

Next, you set this equal to what you know. You know that we have traveled 5 km in 7 minutes. So, we’re going to solve the equation:

     X mile     5 km
     ------  =  -----
       hr       7 min

The main trick here is that you want to make sure that you get the same type of units on the top and bottom on both sides of the equation. On the left side we have a unit of distance on the top and time on the bottom, so make sure you get distance on top on the right side and time on the bottom.

Now this equation IS valid… but you can’t solve it yet because the units are different on the left and right side. So what we need to do is apply some conversion factors.

First off, we all know how hours and minutes are related, and you can ask Google how many km are in a mile, and you’ll get two conversion factors:

     1 hr = 60 min
     1 mile = 1.6 km

Now what we want to do is go back to our equation and we want to multiply the right side by a conversion factor that will change km to miles. Since we have km in the top (and we want to cancel it out), multiply it by the conversion factor that has km in the bottom:

     X mile     5 km    60 min   1 mile
     ------  =  ----- x ------ x ------
       hr       7 min    1 hr    1.6 km

After canceling the units, and multiplying and dividing the numbers, you’re left with:

     X mile     26.6 mile
     ------  =  ---------
       hr         hr

So now you know how fast you were going in units more familiar to you. Apparently, it was a very slow road!

MORE COMPLEX EXAMPLE

Let’s do something more complicated. You own a swimming pool and you just drained it and refilled it, and the pool manual says that you should have 2 ppm (parts per million) of chlorine. Your pool is a rectangle 12 feet long, 15 feet long, and 4 feet deep. You go to Walmart and buy some muriatic acid that says it’s 30% chlorine (HCl). How many cups of the muriatic acid should you add to your pool?

So, the first thing to do is to understand what these number all mean. That’s actually often the hardest part of a story problem. First off, when it says that we need 2 ppm chlorine, what it’s saying is that we want 2 units of chlorine in a million units of the pool. Since the pool is measured in gallons, we want to add 2 gallons of chlorine for every million gallons in the pool.

     2 gal chl = 1,000,000 gal pool

Notice that we want to be very specific in what we’re measuring because we’ve got gallons of chlorine, gallons of pool water, and gallons of muriatic acid, so we’ll always fully specify what we’re measuring.

The muriatic acid is 30% chlorine (which is saying there is 0.30 gallons of chlorine in a gallon of muriatic acid) so we have the factor:

     0.30 gal chl = 1 gal mur

We have that the pool is 12 ft by 15 ft by 4 ft (and we’ll assume that it’s full of water), so we have a pool that is 720 cubic foot. We’ll google how many gallons are in a cubic foot (abbreviated ft^3) and we find:

     1 ft^3 = 7.48 gal

So, what we want to know is the number of cups of muriatic acid to add to a 720 cubic foot pool. Our equation then is:

     X cup mur = 720 ft^3 pool

So, let’s start applying conversion factors to convert the right side to cups of muriatic acid:

     X cup mur = 720 ft^3 pool   7.48 gal pool       2 gal chl
                 ------------- x ------------- x ------------------
                                 1 ft^3 pool     1,000,000 gal pool

                   1 gal mur     4 qt mur    2 pint mur   2 cup mur
                 x ----------- x --------- x ---------- x ----------
                   0.3 gal chl   1 gal mur   1 qt mur     1 pint mur

All the units cancel out. Multiply all of the numbers and you get 0.574 cup.

So, you need to add just a bit more than half a cup of the muriatic acid to pool.

Some things to note: note the conversion factors where I’m converting volumes of muriatic acid. Since I know the generic conversion factor:

     1 pint = 2 cup

Then I also know the same conversion factor for any thing I might be measuring. 1 pint of milk = 2 cup milk. Or 1 pint muriatic acid = 2 cup muriatic acid.

Just be careful that you do not mix conversion factor units, so there is NO conversion factor:

     1 pint muriatic acid = 2 cup pool

Using dimensional analysis, you can ALWAYS solve conversion problems, no matter how complex they might seem.

UNDERSTANDING REAL-LIFE PROBLEMS

So one more problem. This time, I’m going to go back to that graduate level chemistry problem. Let’s see if we can use dimensional analysis to understand a graduate level chemistry problem.

Spectrophotometry is a method to measure how much of a chemical exists in a solution. The basic spectrophotometry equation is:

     A = b * c * e

In order to understand this equation, we need to understand the units associated with each value. I’m going to put the units in brackets as we figure them out.

In this equation, A is the amount of light that gets absorbed and it is measured in absorbance units. So, we’re going to add units to the left side.

     A [abs units] = b [?] * c [?] * e [?]

Clearly, the units on the right side are going to have to be absorbance units too in order to make this work, so let’s figure out the units for each of the things on the right.

The way spectrophotometry works is that you have a glass tube (called a cuvette) and you shine a light through it. The cuvette is made of a totally transparent material, so all of the light goes through. Then, you fill the cuvette with some solution which has a chemical in it which absorbs some of the light. Now, when you shine the light through the cuvette, only some of the light makes it through and some is absorbed.

You can easily imagine this absorption if you have ever looked at a piece of colored glass. The colored glass absorbs some of the light going through it. If you look through a single thin pane of this glass, you can see what is behind the glass fairly well, but it’s a little dimmer because some of the light got absorbed. If you then stack two panes of glass on top of each other and look through both of them, you can still see through them to what is behind them, but now the image is even dimmer because twice as much light got absorbed. Another example is looking through an aquarium. If you look through a small aquarium (one only a couple inches wide for example), you can see whatever is behind the aquarium fairly brightly. But if you then switch to a wider aquarium, say one that is a foot wide, the light that comes through is dimmer because the water absorbed some of it. The wider the aquarium, the dimmer the light that gets through. By the way, you should ignore the fact that the image gets bent. That’s related to a funny property that light has when it goes through different types of matter (so light passing through an aquarium goes from air to glass to water and back to glass and back to air). These changes cause the light to bend, but they don’t affect absorption, so even though the image may appear warped, the thing we’re interested in here is how dim the image is.

So, back to the spectrophotometer. The amount of light that gets absorbed depends on how wide the cuvette is (or in other words, how much distance the light has to travel though the solution). Since the wider the cuvette, the more light gets absorbed, the equation has to take into account the width of the cuvette. The b in the above equation is the width of the cuvette, typically measured in centimeters. So now the equation becomes:

     A [abs units] = b [cm] * c [?] * e [?]

The next thing is that you need to know how much of the chemical there is. Here’s another easy example. Say you have a clear glass of water. It doesn’t absorb much light so the image you see is almost as bright as if there were no glass of water. Now, add something that absorbs some of the light. For example, add a drop of food coloring. Now, some of the light is absorbed and it’s a bit harder to see. Add a second and third drop, and the water gets darker and less light gets through.

The more concentrated the chemical is that is absorbing the light, the more light gets absorbed. So, the c in the equation above is the concentration of the chemical and is typically reported as moles per liter. A mole is an amount of a chemical, so the concentration is how many moles are dissolved in a liter of the liquid that you have in the cuvette. So, we’ll add these units and the equation now is:

     A [abs units] = b [cm] * c [mole/l] * e [?]

Now the e in the equation above is simply a number that you’ll need to look up. Basically, different chemicals absorb lights differently, so chemists who are studying a chemical just put an accurately known amount of the chemical in the cuvette and then they measure how much light is absorbed, and they just store that value in a table somewhere. So the value of e isn’t interesting. It’s dimensions ARE interesting. And using dimensional analysis, you can figure out what they are. So, here’s a homework problem. What are the dimensions of e? Try to figure them out.

You know that you need to get units of absorbance units on the right to match the ones on the left. Also, you have units of cm and mole in the top, and l on the bottom, so you need to get rid of them. So, the units of e are:

     abs units * l
     -------------
      cm * mole

Pretty complex! So, simply by understanding the dimensions, you now have at least a rudimentary understanding of a relatively complex chemical property! That’s how powerful dimensional analysis is.

I took a graduate level applied math class in college which covered dimensional analysis. The math professor took a published journal article which was explaining why a chemical in the body was being gathered in one place using only osmosis. Osmosis typically means that if you have a place where a chemical is more concentrated, it will basically ‘spread out’ so it’s evenly concentrated everywhere. But in a situation described in the article, the chemical was actually going from a place of lower concentration to a higher concentration.

The professor took all of the facts out of the article, and looked at their units in order to understand them, and in the end, we were able to understand the results that were published. It was an amazing example of understanding an enormously complex problem using nothing more than dimensions.

In the same way, you now have at least a basic understanding of spectrophotometry based only on dimensional analysis.

I would encourage you to google dimensional analysis and play with it. Yes, I’m weird and actually find it fun, but even if that’s not the case, you’d be amazed at how much light it will shed on any math problem involving units.