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"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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"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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"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 | |||
"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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"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! | |||
"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? | |||
"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 | |||
"Does anyone know the average airspeed velocity of an unladen swallow? "African or European? | |||
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"85% of single bi fems dont exist!" Don't spoil the dream. Bet you think Santa ain't real as well.  | |||
"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 | |||
"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!! | |||
"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. | |||
"78% of single men on FAB are honest, except to themselves and their wives. lol" | |||
"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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