## 0563: "Fermirotica"

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Iridos
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### Re: "Fermirotica" Discussion

St_Fred wrote:
sabik wrote:
Sk25 wrote:
St_Fred wrote:[...]wouldn't the whole "quantum uncertainty/Schrödinger's cat" issue affect [...]

[Yes] but until we see someone actually having sex they may or may not have it. [...]

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

η

Ah but with quantum uncertainty, the girl isn't pregnant until you test

Think that's what he meant... apropos - am I the only one who feel this would have been a much nicer, more realistic and much much more humane example to use for explaining the problem, instead of killing half a cat?

Although a bit unrelated - may I also point out: Heisenbergs uncertainty principle clearly states that you cannot exactly know how much the person you have sex with enjoys it AND how long your relationship to said person will last. Actually you might - the situation is a bit more complex than with particles - there is a possibility discussed you might be able to learn both facts, but if and only if you don't know the persons name.

neoliminal wrote:
Katieesq wrote:80 times per year? I think I'm a couple standard deviations away from that mean.

Which way?

Indeed, which way... and how on earth did you figure? As this is probably the value that incorporates the average over the whole lifetime, it should be really hard to estimate... until you're sure you've come sufficiently close to the end of your lifetime. (Which I feel makes the figure generally a very dubious one)

tentative wrote:[...] the calculation is dead wrong? [...]

No, it's not "wrong" it is a valid model, just not a very good one - and this just means that the standard deviation is so big as to make the result useless... which really surprises nobody and is probably the reason it wasn't mentioned.... and as such demonstrates the uselessness of statistics if taken to extremes, which was kind of the point (or one of them) of the comic.

I.

aleflamedyud
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### Re: "Fermirotica" Discussion

This only works if the frequency is counted in sexual events per year per person. Otherwise the units don't work out.
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Det.
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### Re: "Fermirotica" Discussion

30 minuets

STOP MAKING ME FEEL INADEQUATE RANDALL

allenb
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### Re: "Fermirotica" Discussion

According to Wikipedia, the Earth has a land surface area of 148,940,000 km^2. Using this figure to calculate the population density term while assuming an Earth population of 2, the result is an average sex distance of 101,929 kilometers. Given that this is roughly 25% of the distance to the moon, this should be proof that in all likelihood, she wouldn't sleep with you, even if you were the last man on Earth (so don't even try, buddy, she's way out of your league).

brunswikstu
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### Re: "Fermirotica" Discussion

Good Comic for April Fool's

na-Fiann
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### Re: "Fermirotica" Discussion

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

Actually, wouldn't Xf mean the average amount of times a single person has sex in a year, making it Lays/(minute*People)? Also, Xd would be in minutes/Lays I think..
Solving the unit gives me a headache though.. but I think I'm going wrong somewhere.

squishycube
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### Re: "Fermirotica" Discussion

allenb wrote:According to Wikipedia, the Earth has a land surface area of 148,940,000 km^2. Using this figure to calculate the population density term while assuming an Earth population of 2, the result is an average sex distance of 101,929 kilometers. Given that this is roughly 25% of the distance to the moon, this should be proof that in all likelihood, she wouldn't sleep with you, even if you were the last man on Earth (so don't even try, buddy, she's way out of your league).

Very nice use of the model, thanks!
My usual approach is useless here

theSAiNT
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### Re: "Fermirotica" Discussion

Why are individuals having sex being treated as independent events? Surely there is a non zero correlation of sex between couples. The current equation would severely underestimate r.

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### Re: "Fermirotica" Discussion

Although the correlation between couples having sex is non-zero it won't be significant when N (the number of couples having sex in a given radius) is close to one, so it won't significantly affect the estimate of r, which is the average radius for one couple to be having sex. The correlation will appear as a second-order term that won't cause an issue until N is large enough that the probability of an threesome, foursome or orgy falling within r becomes significant.

On the other hand, both Xf and Xd will be strongly time-dependent, and possibly seasonally dependent as well, so r will vary with time of day and possibly season quite a lot. But the whole point of a Fermi calculation is to get a first-order estimate based on known or reasonably presumed causal relationships and approximate input parameters.
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kaiso1
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### Re: "Fermirotica" Discussion

I've often thought about this, but in terms of how many people within a certain radius are... ahem... enjoying the climax of such an event (solo or with partner(s)).

The average frequency of sex ought to be much lower in places like Boca Raton, where the median age is skewed towards the elderly... and much higher in small college towns where the population is younger and drunker.

JohnTheWysard
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### Re: "Fermirotica" Discussion

Well, that's good enough Fermi.

The only problem is that I live in very rural and very conservative Idaho so the function's value is pretty close to zero.

meat.paste
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### Re: "Fermirotica" Discussion

So, when I plug in the numbers listed in the comic into the formula, I get an average distance of 13.9 m (sqrt[2/(pi*18600*80*5.7E-5)] = 0.0866 miles = 13.9 m). Where does the 150 m number come from (and where the hell is the error on my math)?

Spoiler:
30 min = 0.5 hr = 0.0208 days = 5.7E-5 years * 80/year = .0046 * 18600/mi2= 84.87/mi2 * pi/2 = 133/mi2. 1/133mi2 = 0.0075mi2. Square root = 0.0866 miles * 5280 ft/mi = 457.29 feet * 12in/ft = 5487.4 in * 2.54cm/in = 13938 cm = 13.9 m

Using similar calculations for Austin, TX (Pop density = 2000 [1], frequency = 80/year, time = 30 min), I get 42 m

[1] http://www.austin-chamber.org/images/charts/PopDensity.pdf
Huh? What?

Louise
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### Re: "Fermirotica" Discussion

kaiso1 wrote:I've often thought about this, but in terms of how many people within a certain radius are... ahem... enjoying the climax of such an event (solo or with partner(s)).

The average frequency of sex ought to be much lower in places like Boca Raton, where the median age is skewed towards the elderly... and much higher in small college towns where the population is younger and drunker.

And much lower in places where people read xkcd.com, cos let's face it, if you're talking about it...

LarrySDonald
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### Re: "Fermirotica" Discussion

"Population density in Kansas" does return a value, but if you attempt to use it in a calculation it screws up - doesn't function like "speed of light" or "pi". Copy/paste says about 3.3km. You've got to be kidding. Baby, I don't care if you've got a headache, we've got to get on this.

UmbralRaptor
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### Re: "Fermirotica" Discussion

Eh, Kansas' population density is pretty variable. Try KC, Lawrence, Wichita, Manhattan, or the like?
(...using my best guesses of 15 minute duration and 60 events per year, I get 500 m. I suspect that I'm overstating quantity)
Why do you assume that I exist?

LarrySDonald
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### Re: "Fermirotica" Discussion

For my area, it's pretty accurate. I'm just saying, it wouldn't kill Google to let you use it in a formula instead of not letting it interact with the math/unit thing. If it can tell me (not give me a link to) of the population density then it'd be nice to be able to say "=(population density in kansas) * pi" and get a correct answer rather then being all confused. And yes, as a matter of fact I am a spoiled brat - I love having the google calc on top of answering "45 cups in liters".

[EDIT] I've edited such a short message enough that I should say "How is babby formed". Sorry.

guyy
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### Re: "Fermirotica" Discussion

I was seriously about to ask where the 2 comes from in that equation. I guess I'm officially better at math that I am at being a mammal.

rubber314chicken
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### Re: "Fermirotica" Discussion

lunarul wrote: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.

does that time average take into account virgins?

and does foreplay count into that time?
kertrats wrote:I wrote this paper up last night. I'd like comments if anyone has any.

That is very scientific. Now please confirm the data.

What we need to do is make a map thing, like the geo-hashing map,where you can click a point at enter a time and it will tell you the radius that at least two (one?) people(person?) are (is) having sex in and what not. We need to take into account varying population density (aka treat a high rise apartment complex separately from the rest of the city), time (when most people have sex) and all the other stuff.

This could be really cool and really creepy.

SecondTalon wrote:Semen! I said semen! tee hee!

Dekoy
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### Re: "Fermirotica" Discussion

I got .58 of a meter :\

This doesn't seem right >.>

dennisw
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### Re: "Fermirotica" Discussion

Dekoy wrote:I got .58 of a meter :\

This doesn't seem right >.>

RTFM

It's in the FAQs.
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InfamousAnarchist
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### Re: "Fermirotica" Discussion

The wikipedia gave me the density for New Hampshire, my state of mind, and I have to actually look a little over a kilometer to find anyone.
1.56942291 kilometers.
EDIT: Actually, my town has a density of 325.8, so the new figure is 1.05312399 kilometers. Well then.

Hurm. Well, the more this is true, the higherthe density, the lower the frequency, eventually, EVERYONE ON EARTH WILL BE HAVING SEX AT THE SAME TIME. AMAZING.
sever every leg

OBloodyHell
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### Re: "Fermirotica" Discussion

> (using the values in the comic, 30 minutes and 80/year, you'd spend about half a percent of your life having sex

Hardly effective reasoning. Sexual activity is not only not constant, it's not even vaguely linear, even taking into account variables like marital state and age of children.

a) One assumes ones sex life is zero until you hit 15 or so... one hopes.
b) Sex life in the first year of marriage is generally argued as greater than the amount of sex you'll have for the rest of the marriage.
c) Presence of small children in the house greatly interferes with opportunity, reducing childmaking activities.
d) Age reduces capacity.

In short, for most people it's a triple lobe "wave" curve (pre-marriage, marriage, post-children -- up down up) which then approaches zero asymptotically until death, after which it is zero. Unless you get involved with someone really, really seriously kinky.

As far as constants, I believe that anyone looking over my shoulder would say, with an almost 100% certainty, "Damn, Y'all are a bunch of fuckin' geeks!!", or some variant thereof.

Otto
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### Re: "Fermirotica" Discussion

meat.paste wrote:So, when I plug in the numbers listed in the comic into the formula, I get an average distance of 13.9 m (sqrt[2/(pi*18600*80*5.7E-5)] = 0.0866 miles = 13.9 m). Where does the 150 m number come from (and where the hell is the error on my math)?
...
13938 cm = 13.9 m

You were doing *so* well right up until that bit.

1 m = 100 cm. Check your division.

SamSam
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### Re: "Fermirotica" Discussion

Why does Google return a result for "1 teaspoon per meter", but not for "1 ml per meter" or "1 liter per meter." Is such a silly conversion just tacked in as an easter egg, and they didn't think of all the possibilities, or what?

hthall
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### Re: "Fermirotica" Discussion

AtRandom wrote:
dln385 wrote:This comic finds the distance at which, on average, one couple within that distance from you is having sex. I want to know what the average distance from you to the nearest couple having sex is.

. . . Here is the formula I got . . .
$d=\sqrt{\frac{1}{2P_d X_fX_d}}$
. . . two-dimensional Poisson point process . . . deduce the expected value . . .

I hope someone found this interesting...

Thank you. I recognized the subtlety, thought of the word "Poisson", and hoped someone would do it right. Which someone did.

I'm also glad Mr. Munroe was careful in the way he stated the claim, with the average taken over the number of instances within a fixed radius rather than an average over radii.
Look at me, still talking when there's Science to do.

hthall
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### Re: "Fermirotica" Discussion

SamSam wrote:Why does Google return a result for "1 teaspoon per meter", but not for "1 ml per meter" or "1 liter per meter." Is such a silly conversion just tacked in as an easter egg, and they didn't think of all the possibilities, or what?

That's curious. "1 ml per inch" or "1 liter per inch" it does calculate, but not "1 liter per meter". It will, however calculate "1 liter per meter in square meters". Similarly, it won't calculate "1 liter" but will calculate "1 liter in cc" or even "1 liter in liters", and it won't calculate "2" but it will "2 plus 2". Apparently your question was too easy for the calculator even to kick in, unless you tell it to explicitly.

Look at me, still talking when there's Science to do.

doctordruid
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### Re: "Fermirotica" Discussion

AtRandom wrote:Actually, there is a slight subtelty here: The distance r in the comic is such that, on average, 1 person whithin this distance is having sex. This is what you prove. However, this is different from the average distance from you to the nearest person having sex.

And I would think it is more interesting to know the average minimum distance. Here is the formula I got: (making the same implicit dubious assumptions as in the comic)
$d=\sqrt{\frac{1}{2P_d X_fX_d}}$
Hence [imath]d=\frac{\sqrt{\pi}}{2}r\simeq 0.887 r[/imath], and for instance r=150m in the comic gives d=133m for the average minimal distance.

I hope someone found this interesting...

I certainly found that result interesting, as I got the same result. My basic explanation for the difference between our formula and Randall's is that the average of a square isn't the square of the average. (If this was so, there would be no variance). So even if one knows the average area of a circle ecompassing the nearest couple, that doesn't give the average distance. The distance can be found by using a 2D Poisson process, for which the area of the smallest circle encompassing the nearest couple is exponentially distributed $f_A(a)=\lambda e^{-\lambda a} ,\lambda \equiv P_d X_f X_d /2$ From there, one can calculate $E\sqrt{A/\pi}=\sqrt{\frac{1}{2P_d X_fX_d}}$ Or, one can find the distribution of the distance, $f_R(r)=2 \pi \lambda r e^{\lambda \pi r^2} , ER=\sqrt{\frac{1}{2P_d X_fX_d}}$ It's also possible to find confidence limits from this.
PS: In order to verify this, I did a simulation of a 2D Poisson process, which gave the same results. Yup, I'm a nerd!

nash1429
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### Re: "Fermirotica" Discussion

it should be square root of ((2 times 365 times 24 times 60)/(PdXfXd))
2 because you need a couple and 365, 24, and 60 to find minutes in a year
this becaomes minutes in a year/minutes of sex

sabik
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### Re: "Fermirotica" Discussion

[quote="nash1429"]it should be square root of ((2 times 365 times 24 times 60)/(PdXfXd))
2 because you need a couple and 365, 24, and 60 to find minutes in a year
this becaomes minutes in a year/minutes of sex[/quote]

You don't need to find minutes in a year, because Google already knows that. That's why you type it in including the units, so that Google converts it for you.

nash1429
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### Re: "Fermirotica" Discussion

sabik wrote:
nash1429 wrote:it should be square root of ((2 times 365 times 24 times 60)/(PdXfXd))
2 because you need a couple and 365, 24, and 60 to find minutes in a year
this becaomes minutes in a year/minutes of sex

You don't need to find minutes in a year, because Google already knows that. That's why you type it in including the units, so that Google converts it for you.

just because google can figure it out does not make it a calculational equation

phlip
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### Re: "Fermirotica" Discussion

What the pants is a "calculational equation", other than a term that doesn't exist?

But anyway, take something simpler... v = s/t. If you travelled 10m in 5s, and you want to find the answer in km/h, then the relevant equation is still just v = s/t. It just that if you take that to mean v = 10 / 5 = 2 km/h, then you fail at maths forever. v = s/t is correct regardless of what units you're using to represent the values involved... if you're using inconsistent units, then you'll have to do some extra conversions outside of the formula, but the same is true for any formula ever. The fact that Google will let you go 10m / 5s in km/h and get the right answer is just a matter of convenience.

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rho421
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### Re: "Fermirotica" Discussion

If they don't exist, then why are there 95 google results?

dennisw
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### Re: "Fermirotica" Discussion

rho421 wrote:If they don't exist, then why are there 95 google results?

Because 95 results on Google is below the threshold of statistical significance.

For comparison: "differential equation"

Results 1 - 10 of about 948,000 for "differential equation"
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nash1429
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### Re: "Fermirotica" Discussion

phlip wrote:What the pants is a "calculational equation", other than a term that doesn't exist?

But anyway, take something simpler... v = s/t. If you travelled 10m in 5s, and you want to find the answer in km/h, then the relevant equation is still just v = s/t. It just that if you take that to mean v = 10 / 5 = 2 km/h, then you fail at maths forever. v = s/t is correct regardless of what units you're using to represent the values involved... if you're using inconsistent units, then you'll have to do some extra conversions outside of the formula, but the same is true for any formula ever. The fact that Google will let you go 10m / 5s in km/h and get the right answer is just a matter of convenience.

word to the wise: use caution when dissing someone about something you are ignorant of. a calculational or computational formula is a form of an equation that is equivalent to the defining equation but is easier to use. they are very common in statistics for finding standard deviation, correlation coefficient, etc. also, if you don't twitch uncontrollably when you see an equation that is not in simplest form, THEN YOU ARE NOT A NERD, and I am hearby revoking your license to be one.

nash1429
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### Re: "Fermirotica" Discussion

dennisw wrote:
rho421 wrote:If they don't exist, then why are there 95 google results?

Because 95 results on Google is below the threshold of statistical significance.

For comparison: "differential equation"

Results 1 - 10 of about 948,000 for "differential equation"

bing has almost 3 mil. results for "calculational equation"

phlip
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### Re: "Fermirotica" Discussion

nash1429 wrote:bing has almost 3 mil. results for "calculational equation"

... until you scroll down to the bottom of the page, and see that it's including results for calculation equation sans quotes - ie, replacing "calculational" with a much more common word, and no longer requiring it to be a contiguous phrase. Click the link for "Show just the results for "calculational equation"." and you get 9 results.

And of those 9 results:
4 use it just on some equation that happens to be used to calculate stuff.
1 uses it in reference to the internal workings of a calculator.
2 are a mirror of two of the above (they're patents - the search found two different patent archiving sites).
1 is a dead link, but the cached version seems to either use it as a non-sequitur, or put it in the first category, I'm not certain... it's a quote from another paper that I don't have access to, and there's not enough context to be sure.

So:
None of them except this thread use "calculational equation" to mean an equation that has extra numerical constants in it to convert between equivalent units, like 24 hours/day, or whatnot.
None of them define what "calculational equation" actually means, as though it were a real term... but rather use it with the face-value definition to mean an equation that happens to be used for calculations... and there's no reason such an equation would need explicit unit conversion factors in the actual equation.

So I stand by my "that term doesn't exist" claim.

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nash1429
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### Re: "Fermirotica" Discussion

phlip wrote:
nash1429 wrote:bing has almost 3 mil. results for "calculational equation"

... until you scroll down to the bottom of the page, and see that it's including results for calculation equation sans quotes - ie, replacing "calculational" with a much more common word, and no longer requiring it to be a contiguous phrase. Click the link for "Show just the results for "calculational equation"." and you get 9 results.

And of those 9 results:
4 use it just on some equation that happens to be used to calculate stuff.
1 uses it in reference to the internal workings of a calculator.
2 are a mirror of two of the above (they're patents - the search found two different patent archiving sites).
1 is a dead link, but the cached version seems to either use it as a non-sequitur, or put it in the first category, I'm not certain... it's a quote from another paper that I don't have access to, and there's not enough context to be sure.

So:
None of them except this thread use "calculational equation" to mean an equation that has extra numerical constants in it to convert between equivalent units, like 24 hours/day, or whatnot.
None of them define what "calculational equation" actually means, as though it were a real term... but rather use it with the face-value definition to mean an equation that happens to be used for calculations... and there's no reason such an equation would need explicit unit conversion factors in the actual equation.

So I stand by my "that term doesn't exist" claim.

it was an origianlly an error: i had meant computational form of an equation, which you obviously missed my correction of in my response. i also notice that you do not mention my point about simplifying. an admission of defeat?

phlip
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### Re: "Fermirotica" Discussion

You're right... I missed that post... mostly because I'd assumed you'd read the rules, and thus wouldn't post two posts in a row, so I didn't look at your previous post when I saw your second one. That's what the edit button is for, after all.

OK, so to respond to that:
nash1429 wrote:a calculational or computational formula is a form of an equation that is equivalent to the defining equation but is easier to use.

OK, but if you want to be taken seriously when you get indignant that we haven't heard of this term, you're going to have to show that people actually use it, and it's not just something that you made up. The fact that none of the search results in a semi-popular search engine, over the entire Internet, for that phrase actually come up with that definition doesn't bode well for you.

And then you'll have to explain why a form of an equation with extra magic numbers thrown in is easier to use than the unit-agnostic formula.

nash1429 wrote:also, if you don't twitch uncontrollably when you see an equation that is not in simplest form, THEN YOU ARE NOT A NERD, and I am hearby revoking your license to be one.

I don't see how, say, "v = s/t * 3.6" or "v = (s / 1000)/(t / 60 / 60)" are a simpler form of the equation than "v = s/t", even if you happen to have meters and seconds and want a km/h result, this one time. One would think that the shorter form, with fewer operations, no magic numbers, and much easier to memorise would be "simpler", but there you go.

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nash1429
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### Re: "Fermirotica" Discussion

if you had done the math you would have noticed that my equation lets you put the numeric values of the variables directly into the equation without accounting for units, which you have to do for the equation given in the strip. google can do it the long way, but my equation saves you a lot of time (solve my equation, without units, and you will see that it has the same answer of 140m as the equation with units). and i have definitely used the term "computational formula" in college math classes, especially stats. this is an example of a textbook being more useful than the internet. and based on what you say about one equation being simpler i am led to believe that you have not taken the basic pre-algebra level math necessary to know what "simplest form" means. my equation cannot be made into your equation without adding factors of 1, but your equation can be made into my equation using only information already in your equation.

phlip
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### Re: "Fermirotica" Discussion

Why the continual baseless attacks on my maths education? I mean, it's pretty irrelevant to the point. I'm not even going to bother to rebut it, wrong as it is, because it'd be pointless.

nash1429 wrote:if you had done the math you would have noticed that my equation lets you put the numeric values of the variables directly into the equation without accounting for units

Oh, I'm sure it does. That doesn't make the form in the comic wrong, and it doesn't make yours better.

Equations that have all the unit conversions built in are only useful if you're using those specific units. "v = s/t * 3.6" is only useful if you're specifically converting seconds and metres into km/h. Now, if you do that a lot, then by all means, remember the relevant conversion rates. But if you have a text that says "v = s/t * 3.6" and you have inches and minutes and want mi/h, then you're pretty stuck. Unless you know (or derive) that the unit-agnostic formula is "v = s/t", and then work from there.

To give another example... a simple formula, to find a certain percentage of a number, could possibly be written "n * p/100". But such a formula is only useful for percentages. If you write the formula as "n * p" and make it known that percentages always represent the number divided by a hundred, then the same formula is useful for percentages, decimals, fractions... because after all, 0.5, 1/2 and 50% are all the same number, there's no reason to have separate equations for them. Just like 1000m and 1km are the same values, and there's no reason to have different velocity formulae to handle the two forms.

nash1429 wrote:but my equation saves you a lot of time

Given the title-text of the comic, half of the point of leaving the unit conversions out is to show off how Google Calculator will do the conversions for you. And it's far from the only one, it's just the most convenient.

nash1429 wrote:this is an example of a textbook being more useful than the internet.

Except, of course, that I've read textbooks where the author of the textbook has invented new terms that, while potentially useful, aren't in common use. And if you try to use them with someone other than the author of the textbook (or someone else who learned from the same text), you'll just get blank stares. And some of the worse textbooks don't even make this clear when they introduce the term.

I think the fact that there doesn't seem to be a usage of the term anywhere on the tubes suggests that this might be the case here.

nash1429 wrote:and i have definitely used the term "computational formula" in college math classes, especially stats.

And the plural of anecote is data.

In other words, my equation has all the information you need, and if your equation includes extra information that can be derived from the same source, then it's redundant. Something that is redundant is not in the simplest form, as it can be made simpler by removing the redundancy.

If you want to claim you're using an arcane definition of "simplest form" such that "v = s /t * 3.6 oh and by the way v has to be in km/h, s has to be in metres, t has to be in seconds" is simpler than "v = s/t", then you're going to have to define that too, because it, too, is not a definition I have come across.

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