DOFPro Team

Is Furlongs Per Fortnight a Thing? is an episode of the Degrees of Freedom Project, Just the Facts videos on chemical and thermal principles. If you want The Full Story version with an in-depth discussion, use the link in the description. Definitions and additional details can be found in the web-page link in the description as well.
Now if \(1\ \mathrm{inch} = 25.4\ \mathrm{mm}\), we can rearrange the equation as:
\(1 = \frac{25.4\ \mathrm{mm}}{1\ \mathrm{inch}}\) or \(1 = \frac{1\ \mathrm{inch}}{25.4\ \mathrm{mm}}\)
Multiplying a quantity by unity does not change its value but does change its units.
\(35.7\ \mathrm{inch} = \frac{35.7\ \mathrm{inch}}{} | \frac{25.4\ \mathrm{mm}}{1\ \mathrm{inch}}=906.78\ \mathrm{mm}\)
An equation such as
\(35.7\ \mathrm{inch} = \frac{35.7\ \mathrm{inch}}{} | \frac{25.4\ \mathrm{mm}}{1\ \mathrm{inch}}=906.78\ \mathrm{mm}\)
is known as a dimensional equation and it will handle most of your unit-conversion issues. Here is a dimensional equation to convert \(344\ \mathrm{m/s}\) to furlongs per fortnight using exact NIST values.
\(\frac{344\ \mathrm{m}}{\mathrm{s}} | \frac{1\ \mathrm{yd}}{0.9144\ \mathrm{m}} | \frac{1\ \mathrm{furlong}}{220\ \mathrm{yd}} | \frac{3600\ \mathrm{s}}{1\ \mathrm{hr}} | \frac{24\ \mathrm{hr}}{1\ \mathrm{day}} | \frac{14\ \mathrm{day}}{\mathrm{fortnight}} |\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2,068,432.36\ \frac{\mathrm{furlong}}{\mathrm{fortnight}}\)
There are precise rules for tracking significant figures, also known as sig figs or significant digits through a set of calculations. If you are curious, check out Wikipedia or the links on the website.
DOFPro will do the following:
Use whatever rounding method you prefer.
The most common web search tool will do unit conversions for you. Try typing the following into your browser url bar or search engine.
Hit return to get a calculator with available conversions and units.
Exact
\(\frac{25.4\ \mathrm{mm}}{1\ \mathrm{inch}}\),
\(\frac{101\ 325\ \mathrm{Pa}}{1\ \mathrm{atm}}\),
\(\frac{4.184\ \mathrm{J}}{1\ \mathrm{cal_{th}}}\),
\(\frac{9\ \mathrm{^\circ R}}{5\ \mathrm{K}}\),
\(\frac{4.184\ \mathrm{J/kg\cdot K}}{1\ \mathrm{Btu_{th}/(lb_m \cdot ^\circ R)}}\).
Close to Exact (just memorize 3 or 4 sig figs)
\(\frac{0.453\ 592\ 4\ \mathrm{kg}}{1\ \mathrm{lb_m}}\),
\(\frac{4.448\ 222\ \mathrm{N}}{1\ \mathrm{lb_f}}\),
\(\frac{0.946\ 352\ 9\ \mathrm{L}}{1\ \mathrm{qt}}\).
Approximate
\(\frac{2.2\ \mathrm{lb_m}}{1\ \mathrm{kg}}\),
\(\frac{1\ \mathrm{Btu/s}}{1\ \mathrm{kW}}\).
Some final words of advice: Whenever possible, work the problem in the units given.
Don’t confuse the pound mass, \(\mathrm{lb_m}\), the pound force, \(\mathrm{lb_f}\), and the pound mole, \(\text{lb-mol}\).
Thanks for watching!
The Full Story companion video is in the link in the upper left. The next video in the series is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.

