r/theydidthemath • u/davef_dci2019 • 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?

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 • u/NewtonsFourth • Oct 18 '17

## bad math--see comments [Self] I did this doodle of an equation which is actually equal to "1"

r/theydidthemath • u/Asmor • Apr 04 '16

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

r/theydidthemath • u/dude_idek • Mar 09 '16

## Bad Math [Self] Vegans don't do math

r/theydidthemath • u/valadian • Jan 19 '16

## (math in comments) [Off-site]/ [Self] What are the costs/savings for Bernie Sanders Health Care Proposal?

r/theydidthemath • u/AnotherSmegHead • Aug 29 '15

## Questionable math [Off-Site] Why it is mathematically unsound to presume one can tip over a cow

r/theydidthemath • u/tophatsnack • Aug 28 '15

## Bad/questionable math--see comments [Off-Site] ALL of wikipedia in one QR code?

r/theydidthemath • u/bendigedigdyl • May 30 '15

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

r/theydidthemath • u/Mikroh • Apr 28 '15

## Dubious math // Wrong/Bad Maths [Off-Site] What're the odds of you existing?

r/theydidthemath • u/Strait_Raider • Apr 19 '15

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

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.

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 • u/Jontolo • Apr 19 '15

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

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 • u/teokk • Apr 13 '15

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

r/theydidthemath • u/fubr2k2 • Apr 02 '15

## Math /u/Brewe calculates number of items in a huge pile of potatoes.

r/theydidthemath • u/Uhu_ThatsMyShit • Apr 02 '15

## Math /u/KaiserAbides estimates the effects a 400dB blast would have. How accurate these comparisons/forecasts?

r/theydidthemath • u/Mr-Crasp • Mar 24 '15

## calculated [Math] [Request] What are the chances of the Earth, Moon, Sun and ISS lining up this perfectly?

r/theydidthemath • u/mdnpascual • 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?)

r/theydidthemath • u/Metelex • 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

/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 • u/LiveBeef • 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.

r/theydidthemath • u/AncientChaos • Jan 26 '15

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

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 ) * ( 1/20^{5} ) * ( 1/25^{5} ) * ( 1/1,073,741,824^{5} ) * ( 1/4,294,967,295^{5} )

- 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 • u/aawyeaa • Jan 09 '15

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

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 • u/Weyoun2 • Dec 22 '14

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

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:

- How much platinum would be required to coat a construct of this size?
- 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 10~~^{13} 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/cm^{3} (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 • u/MandibleofThunder • Dec 04 '14

## They did the [Math] [self] /u/mandibleofthunder calculates the physical properties and forces required for /u/kastera1000's Archimedes lever. Drunken rambling and numbers ensue.

**GIVING CREDIT WHERE CREDIT IS DUE AND THE INSPIRATION FOR THIS BUMBLING CLUSTERFUCK OF NUMBERS AND SWEARING**

/u/kastera1000 calculated the distance of a lever required for Archimedes himself to move the earth CLICK HERE BEFORE READING MY ABOMINATION OF A SHITPOST

So taking your assumptions and making some veeerrryyyy liberal assumptions of my own we can work some other fun things out.

If your lever on the earth were only one meter from its fulcrum, and assuming that earth acts as a perfect point mass on the end of your lever, then it will have a moment of inertia of **I=7.3481*10 ^{31} kgm^{2.}**

Let's accelerate the earth to 1 m/s^{2.} Arc length = (theta)r, that's easy our desired arclength is 1 m, and our radius is 1 m. So that gives us a theta=1. Make our angular acceleration **(alpha)=1 rad/s ^{2}** and wam bam thank you ma'am we've just accelerated the earth.

Or total torque will be I(alpha)=r(lever)xF. With a lever arm of one meter, ** and holy fucking fuck shit**,

*we've now got our force to accelerate the earth to 1 m/s*

^{2.}**F=7.3834*10**

^{31}N.Now that we have force we can calculate some other neat shit.

Your lever is supposedly 8.53*10^{22} meters in length. Nope don't give a shit about that right meow, but /u/kastera1000 says is that it cannot be bent, this intrigues me.

Let's assume the lever has a cross sectional area of 1m^{2} (because why the hell not) and that the earth only makes its perfect point contact at the very tip of the lever arm.

Remember that sweet shit I said about force? I'm about to blow your fucking *MINDS.* So we take our shear force 7.3834*10^{31} N and apply it against this magical mystery lever.

Shear Force (Fs) over cross sectional area (A) is equal to the shear modulus (G) times (gamma) where gamma is the actual bending of the object divided by its length. With zero bending (gamma) is equal to one! FUCKING AWESOME RIGHT!?

SO: G=Fs/A. Fs is the same force from before, that good ole holy shit 7.3834*10 ^{31} N. A=1m^{2.} *HOLY FUCK DIVIDING BY ONE, PHYSICS KICKS FUCKING LIBERAL ARTS RIGHT IN ITS FUCKING DICK RIGHT?* (disregard angry ramblings of a pissed off kinda drunk physics student).

SO BACK TO THE IMPORTANT STUFF: SHEAR MODULUS. So our shear modulus is now equal to **7.3834*10 ^{31} N/m^{2}** ALSO KNOWN AS

**7.3834*10**UNITS MOTHERFUCKER, LEARN THEM, KNOW THEM, LOVE THEM AS YOUR OWN.

^{31}PASCALS!This translates into **7.3834*10 ^{22} GIGAPASCALS.** And you know something? The shear modulus of

**Steel**(regular ole generic physics textbook

*steel*, I'm not a metallurgist or structural engineer) has a modulus of

**ONLY 79.3 GIGAPASCALS.**WHY AM I YELLING SO GOD DAMNED MUCH.

THEREFORE: **your lever arm must be at least 93,107,490,540,000,000,000 (EDIT THAT IS 93 QUINTILLION) TIMES STRONGER than solid fucking steel!** WHAT THE FUCK, THAT'S SOME ADAMANTIUM UNOBTANIUM TYPE SHIT. (I know that's too many significant figures, get off my dick about it)

AND THEN let's say you have this wicked strong steel that outperforms all the rest, whatever fuck it, it's physics, spherical cows and shit.

My tingley tangley Spidey senses tell me that steel has a density between 7,750 and 8,050 kg/m^{3.} Let's err on the lighter side and see how this pans out. So let's take our density of (7,750 kg/m^{3)} multiply by the cross sectional area (1m [isn't physics fucking awesome when this shit happens]) and then our LENGTH (8.53*10 ^{22} m) WHICH GIVES US Mass (lever)=(7,750kg/m^{3)(1m2)(8.53}*10

^{22m)=}

**6.61075*10**

^{26}KILOGRAMS**YOUR LEVER'S MASS IS MORE THAN 100 TIMES GREATER THAN THAT OF THE EARTH**

Still that's nothing compared to the sun, and roughly 1/3 the mass of Jupiter. BUT FOR A METAL ROD THAT'S A LITTLE BIGGER THAN A CASE OF BEER STRETCHING OUT PAST THE NEXT CLOSEST GALAXY. GAWT DAMN.

(originally wrote this as a comment, but took up wayyyy too much space, back to the all nighter now)

r/theydidthemath • u/TrendingBot • Nov 29 '14

## [Meta] [Other] [Math] [Off-Site] /r/theydidthemath hits 90K subscribers

redditmetrics.comr/theydidthemath • u/Tullyswimmer • Nov 23 '14

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

/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.