Sticky Situation

This one's stolen from Guesstimation.

In the second Spider-man movie Spidey stops a six car subway train by attaching his webs to it and pulling it to a stop in 10 or 20 blocks.

How much force does he need to do this?

Food for Thought

Let's answer this in three parts. How much energy does the average American consume, how much does that cost, how much does each calorie cost and how much energy would the Flash need.

There's a number of ways we could estimate how much one person spends per month for food. You could say that you eat out every meal (3 times a day) at some fast food joint for $10 a meal. That kind of life style has a monthly bill of 30 * 3 * 10=$1000 per month. I think that's high at $30 per day. What about $1 per day? At that price food would only cost you $365 each year. Ya right. So the easy numbers between $1 and $30 are about $10 per day or $3 per day. Let's be nice to Wally and say $3 per day or about $100 per month.

So we've guessed it can costs about $3 a day to eat (I know it seems low). Now we must guess how many calories you consume a day and so find the cost of every calorie.

How many calories do you eat per day? Maybe you know this maybe you don't. If not, maybe you know how many calories are in a McDonald's Value Meal (1167 cal) or in a can of coke (200 cal) or in a steak (500 cal). Therefore, you eat more than 1000 cal and probably not 10,000 cal. So then we guess the geometric mean which is 10^{7/2} = 3000 cal per day. So let's convert from calories to Joules (there's 4200 Joules in every calorie). 3000 cal/day * 4000 J/cal = 12 * 10^6 J/day. A physicist like Ray Plamer or Reed Richards would be happier if we turned this to Watts. Watts are Joules per second. So as a side note the power usage of a human being is 12 * 10^6 J/day * day/24hr * hr/60min * min/60sec = 150 W. Oh good - more power than a lightbulb.

So anyway that was a sidequest. The important thing is that a person consumes 3000 cal or 10^7 J everyday. Plus spends $3 a day for that food. That means that $1 will buy you 1000 cal or 4 * 10^6 J.

Next we have to estimate how much energy the Flash needs. Let's guess that most of his energy goes to running (that way if our answer is small compared to a normal person we can add the energy of a normal person to it or if our answer is big compared to a normal person we know that it is insignificant). The energy of him running is kinetic. We already used the equation for kinetic energy KE = 0.5 * m * v^2. If we allow for fictitious elimination of frictitous forces then if we can guess his speed, we will know the energy needed to get him to that speed. He needs that much energy every time he runs to that speed. Remember we are guessing for an average day, not at his fastest nor his slowest.

The speed of light (3 * 10^6m/s)?. %1 the speed of light? The speed of sound (300 m/s)? The speed of thought (only about 100m/s since it depends transport mechanisms of ions crossing membranes)?

Let's guess that he goes 10 times the speed of sound for an average run. The energy he needs to reach that speed is KE = 0.5 *m * v^2 = 0.5 * 100kg * (10 * 300m/s)^2= 0.5 * 100 * 9 * 10^6J=5 * 10^8J.

Since he needs that much energy every time he gets to those speeds we need to ask how often does he go that fast? We are talking about a cocky jock with a fairly large rogue gallery. So let's say he goes more and once a day but less than 100 times a day. The geometric mean: 10 times.

Then the energy the Flash needs for running in a day is 5 * 10^9J. At a price of $1 for every 4 * 10^6 J, the Flash must spend 5 * 10^9J / (4 * 10^6 J/$1) = $10^3 = $1000 more per day than muggles.

We said that a single person could eat for $100 per month and based on that the Flash's monthly grocery bill is $30,000. That's more than some of use make in a year! Notice that if we had said that Wally ate out for every meal his bill would be 10 times greater. A bright supervillain interested in discovering the Scarlet Speedster's secret identity might begin by looking for abnormally large grocery bills.

Like a Bat Out of Hell

I happened to be in the western part of Canada this weekend and thought it would be really cool to do some bouldering in the rockies. I'm here for a really quick trip and so climbing was just going to be something I squeezed in as I drove into the eastern part of the mountains. I packed my shoes and everything seemed to be going really well. The guy I was going with was excited, the people I'm travelling with didn't care if we stopped for a bit and the weather was beautiful. Near the town of Coalman we spotted what looked like a great face. We pulled over, changed and started hiking. We just arrived at the site when suddenly a huge dark cloud lumbered over the mountain. It didn't waste anytime looming. It came fast and hard. We got stuck in a hail storm and didn't get any climbing done. The rain didn't let up until we were out of the mountains again. It was too bad. I don't know when I'll be back there.

The Cost of Being Super

How big is the Flash's grocery bill?

For this superheroic Fermi Question you'd might like to know that every Calorie that you eat has about 4200 Joules of energy. Also guessing how fast he goes is tough. Consider average days for the Flash and think to yourself that he's much faster than the speed of sound but not any where near as fast as the speed of light. (Just an average day. That's all I'm saying.)

Skeletal

How heavy is he without his skeleton? I'm about 70 kg. Logan's pretty buff. Maybe say 100 kg.

How much of that is bone? 1%? Nah. 100%? 30%? 10%? Ya, 10% seems like the best answer. 10% of 100kg is 10kg. So that's 10 kg of bone and 90 kg of other stuff in your body.

Well, let's guess that bone and muscle and everything inside of us is all the same density. This is really not so bad of a guess because nothing inside of us is as heavy as rocks or as light as air. We pretty much have the same density as water.

The density of water is 1 kg/L = 1000 kg/m^3. So then 10kg of bone has a volume of about 10 kg / 1000 kg/m^3 = 10^{-2} m^3.

If that normal skeleton is removed and replaced with an adamantium one. If we can guess the density of adamantium then we can multiply it to the volume of a skeleton that we just found and we will know the mass which we can then ad to the 90 kg of left over non-skeletal stuff.

What's the density of adamantium? Well it's got to be heavier than water. 100 times heavier? No way. 10 times heavier seems like a good guess which says it has a density of 10 * 1000 kg/m^3 = 10^4 kg/m^3. Another way we could find this is we could look up the densities of some metals and notice that tungsten is pretty heavy at 19.25 g/cm^3 and that iron and silver and all of those are all pretty close at 7.9 or 10.5 g/cm^3 which are both 10 g/cm^3 = 10^4 kg/m^3 as far as we are concerned.

So using this density we find that the mass of an adamantium skeleton is about 10^{-2} m^3 * 10^4 kg/m^3 = 100 kg. So now with his new shiny skeleton Dark Claw or Wolverine weighs 100 kg + 90 kg = 200 kg or about twice as much. This isn't a shockingly high value. A person had an adamantium skeleton wouldn't sink in and leave foot prints in concrete or be too heavy to use the elevator.

Dark Claw's Adamantium Skeleton

In the 1990's DC and Marvel did a cross over event and mashed their characters together into "amalgams". The character they combine with Batman was Wolverine to create: Dark Claw.

At that time Wolverine had an adamantium skeleton which was one of the things that helped make him invincible (Batman often suffers from the same problem: invincibility by popularity). I can imagine some side effects to having a metal skeleton. Is weight one of them?

How heavy is Wolverine/Dark Claw with his adamantium skeleton?

Tons of Kryptonite

Let's assume that the entire planet of Krypton is made of Kryptonite. With the wide variety of kryptonite this doesn't even feel like such a stretch of the imagination.

So the planet explodes causing the mass that was the planet to shoot outwards and expand as a shell of debre. If we can figure out the fraction of that shell to hit the earth than we know the amount of kryptonite to hit earth (if we know the original mass of the planet krypton. Let's go with the 10 times greater than earth that Mohitparikh guessed instead of my 25 times (although for estimating it doesn't really matter!)).

We need to know the area of the shell. That depends on how far it's expanded. So we have to ask ourselves: How far is Krypton from Earth?

Well, how long did it take Kal-El to get to Earth? The fastest his rocket (it's usually a rocket) could possibly travel is the speed of light. You don't have to know the speed of light to know that a light-year is a unit of distance. It is the distance that light can travel in 1 year. So if Kal-El was put in the rocket (that goes the speed of light) as a new born and he arrived on earth as a yearling then Krypton is 1 light-year away. If he arrives as a 10 year old then Krypton is 10 light-years away. Here's a fact: No solar system is as close as 1 light-year away from us. Usually Kal-El is not 10 years old when the Kents find him so let's take the geometric mean and say that he was a toddler of 3 when he crashed on Earth and therefore Krypton is 3 light-years away.

Some more facts: Proxima Centauri is 4.2 light-years from us. Sirius A and B are 8.6 light-years. Vega is 25 light-years. So by guessing 3 light-years we are saying that Krypton was one of Earth's closest neighbors in a really big neighborhood.

Let's find out what 3 light-years are in useful units. Everyone rememembers the speed of light? It's 3 * 10^8 m/s. Here's the coolest trick I learnt from the Guesstimation book: You can either to all the slow multiplying of 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, 365 days in a year --- OR --- you can memorize that there are pi * 10^7 s in every year. It's a really good estimate. It's easy to remember that pi is about 3.14 and there are 3.15 * 10^7 sec / year. Cool, eh?

Anyway, how many meters to Krypton? 3 light-years times 3 * 10^8 m/s * pi * 10^7 s/year equals 27 * 10^{15} m (cuz pi=3, remember?). Which we round to 3 * 10^{16} m.

So what is the area of the shell of the exploded Krypton when it reaches earth? Well, we just found the radius of the shell (3 * 10^{16} m) and the total area of a shell is A=4 * pi * R^2. Which means the area is 4 * 3 * (3 * 10^{16})^2 = 4 * 3 * 9 * 10^{32} = 100 * 10^{32} = 10^{34} m^2.

We only need one more piece of information: How much of that hits earth?

All the Kryptonite is kind of evenly spread over the sphere so the density is d = M / A = 6* 10^{25}kg / 10^{34} m^2 = 6 * 10^{-9} kg/m^2. Now we know how many kg pre square meter and all we need is the area of the Earth that this hits. It's really the silhouette area or the cross sectional area A = pi * r^2. To do this you need to know or guess the radius of the Earth. I happen to know that it is r = 6000 km = 6 * 10^6 m.

That means that the cross section of earth is A = 3 * (6 * 10^6 m)^2 = 3 * 6 * 6 * 10^{12} = 100 * 10^{12} = 10^{14} m^2. When we multiply the area by the density we will have the mass of Kryptonite that hit the Earth. 6 * 10^{-9} kg/m^2 * 10^{14} m^2 = 6 * 10^5 kg. Holy Rao, Superman. That's 600 tons! But if you think about spread over the entire earth most of it would be dust and really we never made any guess about what fraction of the mass on Krypton was Kryptonite.

Notice, the Wikipedia page on Krypton says that Superman's home planet is 50 light-years away. If we ignore what this would mean about how fast Kal-El's rocket would travel (not to mention the Kryptonite rocks themselves) and redo the calculation then there's only 2 tons of Kryptonite on Earth. BUT even if it moved at 10% of the speed of light, it would take the Kryptonite 500 years to get here.

How much kryptonite is on Earth?

Batman's best friend Superman is a allergic to green chunks of his home planet. If Batman wants to find all of the Kryptonite on earth and destroy it or lock it away to keep Superman safe, how much should he look for?

OR

Batman doesn't like the Big-Blue-Boyscout who hasn't even figured out he can fly yet. So Batman's going to collect all the Kryptonite in the world to line the Batcave with thus making it impossible to Superman to attack Batman's child kidnapping and government over-throwing operations. How much Kryptonite is available?

This Fermi problem was posed by the book Guesstimation.

Crazy Krypton

The mass of a planet creates gravity. If two planets are the same size and one has twice the mass as the other than it will also have a gravitational acceleration that is twice as big. On earth the acceleration of gravity is g=9.8ish m/s^2. For guesstimating 9.8 is rounded to a nice g=10 m/s^2.

How high is a tall building? 1 story? Nope. 10 stories. No. 100 stories. Yes 100 stories (the Chrystler Building is 77 stories and the Empire State Building is 102 stories. How high is each story? 9ft.

So that means that the total height of a tall building is 900ft. To make ft into meters we multiply by 0.3 making a tall building 300m tall.

Relative to the ground, the potential energy at the top of the building is PE = m*g*h. All we need to do is guess the mass of Superman. Let's use a nice round 100kg (He's not 10 kg and he's definitely not 1000kg (that's a ton) so the geometric mean is 100kg). So at the top of his jump Supes' potential energy is 100kg * 10m/s^2 * 300m = 3 * 10^5 J (J=joules which we can think of as shorthand for all those more basic units which went into the calculation or we can just leave them whatever).

So then in order to get that high he must start out with a kinetic energy which is that big: KE=PE. 1 / 2 m*v^2 = 3*10^5kg m^2 / s^2. We said Superman's mass was 100kg so the velocity is v=sqrt(6*10^3). The square root of 10^3 is 30 because we divide the exponent 3 in half and multiply the coefficient by 3 for the 0.5 that is left in the exponent. The square root of 6? I don't know but I know that the square root of 4 is 2 and the square root of 9 is 3 and 6 is between 4 and 9 so I'll say 2.5.

So Superman's speed when he jumps must be v=2.5 * 30 = 75m/s. That's pretty fast but not nearly as fast as sound (340m/s) so he doesn't have to worry about causing sonic booms. How about as fast as a bullet? Not a chance Check out the Orders of Magnitude wikipedia page. A bullet moves at like 3 times the speed of sound or 1000 m/s.

So if he jumps at 75m/s, what kind of force must his legs exert to get that fast? F = m * a and he starts from standing still and gets to 75m/s. The acceleration is a = v / t. We need some guess at how long he jumps for.

The golden age Superman wasn't really quicker than the average person. He could run and move fast but he sort of jumped normal so let's estimate how long a person jumps for before their feet leave the ground. 1 second? No way. That's a pretty long time. 0.1 second? That seems a little short. Let's take the geometric mean and say 0.3 seconds. Then the acceleration is a = 75m/s / 0.3s = 250m/s^2. And then the force his legs exert is F=ma=100kg*250m/s^2 = 2.5*10^4N (N is newton's which is the unit for force or weight).

That seems big but what force does the average person exert? 1ft? Ya... It seems small but 1ft * 3 = 0.3m and an average person definitely can't jump 1m so let's just pick a number in between 50cm seems like a good guess. So the average person's potential energy at the top of their jump is PE = mgh= 100kg * 10m/s * 0.5m = 500J which means they need an initial velocity:
KE = 0.5 mv^2 = 500J ---> v^2 = 10m^2/s^2 ---> v=3m/s.

So the average person's acceleration is a = v/t = 3 / 0.3 = 10m/s^2. And the total force they jumped with is then F = ma = 100*10 = 1000N = 10^3N.

Superman's jump was 2.5 * 10^4N and and average person's was 10^3N. So Superman's jump is 25 times stronger than an average person's.

So to make Superman jump like a normal person on Krypton, the gravity must be 25 times greater. The gravitational acceleration on Krypton would then be 25 * 10m/s^2 = 250m/s^2. To get a gravitational acceleration that is 25 times greater on a planet the same size as earth, the mass of Krypton would have had to have been 25 times greater too. The mass of Earth is about 6*10^24kg which means the mass of Krypton would have been 1.5*10^26.

You should really read Physics of Superheroes to see what this means about what must have caused Krypton's destruction.

On that note, I think the Fermi Problem I pose tomorrow will be about Krypton's explosion.

What was the mass of Krypton?

Back in the dawning heyday of the superhero, writers were still determining the extent of Superman's powers. There was a time when Superman's powers didn't come from our yellow sun. One important distinction between then and now was that Clark couldn't fly. He was only "able to leap tall buildings in a single bound". It was explained that his super strength was a result of Krypton's larger gravity.

So, Detective, based only on the statement that he is "able to leap tall buildings in a single bound", what was the mass of Krypton? Some help for this Fermi question: It's a good idea to guess that Krypton is about the same size as Earth. You might be interested in remembering some high school physics that to jump to a height (h) you initially need the same kinetic energy (KE = 0.5 * m*v^2) as the potential energy at the top of your jump (PE = m*g*h).

Batty About Buses (a superhero Fermi question)

Ok so the Joker has planted a bomb on one city bus and Batman must track it down before the end of rush hour. How many buses could it possibly be hidden on?

We might have some idea of how many blocks we would have to walk to catch a bus in a dense city like Gotham and how long we would have to wait until the bus would come. So then if we can guess how big Gotham city is we could make an estimated guess at how many buses it would need.

So how large an area is Gotham city? 1km^2? No idea? Ok, well let's say Gotham city is about the same size as NYC. So what's the size of NYC? I have no clue.

So what do we do? We compare to the size of something we know. I said that the area of the earth is 5*10^14 m^2. And how many NYC fit on earth? Ok, bad question but what's the size of the USA? Well how many USAs would fit in the world?

1? Bad guess must be bigger. 10? No still bigger. 100? Ya that's a good guess. But seems a little big.

So we take the geometric mean of 10 and 100 which is 10^((1+2)/2) = 30.

Ok so if the area of earth is 5*10^14 m^2 and 30 USAs fit on earth than the area of the USA is about... well 5/3 is about 1 ... come on! Remember we're estimating. So we say, ``who cares?'' Ok so 5/3=1 and 10^14 / 10 means we subtract the exponents 14-1=13 so (5/3) * 10^(14-1) m^2 = 10^13 m^2. Check out the "actual"' answer. So how many New Yorks fit in the USA? I don't know. But I do know how many states do. Assume all the states are about the same size and say there's a nice round number of them: 50 (if there weren't a nice round number we would round it. Like say 48 to 50). So the average area of each state is 10^13 / 50 = 2*10^11 m^2. So we estimate that New York state has an area of 2 * 10^11 m^2 and the official area of NY is 1.4*10^11 m^2. Not bad.

And finally to estimate the area of NYC: how many NYCs fit in NY state? 1? nah. 100? Maybe? 1000? Somewhere between 100 and 1000. So the geometric mean is 300. Then the area of NYC is about 2*10^11 m^2 / 300. 11-2=9 and 2/3 is about... oh I don't know 1. So then the area of NYC is 10^9 m^2. Actual area of NYC (drumroll please): 1.2144 * 10^9 m^2. Wow pretty damn close.

And Gotham is about the same size as New York. Ok now. How many buses? Let's guess how many roads are in Gotham. First let's say Gotham City is square. Then it has a width or length of the square root of the area. Remember from the last post that we square root by halving the exponent: =10^{3/2} km = 10 * 3km =30km. Ok and how many blocks are in 1km? 1? no. 100? no. 10? ya that sounds about right. 10 blocks for every 1km. That's like 20 city blocks in a mile. So if Gotham is 30km by 30km and there's 10 blocks per km then Gotham is 300 blocks by 300 blocks. The roads make grids around the blocks so there are 600 roads in Gotham city.

How many roads have buses on them? Every road? No way. Every 10 blocks? Too big. Geometric mean = 3. We guess that in a densely populated city like Gotham or New York there is a bus route every 3 blocks. So that means that of the 600 streets 200 have bus routes on them which each run for about 30 km (about the length of the city).

And how often do they come? Never fast enough, right? I don't know every 10 minutes seems fast but every 30 minutes seems slow. Let's guess every 20 minutes.

And how fast do they drive? Not 1km/hr that's slower than you walk (10 blocks in 1 hr? Sloooow) but 100km/hr which is like freeway speed. So Geometric mean = 30km/hr.

Ok so if the buses go 30km/hr and come every 20minutes then we can guess the distance between buses: 30 km / hr times (1 / 3 )hr = 1km.

We've got it!!! On each route there must be one bus every km over a 30 km distance and there are 200 routes. (1bus/km) times (30km/route) times (200routes) = 6000 buses in Gotham City.

That took a little while but the math was easy and we could have scribbled it on a napkin or maybe done it in our head. But how did we do?

Well... (and remember as long as we're within ten times bigger or ten times smaller we're pretty happy) according to the wikipedia article on the MTA fleet there are 6,251 buses in fixed route service. The article on the NYC transit buses says about 4500 buses. Either way, 6000 is a pretty good guess.

How many buses are in Gotham City?

You don't need any information. Just answer it. You do need some idea of the area of something (ANYTHING) to compare to. Any area. The size of your state? Your city? Your country? The planet earth? For example, I happen to know that the planet earth has a surface area of 5 * 10^14 square meters (m^2). Don't ask me why. It's just something I picked up.

Oh and I know I haven't really shown you how. But let this be our learning example.
AND remember no looking up answers and no calculators. Do it in your head or on the back of a napkin or envelope or any scrap of paper.

Feel free to leave answers or guesses or questions or thoughts.

Dealing with Big and Small Numbers

How are we going to learn to guess? It's actually something we can learn. For example, say you ask me, "How many people are on earth?"
So I look on Wikipedia and find an estimate that on May 31, 2009 the earth's population was 6792,467,727. Wow down to the person: 6 billion blah blah blah and TWENTY SEVEN! Good job Census Bureau. But really now, that's a pretty useless number. Most people would have guessed 6 billion and really 7 billion would have been a better guess BUT and this is the point I'm trying to get to: the twenty seven people are a useless detail when estimating.

In fact! The 6 or 7 is pretty useless. When estimating the only important word is BILLION.

Don't get me wrong there will be times when we'll keep the 6 or 7 but remember the point of these Fermi questions is to get good estimates without looking things up and without using a calculator. So say we choose 7 billion and then we have to divide it by 2, ok? 7 / 2 is ... I don't know (actually I do but humor me) that's tougher math than 6 billion / 2. 6 / 2 = 3 SO THREE IS CLOSE ENOUGH. We will always chose the easy math or just drop numbers whenever we want.

Ok? I know that this will take some getting used to for anyone in the audience who wrongly thinks that the details matter. Another example: we were all in some high school math class and the teach asked someone to put some number through their calculator. Say you were calculating the area of a circle of radius 2m. The formula for area is
.
So some kid puts it through their calculator and says 12.56meters squared. The teacher says that's correct BUT then some annoying kid (that kid was usually me, by the way) shouts "No, it's actually 12.56637061...blah blah blah . Ok so maybe the annoying kid was more accurate but come on! The amount of time it takes to write those useless numbers out and he had to have used his calculator. There's no way he did it in his head.Us, here at Building Batman however, will always say things like
.
Who cares about the details. We are just estimating to get a guess. So then we'd say the area of the circle is 3*2*2 = 12. Close enough for the girls we go out with, ok?

And back to the world population of 6 or 7 billion.
How many zeros are in a billion? There's nine. So then the world's population is 7000,000,000. That's a lot of zero's. From now on we're going to use something called scientific notation. Don't be scared of it. It's actually really easy. How many times would you need to multiply 10 to the 7 to get 7000,000,000? Just count the zeros. Nine times. So we write this as 7 * (10 nine times) and the "nine times" is just taking the power. So 7 * (10 to the power 9) =
.
Because I'm writing this on a blog I'm going to write taking

as 10^9, ok?Why this "trouble"? Well, for two reasons: 1) if we wrote out all the zeros sometime we'd miss count or forget a zero and then we'd be screwed because 2) it tells us the size. Really the 7 in 7 * 10^9 isn't so important the important number is the 9 because the 9 tells us the size or the order of magnitude. If we change the 7 to a 6 so what? Our guess would be off by a little bit but it's only a guess so little bit here, a little bit there doesn't matter. We aren't looking for the "real" answer. We're looking for a good estimate. On the other hand the 9 makes a big deal. Turn the 9 to and 8 and only 10% of the population survives!!!

We can do the exact same thing with really small numbers. Think about 0.00097. Again we just count the zeroes (4 not 3). 0.00097 = 9.7 * 10^-4.

Good so far? If I did this too confusedly and you need something explained better just leave a comment.

Ok so three last things:

1) The rules for multiplying and dividing with scientific notation. When you multiply the coefficients and add the exponents.
For example (4*10^8) * (9*10^3) = (4*9) * 10^(8+3) = 36 * 10^11 = 3.6 * 10^12 is about 4*10^12. Cool?
And for dividing we divide the coefficients and subtract the exponents. Like (9*10^13) / (3*10^7) = (9/3) * 10^(13-7) = 3*10^6.

2) 10^0 = 1. Anything to the power zero is 1.
3) We're going to be taking alot of averages. We'll be guessing things and be saying things like, "well... 10,000 is definitely too big but 1 is way too small." So we will want to take the average between them. BUT notice the average of 10,000 and 1 is (10,000 + 1)/2 = 5,000.5. Well that's stupid. If you round 5,000.5 it's just 10,000. That's just the big number!!! 5000.5 is like 5000 times bigger than our small lower bound and only a factor of two smaller than our upper bound. Averaging isn't going to work.

When dealing with really big numbers and really small numbers taking the normal/everyday average isn't the right thing to do. What we really have to do is take the geometric mean. This is a good average for when you are dealing with big and small numbers. The actual definition of the geometric mean is that you multiply the two numbers and take the square root. But square roots are hard and we are doing this all in our heads so the greatest trick in the whole world which I learned from Guesstimation is how to take the approximate geometric mean. To take the approximate geometric mean just average the coefficients and average the exponents.
The geometric mean of 1 (10^0) and 10,000 (10^4) is ((1+1)/2) * 10^((0+4)/2) = 1*10^2 = 100. 100 is a good in between number if 10,000 is your high guess and 1 is your low guess.
Another example, the geometric mean of 4*10^17 and 8*10^5 is about 6*10^11 (because 6 is the average of 4 and 8 and 11 is the average of 17 and 5).
If the sum of the exponents is odd. There is a rule. You must decrease the exponent by 1 making it even and multiply the coefficients by 3. For example, the geometric mean of 1 and 10^3 is 3*10^1 = 30. The reason for this is because when you average the exponents you get a something and a half. An exponential of 0.5 or 1/2 is exactly the same as square root. And the square root of 10 is 3.16 blah blah blah or just 3. If that's confusing, don't worry about why. Just follow the simple rules for geometric mean and we will get some really magical answers from very little information at all.

That was a long post but we need to know all that. Tomorrow I'm going to do a non-superhero one with you and when you see this in action it will make a lot of sense and you will probably be wowed (maybe not but I was).

Fermi Problems and Batman

Batman is the world's greatest detective. And so it's time that I start to develop detective skills but how to start? Do I start with a criminology text book or try to master forensics or just buy a fingerprint kit?

I'm not sure but I think I need more basic skills than that first. Just logic skills. Like Sherlock Holmes. Holmes once said, "From a drop of water a logician could infer the possibility of an Atlantic or a Niagara without having seen or heard of one or the other." Moving forward from this thought I'm going to take you through a couple of Fermi Problems.

Fermi problems aren't Holmesian deduction but they are a method of "scientific guessing". By making good estimates of things that we already know (things that are common knowledge) we will realize that if we think about it we can have a good estimate about something we thought would be impossible to guess.

It's an estimating technique that will allow us to make good "guesses" very quickly and with very little information about things which would seem impossible to compute by people who have never learned this.

For someone like Batman this kind of skill would be invaluable. I have no doubt that Batman has wired his mind to operate this way.

I practiced this from a book called Guesstimation by Lawrence Weinstein (click on that link) and John Adam. Some of the stuff we'll look at are inspired by a question or two in their book but with a superhero twist. It's a great book by the way. If you have fun with these Fermi problems then I'd highly recommend it. The other place of inspiration will be from a book called The Physics of Superheroes by James Kakalios. Some of these problems will definitely involve physics but it will be the most minor physics. Kakalios uses these types of questions to teach physics but really it's not such hard stuff so I'm going to assume that if you haven't taken a high school physics course that I'll be able to teach it to you as we go.

BUT the pro at this, the man whom the entire method is named after is Enrico Fermi. This man was the prototypical physicist. I don't know what to say about him in a half a paragraph but... wow. He was legendary for being able to figure things out in his head from what other people thought was no information at all. Classic Fermi questions he estimated by this method that we are going to learn include:

How many piano tuners are there in Chicago?
How many blades of grass are there on your front lawn?
What is the ratio of paved to unpaved surface area in California?

I'll post a blurb on the how us would-be-Batmans will solve Fermi problems, and then post questions every other day and a solutions the following day. I hope that you guys have as much fun with these as I did (they're all done but scribbled on napkins and notes and stuff but not typed).

PS Once you start to get the hang of these throw out your own comic book based ones and let's see if we can solve them together.

The Brave and the Bould

WOW. Bouldering was amazing. I had so much fun. And my toes hurt so much.

We drove out and hiked in. At first we found many rock formations that weren’t tall enough or were just too boring. But eventually we found this boulder that looked pretty good. Unfortunately, the rock wasn’t very great on the one side and the other was very damp (it was a pretty moist day). But we’d found the right area. So we hiked around a bit looking for the perfect face. We didn’t ever find the perfect one but we did eventually settle on a wall. It needed cleaning and neither of us brought a brush (you use a metal brush like you would use on a BBQ to clean dirt, moss and other slippery substances from the visage) but we decided that that would just have to be part of the challenge today.

So we tied up our shoes and positioned the drop mat. This is my only complaint about the shoes I chose. Lace-up shoes take a significant amount of time to tie-up and untie if you ever want to give your poor feet a break. If I had chosen slip-on type shoes, I could have taken them off while I was spotting.

There were two routes up. The easy one was more than doable by me on my first time out but it did take me a couple of tries. Falling off the face is not nearly as bad as I expected. Your partner watches you as you climb and if you fall he’ll break your descent by catching you. This isn’t a falling-damsel-Superman-type save but rather you let the person’s legs land on the mat and get your hands under their arms. By catching their upper body you make sure that they don’t hit their head on any nearby rocks.

The second route was far more challenging. Although my partner made it up a few times, I could never get past a certain point. I tried and tried but I think it was just a touch too challenging. Luckily, the guy I was climbing with is very patient and a great teacher. I’m determined to make it to the top next time we go out there.

I was in pretty good shape after the climb (sore toes and feet and one bloody knee) so we continued to explore the area. It turns out that there are a couple of other pretty good looking climbs nearby and one amazing looking overhang.

I’m excited to get back out there and my climbing partner has promised to show me some of the other spots in the city’s immediate vicinity.

PS Apparently climber culture is filled with lingo and jargon.

Bouldering Batman

It’s only been one day since I bought my shoes but this evening a more experienced friend and I are going to drive out to the area I explored on my bike last weekend and see what’s out there for climbing.

Bat-Boots

This afternoon I biked across town to the largest sporting store in the city and bought a pair of rock climbing shoes. Oh, man are they uncomfortable. They are supposed to be really tight and mine are (I hope tight enough). But it’s a good thing that I knew exactly what I wanted because the service was really weak and I didn’t get much help.

So what exactly did I look for in a shoe? Price. Since I’m brand new to this, I don’t need the best gear. Plus, I need to have enough money for any other projects that I have on the go for this Building Batman. I expect Jiu Jitsu to cost a bit. The shoes that I settled on are a pair of Mad Rock. I hope they’re what the doctor ordered.

Let There be Sanity

The Source has posted another image of Batgirl. Am I the only one praying it's not Babs? I hear a lot of people who would or wouldn't want it to be Cassandra but please God let it not be Barbara. Yes, she was the best but her crippling was such a daring development of the character. Please don't let them just erase all that. She works so well as Oracle both for her own character and as a part of the Bat-family.

That's a lot of miles...

This weekend I spent a lot of time on my bike. Friday I did a 16 mile bike ride, Saturday a hard 25 mile ride and then on Sunday a final 25 miles but at a much easier pace. This wasn't just to get in better shape but also to explore potential bouldering sites outside of town. One of the locations has real potential but I think that before I can start climbing there I will have to be less exhausted from the bike trip by the time I make it out.

P.S. Batman doesn't bike. Batman flies. In a flying car. Everyone knows that - except The History of the Batmobile site which is uncharacteristically days behind the the current Batmobile (Batmobile iterations must be recorded on a day by day basis since it's changed so often).

Batman and Robin #1

I don't usually do comic book reviews. Hell, lately I don't usually post at all. Maybe you couldn't tell but I do actually read Batman comics. I try my best to buy bound editions after I have read enough reviews rather than risk getting stuck in a crummy story but I often have one or two serial titles that I'm following. Lately, it was Gotham After Midnight and RIP. RIP was well... was... despicable and Gotham After Midnight was ok but I really wish it had been more of a mystery. The story gimmick was to present an unknown character - a new villain who's secret identity was most obviously closely tied to Batman but who was unknown to all the story's characters and to the reader. But then instead of making it a mystery story with clues throughout the title, writer Steve Niles produced only one single character the villain could possibly be, left no clues at all and then had Batman deduce the entire thing in the final issue. It was kind of a disappointment.

Anyway, none of the new Batman titles on DC's lineup really interested me this year except Detective Comics which will be headlining Kate Kane. I have high hopes for this run. The promotional material is stunning.

And then there's Batman and Robin. Batman and Robin. Batman and Robin by Grant Morrison. Batman and Robin by the man who killed Batman (and I don't mean wrote the story in which Batman died) the last time he touched the character. Batman and Robin starring Damian Wayne as the new Boy Wonder.

Oh God. Just let me die now.

BUT drawn by Frank Quitely. ...

...

Oh man. Now I have a conflict. All Star Superman was so good. I even liked JLA: Earth 2. When Grant Morrison and Frank Quitely work together it's such magic. Now I know that some people don't like Frank Quitely's work. I've heard it said that it's distracting (whatever that means for a comic book) and also that it's just uninteresting but personally I'm a fan.

So I said to myself "I'll buy issue #1. Issue #1 only. If I'm not immediately won over that's it. No more chances, Mr. Morrison."

Conclusion: I have a new conflict: This title is so great that I must decide if I'm going to keep buying the monthly issues or wait until the run is done and buy the bound edition that is sure to follow?

So to actually review the issue:

BOOM BOOM. The tone is effectively set by the very first panel. Sound effects are written out BUT this isn't the POW SLAM BEEBOB sound effects of the campy Batman television series. In a very neat manner Quitely has magically incorporated them into the art. The BOOMs for explosions are formed by the explosion itself. The BWKSSSssssss of a rocket is in the rocket's exhaust. SPLASH is spelt out in the waves. This isn't new (Paul Pope's Batman: Year 100 immediately comes to mind) but it is the perfect fit for this crazy Batman title. Another reviewer summed it up as follows: "Silver age wackiness with modern sensibility." Sure we have a frogman as the villain but he's a drug dealer. Although, I think Batman should be as real world and gritty as possible, I do have a soft spot for things like this series or the Batman: The Brave and the Bold cartoon.

The writing is just as perfect. The otherworldlyness of an upside-down dynamic duo is wonderful to read. Nothing is as it should be. Batman is encouraging and optimistic while Robin is curt, confident and arrogant. Although Quitely's visualization of Dick Grayson in Batman's costume does not vary significantly enough from Bruce's physique, Morrison's dialog is extremely differentiating. He never lets you forget that it is Dick under the cowl not Bruce. Even more so with Damien. In fact, it appears as if Morrison is on a crusade to redeem Damien in the eyes of readers. That's not to say that Damian isn't still a completely spoiled brat but with Bruce out of the picture it is apparent that Damien and Bruce possess very similar characters. Morrison even has Alfred comment on the similarity while he and Dick close down shop at Wayne manor and move operations to the Wayne Foundation.

A couple of details here. (1) Getting them out of the Batcave is a good idea. It distances this team from the original. (2) When they pass the Wayne family cemetery, is that an unmarked grave stone in the shape of the Batsymbol? (3) Have you ever googled Building Batman before? Try it. Notice the second site listed: An AT&T building that is often called the Batman Building. Now look again at Quitely's Wayne Foundation. Am I the only one seeing similarities here?

A couple of last things. Before the issue ends we are reintroduced to a new villain: a butcher in a pigs mask (It's ironic. Yes, I get it, Mr. Morrison). This closing scene is quite disturbing which hearkens back to the "modern sensibilities" comment. I really didn't expect a story as brightly colored as this to be able to capture the darkness of the Batman universe but this is one creepy villain.

We then get two sets of images from upcoming issues which tell us some important things:
(1) That funny black thing on Robin's cape is actually a hood.
(2) Arrogant Robin is going to get a humiliating moment of failure.
(3) Damien gets an "I quit" / "run away from home" style story.
(4) The daughter at the end of this issue becomes the Robin-esque equivalent for Red Hood.
(5) The Batman and Robin and Detective Comics stories eventually intersect.
(6) Wayne manor only has only two locks. One of which is a simple tumbler. The other is only a warded lock. (Oh and Dr. Hurt survived the helicopter crash).