How to Convert Knots to MPH, and More Unit Magic
using conversion factors to change units and confirm results

The boring way to convert knots to mph is to multiply knots by 1.15.((This is actually more like 1.150779, but we’ll use the rounded version.)) But this skips over some awesome unit magic.

Specifically, we can use conversion factors to change between related units. Let’s look again at how to convert knots to mph.

Suppose you’re traveling at v\ \text{knots}=v\ \frac{\text{nautical miles}}{\text{hour}}. Then,

    \[ v\ \frac{\cancel{\text{nautical miles}}}{\text{hour}} * \frac{1.15\ \text{miles}}{1\ \cancel{\text{nautical miles}}} = v * 1.15\ \frac{\text{miles}}{\text{hour}}} = v * 1.15\ \text{mph} \]

Did You Just Cross Out the Units?

I did! This is called a conversion factor,((More precisely, you may hear this called dimensional analysis or the factor-label method.)) and is based on the principle that anything multiplied by 1 equals itself. It is an easy way to convert between units, and allows us to solve a variety of aviation problems.

For instance, we know

1.15\ \text{miles} = 1\ \text{nautical miles}

and so

\frac{1.15\ \text{miles}}{1\ \text{nautical miles}} = 1

To convert knots to mph, we multiple our original speed by this conversion factor. We then cancel out common units on the top and bottom of the equation.((We can cancel these out because \frac{\text{miles}}{\text{miles}} = 1.)) So,

    \[ v\ \frac{\cancel{\text{nautical miles}}}{\text{hour}} * \frac{1.15\ \text{miles}}{1\ \cancel{\text{nautical miles}}} = v * 1.15\ \frac{\text{miles}}{\text{hour}}} \]

Practical Uses

Complex Conversions

This approach is particularly handy for complex unit conversions. For instance, suppose your fuel burn is 10 gallons per hour, and you want to know how much weight you’ll lose after flying for 30 minutes.

To find the fuel consumed, convert \frac{\text{gallons}}{\text{hour}} to \frac{\text{gallons}}{\text{minute}}, then multiple by 30 minutes. This looks like,

    \begin{align*}\frac{10\ \text{gallons AvGas}}{1\ \cancel{\text{hour}}} * \frac{1\ \cancel{\text{hour}}}{60\ \cancel{\text{minutes}}} * 30\ \cancel{\text{minutes}} &= \frac{10 * 1 * 30}{1 * 60}\ \text{gallons AvGas} \\&= 5\ \text{gallons AvGas}\end{align*}

To find the weight equivalent, we know 1\ \text{gallon AvGas} = 6\ \text{lbs}, so

    \[5\ \cancel{\text{gallons AvGas}} * \frac{6\ \text{lbs}}{1\ \cancel{\text{gallons AvGas}}} = 30\ \text{lbs}\]

If you’re feeling ambitious, you can do the whole conversion in one equation:

    \[\frac{10\ \cancel{\text{gallons AvGas}}}{1\ \cancel{\text{hour}}} * \frac{1\ \cancel{\text{hour}}}{60\ \cancel{\text{minutes}}} * 30\ \cancel{\text{minutes}} * \frac{6\ \text{lbs}}{1\ \cancel{\text{gallons AvGas}}} = 30\ \text{lbs}\]

Plausibly Useful Idea: Canceling Terms After Factorization

A related idea is using factorization to simplify the math of equations like the one above.

To do this, we first factor out((Meaning break the number into smaller numbers that multiply back to the original number, such as 60 = 10 * 6 = 10*3*2.)) common terms in both the numerator and denominator. We can then cancel those common terms.

Using the example above,

    \begin{align*}\frac{10}{1} * \frac{1}{60} * 30 * \frac{6}{1}\ \text{lbs} &= \frac{\cancel{10}}{1} * \frac{1}{\cancel{10} * \cancel{6}} * 30 * \frac{\cancel{6}}{1}\ \text{lbs} \\&= 30\ \text{lbs}\end{align*}

Unusual Unit Types

Conversion factors also make handling uncommon unit types easier. One unit we encounter in weight & balance is \text{lb-inches}, the unit of a moment.

To find the center of gravity (\text{CG}), we divide the total moment M_{total} by the total weight w_{total}. We can sanity check this logic by canceling units and confirming we end with \text{inches}.

    \[CG = \frac{M_{total}\ \text{\cancel{lb}-inches}}{w_{total}\ \cancel{\text{lb}}}=\frac{M_{total}}{w_{total}}\ \text{inches}\]

Knowledge Test Questions

Another powerful use for conversation factors is reasoning through knowledge test questions. For instance, if you encounter this problem,

If fuel consumption is 75 pounds per hour and groundspeed is 110 knots, how many gallons of fuel are required for an airplane to travel 380 NM?

Based on King School’s FAA Commercial Prep Questions

you can find a solution with conversation factors:((We can use the factorization method here as well:

    \begin{align*}\frac{75 * 380}{110 * 6} &= \frac{(3 * 25) * (19 * 2 * 10)}{(11 * 10) * (2 * 3)} \\ &= \frac{(\cancel{3} * 25) * (19 * \cancel{2} * \cancel{10})}{(11 * \cancel{10}) * (\cancel{2} * \cancel{3})} \\ &= \frac{(25) * (19)}{(11) * (1)} \\ &= 43.18\end{align*}


    \[\frac{75\ \cancel{\text{lbs}}}{1\ \cancel{\text{hour}}} * \frac{1\ \cancel{\text{hour}}}{110\ \cancel{\text{nautical miles}}} * 380\ \cancel{\text{nautical miles}} * \frac{1\ \text{gallons AvGas}}{6\ \cancel{\text{lbs}}} = 43.18\ \text{gallons AvGas}\]

I find myself using this trick nearly every time I’m working with units — I hope you find it useful as well!

And of course, if you’re in the Bay Area and interested in learning to fly, shoot me a message and I’d be happy to meet you!


P.s. This post’s LaTeX functions were made possible by QuickLaTeX.

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