## 0563: "Fermirotica"

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josiahstevenson
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### Re: "Fermirotica" Discussion

HiPhil wrote:I wonder if 18600 people / square mile is already adjusted for little children and senior citizens, who are doubtful to have sex 80 times a year, for 30 minutes.

it's a comprehensive average, i think. Young swingers are supposed to cancel out seniors and children.

Quantumplation
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### Re: "Fermirotica" Discussion

Ah, woops, thought I took care of my units. >_< Shame on me.

dennisw
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### Re: "Fermirotica" Discussion

LapsesofSanity wrote:
coreyjkelly wrote:Anybody know where the 2 in the numerator comes from? I assume it has to do with sex involving two people? Other than that, seems like a pretty straightforward derivation... circle area and population density tell you how many people are within "r" of you, and the frequency tells you what the probability is that any given one of them is having sex.
Dimensional analysis FTW

yeah, with such a small number on top, and a large number on the bottom, wouldnt r always come out to be a very small number? itd be weird having people having sex within 6 inches of you...

So what you're saying is size really matters?

Or is it really all about the units.
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TwilightNecrosis
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### Re: "Fermirotica" Discussion

The formula doesn't take into account multiple-partner intercourse...
I wonder how we could estimate the number of threesomes/orgies within r.
Any ideas?

flguy1980
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### Re: "Fermirotica" Discussion

hthall wrote:Make that 140 meters:

(The title text refers to the fact that Google understands the units and does the adjustment for you. It also converts the miles to SI standard meters.)

I got the same result with my TI-89. The density where I live is 360/mi^2, so my r is 1000 m.

psion
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### Re: "Fermirotica" Discussion

I was actually thinking of something similar a few days ago and I asked my friend how many people does she think are having sex within a mile from us. She answered three. I was intrigued.
She was probably right though, considering it was 4am in a dry county with heavy religious influence and a population density under 1500/sq mi
Last edited by psion on Wed Apr 01, 2009 7:13 am UTC, edited 1 time in total.

SocialSceneRepairman
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### Re: "Fermirotica" Discussion

HiPhil wrote:I wonder if 18600 people / square mile is already adjusted for little children and senior citizens, who are doubtful to have sex 80 times a year, for 30 minutes.

It shouldn't matter, since they're already counted in the average, aren't they?

(Also, LapseOfSanity, keep in mind that duration * frequency is basically the portion of your life you spend actively having sex, and for all the people there, a square mile is tens of millions of square meters, so you're going to get a pretty small denominator if you pull a reasonable distance unit outside. Remember, the figures given in the comic do work out to 150 m, more or less.)

And I imagine multiple-partner intercourse would work by using weighting it against the average as 2/n.

punto
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### Re: "Fermirotica" Discussion

"average persons frequency of sex" is nice, but it'd be nice to have some granularity, specific groups averaged with the general population. Like lesbian couples, couples where the female[s] are 16-50 years old, <60kg, have a penis, etc.

How close is the closest theoretical lesbian-milf-shemale orgy?

iamagloworm
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### Re: "Fermirotica" Discussion

Denyvem wrote:I did the calculations for the world's densest city (Dhaka, Bangladesh) which has the population density of 43,752 Pplz KM^2 or 70003.2 Pplz Mi^2 and I got 71.8449256 meters WooHOOO!

i live in dhaka, this is probably about right if you allow some latitude in your definition of sex...

Chuff
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### Re: "Fermirotica" Discussion

punto wrote:"average persons frequency of sex" is nice, but it'd be nice to have some granularity, specific groups averaged with the general population. Like lesbian couples, couples where the female[s] are 16-50 years old, <60kg, have a penis, etc.

How close is the closest theoretical lesbian-milf-shemale orgy?

Wait, females have a penis?
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Guelphia
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### Re: "Fermirotica" Discussion

2 / (3.14)(3342)(50)(15) = 1,264m

I can imagine them now.

Sunny_314
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### "Fermirotica" Title

Doesn't the Fermiotica title a reference to Fermi Problem of doing back of the envelope calculations based on averages and statistical modeling?
http://en.wikipedia.org/wiki/Fermi_problem

themuffinking
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### Re: "Fermirotica" Discussion

As a resident of a college dorm holding over 600 people, on a floor containing six of the biggest "playahs" (or, as my black friends call them, WHORES) on earth, the odds that someone is having sex within fifty feet of me are roughly 100%. Unless, of course, Family Guy is on.

TheSkyMovesSideways
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### Re: "Fermirotica" Discussion

TwilightNecrosis wrote:The formula doesn't take into account multiple-partner intercourse...
I wonder how we could estimate the number of threesomes/orgies within r.
Any ideas?

Replace Xf with the average person's frequency of having a threesome, and 2 with 3?

I think the biggest problem with this equation is that it assumes that occurrences of sex are evenly distributes through time, rather than tending to occur at certain times throughout the day, week, and year. (Such as when your mum is in town.)

Obviously, we should attempt to accumulate a statistically significant selection of experimental data, and use that to validate the formula. *starts kicking down bedroom doors*
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St_Fred
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### Re: "Fermirotica" Discussion

I am no math/physics major, but wouldn't the whole "quantum uncertainty/Schrödinger's cat" issue affect any statistical voyeurism?

Vempele
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### Re: "Fermirotica" Discussion

psion wrote:I was actually thinking of something similar a few days ago and I asked my friend how many people does she think are having sex within a mile from us. She answered three. I was intrigued.
She was probably right though, considering it was 4am in a dry county with heavy religious influence and a population density under 1500/sq mi

How does that make a threesome more likely than two-person sex?
const int ALMOST_FIFTY = 80;

UmbralRaptor
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### Re: "Fermirotica" Discussion

*goes off to cause a divide by zero error*
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Sk25
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### Re: "Fermirotica" Discussion

St_Fred wrote:I am no math/physics major, but wouldn't the whole "quantum uncertainty/Schrödinger's cat" issue affect any statistical voyeurism?

Well, to go with your reasoning yes, we are just estimating the probability, but until we see someone actually having sex they may or may not have it. However, I refute the applicability of quantum mechanics to sex. Have you ever tried going with a girl who is as wide as a few subatomic particles?
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Knightshire
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### Re: "Fermirotica" Discussion

poxic wrote:
Delass wrote:Wouldnt two people be required to have sex?

That's the generally agreed minimum, yes.

The formula would be a lot scarier if it considered the number of people jacking off in your neighbourhood.

UmbralRaptor
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### Re: "Fermirotica" Discussion

Knightshire wrote:
poxic wrote:
Delass wrote:Wouldnt two people be required to have sex?

That's the generally agreed minimum, yes.

The formula would be a lot scarier if it considered the number of people jacking off in your neighbourhood.

It just says "sex." There's no reason you have to exclude solo sex...
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martin878
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### Re: "Fermirotica" Discussion

pgn674 wrote:If I recall correctly, Google handles unicorn horns and light years in a tablespoon, too.

Best units ever.

1 Imperial teaspoon per light year = 6.2569343 × 10-22 m2

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### Re: "Fermirotica" Discussion

phlip wrote:(…) So PdXfXd is the expected local density of people having sex. Call this D, for brevity. Then you'd expect one couple (two people) per 2/D area... and since we're talking circles, our area is is pi*r2. Solve for r, and you get the formula in the comic.

That factor 2 may not be exactly 2.

Maybe it should be changed into S, the regional average number of partners in one sexual act.

TheSkyMovesSideways
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### Re: "Fermirotica" Discussion

I think there's too much conjecture in this equation.
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### Re: "Fermirotica" Discussion

KyleOwens wrote:The title is a reference to Enrico Fermi. He was quite well known for his ability to estimate just about anything. The most famous example is his estimation of the number of piano tuners in Chicago based only on approximations of population, piano ownership, average time between tunings, etc.

Ah, that makes more sense. I was thinking it was a reference more specifically to Fermi-Dirac statistics, and was trying to figure out what the joke was in making copulating couples obey the Pauli exclusion principle. I kept thinking that Maxwell–Boltzmann statistics would be more appropriate.
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### Re: "Fermirotica" Discussion

Aww?! 250 meters!

I fudged the variables a bit to account for the fact that it's almost only students living here, with more libido and stamina, and got it down to 150 m.
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chrono
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### Re: "Fermirotica" Discussion

Am I the first to notice the worrying implication of the data?

First, the 2 under the square root. That's two people, right?
Also, eighty times per year is far too little for, well, solo action.

See where I'm going with this?

lunarul
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### Re: "Fermirotica" Discussion

Adjusting for my city's density of 22040 / sq mi and to the real average duration of sex according to studies I read recently (6 minutes), then there's someone within 286 m from me having sex right now... I should find a correct estimation of instances per year, too.
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keanebean86
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### Re: "Fermirotica" Discussion

sabik
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### Re: "Fermirotica" Discussion

Sk25 wrote:
St_Fred wrote:I am no math/physics major, but wouldn't the whole "quantum uncertainty/Schrödinger's cat" issue affect any statistical voyeurism?

Well, to go with your reasoning yes, we are just estimating the probability, but until we see someone actually having sex they may or may not have it. However, I refute the applicability of quantum mechanics to sex. Have you ever tried going with a girl who is as wide as a few subatomic particles?

The other problem is that it tends to end up with the girl a little bit pregnant.

η

carls
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### Re: "Fermirotica" Discussion

Am I doing something wrong or why do I work the distance out to be like, 19 centimeters?

EDIT : Unit issues, silly me.

realyze
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### Re: "Fermirotica" Discussion

Phi*r^2*Pd*Fx*Xd == 2? Really?

Stigg
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### Re: "Fermirotica" Discussion

~53 meters for Annapolis, MD

ThemePark
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### Re: "Fermirotica" Discussion

phlip wrote:
madhollywood wrote:Also, I wouldn't necessarily thing that the duration of sex decreases the radius in which it was occuring.

Well, XfXd is the probability that a given person is having sex at a random point in time (using the values in the comic, 30 minutes and 80/year, you'd spend about half a percent of your life having sex... so by the frequentist definition of probability, if we picked a time at random, there'd be a 0.5% chance you'd be having sex then). So PdXfXd is the expected local density of people having sex. Call this D, for brevity. Then you'd expect one couple (two people) per 2/D area... and since we're talking circles, our area is is pi*r2. Solve for r, and you get the formula in the comic.

Nice. Now it all makes sense! Thanks for that explanation.
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chaitanyag
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### Re: "Fermirotica" Discussion

Isn't the equation missing a constant? The Fermotic constant, "F"? The units don't actually cancel out exactly. the 18000 for example is people, the 80 is number of errrrm.... "events", shall we call them Lays. They are two completely different units and don't cancel each other out. If you have [imath]x = 20 people/mi^2[/imath] and [imath]y = 10 apples/mi^2[/imath] , then [imath]x/y = 2 people/apple.[/imath]

So in the current equation, we have [imath]1/ ( (People/mi^2) * (Lays/minute) * (minute))[/imath] we end up with the final units of [imath]mi^2 / (People * Lays)[/imath] . So shouldn't we have F as a constant of [imath](People * Lays)/Mi[/imath] to end up with r as the distance?

Mysidic
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### Re: "Fermirotica" Discussion

Wait, the radius is in meters but the population density is in square miles? Shame on you randall!
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lunarul
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### Re: "Fermirotica" Discussion

chaitanyag wrote:Isn't the equation missing a constant? The Fermotic constant, "F"? The units don't actually cancel out exactly. the 18000 for example is people, the 80 is number of errrrm.... "events", shall we call them Lays. They are two completely different units and don't cancel each other out. If you have [imath]x = 20 people/mi^2[/imath] and [imath]y = 10 apples/mi^2[/imath] , then [imath]x/y = 2 people/apple.[/imath]

So in the current equation, we have [imath]1/ ( (People/mi^2) * (Lays/minute) * (minute))[/imath] we end up with the final units of [imath]mi^2 / (People * Lays)[/imath] . So shouldn't we have F as a constant of [imath](People * Lays)/Mi[/imath] to end up with r as the distance?

Actually the 2 also has a unit of People/Lay... so you have:
[imath](People/Lays) / ( (People/mi^2) * (Lays/minute) * (minute))[/imath]
and you end up with:
[imath]mi^2 / Lays^2[/imath]
which is then reduced to mi/Lays

sabik
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### Re: "Fermirotica" Discussion

Mysidic wrote:Wait, the radius is in meters but the population density is in square miles? Shame on you randall!
Why? That's what a unit-tracking calculator is for; either google, or the units(1) program.

You type in whatever units you have and it converts them as necessary.

η

tetracycloide
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### Re: "Fermirotica" Discussion

So I calculated the distance based on my local stats:

627 people per KM^2
2.5 sex per week
30 min average sexual duration

sqrt(2 / (((((pi * 627) / (km^2)) * 2.5) / week) * (30 minutes))) = 369.407684 meters

Now what are the odds of that?

chaitanyag
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### Re: "Fermirotica" Discussion

lunarul wrote:
Actually the 2 also has a unit of People/Lay... so you have:
[imath](People/Lays) / ( (People/mi^2) * (Lays/minute) * (minute))[/imath]
and you end up with:
[imath]mi^2 / Lays^2[/imath]
which is then reduced to mi/Lays

Ah, of course ! That makes it all better now. Thank you kind sir.

Mr.Randall, my faith in you has been restored.

martin878
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### Re: "Fermirotica" Discussion

tetracycloide wrote:So I calculated the distance based on my local stats:

627 people per KM^2
2.5 sex per week
30 min average sexual duration

sqrt(2 / (((((pi * 627) / (km^2)) * 2.5) / week) * (30 minutes))) = 369.407684 meters

Now what are the odds of that?

A priori, 0.01. A posteriori, 1. Classic seeing the question in the data fallacy.