Calculating terminal speed

Bookmark and Share

In the 22,000 Foot Fall episode of Mythbusters, Adam did a calculation of how long it should take for a falling person to reach terminal speed. It occurred to me that there’s a wrong (but simple) way and a right (but complicated) way to do this calculation — I wonder which one was used on the show to come up with 487 feet?

First, the simple way. From numerous tests in previous episodes, the Mythbusters know that the terminal speed of a person falling through air is about 120 miles per hour. Based on that, you could try to figure out the fall height it takes to achieve that speed using the formula

$$h_0 = \frac{v_T^2}{2g}$$

(It’ll become clear later on why I’m calling the height \(h_0\)). This formula comes from the kinematic equation \(v^2 = v_0^2 + 2ad\) with initial velocity \(v_0 = 0\), final velocity \(v = v_T = \unit{120}{\mileperhour}\), and acceleration \(a = g = \unit{9.8}{\frac{\meter}{\second^2}}\). Plugging in the numbers and calculating gives

$$h_0 = \unit{147}{\meter} = \unit{482}{\foot}$$

That’s reasonably close to what Adam got on the show, so I’m guessing this is the method he used. (In other news, Google actually makes a pretty awesome calculator)

Now, what’s wrong with that, and how do you correct it? Well, for starters, a falling body never really reaches terminal speed. What actually happens is that, as the body’s speed gradually increases, the force of air resistance also gradually increases. This, in turn, reduces the net acceleration, meaning that the speed changes progressively less and less as the body falls. This means that the speed follows what we call an asymptotic curve, getting closer and closer to some particular value but never quite reaching it. For a falling object, that particular value is the terminal speed.

The real situation makes it a bit tricky to calculate just how far off the mark Adam’s calculation was. The thing is, we’re trying to calculate how far the body takes to reach a speed which it never reaches. So technically, the correct answer is “an infinite distance.” What we can do, on the other hand, is calculate what the body’s speed would be after falling that \(\unit{482}{\foot}\) distance, which is probably what’s important for testing the myth. Adam and Jamie simply want to ensure that after falling \(\unit{500}{\foot}\), their dummy’s speed is a reasonable approximation to what its speed would be after falling \(\unit{22000}{\foot}\).

Now, on to the math. To calculate what happens to the falling body, we’ll need Newton’s second law,

$$\frac{1}{2}CA\rho\dot{y}^2 - mg = m\ddot{y}$$

where the dot indicates the time derivative (\(\dot{y} = \ud{y}{t}\) and \(\ddot{y} = \udd{y}{t}\)). This can be rearranged to

$$\udc t = \frac{m\,\udc{\dot{y}}}{\frac{1}{2}CA\rho\dot{y}^2 - mg}$$

and integrated to give

$$t = -\sqrt{\frac{2m}{gCA\rho}}\atanh\Biggl(\sqrt{\frac{CA\rho}{2mg}}\dot{y}\Biggr)$$

(I did it in Mathematica). Now we can rearrange this to solve for velocity,

$$\dot{y} = -\sqrt{\frac{2mg}{CA\rho}}\tanh\Biggl(\sqrt{\frac{gCA\rho}{2m}}t\Biggr)$$

and integrate one more time to get the position,

$$y = h - \frac{2m}{CA\rho}\log\cosh\Biggl(\sqrt{\frac{gCA\rho}{2m}}t\Biggr)$$

In order to compare the speeds when a body is dropped from \(\unit{500}{\foot}\) vs. \(\unit{22000}{\foot}\), we’ll need to solve this for time,

$$t = \sqrt{\frac{2m}{gCA\rho}}\acosh\exp\Biggl[-\frac{CA\rho}{2m}(y - h)\Biggr]$$

and plug it back into the expression for \(\dot{y}\) above. When we do that, notice that we’re going to wind up taking the hyperbolic tangent of an inverse hyperbolic cosine, so we can use the identity

$$\tanh(\acosh x) = \frac{\sqrt{x^2 -1}}{x} = \sqrt{1 - x^{-2}}$$

(valid for \(x > 0\)). In this case, \(x\) is that exponential,

$$x = \exp\Biggl[-\frac{CA\rho}{2m}(y - h)\Biggr]$$

so substituting in, I wind up with

$$\dot{y} = -\sqrt{\frac{2mg}{CA\rho}\bigl[1 - \exp(CA\rho(y - h)/m)\bigr]}$$

That’s a wacky-looking equation… but at last it’s time to plug in some numbers.

First of all, we know (or are assuming) that terminal speed is \(\unit{120}{\mileperhour}\). To make use of that, we can go all the way back to Newton’s second law,

$$\frac{1}{2}CA\rho\dot{y}^2 - mg = m\ddot{y}$$

By definition, at terminal speed (when falling), \(\ddot{y} = 0\). So we can start by plugging that in, and after a little rearrangement we get

$$v_T^2 = \frac{2mg}{CA\rho}$$

This is awfully convenient because the combination of variables on the right occurs as-is in the equation for speed. Now we know its value is just the terminal speed squared, \(v_T^2 = (\unit{120}{\mileperhour})^2 = \unit{2880}{\bigl(\frac{meter}{\second}\bigr)^2}\). And in fact, now we can simplify the speed equation to

$$\dot{y} = -v_T\sqrt{1 - \exp[CA\rho(y - h)/m]}$$

which makes it pretty obvious that the velocity is closely related to the terminal speed — as it should be.

At this point the only combination of unknown variables left is \(\frac{CA\rho}{m}\), which, according that last manipulation of Newton’s law, is equal to \(\frac{2g}{v_T^2}\). Recognize that formula? It’s just \(1/h_0\), the reciprocal of the height we (like Adam) would have calculated the naive way. So we can rewrite the speed equation in an even cleaner form,

$$\dot{y} = -v_T\sqrt{1 - e^{(y - h)/h_0}}$$

Now all that’s left in this equation is a couple of simple, physically meaningful constants, namely \(v_T\) the terminal velocity and \(h_0\) the “vacuum drop height” (or whatever you want to call it), and the variables we want to deal with: final height \(y\), initial height \(h\), and the speed \(\dot{y}\). When you can pull it off, this is the best way to express a physics equation because it’s really clear what each piece means, and that makes it easy to understand what the equation expresses about the world.

Anyway, back to the problem at hand. First, we want to find the speed that the actual gunner in the myth would have had after falling from his plane. Plugging in \(y = 0\) and \(h = \unit{22000}{\foot}\), we get

$$\dot{y} = -\unit{120}{\mileperhour}$$

As expected, it’s pretty much exactly terminal velocity. In fact, you can calculate just how tiny the deviation is by using a Taylor series:

$$v_T - \abs{\dot{y}} \approx \frac{v_T}{2}e^{\frac{y - h}{h_0}}$$

That exponent \(\frac{y - h}{h_0}\) is about \(-44\), so we get a deviation of about \(\snunit{1}{-18}{\mileperhour}\). No measuring device on Earth could detect that! But of course we don’t even know the terminal velocity of a person that precisely — it probably changes by more than that anyway when a fold of his jacket flips up or down.

Finally, though, let’s figure out how quickly the Mythbusters’ dummy would have been falling at ground level if he fell 500 feet (without a guide wire). Just plug in \(y = 0\) and \(h = \unit{500}{\foot}\) and you get

$$\dot{y} = -\unit{96.4}{\mileperhour}$$

That’s a good way off from the \(\unit{120}{\mileperhour}\) they were aiming for — about a 20% difference, actually! It just goes to show that you can’t always trust the simple calculations to be accurate approximations. But in this case, I really don’t think it matters. It seemed pretty clear from the actual test that the dummy, had it been a real person, wouldn’t have survived at any speed, with all the shrapnel flying around. And besides, I think that the presence of the shockwave would have actually made the impact worse, rather than cushioning the fall… but that’s perhaps a question for another day and another post.