An Analytical Approach to Pamphlet Folding

Most of us can say we’ve endured the frustration of folding letters or pamphlets, trying as much as possible to align the corners for a straight fold only to have them wiggle freely as the crease is made. And in an office like the one I work at, half-folding pamphlets is a fairly procedural task. Whenever this work is entrusted with me, I like to build a rig like the one shown below to minimize the chances of me pulling my hair out.

Fig. 1: Makeshift pamphlet folding rig

It's easy to arrive at a tool like this based on common-sense, but we can also take a more analytical approach as to why a simple 90° corner can make paper folding so much easier. We need to start with a surface-level premise about the universe we live in; as far as we can tell, our universe is comprised of three spatial dimensions, mathematically attributed as x, y and z and experienced as length, width and height, and one temporal dimension. The temporal dimension and the other however many must exist according to string theory or what have you aren’t as important in this task as the three spatial dimensions with which we interact on a daily basis. An unfolded pamphlet can move freely through these three spatial dimensions, as throwing them out the office window will demonstrate when they flutter down to the street below. But while they are sitting on my desk waiting to be folded, they don’t budge. This is because the desk on which they sit is applying an upward force on the pamphlet stack, counteracting gravity and preventing them from moving towards the center of the earth. Balancing forces like this is the foundation of mechanics and is the way that the following discussion manifests in the physical world we live in.

So what else is there to consider? Well, three free dimensions can also be referred to as three “degrees of freedom,” a fancy term that just says a variable can change as necessary. The more degrees of freedom that exist, the harder it is to pinpoint the position of an object that moves freely, like the corner of a pamphlet. What’s also cool about the three spatial dimensions is that even though by default we orient the basis of the defined dimensions to align one direction with gravity because this simplifies calculations, linear algebra tells us that the basis vectors x, y and z are easily translatable to different positions. Now let’s take my desk, oriented by default in the x-y plane, and translate it so that it now exists in the y-z plane. This is kind of similar to if I were floating with my body parallel to the desk, which would make more sense if I were in outer space since there is not as strong a default frame of reference when gravity is not apparent. What we see now is that my desk has somehow become like a wall! Amazing! This tells us that floors, walls and ceilings are not very dissimilar from each other, the common theme being that if I beamed a ball in a closed room, it would bounce off of all surfaces because its free motion in the three spatial directions has been truncated.

Fig. 2: Designation of three spatial dimensions adjusted to rig

Let’s return to the pamphlet folding rig. The rig is comprised simply of three walls: the desk and two others constructed from a box and a paper tray. Additionally, while folding gravity acts on the pamphlets to further constrict movement in the common z dimension, or up, while keeping the dimension still partially accessible so as to get the right side of the pamphlet over the left. My hands as well work to apply force to keep the pamphlet to the left against the box and away from me towards the paper tray while creasing, limiting the x and y dimensions. If we look at the two corners of the pamphlet tucked into the rig while half-folding, all three degrees of freedom have been collapsed, the pamphlets sandwiched between one surface and my hands along all three axes. Not only are these corners restrained, but the geometry of the pamphlet then guarantees that all other corners of the pamphlet should be aligned as well. So while in practice errors on my part and in my shabby rig’s construction limit the effectiveness of this folding method, in theory restricting the three degrees of spatial freedom of the pamphlet should produce a perfect fold every time without the frustration of any dancing corners.

Fig. 3: The final folded pamphlets

While some people may not appreciate such a lengthy analysis of as simple a contraption as a corner for folding pamphlets, I think really delving into why something so common-sense works from a physics standpoint demonstrates how approaching the world from an analytical perspective provides insights that can help everyday people live better lives. I hope you enjoyed this article, and if you have any feedback or questions, leave me a comment below. I love hearing from you guys. Thanks!

Iodine Salt to Treat Radiation?

Fig. 1: Next, movie cover (derricklferguson)

Over winter break I watched the movie Next starring Nicolas Cage with my dad. In the movie, the FBI was able to link a dead woman knifed in her room to a Russian nuclear attack plan because of a few potassium iodide (KI) pills found at the crime scene. The star FBI policewoman (who is also President Coin from the Hunger Games movies!) was quick in realizing that the only reason someone would take potassium iodide pills was to combat radiation poisoning. Before this movie, I had never heard of KI being used for this purpose, and expectedly I was skeptical. Before you judge me, imagine if someone had told you that a sugar pill could prevent you from dying of stomach cancer. That is the same magnitude of ridiculousness that I felt the whole KI pill thing had to be.

But, I was wrong of course. KI supplements are an established treatment for preventing thyroid cancer, one of the biggest health impacts observed after the Chernobyl meltdown [1]. So how is it that something so simple as a salt pill, because that’s what it is, can prevent one of the most odious conditions of modern times caused by a technology that took humans thousands of years to create? Turns out it’s by inhibition [2]. KI pills for thyroid cancer prevention aren’t made up of just any iodine; they are made of the ¹²⁷I isotope, which is iodine’s only stable form [3]. Ingested iodine is taken up by the thyroid gland, and if the iodine is of a radioactive isotope the subsequently produced radiation can cause thyroid cancer. KI pills work by dumping stable ¹²⁷I into the person’s blood stream, flooding the thyroid and reducing uptake of other radioactive iodine isotopes. KI pills only work in preventing thyroid cancer caused by radioactive iodine exposure, however, not other conditions caused by general radiation exposure.

Fig. 2: ²³⁵U fission product properties (Hochel, R. C.)

So, a few other questions obviously arise from this talk of iodide pills, one of which being where do the radioactive iodine isotopes come from in the context of nuclear fission? Fission of heavy atoms results in atoms of lighter weight and free neutrons that propagate the nuclear fission reaction. Some of the fission products of ²³⁵U are various iodine isotopes, including ¹³⁵I (6.33% yield), ¹³¹I (2.83% yield) and ¹²⁹I (0.9% yield) [4]. These are clearly not the main fission products of ²³⁵U, but they can still accumulate in contaminated environments, especially where large-scale nuclear fission reactions were involved such as nuclear meltdowns and atomic bomb testing sites. Another question to answer is how do the radioactive isotopes end up being ingested by people in contaminated regions? Scientists at Dartmouth, New Hampshire were able to measure increases in ¹³⁵I concentration, an indicator also of the presence of undetectable ¹²⁹I, on land but especially in local streams a year after the 2011 Fukushima meltdown in Japan [5]. They cited the increase as due to nuclear fallout from the Fukushima incident that blew across the continent and deposited itself in groundwater sources. This implies that the radioactive iodine isotopes can be both airborne and waterborne. If everything is coated in radioactive iodine, ingestion is believably imminent. To bring us full circle, there was also a run on KI pills in 2011 on the American West Coast due to fears of radioactive iodine finding its way into homes and food supplies there as well [6]. It’s funny how analyzing a simple movie premise can take us all the way to a not-so-late nuclear disaster.

Yesterday was the first day of classes, and soon enough school will be back in full swing. I've based my schedule this semester off of a google calendar with the idea that better organization will make hectic school life just a bit easier, so we'll see how that goes. My course load is two materials science classes, one materials science lab, orgo 2 and an anthropology class on modern culture. I'm hopeful that this semester will go better than last, and I'll keep you guys updated on what goes on. If you like this article or have ideas for another, be sure to leave me a comment below. Thanks for reading!

Can the Earth's Magnetic Field Support a Space Railgun?

Fig. 1: Railgun diagram (HowStuffWorks)

In physics class, you’ve probably learned about railguns and how they use magnetic fields and current to generate a propulsive force. And if you’re anything like me, you’ve probably wondered if we can launch a person or spaceship with one, because why not? What would be even cooler is if we could do it with the earth’s ambient magnetic field. Well, let’s test this idea a little bit.

We know that the force generated by a magnetic field on a charged particle is

1.       F = d/dt(L)q x B = qv x B (L is displacement, q is charge, B is magnetic field, v is velocity)

An equivalent statement more applicable to railguns is 

2.       F = d/dt(q)L x B = IL x B          (I is current)

This change of derivative position is acceptable because it represents a change in reference frame, the displacement derivative in the frame of the charged particle and the charge derivative from a fixed external frame of reference, watching the charges flow by. In the setup of a railgun, B and I are orthogonal and the resultant vector of the cross product is in the direction of propulsion. Solving the cross product yields

3.       F = ILB 

We also know the induced current through the railgun circuit to be related to the generated emf, 

4.       Ɛ = - d/dt(Φ_B) = - d/dt(BA) = - Bd/dt(A)          (Φ_B is magnetic flux, A is area)

By geometric analysis, we know the change in area to be the fixed base length/bar length times the change in height, or

5.       d/dt(A) = L d/dt(h) = Lv          (h is height)

The full expression is

6.       Ɛ = IR = - BLv

7.       I = - BLv/R

Appending our force equation to account for induced current that works against the initial current, we get

8.       F = (I - BLv/R)LB = LBI – (L²B²/R)v

This equation is useful because by setting the magnetic force to zero we can solve for a terminal velocity,

9.       v = IR/LB

This is no good! If we’re to launch someone high into the sky, we need a constant force to produce a constant acceleration to exceed that of gravity. So what if we keep a constant effective current to counteract the current induced by propulsion? Looking at equation 8, we see that this is possible if we establish current as a time-dependent function given by

      10.   I(t) = I₀ + BLv/R                                                 (R is resistance)

      11.   I(t) = I₀ + BL(at)/R = I₀ + BL(Ft/m)/R               (from kinematic equations v=at and F=ma)

The F term here should be substituted with the constant force we predict to be generated by the current adjustment

12.   F = I₀LB

Substitution and regrouping produces

13.   I(t) = I₀(1 + (B²L²/Rm)t)          (m is mass)

This equation can be verified by substitution into equation 8. Now that our force is constant, let’s calculate what current would accelerate the average person [1] on a weightless conductive beam unaffected by air drag 1m in length enough to counteract gravity (a pretty low bar for testing viability of concept). Since the magnetic field generated by the circuit would probably diminish too close to the walls of a 1m launcher, an external field is necessary. Let’s try using the upper end of the earth’s magnetic field strength range [2].

                     14.   F = mg = I₀LB

                     15.   I₀ = mg/LB = (62kg)(9.8m/s²)/((1m)(65x10^-6 N/Am)) = 9.3x10⁶ A

That’s a lot of current, but perhaps possible? Adjusting for a spaceship weighing 1,000 times the average person 10m wide with an acceleration of 1m/s² [3], we get a figure on the scale of 1x10⁷ A. In 2014, the CERN Superconductors team was able to pass 20x10³ A through an MgB₂ superconducting wire, a world record at the time [4]. This record value is clearly magnitudes less than even the current necessary to lift a single person, let alone a spaceship. Looks like the earth's ambient field is a no-go. But! If the person were in a magnetic field of an achievable 1T even, then with currently achievable currents it should be possible to throw them at least into the sky if not into space. Hooray for space railguns!

If you liked this post or have any ideas for another, let me know by leaving a comment. Thanks for reading!

Why Cold Drinks "Sweat"

With a horrible heat wave hitting the Philadelphia area, it’s good to think cool thoughts. Already feeling the heat last night, I left a coconut water in the freezer with the intent to drink it but forgot and so took it to work this morning frozen solid. I figured since it’s so hot outside and the metal can is a good conductor, it’d probably melt pretty quickly. And while the ice in immediate contact did melt, the inside remained frozen and I had to cut the top open with scissors to eat it. Before I figured this out, the can had already shed a puddle at my desk. Have you ever wondered why it is that cold things "sweat."

Fig. 1: My favorite coconut juice brand, Foco (pinstopin.com)

Most of us are familiar with the concept of condensation, having learned about the water cycle in elementary school. We are commonly taught in elementary that water exists as vapor at hot temperatures, condenses to liquid as the temperature drops and eventually expands (not condenses, as ice has a lower density than water due to hydrogen bonding) into ice as the temperature drops further. In high school, we learn about the ideal gas law and how pressure also affects phase transitions, yielding the phase diagram.

Fig. 2: Phase diagram for water (myhomeimprovement.org)

So from this standpoint, we are all familiar with why cold things "sweat." What else is there to it? While the basic principles stand, there are some other viewpoints from which we can view this phenomenon.

Phase transitions can be viewed as being an equilibrium process, as is demonstrated by the fact that an ice and water mix maintains a 0°C temperature. In such a mix, the ice melting and the water freezing are competing processes that are controlled by environmental factors; if you cool the mix the ice expands, but if heated the ice melts. Additionally, the entire mix must either become ice or water only before the temperature can deviate significantly from the equilibrium temperature of 0°C. What’s cool about this process is that if you track the energy entering the ice and water mix, say a glass of iced coconut water (let's treat this as an ideal glass of pure iced water), we can predict the corresponding phase transitions based on molecular kinetics.

When bonds are formed, whether strong or weak, we know that energy is released as heat. The reverse is true as well, breaking bonds requiring energy. The direction of bond energy transfer can be simplistically remembered taking into account the conservation of energy in a two molecule one-dimensional collision. Say two water molecules are moving towards each other and stick together upon impact. Where did the kinetic energy go? Ignoring molecular vibrations, the energy had to have been released as work, or heat. In order to separate the water molecules, we need to get them to move apart, a.k.a. add work, or heat, to yield kinetic energy. In our glass of iced water, this sort of energy transfer is happening extremely fast and on a large scale, one that can be described by Le Chatelier’s principle since the ice and water form an equilibrium.

Fig. 3: Ice-water equilibrium state (JVC's Science Fun)

Now let’s put the iced water outside on a hot summer Philadelphia day. From experience, we know that the ice will melt and the water will become unappealingly warm. If we track the direction of energy transfer, the higher energy hot air must be donating energy to the lower energy iced water simply because this is the default direction of energy transfer in our universe according to the Second Law of Thermodynamics. The added energy must translate into kinetic energy as temperature is positively correlated to molecular kinetic energy. From our two water molecule system we know that a decrease in water molecule association is predicted, favoring water over ice and vapor over water. This manifests as the ice melting and the water warming and eventually evaporating.  

What has been so far described, however, is only focused on the iced water itself. Let’s change our basis to focus on the hot air along the iced water glass instead (assume the water glass does not hamper kinetic energy exchange between air and iced water). Hot air carries a lot of water since at higher temperatures water enters the vapor phase preferentially according to Le Chatelier’s principle. From the viewpoint of the air, the cold iced water is pulling kinetic energy from it, accordingly cooling the air within a certain range of the glass. Plugging this information back into our two molecule system, the energy must be afforded by reducing the molecular kinetic energy of the air, increasing the probability of water existing in associated groups, i.e. water. And this is why a cold drink sweats in the summer.

Since school is starting up again, I will not be able to post as frequently as I have been during the summer. I will try to post at least once a week, and will probably be doing so during the weekend since this is when time is most available. Please have patience with me on this, and as always thanks for reading!

The Fun of Latin and Ballroom Dancing Explained with Mechanics

Last semester, I picked up Latin and ballroom dancing as a hobby. All through high school I didn’t dance, but thanks to the recommendation of a friend I decided to go to free introductory dance lessons. At first it was just fun to socialize, learn some new steps and practice body coordination, but I soon grew to love it. Our Latin dance teacher has a particular habit of describing dance movements in terms of coordinates and physics; rumba walks require your center of mass over your front foot for balance and look best if you extend the axis running from your shoulder to opposite leg as far as possible. This mentality has slowly creeped its way into my brain, and eventually I realized that, hey, dance really is about physics! You don’t know it while dancing, but every movement is purposeful when viewed through a pair physics-colored safety glasses.

When first starting to dance, everything is about mechanical control of your body and constantly asking the question, where is my center of mass (let's abbreviate it as COM) right now? Of course you wouldn’t consciously ask yourself that, but even without asking your body will give the answer. While posing still, if your COM is not somewhere well supported by your limbs, you’ll fall down because relative to the point of floor contact your bodyweight is generating torque on your COM. On a single limb, your COM needs to be over the supporting limb to stay balanced. On more than one limb, your COM should be somewhere between the limbs based on how your bodyweight is distributed towards each (unless the floor is slippery. Then it’s best to choose a limb to center over otherwise your supports will simply fall apart and you’ll hit the floor). When you begin to move horizontally, core muscles contract in order to support your center of mass on its journey from one place to another through fluid motion.

Spinning generates torque along other axes and accordingly makes you conscious of where your body axes are. Since torque acts along levers originating at your COM, all body axes will include your COM. Torque is defined in parallel to Newton’s second law by the equation

                                                              1.       τ=Iα    (torque equation)

Where τ is torque, I is inertia and a is angular acceleration. Inertia is related to the distribution of your body mass in space around a rotational axis. While spinning on an axis perpendicular to the ground like a ballerina or ice skater, contracting your arms from an extended position can reduce your moment of inertia and increase your angular acceleration based on the existing torque. And expanding your body to increase inertia while tilting your axis slightly can slow your spin and generate a simultaneous forward momentum to get you out of the spin and into horizontal motion. Manipulating the position of your center of mass so that it remains relatively stationary while moving around it by torque is what allows break-dancers to do such complicated flips. Dancing is all a game of mechanics.

Fig. 1: Dancing as an act of balancing forces (University of Illinois at Urbana-Champaign)

What makes Latin and ballroom dance special to me is that all of the dances are partner dances. This means double the complication, but also double the fun! Instead of your COM, your tandem COM is what counts. And all movements and spins are done together accordingly. What’s even more exhilarating is that your tandem COM is without your body, so when you go for a partner spin you’ll both be spinning about an axis between the two of you (assuming equal mass). In Latin and ballroom dance, a concept that is stressed is partner connection. This means making sure both of your bodies push against each other at the point of contact in a way that conveys information about how each of you is moving. What this also does is make the tandem COM more stable. When partner connection is weak, your tandem COM can constantly dissolve and reform. This makes both partners unstable and rely on knowing their own COMs to maintain balance. A strong connection means both partners can fully commit to the unwavering tandem COM and structure each motion together. And believe me, you can feel this. It’s really exciting when you and your partner start to move as one, and not having to worry about your own balance allows you to focus on the art of the dance.

This post is a little less reference and math heavy than some of the past ones, which I thought would be a nice break. I just wanted to share what makes Latin and ballroom dance such a special hobby to me. Dancers out there, please let me know if you think my description does dance justice in the comments below. Other Thinkers, I wanna hear your thoughts too. And check out a few minutes of this video of world-class Latin dancers. Thanks guys!

Wind Turbine Efficiency, Part 3: The 40.3% That Got Away

You made it to Part 3! If you haven’t already please go back to read Part 1 and Part 2.

Let's continue. From calculus, we know that to find a maximum of a given single-variable function, we take the derivative, or the slope of the function, with respect to the variable and find where the derivative equals to zero. This is because the slope of a function is zero where the function changes from increasing to decreasing or decreasing to increasing output values (aka peaks and valleys). Equation 24 from Part 2 is rather ugly, however, so let’s set the derivative with respect to k=v2/v1 (the ratio of final and initial velocity) as our plan of attack (reminder: ve, v1 and v2 are all known constants in this expression). Let’s do this now.

            25.     P=1/4rAv1³(1-k²+k-k³) (substitution of k=v2/v1 into equation 24)

            26.     dP/dk=1/4rAv1³(0-2k+1-3k²)=1/4rAv1³(-2k+1-3k²)=0

            27.     -2k+1-3k²=0 ((1/4rAv1³) is a giant multiplicative non-zero constant term, so it can be
             dumped into the zero never to emerge again (except by integration))

            28.     k=1/3, -1 (solutions to equation 27, obtained by factoring the second-degree polynomial and
             letting each set of terms equal to 0 since if ab=0, either a=0 or b=0 makes the statement true)

Obviously, a negative value of k=v2/v1 does not make sense since this requires that the wind hit the turbine and flow backwards, so we will only keep the k=1/3 value. Plugging the value for k into equation 25, we get:

                                           29.     Pmax=1/4rAv1³(1-1/9+1/3-1/27)=16/27(1/2rAv1³)

To interpret this result, all that must be remembered is that the total power in the original cylinder of wind was found to be Pwind=1/2rAv1³ in equation 7. Substituting in this expression reveals the following relationship:

                                                                      30.     Pmax=16/27Pwind

In other words, only 16/27, or 59.3%, of the total power in a column of wind can ever be extracted by wind turbines (this is assuming 100% turbine internal efficiency relative to the 59.3% limitation). This means that there is an inherent limitation in the production efficiency of wind power relative to available energy to be harvested.

Fig. 4: Turbine efficiency cartoon (Wikimedia)

And there you have it, Betz's Law in only 30 steps.

So, knowing this, should we abandon wind power? The answer is a definite no. Wind turbines have an installation cost in both money and materials and also severe limitations as to practical operation, such as a decreasing efficiency in high wind velocities to prevent structural damage, but they are an important member of a small set of what I would consider to be true renewable energy strategies. I won’t talk about this now since I feel guilty about the density of the topic I just covered, but I will make this the topic of my next post.

If you have any questions, leave them in the comments section below and I’ll try to answer them as best I can.

Wind Turbine Efficiency, Part 2: Too Many V Terms

Welcome back! If you haven’t already, please read Wind Turbine Efficiency, Part 1: A Windy Cylinder of Power.

All calculations thus far have dealt with air in front of the wind turbine. We should take a look at what happens behind it as well. From fluid dynamics we know the following equations to be true through logical progression:

                               8.     V1=V2 (V1 is air volume before turbine, V2 is air volume after)

                               9.     dV1/dt=dV2/dt

                               10.     dV/dt=Av (equation 5)

                               11.     A1v1=A2v2

Equations 9-11 simply state that the volume of wind entering our cylinder from Part 1 must change at the same rate as the volume of air exiting the turbine from behind. This is true if we assume no changes in temperature or atmospheric pressure before to after the turbine (beyond those induced by the slowing wind, which we are calculating for now). From energy conservation, we know that if work is performed on the wind turbine, then the kinetic energy of the wind after the turbine, and accordingly the wind velocity, must be less than that before. In other words, v1>v2. The result is that A2>A1, so the wind after the turbine must effect a cone of area expanding away form the turbine blades.

Fig. 3: Air flow through a wind turbine (Danish Wind Industry Association)

Let’s try to piece apart the maximum power obtainable from the wind cylinder calculated above. Wind force is defined as follows:

                                     12.     F=ma=d(mv)/dt=(dm/dt)v+m(dv/dt) (force equation)

For simplicity’s sake, we are assuming that the wind before the turbine has velocity v1 and slows to velocity v2 after performing work on the turbine, both of which are know constant values. When wind blows through our wind cylinder, the mass of air, and as a result the effective length of the cylinder since base A is constant, changes with time. Therefore, we are not concerned with how force changes with velocity, so the m(dv/dt) term can go to zero and we can evaluate the change in wind force over the turbine as follows:

                                  13.     ΔF=(dm/dt)Δv=(dm/dt)(v1-v2)

                                  14.     ΔF=r(dV/dt)(v1-v2)=rAve(v1-v2) (from equations 4 and 5)

A new variable has been introduced in equation 14, namely the effective velocity ve, which corresponds to the wind velocity that the turbine experiences as part of work production (in other words, if the wind were to lose all kinetic energy after contacting the turbine, a wind velocity of ve would generate a quantity of power the same as from wind with velocity v1 slowing to v2). Concerning the power obtained from the change in force, we know that since the change in wind force is equal to the force exerted on the turbine, power can be calculated as follows concerning the wind contacting the turbine:

                         15.     W=FL (instantaneous work equation, L is analogous displacement here)

                         16.     dW/dL=F, therefore dW=FdL

                         17.     P=dW/dt=d/dt(FdL)=F(dL/dt)=Fve (power equation)

                         18.     P=rAve²(v1-v2) (from equation 14)

From expanding equation 4 to accept that the wind velocity after the turbine is not zero, we obtain the equation:

               19.    P=1/2(dm/dt)(v1²-v2²) (this is just the difference in kinetic energy with changing mass)

               20.    P=1/2rAve(v1²-v2²) (from equations 5 and 6, parallel to equation 7)

Now we have two different equations for power, one from the work equation and the other from the kinetic energy equation! Setting these equal to each other, we obtain the expression:

                   21.     P=1/2rAve(v1²-v2²)=rAve²(v1-v2)

                   22.   v=1/2(v1+v2) (simplification of equation 21 using (v1²-v2²)=(v1+v2)(v1-v2) identity)

This equation tells us what the relationship between the effective velocity ve and the initial and final wind velocities v1 and v2 are. Specifically, the relationship is that the effective velocity is the average of the initial and final wind velocities. Using this relationship, let’s redefine the kinetic energy-derived power equation in terms of v1 and v2 in preparation for calculating maximum power.

                            23.    P=1/2rAv(v1²-v2²)=1/2rA(1/2(v1+v2))(v1²-v2²) (from equation 20)

                            24.    P=1/4rAv1³(1-(v2/v1)²+(v2/v1)-(v2/v1)³) (simplification of equation 23)

Things are starting to get exciting! Read on to Part 3 to finally start calculating the external efficiency limitation of wind turbines, and let me know any questions in the comments below.

Wind Turbine Efficiency, Part 1: A Windy Cylinder of Power

When people think of renewable energy, probably what comes to mind is a mixture of solar panels, wave motion generators, geothermal plants, hydroelectric dams and wind turbines. Of these, the two most associated with the image of a sustainable society are probably solar panels and wind turbines. That’s why when I caught a glimpse of this wind farm while at the Jersey shore, I felt like I had been transported into some sort of futuristic society apart from our own. 

Fig. 1: Wind farm near Atlantic City (personal photo, hence the quality)

Assumedly, we all know that every energy-harvesting method has efficiency limitations, even though some people like to pretend they don’t. With solar panels, calculating efficiency can be complicated as it comes down to a matter of quantum efficiency, but the efficiency limitations of wind turbines, also called Betz’s Law, can be readily calculated with some physics. Let’s explore this.

(Note: though not particularly difficult conceptually, this derivation is math-heavy. I will do my best to explain the underlying logic as we go along. Also, this explanation will be separated into three parts so that your eyes don’t bleed from the endless stretch of math. Backbone calculations for Betz’s Law come from the Wikipedia article and this MIT presentation, but I hope you will find that my explanation, though lengthy, is more intuitive than these sources. Enjoy!)

Let’s start with calculating how much power there exists in a cylinder of wind with base area A corresponding to the arm span of the wind turbine.

Fig. 2: Cylinder with base A and wind influx at velocity v (MIT)

                          1.  KE=1/2 mv²  (kinetic energy equation)
                   2.   KE=W (work-energy equation)
                   3.   P=d/dt(W)=1/2(dm/dt)v²+mv(dv/dt)  (power equation, product rule)

For equation two, let’s assume all work is converted to electrical energy, not friction or blade deformation or other stuff. Now in equation three, what normally happens is that the dm/dt term, or the change in mass over time, is assumed to be zero, producing the familiar mv(dv/dt) or (d(mv)/dt)v or Fv term associated with power. Instead, let’s assume that the air cylinder has a constant known speed v1. Therefore, the term mv(dv/dt) goes to zero, and we are left with equation 4:

                                                                   4.  P=1/2(dm/dt)v1²

From fluid dynamics, we know that

                                                         5.  dm/dt=(m/V)(dV/dt)=r(dV/dt)

or that the mass per volume, density, multiplied by the change in volume over time is equal to the change in mass over time. Makes sense, right? This is essentially just dimensional analysis. With constant wind cylinder base A and changing length L, change in volume over time is defined as

                                                         6.  dV/dt=d/dt(AL)=A(dL/dt)=Av1

Combining equations 4, 5 and 6, we get the general power equation for a cylinder of wind to be

                                                                       7.  P=1/2rAv1³

where r is density, A is cylinder base area (also area of turbine span) and v is a constant wind velocity. 

I’ll stop the first post here since it’s a good spot to take a break. Read on to the next post to learn how to calculate the effective wind speed acting on a wind turbine, and let me know in the comments below if you have any questions.

Party Science, Part 1: The Beats

Who doesn’t like to party, am I right? Going to your favorite concerts, dancing at clubs, or just chilling with friends, partying helps us all to relieve the stresses of daily life and maybe make a few lasting memories. Living in the time we do, we have the technology to throw some epic shindigs, and not to get preachy, but from the flashing lights to the music beats, a lot of what allows us to party in the way we do is realized through science. I want to take the next couple posts or so to talk about some of the basic aspects of party science. Today we’ll cover the beats.

What is sound anyway? In its basic form, sound is the perception of longitudinal waves propagating in the air around us. Take the drum for instance. A stereotypical drum consists of some sort of membrane adjacent to an echo chamber. When struck, the membrane implodes at the point of compression, but nearby surface immediately works to restore equilibrium by pulling up on the impact point while the impact point pulls the surrounding membrane down [1]. The result is a wave propagating across the membrane. What does this mean for the air around the drum membrane? when the membrane implodes, air under the membrane is put into pressure and triggers movement away from the drum surface, only to feel the opposite reaction when the membrane at that point fluctuates outward.

Fig. 1: Production of longitudinal waves in a drum (University of Leicester)

This air movement creates a series of propagating low- and high-pressure zones comprising the longitudinal sound waves that have communicated human culture for thousands of years. On string instruments, the concept is the same. When plucked, the string oscillates and sends those vibrations to the instrument belly, causing air fluctuations by the face of the string instrument, say, a guitar, to produce music that has colored many cultures with Spanish influence [2]. However, the way this effect is experienced differs with modern music like dubstep, electric, house or other computer-altered music types that don’t come from physical musical instruments per se. With this sort of music, the component properties of previously recorded sounds are altered with the help of computer software. For example, a gradual increase in sound frequency and tempo is the sensation often used to foreshadow a dubstep drop. How do you send these synthetic beats out to awaiting ears? The answer is with a speaker, what some would consider a crucial part of any party. The basic speaker consists of a membrane or cone attached to an electromagnetic coil (solenoid) next to a permanent magnet. When current is sent through the solenoid, a magnetic field is produce down the length of the coil oriented towards or away from the electromagnet at a given moment if the field is envisioned as a vector pointing down the coil center. As the current fluctuates, the membrane is moved towards and away from the magnet, creating sound waves. When worked in reverse, with sound waves generating current, the device is called a microphone. The single waveform current signal generated can then be broken down using Fourier analysis into component frequencies with corresponding amplitudes (aka that bar diagram widget that moves when your music plays). To generate music, the speaker current is controlled by programs that mathematically synthesize waveforms from music files and divvy up the responsibilities of producing a net sound effect between available speakers. In the end, if executed properly with a location that offers the right amount of acoustic bounce back, say in a closed room among friends or in the middle of Miami skyscrapers like in the video of Hardwell below, you’ve got the basis of a good time.

When you watch this video, think about how the sounds are being synthesized. Is wave frequency/amplitude being altered? Are the sounds natural or computer generated? Let me know your thoughts in the comments (and don’t feel obligated to watch the entire video, it’s long).