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 long at a slope, which then curves upward to a launch ramp that terminates 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.

(here's a full-size version)

There are two parts to this problem:

1. The slide
2. 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

and plugging in, term by term.

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

where is going to be the angle of the acceleration ramp. This frictional force acts antiparallel to the motion over a distance of (plus a little correction that I'll leave for later), so we get

The next term, , is the final energy, at the end of the slide when the person flies off. This is a combination of gravitational potential energy, , and kinetic energy, . Accordingly, I'll write it

using 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,

Putting it all together,

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

By the way, this factor of 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 at an angle to a surface at height ; how far does it travel?

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

Plugging in the relevant expressions, we get

which has the solution

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

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

At this point we could substitute in 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 , the height of the end of the slide above that lowest point to be , the height of the water below the lowest point of the slide to be , and the angles of the slide and the launch ramp respectively to be and . Plugging all these in to the expression for velocity, I get

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 . The value , 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 and use Mathematica to solve for the coefficient of friction, . It comes out to be . Seems reasonable.

Let's try this another way, though. Again using Mathematica, I can substitute my expression for into the expression for . Both Jamie and Adam flew about horizontally, so that's my value for . Then I can plug in all the other distances and angles taken from the show and get another equation for :

That one takes a bit more algebra — of course, I'll just get Mathematica to do it ;-) The solution works out to... ! 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 , which was first a half, then three quarters, of the total length. So if I plug in , I get

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 ?

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 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 to narrow in on the one that yields a distance of . It works out to , or , 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 slide to achieve a flight.

Interesting, isn't it?

The latest episode of Mythbusters features a myth with a deep physical explanation... no pun intended! Well, maybe. Anyway, the myth is that by diving under the water, you can escape injury from an explosion occurring above the surface. Adam and Jamie tried to solve this puzzle by experiment (what else), and their results seemed to show that the myth might actually be true, but I want to look at it from the theoretical standpoint: why might being underwater protect you from an explosion?

There is actually a not-too-obscure answer to this puzzle, and it has to do with refraction and reflection. These are phenomena that occur when a wave (of any sort — light, sound, or whatever) crosses a boundary between two media in which it has different speeds. Part of the wave bounces back (that's reflection) and part of it continues through, but in a different direction (that's refraction). Exactly how much of the wave's power is reflected and how much is transmitted through, as well as the new direction of the transmitted part, depends on the angle of the incoming wave with respect to the surface, and also on the relative speed of the wave in the two materials.

Reflection and refraction are most often discussed in connection with light. Lenses, for example, work by refracting incoming light, since light travels slower in glass (or plastic) than in air. Because the surface of the lens is curved, it bends different light rays in different locations by different amounts, and this can be used to focus an image. But the same thing happens with pressure waves, like sound waves and explosive shockwaves, at the surface of a lake, because pressure waves travel faster in water than they do in air. It stands to reason, then, that part of the energy in a blast wave would be reflected off the surface of the lake, reducing the energy available to damage anyone (or anything) below the water.

Unfortunately, calculating how much energy is reflected and how much is transmitted when a shock wave hits water turns out to be a very involved problem, not something that can be learned and solved in a week ;-) Still, there are some qualitative things I can say about the process. Consider what happens to a pressure wave in air when it hits a boundary, such as a wall: the wall will vibrate a bit, but the wave mostly just bounces back. You'd expect that a similar sort of thing would happen if the wave hits a water surface instead, since compared to air, water is very "hard" — technically speaking, it has a very low compressibility. This is related to a material property called the bulk modulus, which basically measures how much the material resists being squashed (compressed). The bulk modulus of water is about 10000 times larger than that of air, so the increase in pressure applied by a wave will have a relatively minor effect on the water. It'll have a much greater effect on the surrounding air, which means that that's where most of the energy goes: back into the air.

If pressure waves don't affect water that much, though, why was there such a huge splash every time the Mythbusters set off one of their explosions? The amount of energy in an explosion shockwave is so large that even "relatively minor" can be enough to throw a lot of water around! Since air is transparent, most of the wave's effect on the air goes unnoticed.

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

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

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 , we consider a molecule of water hitting the stone with the same speed but in the opposite direction, .

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,

and then the difference

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

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

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, , times the volume the surface passes through in a unit time , which is . The dot product takes the component of velocity parallel to the area's normal vector, and it just works out to , or . Multiplying this by the change in momentum per molecule, we get

But notice that , the mass density of the water. Dividing both sides by , we get the force,

Hmm, this factor 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 points directly along the 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 , use , and instead of , I'll invent a vector which points perpendicular to by definition. Unlike and , these vectors won't arbitrarily change their orientation when we switch coordinate systems. I can write the force as

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

and

Plugging in and simplifying,

which conveniently reduces down to

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,

where 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 and 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 , that's just the speed of the car, and 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 . 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 . We can get an approximate measurement of 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 . is a bit harder to determine, but it has to be pretty small, probably not more than . With these numbers, the force works out to , or . 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, , 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 and , but knowing that they're both really small,

That amplitude is about , so even with as small as about , it'll be enough force to completely counteract the weight of the car and bounce it back up into the air.

Following up on my calculation of the lifting power of helium balloons, it's time to see how the same argument applies to ping-pong balls being used to raise a sunken ship.

Raising a ship with ping-pong balls is, in fact, nearly the same situation as raising a child with helium balloons. All you have to do is replace the air with water, the helium with air, the rubber balloons with plastic balls, and the child and harness with a boat (though preferably not in that order). The physical principle at work (Archimedes' Principle) is exactly the same, and so the same equation I used last time is equally applicable here: the buoyant force on an object (ping-pong ball) immersed in a fluid (water) is equal to the weight of the water displaced by the fluid,

Let's see what this says about how many ping-pong balls it would take to raise the Mythtanic II, which weighs about according to the show. We can start by figuring out how much mass it takes to balance out the buoyant force on a single ping-pong ball, using as we did last time:

• Water has a density of about . What about the volume of a ping-pong ball? These balls come in two standard diameters, and . The ones that Adam and Jamie used look like low-quality practice balls I've seen in stores, so let's use as the diameter, giving a volume of . Multiplying that by the density of water gives .
• From that quantity, we need to subtract the mass of the ping-pong ball and the air inside it. Standard ping-pong balls weigh , easy enough. As for the air, I'm just going to assume that it's at normal atmospheric pressure, so we can multiply the normal density of air by the volume of the ball to get . That's a tiny amount of mass, about 2% of the mass of the ball itself, and our numbers just aren't precise enough to bother caring about it. So the net lifting power of a single ping-pong ball in water is , or .

Given that figure, we can easily calculate how many balls it should take to lift :

That's a lot of balls.

But wait! We're forgetting about the same thing that Adam and Jamie forgot about in their calculations — the buoyant force on the boat itself. It's an easy mistake to make, because we're used to dealing with objects in air, where there's so little buoyant force that we usually just ignore it (except for things that blatantly "advertise" their buoyancy by floating away). Not so in the water, though; remember that buoyant force is proportional to the density of the fluid, and since water is so much denser than air, the buoyant forces are correspondingly larger.

Now, without knowing exactly what the boat is made of, there's no way to tell just how much that force is, but let's make a guess. Pretend the boat is made entirely of fiberglass. Now, the density of fiberglass) is about 1.5 to 2 times that of water, so if the boat is made of of the stuff, its volume would be

(In case you're wondering, yes a cubic meter is actually pretty sizable. Think about it.) With the Mythtanic underwater, all this volume of fiberglass is displacing water and contributing to the buoyant force along with the ping-pong balls. But of course, we know how to calculate how much mass this buoyant force can support: , or in this case,

and we also know how to calculate how many ping-pong balls it would take to lift that amount of mass in water:

That means that the boat itself provides the lifting power of about 26000 ping-pong balls, which in turn means that we only need an additional 26000 real ping-pong balls to achieve the target of 52000, which should bring the Mythtanic II back to the surface. And guess what? If you were watching the show, you would notice that it took Adam and Jamie just about 27000 balls to bring their boat up — practically right on target. Whaddya know, physics works!