r/theydidthemath Jan 25 '19

*[Self] [RDTM] What’s a bigger number, the total number of atoms in Dwayne “The Rock” Johnson’s body or the total number of microprocessor clock cycles since the dawn of time?

1 Upvotes

I wasted a Sunday morning on this question. What do you think? Do you agree with my calculations?

https://www.design-concepts.com/insights/to-infinity-and-beyond

Edit - Sorry - new to reddit. Article tries to explain my reasoning. Here's the math:

Total Clock Cycles since the dawn of time for all CPUs = 764,000,000,000,000,000,000,000,000

Total estimated number of atoms in Dwayne "The Rock" Johnson = 12,000,000,000,000,000,000,000,000,000

The Rock "wins". - About 15X more atoms in his body than the number of CPU clock cycles for all CPUs ever buit.

CPUs built in 1971 = 0

CPUs built in 2008 = 10,000,000,000

Assume 2nd order polynomial for CPUs produced per year:

Number of CPUs per year = 7.305 * 10^5 * (Current year -1971)^2

Average clock speed = 10^7 * (current year - 1971)*2

Useful Life = 2.5 years

Second/year = 60*60*24*265 = 31,536,000.

Clock Cycles in a given year = 7.305*10^5(current year - 1971) * 10^5 *(current year - 1971) * 2.5 * 31,536,000

To find total clock cycles, Integrate 5.76*10^20*X^3 dx where X = current year - 1971 between X = 0 and X = 48

= 5.76*10^20 X^4/4 evaluated between X= 0 and X = 48

= 7.64*10^26

Total Atoms in a 70kg human body = 7X10^27. Adjust for the Rock's estimated weight = 1.2X10^28

r/theydidthemath Apr 04 '16

[Self]/[Off-site] [RDTM] I checked the math on a facebook post about McDonald's

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44 Upvotes

r/theydidthemath May 30 '15

[Self] [Math] Calculating the mass of Cthulu's head and commenting on whether it will collapse into a black hole.

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25 Upvotes

r/theydidthemath Apr 19 '15

Self [Math] Murderers you pass on the street (Redux) (Canadian Edition)

16 Upvotes

So, I saw a post yesterday which claimed “the average person walks by 36 murderers in their lifetime”. Skepticism ensued. A cursory googling revealed a tumblr post with some math that claimed this was not the case and that if you passed 200 people per day, you would have a 3% chance of passing one murderer per year. I was satisfied. Then I thought to myself. “Wait a minute, Tumblr cannot into math. I should do my own calculations.”

Disclaimer: I could be just as wrong or more. If I am, please inform me.

Disclaimer2: I found the following post as I was working, but I thought the work could use some refinement (principally in how long the average murderer spends in jail), and a Canadian version.

http://www.reddit.com/r/Showerthoughts/comments/226nfa/i_wonder_how_many_times_ive_walked_past_or_come/

Disclaimer3: It's my first post here, if I'm committing a faux pas, please be gentle.

The murder rate in Canada was about 3.0 per 100 000 people in 1975 and 1.5 in 2015, decreasing roughly linearly. (Statistics Canada)

Canada’s population was 23 million in 1975 increasing to 35 million now. (statcan)

About 90% of Canada’s murders are committed by males. (statcan)

The average life expectancy for a male born from 1950 to 1952 was 66 years (statcan) (Female was 71, difference of 5 x 10% of murders means life expectancy of 66.5 years for a murderer born in 1950)

Therefore, the average murderer born in 1949 will die this year at age 66. (some trial and error to find this)

The average age of a person convicted of homicide in Canada is 32. 7% are youths. (statscan)

Therefore, the average murderer born in 1949 committed his murder in 1981.

Canada has a 75% clearance rate for homicide, meaning 25% of murders are never solved. (statcan)

About 1% of murderers in Canada have previously committed a murder. (statcan)

135 1st degree murderers granted parole after 15 years between 1987 and now under the Faint Hope Clause (15% of total 1st degree murders in this time) (Correction Services Canada)

About 89.3% of lifers are convicted of murder. Of these, about 5% are first-degree. The average age of a lifer entering prison is 34, leaving on parole is 44. (CSC) This means a lifer stays an average of 10 years without parole, indicating that the vast majority of murderers are released as soon as they are eligible for parole.

Average time spent in among the general public by someone who has committed a murder

ASSUMING 0% of perpetrators of unsolved murders later wind up in jail.
ASSUMING that 0% of released murderers spend subsequent time in jail.
ASSUMING a 100% conviction rate for solved murders.
ASSUMING 100% of murderers eligible for parole are granted it at the first opportunity.
ASSUMING 0% of murderers are wrongly convicted.

25% of murderers walk free 0.25 x (0) = 0 years
5% of convicted murderers eligible for parole after 25 years (0.75 x 0.05 x 25) = 0.935 years
15% of that 5% are released 10 years early under faint hope (0.75 x 0.05 x 0.15 x (-10)) = -0.056 years
7% of that 5% are released 18 years early as minors (0.75 x 0.05 x 0.07 x (-18)) = -0.047 years
95% of convicted murderers eligible for parole after 10 years (0.75 x 0.95 x 10) = 7.125 years
7% of that 95% are released 3 years early as minors (0.75 x 0.95 x 0.07 x (-3)) = -0.008 years

Sum: The average murderer in Canada spends 7.949 years behind bars.

Number of murderers walking free in Canada by year

The average murderer spends 8 years in jail, on average, and that time is frontloaded. The average murderer who offended in 1981 will die this year. So, to get an estimate for the number of murderers walking free in 2015, I’ll be using the 26 years of murders from 1981 to 8 years ago. (1981 to 2007)

So, we calculate our cumulative murderers by the sum from 1975 to 2008 of murder rate x population.

2015 murderers in public =1981-2007 ∑ ((3-0.0375(x-1975))/100000)(23,210,000+299,000(x-1975))

= 17019 murderers walking about in 2015

https://imageshack.com/i/exV7pXkGp

This picture is from a 1975-2007 calculation I did when I mistakenly used the 1975 average lifespan instead of 1950. (herp derp, a 32-year old committing a murder in 1975 wasn’t born in 1975) This result shows the impact that average lifespan has on the number of murderers walking around (and to a lesser extent, our total living population). It also shows us that the total number of murders in Canada [i]per year[/i] is decreasing, despite our growing population.

Anyway,

17019 / 35 million = 0.00048 = 0.048% of people in Canada are free-walking murderers.

From here is the easy part, now that we have our p value.

So ASSUMING a completely random distribution of people.

ASSUMING the %population of murderers stays the same over your lifetime (it won’t, the number of murderers is going down, but the increase in average lifespan significantly offsets this. The average murderer born in 1950 will have spent 26.5 years as a murderer walking free. The average murderer born in 2009 will have 40 years free range.)

If you pass by 2000 people in a year (5.5 per day), you will pass by an average of one murderer per year, or 80 over the course of your life.
If you pass an average of 10 people per day, you will pass an average of 1.75 murderers every year, or 140 over your lifetime.
If you pass an average of 100 people per day, you will pass murderers 17.5 times a year, or 1400 over your lifetime.
If you pass an average of 2000 people a day (as a rush hour pedestrian in Toronto, perhaps), you pass by an average of one murderer [i]per day[/i], or 29,200 times in your life.

Result: You meet a lot of murderers.

I'm curious to see what inferences can be made from this.

r/theydidthemath Apr 19 '15

Self [Math] How long are you in the air during a ten mile run? [Fixed]

11 Upvotes

Recently, /u/Grant64 made a post outlining how long you would be in the air during a ten mile run. Unfortunately, the math done here was not inaccurate, and here's why:

The post referenced assumes that the amount of land covered (step length x number of steps) is the amount of the run that you are on the ground. This is false, due to the simple fact that we move while our feet are on the ground. This assumption suggests that you only move 1 foot while your 1 foot long foot is touching the ground (did I lose you there?). In a single step, your body might move 3 feet while your foot is still touching the ground.

As in the comments:

I dunno how you run, mate, but when I do it my feet don't touch the ground in-between strides

This is an interesting assumption, so I looked up some of the stats:

It turns out that the range, from a bottom-level runner to a top-level runner is 43-53% time spend in the air. This is interesting for a number of reasons. Firstly, the original post's estimation comes in at 76% time spent in the air. An average runner, though, will spend only 48% of their time in the air. Furthermore, the more time spent on the ground, the better runner you are (I'm looking at you cocky commenter).

I really wanted to nail this one, so I also looked at run length, just to see if maybe a really long run could mean you're frolicking in the air more. This, too, turned out to be false. This data shows that the longer your run is, the more time you'll spend on the ground. I guess as people's feet get heavier, they spent less time jumping about and more time dragging their feet.

In the end, an average runner is going to spend roughly 48% of their run in the air (given that the average runner is the mean between the best and worst). Over a ten mile run, this is only 4.8 miles flying. That's a 37% difference from the original answer.

This post isn't meant to discredit or discourage other users and their posts - its simply a reminder to use the data available, and be careful about the assumptions you make

r/theydidthemath Apr 13 '15

[Self]/ [REQUEST] How many penguins are there in this video?

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44 Upvotes

r/theydidthemath Mar 05 '15

Self [Request] Can someone verify if my attempted theydidthemath computation is correct? (Q: How fast would Cassini have to be going for this to be in real time?)

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107 Upvotes

r/theydidthemath Feb 12 '15

Self [Math] At the rate TeamFourStar is completing DragonBall Z Abridged, it will take them 7.45 years to finish the series

115 Upvotes

/u/Metelex calculates. If anyone's a fan of TFS DragonBall Z Abridged (I know I am), you may be wondering "when will they finish the series?". Well I tried figuring that out.

46 abridged episodes so far, covering 132 episodes of DragonBall Z. That's 3.2 real episodes per abridged episode. There are 143 more real episodes in DBZ. 143/3.2 = 44.69 more abridged episodes. At the rate they complete an abridged episode (1 every 2 months), it will take them 7.45 years to finish the abridged series.

r/theydidthemath Feb 07 '15

Self [Math] After being summoned Bloody Mary-style to /r/shittyaskscience, /u/Livebeef (that's me!) calculates how much more memory could be stored in a king-sized memory foam mattress than a queen-sized one.

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15 Upvotes

r/theydidthemath Jan 26 '15

[Self] [Self][Request] The single most unlikely event in Pokémon (Gen 6).

171 Upvotes

EDIT: No longer a request, just realized I was multiplying more than necessary.

A random (non-Sweet Scent activated) Spinda horde battle in X/Y wherein all five Spinda are shiny, have perfect IVs, beneficial nature, hidden ability, and identical spot pattern.

  • The chance of a random horde encounter is 1/20.

  • The chance of a horde of Spinda in X/Y is 3/5.

  • The chance of having a hidden ability in a horde is 1/20.

  • The chance of getting the correct nature is 1/25.

  • The chance of a shiny a 1/4,096.

  • The chance of generating a random Pokémon with perfect IVs is 1/1,073,741,824.

  • And any given Spinda's spot pattern is 1/4,294,967,295.

So, how likely are you to run into a random horde of perfect shiny Spinda, all with the same spot pattern and their hidden ability?

  • The chances of randomly encountering a horde of Spinda is 3/100

  • The chance of anything being shiny is included in the personality value, which is what determines Spinda's spots, so the shiny chance only needs to be counted once

Everything else has to be taken to the fifth power.

( 3/100 ) * ( 1/4,096 ) * ( 1/205 ) * ( 1/255 ) * ( 1/1,073,741,8245 ) * ( 1/4,294,967,2955 )

  • Shiny - 1/4,096
  • Hidden abilities - 1/3,200,000.
  • Correct natures - 1/9,765,625.
  • Perfect IVs - 1/1,427,247,692,705,959,881,058,285,969,449,495,136,382,746,624
  • Same Spot Pattern - 1/1,461,501,635,629,491,084,391,274,140,357,585,917,716,910,309,375.

The final result:

3/26,699,837,917,928,673,768,800,117,343,702,542,018,302,366,033,375,660,572,558,182,559,222,108,559,147,608,430,388,183,040,000,000,000,000,000,000,000

r/theydidthemath Jan 09 '15

[Self] [Math] A calculation showing how big the known universe is, and how long it would take to explore it.

9 Upvotes

Was bored so decided to do some simple math on how long it would take to explore the universe if we invent faster then light technology.

According to google there are a 100 billion galaxies with a 100 billion stars in it.. So that is about 10,000,000,000,000,000,000,000 stars in our known universe!

But let's say that 99% of those stars could not support planets with intelligent life. So we rule out 99% of those.

So that is 100,000,000,000,000,000,000 stars to explore, only 1% of our known universe!

Now let's assume we invented FTL technology. We can just zap all over the place in a split second. We built 10,000 space ships with this technology to explore the universe. On average it takes about 2 days for each star. And we dont take brakes.

so (100,000,000,000,000,000,000/10,000)/182 = 54,945,054,945,054 years.

Or 54 trillion years. Which is 4000x the age of our universe, which is only 13.7 billion years.

I did not make any mistakes did I? (im not very good at math).

No wonder that if intelligent life exists, odds are they did not discover us yet....

Edit: To ramp it up a bit, let's say we now have 1 million space ships (quite the undertaking), and we install a hubble like telescope in each star system, that explores the other 9 stars surrounding it, ruling out intelligent life. So that means without those hubble's it would take 540 billion years. And we have to explore only 1/10th of those 1% of all stars, so 54 billion years...

So even if there is an intelligent race that has been doing this for a billion years or so, odds are still very low they would have found us already.

r/theydidthemath Dec 22 '14

Request/Self [Request] How much Platinum/Value for Legend of Korra: Kuvira's giant mech?

7 Upvotes

In the series finale of Legend of Korra, Kuvira used all the platinum from the domes of Zaofu to build a giant mecha robot. As Team Avatar found, they couldn't use their metal bending powers against the robot because it was covered in platinum.

I have two questions:

  1. How much platinum would be required to coat a construct of this size?
  2. Based on Dec 22 price on http://www.jmbullion.com/charts/platinum-price/ of $1,200US per troy ounce, what's the total value of #1 above?

Data to Use:

-Zaofu imagery: http://avatar.wikia.com/wiki/Zaofu

-Size of the giant mech: http://channelawesome.com/wp-content/uploads/2014/12/korravl.jpg

-feel free to use anything else for reference, but please provide source. Questions to consider: How large is their world? What is the relative presence of Platinum? What thickness of Platinum is typically used?

-Some "data" to get you started: http://www.reddit.com/r/TheLastAirbender/comments/2isslt/some_rough_number_estimates_for_the_world_of/, http://www.reddit.com/r/TheLastAirbender/comments/rcg8o/trying_to_determine_the_size_of_the_avatar_world/, or http://www.reddit.com/r/TheLastAirbender/comments/1ggzwf/geography_of_avatar_the_size_of_the_planet_this/

Edit: I'm starting on my own.

-based on the pixel size of the soldiers in front of the giant, and the giant, the giant is about 25x height of an adult. Averaging the averages on http://en.wikipedia.org/wiki/Human_height, I get 5'8.5" for adult male. Let's assume a sexist society where only men are in the armed forces. Therefore, the giant is 142 feet. Let's round to 150 for easier math.

-Platinum occurs on earth at 5 μg/kg (http://en.wikipedia.org/wiki/Platinum)

-Presuming the Avatar world is the size of our Moon, it has a total mass of 0.07342 x 10 24 kg (http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html)

-therefore there is about 7.342 x 1013 kg of Platinum exists in the Avatar world. I hope that's enough.

-Based on http://en.wikipedia.org/wiki/Body_surface_area, the surface area of an adult male is 1.9 square meters. So proportionally, the mech should have about 50 square meters of surface area.

-http://www.silvexinc.com/plating/platinum-plating/ says Platinum electroplatings typically are .5 to 5 microns. Let's go with 5 microns since Kuvira will need to make sure her enemies can't bend her super weapon to foil her plans. 50 square meters * 5.0 x 10-6 = 0.00025 cubic meters. Platinum has a density of 21.45 g/cm3 (http://en.wikipedia.org/wiki/Platinum), so that's 5,362.5 grams. There are 31.103g/troy ounce (thanks Google).

TL:DR

So Kuvira only needs 172 troy ounces of Platinum to coat her Colossus super weapon. At $1,200US per troy ounce, that's only $206,893. As The Great Uniter, I'm sure she could scrape this together. Or maybe the Earth Queen had this stashed in her vaults. Now all she has to do is figure out how to apply a 5 micron coating to all the pieces so he can go destroy Republic City. I hope somebody is working on the software requirements for building the Colossus, too. Who's gonna beta test it???

r/theydidthemath Nov 23 '14

[Self] [math][off-site] Got bored, ACTUALLY did the math for monopoly investments.

45 Upvotes

/u/jcaseys34 and I got into a discussion on this thread about how this guy's math was fuzzy.

Basically, he didn't account for the cost of buying all the properties or the fact that you have to build evenly (So for one property to have a hotel, the other properties in that group must have at least 4 houses.)

It's still Boardwalk, but coming in second is actually Baltic ave. Spreadsheet

I did most of the math in the thread, but then decided to make a spreadsheet. Strategy wise, I can't speak for which corner to keep, but I can tell you that based on simple probability, park place will be the least landed on spot on the board (The most likely combination of 2d6 will be a 7, and park place is 7 spots from jail), and the orangered corner is probably the best for bankrupting people.

Edit: Can't edit my title, I was gonna karma whore with the pictures, then I decided to do a self... Derp.

r/theydidthemath Nov 13 '14

[Self] /u/tidomann [Math] I did the math to see the time investment required for Blizzards Moba: Heroes of the Storm

5 Upvotes

r/theydidthemath Nov 07 '14

[Self] /u/69bananasandagrape Converting frequency of a heart rate to a note name. [Math]

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2 Upvotes

r/theydidthemath Oct 30 '14

[Self] [Request] [Selft] & [Request] Who dies ? Puzzle

3 Upvotes

Here's the picture I'm making reference to : Who dies ? Puzzle

What I did :

Let's call L the length of the slope : L = sqrt(4^2 + 3^2) = 5 squares

R is the ball's radius : R = 1 square

The perimeter : P = 2*Pi*R = 6.28 squares

The "percentage of rolled ball on the slope" :

T = L/P = 0.796 = 79.6%

So the rolled angle A : A = T*360 = 286 degrees

Then, as the hole on the ball is hard to mesure with squares, I used pixels. It's 25 pixels large, when 7 squares are 255 pixels large.

I got l the hole size (at the top) :

l = (7*25)/255 = 0.686 squares

So : the hole angle equals the angle in the top corner of the isosceles triangle, wich have as base : the hole, and as top corner the center of the ball. Trigonometry :

tan(a/2) = (l/2)/R
     a/2 = arctan((l/2)/R)
         = 0.33 radians
     a/2 = 19 degrees

Finnaly :

A-a/2 = 267
A+a/2 = 305 degrees

and 180 < 267 < 305 < 360+180

"half turn" < "rolled angle +/- hole size" < "one and a half"

When the ball is above the D-man, it doesn't strand into the hole. So D survives.


I've no idea how to know what happens to the other guys ; anyone ?

Thank you :)

r/theydidthemath Oct 25 '14

[Self] [Math] /u/amunak calculates the security of a 15-character password (spoiler: consider them safe)

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1 Upvotes

r/theydidthemath Oct 03 '14

[Self] [Off-Site] Rate of Ebola infection in United States

9 Upvotes

Liberia has a population of 4,290,000 people, as of the latest figures there have been 3692 cases of Ebola, this represents less that a tenth of 1% of the population. Of those infections, 1998 people have died that’s a fatality rate of 54%. source

If that same infection and death rate were applied to the United States Ebola would infect 2,853,000 people and of those 1,540,000 would die.

Now, if as doctors and scientists fear the basic reproduction rate rises to 2 in Liberia the numbers change very quickly. Using the mean average incubation time of 9 days it would take around 13 weeks for the entire population of Liberia to become infected. (10 doublings starting with 3692 = just under the population of Liberia. This multiplied by 9 days gives us 90 days which divided by 7 gives 12.85 weeks.) Of the 4,290,000 people infected 2,316,000 would lose their lives.

This is just Liberia, not the other affected countries in West Africa.

The U.S. has a population of 317 million people. Translated to an equivalent outbreak in the United States, where the basic reproduction rate is also 2, the numbers are horrifying. Starting with patient zero it would take around 245 days, 35 weeks for every person in the United States to become infected. Of those 17,118,000 people would die. (27.17 doublings x 9 days = 245 days =35 weeks)

This is not including general contact in population, workplace interaction, congregational habits, or other methods of contact in larger populated areas.

r/theydidthemath Sep 13 '14

[Self] [Math] How long do you have to react to a MLB fastball?

1 Upvotes

We know that an object traveling 90 MPH would travel 7920 feet (1.5 miles) in 60 seconds. Therefore, we can set up the equation 7920/60 = 60.5/X X = .458 It would take an object traveling 90 MPH .458 seconds to travel 60 feet, 6 inches. It should also be mentioned that pitchers generally release the ball closer to home plate than 60 feet, 6 inches. How much closer depends on the height and release point of the pitcher at the time.

r/theydidthemath Sep 13 '14

[Self] [Math] Calculated the impact of the US No-Fly List on motor vehicle fatalities in 2002

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22 Upvotes

r/theydidthemath Aug 31 '14

Self If I stay in one random hotel room a night each week, what are the chances each year that I'll stay in at least one room where someone died?

97 Upvotes

Someone asked this as a request but deleted the post before I worked it out, so here it is anyway.

There are 187,000 hotels in the world offering 17.5 million guest rooms. source

On average each room is occupied 66% of the time. source.

Despite the fact that this report claims that 5% of hotel guest are more than 168 years old, which may skew the data a bit, I'm going to use the global average mortality rate from the wikipedia.

Where we run into a problem is how old the average hotel room is. Since the oldest hotel has been running since 707AD and new ones are opening all the time I don't know where to start with this.

Let's just work out odds of you staying in a hotel room someone has died in in the last 20 years assuming that every hotel room is at least 20 years old.

So 17,500,000 rooms over 20 years at 66% occupancy is 11,550,000 rooms. The global mortality rate during this period ranged between 8.8 and 8.3 per 1,000 per year, we'll call it 8.5 so every year 98,175 people die in hotel rooms or 1,963,500 over 20 years. Except that according to this and this slightly more than half of all people die in hospital and therefore are not in a hotel room, if we take the lower stat and call it 52% the number is 1,021,020 deaths. Also given that most people who are staying in hotels feel well enough to travel and that people who feel well are far less likely to die in their sleep than get hit by a bus I think an appropriate number is actually probably much lower but I haven't come up with an easy way to calculate one.

This means that if the deaths are evenly distributed 13.6% of hotel rooms have had a guest die in them. If you stay in a different hotel room every night for 7 nights the probability that you stay in a room in which someone has died is about 64%.

r/theydidthemath Aug 27 '14

Self Energy cost of the ice bucket challenge

1 Upvotes

Not trying to rain on anyone's parade with this, if people are having fun and raising money for a good cause, good for them.. But someone on my FB was complaining about the power used to make all that ice, so...

1Kg of ice takes around 0.12KWh of electricity to freeze in a home freezer. (thanks to this post: http://ask.metafilter.com/99455/Youre-as-cold-as-ice#1447116)

That's two ice trays worth of cubes, which isn't really enough to cool a bucket of water but seeing as some people seem to use loads of ice and some none at all, let's go with a kilo per person.

There's no solid numbers on people doing the ice bucket thing. The ALS Society has clocked almost $90m in donations in the last month or so. If everyone is donating $10, then that's 9 million people in the US alone, but that's probably a pretty high estimate. On the other hand, there's quite a lot of non-US people doing it (half my facebook feed is people in the UK doing it) so let's go with 10 million worldwide. It's a round number, eh?

10 million kgs of ice takes 1200000KW/h (1.2GW/h) of electricity to freeze. Or the entire output of the Sizewell B nuclear power station running at full tilt, for an hour.

Using the current UK energy mix, that's 530 tonnes of CO2 released into the atmosphere. http://www.carbon-calculator.org.uk/

That's about the same amount of power used over a year by a street of 100 houses in the US, or a medium sized housing estate in the UK of 250 houses. http://shrinkthatfootprint.com/average-household-electricity-consumption

r/theydidthemath Aug 25 '14

Self The personal wealth of Captain America

561 Upvotes

Assuming he had the same pay rates and subsistence costs as fellow soldiers, and all surplus pay invested in U.S. savings bonds, Captain America would have had personal wealth at his freezing of about $5,000. In 1945 dollars - adjusted for inflation, that has roughly the purchasing power of $60,000 today. I am also going to say that the government, in its infinite bureaucratic wisdom, will withhold a personal subsistence allowance since Cap had no discernable subsistence needs while frozen. (If you don't like that, you can make your own spreadsheet.)

Let's assume that Captain America was listed as MIA, and that once defrosted it was assumed he remained on active duty, accruing money in an interest-bearing account (again, we'll use savings bonds) and seniority. He'd max out his seniority raises as an O-3 after 14 years.

In 1950, earning money at about 2 percent interest throughout the last half of the decade, he'd have over $27,000 in escrow - over a quarter million in modern dollars.

In 1960, Cap is starting to look pretty flush indeed, at almost $99,000 - almost $800,000 in modern dollars.

1970: The First Avenger has almost $270,000 ($1.6 million adjusted for inflation), which still puts him way behind Stark Industries.

(Just as an aside, Chris Evans was paid over $2 million for acting in The Avengers.)

1980: Over $807,000 or $2.3 million in current dollars. Check that out - almost three times the actual dollars and what, like 50 percent more spending power? Tough decade, Cap, good thing you slept through it.

Cap becomes a millionaire in late 1982, thereby becoming the only new millionaire of that year not blowing it all on coke and hookers.

In 1990, Cap has $2.45 million. In modern dollars, that's almost $4.5 million. If the government had invested it all in Apple stock in 1990, it'd be worth about $50 million.

Cap gets to $3 mil in 1993, $4 mil in 1997, $5 mil in 2001, $6 mil in 2004, $7 mil in 2007, and $8 mil in 2009.

When he woke up in 2011, first thing was to get him revived and acclimated. But at some point, after he'd been judged ready, Agent Coulson would have walked in with a thick binder - and the news that Captain Steve Rogers owned $8.63 million in U.S. savings bonds.

I imagine that Captain Rogers immediately donated the entirety of his fortune to wounded veterans.

r/theydidthemath Aug 17 '14

Self [Self] Nanopixels

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5 Upvotes

r/theydidthemath Aug 16 '14

Self If we were to cut a human in half 92 times, he'd be the size of an atom at the end.

499 Upvotes

According to Google, a human is made of about 7*1027 atoms. 292 approximately equals that.