Warning: Science Content

Hidden Nasties

October 3rd, 2010 No Comments

When we go to full scale… who knows what’s going to happen?
Tori Belacci

Well, I’ll mostly ignore the gross biology stuff from this episode, as it’s not my area of expertise. I definitely prefer to pay attention to the destruction of automobiles.

I enjoy it even more when I get to talk about the Buckingham Pi Theorem. Non-dimensional numbers are my favorite kind of numbers.

“But Ryan,” you say, “what’s all of this non-dimensional non-sense?”

Well, after their first test failed, you’ll kindly recall that Tori did some small scale tests with a 1:12 model of the Lamborghini. However, instead of using a model that weighed 1/12 of the actual car, they went all the way down to two pounds. And they scaled 50 mph down to 14 mph. Where did those numbers come from?

This is where non-dimensional numbers come in. If an event has the same non-dimensional numbers as another situation, they two situations should be indistiguishable. This is the foundation of scale modeling. If you can reproduce the relevant non-dimensional numbers, a small scale test should replicate a large scale test, only much cheaper and maybe safer.

For more details, I’ll refer you to the Wikipedia articles on dimensionless numbers and the Buckingham Pi Theorem.

Okay, so let’s think about the important characteristics of the car’s motion and interaction with the water. I’ll go ahead and list what I would consider the five most relevant physical parameters of the problem:

$L_{c}$ – length of the car
$V_{c}$ – velocity of the car
g – gravitational acceleration
$\rho_{liquid}$ – density of water (or whatever liquid we’re trying to skip)
$m_{c}$ – mass of the car (assume for simplicity, it’s distributed consistently throughout the car)

We expect there to be a couple of dimensionless numbers: $\frac{V_{c}^{2}}{L_{c} g}$ (square of the Froude number) and $\frac{m_{c}}{\rho_{liquid}L_{c}^{3}}$ (sort of a specific gravity). We want them to be the same in both the large and small scale tests. There are a couple things we’re stuck with. Since the car is scaled 1:12, it is, let’s say 1 ft (0.3 m) long instead of 12 ft long. And gravity isn’t going anywhere, so we’re stuck with $g=9.81\:m/s^{2}$ . Maybe we could use a different liquid, but we’ll stick with water for now, so $\rho_{liquid}$ . So that only leaves our new speed and new mass (note, 50 mph = 22 m/s and 2600 lbm = 1200 kg):

$\left(\frac{(22 \: m/s)^{2}}{(3.6 \: m)(9.81\: m/s)}\right)_{full\: scale}$ = $\left(\frac{V_{c}^{2}}{(0.3 \: m)(9.81\: m/s)}\right)_{small\: scale}$

and

$\left(\frac{1200\: kg}{(1000\: kg/m^{3})(3.6\: m)^{3}}\right)_{full\: scale}$ = $\left(\frac{m_{c}}{(1000\: kg/m^{3})(0.3\: m)^{3}}\right)_{small\: scale}$

Solving for the small-scale values car gives $V_{c} = 6.4 \:m/s =14 \:mph$ and $m_{c}=0.7\:kg=1.5\: lbs\approx 2\:lbs$ . What do you know? That was the velocity and mass they used! That means they used science! And since they had no explanation of why they did what they did, they left me something to talk about. FTW!

Now, I must admit, I was slightly surprised/impressed that the team got as much right (mass/speed) as they did. But considering they were in the business of special effects, it is a no-brainer that they know these ins and outs. If you want to make the filming of a miniature disaster look like a full scale one (before the miracles of CGI came along), you better know a thing or two about dimensional analysis.

On the other hand, there are a few more potentially important variables that go into the physics of this particular problem, when we talk about the car skipping on the water. If we included the viscosity and surface tension of the water, or air resistance, etc, we would find it almost impossible to duplicate all of the dimensionless numbers in the small-scale (it would have to involve some exotic liquids and gases). But I think we hit the most important numbers for this situation, which is probably why the small-scale tests looked so similar to the full-sized tests.

Mythssion Control

September 24th, 2010 No Comments

Satisfying ding isn’t it?
Adam Savage

It is, it is. But not as satisfying as them crashing cars to debunk Jamie’s reckless statements about car smashing from a few (episodes? years?) back. Ok, it wasn’t that bad. However, I think it doesn’t take much reflection to realize that the collision of two cars going 50 mph in opposite directions is the same as one car crashing into a stationary car at 100 mph, or one car crashing into a rigid wall at 50 mph. As much as I liked the follow-up, I got kind of tired of Adam expressing his amazement and confusion of this fact.

Although they didn’t wade too deep into the math, I was also impressed that they noted that when a car doubles its speed, it quadruples its kinetic energy, thanks to $(\frac{1}{2}mV^{2}$ ). I also wish they dipped into the math a little more when they did their very clever small scale test with pendulums of clay sandwiches. Then again, if they did, I’d have nothing left to do…

So since they are released from a height with zero velocity and swing down to the zero height from which we are calculating potential energy, it is safe to say that the pendulum cars are converting all of their potential energy into kinetic energy:

$mgh = \frac{1}{2}mV^{2}$

When doing these sorts of calculations, especially when comparing relative magnitudes with no need for precise quantification of the actual velocity, it’s useful to use unity as a value for unknowns that both objects have in common. For instance, I’ve gone ahead and deemed the height of the 2X velocity car is unity (making the height of the speed X car $\frac{\sqrt{2}-1}{\sqrt{2}}$ ). Once we cancel out the masses, we have for speed X:

$V_{X} = \sqrt{2g\frac{\sqrt{2}-1}{\sqrt{2}}} = 0.77\sqrt{g} = X$

and speed 2X:

$V_{2X} = 1.41\sqrt{g}$

So the resulting 2X is in fact:

$V_{2X} = 1.41\sqrt{g} \frac{X}{0.77\sqrt{g}} = 1.85X$

Not quite twice as fast as the lower pendulum, but we’ll let them slide.

Knocking Socks

Again, I have to commend the team for tackling an old issue and its associated lingering questions. However, while they were thorough in many respects, I don’t think they were thorough in one of the more critical respects. The force of removal tests on Tori, for instance, were really neat. They appeared well-conceived and executed. And I have to admit, I think they’re right about knocking the socks off someone just wearing socks. However, in my experience, most boxers do not just wear socks. They also wear a little something I’ll call “shoes”.

The problem is that the team did not think hard enough about the mechanism by which socks are removed. To the extent I can put it in plain English: The removal of socks hinges on the inability of the accelerating foot to apply sufficient force to the socks to accelerate them at the same rate.

Imagine a worst-case scenario world, where while we’ve minimized friction force between the foot and sock (maybe some sweat would help in this respect), the friction between the sock and shoe is infinite. So in order for the shoe (and sock) to stay on the accelerating foot, the mass of the sock and the shoe must be accelerated at a comparable pace. This would take much more force (seeing as shoes weigh much more than socks, and all). So, perhaps a bit counter-intuitively, as long as the shoes weren’t tied on overly tight, and there was considerable friction between the socks and shoes, they would have had a much easier time knocking Buster’s socks off if they had just given him some appropriate footwear.

Using a Thermistor

September 20th, 2010 No Comments

A thermistor, as the name indicates, is nothing more than a resistor, whose resistance value is sensitive to temperature. Like a normal resistor, it has no polarity, that is, you can’t hook it up backwards. Depending on the type of thermistor, the resistance may increase or decrease with increasing temperature.

To use a thermistor to measure temperature in a project, you must monitor the resistance across the thermistor. Generally, this is done by connecting a resistor of a known value in series with the thermistor, bridging a higher voltage (say, + 5V) to ground. To eliminate confusion, let’s just say that the thermistor is the component connected to ground. Then, an analog input can be connected to the middle of the thermistor and resistor, measuring the voltage across the thermistor.

Knowing that $V_{therm}/R_{therm} = V_{res}/R_{res}$ , the resistance of the thermistor can be measured to be $R_{therm} = V_{therm}R_{res}/V_{res}$ . If the supply voltage is indeed 5 volts, then $R_{therm} = V_{therm}R_{res}/(5-V_{therm})$ .

That just leaves converting resistance to temperature. This conversion will depend on the particular thermistor and the information supplied. If given a B or $\beta$ and $R_{0}$ value (usually the resistance at 25 C), the equation for temperature as a function of resistance can be written as:

$T = \frac{\beta T_{0}}{\beta+T_{0}ln\frac{R}{R_{0}}}$

For instance, if you have a $10 k \Omega$ , $\beta=3977 K$ thermistor, the hard-coded equation for temperature (in Kelvin) would become:

$T = \frac{1}{0.003354+0.0002502ln\frac{R}{10,000 \Omega}}$

You may find yourself asking what value of resistor to use. Well, nearly anything will work, but you can make some smart decisions based on the thermistor and the temperature range you expect to be operating at.

I’ll get back to you on how to be smart about it at a later date; be sure to check back…

“Informed” Consent

September 19th, 2010 No Comments

The best argument against democracy is a five-minute conversation with the average voter.
Winston Churchill

I really try not to be a cynical person, but I’m afraid it’s in my nature. I’ve either come by it genetically, or I’ve cultured a cynic’s attitude over the years. I think, if channeled in a healthy manner, cynicism manifests itself as skepticism, which is a valuable tool in a scientist/engineer’s toolbox.

Unfortunately, every time I think I have managed to rehabilitate myself just a skosh, I come across something like this editorial by Dr. Alicia Fernandez (Annals of Internal Medicine, Vol. 153, No. 5), writing about one of the short-comings of informed consent:

The consent form used in this study was written at a 12th-grade level — far above the 8th-grade mean of both the population at large and recommended standards.

As the esteemed Dave Chappelle, impersonating Lil John would say: What!?

Did she just state that the mean education level is at the junior high level? In America? What about the mode? The median?

Obviously, if I’m writing in a blog like this, I care about education and learning. Am I supposed to be motivated by a statistic like that? It tends more to cause despair. Is the problem in our schools? Is it a defect in the social fabric? Is it even a real problem? 100 years ago, I’ll bet the mean education level was much lower.

I really don’t know what to think about it. I do know that I am sadder for learning that. And a little despondent…

Dive to Survive

September 10th, 2010 No Comments

That’s crazy! I mean, it’s the same amount of material that it’s passing through.
Tory Bellaci

There was a lot to like about this episode. Anytime you can get a remote-controlled car to drive through a hail of gunfire, I’m watching. Unfortunately, some of the groundwork they laid with their terminal ballistic research was a bit shoddy.

Terminal Ballistics

My biggest problem is that for some reason, they think you can evaluate armor by looking at a projectile’s depth of penetration (DOP) into an essentially semi-infinite target, and then use that distance to define the final armor thickness.

They repeated this mistake time and time again throughout the phone book part of this episode. First they shot through their modified car door, backed by 10 phone books. They saw that the M14 almost made it to the back of the first phone book of their stack. Then they jumped to the conclusion that it would have been stopped without the other nine books backing it up. Then they shot through a bunch of phone books backing a car window, and they observed that it only went through three phone books. So they only need three phone books!

Wrong on both accounts.

While the observation that the glass does some work in blunting the projectile is probably true*, I’d bet a pound to pennies that if they had shot the window backed by only three phone books (as opposed to their setup where the window was sandwiched), they would have also seen the bullet fly straight through. This is because the integrity of the back of the target is enhanced by additional material behind it. You are almost never going to “just stop” a projectile; you will always need some amount of armor behind it with enough strength left to “catch” the projectile. Would it have killed them to actually try this to confirm their speculation?! They could have busted the myth they just created!

If I think of a better example, I’ll give it, but for now, think of the simple case of trying to stop a baseball with paper. If I stack a bunch of phone books up and hurl a baseball at them, it will bounce right off. Following the Mythbuster’s logic, that means I’d only need one sheet of paper to stop it. Of course, that’s not true; a baseball would rip a hole right trough a piece of paper without the help of all the pieces of paper behind it.

Let’s use a quick diagram to drive the point home. Here, the three phone books are contributing to stop the bullet. In fact, the bullet stops in the second phone book. However, removing the green phone book means there’s not enough structural integrity behind the yellow phone book anymore. Repeating the test, the bullet would have no problem tearing through that last bit of yellow book and going along its merry way.

Really quick aside: I don’t want to mislead anyone into thinking that the solution is to add one more phone book than you think you need and you’ll be fine. It depends on a lot of variables, not least of which is the thickness of the phone book in question. It’s hard to know a priori what you need; you mostly just have to shoot it multiple times to see how it performs.

When evaluating a specific armor (let’s concentrate on homogeneous armor, made from a single material), engineers define V50 for a given projectile. That is, the speed at which 50 percent of the projectiles will be stopped, and 50 percent will defeat the armor. And the funny thing is, when a projectile defeats armor, it will virtually always be carrying a head of steam behind it (as we saw when they shot their sandwiched window). The projectiles that go through at V50 are still carrying a significant portion of their velocity with them. In the same vein, projectiles that are stopped will only go through a fraction of the armor (maybe a large fraction, but not all the way to the back of the armor).

* Armor designers are actually acutely aware of the power blunting a projectile can have on stopping power. So are projectile designers, which is why we have armor-piercing rounds, which aim to retain a streamlined profile.

Transmission of Shock Waves

I mostly liked what they did with the experiments to see whether it was safer to be underwater than not when an explosion goes off. But I think they may have come to the right conclusion for the wrong reasons. Not that their full-scale experiment was terrible, but I don’t think it was very fair.

When they set up their burst-discs, they spaced them from the explosive at intervals (5, 10, 20, 50 feet) and above and below the water (5 feet above, 5, 10 feet below the water line). However, their explosives all seemed to be located approximately 5 feet above the water themselves. A little geometry shows how problematic this can be. When the above water distance from the explosion was 5 feet, the below water discs are actually 11 and 16 feet respectively. The disparity is less dramatic the further you get away, so I think their last test with TNT is pretty indicative of the protection water affords.

But why does water afford protection at all? Well, it’s actually pretty simple. Anytime a wave front reaches a transition of materials (media), two things happen. Part of the wave is transmitted, and part is reflected (and of course, energy is conserved!). The degree to which the wave is transmitted has to do with how similar the two materials are (or aren’t). So there’s nothing special about water. It’s just not air. If the explosion occurred underwater, the reverse would be true, and you’d be exposed to less overpressure if you were out of the water!

SKS Rifles and Harmonics

August 18th, 2010 No Comments

Resonant frequency is the frequency at which an object will begin to vibrate in excess of the energy that you put into it.
Grant Imahara

Ugh… just typing that makes me cringe. Grant said it on the Fire vs. Ice episode (no. 121) that aired May 27, 2009. Hopefully, any statement about energy being created out of nowhere should automatically activate the B.S. detector in your brain. If you don’t have a B.S. detector, let me know and I’ll send you some blueprints.

Anyway, of course, an object vibrating at its resonant frequency will not vibrate in excess of the energy you put into it. That would violate all sorts of physical laws. What it will do is accumulate the energy being put into it, amplifying its response. Many times, if the forced input continues to match the natural frequency for a good length of time, the energy will tear the object apart; just like Grant’s example of the wine glass.

I wanted to do a swing-set example, but I saw Wikipedia already took care of it, so I won’t duplicate their effort.

SKS rifle pin orientation The other problem with their experiments is that they often aligned their guns perpendicular to the sound waves that were hitting them. Acoustic waves are longitudinal waves, which means they transmit energy in the same direction in which they move. In order to “accidentally” fire a gun, the pin needs to line up with the wave propagation direction. Oftentimes (most egregiously, in the sound chamber they used), the gun in question is lined perpendicular to the acoustic wave propagation, meaning there is no impetus for the pin to move along its axis. Think about getting a Shake Weight going… you shake it along the axis, right? Perpendicular shaking would be all sorts of counter-productive.

What I would have liked to see is for the ‘busters to mount the gun to some sort of programmable shaking device that would vibrate the gun along the firing pin’s axis (i.e. its barrel). Then they could have figured out what frequency/amplitude it took to get a misfire. Then, the final step would be to figure out if they could replicate those circumstances with sounds waves in a car. Done and done.

22,000-Foot Fall

August 4th, 2010 No Comments

Eh..you know, it’s not exactly an equal and opposite reaction… so what do you think?
Kari Byron dropping knowledge on our heads

I have to say, I was a little disappointed by this experiment (22,000-Foot Fall, December 13, 2006). I appreciate that it was very hard to execute, and there are no second chances when dealing with such an elaborate setup, but I would have liked them to apply a slightly more critical eye to themselves. For instance, they spend the whole show assuming they wanted the explosion to go off just as the poor gunner reached the top of the roof.

But why? Maybe the pressure wave would been just the right amount to slow him down without killing him a little further above the “train station”. There are codes out there (I’m thinking BlastX, in particular) that will accurately predict pressure/impulse as a function of distance and explosive type/mass (which I would do, but it’s a restricted code). With this info, you could not only estimate a safe distance to hit the blast wave, but also how much deceleration to expect $(\Delta V=I/m)$ . Solving the mystery of the airman’s survival needs to start there, since nobody can afford to repeat a bunch of drop tests with 500 lb bombs.

Terminal Velocity

Not sure how Adam figured out how long it takes to reach terminal velocity (he reported 5.5 seconds) or how high he needed to be to make that happen (500 ft). Typically, terminal velocity simply means that a falling object’s drag force equals the force of gravity:

$mg = 0.5AC_{d} \rho V^{2}$

which is all fine and well, and almost looks like it would help us get to 5.5 seconds (or whatever the time actually is). The only problem is that the non-dimensional drag coefficient, $C_{d}$ , is not only a function of how the paratrooper falls (I doubt he will naturally assume a typical paratrooper’s position), but also his velocity. And it is a function in unknown ways that can only be determined experimentally (or computationally, I suppose). So we can’t even do an integration without more data or assumptions. I’d be fascinated to know if Adam had something up his sleeve, or was in over his head.

Ping Pong Rescue

August 3rd, 2010 5 Comments

We never believe the math on Mythbusters.
Adam Savage

Well, Adam, that’s what we’re here for. After just a little explanation, the most difficult thing about buoyancy is spelling it correctly. The Mythbusters gave a couple great examples in a single episode (Ping Pong Rescue; November 3rd, 2004) with ping pong balls and helium-filled balloons. The show was, as usual a little short on explanation; we’ll try to remedy that here and now.

Buoyancy Review

If you’re going to take away one thing from this post, it should be that it’s really quite straightforward to think about floating vs. sinking. Just visualize the object in question. Then imagine the amount of water it is going to displace (generally equal to the object’s volume). If the object weights less than an equivalent volume of water, then the object will float; if it is heavier, it will sink. This also applies to objects in air, it’s just that

there aren’t a whole lot of objects that are “lighter than air” and
the “weight” of something becomes a little hazier in that regime (negative weights are sort of counter-intuitive)

Hence, both the ping pong and balloon myths are both myths about buoyancy!

For an object that sinks, its effective underwater weight (if you were to put it on a scale while submerged) would be its above water weight minus the weight of displaced water. This is why the guys ended up buying too many ping pong balls. Not that it was a bad way to cover their butts, because by buying enough ping pong balls to lift the “dry” weight of the boat, they were addressing the upper bound of their ping pong needs. If the boat had been crushed to an very small volume, you would have needed a lot more ping pong balls. Because it displaced quite a bit of water, it had a relatively low “wet” weight which the ping pong balls needed to offset.

Not to spend to much time on this point, but how about a quick example? For instance, if you pulled an iceberg out of the ocean and weighed it, with Adam’s method you would estimate that you’d need a gaggle of ping pong balls to float it. Of course, it will float without a single ping pong ball, which hopefully stresses that buoyancy should be taken into account for any such calculation. In fact, to determine if any assemblage of objects will float or sink, simply add the volumes and weights of the objects and see if that is heavier than the total equivalent volume of water.

To the math!

Ping Pong Balls

Suppose a ping pong ball has a mass of 2.7 grams (0.006 lbs) and a diameter of 40 mm (1.6 in). Then an equivalent volume of water would weigh $V\rho g=(1.96\text{ in}^{3})(0.036\text{ lb/in}^{3})=0.071\text{ lbs}$ , for a net buoyancy of 0.071 – 0.006 = 0.065 lbs. Just as Adam said, it would take $\frac{1\text{ lb}}{0.065\text{ lbs/ball}}\approx 15\text{ balls}$ to buoy the 1 pound of shot. The difference is, I trust the math.

For their 3,500 lb boat, Mythtanic II, let’s take a stab at how much water it displaced. Since it took 26,000 ping pong balls:

$\begin{array}{rcl} m_{total}g &=& V_{total}\rho_{water}g \\ m_{boat}+m_{balls} &=& ({V_{boat}+V_{balls}})\rho_{water} \\ 3500\text{ lbm}+(26000\text{ balls})(.006\text{ lbm/ball}) &=& ({V_{boat}+(26000\text{ balls})(1.96\text{ in}^{3}/\text{ball})})(.036\text{ lbm/in}^{3}) \\ V_{boat} &=& 50600\text{ in}^{3}=29 \text{ ft}^{3} \end{array}$

So, the boat displaces about 29 cubic feet of water (whether floating or sinking). Any boaters out there who can tell me if that number seems reasonable?

Packing Spheres

Adam spends a little bit of time talking about how efficiency ping pong balls pack together. Of course, he doesn’t ask how efficiency can ping pong balls pack together, but let’s answer that question, anyway. Unfortunately, while they show him measuring up the aquarium and punching into his calculator, they don’t actually shed any light onto the numbers he got, except to say that he got a 51% packing efficiency.

A great number of people have looked at this problem before us, and when it comes to packing spheres, there are a couple of methods of arranging them that will net 74% packing efficiency (see Wikipedia: Close-packing of spheres). Of course, one has to be careful, and consistent on how the spheres are packed, and surely, when the boat is floated, we cannot count on any sort of optimal packing efficiency. So in that sense, Adam’s more empirical method probably gives a better sense of what the actual packing efficiency will end up being (and we’ve already checked its reasonableness by comparing it to the maximum possible packing efficiency).

Balloons

While the crew takes a reasonable approach to trying to figure out how many balloons they’ll need, they are at a disadvantage to Adam and Jaime inasmuch as the volume (and therefore buoyancy) of a ping pong ball is very consistent and predictable, a balloons buoyancy will vary quite a bit depending on how well its inflated. That is, as the volume of each balloon varies (depending on how “full” its filled), its buoyancy will vary. So it is no surprise that they do not have a lot of success predicting how many balloons they would end up needing. Furthermore, the harness and rope needed to connect the balloons together and to the child are sure to increase the final balloon count. Judging that ultimately, they had the balloons and time needed, they were aware that their estimate may have undershot the eventual requirements.

Flowing Ping Pong Balls Down

This could be interested to look at, but I’m running out of steam for this one show. Anyone interested in examining the physics of moving the balls down to the boat just needs to estimate the velocity of water needed to provide a force that will cancel (or exceed) the force of buoyancy (hint: both of these forces are essentially independent of depth).

Even More Science

Objects experience buoyancy when they are immersed in a fluid subject to hydrostatic pressure. That means, as you move down in the fluid, the pressure increases, due to the extra weight of all the fluid above it. Water is a really straightforward example of this, because it’s essentially incompressible, so it’s easy to calculate pressure as a function of depth; it increases linearly as depth increases: $p=\rho gd$ . The atmosphere, being compressible, can be a little more complicated, especially since we start at the “bottom”, so we’ll concentrate on buoyancy in water.

Of course, when Adam and Jaimie imploded the ping pong balls in their pressure chamber, the balls didn’t start floating (because ping pong balls don’t float in “regular” air). So it’s not just increasing the static pressure that makes objects float. It’s the incremental increase of pressure with depth that exerts an upward force. Basically the surface on the bottom of the object is experiencing higher pressure than the surface on the top, leading to a net upward force.

Look at the simple example of a submerged cylinder. The top surface of the cylinder is experiencing a pressure of $\rho g d$ , while the bottom surface is experiencing a pressure of $\rho g(d+h)$ (the sides of the cylinder are irrelevant, as the pressure on them acts perpendicular to gravity and ultimately cancel each other out). Or, doing a force balance in the z-direction, the net force on the cylinder is

$\begin{array}{rcl}F_{net} &=& F_{bottom}-F_{top}-F_{weight} \\ &=& A \rho g (d+h)-A \rho g d-mg \\ &=& A\rho gh-mg\\ &=& V\rho g-mg\end{array}$

A-ha! We have a net force that depends on the weight of the cylinder, the weight of an equivalent volume of water, and it is independent of the depth!

If you care to imagine it, this can be generalized to an arbitrarily shaped object simply by building the object out of a very large number of very small diameter cylinders (better yet, square prisms so there are no gaps in between!) that extend from the top of the object to the bottom. The sum of the forces will come out the same (assuming you use skinny enough cylinders).

Hidden Nasties