You’re right, we’re asking for reader assistance again. This time we are trying to quantify just how wrong is the IPCC’s magically magnified greenhouse effect.

All that’s required are listings of tropical places (latitudes ≤ 23°27’N or S) with their solstice average temperature and humidity. You can leave them in comments or send them directly to Ed. Ideally some will be in the wet tropics, some wet-dry and some dry tropics.

So, what’s the purpose of this inane exercise?

We’re trying to demonstrate *simply* that greenhouse effect is a self-limiting process.

The whole argument over enhanced greenhouse is about positive feedback – whether the atmosphere warmed slightly by addition of the minor greenhouse gas carbon dioxide (CO_{2}) and therefore capable of holding more water vapor (H_{2}O), the main greenhouse gas by volume and effect and thus further enhancing greenhouse effect, increasing atmospheric temperature, enabling more H_{2}O content…

The coarse calculations are apparently a bit difficult for some people to follow, hence the attempt to show actual locations.

At this point we might as well lay out the calculations again, since some correspondents still ask how we can tell what temperature the earth “should be”. Skip this

**Sidebar: calculating earth’s expected temperature**

Note that various references use different rounded values for effective emission temperature of the sun, solar radius, distance from earth, bond albedo etc., so your mileage may vary, as they say.

We’ll use:

Solar effective emission temperature (Ts): 5,780K (that’s in kelvins which are the same as °C + 273.15)

Solar radius (Rs): 6.955 x 10^{5}

Radius Earth-Sol (Res or 1AU): 1.496 x 10^{8}

Sigma σ (Stefan–Boltzmann constant): 6.704 x 10^{-8}

To find the expected Top Of Atmosphere radiation from the sun then our calculation is σ x Ts^{4} x (Rs/Res)^{2} or 1,368 Watts per meter squared (W/m^{2}), which is about the amount we measure by satellite.

Now, because earth rotates and is roughly spherical the average surface insolation received pole to pole is one-fourth* that of the Top Of Atmosphere, 342 W/m^{2}, so converting that to temperature (342/σ)^{1/4} yields approximately 279 K or about 6 °C.

Now we need to take into account the proportion of incoming solar radiation reflected (called a planet’s Bond albedo), which in earth’s case is just a tad over 0.3 but we aren’t worried about greater precision than 0.3 at the moment. That reduces our surface insolation to about 239 W/m^{2} and surface temperature to the oft-listed 255 K (-18 °C) for earth’s no-greenhouse state.

The final part of the calculation then is to add back the feedback from the atmosphere because the atmosphere is composed of molecules of non-zero temperature and all bodies of non-zero temperature emit electromagnetic radiation as a function of spontaneous entropy. The frequencies (or wavebands) of these emissions can be determined using Planck’s Law but are not important to this calculation.

This feedback is approximately 38.5% so, 239/(1-0.385) yields approximately 389 W/m^{2}, giving us our surface temperature (389/σ)^{1/4} or ~288 K (15 °C), just slightly warmer than we believe the planet to be.

* Some people have a little difficulty with why it’s one-fourth and not simply one half to account for night/day rotation. The answer is that as you travel to higher latitudes the sun is lower to the horizon – sunlight is arriving at an angle. At the equator that angle is 90° – the sun is overhead (at equinox) and so the day half is the correct amount. At latitudes of 50° (about that of Prague or Winnipeg) a square meter of land surface receives only 64% of that at the equator, 60° (Oslo or Anchorage), 50%, Polar Circles, 40% and so on, varying slightly with solstice.

To return to the original problem, we are trying to show that this feedback mechanism of “greenhouse effect” is actually self-limiting – the atmosphere does not warm indefinitely according to greenhouse gas constituents constantly enabling more greenhouse content and greenhouse effect in a “runaway” endless loop.

We can do it with the calculation above but that’s too tedious and a little intimidating for some.

The simplest way we can think to do so is by demonstrating that earth regions with the greatest insolation, i.e. the tropics, do not have a feedback similar to the planet as a whole.

At the equator insolation is perpendicular, so we need only reduce insolation for rotation, not earth surface curvature as we do for higher latitudes. Thus 1368/2 or 684 W/m^{2}.

If we simply reduce that by Bond albedo we have ~475 W/m^{2}, which resolves to 303 K or ~30 °C, all without any greenhouse effect whatsoever and consequently far too high a surface temperature. Since the atmosphere has temperature in the tropics it *must* emit electromagnetic radiation, some of which must be absorbed by the earth. If we use the global average calculation then the tropics must be approximately 342 K or almost 70 °C, an obvious nonsense.

So there we have it. The tropics demonstrate that negative feedbacks are operating, have overwhelmed the increased-moisture positive feedback and deplete additional greenhouse effect.

Where to from here?

It seems to us that we should probe temperature/moisture pairs to try to establish an empirical “break even” point. How far from the equator does this net negative feedback zone extend? Are there portions of the planet which could be seriously affected by remaining greenhouse effect or are they too cold to radiate important quantities of energy in wavebands absorbed by GHGs?

There are still plenty of things left to determine and there actually might be places negatively affected by increasing atmospheric CO_{2}, although this appears increasingly unlikely.

One thing is for sure. The IPCC’s simplistic marvelous magical multipliers for enhanced greenhouse effect are foolish, destructive and wrong at least in the most energetic portions of the atmosphere and planet.

Too far off-topic

Oliver– this thread is specifically for enhanced greenhouse. –Ed.Do you mean like

June 20th 2011

Quito Ecuador

Mean temperature 13°C

Average humidity 69

Elev: 2812 m

Lon: 78.5° W

Lat: 0.1° S

http://www.wunderground.com/history/airport/SEQU/2012/6/20/DailyHistory.html

Do you want every year or just this year and does elivation matter? Doe you want temps in centigrade?

Should read June 20th 2012

I would have looked up the elevation but if it comes supplied that’s even better. The monthly mean temperature for June or December is fine for tropical sites. Either temperature scale is fine.

Ok

http://www.wunderground.com/history/airport/SEQU/2012/6/1/MonthlyHistory.html

If I look up the June monthly stats and average the averages (using Excel) I get 13.83°C and humidity 61.07 but if I average the high and the low values, add them together and divide by two I get 14.04°C and 59.07

I also note that the original numbers are all whole numbers. Are these numbers suitable?

Yes

Tiny, these are fine – I’ll need to locate weather stations for sites eventually selected so there’s no need initially to perform any averaging.Initially all that are required are coarse items like this Galapagos Weather because we are simply locating interesting triplets or pairs (equatorial sites with similar locations and different latitudes).

Once we have a range of wet, wet-dry and dry tropical locations we can get down to the compare and contrast.