# A Sense of Scale (in Dollars and Cents)

I hate politics, but for some reason I obsessively read about it. I don't know why I do this, but I assume it's the same reason people slow down for car wrecks and pay to see the geek [1]. Anyway, the big thing in political news now is that if Congress can't pass a budget [2] by the end of the day Friday, the government will shut down. Shutting down the government means that 800,000 federal employees will go without pay [3], lots of services will be put on hold and you won't be able to go to the Smithsonian or the Grand Canyon. So it's kind of a big deal. Since the ramifications of a government shutdown are so serious, there must be some really important disagreements holding it up, right? Right? A quick search (for example, here), shows that the big hold-up in passing the budget comes over a disagreement on how much money should be cut from the budget. Republicans want to cut $40 billion dollars and Democrats are willing to cut$34.5 billion dollars. So the hold-up is over $5.5 billion. Let's consider how utterly and stupidly insignificant this is. We shall take as our inspiration Alemi's post a while back offering several ways to help visualize the large and scary sounding numbers thrown around in government budgets. The first thing we'd like to do is find out what fraction this$5.5 billion discrepancy is compared to the total budget. Alemi cites the total budget of the U.S. government in the 2010 fiscal year to be $3.55 trillion. Hot dog! So we see that: $$\frac{\mbox{Disputed Difference}}{\mbox{Total Budget}} = \frac{\5.5 \times 10^9}{\3.55 \times 10^{12}} = 0.0015$$ So the disputed part amounts to 0.15% of the total budget or one and half parts in a thousand. Let's compare this to some things for which we have a better sense of scale. Let's pretend we are looking to buy a car. We need it to get to work and, you know, get stuff done. If our car costs$10,000 dollars then 0.15% of our total costs will be $$\mbox{\10,000} \times 0.0015 = \15$$ So the current budget situation is like arguing for 6 months over a $15 charge on your$10,000 car. Sounds reasonable! Now let's switch gears and consider a timescale. Let's consider a 40 hour work week. What's 0.15% of 40 hours? This comes to $$40 \mbox{ hr} \times 0.0015 = 0.06 \mbox{ hr}$$ or, if you prefer, $$0.06 \mbox{ hr} \times \frac{60 \mbox{ min}}{1 \mbox{ hr}} = 3.6 \mbox{ minutes}$$ So the current budget problem is like arguing for six months over about three and a half extra minutes to your work week. Now let's consider a length scale. Consider the US \$1 bill. According to Wikipedia, the dollar bill is 6.14 inches long, 2.61 inches wide and 0.0043 inches thick. That means that the ratio of the thickness to the width of a dollar bill is $$\frac{\mbox{Thickness}}{\mbox{Width}} = \frac{0.0043 \mbox{ inches}}{2.61 \mbox{ inches}} = 0.0016$$ or slightly more than the disputed fraction of the budget. Hooray! "That's all well and nice," you say, lowering your voice and leaning over in a way that makes me uncomfortable, "but who is to blame?" That's a great question! My answer will take the form of an experiment. First, get a coin. Got it? Great. Now if you're Republican, let heads be "The Republicans" and tails be "The Democrats." If you're a Democrat, let heads be "The Democrats" and tails be "The Republicans." If you are neither, then randomly assign a party to heads and allow the other to be tails. Ready? The coin is to blame. [1] I use this in its original meaning. That is, the guy whose job it was to do horribly gross things at a carnival for money, not the guy whose main form of social interaction is debating scenes in Star Wars on internet forums. A more up-to-date comparison would have been to say "for the same reason that people watch Jersey Shore." [2] That is, the budget for the 2011 fiscal year, which started on October 1st 2010. [3] Don't worry, all members of Congress would be exempt from this and would still pick up paychecks!

# Collective Wanderings

Update [04/04/11]: It seems Clicky got too popular for our bandwidth limits on the physics servers. Hopefully we'll be able to fix this sometime soon... Update [04/02/11]: We made the code faster, so check out the new and improved Clicky! Hey, kids! Would you like to be a bit player in a grand experiment with poorly thought out objectives? If so, then check this out. It's a little interactive "game" [1] that Alemi and Matt coded up. When you click on the link, you will be redirected to a page showing something like this: Using the arrow keys on your keyboard or by clicking on the arrow buttons on the screen, you can move the dot around. Sound fun yet? Well, the interesting bit comes in with the fact that anyone who wants to can move the same dot at the same time. The plot will update automatically, so if you're really bored you can sit around and watch someone else move the dot around. Heck, you could even give the dot a cute name like Mr. Dottington and tell it how your day was. We won't judge you. [2] Anyway, check it out if you've got a bit of time to kill. Explore the space and we'll get back to you with the "results" of this "experiment" sometime in the "future." [1] There are no objectives and "winning" is undefined (for a contrary argument, however, see Sheen et. al (2011)) [2] But Mr. Dottington will.

# Special Virtuosi Book Announcement!

Coming soon to a Kindle near you!

We've been at this whole blogging thing for about a year now and I think we've amassed a large and dedicated enough fanbase to finally release a book! The track record so far for physicist-writers has been quite good of late, so we figured why not us? Well, lots of reasons actually. For one thing, it's really hard. Books are, like, hundreds of pages long. I barely stay coherent and on-topic in a one page blog post. For another thing, it takes lots of time. I hardly have enough time to do my laundry in time scales deemed "socially acceptable." How could I ever find the time to write a book? Despite these potential setbacks, the millions and millions of dollars that writers make still seems really appealing. Who wouldn't want to be rich and popular forever. I mean, just look at Oscar Wilde, Edgar Allan Poe and Herman Melville! Luckily, a solution presented itself. I don't have time to write a book now, but I found an old copy of my novel Blue Dragon laying around the house that I was able to sell using the immense popularity of the Virtuosi brand. The book will be published this summer by Clark Hall Publishing. Here are a few advance reviews: "Audacious, bodacious, entropic, synoptic, electric, eclectic, entertaining, hyperbraining, high-roller, tripolar... BuyAgainst the Day Blue Dragon." -- THE PHILADELPHIA INQUIRER "Corky isn't easy to 'get'. His melodies and chords I would describe as complicated. But, behind all the disharmony and 11ths and so forth.. it swings. This is my favorite Virtuosi book. It's an amazing book and that's all you can say." -- J.D. Jackson "This is a book rich in allegory and poignant social commentary. Also, you should call home more" -- Corky's Mom So these unbiased critics think it's totally rad. Should you just take their word for it? Absolutely! But if you really want see the book for yourself, here it is:

# Nickel Gnomes

Perhaps Step 2 was to steal copper?

While flipping through a CRC Handbook whose days in the United States are dwindling, I came across a section that described the naming conventions of each chemical element. Most of the names made sense to me. For example, Nobelium is named after Alfred Nobel (surprise!). However, the Nickel entry was the following: Nickel: Named after Satanor Old Nick This confused me greatly. What the heck is Santa Claus doing hanging out with Satan [1]? After a bit of poking around on the internet, I found an article from 1931 by a guy named William Baldwin called The Story of Nickel, How "Old Nick's" Gnomes were Outwitted. Needless to say, this did not allay my confusion. Here is the relevant passage from the article: In the early part of the eighteenth century fresh lodes of ore were laid open in Saxony where from times immemorial silver and copper mines had been worked. This new ore was so glittering and full of promise as to cause the greatest excitement, but after innumerable trials and endless labor all that could he obtained from the ore was a worthless metal. In disgust the superstitious miners named the ore kupfer-nickel after "Old Nick" and his mischievous gnomes who were charged with plaguing the miners and bewitching the ore. This "worthless" metal is nickel. Kupfernickel is the German name for the nickel ore (see image below) which comes from the words kupfer, meaning "copper" and nickel, meaning "demon."

Ore that, if not for gnomes, would contain that sweet sweet copper

And that's the story of Nickel! Tune in next time [2] for our ongoing 118-part series, Better Know an Element! [1] Turns out "Old Nick" is an English name for the Devil. Go figure. But those of you named Nicholas should not be upset, as your name comes from the Greek Nikholaos, which literally means "victory-people." So you've got that going for you. [2] I have no plans of ever doing this again.

# Blown Away

I was reading a discussion on green energy recently, in particular wind power, where the following claim was made

enough wind turbines to power the world would cover the surface of the world.

Now, this was quickly decried by supporters of wind power, but the claim has stuck with me. The question on my mind today is: How much of the earth's surface would have to be covered to power the earth with wind turbines? We can't hope to put an exact number on this, the best we'll be able to do is an order of magnitude. I also don't know much about wind turbines, so I'll be making liberal use of wikipedia as I go. Let's start with the size of the wind turbine. According to wikipedia the largest wind turbine has a rotor sweep diameter of 128 m. To an order of magnitude, we'll say that our average wind turbine has a diameter of 100 m. Next we need to know how much power this puts out. The maximum power of this turbine is \~8 MW. However, it certainly wouldn't be producing that at all times. Current wind farms produce around 20-30% maximum capacity. However, these turbines are careful placed in areas of high wind. We're not going to get that lucky with our wind dose when we place our turbines haphazardly, so we'll assume they produce at 1% maximum capacity. According to wikipedia, the world energy consumption in 2008 was 474 EJ (exajoules), or an average power use in 2008 of 15 TW. To an order of magnitude then, the area we'd have to occupy with wind turbines to power the world would be: $$\left( \frac{(100\text{ m})^2}{1\text{ turbine}}\right)\left(\frac{1\text{ turbine}}{.018 \text{ MW}}\right)15 \text{TW} = 2\cdot 10^{12}\text{ m}^2$$ That's 210^6 km^2, or, in english, 2 million square kilometers. For comparison, the land area of the united states is roughly 10 millon square kilometers. So we'd only have to cover 1/5th of the united states with wind turbines to power the entire world (in 2008, no doubt power use has risen since then)! While that is a lot of space taken up, it is nowhere near the entire surface of the world. There are, of course, other concerns about wind power. Note: maybe the wind turbines are less efficient overall. Also, I assumed that the footprint was just the square area of the turbine diameter. I know this is the size of the face of the turbine, but to an order of magnitude I imagine it is correct for the space occupied on the ground.

# Physics Challenge Award Show

In an emergency, Richard Dean Anderson's mullet can be used as a flotation device and/or standard kilogram.

Welcome to the First Physics Challenge Problem Award Show! We received an integer number of solutions to our challenge problem and at long last and after much deliberation, we have chosen our winner. We had before indicated vaguely that there may be some sort of prizes involved in this competition. After consultation with our financial advisors and breaking Alemi's piggy bank, we have decided on the following prizes: First Place: A brand new CRC Handbook! Second Place: An autographed [1] picture of Scott Bakula! Honorable Mention: Nothing! [2] So before we officially announce our winner, let's backtrack and build up some suspense. The challenge was to come up with a bunch of MacGyveresque experiments to determine as closely as possible the standard second, meter and kilogram using only the materials handy to you on a desert [3] island. Just about every response we got successfully answered the question, so we had to base our final result on robustness and uncertainties. We also tended to favor those that did not rest on precise knowledge of one's own height, weight, etc (though there is nothing wrong with these approaches). So without further ado, the winner is.... [pause] [unnecessarily long secondary pause] George from Australia! George provided a list of no fewer than 11 different methods for determining the second, meter and kilogram. The second was found by timing (by means of a pendulum) the time it takes for the sun to move through an a given angle measure. This measurement can be made each day for a year to get the best results (remember, you've got all the time in the world!). Now we have the time it takes the sun to go through a given angle measured in "swings of a pendulum." We can calculate the transit time of the sun into seconds and then we can find the number of seconds per swing of the pendulum. So now we have a second. But since the period of a pendulum is given by: $$T = 2\pi \sqrt{l/g},$$ we also have the length of the pendulum in meters. From this we can make a standard meter. Now all that is left is to find the kilogram. This can be done by making a water-tight enclosure of volume 1000 cm^3 and filling it with water, giving a kilogram that can be compared to other objects to get a longer lasting standard kilogram. Any other knowledge you may have (i.e. known height and weight) may be used to check the above measurements. A very close second place goes to Alireza, who provided a very strong submission and perhaps one of the best uses of snail mucus in the entire competition! Alireza's solution was to use one's own known height to construct the standard meter. From this you can make a pendulum to get the second and a box o' water (sealed with tar, glue and snail mucus) to get the kilogram. Both George and Alireza provided very detailed responses with special attention paid toward reducing uncertainty and checking their answers through several independent measurements. They also both carried out parts of their experiments. Congratulations to you both! In addition to the winning responses, we also received several other submissions that merit honorable mention. They are.... Oliver, who provided several acoustical experiments using his SUPERPOWER of perfect pitch. Nicole, who noted that we were not correct in giving the Titanic the designation "HMS" as this is reserved for ships in the British Royal Navy. In fact, the Titanic was designated "RMS" since it carried mail. Thanks, Nicole! Gary, who completely ignored the premise, used arbitrary units and left a very amusing note on his deathbed giving all his arbitrary units to be found and converted by the first enterprising explorer who lands on his island. I know you had asked for your prize in cash and while I cannot directly accomodate that wish, do know that Richard Dean Anderson smiles can be cashed in most banks nationwide (see [2]). So thanks to everyone who submitted a response to this Challenge. We enjoyed reading all of the solutions and we hope you had fun thinking about it. And next time you go on a boat, don't forget that CRC! [1] Autograph will be signed by me on an 8.5" x 11" picture of Scott Bakula printed by a laser printer on regular copy paper. [2] Well, you receive nothing of monetary value but, know deep down in your hearts that in the picture at the top of this post, Richard Dean Anderson is smiling at you. So you've got that going for you, which is nice. [3] Sadly, we did not choose a dessert island. That would have been much more fun.

# Japan Nuclear Crisis

Though I know that two posts in one day is recently unprecedented, I've been meaning to post about the Japan nuclear crisis for a few days. The various major news outlets are doing a good job, or so it seems, of keeping us informed of the events going on over there. However, I found myself rather puzzled over the physics of what was happening. From the news articles I was unable to figure out what was actually causing the meltdown, beyond some problem with the cooling. As a postdoc in my lab asked, "Isn't all they have to do drop the control rods and the reaction ends?" So I decided to do a little digging. I've found a couple of places that do a nice job of explain some of the physics of what is actually happening, nature news (not sure if the nature blogs are behind a paywall), and scientific american (not up on current events, but a nice summary of what can/might go wrong). I'm sure there are many other places doing a good job of explaining things, but these are the ones and I found, and hopefully they help clarify what is actually happening.

# Apologies + Saturn!

Hello, again! Remember us? I don't. Anyway, apologies for the lack of activity here. There are plenty of people* to blame for this lack of activity, but I don't want to name names. The real purpose of this pseudo-update is to SUPER DUPER promise that the winners of our first Monthly Physics Challenge problem will be announced tomorrow. Thanks for your patience! In the meantime, there's a totally rad video-ification of photos of Saturn (and moons) taken by the Cassini spacecraft that was Astronomy Picture of the Day yesterday and will (presumably) be part of an IMAX movie in the future. You can check it out here.

Saturn. It's a planet!

See you tomorrow! * Everyone. Especially Matt.

# Fun Fact: Lebron James Plays Basketball

Between building airplanes and playfully destroying everyone else in my apartment at Super Smash Brothers, my roommate Nathan brought up an interesting recent fact about LeBron James. He told me that LeBron scored 11 consecutive field goals (not in football... you know who you are) in one game. Apparently this was a pretty special event, but how rare is it for a player of LeBron's caliber? TO THE SCIENCE-MOBILE! The Problem! ESPN 8, The Ocho tells me that LeBron's career field goal percentage is 47.5%. Considering the number of shots he takes, this is a pretty good number. To compare, the highest field goal percentage for a single season was Wilt Chamberlin with 72.7%, but eye witness testimony says he was around 10 feet tall and would wait in the offensive paint all game. Let's see how improbable this 11 in a row streak is. The generic question we are going to need to answer is as follows: If a basketball player takes N shots in one game, with a shooting probability of q, what is the probability that the player will make AT LEAST k shots in a row? We'll call this probability P(N) This turns out to be a tricky problem, but let's take a shot (awful pun... I sincerely apologize). We can take care of simple cases: If N < k, then P(N) = 0. This tells us you can't have a streak of k if you don't take k shots! If N = k, P(N) = q^k. This is the probability of getting k in a row if you take k shots, not too surprising yet. When N > k, things get more interesting. Finding the Recurrence Relation Our goal is to write a relationship that has this form: P(N) = P(N-1) + blank What's blank? "You don't worry about blank... let me worry about blank!" We'll need to look at the inclusion-exclusion principle. This principle basically says that when we want to take all DISTINCT items in two sets, we need to take all of the elements in one set, and add all elements in the second set which are DISTINCT from the first. For example, if A = {0, 1, 2, 3, 4} and B = {3, 4, 5, 6, 7, 8}, then the union of A and B is {0, 1, 2, 3, 4, 5, 6, 7, 8}. Note that I did not include 3 and 4 twice. Let's take a look at the expertly designed (5 minutes before class) google docs drawing below: The entire line represents N shots being taken. Each shot gets its own little column (not all columns shown). Using the inclusion-exclusion principle with the following sets will give us the answer. Choose A to be the first N-1 shots, and B to be all N shots. The principle tells us first to take everything from A, which is the probability P(N-1) shown in red. B will be the entire line, but the principle tells us to only add DISTINCT chances from B. Since the only difference in B is one more shot than A, the only distinct chance for a streak of k shots will be in the last k shots, shown in yellow as P(k). This is only distinct if the (k+1)th to last shot shown in green is missed! Otherwise a streak of k would have been included in A already. There is one more place for a streak to be already included in A. If there was a streak in the blue section, we must not include the B streak so we don't double count. Phew... Let's put this all together by multiplying the probabilities of each of those events: P(N) = P(N-1) + (probability of yellow streak)(probability we miss green)(probability of no streak in blue) $$P(N) = P(N-1) + q^k \times (1-q) \times (1 - P(N-k-1))$$ This gives us a recurrence relation for the probabilities! This is a general statement about the probability of at least one streak of length k out of N chances, given each has a probability q. Since I'm just going to plug this into Python anyway to handle the data, this equation is good enough. The expectation value of an event is the probability multiplied by the number of chances. For example, the expectation value of getting heads with 2 tosses is just (1/2)*2 = 1. The plan is to compile a list of his field goal attempts in every game LeBron has played in the NBA, and sum the expectation values for each N. $$\mbox{Expectation} = \sum_i P(i) \times \mbox{(number of games with i shots)}$$ Using LeBron's actual field goal attempt data for each game (up to February 4, 2011), we find that LeBron is expected number of games with at least a streak of 11 in a row is 1.128. This is a higher expectation value than the number of heads in 2 coin flips! So this is MORE expected than the number of heads we would see with 2 coin flips. This isn't very exciting given the number of shots he has taken and his shooting percentage. Data Tables

Consecutive Shots

Expected out of 667

Percent of Games

1

666.9

99.99

2

643.6

96.49

3

489.0

73.32

4

281.8

42.25

5

140.0

21.00

6

65.23

9.780

7

29.55

4.431

8

13.20

1.980

9

5.854

0.8778

10

2.578

0.3866

11

1.128

0.1691

12

0.4898

0.07344

13

0.2110

0.03164

14

0.09011

0.01351

15

0.03807

0.005709

16

0.01592

0.002387

17

0.006560

0.0009839

18

0.002660

0.0003995

19

0.001067

0.0001600

20

0.0004180

6.267E-05

21

0.0001599

2.397E-05

22

6.03E-05

9.033E-06

23

2.22E-05

3.328E-06

24

8.06E-06

1.208E-06

25

2.84E-06

4.259E-07

26

9.98E-07

1.495E-07

27

3.51E-07

5.255E-08

28

1.20E-07

1.797E-08

29

3.92E-08

5.870E-09

30

1.15E-08

1.717E-09

31

3.45E-09

5.171E-10

32

1.06E-09

1.589E-10

33

3.37E-10

5.050E-11

34

7.23E-11

1.083E-11

35

1.70E-11

2.554E-12

36

2.30E-12

3.442E-13

The streak in question is highlighted in red, so it appears we expect it to happen 0.169% of his games. The Realization Of course I did all of this before looking up the actual article. I'll quote the blurb here:

LeBron James set a personal record by making his first 11 field goals to start the game. His previous career-high was 10 straight field goals after tip-off, recorded against Chicago in 2008. After hitting his first 11 field goal attempts on Thursday night, James shot 6-for-14 thereafter.

Well... this calculation just got a bit easier. He has played 667 games, and the probability of getting 11 straight off the bat is q^k = 0.475^11 = 0.0002777. Multiply this by 667 games to get the expected value of 0.185. Sure this is 6 times smaller than our previous calculation; however it's still not statistically that impressive. How would LeBron's expected number change if he shot the same percentage (72.7%) as Wilt for his record breaking season? The expected number of games with 11 in a row during the game would be 72.38 games!! So this is incredibly dependent on the shooting percentage. We have a factor of q^k everywhere! Certainly it's dependent on the number of shots taken in a game too. The probability P(N) is a monotonically increasing function! Moral Given LeBron's shooting percentage and high number of shots per game, we expect that he would have at least 1 of these streak of 11 games so far in his career. This is certainly not to diminish this feat though. You still need to take 20 some shots a game in the NBA with nearly 50% shooting accuracy! We also have a nice formula to apply to more sports streaks! More to come...

# Problem of the Month: Gilligan Physics

So, some of us over here at Virtuosi Central have organized a challenge problem for the physics community here at Cornell. Well, we thought we would open up the challenge to the great wide world. The more submissions the merrier. The deadline is March 1, and submissions can be sent to our email. Details can be found at bit.ly/physicschallenge. Good luck and happy hunting.