2012

Apr

Square wheels

Posted by David on April 14, 2012 — Comments

It hasn't escaped my notice that Mythbusters is back with a new season! Actually, it's not really that new, since we're now three (well, now four) weeks in, but I missed the first two episodes since I was out of the country. But it works out because this (actually last) week's myth is full of interesting physics to analyze!

This past Sunday, Adam and Jamie tested the myth that if you're driving fast enough, square wheels can actually provide a surprisingly smooth ride. At first, the idea of square wheels working at all, much less actually being smooth, can seem a little wacky, but with a bit of physical intuition, it's not hard to convince yourself that it's actually pretty plausible. As they explained in the show, the reason a square wheel is expected to bounce you up and down is that the distance from the axle to the bottom of the wheel changes as it turns. If you're going slowly, every time the wheel tips over another corner, it's going to fall down until its side is resting against the ground, taking you with it. But if you speed up enough, the wheel won't have time to fall very far before it rotates through a quarter turn, and the next corner gets under it to hold it up.

Simple model: a slowly turning wheel

You can actually calculate about how fast you would need to go to do this. Let's consider just a single square wheel, and at first, suppose it's going really slowly. That way, the wheel is going to constantly stay in contact with the ground. There are basically two "phases" in the cycle of a slowly turning square wheel:

Starting from a position where the wheel is "perched" on its corner with the corner pointing down, it's going to first just pivot forward around that corner, and fall down on its side.
After that, it'll pivot up around the next corner, until that next corner is now pointing down.

Diagram of rotating square wheel

Suppose the wheel has a side length of $2r$ and is rotating at angular speed $\omega$ , which I'm going to assume is constant for simplicity. Based on geometry, the height of the first corner relative to the wheel's center is

$y_1(t) = -\sqrt{2}r\cos(\omega t)$

and since the second corner trails by an angle of $\frac{pi}{2}$ (a quarter circle), its height relative to the center is going to be

$y_2(t) = -\sqrt{2}r\cos\biggl(\omega t - \frac{\pi}{2}\biggr)$

At any given time, the height of the lowest point on the wheel relative to its center will be the lesser of these two expressions: $y_1$ for the first eighth turn, and $y_2$ for the next eighth. But what we really want is the height of the center of the wheel above the ground, which will be the negative of that minimum:

$y_\text{slow}(t) = \begin{cases}\sqrt{2}r\cos(\omega t) & 0 \le t < \frac{\pi}{4\omega} \\ \sqrt{2}r\cos\biggl(\omega t - \dfrac{\pi}{2}\biggr) & \frac{\pi}{4\omega} \le t < \frac{\pi}{2\omega}\end{cases}$

This function tells us the height of the truck as a function of time as one quarter turn of the wheel elapses. It looks like this:

Height function for slow wheel

From this, we should be able to figure out how bumpy the ride on this slowly turning wheel would be. But that brings up another question: how exactly do you measure bumpiness?

Think back to the last time you were on a car driving on a road with a lot of potholes, or any other rough surface. What makes it unpleasant is that you get shaken up and down a lot. The larger the vibrations, the rougher the ride. So it makes sense to say that our measure of bumpiness should be related to the distance by which the car bounces up and down in a cycle — in other words, the maximum height minus the minimum height, which is often called peak-to-peak amplitude.

But if you think about it, the time scale over which these oscillations occur is also important. When you drive up and down a mountain, that's a huge bounce, but it doesn't feel like it because it's so slow. So the "bumpiness metric" should also be anti-correlated to the cycle time: quicker bumps at the same amplitude have more of an effect. Accordingly, I'm going to define a simple measure of bumpiness as the ratio of the peak-to-peak amplitude to the period for one up-and-down cycle of oscillation (which is actually a quarter cycle of the wheel). I'm sure there are more complicated (and more realistic) ways to define bumpiness, but this one should be good enough to make my point here.

For the model of a slowly rotating square wheel, we can find the peak-to-peak amplitude using the maximum value of $y(t)$ , which occurs at $t = 0$ (and again at $t = \pi/2\omega$ ), and the minimum value, which occurs at $t = \pi/4\omega$ .

$B = \frac{\sqrt{2}r\cos(0) - \sqrt{2}r\cos\bigl(\omega\times\frac{\pi}{4\omega}\bigr)}{\frac{\pi}{2\omega}} = \frac{2}{\pi}(\sqrt{2} - 1)r\omega$

Speeding it up

With the simple slow wheel model out of the way, let's see what happens when you speed the wheel's rotation up. The most important change comes from the fact that there is nothing actually holding the wheel's surface to the ground. In the first part of the cycle, the only thing pulling the wheel down is gravity, and gravity can't accelerate it any faster than $\SI{9.8}{m/s^2}$ . So if our slow-wheel model says that the wheel should be moving downward at faster than $g = \SI{9.8}{m/s^2}$ , that is if $y''(t) < -g$ , then we've got a problem.

Of course, it's not hard to figure out when this actually does happen, using basic Newtonian mechanics. There are two relevant forces acting on the wheel, gravity and the normal force from the ground. Their relationship is given by Newton's second law, $\sum F = ma$ , or in this case:

$-mg + F_N = my'' = -\sqrt{2}mr\omega^2\cos(\omega t)$

For a slowly rotating wheel, $\omega$ is small, and thus $F_N$ will need to be positive to make this equation true. Once $\omega$ gets large enough that $-mg > -\sqrt{2}mr\omega^2\cos(\omega t)$ , though, there will be no zero or positive value of $F_N$ that can make the equation true. That's when the wheel is going to leave the ground. This will happen at $t = 0$ as long as

$\omega^2 > \frac{g}{\sqrt{2}r}$

If the normal force is going to be zero, the wheel won't be touching the ground as it rotates. Instead, it's going to be in free fall for some amount of time. That's an easy situation to analyze; the height of an object in free fall is just $y = y_0 + v_0 (t - t_0) - g(t - t_0)^2/2$ , and since in this case the free-fall phase starts at time 0 with zero vertical velocity, that just reduces to

$y_\text{free}(t) = \sqrt{2}r - \frac{gt^2}{2}$

The wheel will remain in free fall as long as this height $y_\text{free}$ is greater than the height difference between the center of the wheel and its lowest point. The latter quantity is something we've already calculated: it's $y_\text{slow}$ . So we need to identify the first nonzero time at which $y_\text{free}(t) = y_\text{slow}(t)$ , the solution to

$\sqrt{2}r - \frac{gt^2}{2} = \begin{cases}\sqrt{2}r\cos(\omega t) & 0 \le t < \frac{\pi}{4\omega} \\ \sqrt{2}r\cos\biggl(\omega t - \dfrac{\pi}{2}\biggr) & \frac{\pi}{4\omega} \le t < \frac{\pi}{2\omega}\end{cases}$

This is a little tricky for a couple of reasons: first, it's a transcendental equation, because it involves both a polynomial in $t$ and a trigonometric function of $t$ . That means you can't write the solution as a symbolic function. You can still solve it numerically, though, and that's what I'm going to do shortly. The other issue is that it has a piecewise function on the right. That's not that hard to deal with, at least not if you have numbers for everything; you can just solve the first case, and see if the solution you get satisfies the condition for that case ( $0 \le t < \frac{\pi}{4\omega}$ ); if not, then the solution comes from the second piece. It turns out that in our situation, because of the requirement $\omega^2 > g/\sqrt{2}r$ , the solution is almost always going to come from the second case; for most values of $\omega$ the first corner of the wheel never touches the ground again after the very beginning of the cycle, and the small region of $\omega$ where that's not the case is basically negligible. (Plus it would take a whole other post to do that analysis properly) So we can reduce this last equation to

$\sqrt{2}r - \frac{gt^2}{2} = \sqrt{2}r\cos\biggl(\omega t - \dfrac{\pi}{2}\biggr)$

The solution to this tells us when the second corner of the wheel is going to hit the ground. Call that solution $t_2$ . Then the height as a function of time for a square wheel which does not have to be moving slowly is

$y_(t) = \begin{cases}\sqrt{2}r\cos(\omega t) & 0 \le t < t_2 \\ \sqrt{2}r\cos\biggl(\omega t - \dfrac{\pi}{2}\biggr) & t_2 \le t < \frac{\pi}{2\omega}\end{cases}$

This function looks like this:

Height function for fast wheel

Notice the difference between this and the equivalent graph for a slow wheel (which is included in the background, for comparison). The peak-to-peak amplitude of this motion is considerably less. This will also be reflected in the formula for the bumpiness, which is

$B = \frac{\sqrt{2}r\cos(0) - \sqrt{2}r\cos\bigl(\omega t_2 - \frac{\pi}{2}\bigr)}{\frac{\pi}{2\omega}} = \frac{2}{\pi}\bigl(1 - \sin(\omega t_2)\bigr)r\omega$

Putting it all together

To recap, our measure of bumpiness over all possible rotational frequencies is given by the following piecewise function:

$B = \begin{cases}\frac{2}{\pi}(\sqrt{2} - 1)r\omega & \omega^2 \le \frac{g}{\sqrt{2}r} \\ \frac{2}{\pi}\bigl(1 - \sin(\omega t_2)\bigr)r\omega & \omega^2 > \frac{g}{\sqrt{2}r}\end{cases}$

For a square wheel with a $\SI{50}{cm}$ side length, a plot of this function looks like this:

Bumpiness plot

The horizontal axis shows speed in meters per second. There's a peak in the graph around $\SI{6.2}{m/s}$ , or $\SI{14}{mph}$ , and after that it starts going down — which means that for this size of wheel, once you pass 14 miles per hour, the ride actually should start getting smoother! And that's pretty close to what Jamie and Adam actually observed in the show.

2011

Oct

Bird Balance

Posted by David on October 11, 2011 — Comments

A lot of the discussion about the physics in the latest Mythbusters episode focused on the giant Newton's cradle. But what about the other myth? While Adam and Jamie were playing around with their giant balls, Kari, Tory and Grant balanced a car on a cliff in order to test whether a bird landing on the hood would be able to send it falling over the edge. Unsurprisingly, it turns out there's some interesting physics to be found there as well.

Potential Energy and Equilibrium

First of all, what does it take to balance a car on a cliff? The easiest way to think about this physically is in terms of gravitational potential energy. Let's say that the top of the cliff, at the point where the car is resting on it, is at height zero. Then the potential energy of the car is $U = mgh_\text{cm}$ , where $h_\text{cm}$ is the height of the car's center of mass. As a rough rule of thumb, the car "seeks out" the configuration of lowest potential energy (there are a bunch of caveats to that statement but it's good enough for now), which means its center of mass is going to fall as low as possible.

Diagram of car teetering on cliff For this car, there are really only two ways it can move: it can rock forward or backward. If the center of mass is behind the edge of the cliff, then rocking forward increases the potential energy, whereas rocking backward decreases it, so the car will tend to tilt backward, keeping it on the cliff. On the other hand, if the center of mass hangs over the edge of the cliff, it's rocking forward that will decrease the potential energy, so the car will fall off. Mathematically, we can parametrize the position of the center of mass as $\vec{r}_\text{cm} = h_\text{cm}(\unitx\sin\phi + \unity\cos\phi)$ and find the torque acting on the car as

$\tau = -\pd{U}{\phi} = -\pd{}{\phi}[mgh_\text{cm}\cos\phi] = mgh_\text{cm}\sin\phi$

(This is an example of a generalized force, for the curious.) The fact that there is no negative sign in this expression means that we have an unstable equilibrium: the car will be balanced at some particular angle ( $\phi = 0$ ), but as soon as it tips the slightest bit in one direction or the other, the torque is going to push it further and further in that direction. If you look at the graph of potential energy as a function of the angle $\phi$ , you'll see that the balancing point at $\phi = 0$ corresponds to a local maximum of the potential energy, which is the signature of an unstable equilibrium.

Graph of potential energy as a function of angle for unstable equilibrium

Diagram of pencil balancing on tip The thing is, if you have an object in true unstable equilibrium, it's impossible to actually balance it. Even the tiniest push in one direction or another, like a breeze or a light touch, or even the random thermal motions of the molecules, will be enough to displace the object and get it falling in one direction or another — for an example, try balancing a pencil on its tip. To get an object to actually balance, you need a stable equilibrium, in which pushing the object in one direction or another produces a torque that tends to push it back. This corresponds to a local minimum in the potential energy, as in this graph:

Graph of potential energy as a function of angle for stable equilibrium

Of course, the potential energy for real-world objects usually isn't just a nice parabola (or cosine curve); in general, it'll have some combination of maxima and minima. In that case, you can actually have an equilibrium which is stable for small displacements, but unstable for larger displacements. Diagram of pencil balancing on eraser For example, consider balancing a pencil on its eraser. (One fresh out of the package, so that the eraser is flat.) If this pencil tips in either direction, its center of mass is going to rise a little bit.

$U(\phi) = \begin{cases}mg(h_\text{cm}\cos\phi + x_R\sin\phi),&\phi\ge 0 \\ mg(h_\text{cm}\cos\phi - x_L\sin\phi),&\phi\le 0\end{cases}$

where $x_R$ is the distance along the base of the eraser from the center of mass to the right pivot point (in other words, when the pencil is perfectly upright, $x_R$ is the horizontal distance between the center of mass and the right pivot point), and $x_L$ is the same for the left pivot point. The potential energy graph in that case looks like this:

Graph of potential energy as a function of angle for locally stable equilibrium

This is basically what's going on with the Mythbusters' car. Even though being balanced on the edge of a cliff is an unstable situation for large pushes, there's a little "dip" in the potential energy function which makes it stable for very small displacements. If the car tilts a little bit to the left or right, it will still tend to return to its original position, as long as it doesn't tilt far enough to escape the dip. This dip could be caused by, say, the bottom of the car getting dented when they were trying to get it into position.

External Forces

Now that we know what it takes to balance a car, let's figure out what it takes — in physics terms — to knock one over.

When a bird lands on the hood of the car, it has two effects:

It shifts the center of mass of the car-bird system by a small distance toward the bird
Any momentum the bird had when it landed is transferred to the car

Both of these can contribute to moving the car out of its stable equilibrium.

Shifting the Center of Mass

The new center of mass can be computed as a weighted average of the centers of mass of the car and bird individually.

$\vec{r}_\text{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$

For this situation, if you treat the horizontal location of the car's original center of mass as $x = 0$ , you can calculate that the new center of mass of the bird-car system shifts by a horizontal distance

$\Delta x_\text{cm} = \frac{m R}{M + m}$

where $m$ is the mass of the bird, $M$ that of the car, and $R$ the distance between the bird's perch and the car's original center of mass.

Why is this important? If you look at the equation for potential energy a few paragraphs back, you'll notice that $U(\phi)$ actually depends on the position of the center of mass, through the variables $x_L$ and $x_R$ . So when the bird shifts the center of mass, say, to the right, it changes $x_L$ and $x_R$ :

$x_L \to x_L + \Delta x_\text{cm} \qquad x_R \to x_R - \Delta x_\text{cm}$

which in turn changes the potential energy function:

Diagrams showing change in potential energy function with shifting center of mass

As the graphs show, this shift doesn't change the height of the equilibrium position at $\phi = 0$ , at least not at first. But it does reduce the height of the "hump" on the right side, which is going to make it easier for the car to fall. It also moves the position of the hump closer to the current tilt of the car, $\phi = 0$ , which means the car doesn't have to tilt as far before it escapes its stable equilibrium. As an extreme case, if the center of mass shifts all the way over to the right pivot point, we'll have $x_R = 0$ , the hump will disappear, and the car will fall right away. You can confirm this with a calculation of the position of the hump, represented by $\phi_c$ (the "critical" angle at which the car will fall), which lies at the local maximum of the potential energy:

$\evalat{\ud{U}{\phi}}{\phi_c \ge 0}{} = -mg(h_\text{cm}\sin\phi_c - x_R\cos\phi_c) = 0$

$\phi_c = \atan\frac{x_R}{h_\text{cm}}$

So when $x_R = 0$ , the critical angle becomes zero. Since the car is already sitting at $\phi = 0$ , it'll be tilted enough to fall, without any wobbling. That's what happened at the end of the episode, for example, when Kari filled up the car's hood with frozen hens.

Momentum Transfer

Even if you don't shift the center of mass all the way past the pivot point, it's still possible to make the car tip. As the Mythbusters pointed out in the later part of the episode, the momentum transferred from the bird as it lands might be enough to push the car over the energy "hump," even if it wouldn't tip based on the position of the center of mass alone.

To figure out what it takes to make this happen, we need to convert the momentum of the bird to the amount of energy it transfers to the car. The bird's landing is basically an inelastic collision in which one of the objects rotates around a fixed pivot, and in cases like this, you use conservation of angular momentum. Coming in, let's say the bird has a landing speed $v$ ; then its angular momentum around the pivot point will be $L \approx mvR$ . Once the bird-car system comes together and acquires this angular momentum, it will have a rotational kinetic energy of

$K_r = \frac{L^2}{2I}$

If the bird lands on the very edge of the hood, that's

$K_r = \frac{(mvR)^2}{2\beta (m + M) R^2} = \frac{m}{\beta (m + M)}\times\frac{1}{2}mv^2$

where $\beta$ is the coefficient that appears in the formula for moment of inertia ( $I = \beta mr^2$ ). It depends on the detailed shape of the car, but since a car is roughly the shape of a rod (which has $\beta = \frac{1}{12}$ ), we can use $\beta \approx 0.1$ as an order-of-magnitude estimate.

When the bird adds this extra energy to the car, it's not going to stay in place at $\phi = 0$ anymore, because the car-bird system now has more energy than just the potential energy associated with that position. The car is going to tilt downwards, and as it does so it will progressively convert that kinetic energy $K_r$ to more potential energy — remember, its center of gravity is rising as it tilts (at first). If it has enough energy, it will eventually reach the critical angle $\phi_c$ , and then the car topples down the cliff; if it doesn't have enough energy, it will stop at some point before that and start rocking back in the other direction.

Graphically, you can see this by comparing the total energy to the height of the hump:

Graph of potential with an example of total energy also plotted

The points at which the total energy line intersects the potential energy line are the points at which the car runs out of energy and rocks back in the other direction — appropriately, they're called turning points. If the total energy is lower than the hump, the car will oscillate (rock) back and forth between the two turning points. but if the total energy is higher than the hump, there will be no intersection (no turning point) on the right and the car just keeps going in that direction.

Mathematically, the amount of kinetic energy needed to reach the critical angle is

$K_{r,c} = U(\phi_c) - U(0) = (m + M)g\biggl(h_\text{cm}\cos\atan\frac{x_R}{h_\text{cm}} + x_R\sin\atan\frac{x_R}{h_\text{cm}}\biggr) - (m + M)gh_\text{cm} = (m + M)g\Bigl(\sqrt{x_R^2 + h_\text{cm}^2} - h_\text{cm}\Bigr)$

Setting this equal to $K_r$ from above shows that the bird needs to land with a speed of at least

$v = \sqrt{2\beta\biggl(\frac{M}{m} + 1\biggr)^2g\Bigl(\sqrt{x_R^2 + h_\text{cm}^2} - h_\text{cm}\Bigr)}$

to tip the car.

Estimating Results

OK, so now that all that math is out of the way, let's plug in some numbers and see whether these formulas make any sense! Remember that there are two different ways for the car to tip: either its center of mass shifts past the pivot point, or it gets enough kinetic energy from the bird to tilt it past the critical angle.

First, let's check the center of mass shift. In order to calculate that, we need to know the masses of the bird and the car, as well as the distance from the pivot point at which the bird lands. Fortunately, they told us in the episode that the car had a mass of 3000 pounds, so $M \approx \SI{1400}{kg}$ , and the heaviest bird they tested was 20 pounds, so $m = \SI{9}{kg}$ . The distance between the landing point and the forward pivot point is a little fuzzier, but we do know that it was roughly 6 and a half feet, or $\SI{2}{m}$ , from the front of the car to the point where the Mythbusters eventually managed to balance it. In the best-case scenario that the bird lands right on the front of the car, that's exactly the distance we need. Plugging the values in gives

$\Delta x_\text{cm} = \SI{13}{mm}$

Is that enough of a shift to make the car tip? Well, obviously we know from the show that it wasn't, but in order to calculate an answer, we would have to compare that distance to the size of the dip in the car's potential energy function — or more precisely, the horizontal distance that the car's original center of mass would have to move before it passes the cliff-side pivot point. This is something we don't have any direct information about.

But let's think about the result another way: in order to allow the car to be tipped by a bird, the Mythbusters would have had to get the center of mass within $\SI{13}{mm}$ of the cliff edge. Does it seem reasonable that they could have managed to do that? I don't think so. They were having trouble positioning the car any more precisely than to within a few inches, and $\SI{13}{mm}$ is a small fraction of that, so they would have had to have gotten rather lucky to get the positioning exactly right.

What we can do is calculate how close they probably did get it, based on the amount of mass they actually had to put on at the end. Kari dropped 80 frozen hens, each weighing about a pound and a half, plus a mechanical turkey, on the hood of the car, so all we have to do is replace the mass of the bird with the mass of the frozen hens (120 pounds) plus that of the turkey (20 pounds). The total comes out to $\SI{63}{kg}$ . Then again, not all the birds were all the way down at the end of the hood, so their "effective mass" would be a bit less than that; I'll estimate $\SI{55}{kg}$ . With this mass, we get almost 6 times as large of a shift:

$\Delta x_\text{cm} = \SI{77}{mm}$

This means the Mythbusters managed to find the position of their car's center of mass to within about 3 inches, less than the width of an average person's hand. Considering the equipment they had to work with, that's not bad.

What about the extra kinetic energy from a bird landing on the hood? Conveniently enough, we have a straightforward formula for this one, which tells us how fast the bird would have to have been traveling when it landed. The numbers that go into it are easy enough to estimate: for the moment of inertia coefficient, $\beta \approx 0.1$ ; we already have $M \approx \SI{1400}{kg}$ and $m = \SI{9}{kg}$ from the show; and gravitational acceleration is of course $g = \SI{9.8}{m s^{-2}}$ . The value we just calculated for how precisely they position the car, namely $\SI{77}{mm}$ , can go in for $x_R$ . The only thing we're left with is $h_\text{cm}$ , the height of the car's center of mass above its pivot points. We don't actually know this, but it has to be a little less than half the height of the car, which means $\SI{40}{cm}$ should be a reasonable estimate. Plugging in those numbers, I get

$v = \SI{18}{m s^{-1}}$

That's pretty fast — over 40 miles per hour! Many birds can fly at that speed or faster (including the condor, which goes up to 55 miles an hour), but no bird would actually land on a hard surface going that fast. Given that, it's clear that the pigeons and hawks and owls never stood a chance. Far from being able to be toppled by a breeze, as Kari speculated during the show, this model suggests that you pretty much would have had to fire a cannonball into the hood for that car to go over the cliff.

2011

Oct

Scattering a bullet off an RPG

Posted by David on October 05, 2011 — Comments

Yep, that's right, a rocket-propelled grenade finally made its way on Mythbusters! Personally I'm surprised it took them so long...

Anyway, let's not get distracted from the science by cool explosions just yet. On last week's season premiere of Mythbusters, Kari, Tory, and Grant tested a myth based on a scene in RED in which two characters are facing off, one (the hero) with a revolver and the other (the villain) with an RPG launcher. In the movie, they both shoot at the same time, the bullet hits the RPG in midair and detonates it, and the resulting explosion kills the villain. Now, in the show, this myth was busted on several counts:

RPGs don't even arm until about 60 feet after launch, about $\frac{3}{4}$ of the way to the target and long after this one would have been hit by the bullet
When an RPG explodes, it sends balls of molten copper flying forward, which wouldn't have been stopped by the bullet
The distance at which the explosion would have taken place, 16 feet from the villain, is quite survivable

Unfortunately, they didn't test what I thought was the most obvious objection to the scene (albeit the one that gives it its "cool factor"): you'd need excellent aim, and some amount of luck, to reliably hit as small a target as an RPG detonator head on with a bullet. Or so it seems. But is this really the case? Even if the RPG was armed, and even if it somehow could lose all its forward momentum and explode in a giant fireball, how likely is it that the collision would actually take place?

head-on view of bullets approaching an RPG As it turns out, shooting two objects at each other and characterizing what happens is a very common thing to do in physics — for instance, this is exactly what happens at particle accelerators like the recently departed Tevatron or the LHC — so the methods used to analyze this sort of situation are quite well studied, especially in a simple case like this one. Imagine the collision from the point of view of the RPG, which "sees" a bullet coming at it. Actually, imagine doing this experiment many times to get a statistical sample, so the RPG actually sees a whole beam of bullets coming at it, as shown on the left. The detonator cap occupies a circle (the solid red one) somewhere in the middle of the beam, and any bullet that hits that circle will trigger an explosion. It's probably safe to say that even a glancing blow will be enough to set the RPG off, so we should expand that circle a little, specifically by half the radius of a bullet, so that any bullets whose centers are within the circle will cause explosions — that's the dashed circle in the diagram. The area of this circle is called the scattering cross section, $\sigma$ .

side view of bullets approaching an RPG In order to figure out the probability of a collision, we need to find the fraction of bullets that will hit the aforementioned circle. The transverse positions of the bullets relative to the RPG form some distribution, called the beam profile $\rho(\mathbf{r}_\perp)$ , so we can figure out the probability by integrating the beam profile over the circle to get the number of bullets (out of this large statistical sample) that will hit.

$P = \iint_\text{circle} \rho(\mathbf{r}_\perp)\uddc\mathbf{r}_\perp$

So if we can get the normalized beam profile $\rho(\mathbf{r}_\perp)$ of a stream of bullets, we'd be all set. This is the tricky part, though, since doing so would require looking up some statistics on the accuracy of handgun aiming, and I'm not sure where to get reliable information about that. So in lieu of that, I turn to the internet. Here's a thread about accuracy at The Firing Line forums which includes some photos of shots made at a specified range. I can extract the coordinates of the bullet holes, compare them to the given scale, and thereby find the beam profile of the person who shot them. Sure, it's not a statistically significant sample, but I'm just going for a rough estimate anyway.

Since this isn't a rigorous experiment, I'm just going to work from this one image.

In order to easily determine the coordinates of the bullet holes, I turned to the first tool I had at hand, Mathematica. This little bit of code (for Mathematica 8) will let you click on each hole (or any point) and print out a list of the coordinates you clicked:

DynamicModule[{p = {}}, 
 EventHandler[
   Dynamic@Show[i, Graphics[{PointSize[Large], Red, Point[p]}]], 
   "MouseClicked" :> {AppendTo[p, 
      MousePosition["GraphicsAbsolute"]]}]
  Button["Clear", {p = {}}]
  Button["Print", Print[p]]
 ]

With this I got the following list of positions:

{{280.5,498.},{236.5,447.},{290.5,407.},{335.5,330.},{391.5,449.},{405.5,471.},
 {516.5,409.},{439.5,292.},{299.5,264.},{373.5,227.},{413.5,198.},{453.5,150.},
 {505.5,256.},{554.5,268.},{531.5,132.}}

I also determined that the distance corresponding to the 5.5 inch line marked at the bottom is 316 pixels, which I can use to convert the pixel positions into centimeters. After that, we have to make a distance correction: the picture shows shots taken at a distance of 150 feet, but the scenario in the movie involved a shot of 64 feet, so (as a simple approximation) we just insert a factor of $\frac{64}{150}$ .

This code graphs the fraction of bullet holes as a function of distance from the average bullet position, rescaled to a 64 foot shot:

Histogram[64./150. (EuclideanDistance[Mean[points], #] & /@ points), 10, "CDF"]

Here's a fancier version:

This graph is an approximation to what I would call the cumulative radial beam profile, which is related to the actual beam profile as

$\rho_\text{cuml}(R) = \int_0^R\int_0^{2\pi}\rho(\mathbf{r}_\perp)r_\perp\udc r_\perp\udc\phi$

Because of the small number of shots I've analyzed, it's hard to tell from the graph exactly what the form of $\rho_\text{cuml}(R)$ is — it could be linear, or quadratic (as you'd expect for a uniform distribution over a circle), or perhaps Gaussian (although I would expect more clustering in the center if that were the case), or something else entirely.

Thankfully, we don't really need to know the form, at least not if we just want a rough estimate of the probability. Just set $R$ to the radius of the scattering cross-section of the detonator cap, and $\rho_\text{cuml}(R)$ is the probability of hitting it. From what I saw in the show, I'd estimate $R \approx \SI{2.5}{cm}$ , which puts the probability at $P \approx 0.5$ . That's actually not a bad chance — not one I'd want to stake my life on, but a lot better than I would have guessed. And according to this article on handgun accuracy from Guns & Ammo magazine, it seems like that's average for a skilled casual shooter. I bet a trained CIA agent, especially one worthy of being played in a movie by John Malkovich, would have no problem making the shot.

2011

Jun

Spinning bullets

Posted by David on June 08, 2011 — Comments

While catching up on some old Mythbusters episodes, I ran across an interesting myth about spinning bullets. Apparently when you shoot a bullet into the surface of a frozen lake, it bounces backward, bounces over the ice a little way, and keeps spinning even after it comes to rest.

This is a very curious result. At first, it kind of seems to make sense. A bullet comes out of the gun spinning at 80000 revolutions per minute, according to the show, which is pretty fast. So it makes sense that it has a sizable amount of angular momentum, which in turn means it'll take quite a bit of torque to stop it. Colliding with the ice conceivably might not be able to exert enough torque to do that, so the bullet would keep spinning.

But if that were the case, the bullet should maintain its orientation — it'd keep pointing in the same direction that it came out of the gun, because angular momentum is a vector quantity and it won't change in either magnitude or direction without an external torque. That clearly wasn't the case; the bullets the Mythbusters shot skittered across the ice, spinning in multiple directions, and landed with an orientation completely different from what they had when they were shot. How does that happen?

I'm not entirely sure what actually goes on with these spinning bullets, but I think the key lies in the direction of the torque exerted. Like angular momentum, torque is a vector quantity. In order for a torque to stop the bullet, it needs to point directly antiparallel to the angular momentum, but when a bullet collides with ice, the collision is quite complicated and there's no reason to expect that the net torque will point neatly along the angular momentum axis.

There are two main sources of torque that I can think of: friction between the edge of the bullet and the ice, and impacts between the bullet and fragments of ice that are flying around in the aftermath of the collision. The friction would certainly contribute to slowing the bullet's rotation, but the impacts would be more or less random, so they could conceivably create the right kind of torque to change the orientation of the spin without necessarily slowing it down. Plus, I can imagine that the torque provided by friction may not be all that much — after all, ice is slippery, and besides, the bullet bounces out (and is no longer touching the ice) after a very short time. If the impacts provide much more torque than the friction, it would account for the tumbling behavior you can see in some of the high-speed excerpts they put in the show, and it would also explain why the bullet keeps spinning as it bounces around the ice.

2011

May

Blow your own sail

Posted by David on May 18, 2011 — Comments

The latest episode of Mythbusters tested a slightly controversial and very physics-related myth: that you can propel a boat forward by putting a fan on the boat and pointing it forward, into the sail. What's going on here?

First of all, why wouldn't you expect this to work? Actually, first of all, why would you expect this to work? Think about the naive explanation for why a sailboat moves: the wind pushes forward on the sail, and the sail pushes forward on the boat. So someone who had never heard of physics might think that putting a fan on the boat and pointing it into the sail just gives you a convenient, portable source of wind. Presto, instant speedboat!

But, as explained on the show, that reasoning doesn't work, because of Newton's laws of motion. There are actually a couple of different ways to apply Newton's laws to this scenario:

Newton's first and second laws (which are kind of the same thing) say that an object maintains its state of motion unless subject to an external force. The key word there is "external": if you want your boat to move, you need something outside of the boat to provide the force. Natural wind can do that, but not a fan on the boat itself; that's an internal force.
For a more detailed description, consider Newton's third law, which says that forces occur in complementary pairs: If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. In this case, the objects are the air and the fan. In order to blow the sail, the fan needs to push the air, which means the air pushes back on the fan. And since the fan and the sail have opposite effects on the air (the fan takes it from still to moving, the sail takes it from moving to still), the forces from the fan and from the sail exactly cancel each other out.

If you watched the episode, of course, you know that blowing your own sail actually does work (a little) in practice. So what's wrong with the preceding argument? Well, the air exerts a forward force on the sail and a backward force on the fan, and I initially assumed that those cancel out. But in reality, they don't, because the sail doesn't just stop the air, it reflects it. It takes more force to bounce something backwards than it does to just stop it — twice as much, in fact, because of Newton's second law in the form $F_\text{avg} = \frac{\Delta p}{\Delta t}$ (reflecting an object takes the momentum from $p$ to $-p$ , which is twice as much change as $p$ to $0$ ). So the force on the sail, which has to make air molecules bounce backwards, is greater than the force on the fan, which just has to make them accelerate from rest.

Or is it? Actually, if you think about it, what happens to that reverse airflow coming off the sail? Yep, it heads right back toward the fan. So the fan isn't just accelerating air from rest; it potentially has to take the air that was moving backward (from the sail) and change its direction to move forward. This represents a change in momentum from $-p$ to $p$ , which exactly cancels out the effect of the sail, assuming the same amount of air is involved.

This is why the overall effect of blowing your own sail depends on how much air is moved by the fan, relative to the amount that's reflected by the sail. As we saw in the show, if the sail is much smaller than the fan, then the fan blows more air than the sail can reflect, so the force on the fan is greater, and the boat moves backward. But if the sail is much larger than the fan, then the fan isn't able to capture all the air reflected by the sail, so the force on the sail is greater, and the boat moves forward. And if the fan and the sail are reasonably close to the same size (or the sail is only slightly bigger than the fan), then the fan can capture essentially all of the air reflected back by the sail, and the forces roughly balance out.

It's interesting to notice that if you treat the combination of the fan and sail (and the air between them) as a "black box," you'll notice that when the boat moves backwards, it's emitting an air stream forwards, and vice versa. So the fan and sail act almost like a rocket: they propel the boat by emitting a stream of exhaust gases in the opposite direction, except that the exhaust gas is air instead of burnt rocket fuel. Still, at best this stream contains only get a fraction of the air being blown by the fan, so for the most efficient propulsion, it really is better to just point the fan backwards to begin with.

Or in true Mythbusters fashion, mount a rocket on the boat. But that's a story for another day.

2010

Dec

Death rays and thermal radiation

Posted by David on December 11, 2010 — Comments

It's been far too long since I did a Mythbusters writeup, but I think it's time to stop stalling and bring this series back. On this week's episode, Adam and Jamie tested the myth of Archimedes' heat ray for a third time — that has to be some kind of record — at the request of President Obama.

The gist of the myth is this: by focusing enough of the sun's rays, using a large number of mirrors, on an enemy ship, the Greeks hoped to heat it up enough to make it catch on fire. So far (spoiler alert), there's no evidence that this thing ever could have worked. All three of the Mythbusters' tests have failed.

But I think I can shed some light (no pun intended) on why. As it happens, I taught a lab on thermal radiation transfer this week, and that (along with an interesting perspective on gravitationally baking a turkey) reminded me that it's fairly straightforward to calculate, at least in a simple model, the amount of radiation it takes to heat something up to a particular temperature. It all stems from the Stefan-Boltzmann equation,

$P = \sigma\epsilon AT^4$

This equation actually serves a dual purpose: it tells you the rate at which an object at temperature $T$ loses energy due to thermal radiation alone (not counting other effects like heat conduction), and it also tells you the rate at which an object in an environment at temperature $T$ gains energy from that environment, again due to thermal radiation alone. In either case, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the object's surface area, and $\epsilon$ is the object's emissivity. Since any real object will be both emitting and absorbing radiation, the two uses of the equation are often combined to give the net power loss due to thermal radiation:

$P = \sigma\epsilon A(T^4 - T_0^4)$

Here I'm using $T_0$ to designate the temperature of the environment, and plain $T$ is the temperature of the object.

Blackbody power equilibrium

So here's the simple model. Suppose that the piece of the ship you're trying to set on fire is effectively insulated, so that it doesn't gain or lose heat by conduction. (The parts of a ship that you'd try to set on fire, the sails or perhaps the wooden hull, generally don't conduct heat very well; besides, the problem gets a lot harder if you relax this assumption.) The piece will lose energy through thermal radiation, at a rate determined by the equation above, and it will gain energy from the sunlight being focused on it. When it's at thermal equilibrium (which it essentially always is), those two will be equal,

$P_\text{sun} = \sigma\epsilon A(T^4 - T_0^4)$

If the equilibrium temperature $T$ is greater than or equal to the flash point of the material, it catches on fire. Simple enough.

The final piece of the calculation involves figuring out the power input from sunlight. This is just going to be the solar irradiance $I$ , representing the power of sunlight per unit area, times the area $A_\text{mirror}$ of the mirrors used to reflect the light, times the cosine of the angle of incidence of the sunlight, times the reflectivity of those mirrors, $\rho$ .

Mirror reflection at an angle

But to get the amount of this sunlight that actually reaches the sail, we need to throw in a couple of extra factors. First, since the aiming of the mirrors is not perfect, we'll need a number $n$ representing the fraction of mirrors that were actually reflecting their light to the right place. Also, since some of the radiation reaching the sail will be reflected, only a fraction of it actually goes towards heating the sail, so we need to multiply in the absorptivity $\mathcal{A}$ of the sail's material. Putting it all together, we come up with this expression,

$P_\text{sun} = In\mathcal{A}\rho\cos\theta A_\text{mirror}$

giving the following for the area of mirrors required:

$A_\text{mirror} = \frac{\sigma\epsilon A_\text{target}}{In\mathcal{A}\rho\cos\theta}(T_f^4 - T_0^4)$

So it's time to plug in some numbers.

Let's start with the Mythbusters' most recent test:

They used a square patch of sail $\unit{50}{\inch}$ on a side, with an area of $A_\text{target} = \unit{1.61}{\meter^2}$ .
We were told in the show that the flash point of the sail was $\unit{410}{^{\circ}F}$ , or $T_f = \unit{483}{\kelvin}$ .
The initial temperature of the sail was reported as $\unit{85}{^{\circ}F}$ , or $\unit{303}{\kelvin}$ (although this number is not going to matter much once you take the fourth power).
We don't know the angle of incidence for the incoming sunlight on the mirror, but it's possible to estimate it by measuring the tilt of the mirrors from the episode, knowing that the rays were being reflected on to a pretty much horizontal path. I come up with an estimate of $\unit{0.4}{\radian}$ (that's around $\unit{23}{\degree}$ ); perhaps it was a little more in reality.
To get the solar irradiance, I integrated the direct ASTM reference spectrum, coming up with a value of just about $\unit{1}{\frac{\kilo\watt}{\meter^2}}$ .
From Wikipedia, I find that the reflectivity of an aluminum mirror, which is a fairly common modern design, is around $\rho = 0.9$ .
Another reference listed on Wikipedia gives the emissivities of various sorts of black paint in the range $\epsilon \sim 0.9$ .
Yet another reference (found by a random Google search) shows that the absorptivities of many black paints are in the vicinity of $\mathcal{A} \sim 0.9$ . (no surprise there...)
Since I've been making optimistic assumptions so far, I might as well continue that trend by going with $n = 0.8$ : 80% of the mirrors are correctly aligned to reflect on to the target.
Finally, the Stefan-Boltzmann constant has a numerical value of $\snunit{5.67}{-8}{\frac{\watt}{\meter^2 \kelvin^4}}$ .

Plugging in all the numbers, I come up with $\unit{6.5}{m^2}$ for the required surface area of the mirrors.

Hm... that's not very much. The 500 mirrors would have had no problem setting the boat on fire. Clearly something is wrong with this calculation. A couple of things come to mind almost immediately: perhaps the absorptivity of the sail was actually lower, so that it reflected more sunlight. Perhaps its emissivity was higher, so that it was more efficient at radiating away excess energy. Perhaps the fraction of mirrors on target was lower than $0.8$ (and I can't really blame the kids, those mirrors were nearly as big as some of them were).

Digging a little deeper, it's possible that the attenuation of sunlight by the atmosphere was more than the ASTM spectrum accounts for, although I wouldn't expect this to have been a big issue. The ASTM spectrum models the radiation received when the sun is about $\unit{42}{\degree}$ above the horizon, and based on my observations from the show, the sun was pretty close to that inclination when they did the experiment. Admittedly the pressure, humidity, etc. at the test site were certainly different from the ASTM reference values, but again, I don't see that making a huge difference.

Another thing to consider is that several of the values I included in the formula are actually wavelength-dependent, namely emissivity, absorptivity, and reflectivity, as well as the spectral irradiance from the sun. So the incoming power should really be expressed as a function of the wavelength of the light and the angle,

$P_\text{sun}(\lambda) = I(\lambda)n\mathcal{A}(\lambda)\rho(\lambda)\cos\theta A_\text{mirror}$

The same goes for the thermal equilibrium condition, although it's still the total power that determines whether the temperature changes,

$\int\frac{P_\text{in}(\lambda)}{\epsilon(\lambda)}\udc\lambda = \sigma A(T^4 - T_0^4)$

Putting these together makes the mirror area formula look much more complicated:

$A_\text{mirror} = \frac{\sigma A_\text{target}\cos\theta}{n}(T_f^4 - T_0^4)\biggl[\int\frac{\epsilon(\lambda)\udc\lambda}{I(\lambda)\mathcal{A}(\lambda)\rho(\lambda)}\biggr]^{-1}$

But usually, it's reasonable to approximate the functions of $\lambda$ as constant over the range of wavelengths we're dealing with, so I doubt that this would account for the two-orders-of-magnitude error in my result. (If you have access to a lot of data on optical properties of various materials, feel free to plug this thing into a computer and find a value for it to see whether that's in fact the case.)

Honestly, though, I think the biggest problem the model doesn't work is that I completely ignored other ways in which heat is removed from the target. Specifically, heat loss to the surrounding air is likely to be a major factor, because convection can carry away a lot of energy. I'm going to say that working out exactly how fast energy is lost this way is "beyond the scope of this post" — that is, either impossible or more work than I have time for ;-) But we can make a quick estimate by modifying our equation to include a generic heat loss term,

$P_\text{sun} - P_Q = \sigma\epsilon A(T^4 - T_0^4)$

Plugging in the other formulas from before, I get

$P_Q = In\mathcal{A}\rho\cos\theta A_\text{mirror} - \sigma\epsilon A_\text{target}(T^4 - T_0^4)$

Now, with the numbers previously estimated and using $\unit{500}{\meter^2}$ for the total area of all the mirrors (should be within a factor of 2, right?) gives

$P_Q = \unit{294}{\kilo\watt}$

That's kind of a lot. Wait — is it a lot? Well, if you calculate the total power that should be coming in from the sun, you get $P_\text{sun} = \unit{298}{\kilo\watt}$ (and I think that formula can be trusted). So if this is correct, the sail is incredibly efficient at radiating away the energy it receives from sunlight; a mere 1% of the incident energy in this situation actually goes into heating it up! No wonder it's such a hard thing to pull off.

2010

Oct

Experimental Design

Posted by David on October 13, 2010 — Comments

Mythbusters is back as of last week with new episodes! Although the season premeire about throwing dogs off the scent isn't directly physics-related, I do have one issue to point out.

Tory, Kari, and Grant did an experiment in which they tried to hide some "contraband" from a dog by sticking it in a container with a strong-smelling material, like coffee or peanut butter. They hid the container somewhere in a 17,000 square-foot warehouse and set the dog loose to find it — which he always did. So the conclusion was that strong scents can't distract a sniffer dog.

But what if the dog was actually following the distraction? I wouldn't be surprised if the scent of e.g. peanut butter itself was a dead giveaway to the location of the container, even if did mask the smell of the fake drugs. It was probably the strongest smell in that warehouse. A better experiment would have entailed filling several containers with the smelly substance, distributing them around the building, and only hiding the fake contraband in one of them, then seeing if the dog identifies the right one.

Of course, they did basically that later in the show, in the luggage store with the hidden meats, and it didn't deter the dog. So I doubt that it would have made a difference in the result. And I know the Mythbusters don't strive for scientific rigor (and I don't blame 'em, because real science is often boring). Still, it wouldn't have taken much for them to upgrade from a for-fun experiment to something more scientific, and I think we could all stand to try for that as often as possible.

2010

Jun

Decapitation: Energy or momentum?

Posted by David on June 11, 2010 — Comments

http://scienceblogs.com/dotphysics/2010/06/collisions_kinetic_energy_or_m.php

Took the words right out of my mouth. Or right off my keyboard. Whatever. I'm just happy to see other people are considering the same questions.

But I've got a couple of things to add. First of all, kinetic energy can be easily related to momentum using the formula

$K = \frac{p^2}{2m}$

which tells you directly that an object with the same energy but larger mass will have a larger momentum.

Also, the big question: is it energy or momentum that gives a collision its decapitating power? My thought is that it actually depends on force. Think about this from the point of view of a particle in the neck. This particle doesn't "know" (as if a particle could "know" anything) how big the sheet of glass hitting it is; it doesn't "know" how much energy or momentum the glass has. The reason is that energy and momentum are what I'm going to call global properties, basically meaning that the total amount of them possessed by some object comes from contributions from all different pieces of the object. To measure a global property of the glass, you need to have access to the entire piece of glass. But a particle in the neck doesn't have that access; it only interacts with a few atoms on the edge, so those are the only ones that can influence it. This means that whatever happens to the neck must be determined by local properties, which can be measured at a point, without having access to the entire piece of glass. And the relevant local property is force. (Think about it: when you feel a force from something pushing on you, you can tell how strong the force is without knowing anything about the object exerting it.)

But of course there's a little more to it than that. The ability to cut through a neck doesn't just depend on the glass being able to exert enough force; it also requires sufficient force to be maintained across the full diameter of the neck. When the glass first hits the neck, it exerts some force on the neck, which cuts through the first layer of skin, but the neck also exerts the same amount of force on the glass, which slows it down. (Newton's third law) Since the glass is moving more slowly, it's going to exert less force on the next layer of skin. The further it goes through the neck, the slower the glass is moving and the less cutting force it exerts. In order to make a clean decapitation, this continual decrease in speed has to leave the glass moving fast enough to cut through a layer of skin after it's made it all the way through the rest of the neck.

I can even apply some math to this, although there's not quite enough information to make a precise calculation. Here's what I do know. The force $F$ on the glass changes the glass's momentum according to Newton's second law, $F = \ud{p}{t}$ (this is only in one dimension, so I don't need to make the variables vectors, but I do need to be careful with signs). Momentum is, of course, $p = mv$ , and I can use that and the chain rule to write

$F = m\ud{v}{t} = m\ud{v}{x}\ud{x}{t} = mv\ud{v}{x}$

So I know how the force affects the glass's speed. That's just general Newtonian dynamics. What I don't know is how the speed affects the force. In other words, how much force does the neck exert on the pane of glass, as a function of their relative speed? If I knew that function, $F(v)$ , I could plug it in to the previous equation and solve the differential equation for $v$ .

To show how that would work, I'm going to make up an example function $F(v)$ , completely out of thin air. Well, not completely out of thin air — I'm going to take the function from... thin air ;-) If you've read my previous posts, you may remember the equation for the aerodynamic drag force,

$F = \frac{1}{2}CA\rho v^2$

This is precisely the kind of equation we'd want: the force of resistance, as a function of speed, on a solid object moving through some stuff. Specifically, a fluid. Of course, a neck is hardly fluid-like, but this is just an example so it'll do the job. (You have a better idea?)

Glass cutting through a neck

In this situation $A$ can be the cross-sectional area of the glass that is in contact with the neck, and $\rho$ will be the volume mass density of neck material (that's mass per unit volume). But $\rho$ depends on the full 3D position, so it's be easier to consider the product of $\lambda = A\rho$ , which is the linear mass density along the direction that the glass moves — in other words, $\lambda$ is the mass per unit $x$ -distance that the glass cuts through. It's a function of only $x$ , not the other coordinates.

If you combine the drag equation with the dynamic equation above it, you get

$-\frac{1}{2}C\lambda v^2 = m_\text{glass}v\ud{v}{x}$

The negative sign is necessary because the force points in the negative direction. This is a separable differential equation, so we can rearrange it to

$-\frac{1}{2m_\text{glass}}C\lambda\udc x = \frac{\udc v}{v}$

and integrate it to get

$-\frac{1}{2m_\text{glass}}\int_0^L C(x)\lambda(x)\udc x = \ln v(L) - \ln v(0)$

$L$ is the total length of the neck that the glass cuts through. (Or you could make $L$ be some distance partway through the neck, if you wanted to find the glass's speed there) I started writing $C$ as $C(x)$ to remind us that it varies with position, and same with $\lambda(x)$ . Now I can solve for $v(L)$ , the final speed, in terms of $v(0)$ , the known initial speed, and I get

$v(L) = v(0)e^{-\int_0^L C(x)\lambda(x)\udc x/2m_\text{glass}}$

So what does this have to say about the original question? Does a larger piece of glass with the same energy have the same decapitating power?

Well, there's no reason to expect this formula to be accurate, given all the unjustified assumptions that went into deriving it. But on the other hand, I'd expect that it would have some tenuous relation to reality. If you increase the mass of the neck (say, by making it denser, increasing $\lambda(x)$ ), the equation says the glass loses more speed, which makes sense. And if you increase the mass of the glass, the equation says it retains its speed better, which also makes sense.

Let's try this out with the Mythbusters' actual experiment. They tried to simulate the effect of the piece of glass Tory threw, with mass $\unit{0.45}{\kilo\gram}$ , if it were moving at a speed of $\unit{300}{\mileperhour}$ , or $\unit{134}{\frac{\meter}{\second}}$ . According to the calculations they did on the show, that has an energy of $\unit{4.04}{\kilo\joule}$ , and the mass required to reproduce that energy at $\unit{80}{\mileperhour}$ is $\unit{6.3}{\kilo\gram}$ . So we have two sets of parameters:

$m_\text{glass} = \unit{0.45}{\kilo\gram},v(0) = \unit{134}{\frac{\meter}{\second}}$

$m_\text{glass} = \unit{6.3}{\kilo\gram},v(0) = \unit{36}{\frac{\meter}{\second}}$

The quantity $\int_0^L C(x)\lambda(x)\udc x$ is going to be the same in both cases. Let me call that $M$ since it has the units of mass. If I plug in the parameters just listed, I get

$v(L) = \unit{134}{\frac{\meter}{\second}}e^{-M/\unit{0.9}{\kilo\gram}}$

$v(L) = \unit{36}{\frac{\meter}{\second}}e^{-M/\unit{12.6}{\kilo\gram}}$

Hmm, so this definitely depends on $M$ . Which I don't know. But if you think about it, $M$ is basically the mass of neck that the glass cuts through, with some scale factor (related to the "drag coefficient" $C$ ) that's probably between $\frac{1}{10}$ and $10$ . (I don't actually know, I'm just guessing here) A window-thin piece of neck probably has a mass of $\unit{100}{\gram}$ or less, which means that $M$ is likely less than $\unit{1}{\kilo\gram}$ . Now look at this plot of $v(L)$ for various values of $M$ in that range:

Plot of final velocity versus M

The top curve is for the smaller piece of glass, the one Tory threw, and the bottom line is for the larger piece they put on the truck. Clearly, for this range of values for $M$ , the smaller piece has a much higher final velocity, and that indicates a much higher decapitating power.

2010

Jun

Firework Man

Posted by David on June 10, 2010 — Comments

More Mythbusters! Yayyy!

Just kidding, I am not that incessantly perky. But I did have an interesting thought while watching last week's episode of Mythbusters. Kari, Tory, and Grant were testing the myth that a man in Germany launched himself 150 feet high using 400 firecrackers and splashed down safely in a lake. First of all, how dumb is this guy? Darwin Award anyone? (I guess he'd technically be ineligible since in the myth, he survived unharmed, but it was certainly a noble effort)

Anyway, here's my physics thought of the week for last week: $v = \sqrt{2gh}$ . That's the formula for the vertical velocity you need to reach a certain height as a ballistic projectile. To get 150 feet of altitude requires $\unit{47}{\mileperhour}$ , and that's if you launch straight up. Coming off an angled ramp, as they did in the show, you'd have to be going even faster. Of course, the guy in the myth isn't quite a ballistic projectile; he gets an extended boost from the rockets. But still, $\unit{150}{\foot}$ of altitude is pushing it, so I would have been inclined (pun? intended? nah) to dismiss this one on that basis alone.

2010

May

Waterslide Wipeout

Posted by David on May 23, 2010 — Comments

Like everyone else, I feel a need to analyze the giant water slide in the latest Mythbusters episode. But honestly, there isn't much left to say. The original video has been around for a while and everybody else who does this kind of analysis has already had the chance to do it. Example 1; example 2 (okay, so I'm only linking to one blog, but there must be more).

I guess I might as well do the obvious calculation, but I'll use the Mythbusters' parameters instead of those from the original video (which are unknown). The slide starts with a downward ramp $\unit{165}{\foot}$ long at a $\unit{24}{\degree}$ slope, which then curves upward to a $\unit{30}{\degree}$ launch ramp that terminates $\unit{12}{\foot}$ above the surface of the lake. They didn't say how long the launch ramp is, but I can work without that information. I'll be trying to calculate two quantities mentioned on the show: how far each Mythbuster flies from the end of the ramp, and his maximum speed.

Diagram of the waterslide

(here's a full-size version)

There are two parts to this problem:

The slide
The flight

The first part is an energy problem. All energy problems can be solved in the same way, starting with the work-energy theorem

$W_\text{ext} = E_f - E_i$

and plugging in, term by term.

$W_\text{ext}$ is the work done by external, nonconservative forces, like friction. The friction is proportional to the normal force, which (for most of the slide) is in turn equal to the component of gravity acting perpendicular to the surface. So

$F_f = \mu_k F_N = \mu_k m g \cos\theta_1$

where $\theta_1$ is going to be the $\unit{24}{\degree}$ angle of the acceleration ramp. This frictional force acts antiparallel to the motion over a distance of $d\equiv\unit{165}{\foot}$ (plus a little correction that I'll leave for later), so we get

$W_\text{ext} = \vec{F}_f\cdot\Delta\vec{x} = -\mu_k m g d \cos\theta_1$

The next term, $E_f$ , is the final energy, at the end of the slide when the person flies off. This is a combination of gravitational potential energy, $U = mgh$ , and kinetic energy, $\frac{1}{2}mv^2$ . Accordingly, I'll write it

$E_f = mgh_1 + \frac{1}{2}mv^2$

using $h_1$ to represent the height of the launch point above the lowest part of the ramp.

Finally, the initial energy is purely gravitational. The height is a lot higher this time; specifically,

$E_i = mgd\sin\theta_1$

Putting it all together,

$-\mu_k mgd\cos\theta_1 = mgh_1 + \frac{1}{2}mv^2 - mgd\sin\theta_1$

The first thing to notice is that mass cancels out, so it shouldn't make a difference how heavy you are. In the show, Jamie flew 2 feet further than Adam, but I suspect that that was just a fluke, not a statistically significant result. I get

$v^2 = 2[gd(\sin\theta_1 - \mu_k\cos\theta_1) - gh_1]$

By the way, this factor of $\sin\theta - \mu_k\cos\theta$ seems to be fairly common — you can often expect to see it when doing a problem that involves a ramp with friction.

Now it's time to move on to part 2, which is a projectile motion problem. I'll use the same height reference as before, the bottom of the curve of the ramp. So we're launching an object from height $h_1$ at an angle $\theta_2$ to a surface at height $-h_2$ ; how far does it travel?

First step: Find the time it takes to hit the water.

$y = y_0 + v_0 t + \frac{1}{2}at^2$

Plugging in the relevant expressions, we get

$-h_2 = h_1 + v\sin\theta_2 t - \frac{1}{2}gt^2$

which has the solution

$t = -\frac{1}{g}\Bigl(-v\sin\theta_2 \pm \sqrt{v^2\sin^2\theta_2 + 2g(h_1 + h_2)}\Bigr)$

Second step: multiply that time by horizontal velocity to get horizontal distance. This one is easy; the horizontal velocity is $v\cos\theta_2$ so we get

$x = \frac{v\cos\theta_2}{g}\Bigl(v\sin\theta_2 \pm \sqrt{v^2\sin^2\theta_2 + 2g(h_1 + h_2)}\Bigr)$

Given that the solution for the previous part came out in terms of $v^2$ (because I dislike ugly square root signs), here's a useful way to rewrite that:

$x = \frac{v^2\sin\theta_2\cos\theta_2}{g} \pm \frac{1}{g}\sqrt{v^2\bigl(v^2\sin^2\theta_2 + 2g(h_1 + h_2)\bigr)\cos^2\theta_2}$

At this point we could substitute in $v^2$ from before and get a symbolic expression, but it's getting pretty complicated. Probably better to plug in numbers.

First, a recap of the numeric values involved. I'm taking the length of the slide from the top down to its lowest point to be $d = \unit{165}{\foot}$ , the height of the end of the slide above that lowest point to be $h_1 = \unit{6}{\foot}$ , the height of the water below the lowest point of the slide to be $h_2 = \unit{6}{\foot}$ , and the angles of the slide and the launch ramp respectively to be $\theta_1 = \unit{24}{\degree}$ and $\theta_2 = \unit{30}{\degree}$ . Plugging all these in to the expression for velocity, I get

$\unit{370}{\frac{\meter^2}{\second^2}} - \unit{900}{\frac{\meter^2}{\second^2}}\mu_k$

Hmm, well that's a problem. Or is it? We don't know the coefficient of friction, but we do know the number this is supposed to work out to, namely $(\unit{30}{\mileperhour})^2$ . The value $\unit{30}{\mileperhour}$ , was given on the show as the maximum speed reached by both Jamie and Adam as they went down the slide. Technically my expression for velocity is for the speed at the end of the launch ramp, which will be a little slower than the maximum speed, but I've made enough approximations that it probably doesn't make a big difference. So I'll go ahead and plug in $v = \unit{30}{\mileperhour}$ and use Mathematica to solve for the coefficient of friction, $\mu_k$ . It comes out to be $\mu_k = 0.2$ . Seems reasonable.

Let's try this another way, though. Again using Mathematica, I can substitute my expression for $v^2$ into the expression for $x$ . Both Jamie and Adam flew about $\unit{70}{\foot}$ horizontally, so that's my value for $x$ . Then I can plug in all the other distances and angles taken from the show and get another equation for $\mu_k$ :

$\unit{21.3}{\meter} = \unit{16.1}{\meter} + \unit{39.8}{\meter}\bigl(\sqrt{(\mu_k - 0.72)(\mu_k - 0.41)} - \mu_k\bigr)$

That one takes a bit more algebra — of course, I'll just get Mathematica to do it ;-) The solution works out to... $\mu_k = 0.2$ ! The fact that these two values are the same is, at least, an encouraging sign that this calculation actually corresponds to reality.

Here's something else we can try: predicting (actually "postdicting") the distance covered on each of the earlier, shorter slides. Everything (including the coefficient of friction) is the same as before, except for the distance $d$ , which was first a half, then three quarters, of the total length. So if I plug in $d = \frac{1}{2}\times\unit{165}{\foot}$ , I get

$v = \unit{8.6}{\frac{\meter}{\second}} = \unit{19}{\mileperhour}$

$x = \unit{11}{\meter} = \unit{35}{\foot}$

Not so great, considering that the actual distance Adam flew on that run was just 4 feet. But the corrections I alluded to earlier might help that out a bit... that's a story for another post, though. What about $d = \frac{3}{4}\times\unit{165}{\foot}$ ?

$v = \unit{11}{\frac{\meter}{\second}} = \unit{25}{\mileperhour}$

$x = \unit{16}{\meter} = \unit{52}{\foot}$

They didn't give measurements for these runs on the show, but from the video I estimate that Adam flew about 40 feet. Still not that great, but we're getting closer.

Even though I know this model doesn't quite work, I've got one last set of numbers to run. How high would the ramp have to be to achieve the hypothetical $\unit{115}{\foot}$ flight in the video? Since I already had the Mathematical machinery set up to compute flight distance for a given ramp length, I decided to do this one by trial and error, plugging in different values for $d$ to narrow in on the one that yields a distance of $\unit{115}{\foot}$ . It works out to $\unit{85}{\meter}$ , or $\unit{280}{\foot}$ , which is really long. But think about this: for shorter runs than the Mythbusters' full-length slide, my model overestimates the distance traveled. Perhaps for longer runs, it'll underestimate the distance traveled, which means that in reality, maybe a shorter slide would do the job. And in fact, if you look at the data points for the slides on the show — keeping in mind that it's dangerous to extrapolate from only three points — they do seem to fall in a line, and that line suggests that you might only need a $\unit{220}{\foot}$ slide to achieve a $\unit{115}{\foot}$ flight.

Graph of flight distance vs. slide length

Interesting, isn't it?

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