How fast was Godzilla running?

A bunch of fossilized dinosaur footprints in the Mideast were recently discovered by Mohammed Al-Daheri who, incidentally, is a journalist (so all you folk who wanted to be paleontologists but didn’t get around to it, it’s not too late to make great discoveries!). The footprints looked like these ones:

but were in Yemen, and not in the Science Museum in Logroño, Spain (where jynus took this picture). National Geographic has a neat article about the footprints with even neater pictures here. Part of the article states, “The paleontologists were also able to infer the size, age, and speed of the sauropods based on their prints.”

Size of the sauropod is presumably related to the size of the footprints, and age might be related to size, but how did paleontologists determine the speed? To demonstrate, we’ll use footprints to determine how fast Godzilla was moving in the picture below:

The first thing we can do from the footprint is to determine the Godzilla’s leg length. From fossils and stuff “we” know that most dinosaur hind feet were ¼ of the size of their leg length, so in theory we should be able to measure the footprint and multiply by 4. It turns out, though, that Godzilla has disproportionally large feet, so we’ll cheat and measure his leg length exactly. Our Godzilla’s leg length is 6.5″, which we’ll translate to 0.1651 meters so we can be all metricy like the scientists.

Next we have to measure the stride length.

Exactly 8 inches, which is 0.2032 meters.

Now we find the relative stride length with the following formula:
Relative Stride Length (RSL) = (Stride Length)/(Leg Length).

Godzilla’s Relative Stride Length is about 1.23.

Now it gets complicated. A famous scientist guy in the Biology department at Leeds University, R.M. Alexander, published something about something called Dimensionless Speed. (See how well I understand it? I think it takes into account that similar animals move in a similar way, regardless of size. ) He came up with the following equation:

Speed = (Dimensionless Speed)√(leg length ·g)

The constant “g” is that familiar gravitational constant 9.80665 meters per second per second. Bet you never expected to see that here, right?

So now we just need to figure out the Dimensionless Speed, and we do that using the Relative Stride Length. There’s a graph in Figure 2 of this article, but it didn’t have a category for “Giant Monsters”. If Godzilla were like a human, then a Relative Stride Length of 1.23 would correspond to a Dimensionless Speed of about 0.3. On the other hand, in this activity (an outline of doing a lesson like this in High School Classes) there is a chart of four dinosaur’s RSL and DS, and plugging those numbers into my calculator gives:

(Dimensionless Speed) = 0.6425(Relative Stride Length)-0.4679

(with a correlation r=.9997, so it’s pretty good).

If we treat Godzilla like a dinosaur, then plugging in his RSL of 1.2308, we get a Dimensionless Speed of 0.3228, which is close to the 0.3 we estimated above. So let’s use that.

Now we plug in DS=0.3228, leg length = 0.1651 meters, and g=9.80665 into
Speed = (Dimensionless Speed)√(leg length·g)

and we find that Godzilla’s speed is a whopping 0.41 meters per second! That’s 41 centimeters per second, which is pretty fast for a guy whose barely a foot tall. To put it into perspective, if you multiply by the scaling factor of 133.3 (from when we calculated Godzilla’s weight), it’s equivalent to the original 50 meter Godzilla running at just over 54 meters per second, or just over 122 miles per hour! Boy, you’d never know that he could run that fast, could you?

Makes you wonder just how fast those Yemeni sauropods were booking, doesn’t it?

I’m thinking that if I’d done the calculations before writing all this up, Godzilla’s footprints would have been placed a little closer together. Thanks to Bill Korth, who demonstrated this method of figuring out dinosaur speeds at a teacher workshop. He found it in Dinosaurs: The Textbook (3rd ed.) by Spencer G. Lucas.

Mathematician of the week: Paul Guldin

Paul Guldin spend much of his teaching career at Jesuit colleges in Rome and Graz, Austria. He is remembered for Guldin’s Theorem: If any plane figure revolve about an external axis in its plane, the volume of the solid body so generated is equal to the product of the area of the figure and the distance traveled by the center of gravity of the figure.

Guldin published this result in his De centro gravitatis (1635-1641), and the result is more widely referred to as Pappus’s Theorem, having been anticipated by Pappus of Alexandria circa 320 CE.

Ivor Bulmer-Thomas writes about the question of whether Guldin had known of Pappus’s work in an article in Isis (vol 75, 1984).  He argues convincingly that Guldin is almost certain to have read Pappus’s work in translation in 1618, and probably unconsciously plagiarized it some 20 years later when completing his own work on centers of gravity.  Bulmer-Thomas’s article debunks a longstanding error in historiography:  Moncula’s 1758 history of mathematics erroneously stated that Pappus’s Theorem had been omitted from the first Latin translation of his work, and consequently Guldin’s discovery must have been arrived at independently.  Moncula corrected this error in the second edition of his history (1799), but the incorrect account has gotten wider dissemination (from Paul Ver Eecke [editor of a modern collection of Pappus's writings], Carl Boyer’s History of Mathematics, and even the Dictionary of Scientific Biography’s entry for Guldin, for example).

Moral:  as much as possible, check and recheck the claims of your secondary sources, even if they are highly regarded and generally known to be sound.

Mathematicians with birth or death anniversaries during the week of June 8 through June 14:

June 8: Birthdays of Giovanni Cassini [1625] (discovered Cassini gap in Saturn’s rings; introduced use of Jovian moon observations to determine longitude), Caspar Wessel [1745] (”Argand” diagram), and Charlotte Angas Scott [1858] (algebraic geometer; first head of math dept at Bryn Mawr)

June 9: Birthday of John Littlewood [1885] (worked with Hardy on theory of functions); death anniversaries of John Machin [1751] (use of series to compute over 100 digits of π) and Vivienne Malone-Mayes [1995] (asymptotic analysis; fifth African-American woman to earn a PhD in mathematics [in 1966])

June 10: birthdays of Mohammad Abu’l-Wafa [940] (introduction of tangent function; mathematical astronomy) and Pierre Duhem [1861] (mathematical physics); deaths of André-Marie Ampère [1836] (PDEs, chemistry, and physics) and Antonio Cremona [1903] (introduced use of graphical methods in study of statics)

June 11: Birthday of Hilda Hudson [1881] (algebraic geometry); death of Wilhelm Meyer [1934] (invariant theory)

June 12: Birthdays of Paul Guldin [1577] (Guldin’s Theorem) and Vladimir Arnold [1937] (solution of Hilbert’s 13th problem; 2001 recipient of the Wolf Prize); death of Jean Frenet [1900] (differential geometry)

June 13: Birthdays of James Clerk Maxwell [1831] (study of electricity and magnetism), Ernst Steinitz [1871] (abstract algebra), William Gosset [1876] (”Student”’s t-distribution), and John Nash [1928] (game theory)

June 14: Birthdays of Nilakantha Somayaji [1444] (mathematical astronomy, series for arctangent), Andrei Andreyevich Markov [1856] (probability theory), Alonzo Church [1903] (computability; lambda calculus), and Atle Selberg [1917] (zeta function); death of Colin Maclaurin [1746] (analysis)

Sources:  Ivor Bulmer-Thomas, “Guldin’s Theorem-Or Pappus’s?”  Isis 75 (1984) 348-352 (for information on Guldin’s Theorem and the controversy over its provenance);   the MacTutor History of Mathematics website (for dates and biographical data generally; for the image of Paul Gulden)

Hard Drive Failure

One of my stats students stopped by this afternoon with an interesting question: she needed a new hard drive, and was looking at this one, which boasted an Annual Failure Rate of 0.34%. What she was wondering was how to calculate the chances that the hard drive would fail within, say, 5 or 10 years. Her first guess had been to multiply 0.34% by that number of years, but she didn’t think that was actually the way to go.

My first thought was to break it down into the chances that the hard drive would fail in one year [(0.0034)] and then add on the chances that it didn’t fail the first year but did the second [(0.9966)(0.0034)] and then add on the chances that it didn’t fail the first two years but did the third [(0.9966)²(0.0034)] etc., all the way up to adding on the chances that it didn’t fail the first (N-1) years but did in the N^th year [(0.9966)^N-1(0.0034)].

Then I added them all up, and factored out the (0.0034), giving (0.0034)[1+(0.9966)+(0.9966)²+...+(0.9966)^N-1]. But 1+x+x²+…+x^N-1=(1-x^N)/(1-x), so that equation simplified to (0.0034)[1-(0.9966)^N]/[1-0.9966],
which further simplified to 1-(0.9966)^N. And when I thought about it, that made sense: (0.9966)^N is the probability that the hard drive does NOT fail anytime in the first N years, so the probability that it does fail would be 1-(0.9966)^N.

Then we started looking at a few actual numbers. The likelihood that the hard drive fails within one year is 0.34%, that it fails within two years is 0.68%, that it fails within three years is 1.02%. Indeed, since 0.9966 is so close to 1 it has a very small effect so the student’s initial thought of N·(0.34%) turns out to be a pretty good estimate for quite a few years [at 10 years out the theoretical probability would be 3.35% compared to 3.4%, and 20 years out the theoretical probability is 6.59% compared to 6.8%.]

This lead me to wondering about another piece of data she’d pointed out: the Mean Time Before Failure (MTBT) was listed as 700,000 hours, which translates to about 79.85 years of continuous play. And where do they get THAT number from, I ask you? Certainly not experimental data. A later thought I had was that it was the number of years before the likelihood of disk failure was 50%, but 1-(0.9966)^79.85≈0.238, so there should be less than a 25% chance that a given hard drive fails within 79.85 years. Indeed, 1-(0.9966)^N=.5 when (0.9966)^N=.5, which happens when N=ln(.5)/ln(0.9966)≈203 years. Hey, why aren’t stores advertising that data? Shouldn’t the Antiques Roadshow have featured a few Revolutionary War hard drives, bought for 25¢ at the neighborhood garage sale?

But then when I was showing a rough draft of this post to TwoPi, he pointed out that that’s the median, not the mean. The mean would be an expected value, which would be the infinite sum ΣN(0.0034)(0.9966)^N-1, which is (0.0034)·ΣN(0.9966)^N-1. Thinking back to Calc II, the geometric series Σx^N converges to 1/(1-x) for |x|<1; taking the derivative of both sides yields ΣNx^N-1 converging to 1/(1-x)² when |x|<1. This means that (0.0034)·ΣN(0.9966)^N-1 just equals (0.0034)·1/(1-0.9966)², or 1/(0.0034)≈294 years.

OK, all that work and it didn’t clear up anything except to indicate that George Washington could have gotten a hard drive for his birthday and it might still be working.

All in all, I think we got a good handle on the theory and absolutely none on the reality. She sent me a link later this afternoon to this article on a study by Carnegie-Mellon that indicated that the Mean Time to Failure Rates were greatly exaggerated by manufacturers (ya think?), up to 15 times as long as they should be, because they failed to take into account that the #1 cause of failure is age: the likelihood that a hard drive will fail in its 10^th year is a lot higher than the likelihood that it will fail in its 2^nd year.

In other words, it doesn’t really matter if they say the drive will last 80 years or 800 years, because it won’t. But it’s still worthwhile to buy a new hard drive when the old one goes belly up.

Thanks to Anya for the idea and for the links in this post!

Carnival Submissions Welcome!

Just a reminder that in two days (Friday May 30), we’ll be hosting the 34^th Carnival of Mathematics!!  Submissions are welcome via the Official Submit Form, via the comments below, and via email to hlewis5 followed by the @ and then ending with naz.edu (put something like “Carnival” in the subject line if you think of it).

In theory we’ll be putting it together tonight and tomorrow.  In reality, we’ll be working on it tomorrow night and Friday morning, so anything submitted anytime Thursday will certainly be included.

How much is enough?

If you notice a pattern, how many times do you have to check that it works before being certain. Six? Twenty? Two thousand?

Two favorite examples of mine that demonstrate that patterns can continue for a long time before going awry:

The first example is the polynomial f(n)=n²+41n+41. If you plug in any whole number for n from 1 to 40, then f(n) is a prime number: f(1)=83, f(2)=127, all the way to f(40)=3281. But f(41)=3403=41·83 is not prime. So something that works 40 times in a row might fail.

The second example are the cyclotomic polynomials. Look at polynomials of the form xⁿ-1 that have been factored:

x-1=(x-1)
x²-1=(x-1)(x+1)
x³-1=(x-1)(x²+x+1)
x⁴-1=(x-1)(x+1)(x²+1)
x⁵-1=(x-1)(x⁴+x³+x²+x+1)
x⁶-1=(x-1)(x+1)(x²+x+1)(x²-x+1)

The last polynomial in each case is the cyclotomic polynomial of order n. [It has a much more technical definition using imaginary numbers and the product of primitive roots of unity]. And at first glance it looks like the coefficients are 0, 1, or -1. Even at second glance or sixth glance, since it’s true for the first 104 cyclotomic polynomials. But not the 105^th.

x¹⁰⁵-1=(x-1)•(x²+x+1)•(x⁴+x³+x²+x+1)•
(x⁶+x⁵+x⁴+x³+x²+x+1)•(x⁸-x⁷+x⁵-x⁴+x³-x+1)•
(x¹²-x¹¹+x⁹-x⁸+x⁶-x⁴+x³-x+1)•
(x²⁴-x²³+x¹⁹-x¹⁸+x¹⁷-x¹⁶+x¹⁴-x¹³+x¹²-x¹¹
+x¹⁰-x⁸+x⁷-x⁶+x⁵-x+1)•
(x⁴⁸+x⁴⁷+x⁴⁶-x⁴³-x⁴²-2x⁴¹-x⁴⁰-x³⁹+x³⁶+x³⁵
+x³⁴+x³³+x³²+x³¹-x²⁸-x²⁶-x²⁴-x²²-x²⁰+x¹⁷+x¹⁶
+x¹⁵+x¹⁴+x¹³+x¹²-x⁹-x⁸-2x⁷-x⁶-x⁵+x²+x+1)

See those two coefficients of 2 in that last polynomial? So the pattern of coefficients being only 0, 1, or -1 fails. Interestingly, the reason for this initial failure occurring so late in the game is that 105 is the smallest number that has three distinct odd prime factors (105=3·5·7). The integer 385 is also a product of three distinct odd primes (385=3·5·11) and the 385^th cyclotomic polynomial is the first one to have a 3 as a coefficient (see Wolfram Mathworld).

Patterns. You just can’t trust them.

Fractured Theorems

The Bovino-Weierstrass Theorem:  In any bounded pasture containing an infinite number of cow pies, you can stand in one location where no matter how close you look, there are an infinite number of cow pies near your feet.

 Surely there must be more such fractured theorems lurking out there in our collective imagination… If you come up with any new ones, share them in the comments.

Math Video Contest

Robert Vallin at the MAA is running a math video contest (MathTube) for undergraduate students. Teams of up to three students can submit videos showing the “entertaining side” of mathematics. He cites the Finite Simple Group (of Order Two) YouTube video as an example of the kind of thing the judges are looking for.

The deadline for submission is December 7, so if you’re going to win this thing, you’d better get busy. (How cool would it be if one of our students did win?)

Music and Light Show All in One

(From Good Math, Bad Math) Apparently you can play music with a Tesla coil. Or two. Watch and enjoy.

They have another one, too.

Mathematician Heads of State?

James A. Garfield, the 20th President of the United States, is perhaps best known (at least among math teachers) for having devised a novel proof of the Pythagorean Theorem. He published the proof in 1876, while a member of congress, in the New England Journal of Education [ Garfield, J. A. "Pons Asinorum." New England J. Educ. 3, 161, 1876. (citation from mathworld.wolfram.com )].

A request, and an open question:

1. Request: If you have access to Garfield’s original article, I’d love to get a paper copy or pdf! Or a hyperlink! Or….

2. Question: Has there been any head-of-state who was by training a mathematician? Which mathematician has risen closest to playing the role of head-of-state of their homeland? [I have a candidate: Isaac Newton, as an MP, and as Master of the Mint in his later years. But I suspect we can do better than that.]

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