## Bullet time

I decided to apply a little physics to my KitchenAid mixer misadventure, in which I claim I was nearly killed by a hunk of metal propelled past my head by spinning mixer blades. When that projectile whizzed past my head, how much danger was I really in compared to, say, a bullet from a sniper’s rifle?

It comes down to a calculation of the kinetic energy of the projectile. Fortunately it’s very easy to approximate by making a few assumptions and by ignoring the effects of air resistance and the spray of cake batter.

The projectile was the mixer’s own removable endcap. As soon as it vibrated loose, fell into the bowl, and was struck by the spinning mixer blades, it was on a ballistic trajectory, arcing up, past my head, and then down behind me. Let’s assume that the highest point of the trajectory was about level with the top of my head, roughly 1.75 meters off the ground, and that this height was attained just as the projectile was passing me, meaning that once it did pass me, it had already started down.

We can decompose the motion of the projectile into a pair of vectors: the one pointing straight down to the floor and the one perpendicular to it, pointing horizontally past my head. The speed in that direction was constant until the endcap hit the floor. The speed in the floorward direction was increasing due to gravity.

Since we’ve assumed the endcap reached its apex as it passed my head, we know that its downward velocity at that moment was zero. We also know that the acceleration in that direction (due to gravity) is 9.8 meters per second per second. Finally we know from high school physics that:

distance = initial-velocity×time + acceleration×time2/2

and since we know initial-velocity is 0, we can rearrange this to say:

time = √distance/acceleration

And since we know “distance” is 1.75 meters and “acceleration” is 9.8 m/s2, we know that it took about 0.4 seconds for the endcap to fall from the height of my head, regardless of its motion in the horizontal direction.

In that 0.4 seconds I estimate (based on where I later found the endcap) that it covered a horizontal distance of 3 meters, giving it a speed of 7.5 meters per second.

I haven’t weighed the endcap but I’m going to guess it’s about 0.25 kilograms (around half a pound). Again, high school physics tells us that kinetic energy is:

mass×velocity2/2

which means the endcap, if aimed just a bit differently, would have struck me with about 7 joules of energy.

How bad would that have been, compared to a bullet? Apparently even the wimpiest guns deliver hundreds of joules to their targets, so we’re not looking at a shearing-off-the-top-of-my-head scenario here. On the other hand, I was there, and I’m not exaggerating when I say that hunk of metal could have dealt me a grievous injury at the very least. If that was what seven joules looks like, I have a whole new appreciation for the stopping power of a bullet.

## How I wonder what you are

Often when I’m gazing at the night sky I will focus on a very dim star and marvel at its ability to shine steadily. No matter how long I look, and no matter whether I move a few inches to the left or right, it keeps right on shining. (It may twinkle a bit, but as you probably know, that’s due to our fluid, shifting atmosphere causing the starlight to refract slightly differently from one moment to the next, not due to anything about the star.) Photons emitted by that star hundreds of years ago poured out in such numbers that, even as they fanned out across unimaginable distances, there are still enough of them landing continuously in the tiny target of my eyes that the star remains a constant point of light.

Knowing the distances involved, it seems implausible that enough photons would be headed in my exact direction — of all the other places in the universe they could have gone! How implausible? Let’s figure it out.

We’ll start by estimating that the area of my dilated, night-sky-gazing pupils is about one square centimeter, or (to keep all measurements using the same units) one ten-thousandth of a square meter.

Thanks to Wikipedia I know that a single photon of visible light carries about 4×10-19 joules of energy which is just enough to excite a photoreceptor in my eye. How many must arrive each second for me to perceive a continuous image? I know that movies, projected at 24 frames per second, are good enough to trigger the persistence-of-vision effect. Since a distant star has far less detail than a frame of movie film, I’m guessing that it can appear steady at even fewer than 24 “frames” per second. Let’s guess 10 photons per second must arrive in my eyes to perceive a steadily shining star. 10 photons delivering 4×10-19 joules every second is 4×10-18 watts (because one watt is equal to one joule per second).

Now, how far is that star? A bit of googling reveals that a typical faintly visible star is 200 light years away (though many are much closer and many are much farther). 200 light years is 1.89×1018 meters.

From the perspective of that star, what fraction of its “sky” is taken up by my eyes? If you imagine a gigantic spherical shell centered on and surrounding that star, with a radius of 200 light years and my eyes on the inner surface, then the total area of that inner surface — the star’s “sky” — is 4×π×200light-years2, which is 503,000 square light-years, which is 4.5×1037 square meters. My pupils, one ten-thousandth of a square meter, comprise 1/2.22×1042 of that “sky.” That’s two millionths of a trillionth of a trillionth of a trillionth — not like that’s going to make my big ego feel small or anything.

So if 4×10-18 watts of visible light is delivered to 1/2.22×1042 of the star’s sky, then the total power delivered to its full sky is that first number divided by that second, which is 1.8×1024 watts, or 1.8 trillion terawatts. Is that a plausible number?

Yes it is. Our own sun has a visible-light output of 3.8×1026 watts. Which means, incidentally, that (using my guesses from above) it would be visible as a steadily shining star to a distance of 2.7×1019 meters, or about 2,900 light-years. Future criminals exiled from our galaxy will lose sight of their home star while barely a thousandth of the way to their new home in Andromeda.

## Do the abortion math

It’s the thirty-fifth anniversary of the famous Supreme Court decision in Roe v. Wade that affirmed the legality of abortion.

Of course abortion remains a fantastically polarizing subject. It seems like the country is divided fairly evenly in favor of choice and opposed to it (and not necessarily along traditional party or ideological lines); and that most opinions are strongly held.

I for one believe that Bill Clinton got it exactly right when he said that abortion should be “safe, legal, and rare,” but why do I believe that? No opinion is worth having if it can’t be examined, challenged, and defended, so here is my defense.

First we must determine whether it’s possible to attack or defend abortion dispassionately, without appeal to emotion. Most people who oppose abortion do so because they believe that it is murder, a naturally emotional subject. We could try to remove or distort the emotional component (e.g., by confining ourselves to a discussion of costs and benefits — as routinely happens in cases of state-sponsored murder such as executions and wars), but I think there’s another way that sidesteps the question of murder altogether.

Whether or not abortion is murder depends on whether or not a life exists to terminate. Indeed that’s what it came down to in the Roe v. Wade decision:

We need not resolve the difficult question of when life begins. When those trained in the respective disciplines of medicine, philosophy, and theology are unable to arrive at any consensus, the judiciary, at this point in the development of man’s knowledge, is not in a position to speculate as to the answer.

Nearly everyone agrees that life does not exist prior to conception. Nearly everyone agrees that life does exist upon birth. So where, in between those two events, is the switch flipped? We don’t need medicine, philosophy, or theology to decide; just simple math.

Life may not exist the entire time, but what does exist is an ever-increasing probability of life — more precisely, a probability of being born alive. The appearance of a fetal heartbeat at about eight weeks after conception improves the odds. “Viability” (a minimal ability to survive outside the womb) at about 22 weeks improves it some more. The chances of being born alive continue slowly to climb as the fetus develops, until finally it reaches 100% at the moment the baby takes its first breath.

Can we scientifically choose a probability threshold before which we say abortion is OK and after which we say it isn’t? Sort of. A threshold very close to 100% would be unacceptable to most people, even pro-choice advocates, for emotional reasons (although this has not been true at all times or in all cultures, some of which routinely disposed of unwanted infants simply by exposing them to the elements). It makes no sense to talk about a threshold of 0% as some extreme anti-abortionists might prefer it, because the probability is higher than zero even before conception — especially if Barry White is playing and the lights are turned low. Any other choice between 0% and 100% would be arbitrary, so the best we can do is to choose the least arbitrary number in that interval: 50%. As it happens, in modern America a 50% chance of a live birth appears to be reached, on average, between the 22nd and 28th week of pregnancy.

So that’s my position: abortion should be legal (and safe and rare) before about 22 weeks, and illegal (with the usual pragmatic exceptions — rape, incest, health of the mother) otherwise. Your emotions aside, I think I’ve shown that no other position on this subject is more rational than that one.

## Numerology, 2008 edition

This is the year 2008. The number 2008 has four prime factors: 2×2×2×251. Last year had three (3×3×223) as does next year (7×7×41).

The year 2000 had 7 prime factors (2×2×2×2×5×5×5), which was the most since 1984, which also had 7 (2×2×2×2×2×2×31). 1944 had 8 (2×2×2×3×3×3×3×3), which we won’t see again until 2016 (2×2×2×2×2×3×3×7). 1920 had 9! (2×2×2×2×2×2×2×3×5.)

Those of us who can make it until 2048 will get to see an 11-factor year (2×2×2×2×2×2×2×2×2×2×2) and the first year since 1024 that’s a power of two, which is a big deal to computer nerds like me.

The most recent prime-number year was 2003 and the next is 2011. The most recent twin-prime years were 1997 and 1999, and the next are 2027 and 2029.

What is the significance of all this? Not a damn thing. But you read it all with interest, didn’t you?

## What brings you here, 2007 edition

Here are some of the top queries from various search engines that resulted in hits on my blog during the past year or so, reproduced verbatim from my server logs. (Last year’s results are here.) Each related family of queries is listed with a main variant in bold and selected other variants, plus the percentage of query-hits represented by that family.

I was at first surprised to see that hits for “James Bond villains” outnumbers hits for “vampire lesbian girl scouts” (etc.) and “sex” (etc.) combined, but then realized: the percentages are a function both of the popularity of that search and of the ranking of my site in the search results. In other words, if you’re looking for anything about vampires or lesbians or sex I regret to say there are a lot of likelier websites for you to visit before mine.

## Don’t be so negative

This week’s Car Talk Puzzler is as follows:

In a certain family each daughter has the same number of brothers and sisters. Each son has twice as many sisters as brothers. How many sons and daughters are there in the family?

This is obviously a very straightforward problem in algebra. Let G be the number of girls in the family and B be the number of boys. Each girl has G-1 sisters and B brothers. Each boy has G sisters and B-1 brothers.

Converting the problem statement to a system of equations gives us:

G-1 = B
2G = B – 1

Now it’s a simple matter to rearrange the terms and solve for G or B. Let’s start by adding 1 to each side of the second equation:

2G + 1 = B

Now let’s substitute G – 1 for B, since the first equation tells us they’re equal:

2G + 1 = G – 1

Now let’s add 1 to each side again.

2G + 2 = G

Now let’s subtract G from each side.

G + 2 = 0

Finally, let’s subtract 2 from each side.

G = -2

…which implies that B = -3. So this particular family has negative two daughters and negative three sons.

Either the family is composed of antimatter, or my ability to do really simple algebra is gone, or there’s some subtle trickery in the wording of this puzzle that I’m missing. For example, maybe the trick is hiding in the difference between the words “and” and “as”: “each daughter has the same numbers of brothers and sisters,” while “each son has twice as many sisters as brothers.” The first sentence could be parsed to mean, “Each daughter has the same number of brothers-and-sisters (i.e., siblings) as the other daughters do,” which is trivially true no matter how many sons and daughters there are. But that reduces the system of equations to just:

2G = B – 1

which has infinitely many solutions for G and B. So where’s the error?

Update: I’m an idiot, as pointed out (gently) in the comments. “Each son has twice as many sisters as brothers” doesn’t mean

2G = B – 1

it means

G = 2(B – 1)

So the family is made of ordinary matter after all.

## No place for common sense

Not that this is especially deserving of a reasoned rebuttal:

Peanut butter disproves evolution

…A (serious) Creationist clip showing how peanut butter disproves the theory of evolution…

The video explains that evolutionists claim that energy plus matter sometimes results in the creation of life. But since no one has ever found spontaneously-generated life in a jar of peanut butter, that means that matter plus energy from the sun couldn’t have caused life on Earth… Link

…but I just happened to have one handy in some old e-mail. An outspoken creationist friend of mine wrote:

there are over 200 million different species on this planet. Since each is (presumably) evolving differently and over time, it seems reasonable to expect that one, only one, just one tiny one, of these 200,000,000 species would have “sprouted wings” in the last 150 years

where I understood “sprouted wings” to mean “underwent a significant, observable evolutionary change.” That may be a common sense outlook, but this is no place for common sense. Common sense breaks down when dealing with fantastically large numbers and fantastically small odds. Here is how I replied:

Let’s say the earth is 4.5 billion years old, and it took all that time to produce 200 million existing species. (We’ll treat the many other species that have come and gone as statistical fluctuations.) That’s 0.044 species per year on average. Over 150 years you should then expect to see the emergence of 6.67 new species on average, which is .00000334% of the total number of species. Easy to miss.

Let’s do it another way: 0.044 species per year is 22.5 years per species — that is, we should expect a new species every 22.5 years. Assuming each of the existing 200 million species is equally likely to spawn that new species, each species must wait an average of 3.12 billion years to have a 50-50 chance of creating a successor.

(That’s

22.5×log1-1/2000000000.5

1-1/200000000 is 0.999999995, which are the odds of a species not spawning a new species in one year. 0.999999995×0.999999995 are the odds of not spawning a new species for two years in a row; 0.999999995×0.999999995×0.999999995 are the odds of not spawning a new species for three years in a row; and so on. How many times must you multiply 0.999999995 by itself to get to odds of 0.5? That’s what log1-1/2000000000.5 tells you.)

That’s by no means a rigorous analysis — it’s full of extremely coarse assumptions, among other things — but it should be at least accurate enough to convey the vastness of the timescales involved, the number of species, and the odds against having any particular evolutionary expectation met.

Or, as Darryl Zero said,

Now a few words on looking for things. When you go looking for something specific, your chances of finding it are very bad, because of all the things in the world, you’re only looking for one of them. When you go looking for anything at all, your chances of finding it are very good, because of all the things in the world, you’re sure to find some of them.

## World widescreen web

Thinking of upgrading your conventional picture-tube TV to a fancy new flat-panel widescreen? But you’re on a budget and don’t want to go overboard? Confused about what size TV to buy? You’ve come to the right place.

The main criterion for choosing a screen size is one that I have not seen described in other TV buying guides: viewing area. The viewing area of a 32″ conventional TV is 492 square inches, whereas the viewing area of a 32″ widescreen TV is a mere 438 square inches! If you’re upgrading from a 32″ conventional TV you’ll want at least a 34″ widescreen to get the same viewing area.

Here’s how I arrived at those figures.

The advertised size of a TV display is the length of the diagonal. If from the diagonal we can determine the height of the display, h, and the width, w, then the viewing area is h×w. Thanks to Pythagoras we know that h2+w2 = 322. But this isn’t enough information to determine the viewing area: we also need the fact that the aspect ratio of most conventional TV displays is 4:3, which means the width of the display is four-thirds the height.

Substituting 4h/3 for w and then simplifying gives us:

h2+(4h/3)2 = 322
h2+16h2/9 = 322
25h2/9 = 322
h = √(9×322/25)
h = 3×32/5 = 19.2

Plugging that into the formula for viewing area (h×w) and recalling that w = 4h/3,

h×4h/3 = 19.2×4×19.2/3 = 491.52 square inches

Knowing that the aspect ratio of widescreen displays is 16:9 and using similar arithmetic gives a result of 438 square inches for a 32″ diagonal.

In fact, the math shows that for a given diagonal, the viewing area of a 16:9 display will always be about 11% less than the viewing area of a 4:3 display.

But wait! It’s not as simple as finding the widescreen TV that has at least the same viewing area as your conventional TV. You should also take into account the kinds of programming you watch.

Do you watch a lot of wide-format movies on your 4:3 TV? If so, you’ve certainly noticed the “letterboxing” needed to fit the wide aspect ratio of the film into the narrow one of the display. You’re not using the entire viewing area; some of it is wasted, as much as 32% of it for very wide format formats such as “CinemaScope.” With a 16:9 TV the need for letterboxing wide-format movies is decreased or eliminated.

Similarly, if you watch a lot of conventional TV programming (sitcoms, newscasts, etc.) on a widescreen TV, you’ll get “reverse letterboxing,” also called pillar boxing, where the black bars appear not on the top and bottom but on the left and right of the image to make the taller aspect ratio fit into a shorter one. Here again you’re wasting some of your viewing area.

So think about the kinds of programming you watch and consult this handy table that shows the true image size (in square inches) for various combinations of TV diagonal size, TV aspect ratio, and programming aspect ratio. Choose a TV that gives you the best image size you can afford for the types of programming you typically watch.

Program aspect ratio
1.33
(4:3)
very common
1.66
(5:3)
some movies
1.77
(16:9)
“widescreen”
1.85
(13:7)
VistaVision
2.35
(33:14)
CinemaScope
4:3
screens
20″ 192 154 144 138 109
27″ 350 280 262 252 199
32″ 492 393 369 354 279
36″ 622 498 467 448 353
42″ 847 677 635 610 480
46″ 1016 813 762 732 576
50″ 1200 960 900 865 681
16:9
screens
20″ 128 160 171 164 129
27″ 234 292 312 299 236
32″ 328 410 438 420 331
36″ 415 519 554 532 419
42″ 565 707 754 724 570
46″ 678 848 904 869 684
50″ 801 1001 1068 1027 808

## e to the i pi plus one equals zero

One of the best teachers I ever had was Mr. Arrigo, for 11th grade pre-calculus. He was young, funny, hip, and energetic. It was almost incongruous that he was a math instructor. He seemed more like the big brother who’d already gone off to college. Of course he wouldn’t be one of my best teachers if he wasn’t also excellent as a teacher, which he was.

Throughout the year we covered topics in trigonometry, complex numbers, transcendentals, and logarithms. Little did we know that Mr. Arrigo was working up to a unification of all four.

One day in class he was particularly animated. We had been discussing the Euler formula, which gives this equivalence:

cosθ + isinθ = eiθ

(Here, e is the natural logarithm constant ≅ 2.718 and i is the imaginary number √−1.) He then asked us to work out the special case where θ is π (the ratio of a circle’s circumference to its diameter ≅ 3.14159). The cosine of π is −1 and the sine of π is 0, so Euler’s formula gives this amazing relation:

eiπ = −1

or, stated just a bit differently,

eiπ + 1 = 0

When Mr. Arrigo derived this result on the blackboard, he literally jumped up and down as he exclaimed, “One simple equation relating the five most important constants in all of mathematics!”

An xkcd comic

His excitement was infectious. Of course the result above is breathtaking. (If you’re not mathematically inclined, you’ll just have to take my word for it. Imagine being a world traveler collecting random antique curios from cities around the globe — then discovering one day that five of your favorite ones just happen to fit together perfectly to make an exquisite and accurate pocket watch. It’s kind of like that.) But Mr. Arrigo’s passionate presentation made it something more — an emotional highlight of my academic career.

A few years later, in college, I shared an office with my engineering friend Steve. One afternoon his actress girlfriend Amy stopped by and, one way or another, the conversation turned to mathematics. Taking turns scribbling on the whiteboard, Steve and I explained to a willing Amy how it’s possible to derive all the familiar rules of numbers and arithmetic from a tiny kernel of laws called Peano’s postulates. Amy seemed interested, so we pressed on into other topics such as geometry and its cognates, trigonometry and set theory. Thrilled that her interest didn’t flag, I mustered Mr. Arrigo’s passion and derived for her the amazing relationship between e, i, π, 1, and 0.

Soon after that, Amy’s drama-major mind finally had as much math as it could handle and our memorable pedagogical session petered out. But I could have kept going all day. I suspect Mr. Arrigo planted in me the desire to teach. I think Jonah and Archer are getting the benefit today of Mr. Arrigo’s passion way back then, and who knows? I may yet heed the urging of friends and family and become a teacher myself some day.