We have a winner for our “Hamster-Powered Mansions” estimation contest. The question: How many buff hamsters would it take to completely power a mansion?
Mansions come in a wide range of sizes, and each will have different energy requirements. According to Factcheck.org, Al Gore’s mansion used 191,000 kilowatt-hours in 2006. Dividing by one year, we can compute his average power consumption to be 22,000 W.
From this MAKE magazine video1, we can see that a hamster is at least powerful enough to light up an LED, but this is only a lower bound since he might be able to power even more LEDs if we connected them in the circuit. According to Otherpower.com, their hamster Skippy had no trouble lighting up 6 LEDs, and they estimate he should be able to power 200. Being fairly conservative, I’ll assume our hamsters can power 50 LEDs. According to Wikipedia, the voltage drop across one LED is anywhere from 1.5-4.5 V and the current should be between 1 and 20 mA. Assuming the LED acts like an Ohmic resistor,2 we can estimate the electrical power created by a wheel-spinning hamster,
power = (# of resistors) · (current) · (voltage drop)
= (50) · (10 mA) · (3.0 V)
= 1.5 W per hamster.
From this and the power consumption of Al Gore’s mansion given above, we can estimate the number of hamster’s you’d need to power a mansion,
# of hamsters = (power per mansion) / (power per hamster)
= (22,000 W) / (1.5 W per hamster)
= 15,000 hamsters.
You’d need about 15,000 hamsters to power a mansion. Congratulations to our winner Bryan Merrill. Bryan will be receiving a free copy of How Many Licks? Keep reading for our next contest.
[1] Speaking of MAKE magazine, check out Maker Faire in Detroit July 31 and August 1.
Anyone who reads Diary of Numbers regularly knows I’m far from a grammar nazi. Despite my best efforts, my grammar is typically atrocious. There is, however, one thing I’m very particular on. Even though both of the following are equally correct, I much prefer the latter:
(1) “Dorothy was afraid of lions, tigers and bears.”
(2) “Dorothy was afraid of lions, tigers, and bears.”
Recently, something struck me: since both usages are perfectly correct, I’ve been wasting printer ink all my life. How many cartridges of ink do you waste in your lifetime by always including the extra comma?
This is a tricky problem because the amount of printer ink a person uses varies considerably depending on his/her career, hobbies, etc. If you’re a novelist, you’ll probably be printing a lot more commas than a rodeo clown. Moreover, the problem’s a little wishy-washy because of how I originally phrased it1. A better way to phrase the question would have been, “How many cartridges of ink would you waste in your lifetime if everyone included the extra comma?”
Some days I print 100 pages, other days I print none. On average, I probably print about one full page per day2. Lists appear about once every 10 pages3. This means I’d be printing one superfluous comma every 10 days. At this rate, I’d produce 2900 needless commas over an 80-year lifespan.
According to HP’s website, the average HP LaserJet Q2612A Black Print Cartridge yields 2000 pages worth of ink. A single comma requires about one-fourth the ink of a letter and according to one of my MS-Word documents, there are about 2000 letters per page. This means one page worth of ink is equivalent to about 8000 commas. From this and the info above, we can estimate the number of printer cartridges one will waste in a lifetime,
# ink cartridges = (# ink cartridges per page) · (# pages per comma) · (# of commas)
= (1 ink cartridge per 2000 pages) · (1 page per 8000 commas) · (2900 commas)
= 0.00018 ink cartridges.
Over the course of your lifetime, you will waste 1/5,000th of an ink cartridge by including the extra comma. Even if everyone in America included the extra comma, we’d waste only about 680 printer cartridges each year.
After posting this problem, a good friend, who had previously worked for Yale University Press as a copy editor, told me that Yale never included the extra comma to save ink!
[1] Sorry about that.
[2] This is more than printing one page every ten days and less than printing 10 pages per day, which seems like reasonably good bounds for and order of magnitude estimate.
[3] Again, this estimate is reasonably good if you consider upper and lower bounds. A list certainly does not appear on every page but you’ll probably see more than one on a hundred pages of text.
I like mime. There. I said it. Just to be clear, I’m not talking about the guys who stand on street corners painted gold pretending to be statues just to get money. Those guys are creepy1. I’m talking about real mime. The kind that requires actual talent. I’m talking about the Marcel Marceau kind of mime.
About ten years ago, an acting teacher suggested I see Marceau perform in town that night. Knowing the type of reputation mimes had, I hesitated, not wanting to associate with their type. Ultimately, I decided to go since Marceau was 77 at the point2, and I was unlikely to get another chance to see him perform if I changed my mind3. I wish I could say I had an instant conversion to becoming a mime-lover during the show, but my reaction was more of a “meh.” It was a cute show, definitely entertaining and at certain points even laugh-out-loud funny. It was even good enough for me to admit that not all mimes were bad. But this was supposed to be the bestmime in the world, and with that kind of reputation, I expected something better than a bunch of funny and occasionally thought-provoking skits.
I went back to work on Monday. I’d told a postdoc about the show beforehand and she’d chided me about it, but she asked me how it went:
Her: How was the show?
Me: It was all right. Pretty funny actually. There was one bit with a mask maker where he gets a smiley mask stuck on his face. There was also one with a kid in a park where he kicks a ball through a window. In the background, there are these two old ladies watching and gossiping. I think my favorite one was this dating service that keeps sending him women who are either too tall or too fa—
Her: Wait. I thought you said it was just him.
Me: [Confused] It was.
That’s when I realized the magic of Marcel Marceau. None of those other people were there. There were no set, no props, no text nor sound of any kind. Not even a lighting change. Just a silent 77-year-old man alone on a stage.
In honor of the great Marcel Marceau, see if you can calculate how many fewer words he said during his lifetime because of his career choice.
People usually say about 3 words per second when talking. The only times mimes don’t speak is during performance. Marceau might have averaged 5 shows per week during his 60 years of performing. Each show is about 1.0 hour long. If we assume only 5% of his time was spent talking, we can estimate the number of extra words he could have spoken in his lifetime,
words = (yrs performed) · (wks per yr) · (shows pr wk) ·(hrs per show)
· (fraction of time talking) · (words per s)
= (60 yrs) · (52 wks per yr) · (5 shows pr wk) ·(1.0 hr per show) · (0.05) · (3 words per s)
= 8.4×106 words.
Marceau would have said about 8.4 million extra words had he not been a mime.
[1] I’m convinced they’re the reason real mimes are listed below jugglers and magicians on the street performer totem pole.
[2] As another member of our class put it, “He’s still alive?!”
“I recently lost my engagement ring. (It's not the first time I lost it, but I think this time it's really gone. Sigh.). But, I’m not alone—as I just learned that Jupiter lost a band recently… How many times bigger is Jupiter's band than my engagement band?”
OBERLIN
Jupiter’s radius is about 71,000 km.1 My sister’s ring’s radius was likely about 1.0 cm. This would make Jupiter’s ring (or band) about 7 billion times bigger if we’re talking about its linear dimension. However, the word “bigger” has several meanings. If we’re talking about how many times larger the volume of the ring is, then we need to cube the linear dimension. In this case, Jupiter’s ring is about 3.6×1029 times bigger. As antoniseb has pointed out, there’s also the issue of density. If we want to talk about how much more massive the ring is we need to know the ratio of Jupiter’s ring’s density to the density of gold. The density of gold is 19.3 g/cm3. Jupiter is a gas giant with a low average density of about 1.33 g/cm3. The upper atmosphere where the ring is located is probably much less dense. Assuming it has the same density as air at the surface Earth (~0.0012 g/cm3), Jupiter’s band will be about 2.2×1025 times bigger.
Sorry about the ring. If you want to make sure you don’t lose your next ring, you can always do what I did.2
[1] Since the ring is not at the equator, it will actually be a little smaller than this.
Happy Father’s Day to all the dads out there. My Dad suggested a problem the other day: How long would it take to wear down a tire by 0.75 inches if you’re driving a 55 mph?
I’ll assume a tire tread starts about an inch thick. By the time 0.75 inches have worn off, it’s time to get them replaced. Tires presumably wear down because of friction between the rubber and the road. When tires rotate, this frictional force from the ground pushes the car forward while an equal and opposite force pushes back on the road. The faster you drive, the greater the frictional force is and this presumably causes more wear and tear. However, if you’re driving at a constant speed, you probably don’t burn as much rubber as you would by slamming on the breaks every five seconds. It’s not immediately clear which of these effects causes greater erosion, but you’re generally advised to change tires roughly every 50,000 miles1. If that’s the case for driving at 55 mph, then it’s easy to compute how long it will take before your tires have worn down by 0.75 inches:
As I was driving the other day, I noticed the little dashboard compass needle spinning whenever the car turned. Perhaps there’s a more energy efficient way to travel with a compass. How fast could you go from Brooklyn to Manhattan by sitting on the point of a giant compass needle?
***********WARNING: Math Ahead***********
Imagine a giant compass needle whose pivot is centered halfway between Manhattan and Brooklyn. Initially the needle points south so that its tip lies in Brooklyn. The needle would have to be about 8.0 km (~5.0 mi.) long. It would need to carry people, so it might look somewhat like a subway car, making it at least 3 m (~9.8 ft) in diameter.1
The torque, or rotational force, on a magnet is given by the equation,
The angle in question is the angle between the magnetic moment and the magnetic field. This angle would change as the needle turns. On average, the sine of the angle would be about 0.63 as the needle passes from Brooklyn to Manhattan. The magnetic field would be the Earth’s magnetic field, which has a magnitude of roughly 5.0×10-5 T.2 The needle could be a large permanent magnet (like the ones on refrigerators), but these are not as strong as electromagnets and aren’t able to flip the north and south poles like an electromagnet can3. An electromagnet can also be turned on and off, so you could control when our magnetic transporter was about to leave the station. We can make an electromagnet by wrapping tons of wire around the needle in a helical fashion and then applying a current through it. The magnetic moment for a cylindrical coil is given by,
magnetic moment = (number of coil turns) · (current) · (area across the cylinder).
From the diameter above, the area would be about 10 m2. If the wire is 1.0 mm thick and there are 10 rows wrapped around each other, then the 8.0 km needle would have about 8 million turns in it. I’ll assume the current in the coils is 1.0 A.4
The time it takes the needle to rotate to Manhattan is given by the equation,
time = [2 π · (moment of inertia) / torque]1/2.
The moment of inertia of the needle can be computed using a formula found here. Plugging in the values for our needle, we can calculate the moment of inertia to be about 3.4×1012 kg·m2.5
By combining equations and plugging all the numbers in from about, we find that it would take about 1.1 days to get from Brooklyn to Manhattan traveling by giant compass needle.
If you made a giant compass needle to harness the Earth’s magnetic field and tried to use this to travel, it would take you a day to get from Brooklyn to Manhattan. Since the subway costs about $2.00 and takes less than an hour, building a giant compass is probably not an efficient use of time or money.
There’s another interesting dilemma in this problem. Nothing in life is free, so you’d suspect that even our poorly functioning magnetic transporter must also sap energy from somewhere. Much like oil, it’s possible the source of Earth’s magnetic field would also eventually run out if we made lots of giant magnetic transporters. How long do you think it would take to sap away the Earth’s magnetic field?
[1] It would also be highly magnetic, so you couldn’t bring your credit cards.
[2] The “T” stands for “Tesla,” a unit of magnetic field.
[3] If you couldn’t flip the poles you’d never be able to get back to Brooklyn in the same way you left.
[4] I don’t have a good reason for assuming this value of current other than the fact that one of my intro physics books uses this value in a problem involving a copper wire. It would be interesting to see how this problem would work with superconducting wires, but you’d have to cool them down to very low temperatures, so the passengers might become a wee bit cranky and hypothermic.
[5] You can estimate this by assuming the needle is a 1.0 ft thick cylindrical shell made of iron.
Today’s question comes from the very funny comedian Maryellen Hooper. Ms. Hooper has appeared on The Tonight Show with Jay Leno and in her own Comedy Central special, Lounge Lizards. In 1998, she was awarded “Female Comedian of the Year” at the American Comedy Awards. She writes,
“I've decided to ask a question on behalf of my son, Nate and his best friend, Noah. They're 5 & 6 years old…How many Legos would it take to build a ladder to the Moon?”
***********WARNING: Math Ahead***********
I’ll assume we’re using the standard 8-peg Lego brick. A single Lego is 9.6 mm tall, 32 mm long, and 16 mm wide1. By multiplying these together, we can calculate the Lego brick’s volume to be about 4.9 cm3. To find the total number of Legos needed, we have to estimate the dimensions of the ladder.
The Moon is about 380,000 km away from the Earth. I’ll assume the sides and rungs of the ladder are both 6.0 cm wide and 6.0 cm thick. I’ll also assume the rungs are 60 cm (~2.0 ft) long and that adjacent rungs are separated by 30 cm (~ 1.0 ft). With this rung separation, there will be about 1.3×109 rungs. From this we can compute the total volume of the ladder.
volume = (# of sides) · (volume of the sides) + (# of rungs) · (volume of the rungs)
= 2 · (6.0 cm · 6.0 cm · 380,000 km) + 1.3×109 · (6.0 cm · 6.0 cm · 60 cm)
= 5.5×109 m3.
To find the number of Legos we’d need to build the ladder, we just need to divide the volume of a single Lego into volume of the ladder,
A Lego ladder to the moon would require 1.1 quadrillion Legos!2 Using prices from Lego’s website, this would cost about $51 trillion dollars. At his richest, Bill Gates was still 500 times too poor to afford this. Thanks for the great question, Maryellen!
You can find Maryellen’s tour dates and clips of her doing standup on her website. You can also buy her comedy CDs Fixer Upper and Dignity Under Duress.
[1] Apparently, the good people at Lego are too good for the metric system, so they actually made up a “Lego Unit”. One Lego Unit is equal to 1.6 mm.
We have a winner for our “Hamster-Powered Mansions” estimation contest. The question: How many buff hamsters would it take to completely power a mansion? 
