# DWise1: Unit Conversions

 "Do the math!" - Sega 16-bit video game-set commercial, c. mid-1980's

### Abstract

Units in physics and chemistry have an interesting property in that they can be treated as algebraic variables which you can multiply and divide and use for grouping. This allows you to perform conversions accurately and with confidence. With this knowledge, you can even derive your own conversion factors easily.

Converting units is such a common task in physics and chemistry that there are several videos on the subject on YouTube.

One property of the various physics constants is that "they make the units come out right" (as one of my teachers told us). For example, the units of the Gravitational Constant in the MKS system (meter-kilogram-second) are (kg–1 m3 s-2). When you calculate the gravitational force between two bodies excluding that constant, the units come out as (kg2 m-2). When you include the constant, the units multiply together thus:

(kg–1 m3 s-2) (kg2 m-2) → (kg(2-1) m(3-2) s-2) → (kg m s-2)
By definition, the MKS unit of force, the Newton, is 1 kg m s-2, so the constant's units straightened out those of the calculation to leave us with newtons. QEF (Quod erat faciendum, "that which had to be done").

This is typically one of the first subjects covered in chemistry and physics and is essential knowledge. This page will demonstrate the technique and also how to derive your own conversion factors.

### Introduction

Transitioning from high school to college was an exciting time in my life during which I learned a lot of really cool things and which basically turned me into a life-long learner.

For example, in junior college I took a "physical science" class which was basically algebraic physics, physics for non-science majors but still a couple steps above "Physics for Poets" (the actual course title for an actual course). Later I took the first semester of real physics, which is based on calculus, such that it is said that Newton had to invent calculus in order to express and work with the ideas he was discovering for physics. There are some subjects in physics, such as the moment of inertia used in angular momentum, which are very simple to express in calculus but which require long tables of formulae in algebra (ie, a different formula for the moment of inertia of each different shape of rotating body and different placement of the axis of rotation). In such cases, the algebra just allows you to calculate the value for that specific case, whereas the calculus expresses the concepts.

In many of the lectures in this "physical science" class, the instructor would start with a few formulae and describe what they tell us of the relationships between its various factors (eg, directly proportional, inversely proportional, inverse-square law), another exciting new idea for me. Then he would solve for one of the factors in one formula and substitute that solution for a factor in another formula in order to come up with a new formula and new relationships. Wow! I mean, it has become common practice since then, but when it was first demonstrated to us ... wow!

But one real mind-blower he showed us (in a manner which seemed like an obvious aside at the time) was that you can treat the units of measurement for each factor as algebraic variables in themselves and thus end up with the correct units in your solution. Yes, that is correct; you can treat units the same as algebraic variables. That was one of the cooler things I learned in that "physical science" class. And then he pointed out that one of the more important functions of constants in physics is to make the units come out right -- I will demonstrate that below with the Constant of Gravitation ( G = 6.674×10-11 m3×kg-1×s-2 ) as it applies in Newton's Law of Universal Gravitation.

Outside of the obvious coolness factor, a very practical use for this fact is in deriving conversion factors. There are many times when we need to convert from one unit to another and we cannot always find the conversion factor(s) we need in a handy reference book. Remember, that was decades before the Internet, so having the right reference book was essential (now we can't even find handy reference books anymore except in used book stores -- in case you go looking for one, they were usually called "Books of Tables" and every scientist, engineer, and technology student had one sitting next to his slide rule). And many times, the conversion we need to perform is too unusual for it to even appear in any reference table. Or else we need to work through a chain of conversions (eg, converting miles down to inches). And maybe also have to account for exponents (eg, converting cubic inches to cubic yards). In such cases, we end up needing to create those conversion factors ourselves and we need to be able to keep everything straight.

Later at my first job as a software engineer (1982-1985) I quickly became the team's go-to guy for conversion factors even though I kept trying to teach them how to do it themselves (ie, what this page is about). Now mind you, these were exotic units (eg, binary angular measurement (BAMs) which were angles encoded directly in binary words) that had been defined specifically for our assembly program, so you could not find them in any reference books (nor could you find them on-line since in 1983 there was no such thing as "on-line").

More recently, I have used these techniques on a couple of my web pages: DWise1: Kent Hovind's Solar Mass Loss Claim and DWise1's Creation/Evolution Page: Earth's Rotation is Slowing. In fact, I created this page in order to replace a rather lengthy footnote in the solar-mass-loss page where I derive the rate at which the sun is losing mass due to hydrogen fusion. In the page on the earth's rotation, I cite several sources but all their rates are given in different units, so I have to convert them all into one common set of units in order to compare them meaningfully. On the latter page, I rely on these conversion techniques to ensure that those units get converted correctly and accurately.

### Mathematical Basis

About a year after having written this page, I decided that I should explain the mathematics behind my presentations on this page. It's all just basic algebra, albeit slightly extended to also apply to the units themselves by treating those units as variables. However, in algebra I've always had a bad habit of combining steps (too lazy to write every single thing down), so some of what I'm doing in the examples may not be readily apparent. Rather than include complete explanations in every single example, I have used key words or phrases or gone through the same kind of grouping and canceling factors. I'll show it here as completely as possible so that you will recognize it in the examples.

I should also add that working with units themselves algebraically is a common practice in chemistry and physics and that there are several videos on YouTube describing that practice.

A note on notation. In algebra, multiplication is indicated by concatenating variables and their coefficients; eg:

xy means x times y

4ab means 4 times a times b

Algebra also has an explicit multiplication sign, which is the center dot (⋅), which is used mainly when factoring numberic values; eg 12 = 2 ⋅ 2 ⋅ 3 . Programming languages use the asterix (*) for this while arithmetic and most people use the times "x" (×) in everyday calculations as well as in scientific notation (eg, 1.234×104). I have chosen to use the × in order to widen my reader audience, though there are some places where I use the algebraic notation.

Also, from learning mathematical proofs in geometry, we have the "ergo" ("therefore") symbol ( ) that is used to indicate your conclusing and which I might slip in now and again from habit.

The primary technique I am using here is multiplying by one, as one of my algebra teachers would describe it. In algebra we do it all the time along with adding zero (which we do whenever we add or subtract the same value to and from both sides of the equation; adding zero does not change the value, same as multiplying by one). Basically, by multiplying by one and by adding/subtracting zero, we can manipulate the form of the expression while at the same time preserving its value.

That technique of multiplying by one is based on these fundamental identities:

For the real numbers a and b:
```a × 1 = a -- any number times one is that number; ie, multiplying any value by one does not change that value

a
--- = 1 -- any value divided by itself equals one
a

a               b                         a     b
if a = b, then:  --- = 1  and  --- = 1  -- Corollary:  --- = ---
b               a                         b     a

```

To illustrate the technique, I'll borrow from the Converting Time Units section below. Let us consider determining how many seconds are in a work week, 5 days.

First, we list the relationships between the various units:
```1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds
```
From the identities above, we know that:
```   1 day        24 hours
----------  =  ----------  =  1
24 hours        1 day

1 hour         60 minutes
------------  =  ------------  =  1
60 minutes         1 hour

1 minute        60 seconds
------------  =  ------------  =  1
60 seconds       1 minute

```
And trivially (but it must still be said):
``` day
-----  =  1
day

minutes
---------  =  1
minutes

seconds
---------  =  1
seconds

```

So then we can now set up our problem for determining how many seconds there are in five days. Not only will I be using multiplication-by-one, but also the commutative property of multiplication in order to change the order of the factors.

```
5 days     24 hours     60 minutes    60 seconds
5 days = -------- × ---------- × ----------- × -----------
1         day          hour          minute

(5 × 24 × 60 × 60) × day × hour × minute × seconds
=  ----------------------------------------------------
day × hour × minute

432,000     day     hour     minute     seconds
=  --------- × ----- × ------ × -------- × ---------
1        day     hour     minute       1

432,000 × seconds
=  ------------------- × 1 × 1 × 1
1

5 days = 432,000 seconds

```
And you will be relieved to know that the examples below will not be so pedantic.

### ADDENDUM 2020 Dec 16: An Alternative Methodological Approach

Instead of working from knowledge of the theory to reliably roll your own conversions, you could use that theoretical knowledge to construct a set of rules that anyone should be able to follow without having to know why it works (a good approach if you're a supervisor or instructor whose subordinates need these rules). Such an approach is presented in a number of unit-conversion YouTube videos by LehrerSchmidt, which are all in German. Na ja, ich kann nichts dafür.

I've already referred to a LehrerSchmidt YouTube math video in which he teaches an "old forgotten multiplication method" which we also know as "Russian Peasant Multiplication" in which in order to multiply any two integers you need know nothing more than how to multiply by two, divide by two, identify a number as being either odd or even, and, of course, how to add -- basically, that's how computers perform integer multiplication in binary; refer to my page, DWise1: Old German Multiplication Method. LehrerSchmidt is a professional teacher in Germany who appears to teach at about USA junior or senior high school levels. His videos cover various topics in mathematics, physics, the German language, and other topics that spark his interest. Yes, all his videos are in German and I have not tested YouTube's translation subtitling feature on them yet. However, if you do know some German he does speak clearly enough and in Standard German so you should have a fighting chance of following him.

Since he presents a methodological approach instead of a theoretical one, he has divided his presentations on measuring unit conversions ("Maßeinheiten umrechnen") into a series of videos for each kind of units (in metric, of course), which are listed here via a YouTube search results page: lehrerschmidt masseinheiten umrechnen. We see the same kind of thing in physics, especially with angular momentum. The fundamental equation for calculating angular momentum is very simple and straight-forward, but it involves integral calculus. Solving that simple straight-forward double integral for all kinds of different shapes and along different axes of rotation results in the generation of an almost infinite number of algebraic formulae for calculating the angular momentum of a body.

Truly, in order to understand physics you need to know calculus -- the standard story is that Sir Isaac Newton invented calculus in order to be able to describe what he was discovering. And if you don't know calculus, then you need someone to derive the algebraic formulae for you to use. And if you do understand the calculus behind angular momentum, but your subordinates don't, then you can use your knowledge of the theory to construct a number of algebraic expressions for the shapes of the particular rotating bodies you're working with. They have something to work with without having to understand what's really going on.

That's kind of what I'm talking about. My goal with this page is to teach you what's really going on so that you should be able to resolve any unit conversion problem. LehrerSchmidt's approach provides a way for your subordinates to perform those conversions. Both approaches have their uses.

Watch some of LehrerSchmidt's videos to see his approach in action. Even if you don't know German, you should be able to deduce how it works (not unlike German and Italian cooking videos just show you how to do it so all you have to translate are the subtitles naming the ingredients). Basically, he builds a chain of the different units in descending order: larger units first, then listing smaller units to the right. For the transition between each adjacent pair of units, he has arrows pointing up and down the chain with conversion factors for each link. To convert from one unit to another, you then perform the successive multiplications or divisions of each step in the links to get there.

For example, here's that diagram from his Maßeinheiten umrechnen | Raummaße | m3, dm3, cm3, mm3 und Liter | Lehrerschmidt - einfach erklärt!, which covers volume conversions involving cubic linear measures as well as tying in liters (refer to the "Merke", remarks to the right). In a far inferior reconstruction of his graphics (he typically draws examples on paper with a Sharpie -- yes, that really is how Germans (and many other Europeans) write ones):

```        <-- ÷ 1000 --       <-- ÷ 1000 --       <-- ÷ 1000 --
m3                  dm3                cm3                 mm3
-- × 1000 -->      -- × 1000 -->       -- × 1000 -->
```
So if you have a volume in mm3, you convert that to cm3 (AKA "cc's" in old US medical practice) by dividing by 1000. In order to convert those mm3 to cubic meters (m3), you work progressively up the chain through cm3 and dm3 to arrive at m3: 1000 × 1000 × 1000 = 109 (an American billion). To convert from m3 to the other units, you simply multiply your way through the chain in the other direction. All you need is to be given that chain.

And as LehrerSchmidt demonstrates with several other such conversion chains, you can do the same with any other form of measurement, though doing it in metric is almost trivial since it's all in powers of ten (though in cubic and square measurements you still have to be careful). You could even do it for units of time, units of angular measurement, and any US customary units; for example:

```         <-- ÷ 1760 --          <-- ÷ 3 --        <-- ÷ 12 --       <-- ÷ 6 --         <-- ÷ 12 --
mile                 yard               foot               inch              pica              point
-- × 1760 -->          -- × 3 -->       -- × 12 -->        -- × 6 -->        -- × 12 -->
```
And for example if you follow this chain from miles through yards to feet, you get 1760 yd/mi × 3 ft/yd to get 5280 ft/mi. So you could supplement your chains with lists of conversion factors derived by combining links in the chain, like I just did to get 5280 ft in a mile.

Of course, the rest of my page does not use LehrerSchmidt's approach, but rather I present the theory so that you can construct any conversion chain you could possibly need. But you might still find his approach useful at some point.

### Practical Application Example: Gravitational Force

Newton's Law of Universal Gravitation is perhaps the classic example for how a constant's units will straighten out the units in the calculation.

First, we must choose our system of units. In physics, those are most commonly:

• Large-scale metric: meters, kilograms, seconds (MKS)
• Small-scale metric: centimeters, grams, seconds (cgs)
• US Conventional: foot-pound-second (fps)
There are a number of humorous systems of units, including the infamous furlong-firkin-fortnight (FFF) system. For this example, we will choose the large-scale metric system, MKS.

Within each system we derive many other units based on those basic three units. Here, we need to derive the units for force. Since force is mass (kg) times acceleration and acceleration is distance divided by time squared (m/s2), the units for the unit of force, newtons (N) in MKS, are:

1 N = 1 kg m/s2
= 1 kg m s-2
BTW, in the cgs system the unit of force is the dyne, which is defined as 1 g cm s-2

Now to define the variables and their units in MKS:

F is the force between the masses in newtons (1 N = 1 kg m s-2)
G is the gravitational constant, 6.674×10-11 m3 kg-1 s-2
m1 is the first mass in kilograms (kg)
m2 is the second mass in kilograms (kg)
r is the distance between the centers of the masses in meters squared (m2).
Please note the low magnitude of the gravitational constant. Of the four forces (gravitational, electro-magnetic, strong nuclear, weak nuclear), gravity is the weakest as is indicated by the gravitational constant. The only reason gravity seems so strong to us is because of how massive the earth (5.97×1024 kg) and the sun ( 1.9885×1030 kg) are.

The formula for gravitational force is:

```     G×m1×m2
F = ----------
r2
```
This gives us the following units calculations (using the times sign (×) to make it explicit):
```   m3          kg × kg          m × kg      kg      m2       m × kg
----------  × ---------    →   --------  × ----- × ----  →  -------- →  kg m s-2
kg × s2        m2                s2        kg      m2        s2

```
kg m s-2 is the definition for Newtons. Hence, the Gravitational Constant, G, has served its secondary purpose of making the units come out right.

QEF

### Practical Application Example: Converting a Constant Between Unit Systems

We know what the gravitational constant is in the MKS system, but what is it in the cgs system? Do we need to Google for that? Or can we just figure it out ourselves? Yeah, Goggle'ing would be what we'd usually do, but figuring it out will illustrate my point -- plus it's more fun.

First, we have G = 6.674×10-11 m3 kg-1 s-2 in MKS. To convert that to cgs, we need to change the meters to centimeters and the kilograms to grams and substitute those values in. Or as my high school algebra teacher would say, we just plug them in and grind it out:

```1 m = 100 cm
1 kg = 1000 g

G = 6.674×10-11 m3 kg-1 s-2
= 6.674×10-11 (100 cm)3 (1000 g)-1 s-2
= 6.674×10-11 × 1003 cm3 × 1000-1 g-1 s-2
= 6.674×10-11 × (102)3 cm3 × (103)-1 g-1 s-2
= 6.674×10-11 × 106 cm3 × 10-3 g-1 s-2
= 6.674×10-11 × 106 × 10-3 cm3 g-1 s-2
= 6.674×10(-11+6-3) cm3 g-1 s-2
= 6.674×10(-8) cm3 g-1 s-2
```
So what we come up with as the value of the gravitational constant in the cgs system is 6.674×10(-8) cm3 g-1 s-2. Did we get it right?

Looking it up in Wikipedia, we find it's equal to about 6.674×10-8 cm3 g-1 s-2. So we nailed it. And we also have gone through an example where the terms being substituted have exponents, so it's a two-fer.

### Practical Application Example with Exponents: Converting Solar Core Density to Other Units

Convert the density of the sun's core from g/cm3 to kg/m3:
```1 kg = 1000 g
1 m = 100 cm
Density of Sun's core = 162.2 g/cm3

1 kg        1       (100 cm)3
= 162.2 × (g × -------) × (---- ×  -----------)
1000 g       cm3      1 m3

g    1 kg         1      1003 cm3
= 162.2 × ( --- × -----) × ( ---- ×  ----------)
g    1000        cm3       1 m3

1 kg       cm3        106
= 162.2 × ( -----) × ( ------ ×  -------)
103        cm3       1 m3

1 kg      106
= 162.2 × ----- ×   -------
103       1 m3

106      1 kg
= 162.2 × ----- × -------
103       1 m3

= 162.2 × (103 kg/m3)

= 162,200 kg/m3
```
QEF

### Converting Time Units

Convert 7 days to seconds:
Strategy: repeatedly multiply the value being converted by 1, which does not change its value. Of course, 1 comes in a great many different forms.

```         24 hours                   day
7 days × --------- = 7 × 24 hours × ---- = 168 hours
day                     day

60 minutes                       hour
168 hours × ----------- = 168 × 60 minutes × ----- = 10,080 minutes
hour                          hour

60 seconds                           minute
10,080 minutes × ----------- = 10,080 × 60 seconds × ------- = 604,800 seconds
minute                              minute

Ergo: 7 days = 604,800 seconds
```
QED
Or, if you will be doing this more than once, you could have created a conversion factor for converting from days to seconds:
```                24 hours   60 minutes    60 seconds
1 day = 1 day × -------- × ----------- × -----------
day        hour          minute

day     hour     minute
= 1 × 24 × 60 × 60 × ----- × ------ × -------- × seconds
day     hour     minute

1 day = 86,400 seconds
```
QED
Conversion factor for converting from days to seconds: multiply by 86,400

### Deriving a Conversion Factor

As I said above, at my first job as a software engineer I was the go-to guy for conversion factors even though I kept trying to teach my co-workers how to do it themselves. Now mind you, these were exotic units (eg, binary angular measurement (BAM)) that you could not find in most reference books (nor could you find them on-line since in 1983 there was no such thing as "on-line").

The procedure is simple: you just set up an equation for the same value using different units and then solve for the unit you want to convert from; eg:

Derive the conversion factor for radians to degrees, given that a full 360° circle is 2π radians:
1 radian = (360 / 2π)°
1 radian = (180 / π)°
QED

Conversion factor for converting from radians to degrees: multiply by 180/π

### Converting Speeds

Converting speeds takes us into having to convert two units at the same time while keeping everything straight.

Convert miles per hour (mph) to kilometers per hour (kph):
Repeat the strategy of multiplying by 1, which comes in a great many different form.

```         1 mile     1.609344 km     1.609344 km
1 mph = -------- × ------------- = -------------
1 hour     1 mile           1 hour

Ergo: 1 mph = 1.609344 kph

Example:  60 mph = 96.56 kph  (nearly 100 kph, country-road speeds in Germany)
```

Convert kilometers per hour (kph) to miles per hour (mph):

Same thing only in reverse.
```         1 km       0.62137 miles     0.62137 km
1 kph = -------- × --------------- = -------------
1 hour     1 km              1 hour

Ergo: 1 kph = 0.62137 mph

Example:  50 kph = 31.0685 mph  (standard major street speed in German towns)
```

Many physics and navigation/piloting problems are in feet per second or meters per second, so you might want to convert to mph in order to get a feeling for how fast that is.
Convert feet per second (fps) to miles per hour (mph):
```         1 foot     1 mile        3600 seconds     3600     miles
1 fps = -------- × ----------- × -------------- = ------ × ------- = 0.68182 mph
1 sec      5280 feet     1 hour           5280     hour

Ergo: 1 fps = 0.68182 mph
```

Convert miles per hour (mph) to feet per second (fps):
```         1 mile     5280 feet       1 hour         5280     feet
1 mph = -------- × ----------- × -------------- = ------ × ------- = 1.467 fps
1 hour     1 mile        3600 seconds     3600     sec

Ergo: 1 mph = 1.467 fps
```

Often in movies and shows like Star Trek: The Next Generation, close maneuvering speeds are given in meters per second.

Convert meters per second (mps) to miles per hour (mph):
```         1 meter     1 mile              3600 seconds     3600         miles
1 mps = --------- × ----------------- × -------------- = ---------- × ------- = 2.237 mph
1 sec       1609.344 meters     1 hour           1609.344     hour

Ergo: 1 mps = 2.237 mph
```
Compared to that, mps to kilometers per hour (kph) would be almost trivial, so I leave it as an exercise for the reader:
1 mps = 3.6 kph

Converting Speeds in Odd Units

You can apply the approach above with any number of units, no matter how odd. Our example here is just such a case.

It turns out that many of the writers on The Simpsons are mathematicians, so a lot of math jokes get snuck in (eg, the Math Club's t-shirts saying "(a / b) ≠ π", which is trivially true for a and b being integers).

This often quoted joke spoofs how so many people turn up their noses at the simple and easy-to-work-with metric system in favor of an antiquated and very cumbersome system filled with arcane units that you have to memorize (in 1821, John Quincy Adams wrote: "Conservative educators have estimated that the use of the metric system . . . would save one to two years of the school life of every child."):

Grampa Simpson rants:
"The metric system is the tool of the devil! My car gets forty rods to the hogshead and that's the way I likes it."

OK, so just exactly what is "forty rods to the hogshead" in miles per gallon (mpg)?

• 1 statute mile (AKA survey mile) = 8 furlongs
• 1 furlong = 10 chains
• 1 chain = 4 rods
• 1 rod = 25 links
• 1 link = 7.92 inches

Real simple, right? Should be real easy to convert, right? So just to stay in keeping with the same conversion format we've been using:

``` 1 mile         1 furlong        1 chain            1          mile       1 mile
------------ × -------------- × ------------ = ------------ × ------- = ----------
8 furlongs     10 chains        4 rods         8 × 10 × 4     rods      320 rods

Ergo: 1 mile = 320 rods

```
Now that we know how many rods are in a mile, we need to discover how many how many gallons are in a hogshead ... or the other way around depending on whichever one is larger. I mean, who doesn't know that conversion off the top of their head? Like, nearly everybody.

From that same article only regarding fluid volume (yes, that's right, dry volume measurements are different -- and ignoring the fact that UK Imperial measurements are different as well), we learn this:

1 hogshead = 63 gallons (US liquid)

Now we know enough to be able to figure out how many miles per gallon Grampa Simpson's car is getting:

Convert rods per hogshead to miles per gallon (mpg):
```                       1 rod         1 mile           1 hogshead                1            mile             1 mile
1 rod per hogshead = ------------ × ---------- × ------------ = ------------- × ------- = ----------------
1 hogshead     320 rods         63 gallons             320 × 63   gallons         20160 gallons

Ergo: 1 rod per hogshead = 4.96×10-5 mpg

Actual Problem:

40 rods per hogshead = 40 × 4.96×10-5 mpg = 0.00198 miles per gallon (2 one-thousandsths of a mile)

0.00198 miles per gallon = 10.476 feet per gallon

Therefore, assuming a 20-gallon gas tank, he'll run out of gas in about 210 feet.

```

Yeah, it definitely looks like Grampa needs to get a different car.

In August 2020, I watched a YouTube video about why the US hasn't gone metric yet (I'm very pro-metric, though I do recognize valid problems with the USA converting over which is the subject of another page I'm working on). In the comments section, somebody repeated Grampa Simpson's quote above and, having already done the math for my own citing of the quote on another page I'm working on (repeated and developed above), I replied to that comment with the results showing how horrible his car's mileage is.

I received a reply to that which asked how many rods to the hogshead a car with decent mileage would get. That made me realize that I had neglected converting speeds, so I added that section. I posted my answer to his question in YouTube for him. And I thank him for having turned me to this task.

First, we need to come up with a conversion factor for mpg to rods per hogshead. Trivially, it's just the reciprocal of rods per hogshead to mpg:

```1 rod per hogshead = 4.96×10-5 mpg

1 mpg = (1/4.96×10-5) rods per hogshead = 20161.29 rods per hogshead

```

But to stay in keeping with our methodology, let's take the rigorous approach:

Convert miles per gallon to rods per hogshead:
```         1 mile       320 rods     63 gallons                rods
1 mpg = ---------- × ---------- × ------------ = 320 × 63 × ---------- = 20160 rods per hogshead

Ergo: 1 mpg = 20160 rods per hogshead
```
The difference of 1.29 rods (16.5 feet) is due to round-off error in the first calculation.

Now as for the question put to me, he suggested using the mileage of a Toyota Camry, but I don't know what that value is. My Honda Accord Hybrid gets about 40 mpg, so I'll use that:

```40 mpg = 40 × 20160 rods per hogshead = 806400 rods per hogshead.
```
806400 rods on one hoghead of gas amounts to 2520 miles on 63 gallons of gas.

As a check on my work, dividing 2520 miles by 63 to get miles per gallon gets us back to 40 mpg.

QED

### How Far Away is That Lightning?

As children we all learned the trick of estimating how far away from us that bolt of lightning is by counting the seconds between seeing the flash and hearing the thunder. But what we don't remember (at least what I couldn't remember) is how to convert number of seconds counted to number of miles away that lightning was.

So one day I looked up the speed of sound and figured it out:

About one-fifth of a mile for each second, so count the seconds and divide by five to get the approximate miles.

Or as an algebraic formula:
For:
d = distance of the lightning in miles
t = time between the flash and the thunder in seconds
d = t / 5

Now for how I figured that out. I chose to use algebra's ubiquitous "rate times time" formula:

distance = rate × time in seconds

Rate would be the speed of sound. One problem is that the speed of sound is variable depending on factors such as the medium it's travelling through, temperature, pressure, etc. I chose air as the medium, air pressure near sea level, and normal air temperature (about 20°C, 68°F), for which is given as:

• 667 kn (nautical miles per hour)
• 767 mph (statute miles per hour)
• 1125 ft/sec
• 1235 kph (kilometers per hour)
• 343 m/sec (meters per second)
So then choosing to use the 767 mph rate:
```          mi
d mi = r ---- × t sec
hr

767 mi     1 hr
d mi =  -------- × ---------- × t sec
hr        3600 sec

767 mi             1              1
d mi =  ---------- × t = ------- × t mi ≈ --- × t mi
3600 sec         4.694            5

∴ d (in miles) ≈ t / 5

```
As a check, now choose the 1125 ft/sec rate:
```        1125 ft
d mi = --------- × t sec
1 sec

1125 ft     1 mi
d mi =  --------- × ---------- × t sec
sec        5280 ft

1125 mi           1              1
d mi =  --------- × t = ------- × t mi ≈ --- × t mi
5280            4.693            5

∴ d (in miles) ≈ t / 5

```

If you use the metric system, you'd want to know how many kilometers away that lightning is:

Using the 1235 kph (kilometers per hour) rate:
```        1235 km
d km = --------- × t sec
1 hr

1235 km     1 hr
d km =  --------- × ---------- × t sec
1 hr        3600 sec

1235 km            1              1
d km =  ---------- × t = ------- × t km ≈ --- × t km
3600 sec         2.915            3

∴ d (in kilometers) ≈ t / 3

```
So the rule for kilometers would be:
About one-third of a kilometer for each second, so count the seconds and divide by three to get the approximate kilometers.

Or as an algebraic formula:
For:
d = distance of the lightning in kilometers
t = time between the flash and the thunder in seconds
d = t / 3

### Gas Prices Overseas

When we drive a rental car in another country, we will need to buy gas for it (AKA "petrol", "gasolina", "Benzin", "benzina", "essence"). So just how expensive is that gas? When you see the prices posted at the pump, how does that compare with the prices back home in the USA?

In these calculations, we have to convert both currency and volume, since of course you'll be using that country's money. In addition, they most likely won't be selling gas by the gallon (US) as in the USA, but rather in whatever they use there. Most other countries sell gas by the liter, but I've been told that Canada used to sell it by the half-gallon (and most likely Imperial half-gallons) though they have since switched to liters.

The example I'll use is from the UK:

• Prices are given in pence per litre
• 100 pence (100 p) = 1 pound (£1)
• the prices cited ranged from 114.4 p/litre (N. Ireland) to 119.5 p/litre (London)

Let's use both prices so that we can get a feel for the range of prices. And to simplify our work, I'll derive a formula which would be a function with respect to price.

The two conversion factors that I will use are:

• £1.00 = \$1.39 US
• 1 liter = 0.264172 gallons (US)

Starting with the price in pence per liter (represented by the variable, n), we will then multiply it successively by one (1):

```   n p        £1.00       \$1.39       1  liter             n × \$1.39           n × \$1.39        n × \$1.00
--------  ×  -------  ×  -------  × --------------  = --------------------  = -------------  = -----------
1 liter       100 p       £1.00      0.264172 gal      100 × 0.264172 gal      26.4172 gal      19 gal
```
So our conversion function, f(n) is:
```        n × \$1.00
f(n) = -----------
19 gal
```
Basically, take the British price in pence and divide it by 19.

Now that we have our conversion function, we can evaluate it for both the low and the high prices I was given:

• f(114.4) = \$6.02 / gal

• f(119.5) = \$6.29 / gal
The last time I got gas, the price was \$3.80 / gal, so the prices in the UK at this time is about 1.58 to 1.66 the price in Southern California.

The only caveat to that function I just derived is that while the conversion between pence and pounds and between liters and US gallons are constant and hence will not change, the same cannot be said for converting pounds to dollars (or Euros to dollars for that matter). Therefore, this conversion function will change as the exchange rate between dollars and the indiginous currency changes.

To offer a solution to that wrinkle (and having a case of COVID-induced cabin fever), I'll convert that single-variable function to a two-variable function which adds to the price variable, n, a currency variable, c (US dollars per unit of indiginous currency -- eg, dollars per pound, dollars per euro):

```   n p        £1.00       c × \$1.00         1  liter         n × c × \$1.00         n × c × \$1.00
--------  ×  -------  ×  -----------  × --------------  = --------------------  = ----------------
1 liter       100 p        £1.00         0.264172 gal      100 × 0.264172 gal      26.4172 gal
```
So our new two-variable conversion function, f(n,c) is:
```          n × c × \$1.00
f(n,c) = ----------------
26.4172 gal
```

Final test will be for us to apply this new conversion formula to see if we get the same results:

• f(114.4,1.39) = 114.4 × 1.39 ÷ 26.4172 (\$ / gal) = \$6.02 / gal

• f(119.5,1.39) = 119.5 × 1.39 ÷ 26.4172 (\$ / gal) = 6.29 / gal
They match!

Of course, if prices on the Continent are given in Euros instead of in cents (100 cents = 1 Euro), then you'd have to adjust the formula by a factor of 100.

Deriving Weird US Customary Unit Conversion Factors Using Metric as the Intermediary

How's that for dealing with odd-ball units?

Here's the story. One night four decades ago while I was on active duty, I received a phone call from a friend while I was on duty. He had a friend who was a farmer and who wanted to build a water trough a certain size (length, width, height in feet), but he wanted to know how much water that trough would contain and how much that water would weigh.

My friend had no idea where to start and neither did I. We had no idea what the conversion factor was from cubic feet to gallons, nor the conversion from gallons of water to weight. All we could think of was to wait until the next day when I could get to the university library where I might possibly have to spend hours research for those conversion factors. For my young readers, I might need to point out that the Internet was still in its infancy with few nodes (mainly major corporations and universities with military contracts) -- the Internet did not begin to become available to the public until the early 1990's, so an on-line search was not an option.

However, I did know that the three basic measurements in the metric system are related to each other through water: a cube that measures one centimeter on each side (ie, one cubic centimeter, cc) contains a volume of one milliliter and the mass of one milliliter of water is one gram. Knowing that, all I needed to do was to convert the trough's dimensions from feet to metric, do all the work in metric, and then convert the results from metric back to gallons and pounds. And the work in metric was very simple:

1. Convert the trough's dimensions from feet to centimeters.
2. Multiply the trough's dimensions in centimeters to get the volume in cubic centimeters (cc).
3. 1 cc = 1 milliliter, so we have volume in milliliters with no calculation needed.
4. Divide the volume in milliliters by 1,000 to get the volume in liters. No calculation needed since we can divide by 1,000 by sight.
5. Since 1 liter of water is 1 kilogram, we know the weight with no calculation needed.
6. Now convert liters to gallons and kilograms to pounds in order to present the answers.
Almost disgustingly simple to do in metric, but very difficult in US customary units. Especially when we do not know the conversion factors needed within the US customary system.

So then just what is the conversion factor from cubic feet to gallons? And the conversion factor from gallons of water to weight? I have no idea of either, but let's derive them! And using metric as our medium.

Here are some basic equalities:

1 foot = 30.48 cm
1 gallon = 3.7854 liters
1 cm3 = 1 ml
1 liter = 1000 ml
So I'll take a more standard algebraic approach here in which we solve for the number of gallons, x, such that:
```x gal = 1 ft3

(30.48 cm)3     1 ml       1 liter        1 gal
x gal = 1 ft3 × ------------- × -------- × ---------- × --------------
1 ft3         1 cm3      1000 ml      3.7854 liter

30.483         ft3     cm3      ml    liter
x gal = --------------- × ----- × ----- × ---- × ------ × gal
1000 × 3.7854     ft3     cm3      ml    liter

28316.84659
x gal = ------------- × gal
3785.4

x = 7.48054

```
Therefore, we arrive at: 1 ft3 = 7.48054 gal.

Google gives it as 7.48052 gallons to the cubic foot and the units conversion app on my phone says 7.48051948. Being off by a couple hundred-thousandths can be accounted for by round-off error, so I claim success in deriving this conversion factor.

I should also note that what we just did was a more-or-less straight volume-to-volume conversion. This next one is not so straight forward.

First, we will use most of the same basic equalities as before, plus a couple more: Here are some basic equalities:

1 gallon = 3.7854 liters
1 kilogram = 2.2046 pounds
1 liter = 1 kilogram (of water)
Now we'll solve for the weight, x, such that:
```x lb = 1 gallon

3.7854 liter     1 kilogram      2.2046 pounds
x lb = 1 gal × -------------- × ------------- × ---------------
1 gal         1 liter         1 kilogram

gal     liter    kilogram
x lb = (3.7854 × 2.2046) × ----- × ------ × ---------- × pounds
gal     liter    kilogram

x lb = 8.3453 lb

x = 8.3453

```
Therefore, we arrive at: 1 gal of water weighs = 8.3453 lb.

Google gives the weight of one gallon of water as 8.345404452 pounds, so we are plenty close enough.

A corollary conversion factor relates cubic linear units between the two systems: eg, between cubic feet and cubic meters. We already know that one cubic centimeter (the "cc's" you know so well from medical movies and TV) is equal to one milliliter, so it normally makes sense to convert from cubic centimeters to liters. After all, the liter is a fundamental unit in the metric system, whereas cubic centimeters and cubic meters are derived units. However, it appears that liters are more commonly used for liquid volume while cubic meters are more commonly used for solids and spaces. Of course, the units get scaled up and down depending on what they're measuring, down to cubic centimeters or up to cubic kilometers. For example, the volume of the earth is given in cubic kilometers (1.08321×1012 km3).

When we deal with units raised to a power, even conversions within the metric system can get tricky and require close attention. Basically, while you can go from centimeters to meters to kilometers just by shifting the decimal point (AKA multiplying and dividing by powers of ten), we must remember that when those units are squared or cubed, then we must also square or cube those decimal-point shifts. For example:

```1 meter = 100 centimeters
1 kilometer = 1000 meters

1 cubic meter = (100 cm)3 = 1003 cm3
Therefore 1 m3 = 1,000,000 cm3

1 cubic kilometer = (1000 m)3 = 10003 m3
Therefore 1 km3 = 1,000,000,000 m3

```

So what's the relationship between liters and cubic meters?

```1 meter = 100 centimeters
1 cubic centimeter = 1 milliliter
1 liter = 1000 milliliters

1 cubic meter = 1,000,000 cubic centimeters = 1,000,000 milliliters

1 liter       1,000,000 liters
1 cubic meter = 1,000,000 ml × ---------  =  ------------------
1000 ml        1000

Therefore, 1 cubic meter = 1000 liters

```

Converting cubic meters to cubic feet follows the same procedure we've been using all along:

```1 meter = 3.28 feet

1 cubic meter = (3.28 ft)3 = 3.283 ft3
Therefore 1 m3 = 35.31467 cubic feet

```

So this section shows that we can derive truly odd conversion factors, especially if we expand our toolkit to include other systems of measurement.

Weighing the Unweighable through Density

This is the logical extension of the ideas in the previous section wherein we found that one liter of water masses at 1 kilogram. But what about one liter of stone, such as granite?

For example, you are at a national or state park or in a museum and the guide or broshure tells you that that big boulder weighs this many tons. If you are like I was as a kid, I wondered who the poor guy was who had to lift that boulder and put it on a scale to weigh it, and then he had to put it back exactly where it had been as if it had never been disturbed. Clearly impossible, but how else could they have done it?

Actually, quite simply by determining its volume and then using its density (ρ) to arrive at its weight. A substance's density is defined as its mass per unit of volume; ie, for mass m and volume V, the density ρ is

```      m
ρ = ---
V
```
Then given the density of a substance and how much of it there is (ie, its volume), you can calculate the mass thus:
``` m = ρ × V
```

Substancekg/m3g/cm3lb/cu ftRD
Hydrogen0.08988.98×10-55.606×10-38.98×10-5
Helium0.1791.79×10-41.11746×10-21.79×10-4
Air (at sea level)1.20.00127.49×10-20.0012
Styrofoam700.074.370.07
Cork2400.2414.980.24
Wood7000.743.70.7
Cooking oil9200.9257.43370.92
Water (fresh)1,0001.062.4281.0
Plastics1,1751.75109.251.75
Sand1,6001.699.8851.6
Concrete2,4002.4149.8272.4
Glass2,5002.5156.072.5
Granite2,7002.7168.55552.7
Aluminium2,7002.7168.55552.7
Limestone2,7502.75171.676892.75
Basalt3,0003.0187.2843.0
The Earth5,5145.514344.22785.514
Iron7,8707.87491.37.87
Steel7,9007.9493.57.9
Copper8,9408.94558.18.94
Silver10,50010.5655.4910.5
Gold19,32019.321206.119.32
Platinum21,45021.451339.0821.45

The units for density depend on which system you are using. In physics we use three basic systems that are named for the basic units from which all other units are derived:

• MKS -- meter, kilogram, second -- used for large scale systems -- density in kg/m3

• cgs -- centimeter, gram, second -- used for small scale systems -- density in g/cm3

• foot-pound-second -- used with US Customary units -- density in pounds per cubic foot

A substance's density is determined empirically by measuring the volume and mass (or weight) of a sample and then applying those measurements to the basic formula. Of course it's not quite that simple in practice, since those measurements must be made under standard conditions (eg, standard pressure and temperature).

Related to density, which has units, is relative density (RD), also called specific gravity. Relative density is the ratio of the density of the material to that of a standard material, usually water. Since you are dividing one density by another, the standard, the units cancel out and you end up with a dimensionless value which simplifies comparisons of densities across different systems of units. Often when densities are given, it will be the relative density.

Another use of relative density is that you can easily determine a substance's buoyancy. More specifically, if the relative density is less than one, then the substance will float in the standard substance (most commonly fresh water at 4°C, the temperature of its maximum density). And if it's greater, then the substance sinks. For example, in the table to the right you can see that the RD of cooking oil is less than that of water, which is why oil mixed with water will separate out and float on the top. The same will happen in air such that a substance whose RD is less that air's will float (eg, hydrogen or helium).

But to work the problem that we want, we need a density that has units to it, like kg/m3. But if you look at that table to the right, you will notice that the Relative Density (RD) is the same as density in g/cm3 since by definition one cm3 (AKA "milliliter") masses at one gram. And then all the other densities can be derived by multiplying the density of water by their relative densities.

OK, so let's try a couple examples just to demonstrate how you would go from an object's measurements to its volume to its mass.

At an archaelogical site, you find a limestone block which had been quarried and transported from a great distance away.
```Dimensions of the limestone block:  127 cm × 127 cm × 71 cm
Volumeblock = 1,145,159 cubic centimeters
ρlimestone = 2.75 g/cm3

Massblock = ρlimestone × Volumeblock = 2.75 g/cm3 × 1,145,159 cm3
Massblock = 3,149,187.25 g = 3.149 metric tonnes

Therefore, that limestone block weighs 3.149 tonnes

```

How much does the earth ( ⊕ ) weigh?
```Volume⊕ = 1.08321×1012 km3
ρ⊕ = 5514 kg/m3

Mass⊕ = ρ⊕ × Volume⊕ = 5514 kg/m3 × 1.08321×1012 km3

kg × km3     (1000 m)3
Mass⊕ = 5.972 819 94×1015 × --------- × ----------
m3            km3

kg × km3 × m3
Mass⊕ = 5.972 819 94×1015 × (103)3 × ---------------
km3 × m3

Mass⊕ = 5.972 819 94×1015 × 109 kg = 5.972 819 94×10(15+9) kg

Mass⊕ = 5.972 819 94×1024 kg

```
Wikipedia reports the earth's mass to be 5.97237×1024 kg. So close enough.

Piloting: The 3-Minute Rule

When piloting a craft, be it a ship or a boat in a waterway or even a car on the freeway, there are a number of rules for performing rapid calculations. One that we've all used on the freeway has been to estimate time to travel the next X miles as being X minutes if we're travelling at 60 miles per hour (ie, one mile per minute).

The 3-Minute Rule is given in the US Navy's Quartermaster Nonresident Training Course (NAVEDTRA 14338, page 8-6):

 Another way of solving problems of distance, speed, and time is by using the 3-minute rule. The 3-minute rule will help solve mathematical computations without a nomogram or calculator. The rule states: The distance traveled in yards over 3 minutes divided by 100 equals the speed in knots. To simplify, just drop two zeros from any distance traveled in yard in any 3 minute period. Example 1: Ship travels 1,600 yd. in 3 min. 1,600/100 = 16 (Speed is 16 knots). Example 2: Ship’s speed is 16 kn for 3 min. 16 x 100 = 1,600 yd.
```1 knot (standard abbreviation, kn, also written as "kt" in aviation) is the speed of one nautical mile per hour.
1 nautical mile (nmi, also M, NM, or nm) is 2000 yards.  It is also one minute (one-sixtieth of a degree) of latitude along any line of longitude.

nmi
Rate times time equals distance in knots, minutes, and yards:   r ----- × t minutes  =  d yd
hr

1 nmi = 2000 yd
1 hour = 60 minutes

nmi     2000 yd     1 hr
r ----- × -------- × -------- × t minutes  =  d yd
hr       1 nmi     60 min

2000 yd
r × --------- × t minutes  =  d yd
60 min

100 yd
r × -------- × t minutes  =  d yd
3 min

t minutes
r × 100 yd × -----------  =  d yd
3 min

t
r × 100 yd × ---  =  d yd
3

```
Thus we have the formula for determining how far you will travel at a given speed for a given number of periods of three minutes. Or to figure out what your speed needs to be to travel a given distance for a given period of time (in terms of periods of three minutes).

It should be possible to use a similar method to this to come up with other such "rules".