I'm working on calculating descent velocity of an object under a parachute, but I'm running into issues when checking my units.

The formula for velocity with a parachute is (at least from my research) $$v = \sqrt {{{2w} \over {\rho {C_D}A}}} $$ In English units, $w$ is the weight of the object and parachute in $\rm{lb}$, $\rho$ is the air density in ${{{\rm{lb}}} \over {{\rm{f}}{{\rm{t}}^{\rm{3}}}}}$, $A$ is the area in $\rm{ft}^\rm{2}$, and $C_D$ is the unitless coefficient of drag.

Showing just the units, you get $$\sqrt {{{{\rm{lb}}} \over {{{{\rm{lb}}} \over {{\rm{f}}{{\rm{t}}^{\rm{3}}}}}{\rm{ \times f}}{{\rm{t}}^{\rm{2}}}}}} {\rm{ = }}\sqrt {{\rm{ft}}} \ne {{{\rm{ft}}} \over {{{\rm{s}}}}}$$

How does this work? Am I making a mistake somewhere?

  • $\begingroup$ Is the density $\frac{lb}{ft^3}$ ? Is it not $\frac{lb*s^2}{ft^4}$? $\endgroup$ – Ben S Feb 6 '17 at 5:22

You're making a mistake, but it's a very subtle one (and indeed part of the reason most scientists don't like to use imperial units). There are actually two different units that go by the name of "pound". One is a unit of mass, and the other is a unit of force, sometimes designated $\mathrm{lbf.}$ when one wants to distinguish between them. They differ by a factor of $g$, the standard gravitational acceleration: $$\mathrm{lbf.} = g\times\mathrm{lb.}$$ Note that $g$ has units of distance per time squared. (It's about $32\ \mathrm{ft./s^2}$, but the specific value isn't important for my explanation.)

In your equation, the pound you use to measure the weight $w$ is the pound of force, but the pound you use to measure the air density is the pound of mass. If you're careful to maintain the distinction between these two, you'll see that the units work out correctly: $$\sqrt{\frac{\mathrm{lbf.}}{\frac{\mathrm{lb.}}{\mathrm{ft.}^3}\times\mathrm{ft.}^2}} \propto \sqrt{\frac{\mathrm{lb.}\times\mathrm{ft./s^2}}{\frac{\mathrm{lb.}}{\mathrm{ft.}^3}\times\mathrm{ft.}^2}} = \sqrt{\frac{\mathrm{ft.}^2}{\mathrm{s}^2}} = \frac{\mathrm{ft.}}{\mathrm{s}}$$

  • $\begingroup$ I don't suppose the US will wake up to the metric system for the next 4 years at least :-( They can Australia as an example. We did it in the 1970s and it was very painless. However, for many years the government was ruthless: you were not allowed to use imperial units anywhere. Now that we've switched, the rules have been relaxed. $\endgroup$ – hdhondt Feb 6 '17 at 9:10
  • $\begingroup$ @hdhondt, I was told in 1973, as a high school physics student, that the U.S. was going to convert to the metric system. If that change hasn't happened in 43 years, it's probably not going to happen. $\endgroup$ – David White Feb 6 '17 at 11:59
  • $\begingroup$ Wow, the imperial system really is awful. I knew there was a distinction between lb and lb-ft, but I had no idea it would come into play in this case. $\endgroup$ – Ian Feb 6 '17 at 16:04
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    $\begingroup$ @Ian And then there's units like slugs to "make things easier"; until you forget what slugs are because you're Canadian and have to work in both units and don't want to spend all your time remembering conversion rules. $\endgroup$ – JMac Feb 6 '17 at 16:57
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    $\begingroup$ @Ian Be careful: this is a distinction between lb. and lbf. The foot-pound, lb.-ft. (or technically lbf.-ft.), is yet another unit entirely; it's a unit of energy. $\endgroup$ – David Z Feb 6 '17 at 20:15

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