Layman struggles with Science

[Sunsettommy]

(05-21-2011 12:02 AM)Climate Realist Wrote:  

(05-20-2011 08:28 PM)Sunsettommy Wrote:  I am not sure where you coming from?

Most INCOMING IR fails to reach the planets surface.

This source shows a very nice chart.That states only two types of solar radiation reaches the in any significant amount.

Visible light and Radio Waves

Solar Energy in Earth's Atmosphere

The chart



A little of the FAR Infrared reaches the surface.

That is true, but it is the Infra red that is warming our atmosphere by DIRECT sunlight- not IR reflected from the surface and certainly not from "back Radiation (IR)"

Thus the direct radiation from the sun ensures all CO2 molecules are already in a perpetual excited state.

From the appearance of the chart.It would seem that most of the INCOMING IR are absorbed in the Stratosphere and upwards.

The Troposphere hardly absorb any incoming IR at all since it has already been absorbed at higher altitudes.

Gamma Rays are highly energetic photons that penetrates to about the same depth as the incoming IR.

Yet the Stratosphere is no warmer than the Troposphere.

Atmospheric Structure



CO2 molecules are mostly in the Troposphere,where they can absorb a FRACTION of the outgoing IR.Because they absorb in three small frequency bands.It is not a big player in absorbing IR in the Infrared window.True it absorbs about 100%,but ONLY in those three small bands in the otherwise large IR window.

The CO2's main absorbing band is partly outside of the main IR energy outflow.A lot of IR energy flies right past the CO2 molecules,because they are not in the right part of the IR frequency to absorb much.

This is shown very well HERE

Two of the three CO2 frequency bands are in the very low energy part of the IR window.

[Richard111]

Exactly SST. I've always pointed out how small the level of IR CO2 can absorb. That simple formula from Wein's Law in relation to peak emission from a black body got my interest. As above, a black body with a peak emission of 15 microns will have a temperature of -80C. At 4.3 microns the temperature will be 400C and at 2.7 microns it will be 800C.

I don't see much interaction with CO2 at 400 and 800 degrees celcius on this planet!
And then again at -80C I suppose Antarctica might be in a condition to absorb a few 15 micron photons.

And that chart you mention shows energy level between 2 and 4 microns is very low anyway. Assuming the habitable areas of the earth, where we take all our measurements is 0C to 30C then IR radiation of interest is from 10.61 microns to 9.56 microns which appear to be in the IR window anyway!

So as far as I can see only IR from CLOUDS can have any effect and then only to reduce the rate of cooling at night but markedly reduce the rate of warming when the sun is shining. I am a Svensmark supporter.

[Richard111]

Below is a graph I copied out of the Handbook of AVIATION METEOROLOGY.
Since it is published by the METEOROLOGICAL OFFICE I assume the shape to be correct. This is a very interesting graph as it is actually showing TWO graphs! Look carefully at the x axis and you will see two scales. One for shortwave radiation(0-2microns) and one for longwave radiation (0-40microns). The shortwave trace represents solar radiation with a peak temperature of 6,000C, and the longwave represents terrestrial radiation at 27C. Both curves MUST be identical in shape as they only show the one curve. That is fine by me. If anyone knows of errors on this please say.

The next point of interest is that the peak of the graphs is at 100% on the y axis. If we concentrate on the terrestrial radiation we will see that if the radiative temperature is reduced the peak of the curve will move to the right and DOWN. Remember this is a blackbody curve and the slope down to the right (low frequency end) will remain the same as the peak reduces. The left slope will move slightly to the right as it follows the peak down the slope on the right. If we know the total area of the curve at 27C we can see the curve area will reduce as the temperature decreases. This reduction in area takes place at the high frequency end of the curve as explained by the "cut off temperature" in Claes Johnson's essay.



Talking of area under the curve reminded me of my quest to find just how much energy will be in the 15micron band as a percentage of the total energy being radiated at a given temperature. This graph gave me the idea. What I did was reproduce the curve on a sheet of fine ruled graph paper. Since I cannot do the math I can at least count the number of little squares under the graph. Then I find the number of squares under the 15micron band and I get 5%. So the terrestrial surface radiating at 27C pushes out 460W/m^2 energy of which CO2 in the atmosphere above can absorb just 23W/m^2. The other CO2 bands are out of range at these temperatures so can be ignored. Warmists keep talking about line doubling increasing the power of CO2 but this is rubbish. I posted a graph from Perry's Chemical Engineers Handbook 8th Edition here which shows line doubling at the 15micron band when CO2 is 50% of the gas mix and the temperature is 1,500K !!! These temperatures and CO2 levels DO NOT EXIST IN THE ATMOSPHERE.

Nasif Nahle has pointed out that on average a photon will travel 1,700 metres before encountering a molecule in the atmosphere. So if I assume a 15micron photon will encounter a CO2 molecule over that distance (which I think is highly optimistic) this implies that those 23watts of energy will attempt to warm up some 1,700 cubic metres of air (nearly two tons!).

But does it? Warmists claim that half that radiation is "backradiated" to the surface AND causes the air temperature to increase! You can't have it both ways. I am still unable to ascertain the amount of thermal heating passed to the atmosphere due to the presence of CO2.

My Handbook of AVIATION METEOROLOGY does not discuss carbon dioxide. It is not even mentioned in the index! There is one mention in Chapter 1 in table 1, carbon dioxide is 0.03% of the atmosphere. That's it!

In Chapter 3 Section 13 TRANSFER OF HEAT BY LONG-WAVE RADIATION it mentions water vapour:

Quote:At any height therefore the air tends to gain or lose heat in proportion to the net radiation absorbed or emitted by the water vapour.

it goes on to say that this is of no great consequence and concludes:

Quote:The rate of fall of temperature with height in the troposphere that would result from a radiative balance is however much greater than that which actually exists, and consequently the temperature in this region is not controlled by radiation alone.

Hmm... again no discussion on HOW MUCH HEATING due to radiation. My own simplified checks show that after the initial absorption the energy becomes a sort of "pass the parcel" up the air column until of course dew point is reached.

Then there is extensive discussion of adiabatic processes and the effects of water vapour and dew points and cloud formation etc. etc. All the sort of stuff you would expect in a meteorological text book but the only mention of gravity is in the Appendix "where g is the acceleration due to gravity" in the derivation of formulae.

I can recomend this book for anyone who wants to get some understanding of weather and climate without having to learn any maths. You will also pick up some interesting stuff that airplane pilots must learn in greater detail.

[Richard111]

The more I learn about energy transfer by radiation the more convinced I become that this whole AGW topic is a monumental scam.

My first suspicion: the dry adiabatic lapse rate does not change with changes of H2O or CO2.

Second suspicion: black bodies or gray bodies of equal temperature do not warm or cool.

Third suspicion: the low frequency intensity of radiation does not change as the body warms, only the high frequency range.

Now both the surface of the earth and most of the atmosphere is above -80C, peak temperature where 15micron radiation will reduce if temperature drops. As temperature climbs above -80C intensity of 15micron band does not increase.

Ergo: the earth surface and the CO2 in the atmosphere are exchanging photons but NOT EXCHANGING ENERGY.
This also applies to that global warming gas water vapour. Once condensation starts the water droplets can absorb and emitt ALL AVAILABLE RADIATION which is why clouds can stop surface cooling.

[Richard111]

Curses! Found a graph over at the Physics Forum that clearly shows both high frequency and low frequency wavebands change with temperature. The "high frequency cutoff" as claimed by Claes Johnson seems a bit weak.

Note the y axis; arbitrary units for radiation intensity but the changes of intensity at say 1micron are large for each 1,000K temperature step.

Obviously this low frequency step change will not be so pronounced at say +15C down to +10C but it must be there if this happens at short wave temperature changes. This makes my claim in the above post as not true.



Must look for plots that show the radiation intensity curves from earth surface temperature ranges.

[Richard111]

A little knowledge is not only dangerous but confusing. I've been happy working out how much radiation from what ever temperature but have only just woken up to the fact that energy has more paths than just radiation when we talk about the earth's surface. Energy passes from the surface to the atmosphere by direct contact. Conduction. Convection then moves that energy away from the surface so more energy can be absorbed. So radiation and conduction must share the total available energy. The question again is how much?

In Claes Johnson's essay Climate Thermodynamics he states:

Quote:The heat is transported by the atmosphere in a combination of thermodynamics (turbulent convection and phase change in evaporation/condensation) and radiation, roughly 2/3 by thermodynamics and 1/3 by radiation.

You have to ask yourself how did the heat get into the atmosphere in the first place? Thus my claim in Post:#163 above that a surface at 27C pushes out 460W/m^2 applies to the TOTAL ENERGY! 306.6W/m^2 (2/3) goes directly in the atmosphere by conduction and 153.4W/m^2 (1/3) is the radiation level. So the 15micron band only has 7.6W/m^2 (5%) radiation for the CO2 to absorb.

Its worse than I thought!

[Scpg02.]

(06-22-2011 11:48 PM)Richard111 Wrote:  A little knowledge is not only dangerous but confusing. I've been happy working out how much radiation from what ever temperature but have only just woken up to the fact that energy has more paths than just radiation when we talk about the earth's surface. Energy passes from the surface to the atmosphere by direct contact. Conduction. Convection then moves that energy away from the surface so more energy can be absorbed. So radiation and conduction must share the total available energy. The question again is how much?

In Claes Johnson's essay Climate Thermodynamics he states:

Quote:The heat is transported by the atmosphere in a combination of thermodynamics (turbulent convection and phase change in evaporation/condensation) and radiation, roughly 2/3 by thermodynamics and 1/3 by radiation.

You have to ask yourself how did the heat get into the atmosphere in the first place? Thus my claim in Post:#163 above that a surface at 27C pushes out 460W/m^2 applies to the TOTAL ENERGY! 306.6W/m^2 (2/3) goes directly in the atmosphere by conduction and 153.4W/m^2 (1/3) is the radiation level. So the 15micron band only has 7.6W/m^2 (5%) radiation for the CO2 to absorb.

Its worse than I thought!

LOL this is why I stick to politics.

[Derek]

This might be of help.

The thermal IR emitted by an object is the temperature signal of the object.
(That's how we measure the temperature of far, far away stars, ie, peak frequency of thermal IR emitted = temperature)
This does not mean, nor should it be used to imply (as it often is), that IR has had the major influence, upon the objects temperature.

Thermal IR emitted is a result of the objects temperature,
which does not necessarily mean thermal IR is the dominant cause of
the objects temperature, or temperature changes of the object observed,
whether they be warming or cooling, but cooling in particular.

I tried to illustrate this with my The (naked) cooling cannonball "thought experiment".
Post 6 in this thread

and in my What does a thermal image of a greenhouse actually show?
Same thread, Post 2.

I think Claes is being (very) conservative within the climate system of earth with 2/3 thermodynamics and 1/3 radiation losses.
- Think of the latent heat of water vapourisation,
and that the earths surface is 71% oceans.
Later addition - Does anyone else remember a point Dr. Richard S Courtney made that tropical oceans never get above a certain temperature,
I think it was 27C. If anyone can, and where it was posted, I think it is relevant to this point.

BTW - re graphs, don't forget the ultraviolet catastrophe.
It is usually not mentioned, but as above, it is clearly depicted.

[Richard111]

Derek, no problem with your cannonball "thought experiments" and others but have found when trying to discuss these things with warmists short snappy questions using numbers they must talk about shows up their lack of commitment to their own credo. Here is a couple of usefull numbers from Nasif:

Determination of Mean Free Path of Quantum/Waves and Total Emissivity of the Carbon Dioxide Considering the Molecular Cross Section.

By Nasif Nahle

University Professor, Scientist, Scientific Research Director at Biology Cabinet©Monterrey, N. L., Mexico.

This article was updated on April 8, 2011. This article has been Peer Reviewed by the Faculty of Physics of the University of Nuevo Leon, Mexico.

Quote:Abstract:

Through the application of astrophysical formulas, the mean free path length of a Quantum/wave stream leaving the surface of the Earth to the outer space before it has collided with a molecule of carbon dioxide and its total emissivity are calculated. The output of this algorithm indicates a value of about 33 meters. Also calculated is the time taken by a Quantum/wave to exit the atmosphere after it has collided with a molecule of carbon dioxide — which is ~4 milliseconds (ms).

Very handy numbers, especially that 33m, much better than 1,700m, and that 4ms means even hurricane winds won't disturb those little photons much.

[Richard111]

Thought I would share this as it blows me away. A highly recommended read.

The Shattered Greenhouse: How Simple Physics Demolishes the "Greenhouse Effect".

Quote:3.1 The Physics of Nitrogen, Oxygen, and Carbon Dioxide
<snip>
If carbon dioxide produced the backradiation claimed by Arrhenius, thermal conductivity measurements of carbon dioxide would be so suppressed by the backradiation of heat conducted into this material, that the correspondingly steep temperature gradient would yield a negative thermal conductivity of carbon dioxide. In reality, a 10,000ppm increase in carbon dioxide could, at most, reduce the conductivity of air by 1%. Given the actual difference between the thermal conductivities of carbon dioxide (0.0168) and zero grade air (0.0260), a 10,000ppm increase in carbon dioxide would lower the thermal conductivity of zero grade air by 0.36%. That would represent a 0.36% increase in thermal gradient, or a surface warming of 0.18% and a ceiling cooling of 0.18% of the total difference in temperature between the top and bottom of the affected air mass. In the case of a tropospheric carbon dioxide increase of 10,000ppm, that would correspond to a warming of 0.125ºC, or one eighth of a degree Celsius at the earth's surface, offset by a cooling of 0.125ºC at the tropopause. On the scale of doubling the troposphere's carbon dioxide, the surface warming predicted by this simple and materialistic thermodynamic approach is on the order of 0.004ºC.

[Richard111]

Do have a look at the Climate Realists web page and read about Joe Postma's new paper discussed in a guest post by John O'Sullivan. This quote below from John O'Sullivan's article supports my original scepticism on AGW as I have lived and worked in Namibia and well remember our drives to the coast to cool off under the moist misty air. I have also lived and worked in Singapore and experienced the effects of bright sunlight being moderated by water vapour during the hottest part of the day. REMEMBER, THIS IS NOT CLOUDS SHADING THE SURFACE, JUST LOTS AND LOTS OF WATER VAPOUR MOLECULES ABSORBING THE NEAR INFRARED FROM THE SUN.

Quote:Water Vapour: the Key Moderator of Earth's Temperature

Schreuder, one of 22 prominent skeptic 'Slayer' scientists had long argued that, “The earth would be warmer if there was to be no water vapour in the atmosphere, and by some margin (but only during the hours of sunshine of course).”

Both scientists refer to the evidence readily seen on a daily basis when comparing maximum temperatures in deserts that have coastal fringes (e.g. Sahara, Namib,

Atacama), where it will be seen that there is a direct link between humidity and maximum as well as minimum daily temperatures.

“Absence of water vapour allows more of the sun's radiation to reach the ground and thus create a warmer earth locally when compared to an atmosphere that holds greater water vapour and is at the same latitude.”

[Richard111]

More thoughts from this struggling layman. The business about the sun having little or no effect on the climate has been troubling me for a while. I've read about the the cosmic ray and cloud forming theory and believe that is very valid and should be accounted for when attempting to predict climate variations.
On reading reports about global heat budgets and how solar irradience only changes by less than 0.5% and therefore cannot effect much in the way of climate change still left me doubtful.

I learnt how global heat budgets were calculated and was shocked at the poor science employed by certain parties. There are many discussions about that elsewhere on this forum so I won't go there. My interest was captured by the following Solar Irradience chart. Note that the vertical scale is in watts per metre squared and only covers a range of just 4 watts of which the trace itself covers 3 watts over the period 1661 to 1991. Not much change indeed!

Look at the modern period for when global "climate change" was being influenced by nasty man made emissions over the period 1921 to 1991. I think it is safe to say the total change in TSI from the sun is just 1 watt per metre squared. I make that a change of 0.07% which is rather less than what the warmists poo-poo as having no influence on global temperatures.



But wait! That 1 watt per metre squared is 1 joule of energy PER SECOND! So what does it add up to? If we take the mean radius of the earth to be 12734.8km [LINK] and calculate the area of the disk the earth is presenting to the sun using piR^2 we get 509,488,200.3square kilometres.

If we multiply that by 1,000,000 we get 5.095 x 10^14 square metres. This is the AREA that absorbs energy from the sun at some 1,366W/m^2. If the sun radiation flux has increased by just 1W/m^2 then we can calculate a figure for the total extra heat to the earth.

So effectively the extra incoming heat from the sun for an increase in 1W/m^2 is 5.095 x 10^14 joules of energy per second. What can we do with that extra energy?

My thought was to see how much water could be heated over just one second. The specific heat capacity of water is 4,187 J/kgK. If we divide our extra energy by that number we get 1.217 x 10^11 kg of water which can be heated 1 degree every second. It is difficult to visualise that number so dividing by 1,000,000,000 we will get a nice number of just 121.8 cubic kilometres of water heated up 1 degree every second!

Now there are an awfull lot of seconds in the period 1921 to 1991 and there has been no letup on that incoming energy from the sun so far. I realise that all that energy is not necessarily being stored in water or whatever, just that this is a very real change over a long period of time and people claim no influence from the sun!

I simply don't believe it.

[Richard111]

Just had a thought --- (doesn't happen often ) this picture shows burning hydrocarbons causes a release of carbon dioxide (CO2).

Plants absorb carbon dioxide and turn it back to hydrocarbons. Why can't we do that? (with all our fancy science?)

Plants are cleverer than us, they do it by sunlight.

[Derek]

(06-26-2011 12:56 AM)Richard111 Wrote:  Derek, no problem with your cannonball "thought experiments" and others but
have found when trying to discuss these things with warmists short snappy questions using numbers they must talk about
shows up their lack of commitment to their own credo.

I am beginning to think exposing their avoidance of grey bodies is the way to go.
Sticking with black bodies and their "figures" is just going round and around in the imaginary "greenhouse physics" land (religion really)
of black bodies, P/4, W/m2, and the failed Greenhouse effect hypothesis.

Reality and grey bodies is where they fall flat on their faces,
(hence my analogies)
because W/m2, P/4, and GH "theory" can not cope with grey bodies.

Yet, "we" are supposed to believe they are scientifically studying the actual and grey bodies relative reality of earth's climate system.
They are not doing such, otherwise they would not be constantly using and invoking "black body", P/4, W/m2, and the failed GH effect hypothesis.
That is why such can not be questioned, "consensus" climate science would be forced back into reality.

[Richard111]

Hmm... I suspect the warmists/greenies have a very good understanding of "grey body radiation". Notice how they have removed access to open country and high altitude station records and now depend almost entirely on temperature readings from town and airports.

[Richard111]

The solar graphic below comes from the Milford Weather Solar page. This post is not about the Milford weather site. It is this layman's attempt to understand what the data means. You will find some discussion over at Milford Haven weather which starts on 23 June, solstice day, or the day after.



The day this picture was recorded, 28/09/11, was a completely cloudless day, no contrails overhead even though jets must have been flying. We see a very smooth bell curve peaking at 13:20 hours (British summer time), if it was GMT the peak would be at 12:20. Why the 20 minutes? Because 360 divided by 24 equals 15 degrees an hour so the sun moves 5 degrees in 20 minutes and Milford Haven is 5 degrees west of the Greenwich Meridian.

My interest is that the smooth curve peak is less that the spikes on a cloudy day. My interpretation is that between the clouds there is less water vapour so that a bit more sunlight gets through. I would say the peak is about 590W/m^2 yet there are three days before showing spikes in excess of 700W/m^2 !
So I have a question: how much effect are the clouds and water vapour having on the local temperature and how to define/calculate this?

In attempting to research this I needed to know how much energy was coming from the sun. 590W/m^2 should warm things up to about 46.4C !!! It most certainly didn't get that hot. So what's happening? Well, Milford Haven is about 51.7 degrees north and the sun is currently over the southern hemisphere so that 590W/m^2 will be spread over a larger surface area. This means math and I'm not too hot in that department so I typed "calculating sun angles" into Bing and up came lots of sites offering solutions.

HERE is a very basic tutorial to get you mind working on the problem. If like me your maths is not up to much have a look at this.

You still need some trig to find out how far the metre is being stretched along the groung, so, I use the lazy man's method:



You just need a 1 metre length of batten, clear sunlight at your local noon, a flat piece of ground and a tape measure. Hold the batten so it is vertical on the ground. Place your opened tape measure along the shadow. Start lowering the batten towards the tape measure and note the shadow as it lengthens. As the batten gets closer to the ground the shadow will get shorter until the batten is flat on the ground where is will show 1 metre. So, raise the batten slowly and note the longest shadow on the tape measure. That is the length of the area being warmed by the sun.

In my case this was 1.66 metres. So that 590W/m^2 was actually going into an area of 1m x 1.66m. So how does that look now? I make it 355.4W/m^2 of heating on the ground, but that is just 8.4C temperature?!? Very confusing. As luck would have it I had grabbed a copy of the temperature graph.



Note the blue trace for the average air temperature. It rises linearly to about 20.7C degrees, peaking at about 15:00 hours, and then descending linearly.

There is no ways the ground surface or the sea surface around the harbour could follow that temperature change. Once again, what is going on?

I can only surmise that we are seeing the effect of water vapour in the lower atmosphere. Those H2O molecules are absorbing NEAR INFRARED and heating the atmosphere directly. I would like to see an instrument that can measure NIR in the 0.7 to 4 micron band where H2O has FIVE active bands and CO2 has ONE active band. They should show as DARK bands on the instrument. The warmists claim half up/half down radiation from GHG molecules in the atmosphere; no argument with that, just they never specify just how much energy went into heating the local atmosphere. The temperature trace above shows that it is quite considerable. That absorbed NIR is from a 5,000K source, the atmosphere over the harbour in Milford Haven reached nearly 300K.

Until anyone can explain where I am going wrong with my reasoning I will continue to believe GREEN HOUSE GASES HELP REDUCE THE INCOMING HEAT DURING DAYLIGHT HOURS.

[Derek]

Amazing find, and a compelling analysis Richard111.

Richard111 wrote:-
" My interpretation is that
between the clouds there is less water vapour so that a bit more sunlight gets through. "

I think you are onto something. A "something" that could be potentially massive.
Congratulations Richard111, is simply not enough.

Richard111 wrote:-
" Until anyone can explain where I am going wrong with my reasoning I will continue to believe
GREEN HOUSE GASES HELP REDUCE THE INCOMING HEAT DURING DAYLIGHT HOURS.

What you have described is far larger than I had imagined so far.

Would such heating help further power convection of latent heat?

[Derek]

BTW - I think this might be more your area than mine Richard111.

Do the often quoted lapse rates, whether they be dry, wet, environmental, normal, etc, etc, etc,
take into account heating throughout the depth of the atmosphere,
(as your above post seems to illustrate to me)
or, do they just assume heat escaping from the surface towards space???

If the lapse rates do "assume" the latter, well, errr, hmmmm.....

[Richard111]

(10-03-2011 12:03 PM)Derek Wrote:  Do the often quoted lapse rates, whether they be dry, wet, environmental, normal, etc, etc, etc,
take into account heating throughout the depth of the atmosphere,
(as your above post seems to illustrate to me)
or, do they just assume heat escaping from the surface towards space???

I reckon I would have a PhD or two if I could answer that. But as I've pointed out before; The midday clear air temperatures on the equator are less than the midday clear air temperatures in a desert region even with the sun over head. This has always puzzled me.

Lapse rates apply to UPWARD temperature reduction (negative) rate from the surface. Only the first metre or so of the air has the same temperature as the surface.

The sensor in Milford Haven is allmost completely surrounded by sea water. The sea water most certainly did not reach 20C !! So that leaves direct sunlight warming the air. The only candidate appears to be H20.

H2O can absorb UPWELLING radiation from the surface but the sea surface was ~10C. (must look into that)

All I can say is the more one looks at the details of the "official" global energy budget the more it looks like a whole pile of cods wallop.

Keep digging, we may find a golden spoon yet.

[Derek]

" The only candidate appears to be H20. "

Having had a little time to ponder further, I am wondering if the effect noticed is "purely" down to water vapour.

Aerosols (natural [sea salt for example], and anthropogenic) could (probably have) have been washed out of the atmosphere inbetween the clouds for instance.
These I assume directly heat the atmosphere when they absorb incoming solar radiation.
The net effect though, would it be to increase the height of convection of latent heat being transported aloft by water vapour.

" The midday clear air temperatures on the equator are less than the midday clear air temperatures in a desert region even with the sun over head. This has always puzzled me. "
Does clear air mean no aerosols? If not, it may be worth considering.

Then there is the Svensmark hypothesis to possibly have to take into account.