Archive for the ‘Math in Pop Culture’ Category

The Scarecrow and the Pythagorean Theorem

January 10, 2010

Ray Bolger, who played the Scarecrow (THE Scarecrow) was born 106 years ago today.  In his honor, here’s a clip from Youtube in which the Scarecrow gets a Doctorate of Thinkology and makes reference to the Pythagorean Theorem — sort of.  (The exact words of the Scarecrow are, “The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh, joy, oh, rapture. I’ve got a brain!”)

Ray, here’s to you!  And remember: 1062=562+902.

An ode to Andreia Chaves

December 14, 2009

(To the tune of “The Christmas Shoes”. Sorry about that.)

It was almost exam time
Sum. Solve. Is this a prime?
Trying to find a good shape or two
Not really in a polygon mood
Then appeared right in front of me
As I surfed somewhat anxiously
The creativity that designers do
But these were mathematical
Artsy pairs of shoes

And these shoes weren’t worn or old
They had V+F=E+2 from heel to toe
Fit for show more than for play
And I couldn’t believe, but they made me say

Yes I wanna show these shoes
To my classes please
There’s origami
Which adds quite a bit of style

I’d better hurry now
My datebook says there’s not much time
You see, classes have met for quite a while
And I know these shoes will make them smile
Cuz those shapes look kind of beautiful
While everyone is studying tonight.

These shoes, designed by Andreia Chaves of São Paulo, Brazil, were featured on yatzer (found via cnet).    All photos are used with permission from the folk at yatzer (Thanks!)

Dieting the MATH Way

May 19, 2009

There are many diets in the world, but most of them fail for the simple reason that they use the wrong kind of mathematics.  Simple logic suggests that food choices based only the arithmetic of

Pounds Lost = (Calories Used – Calories Eaten)/3200

are insufficient for good health.  The math is just too elementary!  If you want an Advanced Diet, you need to use Advanced Mathematics!  Now with the power of Theorems.

The Harmonic Diet
If you’re a person who starts off strong but has trouble sticking to a plan, this is the diet for you!  On the first day, lose 1 pound.  Tough, but possible with that initial surge of motivation.  On the second day, lose 1/2 pound.  On the third day, lose 1/3 pound.  On the fourth day, lose 1/4 pound, etc.   As your motivation wanes, so will your weight loss; however, thanks to the power of Calculus you are guaranteed to eventually reach your goal weight.  One of the amazing features of this diet is that it will still work even if you don’t start with the 1-pound loss:   you can just as easily jump in on Day 3 or whichever day suits your fancy.

The Zeno Diet
Endorsed by Ancient Greek Philosophers everywhere, this plan is perfect for the daytime snacker!  Start by choosing a total number of calories that you’d like to consume in a day:  say, 1000 calories.  Your first meal should be 500 calories; after all, breakfast is the most important meal of the day, and skimping is unwise.  Your second meal can be whatever you want, whenever you want:  just be sure to limit it to 250 calories.  Want another snack?  No problem!  Treat yourself to any 125 calorie food.   Your next tidbit can be 62.5 calories, the next 31.2 [no rounding up!] and so on.  With this diet plan you can snack as many times as you want, and you’ll never exceed (or even reach!) your total allotted calories.

The Banach-Tarski Diet
This diet uses the power of Set Theoretic Geometry to help those people who want larger portion sizes.   Start with small amounts of your favorite foods.  Roll each item into a ball and grab a sharp knife.  Then cut each ball into 5 pieces and reassemble into two balls OF THE SAME SIZE!  Repeat as desired.

The Fibonacci Diet
Is your focus more on healthy eating than actual calories?  The Fibonacci Sequence 1, 1, 2, 3, 5, 8, 13, … is more than just a fancy way to convert between miles and kilometers:  you can arrange your entire plan around these special numbers.  Break your eating into 3 meals with 2 snacks.  Make sure each meal is made up of 1/2 carbohydrates,1/3 protein, and1/5 fat.

It’s OK that the amounts add up to more than 1, becausDaVincie THIS diet was based on a Bestselling Novel and is featured both in Women’s World Magazine in 2006 and in the book  The Diet Code:  Revolutionary Weight Loss Secrets from Da Vinci and the Golden Ratio by Stephen Lanzalotta, which you can buy for as little as 13¢.  Now that’s value.

In a fortunate coincidence, Walking Randomly makes all of this even easier by showing show to use Wolfram Alpha to compute calories.

Math in the Movies

May 6, 2009

120px-enterpriseA recently ran across a great resource page of  “Math in the Movies” from MathBits.  This page lists clips of movies (with links if they’re available online) and worksheets that can be used in the classroom.

Two of the examples relate to Star Trek, which as you all know (right?) will be in theaters starting this weekend.  In Episode 20 (“Court Martial”) of The Original Series, James T. Kirk is accused of murdering Benjamin Finney, the Records Officer.  At one point late into the episode, Kirk uses the computer to hear the heartbeats of everyone on The Enterprise.  As he explains:

Gentlemen, this computer has an auditory sensor.  It can, in effect, hear sounds.  By installing a booster we can increase that capability on the order of one to the fourth power.  The computer should be able to bring us every sound occurring on the ship.

One to the fourth power?  Not so impressive (and was I the only one thinking that you’d hear a lot more than heartbeats if every sound was magnified?  Wouldn’t breathing and moving be really LOUD?  But I’d better be careful with my critiques, because I’m a big fan of The Next Generation and a recent viewing revealed that it isn’t immune to problems either.)

The MathBits worksheet to accompany the scene is here.  They don’t link to a video clip, but you can see it on YouTube.  Actually, you can see the entire episode on YouTube:  the mistake starts at 38:10.

(Edit 5/9:  the embedding no longer works, but you can see the episode at this link.)

Tetrahedral Pyramids (part 2 of 3)

April 24, 2009

The first post in this series, Pascal’s Pyramid (part 1 of 3), explored number patterns that arise in a 3 dimensional version of Pascal’s Triangle: a pyramid with square cross sections.  Another way to do a 3D version of Pascal’s Triangle would use triangular cross sections.   If you think of building a tetrahedron by stacking oranges in a pile, each orange is in contact with (up to) three oranges that lie in the level above it.

As before, we set the top number in the pyramid to be 1, and assume that at the lower levels, each number will be the sum of the (up to) three adjacent numbers on the previous level.

Also as before, we know to expect the binomial coefficients to appear on each of the three triangular faces of our new pyramid.

In the interior, the numbers generated satisfy the recursion

f(n+1, a+1, b+1) = f(n, a, b) + f(n, a, b+1)+ f(n, a+1, b+1)

where f(0,0,0) = 1, and f(n,a,b) is understood to be zero if any of the parameters go out of range (that is, if a or b is either negative or greater than n, or if b>a).

The resulting pyramid:


And for example, the numbers on the seventh level from the top are:


Examining this triangle, we recognize an odd variation on Pascal’s triangle:  each row is a multiple of a row of the usual Pascal triangle, where the multiplier is the left-most entry in the row.  Since those terms are also terms on an outer face of the pyramid, we know they too are binomial coefficients, and we are led to conjecture that  f(n, a, b) = { n \choose a } {a \choose b}.

As before, we can prove this result by induction on n.  The base case, when n=0, works, since {0 \choose 0} {0 \choose 0} = 1.

If we assume (for some specific value of n)  that each f(n, a, b) = {n \choose a} { a \choose b}, then it follows that

f(n+1, a+1, b+1) = {n \choose a}{a \choose b}  + {n \choose a}{a \choose {b+1}}+ {n \choose {a+1}}{{a+1} \choose {b+1}}

= {n \choose a} \bigg( {a \choose b} + {a \choose {b+1}} \bigg) + {n \choose {a+1}}{{a+1}\choose {b+1}}

= {n \choose a}{{a+1} \choose {b+1}} + {n \choose {a+1}}{{a+1} \choose {b+1}}

= {{n+1} \choose {a+1}}{{a+1}\choose {b+1}}

And thus we have our identity by induction on n.

This suggests a direction for exploration:  what interesting things do we know about Pascal’s Triangle, that might generalize to three dimensions?

Coming up:  Sierpinski’s Pascal Pyramid (3/3)

Pascal’s Pyramid (part 1 of 3)

April 20, 2009

q138In an earlier post, Ξ had mentioned a question from Who Wants To Be A Millionaire that made reference to “a famous “pyramid” of numbers that starts with the number one on top”.  The intended answer was “Pascal’s Triangle” (my emphasis).

That got me wondering what a three dimensional analogue of the Pascal Triangle might look like.

Today we’ll explore a Pascal-like construction of a square pyramid.   The top-most point on the pyramid will be assigned the value 1.  As we move down the pyramid,   let’s assign each point a value  by taking the sum of the 4 terms directly above and adjacent to each point.

If you think about the 4 outer faces, each point along a face only has two points that lie above it.  Thus if we restrict our attention just to one outer face, our construction is identical to the two dimensional Pascal triangle, and for  each face we should get the familiar binomial coefficients.

One can describe  how a given entry depends on those above it:  If we let f(n, a, b) be the entry in the nth layer down from the top, in row a and column b, then the entries that lie above will be in layer n-1, their row will be a or a-1, and their column will be b or b-1.  Thus

f (n, a, b) = f(n-1, a-1,b-1) + f(n-1, a, b-1) + f(n-1, a-1, b) + f(n-1, a, b)

with f(0,0,0)=1, and f understood to be zero if any of the parameters get out of range (e.g. if a or b > n, or is negative).

Numerically, we get the following pattern (for 0 ≤ n ≤ 6):


For example, the entries in the fifth layer are:


Each entry appears to be the product of the first entry in its row, with the first entry in its column.  Since we know the row and column values are binomial coefficients, we conjecture that f(n,a,b) = {n \choose a} {n \choose b}.

To prove this, we start by verifying that when n=0, the product of the binomial coefficients is equal to 1.  Proceeding inductively, we assume that the result holds for n-1, and we compute f(n,a,b):

{ {n-1} \choose {a-1}} {{n-1} \choose {b-1}} + {{n-1} \choose {a}} { {n-1} \choose {b-1}} + {{n-1} \choose {a-1}}{{n-1} \choose {b}} + {{n-1} \choose {a}} {{n-1} \choose {b}}

= \bigg( {{n-1} \choose {a-1} } + {{n-1} \choose {a}} \bigg) \bigg( {{n-1} \choose {b-1}} + {{n-1} \choose {b}} \bigg)

= { {n}\choose {a}} {{n} \choose {b}}

QED by induction; each entry in Pascal’s Square Pyramid is a product of two binomial coefficients, and is equal to the sum of the four terms immediately above it.

Coming upTetrahedral pyramids (2/3);  Sierpinski’s Pascal Pyramid (3/3)

Martha Math

March 21, 2009

I just learned about the show Whatever, Martha! from a friend of mine.  The show consists of Alexis Stewart (Martha Stewart’s daughter) and Jennifer Koppelman Hutt watching and commenting on old episodes of Martha Stewart Living.  Like a domestic version of Mystery Science Theater 3000.

I prefer the one about making s’mores (mostly because, Seriously?  Homemade sticks?) but in an effort to keep it math related I’ll post a 5-minute snippet that has some math.  Here’s the word problem:  if the diagonal of the square is 10½″, what is the side of the square?

(Apparently Martha and her mom have a better relationship than this video would imply.  If Martha Stewart is OK with this, my respect for her has just gone way up.)

NUMB3RS Puzzles

February 26, 2009

puzzle6-numb3rsThere’s a new addition to the math-fights-crime TV show NUMB3RS.  This season, the folk at Wolfram have created a math puzzle that goes along with each episode of the show.

For example, in Scan Man, the passage

Charlie:  I’m not sure an Error Correcting Code is gonna get you there — That is what  you’re using, right?

Amita:  …And have been for weeks.

inspired the following puzzle:

A spy captures a code key (first block) and two 17-character mathematical messages. Unfortunately, almost nothing seems to match the key. Can you decipher the two messages, and also find the third hidden 17-letter phrase?

Image used with permission from Wolfram Research, Inc.

The puzzle is here with tabs for a hint (my experience is that the hints are pretty useful for solving the puzzle) and for the quote in the show that inspired the puzzle.  There’s also a link to the solution.

If you’re feeling bad because you love the puzzles but you missed one of the inspiring episodes, fear not!  You can watch the entire season online!  Hooray for online television!   (I don’t know how long they’ll be up, so you probably should get watching while you work on those puzzles.  Clearly this is more important than doing/grading the homework or working on the project you were about to get to.)

There’s a little more background information (including a correction to one of the solutions) in Tuesday’s post on Wolfram’s blog.

Happy Solving!

Juxtaposition: Millionaire Triangles

February 24, 2009

There was a math question on Who Wants to Be a Millionaire yesterday!  According to this transcript of the February 23 game, this was the $50,000 question:

Named for the mathematician who designed it, a famous “pyramid” of numbers that starts with the number one on top is called what?
A. Fibonacci’s triangle
B. Pythagoras’s triangle
C. Pascal’s triangle
D. Fermat’s triangle

Fibonacci's TriangleThe correct answer is Pascal’s triangle.  But it got me thinking, what would the other triangles look like?  We’ve actually posted before on Fibonacci’s Triangle:  as envisioned by Doug Ensley, it’s a triangle with Fibonacci numbers down each side and with each interior number equal to the sum of the two numbers above it (as with Pascal’s triangle).  This is also called Hosoya’s triangle, according to wikipedia (named after Haruo Hosoya, who wrote about it in The Fibonacci Quarterly in 1976).

fermat-triangleSo what about Fermat’s triangle — what would that look like?  Since Fermat numbers are numbers of the form 2^{2^n} + 1, a natural adaptation to Fibonacci’s triangle would be to have the Fermat numbers on the outside, and to use sums of numbers to fill in the inside.  This looks like of funny, though — the numbers on the outside grow much faster than those on the inside.

What has me stumped is the notion of Pythagoras’s Triangle.  Would it have something to do with Pythagorean triples?  Maybe it would be a tetrahedron rather than a flat triangle, with a 1 on top, and then a Pythagorean triple beneath it — say the iconic 3, 4, 5 — and then…then I’m stumped.  What would come next?  Is there even an ordering for Pythagorean triples?  (Well, maybe:  TwoPi wrote earlier about how every Pythagorean triple can be written as (2uv, u^2-v^2, u^2+v^2), for positive integers u and v, so we could put an ordering on u and v and retrict ourselves to the primitive triples.)  But I’m not sure how to generate the entire tetrahedron.

Then again, for a Pythagorean triangle, perhaps we could just draw a right triangle and call it a day.

Incidentally, the link to the transcript also has commentary about whether Pascal’s triangle was the only reasonable answer.  The consensus seems to be YES, if only because the question referred to a “famous” triangle “pyramid”.

Yet more math in bones: calculating weight

January 21, 2009

bathroom_in_the_beamish_museumYes, more math in Bones!  This one from “The Truth in the Lye”, Season 2 Episode 5.  In this episode, a body was found in a bathtub that had been filled with lye, so the body was in an advanced state of decomposition.  (My apologies if you’re reading this over a meal.  It only gets more descriptive, so you might want to skip this post.)

Here Camille Saroyen is talking to the team at the Jeffersonian about how much the person would have weighed.  (I switched to first names for the dialogue, because I had no idea what Camille’s last name was until I looked it up.  Speaking of looking this up, after trying to transcribe it from the DVD I discovered that the whole thing had been transcribed for me here).


Camille: What’s our starting weight, Zack?

Zack: Starting weight is 542.13. [lbs]

“Bones”: The tub itself weighs about 200 pounds. Capacity is 34 gallons.

Camille: Which at about 8.3 pounds a gallon comes to 270-275.

Jack: And two-thirds full makes it about 180, putting this guy somewhere in the 160-pound weight class.

[Brennan nods]

[Cam is stirring the tub, where orange is starting to appear]

Camille:The cream always rises. Or in this case, melted body fat. [raises tong, melted body fat drips off] I’ll measure its volume to determine body type.


My initial thought was to figure out if this was good science or not, assuming that it must be because, hey, these people work at the Jeffersonian.  I couldn’t find the weight of a bathtub to check their very first calculation exactly; however, the estimate I found here suggested that clawfoot tubs were 250-350lbs, and modern tubs weigh less, so that 200lbs might well be in the ballpark.

But I started to worry about all the rounding.  Does the tub weigh 200 lbs exactly, or could that be off by 10 lbs?  Then there’s that whole “about two-thirds full”.  Close, but again I suspect a wide margin of error, especially since they rounded when they said that 34 gallons times 8.3 pounds/gallon was 270-275 (It’s 282.2, which is  10 pounds off in a situation that implies accuracy to more than 10 pounds.)

All in all, I think their final answer of 160 lbs (from 542.13-200-180, which equals 162.13) is really pretty rough.  Ballpark, maybe, but it could reasonably be anywhere from 140 to 180 lbs or more.  I want to believe in their good mathematics, but I fear that Barry Leiba was right when he implied in his comment earlier that they’re not really trying to show good science.*

*Not that that will stop me from continuing to talk about it.

Quick: what’s 500+500?

January 16, 2009

We’ve seen Joey discuss the value of long division (when wanting to portray someone who has just gotten bad news);  now enjoy a Friends excerpt where his addition skills get a work-out:

(You can find the entire scene here.)

Watching this video clip, I’m struck by a number of mutually tangential thoughts:

  • I see a LOT of my students pull out their calculators to do basic arithmetic these days (multiplying a two digit number by 2 or 3, or adding two 2-digit numbers).  I’m often surprised during a calculus quiz to see someone use a calculator to check their integer arithmetic (in combining like terms of an algebraic equation, say).
  • I’m heartened to hear laughter in the video, and wonder just how long it will be before American audiences see nothing uncomfortable or amusing in an adult needing a calculator to find 500+500.
  • I’m reminded that in laboratory studies of human responses to stress, one of the standard ways to ethically induce stress on test subjects is to have them perform multidigit subtraction in their head (e.g. count down from 483 by sevens) while the experimenter urges them to work more quickly.

But most of all, I’m reminded of how much I miss watching television, and wonder how I ever used to find the time to do it.  (Oh yeah, that’s right, that was life before parenthood.)

Thank you to Ionica at wiskundemeisjes, who brought this Friends scene to our attention, and suggested that we both post about it on our blogs simultaneously!

If you can’t trust celebrities, who can you trust?

January 4, 2009

Mariah Carey has been in the science news lately, amazingly enough.  No, not for a discussion of her five-octave vocal range, but rather for a discussion of Einstein’s Theory of Special Relativity.   Her most recent album appears to be named for Einstein’s mass-energy equivalence formula:  E=mc^2, as you can see from the album cover art seen above at this link. [I’m not including the image here, for copyright reasons.]

Apparently Mariah Carey was recently asked about the album title; she has been quoted as having said that it stands for “emancipation equals Mariah Carey times two”.

I suppose “Mariah Carey times two” might be loosely translated as “buy my two most recent albums”, as her previous album specifically referred to emancipation in the title.  But regardless of her intent, the algebraic notation does not refer to multiplying by two, but rather multiplying by a second copy of “c”.  (Perhaps the album should have been a series of duets by Mariah Carey and Charo….)

A British nonprofit organization, Sense About Science, recently called attention to Mariah Carey’s misinterpretation of exponent notation and other celebrity gaffes when speaking about science in public, in their “Celeb Audit 2008“.  While Mariah’s quote is the only one that is explicitly mathematical, many of the others could be useful fodder for discussion in a statistics classroom environment, and in many cases (as in the quotes from American politicians by the names of McCain, Obama, and Palin) involve far more weighty matters, and in all honesty far more disturbing misrepresentations.

How tall was the murderer?

December 13, 2008

bones-leashIn a recent episode (“The Bone that Blew”) of the TV series Bones — and yes, I am thinking of creating an entire category on this blog just for brief mentions of math in Bones — a body was discovered and it turned out to have been dragged by a choke chain. Temperance Brennan points out that the angle of the fracture was 18°, indicating that the leash was being pulled at that angle. The blond assistant guy then added “Assuming a standard 4 foot leash, the person who dragged the victim is at most 5’5″.” They showed a picture like the following, only it had a skull and broken neck at the bottom. I spared you that.


How did they come up with the 5’5″ height? Since the leash was 4 feet long and being held at 18°, the vertical height from the body to the hand that was dragging it must have been 4·sin(18°)=1.24 feet. The leash probably didn’t touch the ground, though: it would have been attached to the neck, and therefore something like 6″ above the ground. Let’s assume the vertical height is therefore 1.74 feet above the ground.

Presumably Blond Guy knows some formula that relates how tall one’s hand is above the ground with their total height, although my extensive searching (i.e. Google) didn’t reveal any such formula. Never one to be deterred by a lack of actual facts, I forged ahead and found that my height is 2.9 times as big as the distance from the floor to my hand. Assuming that ratio holds constant for everyone [which is certainly doesn’t], this would mean that the person dragging the body was (1.74·2.9=5.05) just over 5 feet tall.

So I still have no idea where the 5’5″ came from, assuming it wasn’t just made up out of thin air (which it wasn’t, right? Because that would make me really sad.) Maybe the person was shorter. Maybe my adding 6″ was an underestimate and it should have been closer to 9″ (which would mean I’d add 0.75 to 1.24, getting 1.99, which yields 5’9″ when multiplied by 2.9). Maybe the 2.9 really varies quite a bit person by person. Or maybe the person was tall, but leaning over, and Blond Guy took that into account.

So alas, I really don’t have an answer. But fortunately for the folk at the Jeffersonian, they found additional evidence (identifying the trajectory of the bullets that killed the man, and used that to get a different height that led them to the killer). If you’re dying to know who did it — so to speak — you can see the episode at least for the time being here on Fox. It’s Season 4, Episode 10, and will probably only be up for a few more weeks.

The Day the Earth Stood Still

December 10, 2008

klaatuandgortdepartThe Earth first stood still over 50 years ago, when from out in space came a Warning and an Ultimatum.   And on December 12, it’s halting again for a remake.

So what does this have to do with math? Two things.  The first (which is the one that initially led to this post) is that one of the people in this Friday’s film is 10-year old Jaden Smith.  And do you know what is favorite subject is?  That’s right, math.  Thus says People magazine, after an interview with him last night at the New York City premiere:

“It’s math,” Jaden told PEOPLE at the New York premiere of Earth on Tuesday. Why math? “Because I’m good at it,” he added.

And the second reason, which I only discovered while trying to find out if anyone else in the movie also had math as a favorite subject, is that there is an Actual Math Equation in this movie.  Apparently there’s a chalkboard scene with John Cleese (who plays Professor Barnhardt), which Director Scott Derrickson referred to in this interview on SciFi Weekly:

Derrickson: It’s a real math equation about a real significant high-physics theory about the universe, and we tried to be truthful to the scientific aspects. I just did an interview with Discover, and the interviewer was really surprised at little things in the movie, but that being the biggest one. We had an astrophysicist who worked with Keanu [Reeves] and John. I still remember watching them for quite a long time. I don’t remember where we were in the production, but for quite a long time in a room working out the back-and-forth of that, and then we added material to make it longer at one point … just to get that kind of flow and rhythm to it. I didn’t have much to do with that. I’m just remembering, because it was really Keanu and the math guy, the theoretical physicist, and John Cleese. The three of them just kind of figured that out, and I saw it and thought it was fantastic.

And fantastic it is.

How Fast Could They Have Gone?

December 9, 2008

I love the show Top Gear. In particular, I am a big fan of the “challenges” they have. Recently (yesterday afternoon, when I should have been grading), there was an episode on BBC America from 2006 in which host Richard Hammond was at London’s ExCeL Centre for the British Motor Show, and before the exhibits were set up, he had The Stig (the show’s “tame racing driver”) test two vehicles to see how fast they could go in the 385-metre hall. The first was a Chevy Lacetti, their “Star in a Reasonably-Priced Car” car, which reached 70 mph. The second was a Toyota F1 car (the TF105, I think), which reached…

only 81 mph?

This led me to wonder what the top speed of a car could be on a 385-metre stretch. Let’s find out.

For simplicity, I will assume constant acceleration a (I said simplicity, not accuracy), and constant deceleration from braking (again, not particularly realistic). Let v_T be the top speed. Then the distance covered in accelerating to top speed is

\displaystyle\frac{1}{2}a \left( \frac{v_T}{a} \right)^2

and the braking distance is

\displaystyle \frac{v_T^2}{2\mu g},

where μ is the coefficient of friction between the tires and the floor, and g is gravity. We then have the following equation:

\displaystyle \frac{v_T^2}{2} \left( \frac{1}{a} + \frac{1}{\mu g} \right) = 385

and thus

v_T = \displaystyle \sqrt{\frac{770a\mu g}{\mu g+a}}.

At this point, I plead ignorance. I tried (not very hard) to find a reasonable coefficient of friction for racing tires, and to find a 0-60 time for a Formula 1 car (the McLaren F1 does it in 3.9 seconds). In the end, I made a spreadsheet for μ between 0.4 and 0.7, and a between 6 and 7 m/s2. Given an ideal setup—starting at one end of the hall and stopping perfectly at the other end—the top speed is somewhere between 95.6 mph (μ=0.4, a=6) and 115.5 mph (μ=0.7, a=7).

What does this mean? I don’t know, but it makes The Stig’s 81 mph sound pretty good, given the initial burnout, the nonconstant acceleration and braking, driver reaction time, and an interest in personal safety (cf. Hammond’s “Formula 1 car-shaped hole” comment).