Convert Units Easily with Factor-Label Unit Conversion

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Converting units is a very common work task, and trying not to get tripped up by it is a real pain at times. I'm sure you've already read the Mars Rover and other trite examples, so what can we do?

We have already discussed some methods to help convert units (like the Google Search Tricks and unit conversion program articles), but today we go beyond programs into what you might call a technique, or a frame of reference. It's a way to think through and record your unit conversions that I find clear remarkably and error-resistant, whether you are working by hand, by computer, or both: the factor-label method.

The factor-label method begins by observing that anything multiplied by itself is one. For example, consider arbitrary variable “x.” We know that x * 1 = x, therefore x * 2/2 = x, x * 325235/325235 = x, etc. So mathematically, we can multiply any value by "one" without changing it. Next, we realize that equivalent amounts of a unit are equal, so they can be considered “one.” For example, 100 cm = 1 m, therefore 100 cm / 1 m = 1, in a sense. 16 ounces / 1 pound = 1, etc.

You also remember that any unit divided by itself can be removed from an equation: 1000 m / km * 1 km / 0.6213 miles = 1000 / 0.6213 = 1000 m/0.6213 miles, which is correct.

You can use a “chain” of these conversions, as long as you like, to get from one number to another. You can also use the “chain” to get yourself to conversions you do know.


30 km/hr = how many ft/sec?

Imagine that you forgot how many seconds are in an hour and you never memorized how many feet are in a kilometer. But you do remember 1 m = 3.28 ft and I hope you recall that 60 s = 1 min and 60 min = 1 hr. Therefore you can solve this problem by multiplying 30 km/hr by "one" a few times and cancelling redundant units:

30 km/hr * 1000 m/km * 3.28 ft / m * 60 hr / min * 60 min / sec = 27.3 ft/sec

Interesting note: this method is handy for checking your work! When I was typing this example the first time I accidentally wrote “* 60 min/hr * 60 sec/min.” Having written it out in this way made it easy for me to go back and see my mistake.


Now let’s try a harder example, where we go beyond conversion: Say gas goes up to $2/gallon, and you live 50 miles away from work. What is your gas cost for a daily commute going to be?

I don’t off the top of my know head what equation I should plug these numbers into to solve the problem, or what other info I need. The factor-label method will help me get the answer even if I don’t know quite where to start:

“Hmm, I want to know $/day. But I don’t know that…I will start with something I do know and go from there. I know every day I have to go to and from work: two trips.”

2 trips / day * 50 miles / trip = 100 miles / day

“I now know miles/day and $/gallon, so I need miles/gallon to connect the two. And I can find that out! I can look EPA stats for up my car model…my car gets 30 miles per gallon in the city.”

"30 miles per gallon" implies 30 miles / 1 gallon is sort of the same as “1/1” in this specific case. It also means I can invert it: 1 gallon / 30 miles.

100 miles / day * 1 gallon / 30 miles * $2 / gallon = $6.67 / day.

  Comparison to other methods

Compared to memorizing conversion equations, or tricks like the Force Triangle and Mole Wheel, I find this a lot simpler: start where you can start, generate “1/1” style equations to get you to the units you need, looking things up when you get stuck. It also lets you use whatever unit conversions you are comfortable with or can remember. (I can always use 1 m = 3.28 ft if I forget 1 ft = 0.305 m). And by writing it out, you get a logical trail of breadcrumbs you or a checker can follow later.

With care, you can also extend the method to stranger problems. For example, consider using “$1 un-taxed / $1.13 with tax” or “$1US 2008 / $1.03 US 2009” to add taxes and inflation to a calculation, or “0.75 kW heat to house / 1.0 kW heat released total” to add a 75% furnace efficiency to you calculation.


One place to beware is any units that have a different “zero” starting point, like temperatures and pressures. For these units you can use the FL method to consider a change in the value, but not the absolute values.

For example, say “the room heated up by 10°C. How many °F did it heat up?” You can answer that 10 °C * (9 °F / 5 °C) = 18 °F change. However, if the question is “the room went from 0°C to 10°C, what is the temperature in °F?” You cannot use the same equation, "10 °C * 9 °F / 5 °C = 18 °F." This is wrong Because the zero point for Celsius is 0°C, but that same temperature is actually 32° Fahrenheit. Their "zero" values don't match.

Or for pressure, 6.9 kPa = 1  psi. So if you say “The pressure increased 30 kPa, how many psi is that?” You can do 30 kpa * 1 psi / 6.9 kPa = 4.35 psi. That’s fine.

But if you say “the pressure is 30 kPa gauge, what is it in psi gauge?” You cannot use 30 kpag * 1 psig / 6.9 kPag = 4.35 psig. Because the “zero” point for your gauge is going to be different for the two systems: usually ~101 kPa or 14.7 psi.

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