While I was writing my last Mythbusters blog post, I realized that it'd be really useful to have a "quick and dirty" plotting program — one that just takes the output of another program and plots it, no questions asked. Actually, let me rephrase that: I knew it would be useful, it just never occurred to me that I don't have one. Most of the common plotting programs people tend to use, like GNUplot, expect to be configured with the format and syntax and display options and all that junk that you don't really care about when you just want to turn numbers into a picture.

If you have GNUplot installed, here's a little script to do just that:

#!/bin/bash

using=""

shopt -s extglob
if [[ "$1" == +([[:digit:]])*(,+([[:digit:]])) ]]; then
        using="using ${1//,/:} "
        shift
fi
shopt -u extglob

plotopts=""

if [[ "$*" == *title* ]]; then
        next_is_title=""
        for word; do
                if [[ -n "$next_is_title" ]]; then
                        plotopts="$plotopts title '$word'"
                        next_is_title=""
                        continue
                fi
                if [[ "$word" == "title" ]]; then
                        next_is_title="1"
                else
                        plotopts="$plotopts $word"
                fi
        done
else
        plotopts="$* title 'STDIN'"
fi

gnuplot -p -e "plot '-' ${using}${plotopts}"

Save it as plot, make it executable, and put it in your $PATH, and then you can run things like this:

calculation | plot

and you get a picture. calculation can be any program that produces output in the form

x1 y1
x2 y2
...

You can also pass any GNUplot options, like

calculation | plot using lp

to plot with lines and points, and there's a special case in the script to give the graph a title, like

calculation | plot title 'My Graph!' using lp

It would probably make sense to add xlabel and ylabel options, but... I didn't. I finished using the script before I needed them.

The Hyneman has a new line to add to his resume:

I kid, of course. Still, the Mythbusters' demonstration of a motorcycle riding on water in last week's episode was seriously cool. I have to admit, I didn't expect it to work! But physics says otherwise — both the experiment, as we saw last week, and as I'm about to show you, the theory.

Underneath all the complicated fluid dynamical effects, a motorcycle can ride on water for much the same reason a speedboat does: it presents an upward slanted surface for water to bounce off of. For a speedboat, that surface is the hull; for a motorcycle, it's the bottom of the wheels, especially the front wheel. But wheels are complicated. It'll make things easier to start by examining the behavior of a speedboat-like slanted flat surface as it moves through water.

Conveniently, I've already done that calculation. It comes from another episode of Mythbusters a few years back, when they tried skipping a car across a lake. The calculation went like this: water molecules hit the lower surface of the car with a horizontal momentum and bounce off in a downward direction, with a final momentum . I'm assuming the collision is elastic (just for simplicity; in reality, it's not) so that the water molecules leave with the same speed they came in. They experience a change in momentum of

In a time interval , the surface pushes through a mass of water equal to — that's density () times width () times height () beneath the water times length ().

Dividing by gives the force exerted on the water,

and the corresponding reaction force on the surface,

In my other blog post, I showed that this force is perpendicular to the surface (but it also makes sense if you just think about it). That takes care of the direction, and we can find the force's magnitude using the Pythagorean theorem and the identity (check it), to write this as

where is the angle between the water's impact direction and the surface.

Forces on a wheel

A motorcycle's wheel differs from the undersurface of a boat in one important aspect: it's curved. That just means the force exerted by the water will be different at different parts of the wheel.

Consider a small wedge of angular width at an angle , as shown in the image above. It experiences a force in the direction

and it spans a height

Putting this into the equation for force on a surface, we get

This we can integrate over the range of where water is hitting the wheel, from , straight down, to , which is the direction to the point where the wheel meets the surface of the water. It works out to the slightly messy expression

representing the upward/backward force on the wheel due to its deflection of water.

Analyzing the motion

In this simplified model I've constructed, the motorcycle is subject to two forces: that of the water bouncing off the front wheel, given by the above formula, and gravity, given by good ol' . (I'm ignoring the effect of the back wheel, so this is going to be more like a motorized unicycle, but it gets the main point across.) So we can use Newton's second law to get some equations of motion for the bike. In the horizontal direction,

and in the vertical direction,

This is a messy enough set of differential equations that they can't be solved analytically. But it's not hard to find a numerical solution. I crunched the numbers using the GSL ODE integrators, so let's look at some results!

Numerical results

Finding a specific numerical solution is a matter of setting values for the constants , , , and , and initial values for , , , and , and just letting the integrator run. Depending on the relationships between these values, the solution can behave in a couple different ways.

If the wheel's initial submersion in the water, , is large enough, or it's dropping quickly enough, and if the wheel's forward velocity is fast enough, we'll get something that looks like this:

This shows the wheel bouncing out of the water, because the force exerted by the water against the wheel is strong enough to overwhelm gravity. Once the submersion depth makes it back to zero, the differential equations no longer apply — that's the end of the solution shown in the graph. But then (since I'm pretending the motorcycle is actually a unicycle), projectile motion takes over and the wheel flies through the air until it hits the water again.

Each time the wheel bounces off the water, it loses some forward velocity. A loss of forward velocity translates to a loss of upward force, so eventually the wheel reaches a point where it no longer experiences enough force from the water to counterbalance gravity. When this happens, we get a graph that looks like this:

This plot is not to be trusted too much, because the whole model of water bouncing off the wheel becomes invalid once the wheel gets too deep, but the main point is clear: the thing sinks. Clearly, while it is possible to ride a motorcycle across water, there's a limit to how far it will go.

Putting it all together, here's a combined plot showing the bounces, free flights, and eventual sinking:

This was generated assuming a total mass of , an initial depth of zero (the wheel is right on the water's surface), a vertical rate of fall of (semi-random choice), and an initial speed of , just under . I'm guessing these values are roughly similar to those in Jamie's actual last run over the lake on the motorcycle. You can see that the graph drops down to the bottom just over in — that's , which is not all that far off from the actual distance Jamie was able to ride in the show, about Considering all the other things that are going on that my model doesn't account for (principally extra drag from the water spray and the back wheel, and rotations of the motorcycle), that's not that far off!

As you might be able to tell by the lack of activity on the blog, I've been pretty busy the past couple weeks. Which makes it kind of hard to write about Mythbusters at my usual level of detail.

Fortunately, Rhett Allain has already done it for me. He analyzed last week's episode to figure out whether air pressure is enough to pick up a manhole cover. And it is: an air pressure of , normal atmospheric pressure, applied to a manhole's surface area of (seriously, click the link), gives a force of — over eight thousand pounds! That's way more than enough to pick up a 300-pound manhole cover.

Of course, that would only happen if the manhole cover had a near-perfect vacuum above it. Perhaps if it were a manhole on a spaceship. But that's not the situation on Mythbusters. (How cool would that be? Mythbusters in space... but I digress.) The manhole cover under an Indy car has air on both sides; the pressure of the air above is reduced, though, due to Bernoulli's principle.

Given that the difference in height, , is probably negligible, and the air underneath the cover isn't moving (), that gives

The Mythbusters only measured 37 pounds of force when they ran their car over the manhole cover at , which corresponds to an air speed of , about 61 miles per hour, if the pressure drop is in fact entirely due to the Bernoulli effect. That's Rhett's blog post in a nutshell.

Of course, one might wonder, is the pressure drop actually due to the Bernoulli effect? I'm not an expert in fluid dynamics, but I suspect there's a lot more going on. When a fluid (like air) flows adjacent to a surface, it has some tendency to "stick" to the surface, measured by its viscosity. The region in which the fluid's flow is strongly affected by its viscosity is called a boundary layer. It's often characterized by reduced speed of the flow near the surface, as in this image from Wikipedia,

but more complicated things, like turbulence, can happen as well.

In the Mythbusters' Indy car experiment, there are actually two surfaces, the car and the road. It stands to reason that the car will drag the air next to itself along at or so, while the road will "drag" the air next to itself along at zero speed — that is, it tends to slow the air down. Each of these two surfaces will attempt to create its own boundary layer of air. If the two boundary layers can both fit in the space between the car and the road, then they will probably do so, but if they don't fit, messy things happen.

Here's the formula for the thickness of the boundary layer for fluid flowing over a flat plate:

where is the kinematic viscosity, is the distance "downstream" from where the surface begins, is the speed of the fluid far away from the surface, and are constants roughly of order 1. Wikipedia gives and , so I'll just run with those values. For air, . If we're simulating the Indy car in the Mythbusters' experiment, it's reasonable to assume is not more than , because nothing makes the air move much faster than that. And finally, maxes out at the length of the car, around .

Plugging the numbers in gives for laminar flow, which fits well within the 7/8 of an inch clearance they mentioned on the show — but for turbulent flow, I get , which does not. So I would expect that the airflow underneath an Indy car is strongly affected by boundary effects.

Now, why is this important again? Bernoulli's principle applies best when viscosity can be neglected; essentially, it works for frictionless fluids. But in a boundary layer, viscosity plays a large role in determining the fluid's behavior. Evidently, viscosity can't be neglected for the air under the Indy car, and that means the application of Bernoulli's principle is somewhat suspect.

I know, I know, this blog post is very late! If I'm going to post about a Mythbusters episode, I usually try to do it before the next one airs. But this topic — calculating the circumstances under which a rope will pull a person's leg — turned out to be pretty complicated. Which of course made it impossible to give up on.

And after days of toil, I think I finally figured it out! If you're the kind of person who finds complicated physical interactions fascinating, you're going to love this post. The math isn't too complicated; if you know what a differential equation is, you'll be fine, but the physical reasoning is something you could probably spend a while wrapping your head around.

The setup

On the Deadliest Catch-themed Mythbusters episode which came out last week, one of the myths being tested was that if a person steps in a coil of rope, and that rope is attached to a crab pot (trap) that gets dropped over the side of the ship, the rope will be pulled tight enough around the person's leg that it will drag them overboard, and down to the ocean floor, along with the trap. Now, whether this works can depend on how the rope is coiled, which direction it's pulled, whether it gets knotted, and so on, but at its core, the myth is about friction. The frictional force between the rope and the leg has to be high enough to keep the rope from slipping while it pulls the person overboard. And friction doesn't need knots. I've managed to develop a model for how friction affects the behavior of a rope coiled around a cylindrical object.

Here's how it works: Imagine a very short, infintesimally short, segment of the rope. It's subject to four forces: tension pulling to the left, tension pulling to the right, the normal force , and the frictional force which opposes the motion of the rope.

Let's decompose these forces in cylindrical coordinates, using the unit vectors and , parallel and perpendicular to the rope respectively. The normal force and friction will be in the radial and tangential directions, respectively, but the tension at each end points slightly off-axis. I've marked out the angles in the diagram to show how to decompose the tensions:

Time to move on to Newton's second law. Here's the radial component of :

and the tangential component:

From here, there are two cases to consider. If the rope slips, we have , and if not, we have an inequality, .

Slipping rope

For simplicity, let's take the equality first, in which we assume that the rope is slipping. We can put all these equations together by solving for and and using substitution. That gives

Since and are the tensions at infinitesimally close positions, their difference will be an infinitesimal quantity, . Also, the mass of the segment of rope will be an infinitesimal , which is equal to where is the radius of the coil and is the rope's mass per unit length. Putting all this in, we get

I've set and then ignored the second-order term proportional to . This leads to the differential equation

which is the final formulation of the model. It's a general model, which can apply to several different situations; to get a specific solution, we just need to characterize the rope's acceleration by postulating some form for the components of the acceleration, and .

Example solution: the fixed cylinder

To get a feel for this model, I'm going to investigate a simple situation where the cylinder stays in place and the rope slips around it. The rope's acceleration will be in the tangential direction, , setting and .

With these conditions, the solution to the differential equation is

where is the total angle by which the rope is wrapped around the leg — for each complete loop. This solution has the form

where is the tension at — where the rope just starts wrapping around the cylinder — and is a constant, the asymptotic value of the tension after it's wound around the cylinder many times. Here's a sample plot:

Wait a minute — a negative asymptotic value? Does it even make sense for tension to be negative? Of course not, and here's where a bit of physical intuition comes into play. We know our function starts out positive, , and, if the rope is wound enough times, it ends up negative, . That means that for some intermediate angle, the function hits zero. Physically, that's the point at which the friction between the cylinder and the rope has "accumulated" enough to cancel out the external force applied to the rope. Any further along than that critical angle, the rope is slack, and the function ceases to be applicable.

But wait! We can actually solve for the critical angle at which the tension hits zero.

Let's do a few sanity checks on this formula:

• Greater initial tension corresponds to a greater critical angle. Which makes sense because the harder you pull, the more of the rope you'll be able to pull taut.
• The critical angle decreases as gets larger — which, again, makes sense, because with a rougher pole, your pull won't be felt as far along the rope.
• In the opposite limit, as goes to zero (a frictionless pole), the critical angle becomes . Now, if you take this criterion and rearrange it, you get , which is just for a rope on a frictionless surface! As it should be. If this complicated model that handles friction gave the wrong answer when there is no friction, it wouldn't be much good at all!

That last item in particular also suggests a physical interpretation of this formula: it's effectively an equation of motion, filling the same role as . Whereas we'd ordinarily use Newton's second law to calculate the motion of a rope subject to a given force, for the rope sliding around a pole (or leg) we can use this formula instead, and it will take care of the friction and the changing tension over the length of the rope. This works because at the end of the rope, where it "stops" winding around the cylinder, its tension is zero — it's not pulling anything. (Except perhaps some additional length of rope, but I'm going to assume that's negligible for simplicity.)

Non-slipping rope

The analysis if the rope doesn't slip is nearly the same, except that the coefficient of friction is instead of , and instead of an equation, we get an inequality from .

Following the same procedure from before, this becomes

OK, this is new. Everybody (well, everybody who has a geeky streak) knows about differential equations, but when have you ever seen a differential inequality? They're not that common. But it's not really that hard to understand. In order for the inequality to be satisfied, the rate at which tension changes as you move along the rope (as in the previous section, is negative) has to be greater than the particular combination of acceleration and tension that appears on the right. Physically, the left side is related to the strength of friction, the right side is related to the other forces acting on one segment of rope, and for the rope not to slip, friction has to be strong enough to cancel out the other forces at every point along the contact surface.

Example solution: the moving cylinder

It'll be easiest to make sense of this differential inequality by choosing a specific type of acceleration, as we did with the slipping rope. Let's say the rope just pulls the cylinder in the direction of the initial tension, namely the direction, without rotating it.

This is not really realistic, since there will actually be some torque on the cylinder that will also rotate it, but it's close enough for demonstration purposes.

The components of acceleration in this case will be , and .

Substituting these in, we get

There's still the little problem that this is an inequality, not an equation. But I have a plan. First, let's solve the corresponding differential equation; it gives

or equivalently

with . This function gives a value of tension that, if we plug it back into the differential equation, represents the minimum possible rate at which tension can change over the length of the rope.

Now, given that the starting value is fixed at , there's no way the actual tension can dip below the value specified by this function without the rate of change being less than the minimum specified by the differential equation. This follows from the mean value theorem. Evidently the tension in the rope can be no less than the function we found from the differential equation; that is,

This is the condition that must be satisfied for the rope not to slip. It corresponds to the shaded region in this plot:

Now that that's cleared up, let's look at the properties of this solution. Again, the asymptotic value (actually, an asymptotic amplitude of sorts) is negative, which means this function crosses zero somewhere. But now there are those pesky sine and cosine factors, and unfortunately they make it impossible to find a symbolic formula for the critical angle where the tension drops to zero. We can still calculate numerically, we'll just need actual values for all the constants involved. As in the last section, the equation for the critical angle where defines some relationship between , , , , , and , which we can use as a "smart" stand-in for .

Will the cylinder move?

Now the key question is, if you have a rope wrapped around a cylinder (like someone's leg) and you pull it, will the rope pull the person, or will it slip? It all comes down to whether the force of friction is strong enough to resist the other forces applied to the rope. Mathematically, that corresponds to whether the differential inequality in the last section is satisfied (at least up to the critical angle where the tension goes to zero). If the inequality is violated at any point along the rope, it'll slip, but otherwise the rope stays pressed against the cylinder and pulls it.

To figure out whether that happens, we'll need values of , , , , , and . We plug them into the formula, then set on the left side (which should be true at the end of the rope, in this simplified model) and see whether the inequality is satisfied. If if is, that puts us in the shaded region of the graph, where friction is strong enough to keep the rope from slipping. Otherwise, we're in the unshaded region where friction is just not strong enough and the rope slips.

There's just one problem: what's the acceleration? The values of , , and are readily available (they would be measured), and the experimenter can control the initial tension , but we don't know . Or do we? Well, look at this from an outside perspective. The initial tension exerted on the rope, , has to pull the rope itself and the person. If you look at the rope and person as one physical system, a "black box" if you will,

you can ignore the effect of friction and just use Newton's second law, .

which gives us

Numbers: do they work?

Even though this model is unrealistic, chiefly because it doesn't take into account the rotation of the cylinder due to the tension being off-axis, I still want to try plugging some numbers in. Here are my estimates:

• Radius of the leg:
• Density of the rope:
• Static friction coefficient: (maybe a little low)
• Total mass of the rope:
• Total mass of the person:
• External force exerted on the rope: , I figure it's roughly the weight of the crab pot, which was stated to be 800 pounds in the show — though if you ask me, they did not look that heavy. (It turns out not to matter anyway.)

Now yes, I could have used these values for the graph in the earlier section, but I didn't want to give the answer away! Here's what the plot actually looks like:

It's hard to tell there, so let's have a closer look:

The tension hits zero for the first time just past 8, which means that with these numbers, it takes about 8 turns of rope around your leg to pull you. Any less, and the rope will just slip off.

That's kind of a lot. But what if we up the coefficient of friction? is still probably reasonable, maybe even low, for a rope (they're made somewhat rough, after all, so that they don't slip through your fingers). Doubling the friction coefficient gives this plot (already zoomed in):

Now it only takes four turns for the rope to grab you. I wouldn't form a conclusion on this myth based on the math alone, but compared to what Adam and Jamie found, it seems not too far off from reality!

shhh, the computers are thinking :-P

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