Subsection 5.1.1 Prep Text: Unit Conversions
You’ve measured the mass of a dragon’s scale in ounces, but your spellbook demands grams. You’re plotting a star map in light-years, but your onboard thrusters calculate in meters. What do you do?
You cast a unit conversion spell.
Unit conversions are the dimensional bridges that connect different lands of measurement. They don’t change what you have, they simply describe it in a new tongue. Whether you’re switching from meters to miles, hours to seconds, or joules to electronvolts, you’re not summoning new quantities, you’re just translating between dialects of the universal languages of science.
To convert between units, you multiply by a conversion factor which is a magical ratio equal to one.
Example 5.1.1.
Suppose a star cruiser travels at \(90 \text{ km/hr}\text{.}\) To express this in meters per second:
\(\frac{90 \text{km}}{\text{hr}} \cdot \frac{1000 \, \text{m}}{1 \, \text{km}} \cdot \frac{1 \, \text{h}}{3600 \, \text{s}} = 25 \, \text{m/s} \)
Both of the two conversion factors are fractions that equal one and stem from the following: \(1000 \, \text{m} = 1 \, \text{km}\) and \(3600 \, \text{s} = 1 \, \text{h}\text{.}\)
For example, start with the expression \(3600 \, \text{s} = 1 \, \text{h} \text{.}\)
If you divide both sides by 3600 s, you get \(1 = \frac{1 \, \text{h}}{3600 \, \text{s}}\text{.}\) This tells us that multiplying any quantity by this fraction is like casting a harmless illusion; it doesn’t change the true value of a quantity, only the form in which it’s expressed. In other words, you’re not altering reality, just viewing it through a different unit’s lens.
It’s like casting a chain of transmutation spells, each one carefully crafted so the old unit cancels and the new one takes its place. As long as your units behave, your math will too.
While there’s no universal conversion for all situations below are some common conversions that will occur during the first part of of our journey through classical physics:
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Length
\(\displaystyle 2.54 \, \text{centimeters} = 1 \, \text{inch}\)
\(\displaystyle 1 \, \text{meter} = 3.28084 \, \text{feet}\)
\(\displaystyle 1 \, \text{meter} = 1.09361 \, \text{yards}\)
\(\displaystyle 1.609 \, \text{kilometers} = 1 \, \text{mile}\)
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Mass
\(\displaystyle 1 \, \text{kilogram} = 2.20462 \, \text{pounds}\)
\(\displaystyle 1 \, \text{kilogram} = 35.274 \, \text{ounces}\)
For derived dimensions like area or volume, you can use the above conversions or use the internet to search for a conversion.
Checkpoint 5.1.2.
Convert 5.30 kilograms to pounds.
Hint.
Use the conversion factor \(1 \, \text{kilogram} = 2.20462 \, \text{pounds}\text{,}\) which can be written as a fraction: \(\frac{2.20462 \, \text{pounds}}{1 \, \text{kilogram}}\text{.}\)
Answer.
\(11.7 \, \text{pounds}\)
Solution.
To convert 5.3 kilograms to pounds, multiply by the conversion factor:
\(5.3 \, \text{kilograms} \times \frac{2.20462 \, \text{pounds}}{1 \, \text{kilogram}} = 11.684 \, \text{pounds}\)
Using the proper significant figures: 5.30 kilograms is equal to 11.7 pounds.
Checkpoint 5.1.3.
Convert \(1,530 \, ft^2\) to \(m^2\text{.}\)
Hint.
ft is the symbol for feet, and m is the symbol for meters. Use the conversion factor \(1 \, \text{meter} = 3.28084 \, \text{feet}\text{,}\) twice!
Answer.
Solution.
Note that our original unit is \(ft^2\) which can be written as \(ft \cdot ft\text{.}\)
So we need to multiply by the conversion between feet and meters twice:
\(1530 \, \text{ft}^2 \times \frac{1 \, \text{meter}}{3.28084 \, \text{feet}} \times \frac{1 \, \text{meter}}{3.28084 \, \text{feet}} = 142.1447 \, \text{m}^2\)
...or...
\(1530 \, \text{ft}^2 \times \left(\frac{1 \, \text{meter}}{3.28084 \, \text{feet}}\right)^2 = 142.1447 \, \text{m}^2\)
Using the proper significant figures: \(1,530 \, \text{ft}^2\) is equal to \(142 \, \text{m}^2\text{.}\)
Why Use Them?
Humans, robots, magic folk, and aliens tend to use units that make sense in their context. When they want to communicate with each other, they need a way to convert thier measurement’s unit to anothers preferred unit. For example, a human might measure the length of a dragon’s tail in feet, while a robot might use meters. If they want to work together, they need to convert between these units.
Humans in different fields of study even use different units. For example, in physics, we often use the International System of Units (SI), which uses meters, kilograms, and seconds as its base units. While some engineers tend to use feet, pounds, and seconds as their base units.
Bonus Trick: Fermi Problems
So you have made it this far, from scientific notation all the way to unit conversions, despite the curse placed upon you. Congratulations! Now, you can use your newfound skills to tackle Fermi problems.
A Fermi problem is a type of estimation problem that requires you to make reasonable assumptions and use unit conversions to arrive at an answer. Named after the physicist Enrico Fermi, these problems often involve estimating quantities that are difficult to measure directly, such as the number of jelly beans in a jar or the amount of energy consumed by a city in a day. The key is to break the problem down into smaller, manageable parts, make reasonable assumptions, and use unit conversions to arrive at an answer.
The answer to a Fermi problem is an estimate, not an exact value. Usually the answer is given in scientific notation with the order of magnitude being the most important part of the answer.
Example 5.1.4.
Fermi Problem: How much tea would it take to fill an airship’s cargo bay?
Initial thought 1: The problem didn’t specify the size of the airship’s cargo bay, so we will estimate the size as a volume based on a typical commercial jet’s passenger volume.
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Height of the cargo bay.
The height of a typical commercial jet is tall enough for a person to stand up, so let’s estimate about 2 meters.
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Width of the cargo bay.
The width of a typical commercial jet is wide enough for 6 people to side shoulder to shoulder, and an aisle for 1 person to walk through. So let’s estimate the width of the jet to be about 7 people widths. Since we are looking for an estimate, lets round up to 10 people widths. A person is probably about 0.5 meters wide, so the total width of the jet would be about 10 times 0.5 = 5 meters.
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Length of the cargo bay.
The length of a typical commercial jet is long enough for 50 seats to fit in front of or behind each other. Let’s estimate the length for each seat to be about the width of a person. Using 0.5 meters as a person width again, that means the total length of the jet’s passenger space is about 50 times 0.5 meters = 25 meters.
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Volume of cargo bay.
The volume is the product of the cargo bay’s height, width, and length. So the total volume is about 2 meters (height) times 5 meters (width) times 25 meters (length) = 250 cubic meters.
Initial thought 2: We now have the volume of the cargo bay in meters cubed, thus we also know much volume of tea we need in meters cubed: 250 cubic meters.
While this is an answer, it might not be the most human friendly answer. In other words, do you immediately have a mental visual of how much 250 cubic meters of tea is? Let’s convert it to a more human friendly unit, cups. We will use cups since is reasonable to assume folks drink tea a cup at a time.
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Cups of tea.
A quick internet search tells us that 1 cubic meter is equal to 4,226.75 cups. But remember, Fermi problems are about estimates, so we will round that to 4,000 cups. Or in other words, \(1 \, \text{m}^3 = 4,000 \, \text{cups}\)
So the total amount of tea needed to fill the airship’s cargo bay is: \(250 \, \text{m}^3 \times \frac{4,000 \, \text{cups}}{1 \, \text{m}^3} = 1,000,000 \, \text{cups}\text{.}\)
That’s \(1 \times 10^6 \, \text{cups}\text{!}\) Or one million cups of tea. Enough to host a tea party for every elf in the eastern forest, twice.
In general, when working on a project, if you start in SI units, and stay in SI units unless you have a solid reason (like a prophecy, or a stubborn client from another dimension).
