Bouncing Bullet

May 13th, 2012 No Comments

Ready for the Naked Avenger!
Tori Bellacci

I was pretty excited that both pieces of this episode actually involved some physics, as opposed to the regrettable (on so many levels) Battle of the Sexes episode.  However, I knew I would have to go to work on this post as soon as Jamie and Adam started equating speed and energy.

They were interested in how a bullet ricochets off of asphalt when fired at different angles.  At a shallow firing angle (16 degrees), they noted that the velocity decreased from 1200 fps to 962 fps.  This is not a huge decrease in speed, but knowing that energy is proportional to the square of velocity, I was not willing to say that it was “not a lot of loss”.  In fact, for a bullet mass M, we can calculate the two energies, before and after impact (note that when you multiple (fps)^2 by mass, you get units of energy):

E_{before}=0.5M(1200 fps)^2=720,000M\: (fps)^2
E_{before}=0.5M(962 fps)^2=462,700M\: (fps)^2

In other words, the energy decreases to 64% of the original value (less than the 75% that Adam later quoted).  Losing more than a third of the energy doesn’t seem like something so easily dismissed.  So I was suspicious our heroes were confusing momentum with energy, and forgetting to square the velocity.

The misconception was cemented by Adam when they moved on to the steeper shot at 32 degrees.  In that case, the velocity dropped from 1200 to 475 fps.  Adam said that this had “lost almost two-thirds of its energy”, but now we know better:

E_{before}=0.5M(1200 fps)^2=720,000M\: (fps)^2
E_{before}=0.5M(475 fps)^2=112,800M\: (fps)^2

It lost 84% of it’s energy!  Quite a bit of energy to give up to ricochet at a symmetric angle!

So while it might seem like nit-picking (and to an extent, it is), there is really an important point here that energy is not linearly related to velocity but to the square of velocity.  This is a concept that is important to keep in mind; energies tend to fluctuate greatly as velocities change, and large velocities result in very large energies.

I just wanted to chime in on the other myth of the blast-aided jump.  I thought they did a pretty good job with this; the nitrogen gun was impressively consistent and the high-speed video was fantastic.  They barely touched on this concept of pressure versus impulse, but going to the anfo was definitely a step in that direction.

Anywho, pressure waves have two important components.  One is the $\Delta P$ increase from atmospheric pressure, and the other is the duration.  The time integral of the $\Delta P$ over the duration of the blast (times the area of the target) is called the impulse.  While a large $\Delta P$ can do damage to a human body, essentially jostling adjacent cells apart from each other, in order to really do damage to object and throw things around, you need a big impulse, delivered, which means the pressure and time need to conspire to impart a lot of momentum.  Hence the slow exploding anfo.

I admit, I was also disappointed, if not surprised, that Buster didn’t get pushed a whole lot.  They sort of addressed my concern about a small exposed area (over which to integrate said \Delta P) by attaching their bed-liner sail.  I guess I’ll just go watch The Avengers to satiate my appetite for unrealistic explosions.

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Oven Redux

May 12th, 2012 No Comments

Now that I had established the mis-calibration of my oven, I wanted to characterize it so that I can dial it in to achieve a desired temperature.  I know that it’s easier to raise the temperature than to wait for the temperature to dissipate, so my plan was to start at a low temperature (200 ^{\circ}F) and periodically (approx. every 20 minutes) increase it by 25 degrees at a time.  Below are the results:

A couple of neat points until I get to the heart of the matter.  Notice how the frequency of the temperature control increases at higher temperatures.  Remember, I had established that the cooling off of the oven occurs similar to exponential decay.  The result is that it cools off relatively fast (deg/min) at higher temperatures, and slower at lower temperatures, because the temperature difference with its surroundings is higher at the higher temperatures.  So, because it’s cooling off faster at higher temperatures, the heating element has to turn on more often, causing the temperature to fluctuate much faster.  In fact, at the lower temperatures, I didn’t even get a full cycle over which to average during the 20 minute duration.

So I pick up the temperature correlation at 275 ^{\circ}F.  My initial inclination was to just take the average value between each period’s peak and trough, but because the curve is not regular (it cools off slower than it heats up), this would bias the temperature higher than what the time averaged temperature actually is.  So I took the time to actually time average each curve segment between peaks (or multiple peaks if available).  The following is those results as well as a linear curve fit:

 

In a perfect world, where the oven works like it’s supposed to, the best fit line would have a slope of 1.0 and an intercept of zero, i.e. f(x)=x, i.e. what you dial up on the oven is what the temperature actually is.  Unfortunately, both values differ from their ideal.  So, not only is there an offset between dialed and measured temperatures, but moving the dial by a degree results in 1.06 degree change in the oven temperature.  As a result, the oven is off more at higher temperatures.   For example, when the dial reads 300, the oven is at 336 ^{\circ}F; 36 degrees off.  But when the dial is at 500, the oven is at 548 ^{\circ}F; 48 degrees off.

I don’t think I’ll be wanting to run that calculation every time something needs cooking.  I’ll probably pick a “best” offset value of, say, 40 degrees and hope that the recipes can tolerate a difference of 10 degrees or less.

I knew that thermocouple that I bought with no solid plans for using would come in handy some day…

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Oven Diagnostics

May 11th, 2012 No Comments

I’ve got a new place with a new oven that has been consistently over-cooking my food.  I could call the super, or turn the oven down or cook everything for less time, but none of that gives me an excuse to bust out my Arduino.  Fortunately, I already had a (K-type) thermocouple and MAX6675 amplifier on hand that have been looking to add purpose to their lives.  Using a handy Arduino library for the MAX6675 made life and coding really straight-forward.

My original intention was to simply display the oven temperature on an LCD and take readings at several oven temperature settings.  I’m glad I decided to be a little more rigorous and actually record the serial data that was sent to my computer (one data point per second).  As it turns out, the temperature fluctuated over a wide range, so the data records were handy.  Not only was there a wide range of temperatures, but they were consistently higher than the oven setting of 400 ^{\circ}F .

This is the data taken after turning the oven on.  You’ll notice a couple of things.  First, the temperature oscillates pretty regularly.  As far as I can tell, the heating element goes from all on to all off, contributing to the bouncing around.  Also, the magnitude of the fluctuations made me suspicious that I was getting a bit of a misleading reading due to the radiation coming off of the heating element.  I’d rather not expound on this too much in this post, but try holding your hand in front of your face when next to a fire.  You will immediately feel cooler; most of the heat you from a fire (an outdoor one, anyway), is radiative.  This temperature does not reflect the air temperature; this is why you put a thermometer in the shade.

Anyway, I thought this might be contributing to an elevated temperature reading, so I put a piece of aluminum foil between the thermocouple and the heating element at the bottom of the oven.  The foil was intended to block the radiation from the red hot heating element.  I tend to believe this worked; it’s not as if the foil got red hot (~1500 ^{\circ} F), but obviously, it made little to no difference in the temperature reading.  So either this radiative heating was never or factor, or I am making a mistake.

After 35 minutes, I turned the oven off, just to see how it cooled off:

This looked like something that maybe I could model…  In fact, recalling some of my heat transfer book-learnin’, I figured the best way to start would be to treat the oven as a lumped capacitance system (insert spherical cow joke here).  I didn’t have my old textbook on hand, so I fell back on Wikipedia, which provided me with a nice equation for the expected exponential decay:

T(t)=T_{env} + (T_0 - T_{env})\exp^{-rt}

A characteristic time of 0.00045 min^{-1} seemed to do a nice job of matching the theory to the model:

Not bad.

I did these tests two different days; I was actually shocked at how consistent the temperature histories are!  The oven may be wrong, but it’s wrong consistently, which means I should be able to learn to use it without burning everything.  I guess a potential next test would be to do this analysis at various oven setting and observe what the results are so that I have a consistent adjustment to cook my food appropriately.

At the end of the day, I’m wondering what my oven is doing wrong. Is the feedback (a thermocouple, I presume) mis-calibrated (assuming mine is accurate…)? Where is it placed? Perhaps it’s measuring temperature in a location not particularly relevant to the oven center. Questions for another day…

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Unchained Reaction: Speed

April 15th, 2012 No Comments

[He] knows everything there is to know about mechanics…
Adam Savage

What prompts Adam to say this?  As far as I can tell, Charles Haine is a film director with little to know science/engineering training…

I’m not sure what I think of this show, Unchained Reaction, yet.  I like the idea; it reminds me of Junkyard Wars, which I loved.  I’m trying to convince myself not to record the show and skip everything but the completed runs at the end.  I haven’t done it yet, but I’m tempted.

Anyway, I watched the “Speed” themed episode and there was one big cringe-inducing moment near the beginning of the show that I had to follow-up on here.  I spent a whole week teaching kids about this very principal in physics lab, so I figured I’d take a post to rant on a simple physics concept.

In the first challenge, teams have to complete the first link in their chain.  The special effects team (“Monster Crew”) wanted to race a tortoise and hare, and to get them started, they were using counterweights to pull the animals across a table.  The falling counterweights pulled ropes connected to the animals.

I think Adam and Jamie’s consultant, Charles Haine, over-thought it a little bit, and in an effort to appear useful and lecture to the team, he sort of threw up all over himself.  After looking up his credentials (or lack thereof), I’m not nearly as surprised about what transpired…

He opened his misguided lecture with the following: “You know that two objects of different weights fall at the same speed if they have the same aerodynamic resistance, right?”


This is not a good start.  We all learned that with no air resistance, objects subjected to the same gravitational field with accelerate at the same rate.  But once any significant air resistance is introduced, objects fall at different rates depending on their weight, size, shape, etc.  I’m not even sure what he meant by aerodynamic resistance (drag coefficient, aerodynamic drag force?), but either way, he’s wrong.

The team explains to him that the different speeds will be due to the different weights of the tortoise and the hare.  Charles seems to go along a little bit: “More drag on the system by having a heavier weighted base will probably create enough for you.”  Which isn’t necessarily patently wrong, but he’s not really being helpful, either.

See, to begin with, the hare wasn’t being pulled quickly enough.  As a planned fix, one of the special effects guys (who isn’t dubbed an “expect in all things mechanical”) sensibly proposes to use a heavier counterweight (i.e. more potatoes).

Charles counters: “Remember, extra weight’s not going to fall any faster.”  Argh!  C’mon, your job is to help them.  This is patently wrong, so it is really less helpful than not saying anything at all.

Let’s begin by looking at free body diagrams of each falling weight (sack of potatoes).  For simplicity, I’ll just set one mass to double the other.  Also, we’ll ignore friction and aerodynamic drag (which, if anything, helps Charles’ case) to keep things simple; rest assured, whether or not we make these assumptions, Charles is still wrong.

We know how gravity is acting on the two potato bags, producing a force of just mass times gravitational acceleration.  What we don’t quite know is what the tension they’re experiencing is.  So let’s look at the rabbit and the hare to try and figure it out:

Pulled massAll right, the normal force from the table definitely cancels out the weight of the block.  But we still don’t know what either tension is.  Although, if we assume a perfect (or near-perfect) pulley, we can say that the tension is constant throughout the rope, so that each pair of blocks experiences the same tension force (either T_1 or T_2).

So let’s first look at the left-hand situation.  We have two equations:

T_1=ma_1 (in horizontal dir. for the block on the table)
mg-T_1=ma_1 (in vertical dir. for the hanging block)

Notice I use the same acceleration, a_1 in both equations; since the blocks are connected by a rope (without slack or elasticity), we can assume they accelerate at the same rate.  So that’s two equations and two unknowns (T_1 and a_1).  We can work with that!

With some simple algebra, we can find a_1 = \frac{1}{2}g.  Following the same procedure with the heavier sack of potatoes, we discover a_2 = \frac{2}{3}g.  For those of you keeping score at home, \frac{2}{3}g>\frac{1}{2}g; in other words, the heavier sack of potatoes (and whatever is connected to it) will accelerate more quickly.

What’s happening, in words, is that due to the rope, you’ve essentially got a single mass moving/accelerating here; however, gravity is only able to accelerate a fraction of it.  So as the suspended mass becomes a larger fraction of the total mass, the acceleration of the system will increase, ever approaching the maximum value of g.

Fortunately, one of the team members doesn’t just take his words at face value, and while he can’t really put into words why he’s right about his idea for extra potatoes, I believe his intuition served him well.  Sure enough, in the final cut, they got the hare to zip along the table.

I think a working knowledge of physics is a great thing to have, but it should be used in conjunction with intuition and common sense.  I’m not saying that this world isn’t full of non-intuitive (but explainable) behaviors, but when you perceive the physics of the problem (as you understand them) to fly in the face of common sense, that’s a sign to take a step back and really think through what’s going on.

The topic sort of invites confusion, which is why it was in the physics lab: so it could challenge students’ grasp of the concepts.  The difference is that Charles is not an undergrad; he’s on the show as an expert.

Okay, that’s fine, we all make mistakes, and he probably wishes he could have that back.  But therein lies my problem with it.  This isn’t live T.V.  How does all of that make it past the editing process?  At the very least, mix in a studio voice-over to correct the misinformation.

I’d say that they should bring me on instead, but I’m guessing Unchained Reaction won’t make it to season two…

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Most Wonderful Time of the Year

March 22nd, 2012 No Comments

Just a quick reminder and refresher (though it’s a few days late)…

The vernal equinox just passed, which means, not only does everyone in the northern hemisphere have at least 12 hours of daylight, but the days are getting longer faster than at any other time of the year.

Combined with Daylight Savings Time, that means we are getting finally getting some bright evenings at northern latitudes (if not warmer temperatures).  So enjoy!

And to our neighbors at the bottom of the map… sorry.  You have to wait six months for a similar thrill!

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Piano Strings

March 21st, 2012 No Comments

There’s nothing remarkable about it. All one has to do is hit the right keys at the right time and the instrument plays itself
Johann Sebastian Bach

I found myself recently staring down into a baby grand piano recently, and for the first time I considered the motivation for its exterior curves.  Noting the varying lengths of the piano wires for the different notes, I naturally assumed that their arrangement dictated the piano’s shape.  A little physics, as well as piano history set me straight.

First off, I used Wikipedia to refresh my memory on how to calculate the fundamental harmonics of strings, which after a little rearranging, allows me to calculate a piano string/wire’s length:

L=\frac{1}{2f}\sqrt{\frac{T}{\mu}}

So, for a given tension, T, and linear density, \mu, the length of a wire is inversely proportional to its frequency.  That rings a bell; now if we (incorrectly) assume all of wires have the same tension and linear density, we can reconstruct the trend of the wire lengths as a function of piano keys, once we know the frequency of the keys:

Note that since the values I used for tension and linear density are arbitrary, the units of length are unimportant, just their relative magnitudes.

Now compare to the inside of a piano:

Not bad, right?  Of course, there are some important differences that should be noted.  First, the piano strings are in fact not of constant linear density or tension.  Apparently, the linear density changes throughout the piano; I read somewhere that approximately every sixth string the wire gauge changes.  The tension is obviously varied in order to get precise tuning.  And the lower notes on the left hand side are rearranged completely differently, significantly changing the shape of that end of the piano.  According to this article on piano history:

Overstringing was invented by Jean Henri Pap during the 1820′s and was first patented for use in grand pianos in the United States by Henry Steinway, Jr. in 1859. When overstringing a piano, the strings are placed in a vertically overlapping slanted arrangement, with two heights of bridges on the soundboard instead of just one. This allowed for larger strings to fit within the piano case.

So to avoid the long tapering spike at the lowest notes, it appears that as a matter of convenience, they are angled to help improve the overall footprint of the piano.  I imagine their linear density is also significantly larger than the higher notes in order to reduce their length.

While I was dubious about it to begin with, it appears that in fact, there’s no justification for the inflection point that a typical piano has in it’s exterior.  It’s simply an artifact of the arrangement of the lower notes, and probably aesthetics.

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Vanishing Head

January 28th, 2012 No Comments

I don’t know if you’ve ever played the game where you hold an outstretched hand and pretend to squeeze people and objects in the distance between your thumb and forefinger.  Well, there’s another way you can make-believe cause someone’s head to disappear.  Now I don’t really want to be the type of blog that just regurgitates other posts from around the web, but I suppose I’ll make the occasional exception for exceptionally neat things.  Check it out:




I already knew about the blind spot; it was something to think about paying attention to while star-gazing.  Sky and Telescope relates it with a little more precision:

As it turns out, your eye is most sensitive to a faint object when that object lies 8° to 16° from the center of vision in the direction of your nose. Almost as good a position is 6° to 12° above your center of view. Avoid placing the object very far on the “ear side” of your center of vision; it may fall on the retina’s blind spot there and vanish altogether.

Sure enough, the YouTube video utilizes the blind spot on the “ear-side”, of your left eye.  I guess I also knew that your brain compensates for it, because it’s not as if you’re conscious of a gaping hole in your vision.  But I was quite surprised that it was sophisticated enough to complete the black bar swept through the blind spot, preserving its continuity.

Wikipedia shared another neat fact; the octopus and other cephalopods do not suffer from blind spots; their optic nerves travel behind the receptors, and so they do not form a bundle in front of the retina.

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Feynman Sprinkler

December 31st, 2011 5 Comments

Perhaps you have heard of the Feynman Sprinkler before; it’s a relatively famous physics conceit, though admittedly, physics mind-benders are not necessarily part of the public consciousness.  Feynman himself was particularly coy when writing about it in his autobiography, abstaining from sharing the result of his experiment, which has led to the typical internet speculation (see: airplane on a treadmill).  It’s such a neat problem because smart, informed people can come to completely different conclusions as to what happens when you run a sprinkler in reverse.

   
Normal Sprinkler Feynman Sprinkler

Obviously, our “normal” sprinkler will spin clockwise.  Feynman wanted to know what direction it would spin when run in reverse.  I think there’s a general consensus that an underwater sprinkler, when run in reverse will not spin, but why?  I’ve heard various explanations that are either confusing, misleading, or wrong, so I wanted to tackle it myself.

What’s the big problem anyway?  Why is this so hard to wrap your head around?  I think the core of the confusion is that we have a problem which appears to be symmetric, and in physics, we expect symmetric situations to behave symmetrically.  But, in fact, the symmetry is an illusion.  Why?  We’ll get to that…

First, I’d like to begin the analysis with our trusty friend, the control volume.  Again, we have to be careful; this analysis has given me considerable trouble due to silly assumptions that I’m used to making, but cannot be made here.

All right, first let’s start with the easier case: that of a sprinkler running in the “correct” direction underwater.  To simplify the problem, we will assume the sprinkler is not moving or accelerating or anything.  Speaking of symmetry, we’re going to use it here, and only examine half of our sprinkler:

Here’s a cross-section of the sprinkler; so let’s set up a control volume around the outside of the sprinkler pipe, across the entrance and exit.  Here’s our scary-looking equation:

\sum\left( \vec{r} \times\vec{F}\right)_{tot}=\frac{\partial}{\partial t}\int_{cv}\left( \vec{r} \times\vec{V}\right) \rho dV+\int_{cs}\left( \vec{r} \times\vec{V}\right) \rho\vec{V}\cdot \hat{n}dA

Okay, let’s start rocking the terms.  Since we only care about the static state, we can drop the time-dependent term:

\sum\left( \vec{r} \times\vec{F}\right)_{tot}=\int_{cs}\left( \vec{r} \times\vec{V}\right) \rho\vec{V}\cdot \hat{n}dA

Because I’ve set up the problem so elegantly, everything happens at right angles and the cross products can simplify so that we are simply balancing torques perpendicular to the page:

\tau_{tot}=\rho A_{in}RV^{2}=\rho QRV=\dot{m}RV

The first term is equivalent to the total torque being applied to the control volume, while the second term comes from evaluating the integral for the flow out of the nozzle (top of the diagram). This is the only flow in our problem happening at any distance from the pivot point.  The inflow is occurring at the axis of rotation and so it doesn’t contribute anything to the fluid’s angular momentum about that point.

Our job is now to equate the total torque with the torques on the control volume’s surface.  Anything left over must be due to the external torque needed to keep the sprinkler still while water is flowing through it.

\vec{\tau}_{tot}=\vec{\tau}_{surf}+\vec{\tau}_{ext}

Generally, the torque is calculated by integrating \vec{r}\times\vec{F} over the control surface, where \vec{F} is determined by p\hat{n}dA.  It so happens for this type of finite control volume analysis, it works best to assume that the nominal fluid pressure is zero.  This “gauge” pressure can be set arbitrarily without affecting our results, so for convenience (and to reduce the number of integrals we have to do), we will set it to zero, and fluid pressures will all be relative to this.  That is, all of the surrounding, unmoving fluid in the problem will have a pressure of zero.  Furthermore, in fluid mechanics, a “free jet” such as the one we have here, has a pressure equivalent to the surrounding fluid (that is, zero).

Soooo…. that means the pressure, and therefore the force all around the control volume is zero; which means \tau_{tot}=\tau_{ext} and we have

\tau_{ext}=\dot{m}RV

So the torque required to keep the nozzle in place is \dot{m}RV (positive, out of the page).  Another way to look at it is that this half of the nozzle apparatus (everything in the control volume) is exerting a torque of -\dot{m}RV on the hub, or whatever is holding it in place.  Doing the same analysis on the other half would yield the same result, with the additive, total torque required to keep the sprinkler still equal to 2\dot{m}RV.

Have I lost you yet?

Okay, then; let’s keep going.

Here we have the case of the reverse (Feynman) sprinkler.  It seems like a straight-forward process to redo the calculations, but with opposite flow conditions, but unfortunately some of our more essential assumptions break down.  I think this is what creates confusion — a lack of symmetry in the flow.

A low-hanging fruit is that the pressure of the inflow is not equivalent to ambient in the reverse case.  This can be worked out using the Bernoulli equation to determine pressure as a function of velocity, where along a streamline, (specific) energy is conserved:

\tfrac{1}{2}\rho V^2_1+\rho g z_1 + p_1=\tfrac{1}{2}\rho V^2_2+\rho g z_2 + p_2

Since the  fluid is being accelerated from the quiescent surroundings (which are at zero pressure), and change in elevation is negligible, the equation becomes:

0=\tfrac{1}{2}\rho V^2_2 + p_2
p_2 =-\tfrac{1}{2}\rho V^2_2=\tfrac{1}{2A_{noz}}\rho Q V_2

All right, I haven’t been entirely upfront with you.  This is not a great application of the Bernoulli equation.

Where did we/I go wrong?

My cartoons have not been doing the flow field justice.  The velocities at the exit probably look closer to this (regular sprinkler on the left, reverse sprinkler on the right):

Another illustration of the non-intuitive asymmetry of the problem

The outflow velocities can more or less be considered to be parallel, and have an either laminar or turbulent profile, resulting in our expression for torque being mostly unchanged.  However, I don’t have any good analytic expression for the velocity profile coming in, making our control volume analysis exceedingly difficult (I know there’s a vena contracta and separated flow as the fluid enters the nozzle).  Since we’re assuming the mass flow rates are the same, though, I’ll take a leap of faith and say that right hand side of the control volume equation is the same in both cases (the velocity is the opposite direction, but the sign stays the same because it is squared).

But let’s use our new information to do a better job with the Bernoulli equation.  When streamlines are bent so severely, some of the Bernoulli assumptions about conserved energy (inviscid-ness) are violated.  This energy loss can be accounted for by introducing a head loss term:

\tfrac{1}{2}\rho V^2_1+\rho g z_1 + p_1=\tfrac{1}{2}\rho V^2_2+\rho g z_2 + p_2+h_L

where

h_L=\tfrac{1}{2}K_L \rho V^2

What of this K_L?  It is a loss coefficient that is a function of the geometry of the loss feature; in our case the nozzle entrance.  Similar “re-entrant” pipes tend to have very high coefficients, approaching the maximum of unity.  Supposing it is unity, let’s recalculate the pressure p_2:

0=\tfrac{1}{2}\rho V^2_2 + p_2+h_L=\tfrac{1}{2}\rho V^2_2 + p_2+\tfrac{1}{2}\rho V^2_2
p_2=\rho V^2_2=\frac{\rho Q V_2}{A_{noz}}

I’m sick of caveats, but here’s another one.  The pressure is not uniform across the entrance, nor do the losses happen at the entrance.  The pressure drop actually occurs due to viscous effects in the separated flow along the entrance of the pipe; but since we’re assuming inviscid flow, this is the correction we need to apply to the upstream pressure (at the entrance) to overcome these viscous losses.

We can then determine the torque from surface forces:

T_{surf}=R p A_{noz}
T_{surf}=R \rho Q V
T_{surf}=\dot{m}RV

And now to work it in to our full control volume equation:

T_{tot}=T_{ext}+T_{surf}=\int_{cs}\left( \vec{r} \times\vec{V}\right) \rho\vec{V}\cdot \hat{n}dA
T_{ext}+\dot{m}RV=\dot{m}RV
T_{ext}=0

So that means there are no outside torques needed to keep the nozzle in place; in other words, the nozzle is not going to spin even if we don’t hold it.

Do I have you convinced?!

Although Feynman never revealed “the answer” or what happened when he tested it out in the lab, the general consensus is that there is no torque on the reverse sprinkler.  However, knowing the answer definitely helped me get here, and I’m not sure that I can be totally confident in my assumptions.  That being said, the flow profile, and therefore the analytical solution, is pretty messy in the reverse case, and requires some bold assumptions to make the problem tractable.  Even if this explanation is not a concrete proof, I hope it helps align the math of the problem with experimental results.  It certainly helped me rationalize the phenomenon.

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Toilet Bomb

December 1st, 2011 No Comments

There’s a lot of sexy data we’ve got going here.
Tori Bellaci

Sorry for the long absence, but science conference season is behind me, and the Mythbusters are doing science!  I thought the Mythbusters did an admirable job of actually trying to explain the science of the fluid dynamics regarding the benefit of flying in the V-formation.  Talking to people who knew what they were talking about definitely helped.

wingtip vortex

Thanks NASA!

I won’t harp on it, because, surprise, surprise, Wikipedia does a really good job of explaining it, but something they don’t spend much time on is the effect of the trailing vorticies on the plane (or bird) generating them.  Seeing as the wingtip vortex is creating an upwash (utilized by the trailing aircraft), it is obvious that if the plane is driving air up at the tip, a corresponding negative lift is acting on the wing (as well as increased drag).  This is obviously not beneficial to flight, nor are the powerful, and stable vorticies that can persist for a long time and disrupt other planes’ flight (esp. smaller ones).

While they can’t be eliminated completely in a three-dimensional world lacking infinite wingspans, aircraft manufacturers have devised some strategies for limiting the effect.  The most noticeable, perhaps, is the winglet, which decreases the pressure difference on either side of the airfoil, by making that edge vertical.

As usual, the Mythbusters’ data logging was less than extensive/thorough, though.  I would have really like to see them use the smoke generators on the plane close to the wing tips to visualize the vorticies.  I’m not sure if it is even possible to do this; maybe you’d have to line up a plane’s wingtip behind another’s smoke trail.  But besides being cool, it would have let the trailing planes know where they needed to be in order to exploit the vortex in whatever formation they were flying.

The toilet stuff was cool; I have also had problems with batteries under-performing under cold conditions…

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Aerodynamics Meets Art

October 19th, 2011 No Comments

I love when art and physics demonstrations collide!

When I was in Paris recently, I stayed very close to the Pompidou and was fortunate enough to see this “kinetic sculpture” titled “Flux” by Zilvinas Kempinas.  Basically, by pointing a fan down at table, he is able to suspend a loop of magnetic tape pretty much indefinitely.


I’d love to break down in greater detail how this works someday. I have a feeling it’s sort of the opposite of the more recognizable, equally cool phenomenon of suspending a ping pong ball with a vertical air jet.  In that case, Bernoulli’s principle explains that when air is flowing around the ping pong ball, if the ball shifts slightly outside the center of the air stream, the faster moving air in the center of the jet “sucks” the ball back in to the middle.

The tape is remarkable in that is suspended and relatively stable (it doesn’t drift off).  As the air deflects back up and away from the table, it pushes the tape up against gravity just as is the case with the ping pong ball. But I believe what keeps the magnetic tape centered has less to do with Bernoulli and more to do with drag.  As the jet spreads out away from the centerline, its speed decay.  In this way, when a segment of the tape gets “too far” from the centerline, the push it’s getting from the jet decreases.  When it gets too close to the center, the velocity is greater, so the drag is greater, pushing the tape more towards the periphery.

I dug around YouTube a bit for more Kempinas and found other cool magnetic tape fun:

And another!

I’d love to come up with something this cool someday; I love when science is art is science. My particular aspiration would involve fun with fountains!

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