Cable Guy Paradox

IdleRich

IdleRich
"Guess you didn't look very hard..."
Well, I didn't find that, that's for sure. Anyway, there are a number of complicating factors in that example which need to remain constant for the rule to apply - for example vapour pressure (which I take to mean air pressure in this instance) which is proportional to depth (ie height below where the atmosphere begins) I believe so will increase as the depth of the liquid in the cone diminishes.

"heating up the now smaller quantity of water which would thus require less energy to achieve a given temperature increase and therefore a given rate of evaporation."
That's what I was just about to say.
 

noel emits

a wonderful wooden reason
That's what I was just about to say.
Would that mean that the evaporation rate would actually remain constant if the surface area was always directly proportional to the overall volume? As I think it would be in an equilateral cone.

I don't think this is dependent on 'real world' conditions by the way.
 

IdleRich

IdleRich
"Would that mean that the evaporation rate would actually remain constant if the surface area was always directly proportional to the overall volume? As I think it would be in an equilateral cone."
I dunno. Not exactly sure what you mean. The volume of a cone is directly proportional to its height but the surface area of the base of this cone would actually vary with the square of the height as well so in fact the volume would decrease with the cube of the height (wouldn't it?). I don't think you could have something with this set-up where the surface area was always directly proportional to the volume if I'm understanding you correctly.
 

Mr. Tea

Let's Talk About Ceps
Would that mean that the evaporation rate would actually remain constant if the surface area was always directly proportional to the overall volume? As I think it would be in an equilateral cone.

I don't think this is dependent on 'real world' conditions by the way.

No, for a cone the volume is proportional to the depth cubed, the area proportional to the depth squared - so area is proportional to volume to the power of 2/3.
 

IdleRich

IdleRich
"No, for a cone the volume is proportional to the depth cubed, the area proportional to the depth squared - so area is proportional to volume to the power of 2/3."
That's true for the cone in your example but it's not true in general is it? I mean supposing the base of the cone was of fixed size but you kept increasing the height then the volume would be directly proportional to the height wouldn't it?
 

noel emits

a wonderful wooden reason
I'd have to check I think. It doesn't matter though, the main point still stands. Less water requires less energy to achieve a given rate of evaporation, so there is a relationship to volume.

Just to be clear, I was taking about the proportional relationship between the volume of water in the cone (which would be a cone itself), and the exposed surface area of the water.*

The other part is also true - liquids are always evaporating. I think this depends on molecular collisions though, so maybe there is a point where you get down to one molecule of water? ;)

* So this was what I was saying - the water forms a cone with its 'base' exposed. If the cone is of equal height and diameter then half the amount of water will have exactly half the exposed surface area.
 
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Mr. Tea

Let's Talk About Ceps
I don't think you could have something with this set-up where the surface area was always directly proportional to the volume if I'm understanding you correctly.

I think if you had an infinitely long 'funnel' formed as the solid of rotation of an exponential decay curve - so that, as depth increases (to infinity) the radius goes asymptotically to zero - the area then goes as the radius squared, and the volume goes as the area integrated wrt to depth, which also goes as the radius squared, because when you integrate an exponential you just get the same exponential scaled by some factor. So in this case, the surface area is proportional to the volume.

I think a shape like this is actually the subject of a classic mathematical paradox, let me just have a quick google...

Edit: http://en.wikipedia.org/wiki/Gabriel's_Horn - a shape with infinite surface area but finite volume. Though that's considering the area of the curved surface, not just the little disc at one end.
 
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IdleRich

IdleRich
"Just to be clear, I was taking about the proportional relationship between the volume of water in the cone (which would be a cone itself), and the exposed surface area of the water."
Well in this case what Ollie said would be correct - the radius of the bottom increases/decreases directly with the height, therefore the area of the bottom increases with the square of the height and the volume increases with the cube of the height ie the volume is proportional to the surface area of the bottom to the power (3/2).
 

IdleRich

IdleRich
"Though that's considering the area of the curved surface, not just the little disc at one end."
I was just going to ask how you calculate the area of the curved surface - I should definitely know that. I gues it's the limit of the area of the sides of a multi-sided "pyramid" as the number of sides approaches infinity...
 

Mr. Tea

Let's Talk About Ceps
If the cone is of equal height and diameter then half the amount of water will have exactly half the exposed surface area.

No, because V^(1/3) is prop. to A^(1/2) - so if you halve the volume of water you decrease the area by a factor of (1/2)^(3/2), or about 0.35.
 

noel emits

a wonderful wooden reason
the radius of the bottom increases/decreases directly with the height, therefore the area of the bottom increases with the square of the height and the volume increases with the cube of the height ie the volume is proportional to the surface area of the bottom to the power (3/2).
It was just a bonus, you understand that if there is less water it requires less energy to achieve a given rate of evaporation - and so the rate at which the rate of evaporation approaches zero is not exponential and so does not reach infinity at least, right? That's the main point there that says that the water can completely evaporate.

But what I was wondering about was the relationship between the exposed surface of the water, the disc, and the volume of water in the cone. For the sake of this I am presuming a cone with equal height and diameter.
 

Mr. Tea

Let's Talk About Ceps
That's true for the cone in your example but it's not true in general is it? I mean supposing the base of the cone was of fixed size but you kept increasing the height then the volume would be directly proportional to the height wouldn't it?

Oh, for sure. I was just talking about similar cones, such as would be formed by different amounts of water filling a conical hollow.
 

IdleRich

IdleRich
"It was just a bonus, you understand that if there is less water it requires less energy to achieve a given rate of evaporation - and so the rate at which the rate of evaporation approaches zero is not exponential and so does not reach infinity at least, right? That's the main point there that says that the water can completely evaporate."
Yeah yeah full agreement here.

"But what I was wondering about was the relationship between the exposed surface of the water, the disc, and the volume of water in the cone. For the sake of this I am presuming a cone with equal height and diameter."
What Mr Tea said abou this is correct - regardless of whether the cone has height equal to it's diameter the volume will vary with the surface area of a disc to the power three over two.
 

noel emits

a wonderful wooden reason
Consider an open-topped conical container, like a funnel only instead of a thin open-ended tube at the bottom, it just ends in a conical vertex. An ice-cream-cone shape, I guess. Now suppose I pour some water in it and leave it out in the sun; clearly, the water will evaporate. It seems reasonable to assume that the rate of evaporation, in unit volume per unit time, is proportional to the surface area of the remaining body of water. Because of the containers' conical shape, the water's surface area is proportional to the square of its depth. Thus the evaporation rate is proportional to the square of the remaining depth - so as the depth of remaining water approaches zero, so the evaporation rate approaches zero -
more over, the evaporation rate approaches zero faster than the remaining depth. So the water never, ever fully evaporates. :cool:
What this says is that the surface area (evaporation rate) of the water approaches zero faster than the remaining depth. That would mean that the water would have zero surface area before there was none left. So that can't really be right, right?

Of course the surface area and the depth reach zero at exactly the same point so the rate at which they approach zero is dependant on the ratio between the two, which depends on the angle of the cone.

But there are two problems here.

The first is in measuring depth rather than remaining volume. As it is a cone and not a cylinder, for instance, they are not equivalent. Depth is not a linear measurement of the remaining water.

The second error is that the evaporation rate is not simply proportional to the surface area of the remaining water, it is also to some extent a function of the remaining volume, as we've said.

Indeed, the evaporation rate is actually somewhat inversely proportional to the remaining volume.

So the evaporation rate does in fact approach zero less quickly than the remaining volume, to a degree dependant on the angle of the cone.
 

Mr. Tea

Let's Talk About Ceps
What this says is that the surface area (evaporation rate) of the water approaches zero faster than the remaining depth. That would mean that the water would have zero surface area before there was none left. So that can't really be right, right?

No, that's not what it means. Of course the surface area and depth both reach exactly zero only when the depth reaches exactly zero. But let's say at some point the surface area is one square cm and the depth is one cm, so in cm units they're numerically the same - of course, they can never be literally the same, since they're quantities with different dimension and therefore incommensurate. What I mean by "the surface area approaches zero faster than the depth" is that for all depths smaller than 1cm, the area in cm^2 will be numerically smaller than the depth in cm. If you square a number smaller than 1, the answer will be smaller still.

Of course the surface area and the depth reach zero at exactly the same point so the rate at which they approach zero is dependant on the ratio between the two, which depends on the angle of the cone.

But there are two problems here.

The first is in measuring depth rather than remaining volume. As it is a cone and not a cylinder, for instance, they are not equivalent. Depth is not a linear measurement of the remaining water.

I don't see how this is a 'problem', since the depth, area and volume are all dependent on each other. Each quantity could be used to uniquely define the amount of water left, and it's all gone when any (i.e. all) of these quantities reaches zero.

I agree about the angle of the cone, but the only difference that makes is the point at which the rate of decrease of area 'overtakes' the rate of decrease of depth, I think. So for a shallow cone the area will initially shrink must faster than the depth and for a very narrow cone it's the reverse, but for all cones the rate of area shrinkage will eventually be faster once the remaining volume goes lower than some threshold value.
The second error is that the evaporation rate is not simply proportional to the surface area of the remaining water, it is also to some extent a function of the remaining volume, as we've said.

Indeed, the evaporation rate is actually somewhat inversely proportional to the remaining volume.

So the evaporation rate does in fact approach zero less quickly than the remaining volume, to a degree dependant on the angle of the cone.

I concede that from a physics point of view, this is where the 'paradox' breaks down; namely, a smaller volume of water will be able to absorb heat from its surroundings faster than a larger volume. Then again, the precise way the rate of heat absorption behaves might be non-trivial to work out, depending possibly on some combination of depth, area and volume.
 

noel emits

a wonderful wooden reason
let's say at some point the surface area is one square cm and the depth is one cm, so in cm units they're numerically the same - of course, they can never be literally the same, since they're quantities with different dimension and therefore incommensurate.
That's right, they are not the same, just numerically the same at that point, which doesn't tell you much. What matters is the ratio, which of course depends on the angle of the cone
What I mean by "the surface area approaches zero faster than the depth" is that for all depths smaller than 1cm, the area in cm^2 will be numerically smaller than the depth in cm.
No it's dependant on the angle, that is to say on the ratio between height and surface area, or just width if you like. There may be a point where the area and height are numerically equivalent, but the two values will continue to converge towards zero at the same rate beyond that point.

Your chosen cone has an angle such that when the height is 1cm the surface area is 1cm^2. That's just a particular case.
If you square a number smaller than 1, the answer will be smaller still.
Well don't do it then! :p

It's not really smaller than 1 is it? It's only said to be so in relation to some point you have arbitrarily defined as being '1' because it happens to coincide with a particular unit of measurement. Again, the rates remain the same.
I don't see how this is a 'problem', since the depth, area and volume are all dependent on each other. Each quantity could be used to uniquely define the amount of water left, and it's all gone when any (i.e. all) of these quantities reaches zero.
Erm, what you really need to know is the height to width ratio, i.e. the angle of the cone.
I agree about the angle of the cone, but the only difference that makes is the point at which the rate of decrease of area 'overtakes' the rate of decrease of depth, I think. So for a shallow cone the area will initially shrink must faster than the depth and for a very narrow cone it's the reverse, but for all cones the rate of area shrinkage will eventually be faster once the remaining volume goes lower than some threshold value.
The ratio between the height and area doesn't actually change though does it? So the ratio between the rates of change stays constant as well.
I concede that from a physics point of view, this is where the 'paradox' breaks down; namely, a smaller volume of water will be able to absorb heat from its surroundings faster than a larger volume. Then again, the precise way the rate of heat absorption behaves might be non-trivial to work out, depending possibly on some combination of depth, area and volume.
I would say that the above argues that the ratio between the rates of change of height and area remains constant, and such that they will reach zero at the same time. Of course.

However the additional factor that less water will evaporate faster given the same amount of energy might mean that the rates themselves also increase as the volume of water decreases. I think the ratio would still stay the same though.
 

Mr. Tea

Let's Talk About Ceps
The ratio between the height and area doesn't actually change though does it?

Yes, it most certainly does! The ratio of the depth (height, whatever) and the *radius* of the circular surface of the water of course remains constant, for a cone of any given angle. But the *area* of the surface depends on the square of the radius, so it is NOT proportional to the water's depth. This is why I've been talking about a 'critical' depth, below which the area numerically decreases faster than the depth itself - you're right that the figures involved will depend on the units used, but the actual units are arbitrary, aren't they? We could be talking about (square) metres or (square) centimetres; it doesn't matter, since the problem is unaffected by changes in scale.

Edit: the rate of decrease of depth with depth, dD/dD, is obviously 1. Area is proportional to radius squared, and radius is proportional to depth, so A is prop. to R^2 is prop. to D^2. Take the derivative w.r.t to D, and dA/dD is prop. to 2D, and therefore prop. to D. So at some point, as D decreases, it will become numerically smaller than 1, no matter what units are used.
 
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noel emits

a wonderful wooden reason
Aw, one small error then?

The ratio of the rates of change of the height and surface area doesn't change, as I said on the next line.

In other words the rates of change remain constant or change at the same rate.
below which the area numerically decreases faster than the depth itself
Yes but this is deceptive. As you've said, they are quantities of 'different dimension and therefore incommensurate'. Height is one-dimensional, area is two-dimensional. So while the numerical representations of these quantities may change at different rates, the ratio between those rates of change remains the same.

As for units, what I mean is try thinking of the apex as the unitary point '1' rather than zero.
 

noel emits

a wonderful wooden reason
But the *area* of the surface depends on the square of the radius, so it is NOT proportional to the water's depth.
No, but the area and the depth are related via the angle of the cone, i.e. the ratio, which remains constant.
 

noel emits

a wonderful wooden reason
Edit: the rate of decrease of depth with depth, dD/dD, is obviously 1. Area is proportional to radius squared, and radius is proportional to depth, so A is prop. to R^2 is prop. to D^2. Take the derivative w.r.t to D, and dA/dD is prop. to 2D, and therefore prop. to D. So at some point, as D decreases, it will become numerically smaller than 1, no matter what units are used.
I propose the use of these ingenious new units of my own devising, then.

1fD = 1 full cone depth

1fA = 1 full cone area

Now they will both become numerically smaller than 1 as soon as the water starts evaporating, which is to say immediately and simultaneously.

Instead of area or depth being numerically defined as proportional to each other they are now numerically defined as proportional to the dimensions of the original full cone.

There is no height reduction without evaporation. There is no area reduction without evaporation. They are the same event.
 
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