Re: Radiative Processes in Planetary Atmospheres

[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?

[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.

[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!

>

[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?

[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.

[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?

[Tom P]

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

[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.

[john]

But can't I tap into geothermal
to heat my surface also?
john

[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.

[john]

What part of your numbers takes into
account how much the interior
of the Earth heats the surface?

john

[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]

> Where did you get those numbers?
>

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.

[RedAcer]

On 25/05/13 14:25, Paul Aubrin wrote:
> 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,

What's a looping energy flux. All things radiate energy to each other.
It's a physics.

> 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

What?!
If there is no radiation to observe then how can that be a proof that
there is some radiation to observe . Talk sense.

> is a
> definite proof that the flux is real?
>

This conversation can't move until you accept that Arno Penzias and
Robert Wilson observed the CMBR.

If you accept that than obviously small amounts of IR/microwave can be
detected and and as IR detectors have improved since then there is no
problem detecting the IR if it is there.

[RedAcer]

And how does it work? Will says reflection is not a property of the
material. Those dielectric mirrors have a special design made of precise
layers. Why go to all that bother if reflection does not depend on the
materials surface. Are you saying that Will is wrong?

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

We are talking about reflecting heat back. An anti reflection coating
wouldn't keep the space-probe warm!

[Will Janoschka]

On Sat, 25 May 2013 16:29:27, Tom P <wero...@freent.dd> wrote:

> 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.
>

you can use whatever you want for your nonsense I hjav measured the
BDRF
for ice snow ant water from 3 to 20 microns. Anyone that thinks that
any surface
can radiate or absorb to pi steradiand is simply making shit up.

[emoneyjoe]

Maybe you are, but the AGW brown shirts keep
saying all that back radiation comes from up above,
as if 210K air can radiate at the same rate as the Sun.

[emoneyjoe]

What is this 6000m that is the top of the
atmosphere?

The lapse rate does no cause the air
at the surface to be warmer, the air at
the surface is warmer, and the dry lapse
rate is the result of gravity, nothing else.

The surface can be hotter than the
air in sunlight, is your physics as loose
as your speaking?
That is a good case for UHI being
umderestimated though.

IR radiation from the surface plays
a minor role in warming the atmosphere,
and air at 100 degrees colder temperature
plays an even more minor role.


Could you please talk physics and
stop pandering AGW?

[emoneyjoe]

The only incoming is from the sun.

>Outgoing 398.2+18.4+86.4 = 503
>Check your arithmetic.

There should be a law against bastardizing
physics and math to support a flawed concept.

You can't argue against the premise that
the atmosphere would be warmer if there were
no water and no GHGs.
You can't specify what that mythical extra
"incoming" is at different latitudes, it would
have to be over 500 at the equator.

All this BS is bound to start causing you
to stutter.

[Paul Aubrin]

On Sat, 25 May 2013 17:46:48 +0100, RedAcer wrote:

> If you accept that than obviously small amounts of IR/microwave can be
> detected and and as IR detectors have improved since then there is no
> problem detecting the IR if it is there.

Small amounts of radiation can indeed be detected. But not a single joule
can be extracted from the permanent back-radiation flux which you suppose
to exist at equilibrium between two black surfaces: it is an
impossibility. Zero energy implies no possible measure. This very
different from the situation of the background radiation of the universe
which is faint but really exists.

[RedAcer]

On 25/05/13 20:19, Paul Aubrin wrote:
> On Sat, 25 May 2013 17:46:48 +0100, RedAcer wrote:
>
>> If you accept that than obviously small amounts of IR/microwave can be
>> detected and and as IR detectors have improved since then there is no
>> problem detecting the IR if it is there.
>
> Small amounts of radiation can indeed be detected. But not a single joule
> can be extracted from the permanent back-radiation flux which you suppose
> to exist at equilibrium between two black surfaces: it is an
> impossibility. Zero energy implies no possible measure.

So you are saying that CO2 can't emit IR when it is warmed up?

> This very
> different from the situation of the background radiation of the universe
> which is faint but really exists.

But is has been measured. Go check the literature.

>

[emoneyjoe]

On Sat, 25 May 2013 20:54:59 +0200, hda <agen...@xs4all.nl.invalid>
wrote:

>On Fri, 24 May 2013 18:17:24 -0400, emoneyjoe <emon...@iglou.com>
>wrote:
>
>>On 24 May 2013 15:46:41 GMT, Paul Aubrin <chu8...@free.fr> wrote:

>>
>>>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?
>>
>> 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.
>>
>
>GF's hardly make a proper energy balance. I think RedAcer and Tom P
>won't do it at all.
>
>You never see them properly boxing or framing a process, so all know
>the exact boundaries of the problem in question. You know:
>
>Input + Production = Output + Accumulation
>
>Furthermore they try to sell us that the total atmospheric HTC
>(heattransfercoefficient) will change disasterously due to CO2
>at the TOA (topofatmosphere). Well I think CO2 and H2O are just a
>transmitter overthere and loose heat to space heatsink. The heat
>container is down at the place where H20 can phase-change (solid,
>liqiud, gas) and even reflective change: opacity.

Yes, and those processes moderate in
both directions.

The real heat sink is the air, the mass of the
N2 and O2 is about equal to the top two feet of
the surface.

There are many more cooling processes
than warming, the only source of thermal
energy is the Sun.

What climate science lacks is the self-
controlling aspects of the cooling processes,
always pushing the temperature toward the
nominal.

[Paul Aubrin]

Zero effect cannot be measured.

>
>

[RedAcer]

You would have to do the measurement to check for zero. Have you done
that? If not you will have to keep quite especially as real scientists
have measured it.

How does the MLI keep the spacecraft warm. You still have answered. How
does a cold layer reflect heat to a warmer layer. If it doesn't work
what keeps MLI-wrapped spacecraft warm *and* cool?

>
>>
>>
>

[Will Janoschka]

On Sat, 25 May 2013 16:02:34, john <johnse...@gmail.com> wrote:
> On Saturday, 25 May 2013 09:42:58 UTC-6, Tom P wrote:
> > 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:

> > > 643øC). 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 620øC, 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

> > > 100øC (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.
>
> What part of your numbers takes into
> account how much the interior
> of the Earth heats the surface?
>

john, TomP is talking radiation here all S-B equation
and black-bodies. No need to bring in stupid things
like the actual world or sensible heat. :)

[Will Janoschka]

All of us claim no back radiation. What is the temperature of the
atmosphere
and emissivity, of what you ar radiating to from the warmer surface.
The S-B equation always has explicitly two temperatures and two
emissive
that are never zero or one.

>
> 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.

There is no flux going in two opposing directions, 2LTD.
None of the spectrum is like that of a black-body. The
"only flux" is that that "between" 288 Kelvin and 275 Kelvin.
with both emissivity way less than one, and a solid angle of
near "one" not near PI. No other flux has ever been detected..

All of your BS, TomP, is Climate Clown BS.

There are no measured values in that diagram!!
Latent heat is calculated from precipitation, The others use radiance

rather that actual radiative heat transfer. then mistakenly use these
radiance numbers in a most childlike manner. Learn some Physics!!!!!
.

[Will Janoschka]

On Sat, 25 May 2013 19:11:33, hda <agen...@xs4all.nl.invalid> wrote:

> Are you dodging a question too ? Where are you getting the 66K or
> 66W/m2 ? Or 167K you write next?

Thanks for your help in getting TomP straightend out!
I doubt that it will work.

[Will Janoschka]

> 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)

Joe NASA has never measured the heat absorbed anywhere.
Such a measurement would destroy NASA. And they know it!

>
> 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?

Yes Joe, You missed the point that NASA makes
up numbers to survive! If those numbers are
wrong, that is OK, we have lots more numbers!!!

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

[Will Janoschka]

Are you now doubling your bets? Where is the money?

[Will Janoschka]

On Sat, 25 May 2013 19:00:56, emoneyjoe <emon...@iglou.com> wrote:

> There should be a law against bastardizing
> physics and math to support a flawed concept.
>
> You can't argue against the premise that
> the atmosphere would be warmer if there were
> no water and no GHGs.
> You can't specify what that mythical extra
> "incoming" is at different latitudes, it would
> have to be over 500 at the equator.
>
> All this BS is bound to start causing you
> to stutter.
>

TomP is a robot impervious to all logic and reason !!

[Will Janoschka]

On Sat, 25 May 2013 18:54:59, hda <agen...@xs4all.nl.invalid> wrote:

> GF's hardly make a proper energy balance. I think RedAcer and Tom P
> won't do it at all.
>
> You never see them properly boxing or framing a process, so all know
> the exact boundaries of the problem in question. You know:
>
> Input + Production = Output + Accumulation

I agree, except this wonderful planet always has something
new up its sleeve, that Earthlings know nothing about.
This causes many Aw Shits, from those that think!

>
> Furthermore they try to sell us that the total atmospheric HTC
> (heattransfercoefficient) will change disasterously due to CO2
> at the TOA (topofatmosphere). Well I think CO2 and H2O are just a
> transmitter overthere and loose heat to space heatsink. The heat
> container is down at the place where H20 can phase-change (solid,
> liqiud, gas) and even reflective change: opacity.

I do not sell to easily!

[Will Janoschka]

On Sat, 25 May 2013 21:37:56, RedAcer <rred...@gmail.com> wrote:

> On 25/05/13 22:33, Paul Aubrin wrote:
> > On Sat, 25 May 2013 20:26:27 +0100, RedAcer wrote:
> >
> >> On 25/05/13 20:19, Paul Aubrin wrote:
> >>> On Sat, 25 May 2013 17:46:48 +0100, RedAcer wrote:
> >>>
> >>>> If you accept that than obviously small amounts of IR/microwave can be
> >>>> detected and and as IR detectors have improved since then there is no
> >>>> problem detecting the IR if it is there.
> >>>
> >>> Small amounts of radiation can indeed be detected. But not a single
> >>> joule can be extracted from the permanent back-radiation flux which you
> >>> suppose to exist at equilibrium between two black surfaces: it is an
> >>> impossibility. Zero energy implies no possible measure.
> >> So you are saying that CO2 can't emit IR when it is warmed up?
> >>> This very different from the situation of the background radiation of
> >>> the universe which is faint but really exists.
> >>
> >> But is has been measured. Go check the literature.
> >
> > Zero effect cannot be measured.
>
> You would have to do the measurement to check for zero. Have you done
> that? If not you will have to keep quite especially as real scientists
> have measured it.

Sorry such has never been measured ever. It has been demonstrated
as not to happen.

>
> How does the MLI keep the spacecraft warm. You still have answered. How
> does a cold layer reflect heat to a warmer layer. If it doesn't work
> what keeps MLI-wrapped spacecraft warm *and* cool?

Why not contact those that use such, like I, They will ignore you as
a child
Like I.
>
> >
> >>
> >>
> >
>

[columbiaaccidentinvestigation]

On May 25, 9:46 am, RedAcer <rreda...@gmail.com> wrote:" What?! If

there is no radiation to observe then how can that be a proof that
there is some radiation to observe . Talk sense"

He has a hard time making sense, and seems be an example of a
perpetual idiocy machine.

[Will Janoschka]

Why do you prattle so child. The surfaces are constructed to reflect
Quite independent of what you deluded mind comes up with.
Such reflection is measured regularly. It is not some ridiculous
mind problem.

>
> >
> > http://en.wikipedia.org/wiki/Anti-reflective_coating
>
> We are talking about reflecting heat back. An anti reflection coating
> wouldn't keep the space-probe warm!

Stop using wiki as a reference, A anti-reflective coating is the same
as a dielectric mirror except for the thickness of the layers.
Please go somewhere and learn some science, before posting again.

[Paul Aubrin]

On Sat, 25 May 2013 17:41:36 +0200, Tom P wrote:

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

I previously answered on this point, directly and to the group.
The energy balance at the level of the ground can accommodate any value
of the back-radiation loop, zero watt by square meter included.
Red Acer positively thinks that this ever looping power is real.

I explained with a though experiment why it can only be an intellectual
convenience: any attempt to extract any energy from this loop will
required to provide it from another, real world source. The loop itself
can provide no actual work thus it is undetectable. For me, things which
intrinsically cannot be put in evidence by any detector like leprechauns
or ghost are not real. I understand your back-radiation concept as a
convenience I don't feel compelled to believe it can have any real world
effect.

So my answer to your question about energy balance can be nothing but the
difference between the incoming solar flux less what is lost by the
ground by other means (conduction or phase change and convection).
As in some frequency bands the atmosphere spectral emittance is
significant, the ground cannot emit the spectrum of a black-body, not
even of a grey body (Wien distribution with a coefficient of emissivity
less than one).

[Paul Aubrin]

On Sat, 25 May 2013 22:37:56 +0100, RedAcer wrote:

>> Zero effect cannot be measured.
>
> You would have to do the measurement to check for zero. Have you done
> that? If not you will have to keep quite especially as real scientists
> have measured it.

Nobody can disprove the non-existence of something, but your back and
forth radiation flux cannot be used to provide any physical work
(joules). I cannot see any instrument of measure which could detect it. I
asked you several times if you could indicate on which principle such an
instrument would be based, without an answer.

PS: Here is again the though experiment which according to me proves that
the looping back-radiation energy flux cannot be converted into work.

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?

All the energy in the laser beam must be provided to the system by an
external source, none can be extracted from the "natural greenhouse back-
radiation" loop.

[emoneyjoe]

There isn't any loop, it is more like loops
in millions of different altitude layers.

It is more like thermal energy in the
N2 and O2, with GHGs trying to get rid
of it, and in the process, maybe delays
the upward flow a little.

>So my answer to your question about energy balance can be nothing but the
>difference between the incoming solar flux less what is lost by the
>ground by other means (conduction or phase change and convection).
>As in some frequency bands the atmosphere spectral emittance is
>significant, the ground cannot emit the spectrum of a black-body, not
>even of a grey body (Wien distribution with a coefficient of emissivity
>less than one).

Everything about water, GHGs and clouds
is a cooling process, helping the T^4 power
process regulate the temperature, keeping
it within certain bounds.

Without the N2 and O2, there would not
be anywhere near enough thermal energy
retained to keep us warm for days without
sun, the GHGs would throw all the energy
they can hold away in hours.

[Will Janoschka]

So true. Remember that the earth has so very,very much
sensible heat, it is never "in" thermodynamic equilibrium.
It is playing catch-up trying hard to get to thermodynamic
equilibrium, a spontaneous process, but with a time
constant of lifetimes. What lifetimes? I do not know.
From crickets to effalumps.

BTW The decease in emissivity with increasing angles
from normal, make the use of the S-B equation nonsense
for surfaces of earth.

[Will Janoschka]

Well yes and no!
The laps rate dry or wet is fixed. The
temperature at all altitudes depends
on how much heat this Earth wants to radiate
to space right now! I doubt that Earthlings
affect that number much.

> The surface can be hotter than the
> air in sunlight, is your physics as loose
> as your speaking?
> That is a good case for UHI being
> umderestimated though.
>
> IR radiation from the surface plays
> a minor role in warming the atmosphere,
> and air at 100 degrees colder temperature
> plays an even more minor role.
>
> Could you please talk physics and
> stop pandering AGW?
>

Pandering AGW is what they are hired to do!

[Will Janoschka]

Paul, Can the insane be helped in any way to become less insane?
especially when they believe you are the one insane. Back radiation
is easy to handle compared to this magnificent insanity.

I think the old timers like Galileo and Newton had the same problem.

[Paul Aubrin]

On Sun, 26 May 2013 02:05:07 -0400, emoneyjoe wrote:

>>I explained with a though experiment why it can only be an intellectual
>>convenience: any attempt to extract any energy from this loop will
>>required to provide it from another, real world source. The loop itself
>>can provide no actual work thus it is undetectable. For me, things which
>>intrinsically cannot be put in evidence by any detector like leprechauns
>>or ghost are not real. I understand your back-radiation concept as a
>>convenience I don't feel compelled to believe it can have any real world
>>effect.
>
> There isn't any loop, it is more like loops
> in millions of different altitude layers.

I agree. But unlike the "natural greenhouse effect" energy loop shown in
the Trenberth diagram, each layer leaks a minute but effective energy
flux upward (generally upward).

[Paul Aubrin]

No one can extract even a single watt from the "natural greenhouse
effect" back-radiation without an external energy source, because it
would be a perpetuum mobile. So the "natural greenhouse effect" can have
no effect in the real world of physics.
Red Acer goes on supposing that he cannot detect it because it is
too faint, like the universe 3K background radiation. He cannot detect it
because it cannot carry any real energy without violating the laws of
physics.
The 3K background radiation carries 0.000005442W/m2; the back-
radiation is supposed to carry 324W/m2 in each direction. The back-
radiation is supposed to have increased the whole mass of the atmosphere
and the upper tire of the oceans by 33K. But when asked how it can be put
in evidence by actual instruments, Red Acer pretends that it is even
fainter than the 3K background radiation when it is 120 millions times
bigger.