Ellipsix Informatics: Blog

2010

Apr

Calculating terminal speed

Posted by David on April 25, 2010 — Comments

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.

2009

Dec

How the Mythbusters skipped a car

Posted by David on December 28, 2009 — Comments

On the last episode before breaking for Christmas, the Mythbusters build team undertook the slightly ambitious project of skipping a car across a pond, as shown in the movie Cannonball Run. At first this probably seems like a ridiculous thing to try — of course, on Mythbusters, what isn't? But this one actually worked. Here's a look at the rather interesting physics behind it.

As Jesse explained on the show, there are basically two physical principles that allow you to skip a stone (or a car) across water: the spin, and the reaction force of the water. This isn't buoyant force, like they've dealt with on previous shows; if buoyancy alone were the only thing pushing up on the stone, it'd float. Stones don't float. (Neither do cars.) The force that keeps a stone skipping across the water is related to its speed. Spin and speed, that's the magic formula.

First, the spin. Any spinning or rotating object has angular momentum, which is like a rotational equivalent of linear momentum: roughly speaking, it measures how difficult it is to change the object's motion. Objects with a lot of momentum are either very massive or moving very fast, or both, and in either case they're not going to change that motion easily. The corresponding formula for linear momentum is

$\vec{F} = \ud{\vec{p}}{t}$

showing that the rate of change of momentum is equal to the force, and for angular momentum,

$\vec{\tau} = \ud{\vec{L}}{t}$

showing that the rate of change of angular momentum is equal to the torque. Unless you have something pushing very hard on it, a rapidly spinning object is not going to change its spin by any significant amount in the short time involved in skipping across a pond. What's especially important here is that spin is a vector; it has a magnitude and a direction, and both of them are going to remain effectively constant. So in addition to continuing to spin at the same rate, a skipped stone will keep its spin axis pointing in the same direction, like a gyroscope. (Just kidding, try this) This is the purpose of spinning a stone when you skip it, so that it maintains the same orientation even as it gets jostled by the water. (It wasn't necessary to spin the car because its large mass keeps the water from flipping it over.)

The big question, of course, is why a stone (or car) is able to bounce off the surface of the water at all. As you might guess, this has to do with the speed. Water bouncing off the bottom of the stone exerts an upward force on it, and the faster the motion, the larger the force. Working out a formula for this force is, if not outright difficult, a little tricky. We need to shift our perspective and imagine ourselves moving along with the stone at some particular instant — basically, instead of thinking about the stone hitting water at a velocity $\vec{v}$ , we consider a molecule of water hitting the stone with the same speed but in the opposite direction, $-\vec{v}$ .

A flat surface striking water

Suppose the molecule bounces off the stone elastically; that is, it keeps the same kinetic energy it had before the collision. That means it's going to have the same speed. Only the direction of the velocity changes, because the component of velocity perpendicular to the surface gets reversed; it's just like light bouncing off a mirror, and just like light, the angle of incidence will equal the angle of reflection. From the diagram, we can work out the final velocity of the water molecule,

$-\vec{v}' = -v\cos(2\theta)\unitx - v\sin(2\theta)\unity$

and then the difference

$\vec{v} - \vec{v}' = v[1 - \cos(2\theta)]\unitx - v\sin(2\theta)\unity$

This quantity is important because when you multiply it by the mass of the water molecule, $\mu_w$ , you get the momentum change of the water,

$\Delta \vec{p}_w = \mu_w\bigl[v\bigl(1 - \cos(2\theta)\bigr)\unitx - v\sin(2\theta)\unity\bigr]$

which is precisely equal and opposite to the change in momentum of the surface caused by this collision,

$\Delta \vec{p} = -\mu_w\bigl[v\bigl(1 - \cos(2\theta)\bigr)\unitx - v\sin(2\theta)\unity\bigr]$

The next step is to figure out how many molecules are going to bounce off the surface per unit time. That's just the number density of molecules, $n$ , times the volume the surface passes through in a unit time $\Delta t$ , which is $\vec{A}\cdot\vec{v}\Delta t$ . The dot product takes the component of velocity parallel to the area's normal vector, and it just works out to $n A v \cos(\pi - \theta) \Delta t$ , or $n A v\sin\theta \Delta t$ . Multiplying this by the change in momentum per molecule, we get

$\Delta \vec{p} = -\mu_w n A v\sin\theta \Delta t\bigl[v\bigl(1 - \cos(2\theta)\bigr)\unitx - v\sin(2\theta)\unity\bigr]$

But notice that $\mu_w n = \rho$ , the mass density of the water. Dividing both sides by $\Delta t$ , we get the force,

$\vec{F} = \frac{\Delta \vec{p}}{\Delta t} = -\rho A v^2\sin\theta\bigl[\bigl(1 - \cos(2\theta)\bigr)\unitx - \sin(2\theta)\unity\bigr]$

Hmm, this factor $\rho A v^2$ looks familiar... we've just more or less rederived the equation for aerodynamic drag that pops up rather frequently in the Mythbusters' experiments.

There's one last thing to do with this expression before we can go on plugging in numbers. In the reference frame where the stone is at rest, we've been working in a rotated coordinate system, chosen so that the velocity $\vec{v}$ points directly along the $-\unitx$ axis. But we want to find the vertical force that holds the stone up — that's vertical in the "stationary" reference frame, where the stone is moving and the water is not. Of course, I could just multiply the force we got by the appropriate rotation matrix. But instead I can use a little trick: I'll express all the vectors in terms of the directions defined by the problem itself. What I mean is, instead of writing $\unitx$ , use $\unitvec{p}$ , and instead of $\unity$ , I'll invent a vector $\unitvec{\eta}$ which points perpendicular to $\unitvec{p}$ by definition. Unlike $\unitx$ and $\unity$ , these vectors won't arbitrarily change their orientation when we switch coordinate systems. I can write the force as

$\vec{F} = -\rho A v^2\sin\theta\bigl[\bigl(1 - \cos(2\theta)\bigr)\unitvec{p} - \sin(2\theta)\unitvec{\eta}\bigr]$

and from a drawing of the overall setup, I can easily tell that

$\unitvec{p} = \cos\phi\unitx - \sin\phi\unity$

and

$\unitvec{\eta} = \sin\phi\unitx + \cos\phi\unity$

Plugging in and simplifying,

$\vec{F} = -\rho A v^2\sin\theta\bigl[\bigl(1 - \cos(2\theta)\bigr)(\cos\phi\unitx - \sin\phi\unity) - \sin(2\theta)(\sin\phi\unitx + \cos\phi\unity)\bigr]$

which conveniently reduces down to

$\vec{F} = 2\rho A v^2\sin^2\theta[\cos(\theta - \phi)\unity - \sin(\theta - \phi)\unitx]$

If you look back at the diagram above, you'll notice that this vector describes a force which points perpendicular to the surface of the object being skipped — not to the surface of the water! I'll note that we could have arrived at the same conclusion using the coordinate-free form of the equation,

$\vec{F} = \rho A (\unitn\cdot\vec{v})^2\unitn$

where $\unitn$ is the unit normal to the surface. The derivation is more or less the same as above, if you make some effort to represent all the vectors in terms of $\vec{v}$ and $\unitn$ from the beginning.

Anyway, now that we have an equation, it's time to calculate some numbers. Unfortunately, this is going to be the difficult part. It's easy enough to estimate $v$ , that's just the speed of the car, and $\rho = \unit{1000}{\frac{\kilo\gram}{\meter^3}}$ is the density of water, but the other values depend on the exact shape and trajectory of the car, which are a completely separate issue from the calculation I'm talking about today. For our purposes, I'm just going to make some guesses based on what I remember seeing in the show.

First, let's consider the Mythbusters' attempt to replicate the circumstances of the movie: a (makeshift) luxury car flying off an 8 foot ramp at $\unit{50}{\mileperhour}$ . This car spun around in the air and hit the water nose-down, so the surface of interest is the front grill, which probably has an area around $\unit{1}{\meter^2}$ . We can get an approximate measurement of $\theta - \phi$ by looking at the video capture of the car (or perhaps by calculating it, but that's a topic for another day). Unfortunately, I don't have this video. So I'm just going to guess. At the point when the car hits the water, it's probably inclined by about $\unit{70}{\degree}$ . $\theta$ is a bit harder to determine, but it has to be pretty small, probably not more than $\unit{20}{\degree}$ . With these numbers, the force works out to $\unit{20000}{\newton}$ , or $\unit{4500}{\pound}$ . That's a lot of force! Enough to support the full weight of the car, in fact. But then again, as the car's speed drops, so does the force, which means that the drag force isn't going to last long enough to make our car float. It just means it slows down fast.

Now consider the second car jump, the one without the ramp where the car actually made it across the pond. This time the speed was higher, $v = \unit{100}{\mileperhour}$ , but more importantly, the car didn't spin around in midair, so it hit the water more or less flat, with a fairly large horizontal velocity. That's significant for two reasons: first of all, the surface of interest here is not the front of the car, but the bottom, and there is a lot more surface area on the bottom of the car that could potentially be contributing to the drag force. But more importantly, since the force acts perpendicular to the surface, in this case it's pointing almost directly upward — not antiparallel to the car's velocity! So instead of slowing the car down, this force pushes it back up out of the water, causing it to skip. It's not particularly easy to estimate the angles $\theta$ and $\phi$ , but knowing that they're both really small,

$\vec{F}\cdot\unity \sim 2(\unit{1000}{\frac{\kilo\gram}{\meter^3}})(\unit{1}{\meter^2})(\unit{100}{\mileperhour})^2\sin^2\theta\cos(\theta - \phi) \approx \snunit{2}{6}{\newton}\sin^2\theta$

That amplitude is about $\unit{450000}{\pound}$ , so even with $\theta$ as small as about $\unit{4}{\degree}$ , it'll be enough force to completely counteract the weight of the car and bounce it back up into the air.

2009

Oct

Dirty vs. Clean Car

Posted by David on October 22, 2009 — Comments

Hot on the heels of their Bullet Fired vs. Bullet Dropped episode, the Mythbusters have another result that's poised to shake up the world of science... well, maybe not. But this week's main myth, Dirty vs. Clean Car, is the kind of neat idea that most of us would never think to test and yet turns out to be surprisingly close to practicality. The myth that Adam and Jamie are testing is that dirt on a car has the same kind of effect as golf ball dimples, increasing the fuel efficiency of the car. To sum up the results (SPOILER ALERT ;-), it doesn't work, at least not with dirt — but putting an actual dimpled coating on a car does increase the fuel efficiency by 11%. (Only on Mythbusters would they dimple a car...)

As with a lot of recent myths, this one deals with fluid dynamics — but not just the simple stuff like drag force, as in the bullet myths. The golf ball effect is based on turbulence, specifically the idea that the rough surface of the ball induces turbulence which disrupts the wake (pocket of still air) that trails behind the ball. That pocket of still air takes energy to travel along with the ball; specifically, the whole combination of ball and wake has a kinetic energy

$K = \frac{1}{2}(m_\text{ball} + m_\text{wake})v^2$

The amount of energy provided by the golf club is, on average, pretty much constant. So if you disrupt the wake, there's less still air and $m_\text{wake}$ goes down, which means that $v^2$ can go up. The ball moves faster and travels further.

Turbulent systems are notoriously difficult to analyze in any detail. So I'm not even going to try to reproduce the result from the show with a calculation or simulation. But I do have a couple of points to pick on:

First of all, when measuring fuel efficiency, Adam and Jamie only ran 5 trials for each configuration. Sure, it takes time and effort to run the car down their 1-mile track, so there's a practical limit on how many times you can do that, but the fact remains that 5 is not a very large sample size. With so few trials, is the improvement they observed from the dimples (11%) really significant, in a statistical sense?

To figure that out, we'd like to calculate the "standard error of the mean" for the data the Mythbusters collected. Standard error of the mean, denoted $\sigma_{\bar{x}}$ , is basically a measure of how precise your average is; there's a 68% chance that the actual value is between $\bar{x} - \sigma_{\bar{x}}$ and $\bar{x} + \sigma_{\bar{x}}$ . The smaller the standard error of the mean, the more precise your measurement. If you assume that your individual measurements are fairly reliable (which you could argue about in this case, but I won't), it can be calculated from the formula

$\sigma_{\bar{x}} = \frac{1}{N}\sqrt{\sum_i (x_i - \bar{x})^2}$

Just one problem, though: we don't have the data! So I'm going to make a guess based on the second test, with the fully clean car, for which Adam reported that half the trials yielded a value of $\addunit{\inch}{in}\unit{3\frac{1}{2}}{\inch}$ and the other half yielded a value of $\addunit{\inch}{in}\unit{3\frac{5}{8}}{\inch}$ . If they ran 4 tests, $\addunit{\inch}{in}\sigma_{\bar{x}} = \unit{0.031}{\inch}$ , corresponding to a relative error $\sigma_{\bar{x}}/\bar{x}$ of 0.9%. And if the relative error for the clay-covered car was on the same order, about 1%, that's much smaller than the 11% improvement they noticed. So yeah, it's definitely statistically significant. (Now I feel kind of silly for going through all that work)

Here's my other picking point (this one positive): Adam makes a good observation in the show about why the fuel efficiency doesn't change when they add on 800 pounds of clay. As he explained, this is due to the clever way they designed their test; they don't count the fuel used to accelerate the car up to 65 mph, only the fuel required to maintain that speed along the mile-long track. On the show they only said that the test didn't simulate real-world driving conditions, so the mass didn't have an effect, but here's the quantitative explanation of why you can say that.

As the car moves down the track, it's subject to the force of the engine (of course), a drag force exerted by the air, and a small amount of rolling friction exerted by the road on the tires. Using Newton's second law,

$F_\text{engine} - F_\text{drag} - F_\text{friction} = ma$

Now, fuel economy is measured in miles per gallon, but each gallon of fuel corresponds to a roughly constant amount of energy. So the reciprocal of fuel economy would be roughly proportional to propulsive energy per unit distance:

$\frac{1}{\text{MPG}} \sim \frac{W}{d} = F_\text{engine}$

Putting these last two equations together,

$\text{MPG} \sim \frac{1}{F_\text{drag} + F_\text{friction} + ma}$

The drag force doesn't depend on the car's mass. The frictional force? It probably does depend on mass, but it's so small that we can basically ignore it (that is, after all, why humans invented wheels in the first place). So the only dependence on mass that's left is the $ma$ term. If the car isn't accelerating, that goes away. By running their tests at constant velocity, the Mythbusters managed to basically remove any effect that the car's mass would have on the fuel economy they measured.

Unfortunately, one thing you may notice about that last equation is that it predicts that when acceleration is not zero, it drives the fuel economy down. People tend to do a lot of accelerating (in the physics sense, which includes braking) in their cars, and I have a feeling that's going to be a much larger effect than anything that could be gained by putting dimples on new car models. But hey, like Jamie said, maybe we'll see it on NASCAR someday...

2009

Jun

Curving bullets

Posted by David on June 11, 2009 — Comments

This week the Mythbusters tackled the question of whether you can make a bullet follow a curved flight path, as in the movie Wanted. The characters in the movie are able to do this using some fancy flick of the wrist as they fire the gun, but is it really possible? Apparently a lot of people were wondering.

The short, simple answer is no. It's an obvious application of Newton's first law of motion: objects moving in a straight line will continue moving in a straight line at constant speed, unless subject to an external force. But there are only a couple of external forces that can act on a bullet: air resistance and gravity. Gravity certainly isn't going to make the bullet curve sideways as we see in the movie, and all air resistance will do is slow it down, not change its direction.

Then again, Kari, Grant, and Tory hit on an important point: bullets are highly symmetric and are typically ejected from the gun barrel with high spin. All this is optimized for motion in a straight line toward whatever you're aiming the gun at. What happens if you use asymmetric, oddly shaped bullets?

In order to follow a curved trajectory, the bullet would need to experience a force perpendicular to its motion. Air resistance is normally antiparallel to the motion, but in principle, say by shaping one side of the bullet into a "ramp" for the oncoming air, you could redirect some of that force in a perpendicular direction to make the bullet curve. The thing is, this effect is pretty small, and any force coming from air resistance would only be able to make a microscopic nudge over the time it takes a bullet to traverse, say, a gun range. Specifically, drag force can be calculated as

$F_D=\frac{1}{2}\rho v^2 A C_D$

where $\rho$ is the density of the fluid, $v$ is the bullet's velocity, $A$ is the cross-sectional area of the bullet, and $C_D$ is a unitless constant on the order of 1. If we could somehow direct all of this drag force into curving the bullet, then the centripetal force would be equal to the drag force,

$\frac{1}{2}\rho v^2 A C_D = \frac{mv^2}{r}$

which means that the bullet would have a radius of curvature of

$r = \frac{2m}{\rho A C_D}$

The radius of curvature is a standard way to quantify curvature in physics; basically, it's what you would get if you took a part of a curved path, extended it into a circle, and measured the radius of that circle. Putting it some reasonable numbers — a bullet mass of about $\unit{5}{\gram}$ , an area of about $\unit{1}{\centi\meter\squared}$ , and the density of air as $\unit{1.3}{\kilo\gram\per\meter\cubed}$ — this comes out to about $\unit{77}{\meter}$ . So the best curve that would be ideally possible is a circle of radius $\unit{77}{\meter}$ , or $\unit{252}{\foot}$ . That's pretty big, but you could probably still notice it over the span of a gun range, if you had a laser pointer as the Mythbusters did. Of course, the best curve you could really get in practice would be far, far straighter, since you couldn't possibly hope to harness all the drag force to curve the bullet.