Needless to say, Cameron’s version of the story is much more embellished. In his rendition, their experiment was a success; as he puts it, the pumpkin flew across the field, “goal post to goal post.”

When I first heard him say this, my initial thought was “Is Cameron telling the truth?”  How likely is it that a pumpkin, launched from a slingshot at one end of a football field, could sail through the air to land on the other side?  Note that his story actually ends with the pumpkin falling through the sun roof of someone’s car – this outcome is, of course, highly implausible, and therefore casts a shadow of doubt upon the entire story.  While the sunroof claim would be difficult to verify, one can at least use some basic math and physics to test the plausibility of the first portion of the story.

We would like to be able to find a formula for the distance Cameron’s pumpkin should travel.  Key to our analysis is the conservation of energy principle, and Hooke’s Law.  In particular, we will assume that the slingshot behaves like a spring, i.e. the potential energy it stores is proportional to the square of the displacement (for example, stretching the slingshot two meters gives you four times the potential energy as stretching it one meter).  So, if Cameron and his friends stretched the slingshot a distance of x meters, the slingshot would then store a potential energy of , where k is the spring constant and depends on the physical properties of the slingshot.

Now let us invoke the conservation of energy principle: when the slingshot is released, suppose all that stored potential energy will be converted into the kinetic energy of the pumpkin.  The formula for the kinetic energy of an object is , where m is the object’s mass, and v is the object’s velocity.  So, when the pumpkin is released, by conservation of energy, this kinetic energy should be the same as the potential energy the system had before we chunked the pumpkin.  In other words,

,

which, with a bit of algebra, tells us that the initial velocity of the pumpkin when it comes out of the slingshot must be .

These poor little guys have no idea what's in store for them.

So we have an equation for the velocity of the pumpkin in terms of the spring constant k, the distance x we pull the slingshot, and the mass m of the pumpkin.  Let’s not forget our main goal, though – what we’re ultimately interested in isn’t a formula for the initial velocity of the pumpkin, but rather the distance the pumpkin travels.  With a little more physics, it’s not hard to get from one of these pieces of information to the other.

More specifically, we have a very good understanding of projectile motion.  As I’ve discussed before, if one ignores air resistance (as we will do for now), all the equations of projectile motion can be derived from the fact that the acceleration due to gravity is a constant, meters per second per second.

Using this fact, suppose you throw an object with an initial velocity at an angle to the horizontal.  Then if one sets the initial position of the object to have coordinates (0,0), the x-coordinate of the object at any time t will be given by , and the y-coordinate of the object at any time t will be given by .  Using these formulas and a bit more algebra, one can determine that the distance d an object will travel in the x direction can be written in terms of the initial velocity, the angle , and g, via the formula

Example trajectory of an object shot with an initial velocity v at an angle A. The distance the object travels in the x direction is given by the above formula.

To recap: we have the distance as a function of the initial velocity, the angle the pumpkin is shot from, and the acceleration due to gravity.  We also have the initial velocity in terms of the distance the slingshot is stretched, the mass of the pumpkin, and the spring constant.  Combining these two formulas, we find that

and we now have the desired formula for the distance the pumpkin travels.

Let’s check Cameron’s story against this formula.  If the pumpkin really went from goal post to goal post, the distance it traveled in the x direction must have been at least 120 yards (100 yards for the field of play, plus 10 yards for each end zone).  This is roughly 109.7 meters.  Therefore, we must have .

On the right side of the equation, the term is maximized when equals 45°, and in this case .  So after making this simplification, we see that in order for Cameron’s story to be true, it must be the case that

The acceleration due to gravity is known, but we need to provide estimates for m, k, and x.  The mass m is the easiest to estimate; let’s say the pumpkin is around 10 pounds (roughly 4.5 kilograms).  To be generous, we’ll even round down to 4 kilograms.  What about the distance the slingshot is stretched?  Based on the slingshot used in the episode, it’s unlikely the slingshot in Cameron’s story would have been stretched more than 2 meters or so, but let’s again be generous and say it’s stretched 3 meters.  Plugging in these values, we would have:

where the units on k are Newtons per meter.  Since one pound is approximately 4.45 newtons, this is saying that the spring constant is about 107 pounds per meter – in other words, for each meter you stretch the slingshot, you need to exert 107 pounds of force.  To put it another way, to stretch the slingshot 3 meters or more, you’d need to exert 321 pounds of force.

This seems like quite a lot, though Cameron is a large fellow.  But recall we were being generous, both in our computation of the spring constant and in our estimate of the pumpkin size.  We also neglected air resistance in this model, but air resistance probably has a non-negligible impact here – not only will it slow down the pumpkin’s forward motion, but it also decreases the optimal angle from 45°.  So in a real world situation, I’d have to remain skeptical of Cameron’s story.  On the other hand, these calculations don’t necessarily rule it out entirely (though a more sophisticated analysis might).

For more on the physics of pumpkin chunkin’, here‘s an article from last year courtesy of Wired.  For related trajectory issues, Angry Birds also provides plenty of fodder.

Of all of these shows, though, the one that most piques my mathematical interest is TLC’s Four Weddings.  Based on a British show with the same name, the premise is as follows: four brides-to-be, unknown to one another, meet and attend each others’ weddings.  When one bride gets married, the other three score various aspects of the wedding, and the bride with the highest score among the four wins a honeymoon (contingencies are in place in the event of a tie, though these are not always explained and seem to vary from season to season).  In order to make for good TV, the show frequently manages to bring out the worst aspects of these women, as they nitpick and pass judgment on everyone else’s wedding.  Here’s a short clip to give you a taste for what this show is all about:

How are the weddings scored?  This process is explained in detail during the course of each episode.  The wedding is broken down into four categories: dress, venue, food, and overall experience.  For overall experience, the other three brides in attendance give a score from 1 to 10 (though I’ve never seen a bride give another wedding a 10).  For the rest of the categories, though, the brides can only rank the weddings as being 1st, 2nd, or 3rd in the given category.  1st place gets 10 points, 2nd place gets 6, and 3rd gets 3.  The total possible number of points a wedding can score is therefore 120.  At the end of each episode, the scores for each bride are broken out in detail.  The wedding budget and headcount are provided as well.

After watching a few episodes, it seemed like the best way to ensure a win was to simply outspend your competitors.  Certainly a large budget can help improve the guest experience or score a hip dress, but I was curious as to what overall trend (if any) could be made between, say, money spent on a wedding and the wedding’s overall score.

With this noble goal in mind, I proceeded to DVR 28 episodes of this show.  After recording the scores for 112 weddings, some of my questions were as follows: is the amount a person spends on a wedding correlated to the score they receive from the other competitors?  What about the amount a person spends per guest?  Finally, did the most expensive wedding win more frequently than would be expected by pure chance?

Let’s look at some data.  Here is a scatterplot of each wedding’s budget, vs. the total points earned.

Click to embiggen!

As the dots suggest, there is a slight positive correlation between the amount one spends on a wedding, and the score one receives from one’s fellow competitors.  The coefficient of correlation here is approximately 0.286, though if we discard the two $150,000 weddings, this improves to around 0.348.

Of course, a $10,000 wedding for 10 people might be a much nicer affair than a $10,000 wedding for 1,000 people.  If you spend more money per guest, does this translate into a higher score as well?  The dots don’t lie; here’s another scatterplot:

Click to embiggen!

There is quite clearly an outlier in this set of data – this corresponds to a bride who spent $150,000 for a 120 person wedding, for a whopping $1,250 spent per guest (this particular bride did end up taking first place).  Eliminating this outlier, though, the correlation here is weaker than the correlation for actual cost, at a meager 0.098.  In other words, there may not be a linear relationship between cost per guest and total score.  This may be because certain fixed costs, such as the dress and the venue, won’t necessarily vary much with the guest total, unlike something like food.

This analysis, though, obscures a key point.  In order to win the honeymoon, you don’t necessarily need a high score; you only need a higher score than your three other competitors.  With this in mind, it may be simpler to just look at the scores episode-by-episode, and see how the amount spent on a wedding compares to the wedding’s relative ranking.

In the graph below, the blue bars count the number of times the most expensive wedding was given a certain rank.  The red bars count the number of times the wedding with the highest cost per guest was given a certain rank.  Note that if cost had no bearing on the rank, we would expect an equal number of weddings in each rank (in this case, about 7 per rank).

Click to embiggen!

As you can see, spending the most money seems to bestow an advantage: fully 50% (14 out of 28) of the most expensive weddings were ranked first.  This would be unlikely if cost had no impact on ranking.

The picture is a little murkier for cost per guest -the frequency for each ranking sticks pretty closely to 7, so it’s not clear that spending more per guest gives any advantage.

In conclusion, there is a small positive correlation between amount spent and score received, though this does not transfer to the amount spent per guest.  Compared only to one’s other three competitors, the amount spent appears to confer an even greater advantage, so if you are on this show and want to show your competitors that you are better than them, my advice would be to simply outspend them.  If all you’re interested in is a nice vacation, though, it may be cheaper to just stick to your budget, and plan a vacation on your own.

As a shining beacon of what is hip, MTV has a responsibility during its movie awards to highlight the most popular films of the year. This is in stark contrast to the priorities of higher brow award shows such as the Oscars, for which artistic achievement is placed on the highest pedestal. This is not to say that these two goals need be mutually exclusive; indeed, since the first MTV Movie Awards was broadcast in 1992, the “Best Film” has agreed with the Academy Award winning best film three times (1997′s Titanic, 2000′s Gladiator, and 2003′s Lord of the Rings: The Return of the King). Even so, a quick glance at the nominated films from these two awards shows each year reveals a fairly small overlap, in general. But if MTV’s goal is to prop up films from an angle focused more on pop culture, it is natural to ask how good of a job they do.

Delicious metal popcorn. (Photo: Jason McDonald/MTV)

This question begets another one: how can we best measure a film’s popularity? My first thought was to consider the rankings on IMDB. There, users can give any film a score from 1 to 10; as a prime example of a range voting system, this seemed like a good place to measure the public’s reception of a film.

The results were mixed. With this metric, comparing the 20 years that both awards shows have been around, the MTV best film scored higher than the Oscar winning best film only 5 times. Oscar trumped 12 times, and the two awards tied three times. The trend is also worth mentioning – after 5 consecutive years of beating or tying the Oscars in IMDB score from 1999-2003, the Oscar winning film has bested the MTV winning film ever since. The disparity has become even larger in recent years (I call this the Twilight effect, as the Twilight films have won best film at the MTV awards for three years running). Here’s a graph of the scores over time (the list of MTV Best film winners is here; Oscar winners can be found here):

While you may argue there isn’t enough data here to draw much of a strong conclusion, the recent trend is fairly convincing. By this metric, it seems like Oscar winning films, at least over the past few years, seem to have been more popular.

Rather than looking at only the winner, though, you might expect to get a better sense of the popularity of films on display by looking at all nominated films, rather than just the winners. If we take the average IMDB score for all nominated best films at each awards show, rather than just the winning film, we get the following picture:

It’s come close, but the average IMDB score of MTV nominated films has never been greater than the average IMDB score for Oscar nominated films. We see the Twilight effect among the averages as well, though it was dampened somewhat this year due to the inclusion of critical darlings Inception, The Social Network, and Black Swan on the MTV nominee list.

“Hold up,” you may be thinking to yourself, “this is all a bunch of hooey.” You may think that IMDB scores are not a very good measure of a film’s popularity. It may be quite likely that IMDB scores are biased towards those same features of a film that make it more likely for Oscar consideration. Perhaps the type of person who goes onto the website to rate films is more likely to be somewhat of a connoisseur, and therefore the tastes reflected by the IMDB community are more likely to reflect the tastes of the Oscars. At the very least, it seems likely that teenage girls are not voting on the website in droves; how else can one explain the Twilight series’ limp average of only 4.9?

What else can we use to measure a film’s popularity? Well, to return to teenage girls, they don’t show their support for the Twilight series by rating it highly on the internet. They show their support by going out and seeing the movie multiple times. So perhaps we should look at box office receipts rather than IMDB score (and, of course, by picking sides in the bitter feud between Edward and Jacob). What sort of picture do we see in this case?

If we only consider the winning film from each awards show, the data looks like this (I’m only considering US box office numbers here):

(I’ve cut the graph so that you can’t see the leap in 1997, when Titanic took top prize at both shows.) Things look a little more erratic, but if you look closely, you’ll see that the MTV award winner has taken in more money than its Oscar winning counterpart 14 times out of 20. The Oscar has favored the larger cash cow only three times.

As with the IMDB rankings, we can try to smooth things out by looking at the average box office returns among nominees, rather than just returns for the winner. This yields the following graph:

These numbers aren’t adjusted for inflation, which may explain in part the growth trend (2009 numbers are bumped up because of Avatar, as well). I’m less interested in the actual numbers than I am the difference between the two graphs. And here we see, in contrast to the IMDB case, that the roles of the two awards shows have flipped. While the average IMDB score of Oscar nominees has always been higher than the average IMDB score of MTV nominees, the average box office return of MTV nominees has always been higher than the average box office return of Oscar nominees.

To return to the original question: do the MTV movie awards highlight more popular films than the Oscars? Well, it depends on how you define “popular.” If popular means highly rated, the claim is somewhat dubious. But if popularity is measured by the almighty dollar, then this seems like a fair conclusion to draw. Whatever your line of thinking, though, I’m fairly confident that current trends will continue, at least until the last of the Twilight films has exited theaters.

From a mathematical standpoint, though, there are a couple of inconsistencies. Michael makes no secret of the fact that he has worked for the company for 19 years.  His employees take this loyalty to heart, and in Michael Scott’s penultimate episode, “Michael’s Last Dundies,” they surprise their boss with a song parody of the Rent song “Seasons of Love,” which pays homage to such a long period of service.  Below is the relevant clip – if you don’t have access to Hulu, you can try this YouTube link, although I imagine it will get pulled eventually.

The song begins with a soulful rendition of the following lyrics, courtesy of Andy Bernard (Ed Helms):

9,986,000 minutes,
We actually sat down, and did the math
9,986,000 minutes,
That’s how may minutes that you’ve worked here.

Here’s a question: how does this number compare to Scott’s claim of 19 years?  Let’s make a few estimates.

For a crude estimate, we can simply take the length of a year to be 365 days, and count the number of minutes in 19 years given this assumption.  This isn’t hard; the answer is simply 19 (number of years) x 365 (number of days in a year) x 24 (number of hours in a day) x 60 (number of minutes in an hour) = 9,986,400.  This is exceptionally close to the value cited in the song – in fact, due to a syllable constraint needed to maintain faithfulness to the original song, it seems likely that this is how the number crunchers in the office actually came up with the value they sang.

Andy has the voice of an angel.

But how accurate is this number?  This depends on a few factors.  First, if Michael Scott really did begin 19 years ago, this estimate neglects a few leap years.  A better estimate would come from taking 365.25 days in a year, which in turn would give an estimate of 9,993,240 minutes (or 9,993,000 minutes, if one is interested in a number befitting of a Rent song parody).  What’s more, it’s unlikely that Michael worked exactly 19 years – the true length of his stay is probably somewhere between 19 and 20 years, or between 9,993,240 and 10,519,200 minutes.

Depending on your interpretation of the lyrics, though, any of the numbers given above may be much too large.  When Michael’s employees tell him “That’s how many minutes that you’ve worked here,” do they mean “that’s how many minutes since you’ve started working here,” or “that’s how many minutes you’ve spent working since you started working here?”  If the former, then these estimates seem more or less reasonable.  But if it’s the latter, we really should only be counting the minutes Scott spent at the office.  If we say that he spent 40 hours a week, roughly, in the office over the past 19 years, then with 365.25/7 weeks per year, this comes out to only 19 x 365.25/7 x 40 x 60, or roughly 2,379,343 minutes, a far cry from our earlier estimates.  To take things even further, based on the evidence provided by the show, the amount of work Michael actually does while in the office seems fairly minimal – if we’re talking strictly about the amount of time he’s spent working in those 19 years, I wouldn’t be surprised to find a much lower value.  Either way, there seem to be some accounting issues at work here that have been ignored.

Perhaps if accounting spent less time playing solitaire, these issues would have been ironed out.

Also, at the risk of sounding like comic book guy, just two years ago an episode of The Office focused on celebrations for Michael’s 15th anniversary with Dunder Mifflin; the subsequent cancellation of the festivities by new manager Stringer Bell Charles Miner drove Michael to quit and start a rival paper company.  One could argue that the time Michael spent not employed by Dunder Mifflin should not count towards his 19 years, but more importantly, 15 + 2 is only 17, not 19.  Before you try to argue that perhaps the show’s timeline is simply moving faster than the normal rate of one year per year (as TV shows are occasionally wont to do), one needs only to consider the timeline of Jim and Pam’s baby to see that this is not possible.  So in fact, all of the above discussion is fairly moot – indeed, we can’t even be sure how long Michael has worked at Dunder Mifflin.

All is not lost, though.  While we can’t accurately apply mathematics to this Office-inspired question, others out there have had more success with different questions.  For example, here is a computer science paper on automated recognition of phrases that can reasonably be followed up by the phrase “that’s what she said.”  What a neat problem!  It looks so hard, but they somehow manage to tame it.

Oh, Michael Scott.  You will be missed.

While the institute seems like a delightful place to work, I regret to inform you that the research coming out of the institute is as bogus as the existence of the institute itself.  The claimed solution to the problem of maximizing meat (or meat flavor, depending on your source) while minimizing cost is contained in the 3 BK Stackers pictured here (image courtesy of foodbeast):

As you can see, the Stacker family of burgers has three members, coming in at price points of $1, $2, and $3, respectively.

Right off the bat you should notice something problematic with their solution.  If the aim is to optimize meat, why are there components that don’t involve meat?  These components would only add to the cost, without adding to the amount of meat.  You should be able to sell just a single beef patty with some bacon for less than you would be able to sell the $1 stacker, for instance.  Right away, then, we see the corrupt influence that the Burger King sponsorship has had on the otherwise (I’m sure) irreproachable standard of academic integrity set by the Meat Mathematics Institute.  Without this sponsorship, I’m sure the mathematicians would’ve discovered a far more optimal meat-delivery system: the (possibly bacon-wrapped) meatball.

Perhaps the Meat Mathematics Institute should look for a more appropriate sponsor.

Even if we restrict our attention to solutions to the “meat/money” problem that are in cheeseburger form, however, it’s still easy to see that the solution proposed by the Meat Mathematics Institute is not correct.  Indeed, a quick stop at the Burger King website reveals the meat differences between the $1, $2, and $3 BK Stackers.  The differences are as follows:

• $1 Stacker: 1 beef patty, 2 half slices of bacon.
• $2 Stacker: 2 beef patties, 3 half slices of bacon.
• $3 Stacker: 3 beef patties, 3 half slices of bacon.

If we divide out by cost, we easily see that per dollar, these three options give you different amounts of meat!  In particular,

• $1 Stacker: 1 beef patty, 2 half slices of bacon per dollar
• $2 Stacker: 1 beef patty, 1.5 half slices of bacon per dollar
• $3 Stacker: 1 beef patty, 1 half slice of bacon.

Clearly these three solutions are mathematically inconsistent.  How can the $2 BK Stacker be as optimal a solution as the $1 stacker, when two $1 stackers yield one additional slice of bacon?  Similarly, how can the $3 stacker be as optimal a solution as the $1 stacker, when three $1 stackers yield three additional slices of bacon?  The only way these three solutions can be consistent is if the cost of bacon is 0, which, unfortunately it is not.  But if it were, then the optimal solution still would not be present here, since one could put as much bacon as one wanted on his or her burger without affecting the cost.

It’s a shame that this institute has been compromised by corporate interests, but this is perhaps not wholly unexpected.  While normally I would advise you to trust your local mathematician, in the case of the meat mathematician, I must regretfully ask you to be on guard.  Especially if he asks you to take him out for a meal.

The joke ran at the end of a muppet-themed segment of the show.  In an homage to Sesame Street, after the segment finished (but before the somewhat racy joke involving a very physical muppet Moe) an announcer stopped to give thanks to the sponsors of the show.  Unlike Sesame Street, however, which is sponsored every day by two letters and a number, this episode of The Simpsons was sponsored by one symbol and one number that looks like a letter:

In case you’re late to the party (since I don’t think that clip will be online forever), let me quote: “Tonight’s Simpsons episode was brought to you by the symbol umlaut, and the number e. Not the letter e, but the number, whose exponential function is the derivative of itself.”

Kudos to the writers for incorporating some choice math humor into the tail end of this episode (I’m willing to overlook some qualms with their wording).  Perhaps Simpsons aficianados would can begin preparations for next year’s e day.

In and of itself, this might not seem like a particularly newsworthy achievement – as any Pi Day aficionado can tell you, there are people who have memorized more digits.  Perhaps what makes it newsworthy is the fact that the student is only twelve years old, or, more perversely, the fact that his accomplishment came in response to the challenge of his math teacher, who asked his students to memorize as many digits of pi as possible.  By far the most depressing part of the video is a brief clip that shows all the students in the classroom mindlessly rattling off successive digits of pi.  The lack of enthusiasm is almost palpable.

Now, I don’t want to come off as too much of a curmudgeon here.  I have no doubt that this student is stoked that he made it on to TV for an academic achievement, regardless of the actual merits of that achievement (at least the student is aware enough to remark that the information he’s memorized will probably never be put to use).  That’s fine – he has every right to be proud of himself for making it onto the local news.  What really gets my goat is the fact that this teacher thought it would be a good idea to make students memorize digits of pi.  I can think of few better ways to dampen a natural enthusiasm for mathematical learning than by asking students to memorize a series of digits that will have no practical value for any of them, ever.  It would be like having an English teacher ask students to memorize a random string of words which, taken collectively, didn’t teach the student anything about vocabulary or grammar.

Is there any benefit to this exercise?  According to the teacher, “The ability to memorize that much stuff has to help tremendously.”  Well, ok.  But aren’t there more important things to learn about in math class?  Is math class really the best venue to discover a talent like this?  I am fairly certain that students in Singapore aren’t spending class time and homework time memorizing digits of pi.  I’m sure this teacher has good intentions, but I fail to see much value in this apparently newsworthy event.  The mystique of the number pi, I suppose, never fails to attract attention.

If this exercise is what gets this sixth grader interested in math, then by all means he should memorize as many digits of pi as he can.  For the vast majority of students, however, such an exercise is probably beyond tedious.  I can only hope that this news story doesn’t inspire other teachers to compel other students to do the same thing.