# Make Units Work for You

How do we use units? You may see one of these speed limit signs, nearly every day. Even though neither of them display units, drivers know they are implied. In the United States, a Speed Limit 25 sign implies that 25 miles per hour is the maximum legal speed. The other type of speed limit sign is common in many other countries. A red circle with a number: the 50 implies that the speed limit is 50 kilometers per hour.

There are many examples where units are implied by the context, and no confusion results.

But when solving problems, often a person will set units aside and run through the calculations. When using a calculator, you enter just the numbers to perform operations on. Sometimes units are put with the final answer, but many times they are implied or just forgotten. Errors in calculation can occur, due to multiplying by a wrong factor (or dividing when multiplying was required). It is difficult to find where the mistake happened.

*Dimensional analysis* is a useful tool in problem solving. Often the correct formula for some process or phenomenon can be figured out (perhaps just needing a dimensionless factor), just by looking at the dimensions involved.

Recognizing the dimensions in a problem is a powerful tool, which can ease many problems. It is the first in a series of tools in the Sanjoy Mahajan’s Street Fighting Mathematics.

Dimensions and units are not the same, but they are related. *“A dimension is general and independent of the system of measurement, whereas the unit is how that dimension is measured in a particular system” *See:

https://www.edx.org/course/street-fighting-math-mitx-6-sfmx

Let’s start with something simple, like converting a measurement from one unit to another. Suppose you use a tape measure to measure something, and it measures 30 inches long. Now suppose you want to communicate the measurement in centimeters. Do you multiply or divide by the conversion factor?

Consider this: we know that 1 inch = 2.54 cm. Now lets divide both sides by (1 inch), and we have:

$$1 = \frac {2.54 ~cm} {1 ~in},\rm{ ~or~the~reciprocal~is:~} 1 =\frac{1 ~in} {2.54 ~cm}$$

Either way, we can multiply our measurement by ##\frac{1 ~in} {2.54 ~cm}## or by ##\frac{2.54 ~cm} {1 ~in} ##, and it is the same thing as multiplying by a dimensionless 1.

*This is important.* We are neither changing dimensions, nor are we scaling the measurement. Multiplying by something equivalent to 1, does not change the measurement. So now we just need to pick the one which “cancels out” the inches, and leaves us centimeters in the numerator. So we should multiply by ##\frac{2.54 ~cm} {1 ~in} ##, and our measurement can be expressed as 76.2 cm.

The following is an example of a real world problem which involves simply multiplying and dividing various rates, to answer a problem.

Bill and Ted carpool to work in Ted’s car. They agree that Bill can help by paying half of the cost of fuel. They want to figure out how much Bill should pay for his half. They will calculate the amount for each week, based on the price of fuel at a local gas station. They live next door to each other, so there is no driving between houses.

Here is information that we have:

– The current price for gas is 2.00 dollars per gallon.

– The distance from Ted’s house to work is 15 miles.

– They commute 5 days per week.

– Ted’s car averages 23 miles per gallon.

In this particular problem, we need to multiply or divide the rates and quantities to arrive at a dollar amount (cost for a week).

How do we know? Here is a method that I learned, while attending tutoring sessions for my first college chemistry class. It is fairly simple, and helps prevent mistakes. It can be confusing to remember whether to divide or multiply by the “miles per gallon” figure. This will help.

The answer we are looking for is cost (dollars), so we will need dollars in the numerator. Start by drawing a horizontal line, which is like a big fraction. Everything above the line is multiplied, then divide by everything below the line. We will draw vertical lines to help separate each item.

2.00 Dollars | gallon |

next we need to ‘cancel’ the gallons, so put the ‘miles/gallon’ such that gallons is on top.

2.00 Dollars | gallon gallon | 23 miles

Each trip is 15 miles. There are 2 trips per day.

2.00 Dollars | gallon | 15 miles | 2 trip gallon | 23 miles | trip | day

Then finally they travel 5 days each week.

2.00 Dollars | gallon | 15 miles | 2 trip | 5 day gallon | 23 miles | trip | day | week

Now if we multiply everything on top, and divide by everything on bottom, and cancel units, we get

(300 / 23 ) dollars/week, or $13.04 per week. This is the total cost of fuel for the week. Since Bill will pay a half, divide by 2, to get $6.52 for the week.

Hopefully this is enough to motivate you to start using units to work for you. They can help you to figure out how to use the available data in solving problems.

**I obtained the speed limit photos from https://pixabay.com which is a nice source of royalty free pictures, which are actually free (they do accept donations, of course).*

After learned about electronics and RADAR in the US Navy, I decided to pursue an engineering degree. At Texas A&M University, I got a job as a helper in the Maple computer lab, while getting my BS in Electrical Engineering. I enjoy helping others to learn about math and science.

I play double bass in two community orchestras in the Houston Texas area.

Such a fundamental and useful trick. I still remember when I learned it in one of my first university basic engineering classes. What an epiphany! :smile:

Great first Insight @scottdave!

Thanks,

I had fun writing it, and learned a little more about LateX, as well.

Thanks @scottdave for a great Insight. Dimensional analysis is in my opinion a underrated method/trick to check complicated expressions.

I have an additional comment or two. I like the way @scottdave shows the conversion by multiplying by 1, as in the example with ## 1=frac{2.54 , cm}{1 , inch} ##. There is one other thing that I find useful in doing some of the more difficult conversions, and that is to write the quantity such as length ## vec{L} ## as a vector. In that case ## vec{L}=L_{cm} hat{cm}=L_m hat{meter} ##. We can solve for ## L_{cm} ## in terms of ## L_m ## if we know that ## 1 hat{meter}=100 hat{cm} ##. This kind of thing comes in handy, e.g., in converting E&M (electricity and magnetism) quantities in c.g.s. to M.KS. and visa versa. This is a more advanced application, but on occasion, it is useful also on simpler conversions.

These techniques can definitely be applied to more complex calculations. It is probably more critical to use these methods in those situations.

I chose to keep my examples fairly simple, so that the article might appeal to a broader readership.

I can't tell how many questions I have seen on PhysicsForums and other sites, where these techniques would have solved the problem for the OP.

I like everything about your article, but there is one point of contention I would like to make. Setting these ratios equal to ##1## is a matter of preference, and I prefer to follow the advice of Arnold Arons and :sedulously avoid” it. I understand the motivation for using it, which is to imply that it’s okay to use it as a multiplicative factor in the same way that it’s okay to use ##1## as a multiplicative factor. But expressions like $$\frac{2.54\ \mathrm{cm}{1\ \mathrm{m}}=1$$ can be a source of confusion for learners. I think it better to simply state that multiplying by the factor $$\frac{2.54\ \mathrm{cm}{1\ \mathrm{m}}$$ is a perfectly acceptable practice. After all, it is not the same thing as multiplying by ##1## because it has the effect of changing the units.

But I realize that this is entirely a matter of preference. I have a colleague who disagrees, stating that students are likely to encounter it presented that way. But I disagree.

Reference https://www.physicsforums.com/insights/make-units-work/

Let’s try getting the LaTeX coding right one more time …

I meant to say that I prefer not using expressions like $$\frac{2.54\ \mathrm{cm}}{1\ \mathrm{m}}=1$$

But instead prefer $$\frac{2.54\ \mathrm{cm}}{1\ \mathrm{m}}$$

Interesting point, @Mister T. I see where you are coming from. The point that I intended was it is easier to figure out if the factor by going through and setting 2.54 cm equal to 1 inch. Then, you can have either (2.54 cm) / (1 in) or (1 in) / (2.54 cm), depending on which way you need to go. I believe this is easier than trying to remember “If I want to convert from centimeters to inches, then divide by 2.54”.

It seems that LaTeX works differently in the comment section than in regular posts? At least on this browser.

As you wrote it would be a source of confusion for anyone.

I think you meant ##frac{2.54 text{ cm}}{1 text{ inch}} = 1##, which to me is perfectly fine. Since 2.54 cm = 1 in, divide both sides by "inch" to produce the equation I wrote, or even ##2.54~frac{text{ cm}}{text{ inch}} = 1##.

Except that 2.54 cm is NOT equal to 1 m.

If you accept, say, that 2.54 cm represents the same lengh as 1 inch, then we can write 2.54 cm = 1 in, recognizing that we're talking about the same length measure, but with different units. It follows that the ratio ##frac{2.54 text{ cm}}{1 text{ inch}}## is dimensionless and equal to 1. Being 1, we can multiply whatever we want by it without changing the underlying value of what it is multiplying.

This is so useful for me. Ive watched a bunch of videos on it and this explains it much better. So are you treating the units like variables essentially?

Yes, I agree completely. Teaching students to multiply by the appropriate ratio is the way to go. My point is that it's not necessary, and adds to student confusion, when you set those ratios equal to 1. While I agree that it's valid in the sense that it's being used, its only utility is to justify the practice of multiplying by ratios. There's no need for a student to ever see an expression like 2.54 cm = 1 in. Some people like to instead write 2.54 cm ##hat{=}## 1 in. to make the distinction clear that the two quantities represent the same length and in that sense are equivalent rather than being equal.

Well, I buggered that up pretty good, didn't I? Especially when I wrote m instead of in.

That would be a way to think about it, or as vectors of certain lengths, as @Charles Link stated in his post.