Radiative Processes in Planetary Atmospheres

[RedAcer]

Some great course notes here for anyone who wants to understand more
about how Atmospheres work. The full course notes are there along with
exercises and solutions so that you can test your knowledge.

<http://nit.colorado.edu/atoc5560/>

Description:
This course will cover the basic physics of, and computational methods
for, the interaction of visible and infrared radiation with the gases
and particles in planetary atmospheres. The applications will be mainly
in the role radiation plays in determining planetary temperature
structure and in climate change. The emphasis will be on the Earth's
atmosphere.

Topics include:
Basic radiative transfer
Line absorption by molecules
Scattering by particles
Approximate and exact radiative transfer methods
Radiative equilibrium models
Trace gases, aerosols, clouds, and climate

*Contents*

Part I:
Introduction to Radiative Transfer
Radiative definitions
Absorption, emission, and Planck functions
Radiative transfer equation
Simple radiative transfer solutions
Part II:
Radiation and Gases
Temperature structure and gas composition profiles
Rotational and vibrational molecular transitions
Absorption line shapes and broadening mechanisms
Absorption line intensities
Atmospheric absorption spectrum
Line-by-line models
Integrated single line transmission; random band models
The k - distribution and correlated k-distribution methods
Clear sky long-wave radiative transfer
Heating rates: calculation and results
Weighting functions and the cooling to space approximation
Part III:
Radiation and Particles
Particle size distributions
Cross sections and phase functions
Polarization of radiation
Rayleigh scattering
Mie scattering (method and results for aerosols and water droplets)
Index of refraction of materials
Discrete Dipole Approximation
Geometric optics scattering
Part IV:
Radiative Transfer with Scattering
Radiative transfer equation with scattering of sunlight
First order scattering
Eddington and two-stream approximate methods
Delta scaling of the radiative transfer equation
Discrete ordinates solution methods
Principle of interaction; doubling-adding method
Surface reflection: Lambertian, Fresnel, general BRDF
Plane-parallel radiative transfer results
3D radiative transfer: Monte Carlo method and results
Part V:
Radiative Equilibrium and Atmospheric Applications
Sun-Earth geometry and solar insolation; solar measurements
Earth's radiative energy budget and measurement
Radiative Equilibrium:
Gray models (single layer and profile)
Non gray models, the atmospheric window
1D radiative convective equilibrium; temperature structure
Radiative forcing and climate feedbacks
Cloud radiation interactions
Aerosols radiative forcing
Greenhouse gas warming
Planetary applications

[emoneyjoe]

On Mon, 20 May 2013 09:56:04 +0100, RedAcer <rred...@gmail.com> wrote:

>Some great course notes here for anyone who wants to understand more
>about how Atmospheres work. The full course notes are there along with
>exercises and solutions so that you can test your knowledge.
>
><http://nit.colorado.edu/atoc5560/>

Yeah, everybody is going to head to
Colorado and take that course.

But not me, Colorado is too cold.

[Paul Aubrin]

It is an online course. As Red Acer promised to study it, we are going to
have holidays for a while. When he returns he will be turned into a
sceptic.
I skimmed over the pdf files and could see no back-radiation concept
there. The energy flux seemed proportional to the difference of radiances
(and thus would flow from warm to cold in the case of black bodies)
everything which appears as normal physics, not as post-normal climate-
science Red Acer's physics.

[RedAcer]

That's not your only problem. You are too stupid to see that those notes
are *online* so it is unlikely that you could read them.

>
>
>
>
>

[RedAcer]

On 20/05/13 17:02, Paul Aubrin wrote:
> On Mon, 20 May 2013 10:59:04 -0400, emoneyjoe wrote:
>
>> On Mon, 20 May 2013 09:56:04 +0100, RedAcer <rred...@gmail.com> wrote:
>>
>>> Some great course notes here for anyone who wants to understand more
>>> about how Atmospheres work. The full course notes are there along with
>>> exercises and solutions so that you can test your knowledge.
>>>
>>> <http://nit.colorado.edu/atoc5560/>
>>
>> Yeah, everybody is going to head to
>> Colorado and take that course.
>>
>> But not me, Colorado is too cold.
>
> It is an online course. As Red Acer promised to study it, we are going to
> have holidays for a while. When he returns he will be turned into a
> sceptic.
> I skimmed over the pdf files and could see no back-radiation concept
> there.

I haven't had time to study these,yet, but this is what you were looking
for:
See page 8 and 9 in week2.pdf
They use the term down welling radiance, and calculate it's value at the
earth's surface, if I read it correctly.

[emoneyjoe]

If any radiation from 10 miles up were
to get to the ground, it would have to be
the difference between the flux of a
warmer gas at 10 miles up and a cooler
surface, which seems impossible.

I see visual proof that the net flow
of energy in winter is from the ground
upward, because frost forms when
the sky is clear, and that happens
when the local temperature is
below about 37 degrees.
That tells me the local air does
radiate to the ground, but this
could be studied by having places
of different heights with a black
surface and a small remote control
camera to see frost and a thermometer
to measure air temperature.

I think it is silly to think that any
model could accurately portray the
complexity of the atmosphere and
the various surface materials at
different temperature and the
phases of water.

But they have an objective,
the slowing of the activity in the
industrialized countries while
the rest of the world catches up.

China burns more coal now
than the rest of the world put
together, and releases more
CO2 than the rest of the world
put together, and China will
laugh at anybody who thinks
that can be changed by words
or political agendas.

Very few things about AGW
make any sense, during WWII
the pollution reached an all time
high, and the global temperature
apparently went up, the pollution
then apparently did not block
the sun, so why does it now.


I don't understand RedAcer
seeming not to understand what
I say about GHGs cooling the
atmosphere, I know he would
understand that the coolant
and the radiator and coolant
pump on my car cools the
engine when it is running,
and slows the cooling of
the engine when I turn the
key off.

So I have to think he is
just playing games while he
gets the script from central
control to argue against
common sense for the AGW
agenda.

[emoneyjoe]

If reading notes resulted in good understanding,
nobody would ever have a failing grade.

My wireless connection is too slow to
try to do anything with that resource, and
that course seems to have little to do with
the wild predictions made by AGW alarmists.

[Paul Aubrin]

On Mon, 20 May 2013 19:44:40 +0100, RedAcer wrote:

> See page 8 and 9 in week2.pdf They use the term down welling radiance,
> and calculate it's value at the earth's surface, if I read it correctly.

Radiance not energy flux.

[RedAcer]

This is why you don't understand the GHE. You don't understand basic
physics.
Radiance *is* energy flux.
From wikipedia:
"Radiance is a measure of the *quantity* of radiation that passes
through or is emitted from a surface and falls within a given solid
angle in a specified direction."

"Radiance characterizes *total emission* or reflection. Radiance is
useful because it indicates how much of the *power* emitted by an
emitting or reflecting surface will be received by an optical system
looking at the surface from some angle of view."

[Paul Aubrin]

On Tue, 21 May 2013 07:35:08 +0100, RedAcer wrote:

> On 21/05/13 06:30, Paul Aubrin wrote:

>> [...]

>
> This is why you don't understand the GHE. You don't understand basic
> physics.
> Radiance *is* energy flux.

The radiance is the energy flux that would flow toward a 0K surface.
The radiative energy flux is proportional to the difference of the
radiances (analogous with the thermal flux is proportional to the
difference of temperatures, the electrical flux is proportional to the
difference of potentials).

[RedAcer]

On 21/05/13 08:41, Paul Aubrin wrote:
> On Tue, 21 May 2013 07:35:08 +0100, RedAcer wrote:
>
>> On 21/05/13 06:30, Paul Aubrin wrote:
>>> [...]
>>
>> This is why you don't understand the GHE. You don't understand basic
>> physics.
>> Radiance *is* energy flux.

...snip wrong stuff.

"Radiance is a measure of the *quantity* of radiation that passes
through or is emitted from a surface and falls within a given solid
angle in a specified direction."

"Radiance characterizes *total emission* or reflection. Radiance is
useful because it indicates how much of the *power* emitted by an
emitting or reflecting surface will be received by an optical system
looking at the surface from some angle of view."

Check the definition here.
http://en.wikipedia.org/wiki/Radiance
If you disagree then please provide a reference. None of your made up
bilge.


Radiance is defined by:-

L = \frac{\mathrm{d}^2 \Phi}{\mathrm{d}A\,\mathrm{d}{\Omega} \cos
\theta} \approx \frac{\Phi}{\Omega A \cos \theta}

where

L is the observed or measured radiance (W·m−2·sr−1), in the direction θ,
d is the differential operator,
Φ is the total radiant flux or power (W) emitted
θ is the angle between the surface normal and the specified direction,
A is the area of the surface (m2), and
{\Omega} is the solid angle (sr) subtended by the observation or
measurement.
The approximation only holds for small A and Ω where cos θ is
approximately constant.

[emoneyjoe]

On 21 May 2013 07:41:13 GMT, Paul Aubrin <chu8...@free.fr> wrote:

>On Tue, 21 May 2013 07:35:08 +0100, RedAcer wrote:
>
>> On 21/05/13 06:30, Paul Aubrin wrote:
>>> [...]
>>
>> This is why you don't understand the GHE. You don't understand basic
>> physics.
>> Radiance *is* energy flux.
>The radiance is the energy flux that would flow toward a 0K surface.

That sounds like the voice of reason, finally.

>The radiative energy flux is proportional to the difference of the
>radiances (analogous with the thermal flux is proportional to the
>difference of temperatures, the electrical flux is proportional to the
>difference of potentials).

No, not proportional to temperatures (did
you write that or quote it), but a function of
the fourth power of absolute temperatures.

[RedAcer]

http://en.wikipedia.org/wiki/Radiance

[Paul Aubrin]

On Tue, 21 May 2013 09:00:48 +0100, RedAcer wrote:

> On 21/05/13 08:41, Paul Aubrin wrote:

>> [...]
> ...snip wrong stuff.
Here are some translated extracts from a radiation course:
Emittance M0 is the total flux emitted per unit of surface.
Some pages below there is the radiative exchanges paragraph:
NetFlux emitted by i = sum for j=1 to n of (Si .Fij.(M0i - M0j)
The net (actual) flux is proportional to the difference of emittance.

Energy is not vectorial, it is always positive. I know that you are
tempted to count negatively some energy flux because it is what you do in
an energy balance equation. But in this case, we don't consider an energy
balance. We are talking of real energy flux, which interacts with real
matter and is always positive. They are not annihilated just because they
reach a black surface. You suppose all those W/m2 are bounced back and
forth between the two black surfaces. Actually, the only real W/m2 are
proportional to the difference of emittance. When the two facing black
surfaces are in radiative equilibrium, the actual energy flux is zero.
No instrument, no experiment could put in evidence the virtual flux which
would be generated by the surface emittance, because the emittance is the
potential to emit, not the real flux.

[RedAcer]

On 21/05/13 09:39, Paul Aubrin wrote:
> On Tue, 21 May 2013 09:00:48 +0100, RedAcer wrote:
>
>> On 21/05/13 08:41, Paul Aubrin wrote:
>>> [...]
>> ...snip wrong stuff.
> Here are some translated extracts from a radiation course:
> Emittance M0 is the total flux emitted per unit of surface.
> Some pages below there is the radiative exchanges paragraph:
> NetFlux emitted by i = sum for j=1 to n of (Si .Fij.(M0i - M0j)
> The net (actual) flux is proportional to the difference of emittance.

So you now accept that radiance represents an energy flow.

>
> Energy is not vectorial, it is always positive. I know that you are
> tempted to count negatively some energy flux because it is what you do in
> an energy balance equation. But in this case, we don't consider an energy
> balance. We are talking of real energy flux, which interacts with real
> matter and is always positive. They are not annihilated just because they
> reach a black surface. You suppose all those W/m2 are bounced back and
> forth between the two black surfaces. Actually, the only real W/m2 are
> proportional to the difference of emittance. When the two facing black
> surfaces are in radiative equilibrium, the actual energy flux is zero.
> No instrument, no experiment could put in evidence the virtual flux which
> would be generated by the surface emittance, because the emittance is the
> potential to emit, not the real flux.

This is just a lot of waffle with no meaning. I showed you the
calculation with two plates. Energy doesn't bounce 'backwards and forwards'.

Your problem is trying to understand physics with words. If you thought
my calculation was wrong, show where.
I pointed to the course notes where radiance from the atmosphere at the
earth's surface was calculated. You have to show where that calculation
was wrong or give us a reference to a published calculation.

You still haven't answered the question about objects in your room. Do
your walls emit EMR commensurate with their temperature and surface
type? Does your cup of coffee emit energy. Does your keyboard emit IR?
Why do most of the passive objects in your room have the same
temperature. How do they get to a 'steady state'.
Think about 'dynamic equilibrium' - it will help you understand thermal
equilibrium. i.e. the equilibrium is not 'static' as in a mechanical system.

[Claudius Denk]

Do they allow non-believers?

[Claudius Denk]

If you need any help with this let me know.

[Tom P]

You may recall from the example with the two plates, when you work out
the equations it makes no difference whether you consider each object
radiating independently of the other, or whether you consider the net
energy flux flowing between the objects. All that happens in the algebra
is that the terms wander from one side of the equation to the other, and
the solutions turn out identical. This means that the argument about
which is right or wrong is rather pointless, even assuming that your
position is the correct one.
The reason why your position is wrong is that as a general principle,
the radiation emitted by an object is a function of its own temperature,
and not of any other object. This was already understood in the 19th
century. Otherwise you would have the absurd conclusion that you have
to take into account every other object in the universe in order to
determine how much your object radiates.

[Tom P]

Please show us where it says that the radiation emitted by a body is
dependent on the temperature of other objects.

[Tom P]

If you look at week 2 page 8 he calculates the back radiation (more
exactly the radiance) from the atmosphere.
" Assume the atmosphere emits as a layer at 285 K.
The surface is a blackbody with temperature of 295 K.
The two Planck radiances needed are:
Bλ (295) = 9.12 W m−2sr−1μm−1 Bλ (285) = 7.69 W m−2sr−1μm−1
..
Downwelling radiance at surface: Iλ = (1 − T )Bλ (Ta)
Iλ (0◦) = 0.237(7.69) = 1.82 W m−2 sr−1 μm−1"

To get the total downward flux you integrate the radiance over
steradians and wavelength.

[RedAcer]

I pointed out the exact same pages to him, to no avail. He now denies
that radiance if a flow of energy.
He's refusing to answer the question, whether ordinary objects like his
keyboard or carpet emit EMR.
His problem is that he just doesn't get the concept of dynamic equilibrium.

[Paul Aubrin]

On Wed, 22 May 2013 07:32:28 +0100, RedAcer wrote:

>> Energy is not vectorial, it is always positive. I know that you are
>> tempted to count negatively some energy flux because it is what you do
>> in an energy balance equation. But in this case, we don't consider an
>> energy balance. We are talking of real energy flux, which interacts
>> with real matter and is always positive. They are not annihilated just
>> because they reach a black surface. You suppose all those W/m2 are
>> bounced back and forth between the two black surfaces. Actually, the
>> only real W/m2 are proportional to the difference of emittance. When
>> the two facing black surfaces are in radiative equilibrium, the actual
>> energy flux is zero. No instrument, no experiment could put in evidence
>> the virtual flux which would be generated by the surface emittance,
>> because the emittance is the potential to emit, not the real flux.
>
> This is just a lot of waffle with no meaning. I showed you the
> calculation with two plates. Energy doesn't bounce 'backwards and
> forwards'.

The two black surface plate calculation show that the energy flux is
driven by the difference of temperature and that no extra energy needs to
be emitted by the warmest plate to be back-radiated toward the colder
one, unlike what is written in Wikipedia and what pretended Poutnik. No
actual physical effect can put in evidence this supposed back-radiation
which is thus virtual (non-existent).

[RedAcer]

Yes it can. Did you check how NASA uses MLI like I suggested?
Here are a few questions to ponder.
How does it keep heat *in and out* ?
Why does it have so many layers ?
Why are there spacers to keep the layers apart?
Why are there holes in the layers?
According to your physics it can't work.
So, how does it work?

Good Luck!
>

[Will Janoschka]

On Wed, 22 May 2013 18:51:59, RedAcer <rred...@gmail.com> wrote:
> On 22/05/13 19:21, Paul Aubrin wrote:
> > On Wed, 22 May 2013 07:32:28 +0100, RedAcer wrote:
> >
> >>> Energy is not vectorial, it is always positive. I know that you are
> >>> tempted to count negatively some energy flux because it is what you do
> >>> in an energy balance equation. But in this case, we don't consider an
> >>> energy balance. We are talking of real energy flux, which interacts
> >>> with real matter and is always positive. They are not annihilated just
> >>> because they reach a black surface. You suppose all those W/m2 are
> >>> bounced back and forth between the two black surfaces. Actually, the
> >>> only real W/m2 are proportional to the difference of emittance. When
> >>> the two facing black surfaces are in radiative equilibrium, the actual
> >>> energy flux is zero. No instrument, no experiment could put in evidence
> >>> the virtual flux which would be generated by the surface emittance,
> >>> because the emittance is the potential to emit, not the real flux.
> >>
> >> This is just a lot of waffle with no meaning. I showed you the
> >> calculation with two plates. Energy doesn't bounce 'backwards and
> >> forwards'.

That is correct just like Paul says.

> >
> > The two black surface plate calculation show that the energy flux is
> > driven by the difference of temperature and that no extra energy needs to
> > be emitted by the warmest plate to be back-radiated toward the colder
> > one, unlike what is written in Wikipedia and what pretended Poutnik. No
> > actual physical effect can put in evidence this supposed back-radiation
> > which is thus virtual (non-existent).
>
> Yes it can. Did you check how NASA uses MLI like I suggested?

Nasa uses superinsulation just like I do!

> Here are a few questions to ponder.
> How does it keep heat *in and out* ?

Multiple layers if high reflectance aluminumized mylar

> Why does it have so many layers ?

the radiative losses are the product of the emissivities
of all the multiple surfaces. each surface has an
emissivity of less than 1 %. Fifty layers have a combined
emissivity of less that a 1/ 10^4. multiply that by your
difference in each T^4.

> Why are there spacers to keep the layers apart?

To prevent thermal conductivity.

> Why are there holes in the layers?

sometimes this prevents outgassing problems

> According to your physics it can't work.

High reflectivity is not against any of Pau'ls remarks

> So, how does it work?

I just explained that.

> Good Luck!

Do you have any other problems with Paul's statements?

[Tom P]

If that was his only problem.

[emoneyjoe]

>>> B? (295) = 9.12 W m-2sr-1µm-1 B? (285) = 7.69 W m-2sr-1µm-1
>>> ..
>>> Downwelling radiance at surface: I? = (1 - T )B? (Ta)
>>> I? (0?) = 0.237(7.69) = 1.82 W m-2 sr-1 µm-1"

>>>
>>> To get the total downward flux you integrate the radiance over
>>> steradians and wavelength.
>>
>> I pointed out the exact same pages to him, to no avail. He now denies
>> that radiance if a flow of energy.
>> He's refusing to answer the question, whether ordinary objects like his
>> keyboard or carpet emit EMR.
>> His problem is that he just doesn't get the concept of dynamic equilibrium.
>>
>If that was his only problem.

How do you find week 2?

[Paul Aubrin]

On Wed, 22 May 2013 17:23:09 -0500, Will Janoschka wrote:

>> Here are a few questions to ponder. How does it keep heat *in and out*
>> ?
>
> Multiple layers if high reflectance aluminumized mylar

Red Acer believes that mirrors behave as black bodies.

[Tom P]

That's a strange question, do you mean, is it to my taste?

Well, it confirms a lot of what Red, Sam, Poutnik and others including
myself have been saying. On page 3 it shows how to calculate how much
radiation CO2 at a concentration of 360ppm absorbs. Page 8 how to
calculate back radiation.
Week 3 page 2 there's a diagram showing that WV becomes a trace gas
above 10km.

[RedAcer]

In the case where it is not facing the sun and you are trying to keep
heat inside the craft each layer is colder than one closer to the craft
so how does it 'reflect' heat back to something that is warmer. Maybe
you are beginning to understand how this works.

>
>> Why are there spacers to keep the layers apart?
>
> To prevent thermal conductivity.
>
>> Why are there holes in the layers?
>
> sometimes this prevents outgassing problems

No. It's to allow out-gassing you want there to be a vacuum between the
layers.

>
>> According to your physics it can't work.
>
> High reflectivity is not against any of Pau'ls remarks

How is heat reflected back to something that is warmer. Paul says that
the laws of physics don't allow it.

>
>> So, how does it work?
>
> I just explained that.

No you haven't. But you were nearly there.

>
>> Good Luck!
>
> Do you have any other problems with Paul's statements?

Yes all of them.

>

[RedAcer]

No I don't.
You use the emissivity of aluminium at IR frequencies. Simples!

>

[RedAcer]

>>>> B? (295) = 9.12 W m-2sr-1�m-1 B? (285) = 7.69 W m-2sr-1�m-1

>>>> ..
>>>> Downwelling radiance at surface: I? = (1 - T )B? (Ta)

>>>> I? (0?) = 0.237(7.69) = 1.82 W m-2 sr-1 �m-1"

>>>>
>>>> To get the total downward flux you integrate the radiance over
>>>> steradians and wavelength.
>>>
>>> I pointed out the exact same pages to him, to no avail. He now denies
>>> that radiance if a flow of energy.
>>> He's refusing to answer the question, whether ordinary objects like his
>>> keyboard or carpet emit EMR.
>>> His problem is that he just doesn't get the concept of dynamic equilibrium.
>>>
>> If that was his only problem.
>
> How do you find week 2?

http://nit.colorado.edu/atoc5560/week2.pdf

>
>
>

[Paul Aubrin]

On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:

>> High reflectivity is not against any of Pau'ls remarks
>
> How is heat reflected back to something that is warmer. Paul says that
> the laws of physics don't allow it.
>

You didn't mention in your "two plates in the void" exercice that one of
them was a mirror, you said: black-body. Mirrors (by definition) reflect,
black surfaces (by definition) absorbs. More straw man fallacies?

[Paul Aubrin]

On Mon, 20 May 2013 16:59:41 -0400, emoneyjoe wrote:

>>That's not your only problem. You are too stupid to see that those notes
>>are *online* so it is unlikely that you could read them.
>
> If reading notes resulted in good understanding,
> nobody would ever have a failing grade.

Even a good understanding of the course would not prevent some false
interpretations. If you discuss with persons who believe in astrology,
they will insist that their "science" can be trusted because they base
their predictions on very accurate positions of stellar objects. And this
will be sufficient for them to reject any other objection. My impression
is that Greens are more prone to believe in astrology (and other "natural
is synonym with good" fads) than the general population.

> My wireless connection is too slow to
> try to do anything with that resource, and that course seems to have
> little to do with the wild predictions made by AGW alarmists.

[emoneyjoe]

On 23 May 2013 10:57:07 GMT, Paul Aubrin <chu8...@free.fr> wrote:

>On Mon, 20 May 2013 16:59:41 -0400, emoneyjoe wrote:
>
>>>That's not your only problem. You are too stupid to see that those notes
>>>are *online* so it is unlikely that you could read them.
>>
>> If reading notes resulted in good understanding,
>> nobody would ever have a failing grade.
>
>Even a good understanding of the course would not prevent some false
>interpretations. If you discuss with persons who believe in astrology,
>they will insist that their "science" can be trusted because they base
>their predictions on very accurate positions of stellar objects. And this
>will be sufficient for them to reject any other objection. My impression
>is that Greens are more prone to believe in astrology (and other "natural
>is synonym with good" fads) than the general population.

Martin Brown has been kind enough to write
a detailed explanation here today, I will study
it after I take a nap.

I have seen a couple of energy budget
diagrams that say only 12% of solar insolation
is radiated directly to space, and only 5% of
the radiation from the surface is absorbed
in the atmosphere.

This web page shows two diagrams, the
first doesn't mention back radiation;
The second does, but the back radiation
number doesn't seem to reference the same
origin percentage?


http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php

[Will Janoschka]

Indeed, but you Paul, can understand highly reflective
black bodies.

BTW My US instruments can measure better than
your French ones. I hope. :) .

>

[Will Janoschka]

Maybe Red Acer, you are beginning to understand how this works.

You Red Acer refuse to admit that that reflectivity Is indeed
One minus emissivity. Reflectivity is the refusal to absorb or
emit.

Make things shiney, and they automagically refuse any thing
about the S-B equation, which requires no refectivity.

>
> >
> >> Why are there spacers to keep the layers apart?
> >
> > To prevent thermal conductivity.
> >
> >> Why are there holes in the layers?
> >
> > sometimes this prevents outgassing problems
>
> No. It's to allow out-gassing you want there
> to be a vacuum between the> layers.

That is the out-gassing problem with no
small holes. -idiot-

> >> According to your physics it can't work.
> >
> > High reflectivity is not against any of Pau'ls remarks
>
> How is heat reflected back to something that is warmer. Paul says that
> the laws of physics don't allow it.
>

Reflection is allowed. Spontaneous heat transfer
to a higherr temperature is not allowed.
> >
The Laws of Physics do not allow the transfer of
heat to a higher temperature because there is
no possible reason to do that.

Unless you have a big honking power source
that wants to do that. Then scamm the peons.

> >> So, how does it work?
> >
> > I just explained that.
>
> No you haven't. But you were nearly there.

Huh?

>
> >
> >> Good Luck!
> >
> > Do you have any other problems with Paul's statements?
>
> Yes all of them.

Can you express them in words, here in this newsgroup ?

[Will Janoschka]

On Thu, 23 May 2013 08:40:36, RedAcer <rred...@gmail.com> wrote:
> On 23/05/13 05:32, Paul Aubrin wrote:
> > On Wed, 22 May 2013 17:23:09 -0500, Will Janoschka wrote:
>
> >>> Here are a few questions to ponder. How does it keep heat *in and out*

> >> Multiple layers if high reflectance aluminumized mylar
> > Red Acer believes that mirrors behave as black bodies.
> No I don't.
> You use the emissivity of aluminium at IR frequencies. Simples!
>

You speak bull shit. You measure the reflectivity of vacuum
deposited aluminium on mylar at IR frequencies. You never
use the numbers the manufacturer claims.

[Will Janoschka]

But which do you have? Both have 0% probability of existing.
So much for thought problems.

[Will Janoschka]

On Thu, 23 May 2013 04:32:21, Paul Aubrin <chu8...@free.fr> wrote:

How about very very light grey bodies :)
>

[RedAcer]

Don't be daft. How do you think photons are reflected. They have to be
absorbed by electrons in the surface layers and re-emitted.

>
> Make things shiney, and they automagically refuse any thing
> about the S-B equation, which requires no refectivity.
>>
>>>
>>>> Why are there spacers to keep the layers apart?
>>>
>>> To prevent thermal conductivity.
>>>
>>>> Why are there holes in the layers?
>>>
>>> sometimes this prevents outgassing problems
>>
>> No. It's to allow out-gassing you want there
>> to be a vacuum between the> layers.
>
> That is the out-gassing problem with no
> small holes. -idiot-

Idiot. They *want* the air to escape from between the layers into space
so that there is a vacuum between them. Go check the NASA web-pages.
Work out for yourself why they want a vacuum there.

[Paul Aubrin]

On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:

>>> Good Luck!
>>
>> Do you have any other problems with Paul's statements?
>
> Yes all of them.

Here are a few points I would like to better understand.

First let us use define a bit more precisely what all this is about.
There are two graphs on the web page below (link from emoneyjoe). The
second graph explains the "natural greenhouse effect".
http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
Quote:
"The natural greenhouse effect raises the Earth’s surface temperature to
about 15 degrees Celsius on average—more than 30 degrees warmer than it
would be if it didn’t have an atmosphere. The amount of heat radiated
from the atmosphere to the surface (sometimes called “back radiation”) is
equivalent to 100 percent of the incoming solar energy."

(1) On this graph,155% of the energy flux emitted by the Sun reach the
ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
from the atmosphere. Do you agree that, according to this diagram, the
ground creates this "back-radiation" energy itself?

(2) do you agree that this natural greenhouse effect "back-radiation"
flux exactly cancels the supplement of energy emitted by the ground in
excess of the energy received by the Sun?

(3) Do you agree that, in the above diagram, it would have been possible
to assign any value to this natural greenhouse effect "back-radiation"
flux without violating the energy balance, since it exactly cancel itself?

(4) Do you agree that there is no known way to capture a part of this
natural greenhouse effect "back-radiation" flux to produce any actual
physical effect (physical work, electrical power, phase change in a
liquid...)? If you disagree, please cite a kind of captor that would do
produce the actual effect.

[RedAcer]

I would guess that NASA buys mylar films built to a standard. In any
case they would obviously test the MLI before attaching it to a billion
dollar probe. Don't you think?

>

[RedAcer]

No material has an emissivity of zero AFAIK. Most bodies are 'grey'
bodies. ie they reflect and also emit according to stefan-boltzman. All
you do is put an emissivity coefficient in the equations.

How does a cold piece of mylar reflect IR photons to something that is
hotter? The photons have to be absorbed by the surface of the colder
body and then re-emitted. You said that a hot body cant absorb IR coming
from a colder body.

Check these out.
http://en.wikipedia.org/wiki/Emissivity
- covers two walls facing each other with differing emissivities

http://en.wikipedia.org/wiki/Thermal_radiation
- explains radiative heat transfer

http://en.wikipedia.org/wiki/Multi-layer_insulation

[RedAcer]

On 24/05/13 07:25, Paul Aubrin wrote:
> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>
>>>> Good Luck!
>>>
>>> Do you have any other problems with Paul's statements?
>>
>> Yes all of them.
>
> Here are a few points I would like to better understand.
>
> First let us use define a bit more precisely what all this is about.
> There are two graphs on the web page below (link from emoneyjoe). The
> second graph explains the "natural greenhouse effect".
> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
> Quote:

> "The natural greenhouse effect raises the Earth�s surface temperature to
> about 15 degrees Celsius on average�more than 30 degrees warmer than it
> would be if it didn�t have an atmosphere. The amount of heat radiated
> from the atmosphere to the surface (sometimes called �back radiation�) is

> equivalent to 100 percent of the incoming solar energy."
>
> (1) On this graph,155% of the energy flux emitted by the Sun reach the
> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
> from the atmosphere. Do you agree that, according to this diagram, the
> ground creates this "back-radiation" energy itself?
>
> (2) do you agree that this natural greenhouse effect "back-radiation"
> flux exactly cancels the supplement of energy emitted by the ground in
> excess of the energy received by the Sun?
>
> (3) Do you agree that, in the above diagram, it would have been possible
> to assign any value to this natural greenhouse effect "back-radiation"
> flux without violating the energy balance, since it exactly cancel itself

At first glance their seem to be a few errors/typos on that page. The
problem with some web-pages are that they are not produced by
specialists in the field. eg 100% of the incoming solar is *not*
reflected as back-radiation as the page says.

If you want to understand this stuff I would start with wikipedia and
then work your way up to those notes I posted.

>
> (4) Do you agree that there is no known way to capture a part of this
> natural greenhouse effect "back-radiation" flux to produce any actual
> physical effect (physical work, electrical power, phase change in a
> liquid...)? If you disagree, please cite a kind of captor that would do
> produce the actual effect.

Why not check out the equipment used to detect the 3K CMBR
<http://en.wikipedia.org/wiki/CMBR>
To measure back-radiation use a shielded IR detector cooled down to
liquid helium temperatures and have a shielded aperture/funnel/aerial
which only collects radiation from the sky.

>

[Paul Aubrin]

On Fri, 24 May 2013 07:35:24 +0100, RedAcer wrote:

>> You didn't mention in your "two plates in the void" exercice that one
>> of them was a mirror, you said: black-body. Mirrors (by definition)
>> reflect,
>> black surfaces (by definition) absorbs. More straw man fallacies?
>
> No material has an emissivity of zero AFAIK. Most bodies are 'grey'
> bodies. ie they reflect and also emit according to stefan-boltzman. All
> you do is put an emissivity coefficient in the equations.
>
> How does a cold piece of mylar reflect IR photons to something that is
> hotter? The photons have to be absorbed by the surface of the colder
> body and then re-emitted. You said that a hot body cant absorb IR coming
> from a colder body.

More straw man fallacies, I see. When you discussed your two plate
mental experiment, they were supposed to be perfect black surfaces. If it
had been an experiment about metallized reflexive mylar films, I would
have considered it in my answer.

[Paul Aubrin]

On Fri, 24 May 2013 07:23:54 +0100, RedAcer wrote:

>>>> sometimes this prevents outgassing problems
>>>
>>> No. It's to allow out-gassing you want there to be a vacuum between
>>> the> layers.
>>
>> That is the out-gassing problem with no small holes. -idiot-
>
> Idiot. They *want* the air to escape from between the layers into space
> so that there is a vacuum between them. Go check the NASA web-pages.
> Work out for yourself why they want a vacuum there.

I detect that, here again, you suppose that he said something that he
didn't meant. When I read what he wrote I understood that he wanted to
mention that holes would prevent problems caused if degassing occurred
which is exactly the same as you say now. If the way he expressed seemed
ambiguous, ask for a clarification.

[Dawlish]

On Monday, May 20, 2013 5:02:31 PM UTC+1, Paul Aubrin wrote:
> On Mon, 20 May 2013 10:59:04 -0400, emoneyjoe wrote:
>
>
>
> > On Mon, 20 May 2013 09:56:04 +0100, RedAcer <rred...@gmail.com> wrote:
>
> >
>
> >>Some great course notes here for anyone who wants to understand more
>
> >>about how Atmospheres work. The full course notes are there along with
>
> >>exercises and solutions so that you can test your knowledge.
>
> >>
>
> >><http://nit.colorado.edu/atoc5560/>
>
> >
>
> > Yeah, everybody is going to head to
>
> > Colorado and take that course.
>
> >
>
> > But not me, Colorado is too cold.
>
>
>
> It is an online course. As Red Acer promised to study it, we are going to
>
> have holidays for a while. When he returns he will be turned into a
>
> sceptic.
>
> I skimmed over the pdf files and could see no back-radiation concept
>
> there. The energy flux seemed proportional to the difference of radiances
>
> (and thus would flow from warm to cold in the case of black bodies)
>
> everything which appears as normal physics, not as post-normal climate-
>
> science Red Acer's physics.

RedAcer is a sceptic now. It's the huge difference between himself and you; an out and out climate denier, that is entertaining

[Tom P]

I don't that is quite correct, strictly speaking. In the classical view
of electromagnetics, the incoming wave induces a dipole moment in the
atoms at the surface. The dipole moment in turn produces radiation.
If the incoming light beam is collimated, then the dipole moments of the
adjacent atoms along the wave front are in phase, and hence the
radiation is perceived as being a collimated reflected light beam.

[Paul Aubrin]

On Fri, 24 May 2013 08:05:49 +0100, RedAcer wrote:

>> (3) Do you agree that, in the above diagram, it would have been
>> possible to assign any value to this natural greenhouse effect
>> "back-radiation" flux without violating the energy balance, since it
>> exactly cancel itself
>
> At first glance their seem to be a few errors/typos on that page. The
> problem with some web-pages are that they are not produced by
> specialists in the field. eg 100% of the incoming solar is *not*
> reflected as back-radiation as the page says.

I see that you didn't take the time to understand that those percentages
were meant to express energy flux using the incoming solar flux as a unit
(100%).

>
> If you want to understand this stuff I would start with wikipedia and
> then work your way up to those notes I posted.

The wikipedia links you posted don't give any answer to the questions
here.

>> (4) Do you agree that there is no known way to capture a part of this
>> natural greenhouse effect "back-radiation" flux to produce any actual
>> physical effect (physical work, electrical power, phase change in a
>> liquid...)? If you disagree, please cite a kind of captor that would do
>> produce the actual effect.
>
> Why not check out the equipment used to detect the 3K CMBR
> <http://en.wikipedia.org/wiki/CMBR>
> To measure back-radiation use a shielded IR detector cooled down to
> liquid helium temperatures and have a shielded aperture/funnel/aerial
> which only collects radiation from the sky.

Cooled to liquid helium temperatures... If you cool the receptor near the
absolute zero, the net radiative flux will be very close to the number
you derive from the emittance.

Liquid helium 4.22^4=317
14°C 287^4=6.8 10^9 >> 317

Solar cells can convert the radiations coming from the Sun in electrical
power. Is there a way to convert the "back-radiation" (324Wm-2) IR flux
into something detectable (without creating a net flux by cooling the
detector).

[RedAcer]

On 24/05/13 08:53, Paul Aubrin wrote:
> On Fri, 24 May 2013 07:35:24 +0100, RedAcer wrote:
>
>>> You didn't mention in your "two plates in the void" exercice that one
>>> of them was a mirror, you said: black-body. Mirrors (by definition)
>>> reflect,
>>> black surfaces (by definition) absorbs. More straw man fallacies?
>>
>> No material has an emissivity of zero AFAIK. Most bodies are 'grey'
>> bodies. ie they reflect and also emit according to stefan-boltzman. All
>> you do is put an emissivity coefficient in the equations.
>>
>> How does a cold piece of mylar reflect IR photons to something that is
>> hotter? The photons have to be absorbed by the surface of the colder
>> body and then re-emitted. You said that a hot body cant absorb IR coming
>> from a colder body.
>
> More straw man fallacies, I see. When you discussed your two plate
> mental experiment, they were supposed to be perfect black surfaces.

No they weren't.If you read thru my many posts, I have a choose them to
be black-bodies for simplicity and to make the maths less complicated. I
have said several times that you can put the emissivities in to treat
the case of grey bodies.

> If it
> had been an experiment about metallized reflexive mylar films, I would
> have considered it in my answer.

OK then consider it. Can a cold MLI layer emit/reflect/produce IR and
have it absorbed by a hotter film. If not then how does the MLI keep the
heat in.
>
>

[RedAcer]

I don't understand what you are trying to say here. Lets get back to
your statement, before we move on.

You said above "there is no known way to capture a part of this

natural greenhouse effect "back-radiation" flux to produce any actual
physical effect"

Do you accept that the IR from GHG's can be detected?
If not how dir Arno Penzias and Robert Wilson detect the 3K radiation
which came though the atmosphere from the universe - for which thye got
the Nobel prize.

>

[RedAcer]

Yes. You're right. Posted in haste to the daft response that it was
"refusal to absorb or emit". Feynman in his popular book "QED...." gives
a great explanation of how a single photon interacts with vast number of
atoms at the surface of a mirror or lense.
In the case of absorption by a crystal the photon effectively transfers
its energy to all the atoms in the crystal when it creates a phonon.

[emoneyjoe]

On Fri, 24 May 2013 08:05:49 +0100, RedAcer <rred...@gmail.com> wrote:

>On 24/05/13 07:25, Paul Aubrin wrote:
>> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>>
>>>>> Good Luck!
>>>>
>>>> Do you have any other problems with Paul's statements?
>>>
>>> Yes all of them.
>>
>> Here are a few points I would like to better understand.
>>
>> First let us use define a bit more precisely what all this is about.
>> There are two graphs on the web page below (link from emoneyjoe). The
>> second graph explains the "natural greenhouse effect".
>> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>> Quote:

>> "The natural greenhouse effect raises the Earth’s surface temperature to
>> about 15 degrees Celsius on average—more than 30 degrees warmer than it
>> would be if it didn’t have an atmosphere. The amount of heat radiated
>> from the atmosphere to the surface (sometimes called “back radiation”) is

Why would you want to measure the
one way thermal emmisivity-radiance?

And how can it be known how far
from the detector the radiating molecules
are?

It is likely only the radiation of a few
hundred meters of the lower troposphere
will be measured, warm air is warm air,
and the stratosphere can be 100 degrees
colder.

Has anybody made the same measurement
inside an unheated domed stadium?

[Paul Aubrin]

On Fri, 24 May 2013 09:44:40 +0100, RedAcer wrote:

>> Cooled to liquid helium temperatures... If you cool the receptor near
>> the absolute zero, the net radiative flux will be very close to the
>> number you derive from the emittance.
>>
>> Liquid helium 4.22^4=317 14°C 287^4=6.8 10^9 >> 317
>>
>> Solar cells can convert the radiations coming from the Sun in
>> electrical power. Is there a way to convert the "back-radiation"
>> (324Wm-2) IR flux into something detectable (without creating a net
>> flux by cooling the detector).
>
> I don't understand what you are trying to say here. Lets get back to
> your statement, before we move on.
>
> You said above "there is no known way to capture a part of this natural
> greenhouse effect "back-radiation" flux to produce any actual physical
> effect"
>
> Do you accept that the IR from GHG's can be detected?
> If not how dir Arno Penzias and Robert Wilson detect the 3K radiation
> which came though the atmosphere from the universe - for which thye got
> the Nobel prize.
>

Could you please stick to the question? I am not interested in the 3K
background radiation. I am interested in the "GHG back-radiation". As you
suppose it is real power, there should be way to convert it, at least
partly, into electrical power or mechanical work.

[emoneyjoe]

My bathroom cove heater (7 inches from ceiling)
is a 451 watt radiant heater. It is about 3 feet long
and about 9 inches wide.

If I supply lower voltage so that it draws about
324 watts per square meter, how warm do you
think it will measure?

As a rough guess, it is about a fifth
of a meter in area, but there is substantial
heat from the back side, so maybe about
a third (150 watts) would be equal to
324 watts/m^2.

It is way too hot to touch at 451 watts.

[Martin Brown]

You have to work very hard to do it since 10um thermal band photons
carry a puny amount of energy but it can and has been done and at room
temperature too. See for example this one based on microcantilevers to
acting as micro calorimeters of the incident thermal band radiation.

http://spie.org/x8622.xml

Signal to noise isn't wonderful but it does demonstrate that you can.

Ditto for measuring the atmospheric emission spectrum against a dark
night sky. There are spectrometers with detectors that can detect
thermal photons hence the back radiation is detectable at the ground.

But you are never going to be able to boil a kettle with them...

--
Regards,
Martin Brown

[RedAcer]

We can get to that later.
There is no point going on unless you accept that the 3K CMBR radiation
was detected and you didn't answer this question from above :-

You said above: "there is no known way to capture a part of this natural
greenhouse effect "back-radiation" flux to produce any actual physical
effect"

Do you now accept that this is wrong and that the IR from GHG's can be
detected?

[RedAcer]

Ahhh. Well then. According to Paul that means they don't exist :)

>

[Paul Aubrin]

On Fri, 24 May 2013 15:31:17 +0100, RedAcer wrote:

>> Could you please stick to the question? I am not interested in the 3K
>> background radiation. I am interested in the "GHG back-radiation". As
>> you suppose it is real power, there should be way to convert it, at
>> least partly, into electrical power or mechanical work.
>
> We can get to that later.

Suppose that you devised a device that can convert into actual energy
some part (say 20%) of the background radiation which, according to the
"natural greenhouse back-radiation" concept flows back and forth between
two black surfaces at the temperature of the your room (15°C). For
example, you devised a CO2 laser powered by the "background radiation".
You place your device between the two blackened plates and measure an
outgoing 324 x 0.2 = 64.8W energy flow emitted by the laser beam. How
will you be able to keep your two plates at the room temperature?

[Tom P]

It would be quite an achievement. The photovoltaic effect depends on the
ability of the photon to excite electrons into the conduction band. You
would probably have to cool the material to substantially below room
temperature, which would defeat the object of extracting usable energy.
Of course, you can make use of the temperature differential between
the surface and the atmosphere - the convection currents get converted
into winds by the coriolis effect, and the wind can be used to generate
power.
Alternatively, you could try the experiment on yourself- your body
radiates a few hundred watts all the time..

[Paul Aubrin]

On Fri, 24 May 2013 19:05:27 +0200, Tom P wrote:

>> Could you please stick to the question? I am not interested in the 3K
>> background radiation. I am interested in the "GHG back-radiation". As
>> you suppose it is real power, there should be way to convert it, at
>> least partly, into electrical power or mechanical work.
>>
>>
> It would be quite an achievement.

It is impossible: not a single Joule can be extracted from an energy
loop.

[Tom P]

On 05/24/2013 08:25 AM, Paul Aubrin wrote:
> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>
>>>> Good Luck!
>>>
>>> Do you have any other problems with Paul's statements?
>>
>> Yes all of them.
>
> Here are a few points I would like to better understand.
>
> First let us use define a bit more precisely what all this is about.
> There are two graphs on the web page below (link from emoneyjoe). The
> second graph explains the "natural greenhouse effect".
> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
> Quote:
> "The natural greenhouse effect raises the Earth’s surface temperature to
> about 15 degrees Celsius on average—more than 30 degrees warmer than it
> would be if it didn’t have an atmosphere. The amount of heat radiated
> from the atmosphere to the surface (sometimes called “back radiation”) is
> equivalent to 100 percent of the incoming solar energy."
>
> (1) On this graph,155% of the energy flux emitted by the Sun reach the
> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
> from the atmosphere. Do you agree that, according to this diagram, the
> ground creates this "back-radiation" energy itself?
>
> (2) do you agree that this natural greenhouse effect "back-radiation"
> flux exactly cancels the supplement of energy emitted by the ground in
> excess of the energy received by the Sun?
>

It's simpler to discuss this in terms of absolute values rather than
percentages as on the NASA page, because you have to keep track of
"percentages of what". You can find Trenberths' diagram on the Liou
presentation http://nit.colorado.edu/atoc5560/week13.pdf page 16.

> (3) Do you agree that, in the above diagram, it would have been possible
> to assign any value to this natural greenhouse effect "back-radiation"
> flux without violating the energy balance, since it exactly cancel itself?
>

Well just look at the numbers at the surface:
Energy incoming: 168+324 = 492 W/m2
Energy outgoing; 24+78+390 = 492 W/m2
What was your problem again with the energy balance? If you really
think that the back radiation is zero, then your supposed energy balance
would look like this:
Energy incoming: 168+0 = 168 W/m2
Energy outgoing; ??+??+?? = 168 W/m2

Now Paul, you have a choice of what numbers you think should go in the
second line. If you accept that thermals and transpiration are 24+78,
then that leaves you with 66 W/m2 emitted by the surface radiation- that
would be a black body temperature of around 66 Kelvin.
Even if the transpiration and thermal transport were zero, you would
need an S-B temperature of around 167 kelvin to balance the equation.

So Paul, tell us what numbers you think should go in the second line.

[Tom P]

So why do you ask stupid questions?

[Desertphile]

On Mon, 20 May 2013 09:56:04 +0100, RedAcer <rred...@gmail.com>
wrote:

> Some great course notes here for anyone who wants to understand more
> about how Atmospheres work. The full course notes are there along with
> exercises and solutions so that you can test your knowledge.
>
> <http://nit.colorado.edu/atoc5560/>
>

> Description:
> This course will cover the basic physics of, and computational methods
> for, the interaction of visible and infrared radiation with the gases
> and particles in planetary atmospheres. The applications will be mainly
> in the role radiation plays in determining planetary temperature
> structure and in climate change. The emphasis will be on the Earth's
> atmosphere.
>
> Topics include:
> Basic radiative transfer
> Line absorption by molecules
> Scattering by particles
> Approximate and exact radiative transfer methods
> Radiative equilibrium models
> Trace gases, aerosols, clouds, and climate
>
> *Contents*
>
> Part I:
> Introduction to Radiative Transfer
> Radiative definitions
> Absorption, emission, and Planck functions
> Radiative transfer equation
> Simple radiative transfer solutions
> Part II:
> Radiation and Gases
> Temperature structure and gas composition profiles
> Rotational and vibrational molecular transitions
> Absorption line shapes and broadening mechanisms
> Absorption line intensities
> Atmospheric absorption spectrum
> Line-by-line models
> Integrated single line transmission; random band models
> The k - distribution and correlated k-distribution methods
> Clear sky long-wave radiative transfer
> Heating rates: calculation and results
> Weighting functions and the cooling to space approximation
> Part III:
> Radiation and Particles
> Particle size distributions
> Cross sections and phase functions
> Polarization of radiation
> Rayleigh scattering
> Mie scattering (method and results for aerosols and water droplets)
> Index of refraction of materials
> Discrete Dipole Approximation
> Geometric optics scattering
> Part IV:
> Radiative Transfer with Scattering
> Radiative transfer equation with scattering of sunlight
> First order scattering
> Eddington and two-stream approximate methods
> Delta scaling of the radiative transfer equation
> Discrete ordinates solution methods
> Principle of interaction; doubling-adding method
> Surface reflection: Lambertian, Fresnel, general BRDF
> Plane-parallel radiative transfer results
> 3D radiative transfer: Monte Carlo method and results
> Part V:
> Radiative Equilibrium and Atmospheric Applications
> Sun-Earth geometry and solar insolation; solar measurements
> Earth's radiative energy budget and measurement
> Radiative Equilibrium:
> Gray models (single layer and profile)
> Non gray models, the atmospheric window
> 1D radiative convective equilibrium; temperature structure
> Radiative forcing and climate feedbacks
> Cloud radiation interactions
> Aerosols radiative forcing
> Greenhouse gas warming
> Planetary applications

Ah, thank you. I covered a few of these in my two videos on the
subject.

--
"God wanted me to be in bed with the maid."

[Desertphile]

On Mon, 20 May 2013 10:59:04 -0400, emoneyjoe <emon...@iglou.com>
wrote:

> Yeah, everybody is going to head to Colorado and take that course.

Huh? Idiot.

[Tom P]

There is no such thing as an "unheated domed stadium". What do you want
to do, measure the temperature of the roof?

[Desertphile]

On Tue, 21 May 2013 23:57:04 -0700 (PDT), Claudius Denk
<claudi...@sbcglobal.net> wrote:

> Do they allow non-believers?

They only allow non-believers: that's what colleges are for. You just
stay in your church where you belong.

[Tom P]

I fail to see the relevance of metalized reflective mylar films. The
topic is Radiative Processes in Planetary Atmospheres.

[Paul Aubrin]

Because it was funny to read your answers: yes the loop is real energy,
but sooooo elusive.... or "it is just like the universe background
radiation" (like saying that 3^4 and 287^4 are approximately of the same
magnitude).

[Tom P]

So how do you balance 492 W/m2 leaving the earth's surface with 168
absorbed by the surface?
Recall:
Leaving the surface:
Thermals 24 W/m2 + transpiration 78 W/m2 + radiation 390 W/m2
= 492 W/m2
Absorbed at the surface from incoming solar radiation:
incoming solar 342 W/m2 - reflected: 107 W/m2 - atmospheric
absorption: 67 W/m2
= 168 W/m2

Paul, it's your turn to provide the numbers.

[Paul Aubrin]

On Fri, 24 May 2013 21:07:15 +0200, Tom P wrote:

> Well just look at the numbers at the surface:
> Energy incoming: 168+324 = 492 W/m2 Energy outgoing; 24+78+390 = 492
> W/m2 What was your problem again with the energy balance? If you really
> think that the back radiation is zero, then your supposed energy balance
> would look like this:
> Energy incoming: 168+0 = 168 W/m2 Energy outgoing; ??+??+?? = 168 W/m2
>
> Now Paul, you have a choice of what numbers you think should go in the
> second line. If you accept that thermals and transpiration are 24+78,
> then that leaves you with 66 W/m2 emitted by the surface radiation- that
> would be a black body temperature of around 66 Kelvin.
> Even if the transpiration and thermal transport were zero, you would
> need an S-B temperature of around 167 kelvin to balance the equation.
>
> So Paul, tell us what numbers you think should go in the second line.

The heating plates of my electrical stove have a surface of 0.03 m2 and
an electrical power of 1200W: 40,000W/m2 (Stefan Boltzman temperature:
643°C). If I put nothing on them, the plates are red hot: the warm air is
quickly replaced by colder air by convection, the difference of
temperature is some 620°C, some 1080W are radiated, some 120W are
evacuated by conduction and convection. If I put a saucepan full of
water on it, the temperature at the surface of the heating plate is only
100°C (SB: 1100W/m2 0.03m2 33W radiated), yet the total energy output is
still 1200W, 33W radiated, 1177W evacuated by conduction. Conclusion: in
the presence of conduction, convection or phase change, there is no need
of a radiative equilibrium.
In the atmosphere, the radiative equilibrium takes place at the top of
the atmosphere (around an altitude of 6000m). The warmer temperature of
the air at the surface results from the lapse rate. In absence of
vegetation or moisture, the temperature of the ground can be hotter than
the temperature of the air (dark coloured sand for example) because it
has to radiate much more, just like the stove.

[Paul Aubrin]

The radiative balance is at the top of the atmosphere.
From there the temperature of the air increases by 0,6°-0,7°C each 100m.

[emoneyjoe]

If there is two black body plates radiating
toward each other (never mind all the other
stuff), and a thin plate is place half way in
between them, what would it's temperature
go to?

If IR radiation worked like the warmie GBs
claim, why wouldn't the thin plate get warmer
than the other two plates, it would be absorbing
thermal energy from both sides.

So doesn't this suggest that somehow
the radiation transfer doesn't take place?

A careful observation of this should
show if the thin plate in the center would
warm above the temperature of the two
outer plates before it begins to radiate
enough to prevent heating up more.


Something has to control the absorption,
or something has to control the emitting,
even though that seems impossible.
The thin plate receiving energy from
both sides has to begin radiating as
fast as it absorbs in order to nor heat up.

But once temperatures equalize,
there is considered no thermal transfer.

[emoneyjoe]

Where did you get those numbers?

http://science-edu.larc.nasa.gov/energy_budget/pdf/Energy_Budget_Poster_04_18_12.pdf

Shows 163.3 absorbed by the surface.
(That is warming energy)

The surface loses 18.4 by conduction,
and 86.4 by latent heat (evaporative cooling),
to the atmosphere.
And the surface loses 40.1 direct to space.

86.4 + 18.4 + 40.1 = 144.9

163.3 - 144.9 = 18.4

That's odd, losing 18.4 both by conduction
and IR radiation!

So the surface needs to lose 18.4 by IR
to the atmosphere.


The back radiation mystery of 340.3 plus
the 18.4 IR should equal the 398.2 emitted,
but it doesn't. 340.3 + 18.4 is only 358.7 .........

The claimed 0.6 absorbed by the surface
and retained makes it worse yet.


Can somebody tell me what I missed?

The Earth can't be cooling by almost 40 .......

[emoneyjoe]

Of course there is, don't turn the heat on
when it is cold out.

>What do you want
>to do, measure the temperature of the roof?

No, observe the reading on a cold night
looking up, and as the sun comes up and
heats the roof, it should be obvious if it
is the air in the stadium being measured,
or the roof.


What do you think the sky should measure,
the average between the surface temperature
and the 100 degree colder lower stratosphere?

[Will Janoschka]

Do the math -idiot-
Reflection is not absorption and-re-radiation it is "reflection", a
surface
property, not a property of the material.

I really do know how this works, unlike you.
>
> >
> >> Why are there spacers to keep the layers apart?
> > To prevent thermal conductivity.
> >> Why are there holes in the layers?
> > sometimes this prevents outgassing problems

> No. It's to allow out-gassing you want there to be a vacuum between the
> layers.
That is indeed "the" outgassing problem, hence the holes!
They must "not" line up between layers.

> >> According to your physics it can't work.
> > High reflectivity is not against any of Pau'ls remarks
> How is heat reflected back to something that is warmer. Paul says that
> the laws of physics don't allow it.

Paul is speaking of emission correctly, not of reflection.
> >
> >> So, how does it work?
> >
> > I just explained that.
>
> No you haven't. But you were nearly there.
>
Read again and try to understand.

> >
> >> Good Luck!
> >
> > Do you have any other problems with Paul's statements?
>
> Yes all of them.

Point out even one.

[Will Janoschka]

So true!!! Why not dress up my 3 sphere radiation demonstration
to prove "once and for all" that back radiation never occurs.

Let me repost this from the beginning of this year

Tom was correct in that anything that inhibits radiation from an
radiating object with a fixed amount of power to radiate will indeed
raise the needed temperature of that object as that is the only way to

overcome that very inhibition and radiate that fixed amount of power.

This is not back radiation nor Greenhouse effect. This is only the
way thermodynamics has been, and must be done.

For this I need to simplify the geometry so there are no weasel words
allowed. The parallel plates Tom uses, allow much weaseling.
The fight between Bill and Svante was all over such words

For this I will use the geometry of concentric spheres, with
a fixed emissivity over any temperature range used. Such
a sphere is an isotropic radiator with one cross sectional
area that always radiates to 4 pi steradians. No leaks.

So we can tell the direction of thermal flux each sphere
has some thermal conductivity between the inner and
outer surfaces. To be able to measure I suggest the
conductivity to be 9 mW/ cm-degree Celsius.and a
thickness. of 1 mm. I only use degree to indicate
"unit" Kelvin, rather than Kelvin. For radiation and
all thermodynamics, a huge difference in meaning(
(an interval rather than a value)t

Three concentric spheres 6 cm dia, 9 cm dia 12 cm dia.
6 cm has a 5 watt resistive element on the inside surface
the outside of the 6cm inside and outside and the 9 cm
and the inside if the 12 cm are constant 95% emissivity at
all significant wavelengths. each supported by threads
by the next larger sphere the 12 cm sphere outside is
chilled by suitable means to 0 Celsius. (For Kelvin add
273.15) The two inner spheres communicate via radio
waves to indicating inner and outer surface temperatures.
The 6 cm sphere has a battery and control to produce exactly
5 watts of thermal power. switchable via radio control. vacuum
between all shells so no transfer except by thermal radiation.also
with 0 conductive heat transfer by any wires With no power on
all surfaces are a the 0 Celsius ice water bath.
Here is what I claim as a result:

Outer sphere 0 Celsius radiance 0. 0300 watts/sq
cm-steradian
delta T inner to outer +1 degree
9 cm sphere 38 Celsius radiance 0.0497 watts/ sq cm-steradian
delta T inner to outer <+ 2 degree
6 cm sphere 93 Celsius radiance 0.0935 watts/ sq cm-steradian
delta T inner to outer +4 degree.

By the always decreasing temperature to the outside the thermal

flux is always outward. and no power is ever returned "to" a
warmer surface. Each surface takes on only that temperature
required to move 5 Watts outward. This is the "dynamics"
of thermodynamics


5 Watts of power always from resistor to 6 cm sphere conduction
5 Watts of power always from 6cm to 9 cm sphere thermal
radiation
5 Watts of power always from 9cm to 12 cm sphere thermal radiation
5 Watts of power always from 12cm sphere to bath conduction

The 5 Watts is the only power you have and none is returned

Radiance is not radiation . Radiation can only be 5 watts that
is all the heat energy you have. The same is true for the 9 cm
sphere. the 12 cm sphere is the eventual receiver of all 5 watts.
From this the temperatures of each surface can be calculated

How much a lower temperature of the 6cm sphere. if the 9cm
sphere were not there. The 6 cm sphere t cools to 68 Celsius

Is this the weird concept of Greenhouse effect caused by CO2?
It is only geometry. not Greenhouses!
On Earth, there is no downward, "flux",fFrom the atmosphere
only potential downward flux to some instrument at 0 Kelvin

[emoneyjoe]

Paul, it should be easy to argue that over 300
watts/m^2 could NOT be an average back radiation,
all the numbers in an Earth Energy Budget diagram
have to be averages.

And the actual "back radiation" at the poles
could not be anywhere close to even the average
claimed, the average is supposed to be one-fourth
the daytime high with the sun at zenith.


So the actual back radiation at the equator
would have to be at least 3 times the claimed
average, or more than 1000 watts/m^2.


Not in any GB alarmist's wildest dreams.

[Will Janoschka]

From Trenberths' Chart? Where are the measured numbers?
No measurements of actual heat transfer? Oh we will measure
radiance and use those. is that the way Tom?

> Energy incoming: 168+324 = 492 W/m2

Where does the 324 come from? Radiance of the sky?
None of that radiance is transfered to the warmer earth just to be
radiated
as part of the 390 below? that is circular, nonsense, and does not
happen..

> Energy outgoing; 24+78+390 = 492 W/m2
> What was your problem again with the energy balance? If you really
> think that the back radiation is zero, then your supposed energy balance
> would look like this:
>
> Energy incoming: 168+0 = 168 W/m2
> Energy outgoing; ??+??+?? = 168 W/m2


> Now Paul, you have a choice of what numbers you think should go in the
> second line. If you accept that thermals and transpiration are 24+78,
> then that leaves you with 66 W/m2 emitted by the surface radiation- that
> would be a black body temperature of around 66 Kelvin.
> Even if the transpiration and thermal transport were zero, you would
> need an S-B temperature of around 167 kelvin to balance the equation.
>
> So Paul, tell us what numbers you think should go in the second line.
>

Energy incoming: inWatts/M^2;
74 sensible heat to surface 72 latent heat not at surface = 156 W/m2
Energy outgoing in Watts/M^2;
24 convection to atmosphere,72 latent heat in atmosphere
38 to direct to space, 22 to atmosphere, then to space. = 156 W/m2

Your Idea of the Earth as anything like a black-body is nonsense.
measure something. No surfaces on the planet radiate to pi steradian.

[emoneyjoe]

On Sat, 25 May 2013 00:12:53 -0500, wil...@nospam.pobox.com (Will
Janoschka) wrote:

>On Fri, 24 May 2013 19:07:15, Tom P <wero...@freent.dd> wrote:
>
>> On 05/24/2013 08:25 AM, Paul Aubrin wrote:
>> > On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>> >
>> >>>> Good Luck!
>> >>>
>> >>> Do you have any other problems with Paul's statements?
>> >>
>> >> Yes all of them.
>> >
>> > Here are a few points I would like to better understand.
>> >
>> > First let us use define a bit more precisely what all this is about.
>> > There are two graphs on the web page below (link from emoneyjoe). The
>> > second graph explains the "natural greenhouse effect".
>> > http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>> > Quote:

>> > "The natural greenhouse effect raises the Earth’s surface temperature to
>> > about 15 degrees Celsius on average—more than 30 degrees warmer than it
>> > would be if it didn’t have an atmosphere. The amount of heat radiated
>> > from the atmosphere to the surface (sometimes called “back radiation†) is

That 324 watts is supposed to be an average,
so if it is a lot less at the poles, and all the numbers
are a fourth of the maximum, like the 1367/4 is
for the noon insolation, what would the maximum
back radiation be at noon at the equator?

[Will Janoschka]

On Fri, 24 May 2013 06:25:40, Paul Aubrin <chu8...@free.fr> wrote:
> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
> >> Do you have any other problems with Paul's statements?
> > Yes all of them.
> Here are a few points I would like to better understand.
>
> First let us use define a bit more precisely what all this is about.
> There are two graphs on the web page below (link from emoneyjoe). The
> second graph explains the "natural greenhouse effect".
> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
> Quote:

> "The natural greenhouse effect raises the Earth’s surface temperature to
> about 15 degrees Celsius on average—more than 30 degrees warmer than it
> would be if it didn’t have an atmosphere. The amount of heat radiated
> from the atmosphere to the surface (sometimes called “back radiation”) is

> equivalent to 100 percent of the incoming solar energy."
>
> (1) On this graph,155% of the energy flux emitted by the Sun reach the
> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
> from the atmosphere. Do you agree that, according to this diagram, the
> ground creates this "back-radiation" energy itself?
>

No such thing as back radiation.

> (2) do you agree that this natural greenhouse effect "back-radiation"
> flux exactly cancels the supplement of energy emitted by the ground in
> excess of the energy received by the Sun?

It would is there were such a thing. Even in that poor graph it must
be subtracted
to come close

>
> (3) Do you agree that, in the above diagram, it would have been possible
> to assign any value to this natural greenhouse effect "back-radiation"
> flux without violating the energy balance, since it exactly cancel itself?

It is probably some sort of whole sky radiance number,
but yes it can be any number.

>
> (4) Do you agree that there is no known way to capture a part of this
> natural greenhouse effect "back-radiation" flux to produce any actual
> physical effect (physical work, electrical power, phase change in a
> liquid...)? If you disagree, please cite a kind of captor that would do
> produce the actual effect.

Since such radiation is "not", it cannot power anything or even be
measured.
what can be measured is the amount of sensible heat absorbed in the
day
and radiated outward from the surface at night. the hard part is in
calibrating
such a device as it would depend on ground moisture. if you use a
plant
hygrometer and calibrate many moisture parameters, you can build and
calibrate your the whole thing by yourself, with cost proportional to
how
much automation you wish to have.

1. 10 cm long plastic pipe with "lotsa" small holes to let moisture
into the dirt.
semi rigid soaker hose may work fine.
2. two back to back thermocouples at each end
3. Some dirt just like you are going to measure, Home made core drill

pipe works fine. and leaves a very nice hole in the ground to
put
your measure device. The dirt goes between the thermocouples.
4. Voltmeter sensitive enough to measure 0.1 degree Celsius
temperature
difference between the two thermocouples. If I were building
this I
would invest in a small recording computer that can store
everything
on a flash drive.for analysis later. Get one that can record
the plant
hydrometer reading also. Remember the voltage will flip between.
absorbing and radiating.

By now you should have figured out the rest by yourself.
You are not really measuring radiative heat transfer. You are
measuring total sensible heat transfer between the surface
and 10 cm lower. If this is what you wish to measure. the stuff
to calibrate is also simple as you can use any measurable heat
source. Small resistors work just fine.

Paul, this is what the Climate Clowns should have started
with if they had "wanted" to learn, rather than preach the
Gospel according to the Church of the Anthropogenic.
Any of the engineers at JPL could show how to do such.

[Will Janoschka]

No such thing as back radiation.

Are you using insolation as a measurement for direct
and indirect solar energy to a calibrated solar panel.
That is direct and indirect radiance from the sun?
That number 324 is nowhere close to what is
absorbed by the surface. Your Climate Clowns
have not bothered to measure the actual absorption
and emission on any part of the Earth. The numbers
I put in are my best estimates, These I trust much more
than anything coming from Jimmy Hansen's playstation-64

I have no idea of where you get the figure that the average
is 1/4 of the maximum.except that the total surface area of the
earth is approximately four times the cross sectional area of a
sphere. Neglecting the fact that the real surface has fractal
dimensions.
>
>
>
>
>

[emoneyjoe]

On Sat, 25 May 2013 02:29:36 -0500, wil...@nospam.pobox.com (Will

Janoschka) wrote:

>On Sat, 25 May 2013 05:28:06, emoneyjoe <emon...@iglou.com> wrote:
>
>> On Sat, 25 May 2013 00:12:53 -0500, wil...@nospam.pobox.com (Will
>> Janoschka) wrote:
>>
>> >On Fri, 24 May 2013 19:07:15, Tom P <wero...@freent.dd> wrote:
>> >
>> >> On 05/24/2013 08:25 AM, Paul Aubrin wrote:
>> >> > On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>> >> >
>> >> >>>> Good Luck!
>> >> >>>
>> >> >>> Do you have any other problems with Paul's statements?
>> >> >>
>> >> >> Yes all of them.
>> >> >
>> >> > Here are a few points I would like to better understand.
>> >> >
>> >> > First let us use define a bit more precisely what all this is about.
>> >> > There are two graphs on the web page below (link from emoneyjoe). The
>> >> > second graph explains the "natural greenhouse effect".
>> >> > http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>> >> > Quote:

>> >> > "The natural greenhouse effect raises the Earth’s surface temperature to
>> >> > about 15 degrees Celsius on average more than 30 degrees warmer than it
>> >> > would be if it didn’t have an atmosphere. The amount of heat radiated
>> >> > from the atmosphere to the surface (sometimes called “back radiation†) is

In the Earth Energy Budget diagrams,
they divide the 1367 watt/m^2 solar insolation
by 2 because of day and night, and then
divide the result by 2 because of the
sun angles at higher latitudes.

So any of the other numbers are
1/4th the maximum at high noon at
the equator too.

I don't get the figure, I am just saying
the climate clowns claim numbers that
are simply not possible.

[RedAcer]

On 25/05/13 02:32, Will Janoschka wrote:
> On Thu, 23 May 2013 08:38:25, RedAcer <rred...@gmail.com> wrote:
>

...

I agree 'absorption and re-radiation' is a bit clumsy, but there is no
point in going into too much detail as Paul has little knowledge of physics.
So you accept that radiation can be 'reflected' off a cold body and be
absorbed by a warmer body.

> property, not a property of the material.
>
> I really do know how this works, unlike you.

Go on then. Please explain how a photon is reflected from a surface.

[RedAcer]

On 25/05/13 07:57, Will Janoschka wrote:
> On Fri, 24 May 2013 06:25:40, Paul Aubrin <chu8...@free.fr> wrote:
>> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>>>> Do you have any other problems with Paul's statements?
>>> Yes all of them.
>> Here are a few points I would like to better understand.
>>
>> First let us use define a bit more precisely what all this is about.
>> There are two graphs on the web page below (link from emoneyjoe). The
>> second graph explains the "natural greenhouse effect".
>> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>> Quote:

>> "The natural greenhouse effect raises the Earth’s surface temperature to
>> about 15 degrees Celsius on average—more than 30 degrees warmer than it
>> would be if it didn’t have an atmosphere. The amount of heat radiated
>> from the atmosphere to the surface (sometimes called “back radiation†) is

>> equivalent to 100 percent of the incoming solar energy."
>>
>> (1) On this graph,155% of the energy flux emitted by the Sun reach the
>> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
>> from the atmosphere. Do you agree that, according to this diagram, the
>> ground creates this "back-radiation" energy itself?
>>
> No such thing as back radiation.

Would it help you if it was called back reflection?

[Will Janoschka]

On Sat, 25 May 2013 09:16:20, RedAcer <rred...@gmail.com> wrote:

> On 25/05/13 07:57, Will Janoschka wrote:
> > On Fri, 24 May 2013 06:25:40, Paul Aubrin <chu8...@free.fr> wrote:
> >> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
> >>>> Do you have any other problems with Paul's statements?
> >>> Yes all of them.
> >> Here are a few points I would like to better understand.
> >>
> >> First let us use define a bit more precisely what all this is about.
> >> There are two graphs on the web page below (link from emoneyjoe). The
> >> second graph explains the "natural greenhouse effect".
> >> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
> >> Quote:

> >> "The natural greenhouse effect raises the Earth’s surface temperature to
> >> about 15 degrees Celsius on average—more than 30 degrees warmer than it
> >> would be if it didn’t have an atmosphere. The amount of heat radiated
> >> from the atmosphere to the surface (sometimes called “back radiation”) is

> >> equivalent to 100 percent of the incoming solar energy."
> >>
> >> (1) On this graph,155% of the energy flux emitted by the Sun reach the
> >> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
> >> from the atmosphere. Do you agree that, according to this diagram, the
> >> ground creates this "back-radiation" energy itself?
> >>
> > No such thing as back radiation.
>
> Would it help you if it was called back reflection?
>

Demonstrate that it can ever happen!
Then you can call what happens anything you like.

Demonstrate also that the atmosphere is warmer
at any altitude or that the total spectral transmissivity
has changed since before the Climate Clown Fraud/.
Do you have some problem with measuring things rather
than fishing your answers from the toilet?

[RedAcer]

On 25/05/13 11:25, Will Janoschka wrote:
> On Sat, 25 May 2013 09:16:20, RedAcer <rred...@gmail.com> wrote:
>
>> On 25/05/13 07:57, Will Janoschka wrote:
>>> On Fri, 24 May 2013 06:25:40, Paul Aubrin <chu8...@free.fr> wrote:
>>>> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>>>>>> Do you have any other problems with Paul's statements?
>>>>> Yes all of them.
>>>> Here are a few points I would like to better understand.
>>>>
>>>> First let us use define a bit more precisely what all this is about.
>>>> There are two graphs on the web page below (link from emoneyjoe). The
>>>> second graph explains the "natural greenhouse effect".
>>>> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>>>> Quote:

>>>> "The natural greenhouse effect raises the Earth’s surface temperature to
>>>> about 15 degrees Celsius on average—more than 30 degrees warmer than it
>>>> would be if it didn’t have an atmosphere. The amount of heat radiated
>>>> from the atmosphere to the surface (sometimes called “back radiation†) is

>>>> equivalent to 100 percent of the incoming solar energy."
>>>>
>>>> (1) On this graph,155% of the energy flux emitted by the Sun reach the
>>>> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
>>>> from the atmosphere. Do you agree that, according to this diagram, the
>>>> ground creates this "back-radiation" energy itself?
>>>>
>>> No such thing as back radiation.
>>
>> Would it help you if it was called back reflection?
>>
>
> Demonstrate that it can ever happen!
> Then you can call what happens anything you like.
>
> Demonstrate also that the atmosphere is warmer
> at any altitude or that the total spectral transmissivity
> has changed since before the Climate Clown Fraud/.
> Do you have some problem with measuring things rather
> than fishing your answers from the toilet?
>

Have you worked out how reflection works yet?

[Will Janoschka]

> > surface property, not a property of the material.

>
> I agree 'absorption and re-radiation' is a bit clumsy, but there is no
> point in going into too much detail as Paul has little knowledge of physics.

Paul has much much more knowledge than you will ever have.

> So you accept that radiation can be 'reflected' off a cold body and be
> absorbed by a warmer body.

Not at all. If the reflectivity is 100% the temperature of such a
surface is
of no importance. However such spontaneous transfer of energy by
thermal
electromagnetic radiation would not happen, except if reflected "to"
a colder
emissive surface. Spontaneous heat transfer to a colder object never
happens, 2LTD. Demonstrated "also" for thermal radiated energy 1906,
again here n 2013. If the transfer is not spontaneous, the energy
source and
mechanism of transfer must be described.. Else you claim perpetual
motion.
2LTD holds whether the system is closed or open.

> >
> > I really do know how this works, unlike you.
>
> Go on then. Please explain how a photon is reflected from a surface.

Photons are not reflected as they only mediate the minimum energy
transfered if such transfer is allowed. Photons always have zero
proper time no matter what the distance. Photons are not bullets
radiated in all directions. Gauge boson.

TomP already explained how the electromagnetic field is reflected.

[Will Janoschka]

On Sat, 25 May 2013 08:12:35, emoneyjoe <emon...@iglou.com> wrote:

> On Sat, 25 May 2013 02:29:36 -0500, wil...@nospam.pobox.com wrote:
> >On Sat, 25 May 2013 05:28:06, emoneyjoe <emon...@iglou.com> wrote:

> >> On Sat, 25 May 2013 00:12:53 -0500, wil...@nospam.pobox.com wrote:
> >> >On Fri, 24 May 2013 19:07:15, Tom P <wero...@freent.dd> wrote:
> >> >> On 05/24/2013 08:25 AM, Paul Aubrin wrote:
> >> >> > On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
> >> >> >>>> Good Luck!
> >> >> >>> Do you have any other problems with Paul's statements?
> >> >> >> Yes all of them.
> >> >> > Here are a few points I would like to better understand.
> >> >> >
> >> >> > First let us use define a bit more precisely what all this is about.
> >> >> > There are two graphs on the web page below (link from emoneyjoe). The
> >> >> > second graph explains the "natural greenhouse effect".
> >> >> > http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
> >> >> > Quote:
> >> >> > "The natural greenhouse effect raises the Earth’s surface

> >> >> > temperature toabout 15 degrees Celsius on average more

> >> >> > than 30 degrees warmer than it
> >> >> > would be if it didn’t have an atmosphere. The amount
> >> >> > of heat radiated from the atmosphere to the surface

> >> >> > of (sometimes called “back radiation†) is

WoW! Now that is a childish prank right from the toilet.

>
> So any of the other numbers are
> 1/4th the maximum at high noon at
> the equator too.
>
> I don't get the figure, I am just saying
> the climate clowns claim numbers that
> are simply not possible.

You are correct and all the real numbers can be
measured ,anywhere if Climate Clowns wished to lean.
I supposed they only wish to preach the Gospel of
the Holy Church of the Anthropogenic.
>
>
>
>
>
>

[Paul Aubrin]

On Sat, 25 May 2013 10:14:27 +0100, RedAcer wrote:

>> Do the math -idiot-
>> Reflection is not absorption and-re-radiation it is "reflection", a
>> surface
>
> I agree 'absorption and re-radiation' is a bit clumsy, but there is no
> point in going into too much detail as Paul has little knowledge of
> physics.
> So you accept that radiation can be 'reflected' off a cold body and be
> absorbed by a warmer body.

Please read your course, week 1 page 11. It says:
Kirchoff law: emissivity=absorptivity.
A black body has emissivity=1.
A mirror has absorbtivity=emissivity=0.

http://nit.colorado.edu/atoc5560/

A metallized mylar film is a mirror (emissivity=0); it is not a black-
body (emissivity=1).

Please show a bit more of absorptivity instead of exhibiting too much the
bad quality of your reflectivity.

[RedAcer]

snip the worst bilge I have seen passed off as physics.
EOC.

[RedAcer]

On 25/05/13 13:03, Paul Aubrin wrote:
> On Sat, 25 May 2013 10:14:27 +0100, RedAcer wrote:
>
>>> Do the math -idiot-
>>> Reflection is not absorption and-re-radiation it is "reflection", a
>>> surface
>>
>> I agree 'absorption and re-radiation' is a bit clumsy, but there is no
>> point in going into too much detail as Paul has little knowledge of
>> physics.
>> So you accept that radiation can be 'reflected' off a cold body and be
>> absorbed by a warmer body.
>
> Please read your course, week 1 page 11. It says:
> Kirchoff law: emissivity=absorptivity.
> A black body has emissivity=1.
> A mirror has absorbtivity=emissivity=0.
>
> http://nit.colorado.edu/atoc5560/
>
> A metallized mylar film is a mirror (emissivity=0); it is not a black-
> body (emissivity=1).

It's rarely zero.

>
> Please show a bit more of absorptivity instead of exhibiting too much the
> bad quality of your reflectivity.

You still haven't explained this.
When a colder layer further away from the space craft 'reflects' the IR
radiation back towards a warmer layer does the warmer layer absorb it.
If not where does the radiation go and how does the MLI work in keeping
the heat inside the craft?

>

[Paul Aubrin]

On Sat, 25 May 2013 13:19:40 +0100, RedAcer wrote:

>> Please show a bit more of absorptivity instead of exhibiting too much
>> the bad quality of your reflectivity.
>
> You still haven't explained this.
> When a colder layer further away from the space craft 'reflects' the IR
> radiation back towards a warmer layer does the warmer layer absorb it.
> If not where does the radiation go and how does the MLI work in keeping
> the heat inside the craft?

You still haven't explained this:
If you cannot extract even a single joule from a looping energy flux, how
does the fine physicist you are, manage to put in evidence the "natural
greenhouse back-radiation" energy loop existence? What instrument of
measure do you use to prove that the absence of anything to observe is a
definite proof that the flux is real?

[emoneyjoe]

Grasping ...........

>> property, not a property of the material.
>>
>> I really do know how this works, unlike you.
>
>Go on then. Please explain how a photon is reflected from a surface.

To get the best reflection,

"Reduced Absorption of Visible Light:"

is the objective.

http://www.spectrum-coatings.com/coatings.htm

http://en.wikipedia.org/wiki/Dielectric_mirrors

http://en.wikipedia.org/wiki/Anti-reflective_coating


You really have to travel far to support
the goofy premise of "trapping" energy and
back radiation.

Without GHGs and without water, the
N2 and O2 would retain heat way too much
for comfort.

[Tom P]

Nice attempt to dodge the subject but we're talking about the energy
balance AT THE SURFACE.

[Tom P]

On 05/24/2013 10:33 PM, Paul Aubrin wrote:

> On Fri, 24 May 2013 21:07:15 +0200, Tom P wrote:
>
>> Well just look at the numbers at the surface:

>> Energy incoming: 168+324 = 492 W/m2 Energy outgoing; 24+78+390 = 492

>> W/m2 What was your problem again with the energy balance? If you really
>> think that the back radiation is zero, then your supposed energy balance
>> would look like this:
>> Energy incoming: 168+0 = 168 W/m2 Energy outgoing; ??+??+?? = 168 W/m2
>>
>> Now Paul, you have a choice of what numbers you think should go in the
>> second line. If you accept that thermals and transpiration are 24+78,
>> then that leaves you with 66 W/m2 emitted by the surface radiation- that
>> would be a black body temperature of around 66 Kelvin.
>> Even if the transpiration and thermal transport were zero, you would
>> need an S-B temperature of around 167 kelvin to balance the equation.
>>
>> So Paul, tell us what numbers you think should go in the second line.
>

> The heating plates of my electrical stove have a surface of 0.03 m2 and
> an electrical power of 1200W: 40,000W/m2 (Stefan Boltzman temperature:
> 643°C). If I put nothing on them, the plates are red hot: the warm air is
> quickly replaced by colder air by convection, the difference of
> temperature is some 620°C, some 1080W are radiated, some 120W are
> evacuated by conduction and convection. If I put a saucepan full of
> water on it, the temperature at the surface of the heating plate is only
> 100°C (SB: 1100W/m2 0.03m2 33W radiated), yet the total energy output is
> still 1200W, 33W radiated, 1177W evacuated by conduction. Conclusion: in
> the presence of conduction, convection or phase change, there is no need
> of a radiative equilibrium.
> In the atmosphere, the radiative equilibrium takes place at the top of
> the atmosphere (around an altitude of 6000m). The warmer temperature of
> the air at the surface results from the lapse rate. In absence of
> vegetation or moisture, the temperature of the ground can be hotter than
> the temperature of the air (dark coloured sand for example) because it
> has to radiate much more, just like the stove.
>

Translation - Paul dodges the question.

[Tom P]

On 05/24/2013 10:45 PM, hda wrote:

> On Fri, 24 May 2013 21:07:15 +0200, Tom P <wero...@freent.dd> wrote:
>
>> On 05/24/2013 08:25 AM, Paul Aubrin wrote:
>>> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>>>
>>>>>> Good Luck!
>>>>>
>>>>> Do you have any other problems with Paul's statements?
>>>>
>>>> Yes all of them.
>>>
>>> Here are a few points I would like to better understand.
>>>
>>> First let us use define a bit more precisely what all this is about.
>>> There are two graphs on the web page below (link from emoneyjoe). The
>>> second graph explains the "natural greenhouse effect".
>>> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>>> Quote:

>>> "The natural greenhouse effect raises the Earth’s surface temperature to
>>> about 15 degrees Celsius on average—more than 30 degrees warmer than it
>>> would be if it didn’t have an atmosphere. The amount of heat radiated
>>> from the atmosphere to the surface (sometimes called “back radiation”) is

>>> equivalent to 100 percent of the incoming solar energy."
>>>
>>> (1) On this graph,155% of the energy flux emitted by the Sun reach the
>>> ground (7%+48%+100%), of which 100% (340Wm-2) is the "back-radiation"
>>> from the atmosphere. Do you agree that, according to this diagram, the
>>> ground creates this "back-radiation" energy itself?
>>>
>>> (2) do you agree that this natural greenhouse effect "back-radiation"
>>> flux exactly cancels the supplement of energy emitted by the ground in
>>> excess of the energy received by the Sun?
>>>
>>
>> It's simpler to discuss this in terms of absolute values rather than
>> percentages as on the NASA page, because you have to keep track of
>> "percentages of what". You can find Trenberths' diagram on the Liou
>> presentation http://nit.colorado.edu/atoc5560/week13.pdf page 16.
>>
>>> (3) Do you agree that, in the above diagram, it would have been possible
>>> to assign any value to this natural greenhouse effect "back-radiation"
>>> flux without violating the energy balance, since it exactly cancel itself?
>>>

>> Well just look at the numbers at the surface:
>> Energy incoming: 168+324 = 492 W/m2
>> Energy outgoing; 24+78+390 = 492 W/m2
>> What was your problem again with the energy balance? If you really
>> think that the back radiation is zero, then your supposed energy balance
>> would look like this:
>> Energy incoming: 168+0 = 168 W/m2
>> Energy outgoing; ??+??+?? = 168 W/m2
>>
>> Now Paul, you have a choice of what numbers you think should go in the
>> second line. If you accept that thermals and transpiration are 24+78,
>> then that leaves you with 66 W/m2 emitted by the surface radiation- that
>> would be a black body temperature of around 66 Kelvin.
>> Even if the transpiration and thermal transport were zero, you would
>> need an S-B temperature of around 167 kelvin to balance the equation.
>>
>> So Paul, tell us what numbers you think should go in the second line.
>

> Tom, according to your above, your gray body, emitting 50W/m2 and
> e=0.3, has a T of 233K...
>
> You can calculate energy flux only on delta-T's. What T2 did you have
> in mind for calculating the other T1, together wit your indication of
> 66 W/m2 ?
>

Paul says that there is no back radiation. If that is the case then
there's no delta T. Which way do you want it? You can't have it both ways.

OTOH if you are talking about the Trenberth diagram then the numbers are
based on measured values. The upward flux of 390 corresponds to a black
body of 288 kelvin, which is not far off the average surface
temperature. A downward flux of 324 would correspond to 275 kelvin,
although in reality the spectrum of the downward flux is not really like
a black body.

[Tom P]

Didn't you see the quote? I gave an O/T reference to the numbers in
Liuo's course notes.
Bear in mind that these numbers are not cast in concrete but are best
estimates based on observations, consequently they can change depending
on how recent the research is.

> http://science-edu.larc.nasa.gov/energy_budget/pdf/Energy_Budget_Poster_04_18_12.pdf
>
> Shows 163.3 absorbed by the surface.
> (That is warming energy)
>
> The surface loses 18.4 by conduction,
> and 86.4 by latent heat (evaporative cooling),
> to the atmosphere.
> And the surface loses 40.1 direct to space.
>
> 86.4 + 18.4 + 40.1 = 144.9
>
> 163.3 - 144.9 = 18.4
>
> That's odd, losing 18.4 both by conduction
> and IR radiation!
>
> So the surface needs to lose 18.4 by IR
> to the atmosphere.
>
>
> The back radiation mystery of 340.3 plus
> the 18.4 IR should equal the 398.2 emitted,
> but it doesn't. 340.3 + 18.4 is only 358.7 .........
>
> The claimed 0.6 absorbed by the surface
> and retained makes it worse yet.
>
>
> Can somebody tell me what I missed?
>
> The Earth can't be cooling by almost 40 .......
>
>

Incoming 163.3+340.3 = 503.6
Outgoing 398.2+18.4+86.4 = 503
Check your arithmetic.

[Tom P]

On 05/25/2013 07:12 AM, Will Janoschka wrote:
> On Fri, 24 May 2013 19:07:15, Tom P <wero...@freent.dd> wrote:
>
>> On 05/24/2013 08:25 AM, Paul Aubrin wrote:
>>> On Thu, 23 May 2013 09:38:25 +0100, RedAcer wrote:
>>>
>>>>>> Good Luck!
>>>>>
>>>>> Do you have any other problems with Paul's statements?
>>>>
>>>> Yes all of them.
>>>
>>> Here are a few points I would like to better understand.
>>>
>>> First let us use define a bit more precisely what all this is about.
>>> There are two graphs on the web page below (link from emoneyjoe). The
>>> second graph explains the "natural greenhouse effect".
>>> http://www.earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
>>> Quote:

>>> "The natural greenhouse effect raises the Earth’s surface temperature to
>>> about 15 degrees Celsius on average—more than 30 degrees warmer than it
>>> would be if it didn’t have an atmosphere. The amount of heat radiated
>>> from the atmosphere to the surface (sometimes called “back radiation†) is

At the wavelengths we're talking about the earth is a good approximation
to a black body.

You're welcome to provide alternative measurements, by which I do not
mean something that you just thought up by yourself.