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Old 2012-11-19, 02:23   #1
Uncwilly
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Default Interesting formulæ

Not sure if this belongs in the Lounge or Misc Math.

We all know several formulae, laws, or equations that relate to everyday life, that are not well documented by science. This thread is intended to be a place to document these.

Submitted as the first is one that most all of us are well aware of: the need to urinate as a function of distance from a known restroom, with compensation for temperature and perceived velocity.

u = need to urinate
p = constant (5.7 in United_States customary units)
d = distance to known restroom
t = ambient temperature, the base formula given is for degrees Fahrenheit
v = perceived velocity (1 = exactly 'appropriate' velocity, 2 = twice 'appropriate' velocity), no units required, auto-adjusts for mode of transit.


u=p^{(1+(1/d))}*(1/v)*(98.6-t)^{0.02*(98.6-t)}

Those of you with medical training may be able to refine this.

Below is a data table:
Code:
d	v	t	u
1.00	1.0	98.00	32
2.00	1.0	95.50	15
0.50	1.0	75.00	823
0.50	2.0	32.00	24858
1.00	2.0	98.55	16
1.00	4.0	98.55	8
0.25	1.0	75.00	26754
0.12	0.5	32.00	6091575835
0.10	1.0	32.00	55397617288
0.25	1.0	70.00	40965
0.12	0.7	50.00	706552439
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Old 2012-11-19, 17:00   #2
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Quote:
Originally Posted by Uncwilly View Post
Not sure if this belongs in the Lounge or Misc Math.

We all know several formulae, laws, or equations that relate to everyday life, that are not well documented by science. This thread is intended to be a place to document these.

Submitted as the first is one that most all of us are well aware of: the need to urinate as a function of distance from a known restroom, with compensation for temperature and perceived velocity.

u = need to urinate
p = constant (5.7 in United_States customary units)
d = distance to known restroom
t = ambient temperature, the base formula given is for degrees Fahrenheit
v = perceived velocity (1 = exactly 'appropriate' velocity, 2 = twice 'appropriate' velocity), no units required, auto-adjusts for mode of transit.


u=p^{(1+(1/d))}*(1/v)*(98.6-t)^{0.02*(98.6-t)}

Those of you with medical training may be able to refine this.

Below is a data table:
Code:
d	v	t	u
1.00	1.0	98.00	32
2.00	1.0	95.50	15
0.50	1.0	75.00	823
0.50	2.0	32.00	24858
1.00	2.0	98.55	16
1.00	4.0	98.55	8
0.25	1.0	75.00	26754
0.12	0.5	32.00	6091575835
0.10	1.0	32.00	55397617288
0.25	1.0	70.00	40965
0.12	0.7	50.00	706552439
I don't have medical training but, I know of a condition that has relevance to urination formulae like this I believe and so:

1/((((b*w)/GFR)/B)*(d/vi)) is my best guess based on what I looked up,B= bladder capacity, b= blood volume, w = % waste by volume , GFR = glomerular filtrate rate, v and d have the same values as in your equation. this is one over percentage of bladder that can be filled in the time perceived to be needed to get to the known restroom.

Last fiddled with by science_man_88 on 2012-11-19 at 17:05
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Old 2012-11-19, 18:21   #3
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Quote:
Originally Posted by Uncwilly View Post
Not sure if this belongs in the Lounge or Misc Math.

We all know several formulae, laws, or equations that relate to everyday life, that are not well documented by science. This thread is intended to be a place to document these.
Said Einstein, "I have an equation,
Which to some may seem Rabelaisian,
Let V be virginity,
Approaching infinity,
Let P be a constant persuasion.

"Let V over P be inverted,
With the square root of Mu inserted,
N times into V ...
The result, Q.E.D.,
Is a relative!" Einstein asserted.


That one has been in the fortune(6) database for over 30 years.

Last fiddled with by ewmayer on 2012-11-19 at 18:56 Reason: "Virginity is like a balloon ... one prick and it's gone."
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Old 2012-12-03, 05:05   #4
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Law of Dirty Rugs/Carpets: The chances of an open-faced jelly sandwich of landing face down on a floor covering are directly correlated to the newness, color and cost of the carpet/rug.

f=chance of jelly/carpet contact
n=newness in months
c=grayscale value of carpet color (0=completely non-reflective black, 1=completely white)
d=cost of carpet in dollars
m=monthly income in dollars

f=(1/n)*c*(d/m)
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Old 2012-12-03, 05:10   #5
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Jilly's third law: The worse the haircut, the slower it grows out.
Jilly's second law (IIRC): Windspeed is proportional to the cost of the hairdo.

Does anyone have the formulae for these?
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Old 2012-12-05, 01:08   #6
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This isn't exactly a formula, perhaps an algorithm.

With a divmod function, turning a total amount of time into certain amounts of smaller units is incredibly easy/beautiful.

Code:
#eta is some large number of seconds
days, eta = divmod(eta, 3600*24)
hours, eta = divmod(eta, 3600)
mins, eta = divmod(eta, 60)
secs = eta
if days > 0:
    print("days, hours, mins, secs")
elif hours > 0:
    print("hours, mins, secs")
else:
    print("mins, secs")
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Old 2012-12-08, 07:20   #7
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This is even less like a formula; I came across it as one of the many corollaries of Murphy's Law. It went something like this:
Quote:
To estimate the time a task will take, start with how long you think it should take, multiply by two and change to the next higher unit. So a one hour task will take two days.
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Old 2012-12-08, 07:51   #8
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Quote:
Originally Posted by Uncwilly View Post
...
p = constant (5.7 in United_States customary units)
...
Erm ... is that supposed to be the average length of some body part (5.7 in. ~= 145mm)?
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Old 2012-12-08, 17:40   #9
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Quote:
Originally Posted by retina View Post
Erm ... is that supposed to be the average length of some body part (5.7 in. ~= 145mm)?
No, just a constant.
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Old 2012-12-10, 05:32   #10
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More gushing about Python.

Code:
def divisors(n):
     # Creates a list of all divisors of n
     # len(this list) == num_divisors(n) (includes 1)

     n = _positive(n, "divisors")
     if not isinstance(n, Factors): n = factor(n)

     # Create list by making a list of lists of prime powers for each prime
     # dividing n
     # Then take a "cartesian product" of all these lists, i.e. take all
     # combinations of one element from each list, i.e. take all combinations
     # of prime powers for each prime dividing n

     biglist = []
     for prime in n: # Create list of all powers of a prime for each prime in n
          biglist.append([prime**power for power in range(n[prime]+1)])

     # For 72 = 2^3 * 3^2, biglist = [[1, 2, 4, 8], [1, 3, 9]]
  # For 1800 = 2^3 * 3^2 * 5^2, biglist = [[1, 2, 4, 8], [1, 3, 9], [1, 5, 25]]

     divs = [1]

     # Now take all combinations of one element from each sub list
     # Or rather, take all combinations of the first list, then all combinations
     # with the second, etc.

     for lst in biglist:
          divs = [x*y for y in divs for x in lst]

     # That list comprehension syntax is *so* cool
     # For 1800: First iter: divs = [1*1, 1*2, 1*4, 1*8] = [1, 2, 4, 8]
     # Second iter: divs = [1*1, 1*3, 1*9, 2*1, 2*3, 2*9, 4*1, 4*3, 4*9, 8*1,
     # 8*3, 8*9] = [1, 3, 9, 2, 6, 18, 4, 12, 36, 8, 24, 72]
     # Third iter: divs = [1*1, 1*5, 1*25, 3*1, 3*5, 3*25, 9*1, 9*5, 9*25,
     # 2*1, 2*5, 2*25, 6*1, 6*5, 6*25, 18*1, 18*5, 18*25, 4*1, 4*5...] 
     # and so on
     # All in one, nice, simple, list comprehension
     # Whew!

     return sorted(divs) # Sorting it is the least we can do :)
Code:
>>> for i in range(50, 101):
...     nt.divisors(i)
...
[1, 2, 5, 10, 25, 50]
[1, 3, 17, 51]
[1, 2, 4, 13, 26, 52]
[1, 53]
[1, 2, 3, 6, 9, 18, 27, 54]
[1, 5, 11, 55]
[1, 2, 4, 7, 8, 14, 28, 56]
[1, 3, 19, 57]
[1, 2, 29, 58]
[1, 59]
[1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60]
[1, 61]
[1, 2, 31, 62]
[1, 3, 7, 9, 21, 63]
[1, 2, 4, 8, 16, 32, 64]
[1, 5, 13, 65]
[1, 2, 3, 6, 11, 22, 33, 66]
[1, 67]
[1, 2, 4, 17, 34, 68]
[1, 3, 23, 69]
[1, 2, 5, 7, 10, 14, 35, 70]
[1, 71]
[1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72]
[1, 73]
[1, 2, 37, 74]
[1, 3, 5, 15, 25, 75]
[1, 2, 4, 19, 38, 76]
[1, 7, 11, 77]
[1, 2, 3, 6, 13, 26, 39, 78]
[1, 79]
[1, 2, 4, 5, 8, 10, 16, 20, 40, 80]
[1, 3, 9, 27, 81]
[1, 2, 41, 82]
[1, 83]
[1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84]
[1, 5, 17, 85]
[1, 2, 43, 86]
[1, 3, 29, 87]
[1, 2, 4, 8, 11, 22, 44, 88]
[1, 89]
[1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90]
[1, 7, 13, 91]
[1, 2, 4, 23, 46, 92]
[1, 3, 31, 93]
[1, 2, 47, 94]
[1, 5, 19, 95]
[1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96]
[1, 97]
[1, 2, 7, 14, 49, 98]
[1, 3, 9, 11, 33, 99]
[1, 2, 4, 5, 10, 20, 25, 50, 100]
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Old 2012-12-10, 06:35   #11
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Dub, please re-read the first few posts. Your's have been off topic.
Paul has gotten the right understanding.
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