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What is dimensional analysis, and how does it help do unit conversion problems?

Dimensional analysis is a technique used to ensure that your conversion from one unit to another is correct. Basically, it is setting up your conversion problem as a fraction so that you can double check your end unit. In dimensional analysis, you write a conversion factor as a fraction. For example, 12 inches is in one foot. The conversion factor would be \(\mathsf{ \frac{12 \text{ inches}}{1 \text{ foot}}} \) to convert from feet to inches. To figure out how many inches are in 3.25 feet, you set up your problem like this:

\(\mathsf{3.25 \cancel{\text{ feet}} \times \frac{12 \text{ inches}}{1 \cancel{\text{ foot}}} = 39 \text{ inches}} \)

If you do not have a conversion factor to make your conversion happen in one step, you can use two (or more) conversion factors to find your answer. Watch the video below to see several worked examples of using dimensional analysis for unit conversions.

PDF Download Converting between units of measure using dimensional analysis is a way that you can make sure that your units are going to turn out the way you want to. So in this problem, we want to find out how many miles are in one point two five kilometers. So what we need to look is a conversion factor between kilometers and miles. And we are actually given two conversion factors, and there’s a little squiggly line there to say that it is approximately because it is not an exact decimal. And so we have one kilometer is approximately zero point six two miles, and we have that one mile is one point six one kilometers. So if I set up my fraction, I can rewrite one point two five kilometers and I need to multiply that so that kilometers cancel, so that kilometers are on the bottom and cancel out with the kilometers on the top next to my original number. And miles is on the top so that when I get the solution, it is equal to miles. So I’m going to look at these two numbers, and I’m going to put one of these conversion factors into fraction form. It doesn’t matter which one; we can do it both ways to show that it doesn’t matter. So I’m going to put one kilometer is equal to point six two miles. Notice, I put the one next to the kilometers and the point six two miles next to the miles. And one point two five times zero point six two would give me zero point seven seven five, and this answer is going to be in miles. So really what you are doing is, this is like one point two five over one, you multiply across the top, and then you divide by what’s multiplied across the bottom. So what you get is one point two five times zero point six two divided by one. I can do this with the other conversion factor as well. I can do one mile is equal to one point six one kilometers. In this case, I get one point two five divided by one point six one. Instead of a multiplication problem, I get a division problem. And when I actually do this division, I get a number that is slightly different but almost exactly the same. So it’s only one hundredth off, and that’s because of these little squiggly lines we have over here which means "approximately." So we can now say that one point two five kilometers is approximately zero point seven seven five miles.

So let’s do this problem--two liters is equal to blank gallons. So I’m going to set up my fraction, and I’m going to put liters on the bottom because I want the liters to cancel out, and I’m going to put gallons on the top. I’m given two different conversion factors between liters and gallons. And one gallon is approximately three point seven eight liters. And as I’ve already established before, it doesn’t matter which one of these that you use, you will get the same, or approximately the same, answer. What I like to do is I like to set up the situation so that the one is on the bottom. So I’m going to use this one because that would make the liter a one, and I have a multiplication problem instead of a division problem. It does not matter what you do. So I get two times point two six, and I get zero point five two gallons.

Five point four five pounds is equal to a unknown amount of kilograms. Here we’re going from a customary unit of measure to a metric unit of measure. And I’m going to use a conversion factor between pounds and kilograms. So to set my dimensional analysis problem, I want to multiply with pounds on the bottom and kilograms on the top, so that the pounds cancel. I have two conversion factors: one kilogram is equal to approximately two point two pounds, and I’m also given that one pound is approximately point four five kilograms. Again, I like to the put that on the bottom, so I’m going to select this one, and I get five point four five times point four five. And that is two point four five two five kilograms.

In this one, I have zero point two four miles to centimeters, and again I am switching from a customary unit to a metric unit, but in this situation, I’m not given a direct conversion factor that goes from miles to centimeters. So I have to figure out how I can get from miles to centimeters. I do have a conversion factor from miles to kilometers. And then I can go from kilometers to centimeters. Another way you might do this is miles to kilometers to meters to centimeters. In this case, you would actually need conversion factors. In this situation, you would need only two, but either way would work. So I’m going to set up my dimensional analysis. I’m going from miles to kilometers first, and then I’m going to go from kilometers to meters, and then meters to centimeters. And I chose this way because the numbers are going to be a little bit easier to work with. My conversion factor between miles to kilometers: one mile is one point six one kilometers. One kilometer is a thousand meters. And one meter is a hundred centimeters. I’m going to make sure all my units cancel: kilometers to kilometers, meters to meters; I’m left with centimeters. I went from miles to centimeters, so that I know it is correct. And now I just have one long multiplication problem: point two four times one point six one times one thousand times one hundred, and I get thirty-eight thousand six hundred and forty centimeters.

In this case, I have three point four days, and I have to go to seconds, and I’m not given a conversion factor that goes directly from days to seconds, so I have to go through some steps. Days to hours, hours to minutes, minutes to seconds. So I need three conversion factors. I get three point four day's times. I put days on the bottom and hours on the top, so one day is twenty-four hours, times one hour into minutes which is sixty minutes. And then one minute is sixty seconds. Double check to make sure your units cancel--days, hours, minutes. I’m left with seconds; that’s what I want. So this once again is a very long multiplication problem: three point four times twenty four times sixty times sixty and that gives you two hundred and ninety-three thousand seven hundred and sixty seconds.

The key to these problems is figuring out the steps between the original unit and the ending unit by going through what you do know. So always think, "If I’m going from large to small?" Think, "Okay, what’s the next step down and another step down?" Sometimes it takes two; sometimes it takes three. And it actually could take more than that, but it's unlikely to take more than two or three steps.

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