<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd">
<article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article">
 <front>
  <journal-meta>
   <journal-id journal-id-type="publisher-id">
    acs
   </journal-id>
   <journal-title-group>
    <journal-title>
     Atmospheric and Climate Sciences
    </journal-title>
   </journal-title-group>
   <issn pub-type="epub">
    2160-0414
   </issn>
   <issn publication-format="print">
    2160-0422
   </issn>
   <publisher>
    <publisher-name>
     Scientific Research Publishing
    </publisher-name>
   </publisher>
  </journal-meta>
  <article-meta>
   <article-id pub-id-type="doi">
    10.4236/acs.2024.144025
   </article-id>
   <article-id pub-id-type="publisher-id">
    acs-136478
   </article-id>
   <article-categories>
    <subj-group subj-group-type="heading">
     <subject>
      Articles
     </subject>
    </subj-group>
    <subj-group subj-group-type="Discipline-v2">
     <subject>
      Earth 
     </subject>
     <subject>
       Environmental Sciences
     </subject>
    </subj-group>
   </article-categories>
   <title-group>
    CO
    <sub>2</sub> Back-Radiation Sensitivity Studies under Laboratory and Field Conditions
   </title-group>
   <contrib-group>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Ernst
      </surname>
      <given-names>
       Hammel
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Martin
      </surname>
      <given-names>
       Steiner
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Christoph
      </surname>
      <given-names>
       Marvan
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Matthias
      </surname>
      <given-names>
       Marvan
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Klaus
      </surname>
      <given-names>
       Retzlaff
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Werner
      </surname>
      <given-names>
       Bergholz
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Axel
      </surname>
      <given-names>
       Jacquine
      </given-names>
     </name>
    </contrib>
   </contrib-group> 
   <aff id="affnull">
    <addr-line>
     aInternational Climate Research, A Citizen Research Group, Vienna, Austria
    </addr-line> 
   </aff> 
   <pub-date pub-type="epub">
    <day>
     30
    </day> 
    <month>
     08
    </month>
    <year>
     2024
    </year>
   </pub-date> 
   <volume>
    14
   </volume> 
   <issue>
    04
   </issue>
   <fpage>
    407
   </fpage>
   <lpage>
    428
   </lpage>
   <history>
    <date date-type="received">
     <day>
      19,
     </day>
     <month>
      July
     </month>
     <year>
      2024
     </year>
    </date>
    <date date-type="published">
     <day>
      6,
     </day>
     <month>
      July
     </month>
     <year>
      2024
     </year> 
    </date> 
    <date date-type="accepted">
     <day>
      6,
     </day>
     <month>
      October
     </month>
     <year>
      2024
     </year> 
    </date>
   </history>
   <permissions>
    <copyright-statement>
     © Copyright 2014 by authors and Scientific Research Publishing Inc. 
    </copyright-statement>
    <copyright-year>
     2014
    </copyright-year>
    <license>
     <license-p>
      This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/
     </license-p>
    </license>
   </permissions>
   <abstract>
    We measured the IR back radiation using a relatively low-cost experimental setup and a test chamber with increasing CO
    <sub>2</sub> concentrations starting with a pure N
    <sub>2</sub> atmosphere against a temperature-controlled black reference background. The results confirm estimations within this work and previous finding about CO
    <sub>2</sub>-induced infrared radiation saturation within realistic atmospheric conditions. We used this setup also to study thermal forcing effects with stronger and rare greenhouse gases against a clear night sky. Our results and their interpretation are another indication for having a more critical approach in climate modelling and against monocausal interpretation of climate indices only caused by anthropogenic greenhouse gas emissions. Basic physics combined with measurements and data taken from the literature allow us to conclude that CO
    <sub>2</sub> induced infrared back-radiation must follow an asymptotic logarithmic-like behavior, which is also widely accepted in the climate-change community. The important question of climate sensitivity by doubling current CO
    <sub>2</sub> concentrations is estimated to be below 1˚C. This value is important when the United Nations consider climate change as an existential threat and many governments intend rigorously to reduce net greenhouse gas emissions, led by an ambitious European Union inspired by IPCC assessments is targeting for more than 55% in 2030 and up to 100% in 2050 [1]. But probably they should also listen to experts [2] [3] who found that all these predictions have considerable flaws in basic models, data and impact scenarios.
   </abstract>
   <kwd-group> 
    <kwd>
     Climate Change
    </kwd> 
    <kwd>
      Greenhouse Gases
    </kwd> 
    <kwd>
      CO
     <sub>2</sub> Backscatter
    </kwd> 
    <kwd>
      IR Radiation
    </kwd>
   </kwd-group>
  </article-meta>
 </front>
 <body>
  <sec id="s1">
   <title>1. Introduction</title>
   <p>Our atmosphere mainly comprises radiative inactive diatomic molecules N<sub>2</sub> 78.08% and O<sub>2</sub> 20.95% by volume. The remaining 0.97% consists primarily in Ar with 0.93%, CO<sub>2</sub> with 0.04% in volume, and traces of hydrogen, helium, and other noble gases, methane, nitrous oxide, and ozone. A variable but radiatively dominant part is water vapor, which is on average about 1% in volume at sea level. Greenhouse (GH) molecules absorb terrestrial IR radiation emitted by the surface as result of warming caused by the incoming solar radiation. Their absorption characteristics allow them to act in the retention of heat within the atmosphere and to ensure that the global mean temperature of the atmosphere supports biological life. This is commonly known as the greenhouse effect. Therefore, the IR active components are water vapor, carbon dioxide, methane, dinitrogen monoxide, and ozone, in decreasing order of effectiveness due to their concentrations.</p>
   <p>The infrared IR transmission spectra of CO<sub>2</sub> over a path length of 100 meters is shown in <xref ref-type="fig" rid="fig1">
     Figure 1
    </xref> against the wavenumber, 1/λ = ν/c (λ = wavelength, ν = frequency, c = velocity of light). The transmission T or transmittance is the extent by which the incident radiation at any wave number is transmitted by the sample. For opaque media T = 0 and complete transparent media T = 1.</p>
   <p>In reflection-free media absorbance, A = 1 − T (transmittance) according to Kirchhoff’s law, which also allows the propagation of the equality of long wave emissivity ε and absorptivity a<sub>lw</sub>.</p>
   <fig id="fig1" position="float">
    <label>Figure 1</label>
    <caption>
     <title>Figure 1. CO<sub>2</sub> IR Transmittance over 100 meters as a function of wavenumber from HITRAN <xref ref-type="bibr" rid="scirp.136478-4">
       [4]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId13.jpeg?20241009013646" />
   </fig>
   <p>The CO<sub>2</sub> spectrum is dominated by the bending vibration, centered at 667 cm<sup>−</sup><sup>1</sup> equivalent to a wavelength of 14.992 µm, and the asymmetrical stretching mode at 2349 cm<sup>−</sup><sup>1</sup> equivalent to 4.257 µm. The extra and very weak bands arise from further excitations and represent very small absorptions that are very often claimed to be significantly inaccurate calculations of the GH effect.</p>
   <p>The surface has a mean global temperature of 288˚K and its emission is approximated by a black body at this temperature, consisting of a continuous radiation spectrum unlike that of the GH gases which is specific for each molecule-species and made up of discrete rotation-vibrational bands. The continuous Planck or black body surface radiance is shown by the gray curve in <xref ref-type="fig" rid="fig2">
     Figure 2
    </xref> compared to the limits of the CO<sub>2</sub> absorption bands in orange and red of the main GH gas of CO<sub>2</sub>.</p>
   <fig id="fig2" position="float">
    <label>Figure 2</label>
    <caption>
     <title>Figure 2. Black body radiator at T = 288˚K and major CO<sub>2</sub> absorption bands.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId14.jpeg?20241009013646" />
   </fig>
   <p>The contribution of the 15 µm and the 4 µm CO<sub>2</sub> band to the total black body IR absorption can be estimated by integration of <xref ref-type="fig" rid="fig1">
     Figure 1
    </xref> with a rough maximum of 25%, where the major contribution comes from the 15 µm band.</p>
   <p>The solar irradiance intercepted by our atmosphere is 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        π 
      </mi> 
      <msup> 
       <mi>
         a 
       </mi> 
       <mn>
         2 
       </mn> 
      </msup> 
      <msub> 
       <mi>
         S 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
     </mrow> 
    </math>, where α is the earth’s radius. A fraction α, called the planetary albedo is reflected to space. The remaining portion is absorbed. Averaging over a long time over the total area of the globe, the absorbed solar radiation must be in balance with the radiation emitted by the atmosphere. In a very simplified 2D model we get according to Stefan-Boltzmann</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mn>
          1 
        </mn> 
        <mo>
          − 
        </mo> 
        <mi>
          α 
        </mi> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           S 
         </mi> 
         <mn>
           0 
         </mn> 
        </msub> 
       </mrow> 
       <mn>
         4 
       </mn> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
     </mrow> 
    </math> (1)</p>
   <p>where T<sub>E</sub> is the theoretical emission temperature without any atmospheric greenhouse effect. Assuming on average α = 0.3, we find T<sub>E</sub> = 255˚K, a temperature much lower than the observed global average surface temperature of T<sub>S</sub> = 288˚K. That means that the natural greenhouse effect results in additional ΔT<sub>GH</sub> = 33˚K, corresponding to</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         S 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
      <mo>
        − 
      </mo> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
      <mo>
        = 
      </mo> 
      <mrow> 
       <mrow> 
        <mn>
          150 
        </mn> 
        <mtext>
          W 
        </mtext> 
       </mrow> 
       <mo>
         / 
       </mo> 
       <mrow> 
        <msup> 
         <mi>
           m 
         </mi> 
         <mn>
           2 
         </mn> 
        </msup> 
       </mrow> 
      </mrow> 
     </mrow> 
    </math> (2)</p>
   <p>Detailed evaluation of the Planck function for T<sub>S</sub> = 15˚C or 288˚K, results in the respective absorption bands.</p>
   <p>13.26 - 17.07 µm (15 µ band), in energy densities of 9.22 × 10<sup>−</sup><sup>7</sup> J/m<sup>3</sup> or 17.6% of the total emission or 6.95 × 10<sup>13</sup> photons per m<sup>3</sup>.</p>
   <p>4.19 - 4.38 µm (4 µ band), in energy densities of 5.72 × 10<sup>−</sup><sup>9</sup> J/m<sup>3</sup>, being 0.11% of the total emission or 1.22 × 10<sup>11</sup> photons per m<sup>3</sup>.</p>
   <p>From above considerations, we see that only the 15 µ band is significant with an IR absorption limited to 0.176 × 391 W/m<sup>2</sup> = 69 W/m<sup>2</sup> and back-radiation to 69/2 = 34.5 W/m<sup>2</sup>. Simplified energy balance considerations in a single atmospheric layer are sketched in the following drawing. See <xref ref-type="fig" rid="fig3">
     Figure 3
    </xref>.</p>
   <p>With ε = 1 we get</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         S 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
      <mo>
        = 
      </mo> 
      <mi>
        U 
      </mi> 
      <mo>
        = 
      </mo> 
      <mn>
        2 
      </mn> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mrow> 
        <mi>
          A 
        </mi> 
        <mo> 
        </mo> 
       </mrow> 
       <mn>
         4 
       </mn> 
      </msubsup> 
      <mo>
        = 
      </mo> 
      <mn>
        2 
      </mn> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
     </mrow> 
    </math> (3)</p>
   <p>and</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         S 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mroot> 
       <mn>
         2 
       </mn> 
       <mn>
         4 
       </mn> 
      </mroot> 
      <mo>
        ⋅ 
      </mo> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
      </msub> 
     </mrow> 
    </math> (4)</p>
   <p>With T<sub>E</sub> = 255˚K we therefore obtain T<sub>S</sub> = 303˚K. This value is considerably higher than the observed 288˚K. The discrepancy between theory and observation is explained by the fact that the atmosphere is not totally opaque to long-wave radiation. The presence of the atmosphere raises the temperature at the Earth’s surface considerably. It is known that this effect is referred to as the greenhouse effect. We assume now that the atmosphere is semi-grey, i.e. it absorbs a constant fraction ε of the long-wave radiation, but is still transparent to solar radiation and emits upwards and downwards at a rate given by</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        B 
      </mi> 
      <mo>
        = 
      </mo> 
      <mi>
        ε 
      </mi> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
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       </mi> 
       <mn>
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       </mn> 
      </msubsup> 
      <mo>
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      </mo> 
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      <mfrac> 
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         ε 
       </mi> 
       <mrow> 
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        </mn> 
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        </mo> 
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          ϵ 
        </mi> 
       </mrow> 
      </mfrac> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
      <mo> 
      </mo> 
     </mrow> 
    </math> (5)</p>
   <p>where 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         A 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
     </mrow> 
    </math> is the radiation energy content of atmosphere. This leads to</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         S 
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      <mo>
        = 
      </mo> 
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      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
      </msub> 
     </mrow> 
    </math> (6)</p>
   <p>With ε = 0.78, we then obtain T<sub>S</sub> = 288˚K. We can conclude that in this simplified model only 22% or 42.8 W/m<sup>2</sup> of the IR radiation gets directly to space and a net power balance of 152 W/m<sup>2</sup>—as expected from (2)—gets permanently recycled as part of the backscattered radiation.</p>
   <p>For more appropriate multilayered atmospheres, we can take over the picture from the American Chemical Society (<xref ref-type="fig" rid="fig4">
     Figure 4
    </xref>).</p>
   <fig id="fig3" position="float">
    <label>Figure 3</label>
    <caption>
     <title>Figure 3. A single-layer model of the atmosphere, where S is the Solar irradiance, α is the cloud albedo and β the surface absorbance. U is the terrestrial radiation, of which a fraction (1 − ε) penetrates directly through the atmosphere to space and B is the radiation emitted by the atmosphere <xref ref-type="bibr" rid="scirp.136478-5">
       [5]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId31.jpeg?20241009013646" />
   </fig>
   <p>Instead of using layered-atmosphere models, there exist excellent semi-empirical formulas to be used. Night skies are shielded by clouds and humidity against radiation loss, while clear and dry conditions cannot compensate for strong radiation losses. This has been studied e.g. with the modified Swinbank model <xref ref-type="bibr" rid="scirp.136478-7">
     [7]
    </xref> <xref ref-type="bibr" rid="scirp.136478-8">
     [8]
    </xref>.</p>
   <fig id="fig4" position="float">
    <label>Figure 4</label>
    <caption>
     <title>Figure 4. A multilayer atmosphere model illustrated here with three layers <xref ref-type="bibr" rid="scirp.136478-6">
       [6]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId32.jpeg?20241009013646" />
   </fig>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         P 
       </mi> 
       <mrow> 
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        </mi> 
        <mi>
          B 
        </mi> 
       </mrow> 
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      </mo> 
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       </mo> 
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        </mn> 
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          + 
        </mo> 
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          K 
        </mi> 
        <msup> 
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           C 
         </mi> 
         <mn>
           2 
         </mn> 
        </msup> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        ⋅ 
      </mo> 
      <mn>
        8.78 
      </mn> 
      <mo>
        ⋅ 
      </mo> 
      <msup> 
       <mrow> 
        <mn>
          10 
        </mn> 
       </mrow> 
       <mrow> 
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          − 
        </mo> 
        <mn>
          13 
        </mn> 
       </mrow> 
      </msup> 
      <mo>
        ⋅ 
      </mo> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         S 
       </mi> 
      </msub> 
      <msup> 
       <mrow></mrow> 
       <mrow> 
        <mn>
          5.852 
        </mn> 
       </mrow> 
      </msup> 
      <mo>
        ⋅ 
      </mo> 
      <mi>
        R 
      </mi> 
      <msup> 
       <mi>
         H 
       </mi> 
       <mrow> 
        <mn>
          0.07195 
        </mn> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> (7)</p>
   <p>with</p>
   <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
    <tr> 
     <td class="aleft" width="9.35%"><p style="text-align:left">P<sub>SB</sub></p></td> 
     <td class="aleft" width="90.65%"><p style="text-align:left">upwards and downwards directed atmospheric radiation in W/m<sup>2</sup></p></td> 
    </tr> 
    <tr> 
     <td class="aleft" width="9.35%"><p style="text-align:left">K</p></td> 
     <td class="aleft" width="90.65%"><p style="text-align:left">0.34 clouds &lt; 2 km, 0.18 for 2 km &lt; altitude &lt; 5 km, 0.06 for &gt;5 km</p></td> 
    </tr> 
    <tr> 
     <td class="aleft" width="9.35%"><p style="text-align:left">C</p></td> 
     <td class="aleft" width="90.65%"><p style="text-align:left">Cloud cover (0 clear skies, 1 covered skies)</p></td> 
    </tr> 
    <tr> 
     <td class="aleft" width="9.35%"><p style="text-align:left">T</p></td> 
     <td class="aleft" width="90.65%"><p style="text-align:left">temperature in ˚K</p></td> 
    </tr> 
    <tr> 
     <td class="aleft" width="9.35%"><p style="text-align:left">RH</p></td> 
     <td class="aleft" width="90.65%"><p style="text-align:left">relative humidity in %</p></td> 
    </tr> 
   </table>
   <fig id="fig5" position="float">
    <label>Figure 5</label>
    <caption>
     <title>Figure 5. The effect of humidity on backward radiation compared between RH = 0.007% and RH = 50%. The blue line is the surface radiation at emissivity ε<sub>S</sub> ≅ 0.95 according to Stefan-Boltzmann.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId35.jpeg?20241009013646" />
   </fig>
   <p>As we can see in <xref ref-type="fig" rid="fig5">
     Figure 5
    </xref>, humidity has a strong impact and cloud coverage has even stronger impact on the back-radiation.</p>
   <fig id="fig6" position="float">
    <label>Figure 6</label>
    <caption>
     <title>Figure 6. Thermal backward radiation characteristics of cloudy skies at RH = 10% and RH = 50%.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId36.jpeg?20241009013646" />
   </fig>
   <p>The last figures (<xref ref-type="fig" rid="fig6">
     Figure 6
    </xref>) suggest indeed that climate predictions depend critically on values like average relative humidity, cloud coverage, and albedo. These values vary considerably over the surface, depending on local temperature and geographic details. Gridded models can account for such values over years of observations to a certain extent only. Instabilities and risks due to erroneous positive feedback loop modeling and/or negligence of external geo-or astrophysical influences are obvious.</p>
   <p>
    <xref ref-type="bibr" rid="scirp.136478-"></xref>Data from ground measurements (<xref ref-type="fig" rid="fig7">
     Figure 7
    </xref>) indicate that the downward (backward) radiation of the atmosphere shows indeed full saturation of the IR CO<sub>2</sub> bands and does not support noticeable additional Thermal Forcing (TF) by increasing CO<sub>2</sub> in the lower atmosphere. It shows almost complete saturation of the 15 µ-central peak and close-to-saturation of the surrounding 15 µ band edges. Early studies <xref ref-type="bibr" rid="scirp.136478-10">
     [10]
    </xref> concluded as well that TF will not be significantly influenced by a further increase of atmospheric CO<sub>2</sub> (<xref ref-type="fig" rid="fig8">
     Figure 8
    </xref>). On the other hand, it is well known that concentrations below 50% of the current level would be detrimental to plant growth and climate.</p>
   <fig id="fig7" position="float">
    <label>Figure 7</label>
    <caption>
     <title>Figure 7. Measurements <xref ref-type="bibr" rid="scirp.136478-9">
       [9]
      </xref> of the infrared emission spectrum of the cloud-free atmosphere at the arctic surface looking upward.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId37.jpeg?20241009013646" />
   </fig>
   <fig id="fig8" position="float">
    <label>Figure 8</label>
    <caption>
     <title>Figure 8. Studies of J. Koch at the Physical Department of Knut Angström <xref ref-type="bibr" rid="scirp.136478-10">
       [10]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId38.jpeg?20241009013646" />
   </fig>
   <p>The wavelength-dependent absorption 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         a 
       </mi> 
       <mi>
         λ 
       </mi> 
      </msub> 
     </mrow> 
    </math> and re-emission is calculated from Beers Law, where transmission is exponentially reduced by increasing absorber thickness L</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
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         a 
       </mi> 
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      <mrow> 
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       <mi>
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      </mn> 
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        − 
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         t 
       </mi> 
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         λ 
       </mi> 
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       </mi> 
       <mo>
         ) 
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        = 
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      <mn>
        1 
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      </mo> 
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       <mrow> 
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         </mi> 
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       <mrow> 
        <msub> 
         <mi>
           I 
         </mi> 
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           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mn>
           0 
         </mn> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mn>
        1 
      </mn> 
      <mo>
        − 
      </mo> 
      <msup> 
       <mi>
         e 
       </mi> 
       <mrow> 
        <mo>
          − 
        </mo> 
        <mi>
          k 
        </mi> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mi>
            λ 
          </mi> 
          <mo>
            , 
          </mo> 
          <mi>
            L 
          </mi> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> (8)</p>
   <p>Kirchhoff’s law requires that absorbed radiation should get emitted again, or in other terms absorptivity should equal emissivity. For a limited wavelength window, Δλ the specific emissivity ε<sub><sub>Δ</sub></sub><sub>λ</sub> (L) is therefore obtained by</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mo> 
      </mo> 
      <msub> 
       <mi>
         ε 
       </mi> 
       <mrow> 
        <mi>
          Δ 
        </mi> 
        <mi>
          λ 
        </mi> 
       </mrow> 
      </msub> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mi>
         L 
       </mi> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <mo> 
      </mo> 
      <mfrac> 
       <mrow> 
        <msubsup> 
         <mstyle mathsize="140%" displaystyle="true"> 
          <mo>
            ∫ 
          </mo> 
         </mstyle> 
         <mrow> 
          <msub> 
           <mi>
             λ 
           </mi> 
           <mn>
             1 
           </mn> 
          </msub> 
         </mrow> 
         <mrow> 
          <msub> 
           <mi>
             λ 
           </mi> 
           <mn>
             2 
           </mn> 
          </msub> 
         </mrow> 
        </msubsup> 
        <msub> 
         <mi>
           I 
         </mi> 
         <mi>
           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mn>
           0 
         </mn> 
         <mo>
           ) 
         </mo> 
        </mrow> 
        <mo>
          ⋅ 
        </mo> 
        <msub> 
         <mi>
           a 
         </mi> 
         <mi>
           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mi>
           L 
         </mi> 
         <mo>
           ) 
         </mo> 
        </mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          λ 
        </mi> 
       </mrow> 
       <mrow> 
        <msubsup> 
         <mstyle mathsize="140%" displaystyle="true"> 
          <mo>
            ∫ 
          </mo> 
         </mstyle> 
         <mn>
           0 
         </mn> 
         <mi>
           ∞ 
         </mi> 
        </msubsup> 
        <msub> 
         <mi>
           I 
         </mi> 
         <mi>
           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mn>
           0 
         </mn> 
         <mo>
           ) 
         </mo> 
        </mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          λ 
        </mi> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (9)</p>
   <p>where the spectral extinction coefficient 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         a 
       </mi> 
       <mi>
         λ 
       </mi> 
      </msub> 
     </mrow> 
    </math> is obtained from HITRAN or similar databases.</p>
   <p>The significance of the unsaturated edges in 15 µ band is highly overestimated, as it can be easily shown that with an extinction &lt; 3 they contribute only 0.17% to the full band when we consider their respective integrals (<xref ref-type="fig" rid="fig9">
     Figure 9
    </xref>). This has also been shown by Howard <xref ref-type="bibr" rid="scirp.136478-11">
     [11]
    </xref> used further in this study.</p>
   <fig id="fig9" position="float">
    <label>Figure 9</label>
    <caption>
     <title>Figure 9. Transmission and Extinction Ratios in the 15 µ band.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId47.jpeg?20241009013645" />
   </fig>
   <p>The infrared (IR) spectra of the four main GH gases over a path length of 100 meters are presented in <xref ref-type="fig" rid="fig10">
     Figure 10
    </xref>, their concentrations being those that pertain to the atmosphere at sea level at 45% relative humidity.</p>
   <fig id="fig10" position="float">
    <label>Figure 10</label>
    <caption>
     <title>Figure 10. Infrared spectra of the main greenhouse gases as calculated using the HITRAN database; transmission is plotted against wavenumber (cm<sup>−</sup><sup>1</sup>) <xref ref-type="bibr" rid="scirp.136478-4">
       [4]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId48.jpeg?20241009013645" />
   </fig>
   <p>The 0 - 500 cm<sup>−</sup><sup>1</sup> band of H<sub>2</sub>O at 45% RH absorbs 66% of the surface IR radiation. The 1300 - 1800 cm<sup>−</sup><sup>1</sup> band absorbs another 1%.</p>
   <p>A 1 bar pressure atmosphere would have a theoretical thickness of 8.2 km. To visualize the effective column height of current CO<sub>2</sub> at different altitudes, we get the following <xref ref-type="table" rid="table1">
     Table 1
    </xref>.</p>
   <table-wrap id="table1">
    <label>
     <xref ref-type="table" rid="table1">
      Table 1
     </xref></label>
    <caption>
     <title>
      <xref ref-type="bibr" rid="scirp.136478-"></xref>Table 1. Equivalent of 400 ppm atmospheric CO<sub>2</sub> in meters at 1 bar atmospheric pressure.</title>
    </caption>
    <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
     <tr> 
      <td class="custom-bottom-td acenter" width="17.85%"><p style="text-align:center">H [m]</p></td> 
      <td class="custom-bottom-td acenter" width="17.86%"><p style="text-align:center">M(z)/M</p></td> 
      <td class="custom-bottom-td acenter" width="17.86%"><p style="text-align:center">1-M(z)/M</p></td> 
      <td class="custom-bottom-td acenter" width="24.10%"><p style="text-align:center"> 
        <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
          <msub> 
           <mstyle mathvariant="bold" mathsize="normal"> 
            <mi>
              L 
            </mi> 
           </mstyle> 
           <mrow> 
            <msub> 
             <mrow> 
              <mtext>
                CO 
              </mtext> 
             </mrow> 
             <mtext>
               2 
             </mtext> 
            </msub> 
           </mrow> 
          </msub> 
          <msup> 
           <mrow></mrow> 
           <mrow> 
            <mtext>
              up 
            </mtext> 
           </mrow> 
          </msup> 
         </mrow> 
        </math> [cm]</p></td> 
      <td class="custom-bottom-td acenter" width="22.32%"><p style="text-align:center"> 
        <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
          <msub> 
           <mstyle mathvariant="bold" mathsize="normal"> 
            <mi>
              L 
            </mi> 
           </mstyle> 
           <mrow> 
            <msub> 
             <mrow> 
              <mtext>
                CO 
              </mtext> 
             </mrow> 
             <mtext>
               2 
             </mtext> 
            </msub> 
           </mrow> 
          </msub> 
          <msup> 
           <mrow></mrow> 
           <mrow> 
            <mtext>
              down 
            </mtext> 
           </mrow> 
          </msup> 
         </mrow> 
        </math> [cm]</p></td> 
     </tr> 
     <tr> 
      <td class="custom-top-td acenter" width="17.85%"><p style="text-align:center">0</p></td> 
      <td class="custom-top-td acenter" width="17.86%"><p style="text-align:center">0</p></td> 
      <td class="custom-top-td acenter" width="17.86%"><p style="text-align:center">1</p></td> 
      <td class="custom-top-td acenter" width="24.10%"><p style="text-align:center">344.83</p></td> 
      <td class="custom-top-td acenter" width="22.32%"><p style="text-align:center">0.00</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">500</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.06893722</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.93106278</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">321.06</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">23.77</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">1000</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.1331221</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.8668779</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">298.93</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">45.90</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">5000</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.51045834</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.48954166</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">168.81</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">176.02</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">10,000</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.76034896</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.23965104</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">82.64</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">262.19</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">20,000</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.94256738</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.05743262</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">19.80</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">325.03</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">30,000</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.98623621</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.01376379</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">4.75</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">340.08</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">40,000</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.99670149</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0.00329851</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">1.14</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">343.69</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="17.85%"><p style="text-align:center">¥</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">1</p></td> 
      <td class="acenter" width="17.86%"><p style="text-align:center">0</p></td> 
      <td class="acenter" width="24.10%"><p style="text-align:center">0.00</p></td> 
      <td class="acenter" width="22.32%"><p style="text-align:center">344.83</p></td> 
     </tr> 
    </table>
   </table-wrap>
   <p>where we obtained the altitude-dependent mass M from the barometric formula</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mfrac> 
       <mrow> 
        <mi>
          M 
        </mi> 
        <mrow> 
         <mo>
           [ 
         </mo> 
         <mi>
           z 
         </mi> 
         <mo>
           ] 
         </mo> 
        </mrow> 
       </mrow> 
       <mi>
         M 
       </mi> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mn>
        1 
      </mn> 
      <mo>
        − 
      </mo> 
      <msup> 
       <mi>
         e 
       </mi> 
       <mrow> 
        <mo>
          − 
        </mo> 
        <mfrac> 
         <mi>
           z 
         </mi> 
         <mi>
           H 
         </mi> 
        </mfrac> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> (10)</p>
   <p>with H = 7 km. The total mass below the altitude z is calculated as</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        M 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mi>
         z 
       </mi> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <munderover> 
       <mstyle mathsize="140%" displaystyle="true"> 
        <mo>
          ∫ 
        </mo> 
       </mstyle> 
       <mn>
         0 
       </mn> 
       <mi>
         z 
       </mi> 
      </munderover> 
      <mi>
        ρ 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <msup> 
        <mi>
          z 
        </mi> 
        <mo>
          ′ 
        </mo> 
       </msup> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mi>
        d 
      </mi> 
      <msup> 
       <mi>
         z 
       </mi> 
       <mo>
         ′ 
       </mo> 
      </msup> 
      <mo>
        = 
      </mo> 
      <mi>
        ρ 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mn>
         0 
       </mn> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mi>
        H 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mn>
          1 
        </mn> 
        <mo>
          − 
        </mo> 
        <msup> 
         <mi>
           e 
         </mi> 
         <mrow> 
          <mo>
            − 
          </mo> 
          <mfrac> 
           <mi>
             z 
           </mi> 
           <mi>
             H 
           </mi> 
          </mfrac> 
         </mrow> 
        </msup> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
     </mrow> 
    </math> (11)</p>
   <p>The thickness of the CO<sub>2</sub>-layer 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         L 
       </mi> 
       <mrow> 
        <msub> 
         <mrow> 
          <mtext>
            CO 
          </mtext> 
         </mrow> 
         <mtext>
           2 
         </mtext> 
        </msub> 
       </mrow> 
      </msub> 
     </mrow> 
    </math> is a calculated value for a standard atmosphere at 1013.25 hPa and a total weight of 5.13 × 10<sup>15</sup> tons of 8.21 km air at current 400 ppm. The table helps to understand the order of magnitudes when studying the atmospheric greenhouse effects of CO<sub>2</sub>. It also illustrates that 75% of the total CO<sub>2</sub> is contained within the troposphere below 10 km altitude and 95% is below 20 km. If we were to concentrate all the CO<sub>2</sub> (at 400 ppm) of an 8.2 km thick atmosphere with a pressure of 1 bar in a single column, the height of the column would be around 3.5 m. The optical path length of CO<sub>2</sub> IR radiation at these conditions is below 1 cm and therefore we can expect full saturation already at current concentrations.</p>
   <p>From thermodynamics we get</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mfrac> 
       <mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          T 
        </mi> 
       </mrow> 
       <mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          z 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mrow> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <mi>
            γ 
          </mi> 
          <mo>
            − 
          </mo> 
          <mn>
            1 
          </mn> 
         </mrow> 
         <mi>
           γ 
         </mi> 
        </mfrac> 
        <mi>
          m 
        </mi> 
        <mi>
          g 
        </mi> 
       </mrow> 
       <mo>
         / 
       </mo> 
       <mi>
         R 
       </mi> 
      </mrow> 
     </mrow> 
    </math> (12)</p>
   <p>where 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        γ 
      </mi> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           c 
         </mi> 
         <mi>
           P 
         </mi> 
        </msub> 
       </mrow> 
       <mrow> 
        <msub> 
         <mi>
           c 
         </mi> 
         <mi>
           v 
         </mi> 
        </msub> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (e.g. 7/5 for dry air), m = 29 g/mol and R is the gas constant. For the International Standard Atmosphere with 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        γ 
      </mi> 
      <mo>
        = 
      </mo> 
      <mn>
        1.26 
      </mn> 
     </mrow> 
    </math> we obtain 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mfrac> 
       <mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          T 
        </mi> 
       </mrow> 
       <mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          z 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mo> 
      </mo> 
      <mo>
        − 
      </mo> 
      <mn>
        6.5 
      </mn> 
      <mrow> 
       <mrow> 
        <mtext>
          dK 
        </mtext> 
       </mrow> 
       <mo>
         / 
       </mo> 
       <mrow> 
        <mtext>
          km 
        </mtext> 
       </mrow> 
      </mrow> 
     </mrow> 
    </math>. Using this formula, we can estimate the LW back-radiation at higher altitudes (<xref ref-type="fig" rid="fig11">
     Figure 11
    </xref>), when substituting the 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         s 
       </mi> 
      </msub> 
      <mo>
        → 
      </mo> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         s 
       </mi> 
      </msub> 
      <mo>
        − 
      </mo> 
      <mfrac> 
       <mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          T 
        </mi> 
       </mrow> 
       <mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          z 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        × 
      </mo> 
      <mi>
        Δ 
      </mi> 
      <mi>
        z 
      </mi> 
     </mrow> 
    </math>.</p>
   <fig id="fig11" position="float">
    <label>Figure 11</label>
    <caption>
     <title>Figure 11. Estimated back-radiation in a standard atmosphere at three different surface temperatures.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId69.jpeg?20241009013645" />
   </fig>
   <p>This might be compared to measurements taken by a group from ETH Zurich in 1998 (<xref ref-type="fig" rid="fig12">
     Figure 12
    </xref>).</p>
   <fig id="fig12" position="float">
    <label>Figure 12</label>
    <caption>
     <title>Figure 12. Altitude dependence of back-radiation with courtesy from C. Fröhlich <xref ref-type="bibr" rid="scirp.136478-12">
       [12]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId70.jpeg?20241009013646" />
   </fig>
   <p>The absorptivity a can be calculated using Beer’s law for an infrared radiator from Equation (8) in a more generalized form.</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        a 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mi>
          c 
        </mi> 
        <mo>
          , 
        </mo> 
        <mi>
          L 
        </mi> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <mn>
        1 
      </mn> 
      <mo>
        − 
      </mo> 
      <mi>
        t 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mi>
          c 
        </mi> 
        <mo>
          , 
        </mo> 
        <mi>
          L 
        </mi> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <mn>
        1 
      </mn> 
      <mo>
        − 
      </mo> 
      <mfrac> 
       <mrow> 
        <mi>
          I 
        </mi> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mi>
            c 
          </mi> 
          <mo>
            , 
          </mo> 
          <mi>
            L 
          </mi> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
       <mrow> 
        <mi>
          I 
        </mi> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mi>
            c 
          </mi> 
          <mo>
            , 
          </mo> 
          <mn>
            0 
          </mn> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mn>
        1 
      </mn> 
      <mo>
        − 
      </mo> 
      <msup> 
       <mi>
         e 
       </mi> 
       <mrow> 
        <mo>
          − 
        </mo> 
        <mi>
          k 
        </mi> 
        <mo>
          . 
        </mo> 
        <mi>
          c 
        </mi> 
        <mo>
          . 
        </mo> 
        <mi>
          L 
        </mi> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> (13)</p>
   <p>From Akram <xref ref-type="bibr" rid="scirp.136478-13">
     [13]
    </xref> we can obtain k = 1.82 × 10<sup>−</sup><sup>6</sup> ppm<sup>−</sup><sup>1</sup>cm<sup>−</sup><sup>1</sup> to calculate a as a function of L.</p>
   <p>
    <xref ref-type="fig" rid="fig13">
     Figure 13
    </xref> shows IR absorption saturation at current CO<sub>2</sub> levels already at air column length below 20 m. From this simple model, we must conclude that mechanisms other than CO<sub>2</sub> increases must explain significant atmospheric thermal enhancement (ATE) in the total energy budget.</p>
   <fig id="fig13" position="float">
    <label>Figure 13</label>
    <caption>
     <title>Figure 13. Absorbance of EMIRS200 IR light source in CO<sub>2</sub> from <xref ref-type="bibr" rid="scirp.136478-13">
       [13]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId73.jpeg?20241009013646" />
   </fig>
   <p>Howards <xref ref-type="bibr" rid="scirp.136478-11">
     [11]
    </xref> parameterizes the total absorption of strong lines at 15 µm and 4.3 µm according to</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mo>
        ∫ 
      </mo> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mi>
         ν 
       </mi> 
      </msub> 
      <mi>
        d 
      </mi> 
      <mi>
        ν 
      </mi> 
      <mo>
        = 
      </mo> 
      <mi>
        C 
      </mi> 
      <mo>
        + 
      </mo> 
      <mi>
        D 
      </mi> 
      <mi>
        log 
      </mi> 
      <mi>
        w 
      </mi> 
      <mo>
        + 
      </mo> 
      <mi>
        K 
      </mi> 
      <mi>
        log 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mi>
          p 
        </mi> 
        <mo>
          + 
        </mo> 
        <mi>
          P 
        </mi> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
     </mrow> 
    </math> (14)</p>
   <p>and</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mover accent="true"> 
       <mi>
         A 
       </mi> 
       <mo>
         ¯ 
       </mo> 
      </mover> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <mo>
          ∫ 
        </mo> 
        <msub> 
         <mi>
           A 
         </mi> 
         <mi>
           ν 
         </mi> 
        </msub> 
        <mi>
          d 
        </mi> 
        <mi>
          ν 
        </mi> 
       </mrow> 
       <mrow> 
        <msub> 
         <mi>
           ν 
         </mi> 
         <mn>
           2 
         </mn> 
        </msub> 
        <mo>
          − 
        </mo> 
        <msub> 
         <mi>
           ν 
         </mi> 
         <mn>
           1 
         </mn> 
        </msub> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (15)</p>
   <p>Substituting these values for 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         ν 
       </mi> 
       <mn>
         2 
       </mn> 
      </msub> 
      <mo>
        − 
      </mo> 
      <msub> 
       <mi>
         ν 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mn>
        250 
      </mn> 
      <msup> 
       <mrow> 
        <mtext>
          cm 
        </mtext> 
       </mrow> 
       <mrow> 
        <mo>
          − 
        </mo> 
        <mn>
          1 
        </mn> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> in the 15 µm band and 
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         ν 
       </mi> 
       <mn>
         2 
       </mn> 
      </msub> 
      <mo>
        − 
      </mo> 
      <msub> 
       <mi>
         ν 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mn>
        340 
      </mn> 
      <msup> 
       <mrow> 
        <mtext>
          cm 
        </mtext> 
       </mrow> 
       <mrow> 
        <mo>
          − 
        </mo> 
        <mn>
          1 
        </mn> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> for the 4.3 µm band we obtain for a standard atmospheric layer of 1 km thickness (<xref ref-type="fig" rid="fig14">
     Figure 14
    </xref>) and for an 8.2 km thick atmosphere (<xref ref-type="fig" rid="fig15">
     Figure 15
    </xref>) as used for <xref ref-type="table" rid="table2">
     Table 2
    </xref>.</p>
   <fig id="fig14" position="float">
    <label>Figure 14</label>
    <caption>
     <title>Figure 14. Band absorption within 1 km standard atmosphere <xref ref-type="bibr" rid="scirp.136478-11">
       [11]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId82.jpeg?20241009013646" />
   </fig>
   <fig id="fig15" position="float">
    <label>Figure 15</label>
    <caption>
     <title>Figure 15. Absorption ratios for the two strong CO<sub>2</sub> absorption bands <xref ref-type="bibr" rid="scirp.136478-11">
       [11]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId83.jpeg?20241009013646" />
   </fig>
   <p>
    <xref ref-type="bibr" rid="scirp.136478-"></xref></p>
   <p>An increase from 400 to 800 ppm in CO<sub>2</sub> shows no measurable increase in IR absorption for the 15 µ and 4.3 µ bands and therefore total saturation. Other elaborate studies <xref ref-type="bibr" rid="scirp.136478-14">
     [14]
    </xref> estimate the total terrestrial long wavelength absorptivity a<sub>LW</sub>.</p>
   <p>
    <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        a 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mi>
         L 
       </mi> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msubsup> 
         <mstyle mathsize="140%" displaystyle="true"> 
          <mo>
            ∫ 
          </mo> 
         </mstyle> 
         <mn>
           0 
         </mn> 
         <mi>
           ∞ 
         </mi> 
        </msubsup> 
        <msub> 
         <mi>
           I 
         </mi> 
         <mi>
           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mn>
           0 
         </mn> 
         <mo>
           ) 
         </mo> 
        </mrow> 
        <mo>
          ⋅ 
        </mo> 
        <msub> 
         <mi>
           a 
         </mi> 
         <mi>
           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mi>
           L 
         </mi> 
         <mo>
           ) 
         </mo> 
        </mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          λ 
        </mi> 
       </mrow> 
       <mrow> 
        <msubsup> 
         <mstyle mathsize="140%" displaystyle="true"> 
          <mo>
            ∫ 
          </mo> 
         </mstyle> 
         <mn>
           0 
         </mn> 
         <mi>
           ∞ 
         </mi> 
        </msubsup> 
        <msub> 
         <mi>
           I 
         </mi> 
         <mi>
           λ 
         </mi> 
        </msub> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mn>
           0 
         </mn> 
         <mo>
           ) 
         </mo> 
        </mrow> 
        <mi>
          d 
        </mi> 
        <mi>
          λ 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        × 
      </mo> 
      <mn>
        100 
      </mn> 
      <mi>
        % 
      </mi> 
     </mrow> 
    </math> (16)</p>
   <p>and obtain the following results for different latitudes.</p>
   <table-wrap id="table2">
    <label>
     <xref ref-type="table" rid="table2">
      Table 2
     </xref></label>
    <caption>
     <title>
      <xref ref-type="bibr" rid="scirp.136478-"></xref>Table 2. Calculated a<sub>LW</sub> from Harde <xref ref-type="bibr" rid="scirp.136478-14">
       [14]
      </xref> for various CO<sub>2</sub> concentrations.</title>
    </caption>
    <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
     <tr> 
      <td rowspan="2" class="acenter" width="15.62%"><p style="text-align:center">CO<sub>2</sub> (ppm)</p></td> 
      <td class="acenter" width="84.38%" colspan="5"><p style="text-align:center">absorptivities ε (%)</p></td> 
     </tr> 
     <tr> 
      <td class="custom-bottom-td custom-top-td acenter" width="15.39%"><p style="text-align:center">tropics</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="18.01%"><p style="text-align:center">mid-latitudes</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.52%"><p style="text-align:center">high-latitudes</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="18.16%"><p style="text-align:center">average 3 zones</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.29%"><p style="text-align:center">global mean</p></td> 
     </tr> 
     <tr> 
      <td class="custom-top-td acenter" width="15.62%"><p style="text-align:center">0</p></td> 
      <td class="custom-top-td acenter" width="15.39%"><p style="text-align:center">81.90</p></td> 
      <td class="custom-top-td acenter" width="18.01%"><p style="text-align:center">69.44</p></td> 
      <td class="custom-top-td acenter" width="16.52%"><p style="text-align:center">58.98</p></td> 
      <td class="custom-top-td acenter" width="18.16%"><p style="text-align:center">74.68</p></td> 
      <td class="custom-top-td acenter" width="16.29%"><p style="text-align:center">77.02</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">35</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">83.80</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">74.48</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">67.04</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">78.43</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">80.08</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">70</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">84.18</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">75.35</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">68.32</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">79.10</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">80.62</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">140</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">84.65</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">76.31</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">69.80</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">79.86</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">81.29</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">210</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">84.99</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">77.00</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">70.77</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">80.40</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">81.76</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">280</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">85.28</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">77. 51</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">71.52</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">80.83</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">82.14</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">350</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">85.53</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">77.95</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">72.14</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">81.19</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">82.45</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">380</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">85.65</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">78.12</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">72.38</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">81.34</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">82.58</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">420</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">85.76</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">78.33</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">72.68</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">81.51</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">82.74</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">490</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">85.91</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">78.67</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">73.16</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">81.80</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">83.00</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">560</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">86.16</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">78.98</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">73.61</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">82.06</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">83.24</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">630</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">86.35</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">79.29</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">74.02</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">82.32</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">83.46</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">700</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">86.52</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">79.58</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">74.41</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">82.56</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">83.68</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="15.62%"><p style="text-align:center">770</p></td> 
      <td class="acenter" width="15.39%"><p style="text-align:center">86.69</p></td> 
      <td class="acenter" width="18.01%"><p style="text-align:center">79.85</p></td> 
      <td class="acenter" width="16.52%"><p style="text-align:center">74.78</p></td> 
      <td class="acenter" width="18.16%"><p style="text-align:center">82.79</p></td> 
      <td class="acenter" width="16.29%"><p style="text-align:center">83.88</p></td> 
     </tr> 
    </table>
   </table-wrap>
   <p>The next figure (<xref ref-type="fig" rid="fig16">
     Figure 16
    </xref>) shows the global mean values of the table above. While our previous model neglected the layered structure by assuming an isobar atmosphere, Harde <xref ref-type="bibr" rid="scirp.136478-14">
     [14]
    </xref> was taking the barometric effects into account.</p>
   <fig id="fig16" position="float">
    <label>Figure 16</label>
    <caption>
     <title>Figure 16. Influence of doubling CO<sub>2 </sub>concentration on total absorptivity <xref ref-type="bibr" rid="scirp.136478-14">
       [14]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId86.jpeg?20241009013645" />
   </fig>
   <p>This study shows an upper limit of additional ATE by 1.5%. Average surface temperatures of T<sub>S</sub> = 288˚K correspond to 390 Wm<sup>−</sup><sup>2</sup> black body radiation. From the foregoing considerations, we learned that doubling CO<sub>2</sub> atmospheric concentrations from 400 to 800 ppm amounts to a maximum 1.5% change of energy absorbed in the atmosphere, i.e. a maximal 3 Wm<sup>−</sup><sup>2</sup> of back-radiation increase. Using the Stefan-Boltzmann formula we then obtain a first estimation of a corresponding temperature increase or GH contribution of 0.5˚K by having 800 ppm atmospheric content compared to the current 400 ppm. This also corresponds very well with the results from Wijngaarden and Happer <xref ref-type="bibr" rid="scirp.136478-2">
     [2]
    </xref>.</p>
   <fig id="fig17" position="float">
    <label>Figure 17</label>
    <caption>
     <title>Figure 17. Total atmospheric absorptivity changes by CO<sub>2</sub> increases calculated by Wijngaarden and Happer <xref ref-type="bibr" rid="scirp.136478-2">
       [2]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId87.jpeg?20241009013645" />
   </fig>
   <p>Ground measurements fall in line with <xref ref-type="fig" rid="fig17">
     Figure 17
    </xref> and indicate almost complete saturation of IR absorption in our atmosphere. We were therefore interested in designing a simple and independent low-cost experimental setup to demonstrate this saturated behavior. This setup and the results obtained will be described in the following section.</p>
  </sec><sec id="s2">
   <title>2. Experimental Methods</title>
   <p>The aim of this work was therefore to demonstrate in a simple manner that an IR-active gas can indeed influence the heat radiation of a body. Another aim was to demonstrate that CO<sub>2</sub> back-radiation is limited by saturation of absorbance. This was realized by two different experimental setups.</p>
   <p>The measurements should allow us to learn if the saturation behavior measured by others and discussed in the foregoing sections of this work are indeed correct and draw conclusions towards the question of CO<sub>2</sub>-induced global warming.</p>
   <p>According to the second law of thermodynamics, heat flows from hotter to colder objects (“downhill”), unless energy in some form is supplied to reverse the direction of heat flow or there is a medium that absorbs part of the energy and remits it isotropic, like greenhouse models describe the radiation household within our atmosphere. Our test atmospheres were studied against a −25˚C cooled black disc (“Lab Mode”) or a cloudless night sky (“Outdoor or Field Mode”) using the natural IR source of surface and atmospheric back radiation.</p>
   <fig id="fig18" position="float">
    <label>Figure 18</label>
    <caption>
     <title>(a) (b)Figure 18. experimental setup for (a) “Lab Mode” using a cooling compressor and (b) “Field Mode”.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="" />
   </fig>
   <fig id="fig18" position="float">
    <label>Figure 18</label>
    <caption>
     <title>(a) (b)Figure 18. experimental setup for (a) “Lab Mode” using a cooling compressor and (b) “Field Mode”.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId88.jpeg?20241009013647" />
   </fig>
   <fig id="fig18" position="float">
    <label>Figure 18</label>
    <caption>
     <title>(a) (b)Figure 18. experimental setup for (a) “Lab Mode” using a cooling compressor and (b) “Field Mode”.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId89.jpeg?20241009013647" />
   </fig>
   <p>Translating these setups into energy flow diagrams like in <xref ref-type="fig" rid="fig3">
     Figure 3
    </xref> &amp; <xref ref-type="fig" rid="fig4">
     Figure 4
    </xref> we first analyze <xref ref-type="fig" rid="fig18(a)">
     Figure 18(a)
    </xref>.</p>
   <p>As shown in <xref ref-type="fig" rid="fig19">
     Figure 19
    </xref> the “Lab Mode” IR sensor response is solely determined by the back-radiation from Test Atmosphere 2 with increasing CO<sub>2</sub> concentrations. The cold plate on the other side of the test cylinder allows thermal flow according to the second law of thermodynamics. Inside the volume of the test cylinder, a CO<sub>2</sub> detector and a low-power ventilator are mounted and run by a combined power pack. The communication with the CO<sub>2</sub> detector is performed via Bluetooth. β is the intensity ratio 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        β 
      </mi> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mi>
         D 
       </mi> 
      </msub> 
      <msup> 
       <mrow> 
        <mi>
          sin 
        </mi> 
       </mrow> 
       <mn>
         2 
       </mn> 
      </msup> 
      <mi>
        θ 
      </mi> 
     </mrow> 
    </math> for an IR detector with numerical aperture θ within the cylindrical test column and the detector area A<sub>D</sub> of 1 cm<sup>2</sup>. ε<sub>2</sub> solely depends on the GH gas concentrations in the test tube and the intensity</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mfrac> 
       <mi>
         β 
       </mi> 
       <mn>
         2 
       </mn> 
      </mfrac> 
      <msub> 
       <mi>
         ε 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
      <msub> 
       <mi>
         ε 
       </mi> 
       <mn>
         2 
       </mn> 
      </msub> 
      <mi>
        σ 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mn>
         1 
       </mn> 
       <mn>
         4 
       </mn> 
      </msubsup> 
     </mrow> 
    </math> is measured by a Thermopile IR detector from Thorlabs<sup><sup>®</sup></sup>.</p>
   <fig id="fig19" position="float">
    <label>Figure 19</label>
    <caption>
     <title>Figure 19. “Lab Mode” test bench with cooled black disc.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId94.jpeg?20241009013648" />
   </fig>
   <p>
    <xref ref-type="bibr" rid="scirp.136478-"></xref>In the “Field Mode (<xref ref-type="fig" rid="fig18(b)">
     Figure 18(b)
    </xref>) the test cylinder is rotated by 180˚, the cooling compressor is removed and replaced by the clear night sky. The temperature gradient in this case is the difference between 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         S 
       </mi> 
      </msub> 
     </mrow> 
    </math> and 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
      </msub> 
     </mrow> 
    </math>, where 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
      </msub> 
      <mo> 
      </mo> 
     </mrow> 
    </math>is derived from (7)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
      </msub> 
      <mo> 
      </mo> 
      <mo>
        = 
      </mo> 
      <mo> 
      </mo> 
      <mn>
        4 
      </mn> 
      <msqrt> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <mrow> 
           <mo>
             ( 
           </mo> 
           <mrow> 
            <mn>
              1 
            </mn> 
            <mo>
              + 
            </mo> 
            <mi>
              K 
            </mi> 
            <msup> 
             <mi>
               C 
             </mi> 
             <mn>
               2 
             </mn> 
            </msup> 
           </mrow> 
           <mo>
             ) 
           </mo> 
          </mrow> 
          <mo>
            ⋅ 
          </mo> 
          <mn>
            8.78 
          </mn> 
          <mo>
            ⋅ 
          </mo> 
          <msup> 
           <mrow> 
            <mn>
              10 
            </mn> 
           </mrow> 
           <mrow> 
            <mo>
              − 
            </mo> 
            <mn>
              13 
            </mn> 
           </mrow> 
          </msup> 
          <mo>
            ⋅ 
          </mo> 
          <msub> 
           <mi>
             T 
           </mi> 
           <mi>
             S 
           </mi> 
          </msub> 
          <msup> 
           <mrow></mrow> 
           <mrow> 
            <mn>
              5.852 
            </mn> 
           </mrow> 
          </msup> 
          <mo>
            ⋅ 
          </mo> 
          <mi>
            R 
          </mi> 
          <msup> 
           <mi>
             H 
           </mi> 
           <mrow> 
            <mn>
              0.07195 
            </mn> 
           </mrow> 
          </msup> 
         </mrow> 
         <mi>
           σ 
         </mi> 
        </mfrac> 
       </mrow> 
      </msqrt> 
     </mrow> 
    </math> (17)</p>
   <p>The down-dwelling radiation through the test chamber into the IR detector is determined by</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        P 
      </mi> 
      <mo>
        = 
      </mo> 
      <mi>
        β 
      </mi> 
      <msub> 
       <mi>
         e 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
      <msub> 
       <mi>
         e 
       </mi> 
       <mn>
         2 
       </mn> 
      </msub> 
      <mi>
        s 
      </mi> 
      <msubsup> 
       <mi>
         T 
       </mi> 
       <mi>
         E 
       </mi> 
       <mn>
         4 
       </mn> 
      </msubsup> 
     </mrow> 
    </math> (18)</p>
   <p>Using the parametrization of Howard <xref ref-type="bibr" rid="scirp.136478-11">
     [11]
    </xref> we obtain for CO<sub>2</sub> test atmospheres power densities below the detection limit of the IR sensor of 10 µW/cm<sup>2</sup> in this setup. Using stronger GH gases would significantly increase 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         ε 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
     </mrow> 
    </math> as we have tested in our experimental series. This would prove that CO<sub>2</sub> is a relatively weak greenhouse gas even when increasing its concentration into the percentage region, compared to other GH gases, which would be detectable already at relatively low concentrations.</p>
  </sec><sec id="s3">
   <title>3. Results</title>
   <p>The 7-liter test cylinder in <xref ref-type="fig" rid="fig18(a)">
     Figure 18(a)
    </xref> was filled up with pure nitrogen. CO<sub>2</sub> was added in 50 ml steps. Simultaneously the CO<sub>2</sub> detector was used to measure the concentration while the ventilator was trying to homogenize the gas volume admixture. In <xref ref-type="fig" rid="fig20">
     Figure 20
    </xref> the obtained power values of the IR detector are plotted.</p>
   <fig id="fig20" position="float">
    <label>Figure 20</label>
    <caption>
     <title>Figure 20. Run 2.2 and Run 2.3 in the Lab Mode mixing CO<sub>2</sub> into a pure nitrogen atmosphere.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId107.jpeg?20241009013648" />
   </fig>
   <p>From previous considerations, we can estimate the back-radiation by the parametrization of Howard <xref ref-type="bibr" rid="scirp.136478-11">
     [11]
    </xref> (Equations (14)-(15)).</p>
   <p>Low values of absorption length are outside the fit and measurement range of Howard but serve as extrapolations to estimate orders of magnitude. The trend in <xref ref-type="fig" rid="fig21">
     Figure 21
    </xref> is indeed very similar to the measured values in <xref ref-type="fig" rid="fig20">
     Figure 20
    </xref> and confirms the logarithmic behavior of the absorption.</p>
   <fig id="fig21" position="float">
    <label>Figure 21</label>
    <caption>
     <title>Figure 21. P calculated from Howard <xref ref-type="bibr" rid="scirp.136478-11">
       [11]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId108.jpeg?20241009013648" />
   </fig>
   <p>Using the values plotted in <xref ref-type="fig" rid="fig20">
     Figure 20
    </xref> we can estimate the fractional absorbance A<sub>f</sub> for varying CO<sub>2</sub> concentrations in N<sub>2</sub> atmospheres</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        ln 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <msub> 
           <mi>
             I 
           </mi> 
           <mn>
             0 
           </mn> 
          </msub> 
         </mrow> 
         <mrow> 
          <msub> 
           <mi>
             I 
           </mi> 
           <mn>
             1 
           </mn> 
          </msub> 
         </mrow> 
        </mfrac> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <mi>
        ε 
      </mi> 
      <mo>
        ⋅ 
      </mo> 
      <mi>
        c 
      </mi> 
      <mo>
        ⋅ 
      </mo> 
      <mi>
        d 
      </mi> 
     </mrow> 
    </math> (19)</p>
   <p>where c is the molar concentration (mol∙m<sup>−</sup><sup>3</sup>) and d air column length (m). We use the Beer-Lambert law (13) function, where ε. is the absorption coefficient of our test atmosphere to obtain</p>
   <p>
    <xref ref-type="bibr" rid="scirp.136478-"></xref> 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mo> 
      </mo> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mi>
         f 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           I 
         </mi> 
         <mi>
           o 
         </mi> 
        </msub> 
        <mo>
          − 
        </mo> 
        <mi>
          I 
        </mi> 
       </mrow> 
       <mi>
         I 
       </mi> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mn>
          1 
        </mn> 
        <mo>
          − 
        </mo> 
        <msup> 
         <mi>
           e 
         </mi> 
         <mrow> 
          <mo>
            − 
          </mo> 
          <mi>
            ε 
          </mi> 
          <mo>
            ⋅ 
          </mo> 
          <mi>
            c 
          </mi> 
          <mo>
            . 
          </mo> 
          <mi>
            d 
          </mi> 
         </mrow> 
        </msup> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
     </mrow> 
    </math> (20)</p>
   <p>Using (20) it is possible to calculate the back-radiation for a given air column length (<xref ref-type="fig" rid="fig22">
     Figure 22
    </xref>) at ground-level conditions for saturation at a given ppm CO<sub>2</sub>.</p>
   <fig id="fig22" position="float">
    <label>Figure 22</label>
    <caption>
     <title>Figure 22. Calculated back-radiation for different long air columns at room temperature (RT).</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId113.jpeg?20241009013649" />
   </fig>
   <fig id="fig23" position="float">
    <label>Figure 23</label>
    <caption>
     <title>Figure 23. Calculated effect of CO<sub>2</sub> doubling on backscattered power in two air columns.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId114.jpeg?20241009013649" />
   </fig>
   <p>In <xref ref-type="fig" rid="fig23">
     Figure 23
    </xref> we calculated the corresponding atmospheric thermal enhancement (ATE) when doubling the actual CO<sub>2</sub> concentration for a homogeneous atmosphere of 100 m thickness at 1 bar. The result shows a rapid reduction in ATE with increasing the length of the air columns. Applying this result to the real world means saturation in CO<sub>2</sub> concentration with respect to measurable thermal effects. This will be discussed in more detail in the next chapter.</p>
   <p>
    <xref ref-type="fig" rid="fig24">
     Figure 24
    </xref> was the result obtained in the so-called Field Measurements against a black cloudless night sky. There is a clear indication that no measurable backscattering is observable when adding CO<sub>2</sub> from 0 to 5000 ppm to the system. Equation (9) with ε<sub>1</sub>-0.2, ε<sub>2</sub> &lt; 0.2 and β-0.02 gives βε<sub>1</sub>ε<sub>2</sub> &lt; 0.0004 and therefore P = βε<sub>1</sub>ε<sub>2</sub>σT<sub>E</sub><sup>4</sup> &lt; 3 × 10<sup>−</sup><sup>2</sup> W/m<sup>2</sup> = 3 µW/cm<sup>2</sup>. The working power of the used Thermopile Thorlabs detector with 1 cm<sup>2</sup> active area starts at 10 µW. For stronger GH gases it would be possible to exceed the detection limits and obtain a response. This was shown by using the Freon gas C<sub>2</sub>H<sub>2</sub>F<sub>4</sub> as shown in the next figure.</p>
   <fig id="fig24" position="float">
    <label>Figure 24</label>
    <caption>
     <title>Figure 24. Field Mode comparative measurement CO<sub>2</sub> and Freon (C<sub>2</sub>H<sub>2</sub>F<sub>4</sub>).</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId115.jpeg?20241009013649" />
   </fig>
   <p>The detector was indeed reacting when C<sub>2</sub>H<sub>2</sub>F<sub>4</sub> Freon was added up to 1400 ppm. The measured absorbance of C<sub>2</sub>H<sub>2</sub>F<sub>4</sub> by NIST <xref ref-type="bibr" rid="scirp.136478-15">
     [15]
    </xref> (<xref ref-type="fig" rid="fig25">
     Figure 25
    </xref>) is indeed, up to 100 times stronger than CO<sub>2</sub> measured by the same institution <xref ref-type="bibr" rid="scirp.136478-16">
     [16]
    </xref> (<xref ref-type="fig" rid="fig26">
     Figure 26
    </xref>).</p>
   <fig id="fig25" position="float">
    <label>Figure 25</label>
    <caption>
     <title>Figure 25. Absorbance of Freon (C<sub>2</sub>H<sub>2</sub>F<sub>4</sub>) <xref ref-type="bibr" rid="scirp.136478-15">
       [15]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId116.jpeg?20241009013649" />
   </fig>
   <fig id="fig26" position="float">
    <label>Figure 26</label>
    <caption>
     <title>Figure 26. Absorbance of carbon dioxide (CO<sub>2</sub>) <xref ref-type="bibr" rid="scirp.136478-16">
       [16]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId117.jpeg?20241009013649" />
   </fig>
  </sec><sec id="s4">
   <title>4. Discussion</title>
   <p>Atmospheric windows, especially the optical and infrared, affect the distribution of energy flows and temperatures within Earth’s energy balance. The windows depend upon clouds, water vapor, trace greenhouse gases, and other components of the atmosphere.</p>
   <p>Out of an average of 341 watts per square meter (W/m<sup>2</sup>) of solar irradiance at the top of the atmosphere, about 161 W/m<sup>2</sup> reaches the surface via atmospheric windows and through clouds (albedo). IR absorption by the atmosphere and corresponding atmospheric heating leads to an equilibrium of 333 W/m<sup>2</sup> of back-radiation and outgoing LW surface radiation of 396 W/m<sup>2</sup> latent heat from evaporation (80 W/m<sup>2</sup>) and other thermal losses (17 W/m<sup>2</sup>). Our measurements align with limitations to an increase of maximum 3W/m<sup>2</sup> back-radiation by doubling the CO<sub>2</sub> content from 400 to 800 ppm. This minor contribution should not exceed a temperature increase of more than 0.5˚K a value, which is not within the range of significant impact for climatic changes and much lower than annual temperature variations in all regions of the earth.</p>
   <fig id="fig27" position="float">
    <label>Figure 27</label>
    <caption>
     <title>Figure 27. “The Atmospheric Window”, NOAA <xref ref-type="bibr" rid="scirp.136478-17">
       [17]
      </xref> <xref ref-type="bibr" rid="scirp.136478-18">
       [18]
      </xref>.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/4701276-rId118.jpeg?20241009013650" />
   </fig>
   <p>From <xref ref-type="fig" rid="fig27">
     Figure 27
    </xref> it should be obvious that only minor contributions can be obtained from the edges of the two prominent CO<sub>2</sub> windows. Water vapor is indeed the major absorber and other GH gases are not relevant at their current concentrations. Increased water vapor should also lead to cloud coverage at constant aerosol concentrations. Slight variations of solar constants and cosmic rays using balanced feedback loops should allow for long-term thermal equilibrium. Fluctuations are often due to unnatural time scales, long-time constants, and statistical noise. In a simplified picture, where we consider an atmosphere with uniform density corresponding to the surface density ρ = 1.2225 kg∙m<sup>−</sup><sup>3</sup> we obtain an atmospheric air column weight of 10,300 kg∙m<sup>−</sup><sup>2</sup> at a length of 8425 m. From our measurements and a simple Beer-Lambert model of such an air column, we found (<xref ref-type="fig" rid="fig13">
     Figure 13
    </xref>) already at 200 m complete saturation at current CO<sub>2</sub> concentrations, without significant contributions to further ATE. Schildknecht <xref ref-type="bibr" rid="scirp.136478-19">
     [19]
    </xref> using similar arguments finds that doubling of the CO<sub>2</sub> content in air from 380 ppm to 760 ppm in one century is an increase of ∆T ≃ 0.5˚C, which corresponds well with the expected values of this and other works.</p>
   <p>Our results should therefore contribute to previously accepted findings which do not indicate any reason to cause climate run-aways by increased CO<sub>2</sub> contents. Ekholm <xref ref-type="bibr" rid="scirp.136478-20">
     [20]
    </xref> introduced the idea of fictitious atmospheric radiation levels due to CO<sub>2</sub> increase. According to him secular cooling of the earth is the principal cause of variation in the quantity of carbon dioxide in the atmosphere. He explained how carbon dioxide is a key player in the greenhouse effect and how his conclusion was based on the earlier work of Fourier, Pouillet, Tyndall, and others. According to their estimations, a tripling of carbon dioxide levels will raise global temperatures by 7˚C to 9˚C. An increase in carbon dioxide should heat high latitudes more than the tropics and create a warmer more uniform climate over the entire Earth. Hansen <xref ref-type="bibr" rid="scirp.136478-21">
     [21]
    </xref> testified this before Congress and raised public awareness of climate change. Hansen was claiming “The most powerful feedback is provided by water vapor”. While this argument seems strong it must be questioned considering paleoclimatic facts e.g. higher temperatures in the Eemian interglacial period 120,000 years ago and the Holstein interglacial period 5,000 to 15,000 years ago did not trigger a tipping point or a galloping greenhouse effect. Sarker <xref ref-type="bibr" rid="scirp.136478-22">
     [22]
    </xref> mentions correctly that 0.24 gigatons of carbon were emitted into the atmosphere for 50,000 years to cause this severe warming known as the PETM. In comparison, humans emit 10 gigatons, or approximately fifty times that amount of carbon annually, into the atmosphere. There is no doubt that a warming climate has negative impacts on various important properties, such as the aridity index for rivers <xref ref-type="bibr" rid="scirp.136478-23">
     [23]
    </xref>. Even just studying rivers there are many other anthropogenic impacts not related to CO<sub>2</sub> which dramatically influence their current and future environmental integrity <xref ref-type="bibr" rid="scirp.136478-24">
     [24]
    </xref> <xref ref-type="bibr" rid="scirp.136478-25">
     [25]
    </xref>. Current global warming can be natural and/or anthropogenic. To be sure, we must understand how it was possible to obtain 100 m higher sea levels or to end up by glaciation over large areas of landmasses without anthropogenic CO<sub>2</sub> emission triggers. Additional energy or energy loss was due to natural forces which must be fully understood before drawing wrong conclusions. Ice core data do not indicate fundamental different temperature variations over the last 10,000 years. We also need a better understanding of the validity of power flow diagrams as published by the IPCC. It would be a great success of our efforts if such questions could be reconsidered also considering the results presented in this work.</p>
  </sec><sec id="s5">
   <title>5. Conclusion</title>
   <p>Experimental evidence in this work confirms earlier work that increasing levels of CO<sub>2</sub> at current levels in the atmosphere cannot significantly contribute to warming by more back-radiation. We also demonstrated that increasing greenhouse spurious gases like Freon show a strong response in back-radiation when added into our atmospheric test chamber. Climate models and their CO<sub>2</sub> forcings should be revised and much more experimental evidence about the IR radiation response of greenhouse gases should be collected before appointing current warming trends and climate change mechanisms monocausal to greenhouse gas theories.</p>
  </sec>
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