Friday, July 20, 2012
Skeptically Speaking
You can hear me talking about Ballparking on the Skeptically Speaking podcast with the always lovely Desirae Schell. You can also catch my earlier interview on How Many Licks? by clicking here.
Thursday, July 19, 2012
Brian Clegg On Giant Balls of Life
Today's question comes from popular science writer Brian Clegg. Brian has written on a variety of topics ranging from the science of flight to time travel to the bizarre physics behind quantum entanglement. His latest book, Gravity: How the Weakest Force in the Universe Shaped Our Lives, came out earlier this year.
Brian writes,
If you took all the living material (animal, plant, bacterial etc.) on the Earth, took it into space and made it into a ball:
1. What would be the gravitational pull on its surface?2. What existing solar system body would it be closest to gravitationally?3. What effect would it have on the gravitational pull of the Earth?
1. Gravitational pull on its surface.
In order to solve this, it helps to know the total amount of biomass present on Earth. Biomass can mean different things, but here I mean the total amount of carbon that makes up a particular species. For example, the average human weighs about 50 kilograms (110 lbs) and there are seven billion of us worldwide, so our total mass would be about 3.5×1011 kilograms. Roughly 18% of the human body is carbon, so the total biomass would be roughly 6.3×1010 kilograms (70 million tons.) To answer Brian's question, we'll need to know the total biomass of all species on Earth.
I remember very little from high school biology, but I vaguely recall the food chain.
In order to solve this, it helps to know the total amount of biomass present on Earth. Biomass can mean different things, but here I mean the total amount of carbon that makes up a particular species. For example, the average human weighs about 50 kilograms (110 lbs) and there are seven billion of us worldwide, so our total mass would be about 3.5×1011 kilograms. Roughly 18% of the human body is carbon, so the total biomass would be roughly 6.3×1010 kilograms (70 million tons.) To answer Brian's question, we'll need to know the total biomass of all species on Earth.
I remember very little from high school biology, but I vaguely recall the food chain.
The food chain can be summarized fairly succinctly: "Big things eat little things."1 However, for this to be sustainable, the total biomass of the little things has to be much greater than the total biomass of the big things or else they would get eaten up too quickly. For example, land plants comprise 1000 times more biomass than land animals. Humans, being close to the top of the food chain,2 will compose a fairly insignificant percentage of the total biomass, so my estimation of the total human biomass will be of little use to us. Fortunately, Wikipedia provides a more convenient starting point:
Assuming other creatures have roughly the same percentage of carbon humans do, the total mass of all non-bacterial creatures would be roughly 3×1015 kilograms. How does this compare with the total mass of bacteria?
Wikipedia lists an estimate for the total number of individual bacteria as being roughly 5×1030. Assuming a mass of 10-12 grams per bacteria, that would put the total bacterial mass at 5×1015 kilograms, which brings the total mass of all living creatures to roughly 8×1015 kilograms (8 trillion tons.)
"Apart from bacteria, the total live biomass on earth is about 560 billion tonnes C...."
Assuming other creatures have roughly the same percentage of carbon humans do, the total mass of all non-bacterial creatures would be roughly 3×1015 kilograms. How does this compare with the total mass of bacteria?
Most living species have a density close to water at around 1.0 gram per cubic centimeter. Molded into a sphere, our biomass ball would have a radius of about 13 kilometers. We can use Newton's law of gravity to find the gravitational pull at the surface of the ball,
g = G M / r2.
Here, G =
6.67×10-11 N•m2/kg2 is the gravitational constant, M =
8×1015 kg is the total biomass, and r = 13 km is the radius of the ball. Plugging in numbers, we find g = 3.2 mm/s2. That's about 3000 times smaller than Earth's gravitational pull.
2. Closest (gravitational) body in the solar system.
Gravitationally, Pluto is the weakest planet (or "planet" according to some) in the solar system with g = 0.61 m/s2, but even this is 200 times larger than the attraction that would be felt by our ball of biomass.
Even moons are going to have a much stronger gravitational pull than our bio-ball. Gravitationally-speaking, the closest object to our little ball o' life would likely by an asteroid or a comet. For example, Halley's comet weighs about 1014 kg where as asteroids can vary anywhere from 1011-1021 kg.
Gravitationally, Pluto is the weakest planet (or "planet" according to some) in the solar system with g = 0.61 m/s2, but even this is 200 times larger than the attraction that would be felt by our ball of biomass.
"I may not be a planet, but I'm still stronger than you, all of life." |
Even moons are going to have a much stronger gravitational pull than our bio-ball. Gravitationally-speaking, the closest object to our little ball o' life would likely by an asteroid or a comet. For example, Halley's comet weighs about 1014 kg where as asteroids can vary anywhere from 1011-1021 kg.
What a ball of humanity might look like |
3. Effect on Earth.
Unless it smashed into the Earth like the asteroid that took out the dinosaurs, our asteroid is going to have very little effect on the Earth. Even if it were touching the Earth, the gravitational pull would only increase by 0.03%. Apparently, life is only a very small part of the world!
Thanks for a great question, Brian! You can find out more about Gravity and Brian's other books on his website and follow him on Twitter.
Unless it smashed into the Earth like the asteroid that took out the dinosaurs, our asteroid is going to have very little effect on the Earth. Even if it were touching the Earth, the gravitational pull would only increase by 0.03%. Apparently, life is only a very small part of the world!
Thanks for a great question, Brian! You can find out more about Gravity and Brian's other books on his website and follow him on Twitter.
[1] Having grown up in the former whaling capitol of the world, I am aware that there are notable exceptions to this rule.
[2] I suspect cannibalistic humans would technically rank slightly higher than non-cannibalistic humans.
[2] I suspect cannibalistic humans would technically rank slightly higher than non-cannibalistic humans.
Tuesday, July 17, 2012
Pauline Chen and The Red Chamber
Today's guest is my friend and author Pauline Chen. Her recently released book The Red Chamber reimagines the provocative love triangle from the Chinese epic Dream of the Red Chamber. Pauline is also the author of the children's novel Peiling and the Chicken-Fried Christmas.
Pauline's questions regard her new book:
Google Maps lists the distance between Beijing and Suzhou as 1146 kilometers (712 miles). Driving nonstop from one to the other would take the better half of a day, about 15 hours, but her 18th century characters certainly weren't using cars. Let's consider the two modes of transportation Pauline lists.
1. Barge. I have to confess, I'm not all that familiar with barges, so I'm going to guess a bit. According to the Wikipedia entry for "water speed record," the fastest boat speed was about 42 kilometers per hours (26 mph) circa 1885. However, Pauline's story takes place about 100 years before this and her characters wouldn't have had access to propeller or steam technology. As such, I'm betting their barge moved appreciably slower. I'll guess 10 kilometers per hour. If the barge was much slower than this, walking would have been faster. Assuming they remained on the barge for the entire trip, it would take about a week to travel from Beijing to Suzhuo. This assumes the distance is a bit longer than the straight line distance since the Grand Canal is about 1700 kilometers long.
2. Horse. A horse's speed depends greatly on how much of a hurry it's in. According to the Wikipedia entry for "horse gait," a walking horse travels about 6.4 kilometers per hour (4.0 mph) where as a galloping horse can move fast as 48 kilometers per hour (30 mph). It's fair to assume the horse won't be sprinting at full tilt for a thousand kilometers in a row, so we should probably use its walking speed. Assuming 10 walking hours per day, a horse would travel about 77 kilometers each day and would complete the trek in about two weeks. This number is a bit misleading though because it assumes the horse is not carrying heavy cargo.
If the horse pulls a wagon, it will travel appreciably less quickly. According to the Wikipedia entry for "Conestoga wagon," horses can travel about 25 kilometers (15 miles) each day, which would make the journey last about a month and a half.
Depending on the amount of time spent on the barge/horse and the amount of baggage brought along for the trip, the journey could take anywhere from one week to over a month. (Though this assumes they're not distracted by various love triangles and such.)
For the second question, I'll need to be careful. I tend to play fast and loose with numbers since I'm only going for order of magnitude accuracy, but I've gotten myself in trouble with populations before. As one reader pointed out in my Pinker post, population growth can be more accurately described by a geometric series:
If we follow his trend line to the year 1775, we get a population given by
Using my reader's values p = 0.782 and m = 1.565, we find that the world population in 1775 was roughly 170 million people.
As you can see from the chart above, the number is a bit off, but certainly within an order of magnitude. A more accurate number would be about 800 million people. At present, China is about 25% of the total world population. Assuming this percentage has remained roughly the same for the last few centuries, the population in China in the 18th century would be about 40 million people if you use my estimated number, though if you use the more accurate world population, you'll find it's closer to 200 million people.
If we want to find the population of Beijing specifically, we could assume the percentage of the Chinese population living there has remained fairly constant over the last few centuries. This is not necessarily a safe assumption because cities are much smaller than countries. The smaller size makes them more prone to fluctuations in population, but I'm hoping the end result won't be too far off. Beijing is currently home to 20 million people or roughly 1.5% of China's population. Using my 40 million people figure approximated above, we could estimate the Beijing population in the 1700s to be about 600,000 people. Using the more realistic population figure would raise the number to about 3 million people. According to the Wikipedia entry for "largest cities throughout history," Beijing grew from a population of 650,000 people in 1700 to 1.1 million in 1800, which would put my numbers very close to the actual value. It would seem that an 18th century Beijing would look like a 20th century Indianapolis, at least in terms of population.
Thanks for great questions, Pauline!
Aaron Santos is a physicist and author of the books How Many Licks? Or How to Estimate Damn Near Anything and Ballparking: Practical Math for Impractical Sports Questions. Follow him on Twitter at @aarontsantos.
Pauline's questions regard her new book:
At several points in my book characters travel to and from the south, either by barge on the Grand Canal or by horseback, and I've just very loosely estimated the time required by their journey. Could you do a more precise estimate of a journey from Beijing to Suzhou, taking into consideration possible routes and stops for the night?
...My other question concerns what it would have been like to live in Beijing at the time. Can you estimate what the approximate population would have been at that time, and what the population density would have been?
Google Maps lists the distance between Beijing and Suzhou as 1146 kilometers (712 miles). Driving nonstop from one to the other would take the better half of a day, about 15 hours, but her 18th century characters certainly weren't using cars. Let's consider the two modes of transportation Pauline lists.
2. Horse. A horse's speed depends greatly on how much of a hurry it's in. According to the Wikipedia entry for "horse gait," a walking horse travels about 6.4 kilometers per hour (4.0 mph) where as a galloping horse can move fast as 48 kilometers per hour (30 mph). It's fair to assume the horse won't be sprinting at full tilt for a thousand kilometers in a row, so we should probably use its walking speed. Assuming 10 walking hours per day, a horse would travel about 77 kilometers each day and would complete the trek in about two weeks. This number is a bit misleading though because it assumes the horse is not carrying heavy cargo.
With a wagon, a horse can't travel nearly as fast. |
If the horse pulls a wagon, it will travel appreciably less quickly. According to the Wikipedia entry for "Conestoga wagon," horses can travel about 25 kilometers (15 miles) each day, which would make the journey last about a month and a half.
Depending on the amount of time spent on the barge/horse and the amount of baggage brought along for the trip, the journey could take anywhere from one week to over a month. (Though this assumes they're not distracted by various love triangles and such.)
For the second question, I'll need to be careful. I tend to play fast and loose with numbers since I'm only going for order of magnitude accuracy, but I've gotten myself in trouble with populations before. As one reader pointed out in my Pinker post, population growth can be more accurately described by a geometric series:
Let's assume the fertile population in 1900 is "p". Let's also assume that a given generation basically dies out after 75 years. This is a realistic life expectancy.
Then the generations of fertile (<25yr old) people looks as follows:
(1850) : p / m^2
(1875) : p / m
(1900) : p
(1925) : p * m
(1950) : p * m * m
(1975) : p * m^3
(2000) : p * m^4
population in 1775 = p / m7 + p / m6 + p / m5.
Using my reader's values p = 0.782 and m = 1.565, we find that the world population in 1775 was roughly 170 million people.
As you can see from the chart above, the number is a bit off, but certainly within an order of magnitude. A more accurate number would be about 800 million people. At present, China is about 25% of the total world population. Assuming this percentage has remained roughly the same for the last few centuries, the population in China in the 18th century would be about 40 million people if you use my estimated number, though if you use the more accurate world population, you'll find it's closer to 200 million people.
If we want to find the population of Beijing specifically, we could assume the percentage of the Chinese population living there has remained fairly constant over the last few centuries. This is not necessarily a safe assumption because cities are much smaller than countries. The smaller size makes them more prone to fluctuations in population, but I'm hoping the end result won't be too far off. Beijing is currently home to 20 million people or roughly 1.5% of China's population. Using my 40 million people figure approximated above, we could estimate the Beijing population in the 1700s to be about 600,000 people. Using the more realistic population figure would raise the number to about 3 million people. According to the Wikipedia entry for "largest cities throughout history," Beijing grew from a population of 650,000 people in 1700 to 1.1 million in 1800, which would put my numbers very close to the actual value. It would seem that an 18th century Beijing would look like a 20th century Indianapolis, at least in terms of population.
Thanks for great questions, Pauline!
Aaron Santos is a physicist and author of the books How Many Licks? Or How to Estimate Damn Near Anything and Ballparking: Practical Math for Impractical Sports Questions. Follow him on Twitter at @aarontsantos.
Monday, July 16, 2012
Still Time to Enter...
Camera Man: Hmm...that puck seems to be getting big awfully fast.
Puck: LEEROY JENKINS!!!!!!!!
Enter my "Look Out!" contest for a chance to win a free signed copy of Ballparking. Practical Math for Impractical Sports Questions.
Operation: Toys and Bacon
Today's invited guest is Jon Haarr. Jon writes one of my favorite blogs, Toys and Bacon. Seriously, can there be a better combination than toys and bacon? I think not! But I digress.
Jon writes,
This is something that's bothered me about Transformers ever since I got Optimus Prime and Megatron toys for Christmas as a kid. In robot form, they're roughly the same size and would seem to be a fair match for each other in a fight. However, problems arise when they transform into disguised mode, after which Optimus becomes a "toy-sized" truck, whereas Megatron transforms into a life-sized handgun. Hasbro's never been a stickler for length scales,1 but this is a bit much. Even as a fairly stupid seven-year-old, I could tell by Toy Land standards Megatron was hideously deformed, and it was all I could do to convince the other toys not to mention it in front of him because he was probably pretty sensitive about it.
Where was I going with this? Ah, yes! Soundwave. Let's consider three cases:
1. Soundwave Maintains His Robot Weight. At first glance, this might seem like a fair assumption. After all, he's a robot first and foremost, so it makes sense that he'd keep his robot weight when disguised. However, as Jon pointed out to me, "It's obvious that he's not that heavy in his 'disguise', because we see people pick him up." Just how heavy would he be? In the cartoons, he looks like he's at least as tall as a two-story building. Conservatively, this would put his height around 20 feet. Assuming he's made of iron with a density of 7.9 g/cm3 and a thickness and width of 4.0 feet and 5.0 feet, respectively, he would weight roughly 100 tons. Even if you had the entire Ukrainian weightlifting team at your disposal, you still couldn't pick him up, so you can kiss those IKEA shelves goodbye. To be fair, this isn't nearly the densest material known to man,2 but he's still going to have a hard time convincing anyone that he's just a plain ol' cassette player. This brings us to option #2....
2. Soundwave Maintains His Cassette Player Weight. Perhaps Shockwave is small-boned—er—alloyed. Just how dense would he be if this were the case? If I assume a cassette player has a mass of 2.0 pounds and use the dimensions above, then in robot form Shockwave would have a density of 80 grams per cubic meter. That's incredibly light. You might think that a light breeze would be enough to knock him over, but it's even worse that that. If he maintained his cassette player weight, he'd be less than half the density of helium, so you'd have to tie a string around him to make sure he doesn't go floating off into the stratosphere. The only way Soundwave is on remotely good scientific ground is if the Transformers discovered some novel unknown physics on Cybertron, which brings us to option #3...
3. Soundwave Maintains a Constant Density While Transforming. Although this is what the cartoons seem to suggest, the change in size would represent a clear violation of mass conservation.
Or would it? I've dealt with the problem of spontaneous weight gain previously in The Hulk Revisited. In that problem, I hypothesized that Marvel's Hulk gains weight by adsorbing air molecules with the result that he would suck out the equivalent of all the air in a 1500 square foot apartment upon transfiguring, instantaneously killing everyone (friend and foe alike) in said apartment with the resulting vacuum. However, Jon's question is unique in that he specifically mentions energy. One could imagine that Soundwave, during his transformation from 1980's cassette player to soulless killing machine, converts some excess energy to mass. From above, we know roughly 100 tons of mass is unaccounted for during the transformation. Perhaps Soundwave converts this mass to some new form of energy via Einstein's mass-energy equivalence equation E = mc2. Here, c = 3.0×108 m/s is the speed of light. Using this equation, we find that the excess mass is equivalent to about 8.2×1021 J of energy. That's about 80 times the energy that the entire United States uses in a year! I suppose it's possible an evil villain like Shockwave has this much extra energy lying around, but if he does, I imagine he'd find a more terrifying use for it than turning himself into an object whose most destructive power consists of playing Chris de Burgh's "Lady in Red" at moderately high volumes.
Thanks for a great question, Jon.
[1] See my earlier critique of G.I. Joe's U.S.S. Flagg aircraft carrier.
[2] A teaspoon of a neutron star weighs about as much as all of humanity.
Jon writes,
...I realized there's something about the Transformers that always made me think. It's how they deal with weight, density and size changes when they transform. Soundwave is a transforming robot. In order to function as a spy for the evil Decepticons, he transforms into a good old fashioned cassette player, perfectly able to blend in among actual cassette players. The kicker is of course that in his natural mode - as a robot - he's huge....
My questions are: What kind of energy would go into these transformations? Let's say that Soundwave does in fact weigh the same in robot mode as he did when he was disguised as a cassette player. What would Soundwave's structural integrity be in robot mode, and how would it affect his daily work as an evil, giant, murdering machine? And then, let us play with the idea that Soundwave is made of traditional metal and that he maintains the weight from his natural, robot mode. How would that affect his disguise mode, and what would happen if that cassette player stood on someone's shelf?
Note the change in size between here... |
...and here. |
Where was I going with this? Ah, yes! Soundwave. Let's consider three cases:
1. Soundwave Maintains His Robot Weight. At first glance, this might seem like a fair assumption. After all, he's a robot first and foremost, so it makes sense that he'd keep his robot weight when disguised. However, as Jon pointed out to me, "It's obvious that he's not that heavy in his 'disguise', because we see people pick him up." Just how heavy would he be? In the cartoons, he looks like he's at least as tall as a two-story building. Conservatively, this would put his height around 20 feet. Assuming he's made of iron with a density of 7.9 g/cm3 and a thickness and width of 4.0 feet and 5.0 feet, respectively, he would weight roughly 100 tons. Even if you had the entire Ukrainian weightlifting team at your disposal, you still couldn't pick him up, so you can kiss those IKEA shelves goodbye. To be fair, this isn't nearly the densest material known to man,2 but he's still going to have a hard time convincing anyone that he's just a plain ol' cassette player. This brings us to option #2....
Soundwave...one bad robot you don't wanna mess with. |
But who said he's just an evil killing machine? |
3. Soundwave Maintains a Constant Density While Transforming. Although this is what the cartoons seem to suggest, the change in size would represent a clear violation of mass conservation.
Or would it? I've dealt with the problem of spontaneous weight gain previously in The Hulk Revisited. In that problem, I hypothesized that Marvel's Hulk gains weight by adsorbing air molecules with the result that he would suck out the equivalent of all the air in a 1500 square foot apartment upon transfiguring, instantaneously killing everyone (friend and foe alike) in said apartment with the resulting vacuum. However, Jon's question is unique in that he specifically mentions energy. One could imagine that Soundwave, during his transformation from 1980's cassette player to soulless killing machine, converts some excess energy to mass. From above, we know roughly 100 tons of mass is unaccounted for during the transformation. Perhaps Soundwave converts this mass to some new form of energy via Einstein's mass-energy equivalence equation E = mc2. Here, c = 3.0×108 m/s is the speed of light. Using this equation, we find that the excess mass is equivalent to about 8.2×1021 J of energy. That's about 80 times the energy that the entire United States uses in a year! I suppose it's possible an evil villain like Shockwave has this much extra energy lying around, but if he does, I imagine he'd find a more terrifying use for it than turning himself into an object whose most destructive power consists of playing Chris de Burgh's "Lady in Red" at moderately high volumes.
Thanks for a great question, Jon.
[1] See my earlier critique of G.I. Joe's U.S.S. Flagg aircraft carrier.
[2] A teaspoon of a neutron star weighs about as much as all of humanity.
Aaron Santos is a physicist and author of the books How Many Licks? Or How to Estimate Damn Near Anything and Ballparking: Practical Math for Impractical Sports Questions. Follow him on Twitter at @aarontsantos.
Tuesday, July 10, 2012
Grumble Grumble
Sorry for the lack of updates. I just finished moving to Minnesota and am still sans internet and personal belongings. I'll have some fun guest questions posted soon. In the meantime, don't forget to enter the Look Out contest for a chance to win a free copy of Ballparking. Also, check me out on the Skeptically Speaking podcast this Friday.
-A
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