Converting Units: Examples
Purplemath
The useful aspect of converting units (or "dimensional analysis") is in doing non-standard conversions. While you can find many standard conversion factors (such as "quarts to pints" or "tablespoons to fluid ounces"), life (and chemistry and physics classes) will throw you curve balls. Learn some basic conversions (like how many feet or yards in a mile), and you'll find yourself able to do many interesting computations.
What are some real-world examples of converting units?
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Suppose an object is moving at 66 ft/sec. How fast would you have to drive a car in order to keep pace with this object?
A car's speedometer doesn't measure feet per second, so I'll have to convert to some other measurement. I choose "miles per hour". I know the following conversions: 1 minute = 60 seconds, 60 minutes = 1 hour, and 5280 feet = 1 mile. (Yes, I've memorized them. You should, too!)
If 1 minute equals 60 seconds (and it does), then
Note: The fact that the conversion can be stated in terms of "1", and that the conversion ratio equals "1" no matter which value or unit is on top, is crucial to the process of cancelling units.
I have a measurment in terms of feet per second; I need a measurement in terms of miles per hour. To convert, I start with the given value with its units (in this case, "feet over seconds") and set up my conversion ratios so that all undesired units are cancelled out, leaving me in the end with only the units I want.
They gave me something with "seconds" underneath so, in my "60 seconds to 1 minute" conversion factor, I'll need the "seconds" on top to cancel off with what they gave me. This will leave "minutes" underneath on my conversion factor so, in my "60 minutes to 1 hour" conversion, I'll need the "minutes" on top to cancel off with the previous factor, forcing the "hour" underneath. Since I want "miles per hour" (that is, miles divided by hours), things are looking good so far.
They gave me something with "feet" on top so, in my "5280 feet to 1 mile" conversion factor, I'll need to put the "feet" underneath so as to cancel with what they gave me, which will force the "mile" up top. This is right where I wanted it, so I'm golden.
Here's what my conversion set-up looks like:
By setting up my conversion factors in this way, I can cancel the units (just like I can cancel duplicated numerical factors when I multiply fractions), leaving me with only the units I want. Then I do the multiplication and division of whatever numbers are left behind, to get my answer:
I would have to drive at 45 miles per hour.
How do you know which units go where?
Short answer: I didn't; instead, I started with the given measurement, wrote it down complete with its units, and then put one conversion ratio after another in line, so that whichever units I didn't want were eventually cancelled out. If the units cancel correctly, then the numbers will take care of themselves.
If, on the other hand, I had done something like, say, the following:
...then nothing would have cancelled, and I would not have gotten the correct answer. By making sure that the units cancelled correctly, I made sure that the numbers were set up correctly too, and I got the right answer. This "setting factors up so the units cancel" is the crucial aspect of this process.
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You are mixing some concrete for a home project, and you've calculated according to the directions that you need six gallons of water for your mix. But your bucket isn't calibrated, so you don't know how much it holds. On the other hand, you just finished a two-liter bottle of soda. If you use the bottle to measure your water, how many times will you need to fill it to get your six gallons?
For this, I take the conversion factor of 1 gallon = 3.785 liters. This gives me:
= (6 × 3.785) liters = 22.71 L
Since my bottle holds two liters, then:
I should fill my bottle completely eleven times, and then once more to about one-third capacity.
On the other hand, I might notice that the bottle also says "67.6 fl.oz", right below where it says "2.0L". Since there are 128 fluid ounces in one (US) gallon, I might do the calculations like this:
= 11.3609467456... bottles
...which, considering the round-off errors in the conversion factors, compares favorably with the answer I got previously.
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You find out that the average household in Mesa, Arizona, uses about 0.86 acre-feet of water every year. You get your drinking water home-delivered in those big five-gallon bottles for the water dispenser. How many of these water bottles would have to be stacked in your driveway to equal 0.86 acre-feet of water?
The conversion ratios are 1 acre = 43,560 ft2, 1ft3 = 7.481 gallons, and five gallons = 1 water bottle. First I have to figure out the volume in one acre-foot. An acre-foot is the amount that it would take to cover one acre of land to a depth of one foot. How big is 0.86 acres, in terms of square feet?
If I then cover this 37,461.6 ft2 area to a depth of one foot, this would give me 0.86 acre-feet of water, or (37,461.6 ft2)(1 ft deep) = 37,461.6 ft3 volume of water. But how many bottles does this equal?
= 56,050.04592.... bottles
...or about 56,000 bottles every year.
This works out to about 150 bottles a day. Can you imagine "living close to nature" and having to lug all that water in a bucket? Thank goodness for modern plumbing!
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You've been watching a highway construction project that you pass on the way home from work. They've been moving an incredible amount of dirt. You call up the information line, and find out that, when all eighty trucks are running with full crews, the project moves about nine thousand cubic yards of dirt each day. You think back to the allegedly "good old days" when work was all done manually, and wonder how many wheelbarrowsful of dirt would be equivalent to nine thousand cubic yards of dirt. You go to your garage, and see that your wheelbarrow is labeled on its side as holding six cubic feet. Since people wouldn't want to overfill their barrows, spill their load, and then have to start over, you assume that this stated capacity is a good measurement. How many wheelbarrow loads would it take to move the same amount of dirt as those eighty trucks?
The conversion ratios are 1 wheelbarrow = 6 ft3 and 1 yd3 = 27 ft3. (If you're not sure about that cubic-yards and cubic-feet equivalence, then use the fact that one yard equals three feet, and then cube everything. The cube of 1 is 1, the cube of 3 is 27, and the units of length will be cubed to be units of volume.) Using these facts, I get:
= 40,500 wheelbarrows
Wow; 40,500 wheelbarrow loads!
Even ignoring the fact the trucks drive faster than people can walk, it would require an amazing number of people just to move the loads those trucks carry. No wonder there weren't many of these big projects back in "the good old days"!
When you get to physics or chemistry and have to do conversion problems, set them up as shown above. If, on the other hand, they just give you lots of information and ask for a certain resulting value, think of the units required by your resulting value, and, working backwards from that, line up the given information so that everything cancels off except what you need for your answer.
For a table of common (and not-so-common) English unit conversions, look here. For metrics, try here. Here is another table of conversion factors.
When I was looking for conversion-factor tables, I found mostly Javascript "cheetz" that do the conversion for you, which isn't much help in learning how to do the conversions yourself. But along with finding the above tables of conversion factors, I also found a table of currencies, a table of months in different calendars, the dots and dashes of Morse Code, how to tell time using ships' bells, and the Beaufort scale for wind speed.
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