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By (user no longer on site) OP   
over a year ago

Did you know that only 11 out of the 91 single women in the CA post area have a veri posted this year.

This is just a pointless silly stat thead if you havnt already guessed. Got any more?

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By *am sampsonMan
over a year ago

cwmbran

That must have taken you ages to analyse

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By (user no longer on site)
over a year ago


"Did you know that only 11 out of the 91 single women in the CA post area have a veri posted this year.

This is just a pointless silly stat thead if you havnt already guessed. Got any more? "

Wonder how many they got hidden though

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By *ovedupstillCouple
over a year ago

mullinwire

78% of single men on FAB are honest, except to themselves and their wives. lol

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By (user no longer on site)
over a year ago

Sex Panther.

60% of the time it works every time

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By (user no longer on site)
over a year ago

100 % of the couples who 'won't meet bi/curious men' or 'won't meet smokers' that message me get a polite 'thanks but no thanks' message rather than a 'get your sodding eyes tested' one - which is probably more appropriate!

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By (user no longer on site)
over a year ago

did you know that 70% of men are pesistent liars and the other 35 % cant count

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By (user no longer on site)
over a year ago


"Did you know that only 11 out of the 91 single women in the CA post area have a veri posted this year.

This is just a pointless silly stat thead if you havnt already guessed. Got any more?

Wonder how many they got hidden though "

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By (user no longer on site)
over a year ago

The last month not to exceed 1000 topics raised in the Forum lounge, was October 2010 with only 893....

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By (user no longer on site)
over a year ago

Does anyone know the average airspeed velocity of an unladen swallow?

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By (user no longer on site)
over a year ago


"Does anyone know the average airspeed velocity of an unladen swallow? "

With some work, it is possible to calculate an approximate answer.

Notice that the formula for gravitational force, F = GmM/d^2, is simplified when G, M, and d are constant, or nearly so (such as near the Earth's surface) and we introduce merely the proportionality constant g, thus F = mg.

Similarly one equation for drag force is F = 1/2 C?Av^2. C (drag coefficient), ? (density of fluid), and A (cross-sectional area of object) may be regarded as constants here (the variation of A with time is really too complicated to even consider, for e.g. a feather.) So we may write merely a = kv^2 where k is a combined constant that includes the mass of the object also.

Thus at any given time the acceleration is a = g - kv^2. This may be written as a differential equation in h (height):

h'' = k(h')^2 - g.

I don't know how to solve this equation, but Mathematica gives the solution as:

h(t) = A - ln(cosh(v(gk))(t-B)))/k

where A and B are constants.

We need two initial conditions. We don't care about the starting point, so we may write h(t) = -ln(cosh(v(gk)(t-B)))/k, setting A to zero. And we want h'(0) = 0 if we initially merely drop the object, hence we have -vg tanh(v(gk)(t-B))/vk = 0, requiring B = 0 as well. So after time t, the velocity should be -vg tanh(v(gk)t)/vk.

As t approaches infinity, this expression approaches -v(g/k), the terminal velocity, what you would expect from setting g and kv^2 equal to each other (because the object stops accelerating.) But since tanh(z) 1 for all finite z, the terminal velocity is, in theory, never reached, only approached asymptotically. In practice, you can require that the fraction of terminal velocity reached - tanh(v(gk)t) - is at least a certain value, like 0.99, and solve for t in this way. However, it requires knowledge of k, which may have to be obtained experimentally if the value of C is not known.

But I could just be talking bollix

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By (user no longer on site)
over a year ago


"Does anyone know the average airspeed velocity of an unladen swallow?

With some work, it is possible to calculate an approximate answer.

Notice that the formula for gravitational force, F = GmM/d^2, is simplified when G, M, and d are constant, or nearly so (such as near the Earth's surface) and we introduce merely the proportionality constant g, thus F = mg.

Similarly one equation for drag force is F = 1/2 C?Av^2. C (drag coefficient), ? (density of fluid), and A (cross-sectional area of object) may be regarded as constants here (the variation of A with time is really too complicated to even consider, for e.g. a feather.) So we may write merely a = kv^2 where k is a combined constant that includes the mass of the object also.

Thus at any given time the acceleration is a = g - kv^2. This may be written as a differential equation in h (height):

h'' = k(h')^2 - g.

I don't know how to solve this equation, but Mathematica gives the solution as:

h(t) = A - ln(cosh(v(gk))(t-B)))/k

where A and B are constants.

We need two initial conditions. We don't care about the starting point, so we may write h(t) = -ln(cosh(v(gk)(t-B)))/k, setting A to zero. And we want h'(0) = 0 if we initially merely drop the object, hence we have -vg tanh(v(gk)(t-B))/vk = 0, requiring B = 0 as well. So after time t, the velocity should be -vg tanh(v(gk)t)/vk.

As t approaches infinity, this expression approaches -v(g/k), the terminal velocity, what you would expect from setting g and kv^2 equal to each other (because the object stops accelerating.) But since tanh(z) 1 for all finite z, the terminal velocity is, in theory, never reached, only approached asymptotically. In practice, you can require that the fraction of terminal velocity reached - tanh(v(gk)t) - is at least a certain value, like 0.99, and solve for t in this way. However, it requires knowledge of k, which may have to be obtained experimentally if the value of C is not known.

But I could just be talking bollix

"

Thanks! Does that work for both African and European?

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By (user no longer on site) OP   
over a year ago

I always thought the speed of a swallow was in direct relation to the thrust of cock in climax.

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By *ovedupstillCouple
over a year ago

mullinwire

if peter piper picked a peck of pickled pepper, and edward woodward would, but micheal knightly might not, if sea shanties tore womens panties and mens briefs stopped knicker thiefs, the government would want more tax off them all

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By (user no longer on site)
over a year ago

A swallow carrying a coconut ?

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By (user no longer on site)
over a year ago


"A swallow carrying a coconut ?"

No - unladen. Couldn't find a piece of string to attach the coconut - plus of course would then muck up the equation by the inclusion of a second swallow to hold the other end!

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By (user no longer on site)
over a year ago


"A swallow carrying a coconut ?

No - unladen. Couldn't find a piece of string to attach the coconut - plus of course would then muck up the equation by the inclusion of a second swallow to hold the other end! "

Oh, so how did you get the coconuts from the kingdom of mercia?

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By (user no longer on site)
over a year ago


"A swallow carrying a coconut ?

No - unladen. Couldn't find a piece of string to attach the coconut - plus of course would then muck up the equation by the inclusion of a second swallow to hold the other end!

Oh, so how did you get the coconuts from the kingdom of mercia?"

I didn't. There's no coconuts. Just a single, solitary swallow. (Undetermined yet as to whether African or European)

What do you think this is? Monty Bloody Python? lol

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By *uckscouple2007Couple
over a year ago

Bucks


"Does anyone know the average airspeed velocity of an unladen swallow? "

African or European?

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By (user no longer on site)
over a year ago

You've got no coconuts, sounds like a chemical castration job. That must have hurt. I've got a lovely bunch of coconuts.

What's this thread about again, oh stats. Only ones I can can think of are from my recent asbestos awareness test

White asbestos is the most common type and is present in 95% of asbestos containing materials (ACM).

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By *obinhood-maidmarionCouple
over a year ago

Yorkshire area

85% of single bi fems dont exist!

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By (user no longer on site) OP   
over a year ago


"85% of single bi fems dont exist!"

Don't spoil the dream. Bet you think Santa ain't real as well.

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By *ovedupstillCouple
over a year ago

mullinwire


"85% of single bi fems dont exist!

Don't spoil the dream. Bet you think Santa ain't real as well. "

nope, he is fake.

the tooth fairy told me

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By *thwalescplCouple
over a year ago

brecon


"A swallow carrying a coconut ?

No - unladen. Couldn't find a piece of string to attach the coconut - plus of course would then muck up the equation by the inclusion of a second swallow to hold the other end!

Oh, so how did you get the coconuts from the kingdom of mercia?

I didn't. There's no coconuts. Just a single, solitary swallow. (Undetermined yet as to whether African or European)

What do you think this is? Monty Bloody Python? lol "

Nobody expects the Spanish Inquisition!!

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By *ushroom7Man
over a year ago

Bradford


"A swallow carrying a coconut ?

No - unladen. Couldn't find a piece of string to attach the coconut - plus of course would then muck up the equation by the inclusion of a second swallow to hold the other end! "

Not to mention the co-efficient of Drag.

Over to you Soxy ....old bean.

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By (user no longer on site)
over a year ago


"78% of single men on FAB are honest, except to themselves and their wives. lol"

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By (user no longer on site)
over a year ago


"Does anyone know the average airspeed velocity of an unladen swallow?

With some work, it is possible to calculate an approximate answer.

Notice that the formula for gravitational force, F = GmM/d^2, is simplified when G, M, and d are constant, or nearly so (such as near the Earth's surface) and we introduce merely the proportionality constant g, thus F = mg.

Similarly one equation for drag force is F = 1/2 C?Av^2. C (drag coefficient), ? (density of fluid), and A (cross-sectional area of object) may be regarded as constants here (the variation of A with time is really too complicated to even consider, for e.g. a feather.) So we may write merely a = kv^2 where k is a combined constant that includes the mass of the object also.

Thus at any given time the acceleration is a = g - kv^2. This may be written as a differential equation in h (height):

h'' = k(h')^2 - g.

I don't know how to solve this equation, but Mathematica gives the solution as:

h(t) = A - ln(cosh(v(gk))(t-B)))/k

where A and B are constants.

We need two initial conditions. We don't care about the starting point, so we may write h(t) = -ln(cosh(v(gk)(t-B)))/k, setting A to zero. And we want h'(0) = 0 if we initially merely drop the object, hence we have -vg tanh(v(gk)(t-B))/vk = 0, requiring B = 0 as well. So after time t, the velocity should be -vg tanh(v(gk)t)/vk.

As t approaches infinity, this expression approaches -v(g/k), the terminal velocity, what you would expect from setting g and kv^2 equal to each other (because the object stops accelerating.) But since tanh(z) 1 for all finite z, the terminal velocity is, in theory, never reached, only approached asymptotically. In practice, you can require that the fraction of terminal velocity reached - tanh(v(gk)t) - is at least a certain value, like 0.99, and solve for t in this way. However, it requires knowledge of k, which may have to be obtained experimentally if the value of C is not known.

...suppose one needs to be good with numbers to be a swinger

But I could just be talking bollix

Thanks! Does that work for both African and European? "

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By (user no longer on site)
over a year ago

At least 1 fab member (possibly 2) has (have) far too much time on his / her / their hands !

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By (user no longer on site)
over a year ago

What are the statistics that there are some famous folks use this site?

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