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.
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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
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.
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:
(The image above is animated on the "live" page.)
...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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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.
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!
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.