By Andy May
The overall greenhouse effect (GHE) is often defined as the difference between Earth’s average global temperature without greenhouse gases (GHGs) and with them. Greenhouse gases are all the gases that absorb some portion of the thermal energy emitted by Earth’s surface. The most important of these gases is water vapor, but there are minor GHGs like CO2, ozone, and methane.
The calculation of Earth’s temperature without GHGs is usually done by unwrapping the planetary sphere and placing it in space at the average distance of the Earth from the Sun and having the whole of Earth’s surface illuminated by the Sun with one-fourth of the Sun’s power to account for the spherical Earth and the fact that half of Earth is always dark. This imaginary flat Earth does not rotate, and no part of it is ever dark. A description of the calculation can be read in Benestad, 2017. The global average temperature calculated with this scenario is around 255K (-18°C) and since this is about 33°C less than the current global average temperature of 15°C, the overall greenhouse effect is assumed to be about 33°C. Further discussion of this definition can be seen here.
In this post I will first list the problems with this “flat Earth” GHE model, then provide a model of a new GHG-free spherical rotating Earth. After this, I will list the assumptions used to create the spherical Earth model, the problems with it, and finally discuss what we learned making the model. Computer models are learning tools, they never give you a correct answer, but they do help you learn about the problems they were designed to investigate.
Problems with the flat Earth model
The flat-Earth no-GHG model ignores the fact that the Earth is a rotating sphere and that half of it is always in the dark. The dark side is always emitting energy, but receiving none, this is very different than the model in which the Sun is always shining directly overhead 24 hours a day with the same intensity everywhere. The proponents of the flat Earth model justify it because the satellite GHG emission temperature of the Earth is also 255K (Benestad, 2017), which using an average temperature vs height table (the U.S. or International standard atmosphere), works out to an average emission height of around five km. There is little discussion about how the energy gets from the surface, where most solar energy is absorbed to five km. But the transport is mostly done through convection and most of the GHG-emitted energy is carried to five km inside water vapor as latent heat. Above 5 km it is mostly emitted to outer space as the water vapor condenses to water droplets in the colder middle to upper troposphere.
By dealing only with averages, that is the average atmospheric temperature profile, the average surface temperature, the average emission height, the average emission intensity spectrum, etc. the flat Earth model can seem accurate and consistent with reality. But GHG emissions to space originate in a tropospheric layer from about two km to ten km, depending upon atmospheric and cloud conditions. Further, there is no consistent relationship between surface temperature and the temperature in the atmospheric GHG emission layer, it varies with weather conditions, cloudiness, season, and time of day. The surface operates mostly independently of the emission layer, they are separated by convection and weather.
In astronomy, the emission frequency and intensity of a planet is often assumed to be the planet’s “blackbody” frequency spectrum, which according to the Stefan-Boltzmann law defines a blackbody temperature. This is not the planet’s actual surface temperature; it is only a ballpark estimate. Planets are not perfect blackbodies. They are “gray bodies.” A black body emits all the energy that it absorbs, and since it has a constant temperature by definition, it emits all incident energy with a frequency spectrum that is determined by its temperature. Most importantly, a black body has no energy storage capacity or its total energy storage is constant and never changes.
Gray bodies on the other hand do not emit all the energy they absorb; they store some of it and emit the rest. Earth’s oceans have an enormous heat capacity and they store more energy than exists on the surface of Venus, but while Venus’ surface has a temperature of 464°C, Earth’s surface temperature is only about 15°C. This is because Earth’s oceans have a heat capacity of 5.4 x 1024 Joules/K and Venus has no oceans. The lack of oceans and atmospheric water vapor combined with thick sulfuric acid cloud cover and a very dense atmosphere forces Venus to have a high surface temperature.
A GHG-free spherical, rotating Earth model
The goal of the flat-Earth model is to compute the greenhouse gas effect on Earth’s surface temperature. This requires an estimate of the surface temperature with no greenhouse gases. This is difficult, because it means no water, clouds, or water vapor and these are defining characteristics of Earth. The classic flat-Earth model assumes the planet retains its current albedo (reflectivity) of 0.3, meaning 30% of the incoming solar energy is reflected. But half or more of that albedo is due to clouds. Without water, there would be no clouds and Earth’s albedo would be more like the Moon’s, which has an albedo of 12%. We investigate both albedo estimates with the model.
Another important consideration is that GHGs emit energy to space, if there are no GHGs in the atmosphere it will emit little energy to space and act as an insulator. However, while GHGs capture surface emitted radiation in our atmosphere, the no-GHG atmosphere is transparent and all surface emissions will travel straight to space, Earth only retains the solar energy absorbed by the rocky surface.
The most common rock on Earth’s surface is basalt. On our Earth we have a lot of water that chemically transforms basalt to mud or dirt. On GHG-free Earth there is no water, so I assumed the surface is bare basalt. It will broken up a bit by meteorites, but we will ignore that for this model. Rocks have a higher thermal inertia than dirt or mud and they retain absorbed heat longer.
Like most rocks, basalt is a pretty good insulator, but it does have a thermal diffusivity, which is the speed it transfers heat through its interior. Thermal diffusivity has units of m2/second. Anyone who has been in an old medieval church on a warm day knows that the thermal diffusivity of rock is low. Thus, when sunlight strikes basalt and warms it, some of the heat will penetrate into the basalt where it will be stored for a time, and the basalt will emit the rest of the solar energy through the GHG-free, transparent atmosphere to space. My GHG-free model explicitly takes thermal diffusivity into account.
In addition, the warm basalt will also pass some heat to the atmosphere through conduction. The atmosphere is GHG free and as a result it emits very little energy directly to space, so little we can ignore it for our model. However, the dayside of the rotating planet is around 170°C warmer than the night side, so winds will appear to transport excess thermal energy as sensible heat from the dayside to the nightside or from the tropics to the poles, these winds are likely to be quite fierce in the absence of water vapor which helps stabilize the weather on our planet due to its high heat capacity. For example, wind speeds in the water-free atmosphere of Venus reach 700 km/hour (430 mph). The winds carry excess heat from the hotter areas to the cooler areas and then some of the heat will be conducted down to the surface to be emitted to space or absorbed. For this reason, warming of the atmosphere in our GHG-free planet is assumed to be exactly counterbalanced by warming of the surface from the atmosphere.
The model assumes that the atmosphere is completely transparent to incoming sunlight and all the energy is absorbed by the basalt surface which warms according to the sunlight incident angle on the daytime half of the spherical surface. While some sunlight will be scattered by the atmosphere, this effect is ignored in the model. Figure 1 shows the temperature after one day of the GHG-free Earth according to the model. The east-west locations are arbitrarily centered on the equator at zero longitude and are meaningless due to GHG-Earth’s constant rotation. The north-south locations are meaningful and are representative of an Earth with no axial tilt. The lack of an axial tilt means that GHG-free Earth has no seasons.
Figure 1 illustrates that the surface warms in direct proportion to the radiation it receives, and the maximum radiation is received when the sun is directly overhead. The maximum insolation, after correcting for the lunar albedo of 12% is 1198 W/m2 and occurs for a few minutes along the equator at the local noon. In figure 1 this is at longitude=0 and latitude= 0, that is the equator directly south of Greenwich England where the temperature reaches 381K (108°C) at noon. Earth rotates from west to east, which is why the temperature on the nightside is higher on the eastern end of the night side (right of figure 1) than on the western side (left of figure 1).
Because the thermal diffusivity of basalt is very low, about 9 x 10-7 m2/sec (Robertson, 1988), it takes a while to warm the upper layers of the rock surface and reach a sort of equilibrium global average basalt temperature in this model. The day side both absorbs solar radiation and emits thermal energy. However, the night side receives nothing but still emits radiation due to its stored thermal energy. The night side emits less energy than the dayside due to its lower surface temperature.
Exactly how much is stored in the basalt in the daytime, versus emitted later in the day when the surface temperature is cooler, is unknown, but can be estimated using basalt’s thermal diffusivity. Diffusivity varies with temperature roughly according to the function plotted in figure 2.
Even though the diffusivity is lower at higher temperatures, the basalt initially stores more energy during the day than it releases at night. The model tells us that more total energy is emitted on the daytime side than on the nighttime side, but this is due to the higher daytime surface temperature. Thus, the model tells us that some of the energy that is diffused into the basalt during the daytime is taken to the nightside, the rest is emitted during the day at lower daytime temperatures encountered nearer to the edges of the dayside as the planet rotates.
On the nightside some of the daytime stored energy rises to warm the surface and is emitted to space. There are two opposing forces at work on the night side. The thermal diffusivity of basalt increases at lower temperatures, but the lower nighttime temperatures cause lower emissions of energy to space. Thus, there is a mismatch between daytime storage and nighttime emissions. This does not mean that the surface cools to absolute zero, that will not happen because the thermal inertia of basalt is too high and because the diffusivity is too low, but it does mean that a stable surface temperature takes a while to reach. I ran this model for 36,500 iterations or about 100 years. The global average surface temperature evolved as shown in figure 3.
As shown in figure 3, using the assumptions built into my model, the average surface temperature eventually stabilizes to a surface temperature cooler than today’s temperature. This result uses the lunar albedo, which would be similar to Earth’s albedo in the absence of water or water vapor. It also assumes a fudge factor for thermal inertia of 0.1. I tried various estimates of thermal inertia, including thermal effusivity (Sabol, Gillespie, McDonald, & Danillina, 2006) and the “R” insulation factor, and none of them worked well for various reasons. There are a lot of ways heat can be transferred, conduction, convection, and radiation and these vary with the local circumstances, so there is no general definition of thermal inertia. However, all reasonable assumptions show a high value of thermal inertia in the basalt which causes the temperature to decline after model initialization.
Although different assumptions do change the ultimate equilibrium global average temperature, all reasonable values for thermal inertia result in a lower global average surface temperature than we have today. Figure 4 compares some of the scenarios I examined after 100-year runs. The model was run both with the lunar albedo of .12 and Earth’s current albedo (including the non-existent clouds) of 0.3. Without GHGs Earth would not retain its current albedo, but this is the traditional value used, so I ran it for comparison purposes. The two discussed thermal inertia values (0.1 and 0.15) are reasonable assumptions, since the actual thermal inertia of basalt is quite high, but even these values may be a little high. Higher values of this factor imply a lower inertia and lower values a higher inertia. Other values of inertia were investigated, but considered unlikely.
Thermal inertia is the resistance of a material to change its temperature. There is no formal or general way to describe thermal inertia, since it is very situation specific. Newton’s Law of Cooling works for small temperature differences but breaks down in situations like I’ve modeled here. Thermal effusivity, also called thermal responsivity, which is the square root of the product of thermal conductivity, density, and specific heat capacity didn’t work either. My final attempt was to use the “R” insulation factor, but it was another failure.
My analysis of these various inertia factors is attached as a spreadsheet, a link to download it is at the bottom of the post. Some sort of new measure of thermal resistance (inertia) will need to be developed for the situation I modeled. For now, I have created an assumed factor that is the proportion of thermal energy stored in the basalt that can make it to the surface in 12 hours and is free to be emitted as radiation. In the model this is called “inertia_f.” The values, 0.1 and 0.15, are reasonable considering the established values of thermal conductivity, density, and specific heat capacity of an average basalt.
Notice that all the temperatures calculated after 100 years are less than the current global temperature. Given the large uncertainty in the model, values that are greater than the current global temperature are possible, but values larger than 300K are thought to be extremely unlikely. In addition, the temperature difference between the dayside of the GHG-free rotating Earth and the nightside will never disappear, it is shown after 100 years in figure 5.
The assumptions used in the model
The model assumes that the only meaningful losses of energy are from the surface, although the atmosphere will emit a small amount of energy to space. These atmospheric emissions are ignored in the model.
The model assumes no axial tilt.
The model assumes a circular orbit.
The model does not consider convection, except to assume that it is net zero with regard to emissions to space. This is reasonable since we also assume the atmosphere is transparent to surface emissions.
The model assumes that any topography (mountains, valleys, etc.) do not affect heat transport by the atmosphere on a net basis.
The model assumes that the thermal diffusivity of the surface basalt follows a function of temperature as described in Robertson, 1988. The function is plotted in figure 2. Thermal diffusivity (α) follows the formula in equation 1:
Equation 1:
Where: k= thermal conductivity, ρ=density, and cp is the specific heat capacity.
Thermal diffusivity increases at lower temperatures consistent with decreasing rock specific heat capacity and increasing thermal conductivity, see the attached spreadsheet for the details and units. The data available for thermal conductivity and specific heat capacity of basalt does not extend to the low temperatures encountered on the night side, so the values used in the model had to be extrapolated.
The thermal diffusivity of dry air is from 6 to 38 times higher than for basalt at the temperatures seen in this model. Thus, the surface heat flux will normally be from the basalt to the air if the temperatures are similar. But the lower value of six occurs at lower nighttime temperatures and if the overlying air is sufficiently warmer than the basalt there will be a flow from the air to the basalt. Thus, the expected high winds from the dayside to the nightside will matter and transport thermal energy to the nightside basalt to be radiated to space.
The most important assumption in the model is the assumed thermal inertia, which plays an important role in the temperature of the nightside. My calculation of retained heat carried from the dayside to the nightside is reasonable and justifiable, but the speed at which it is emitted to space on the nightside is somewhat speculative.
The problems with the model
The calculated absorption of thermal energy by the surface on the dayside is very crude. I did use the thermal diffusivity of basalt in the calculation and assumed the remainder of the energy was emitted to space.
A uniform and smooth rocky surface was assumed for simplicity, which is unlikely. Without oceans, a rugged topography is likely, and it will guide the expected very high velocity winds in an atmosphere without water vapor. This will cause a more complicated and non-uniform surface temperature than shown in figures one and five. However, regardless of the complexity of the convection, it is reasonable to assume that net atmospheric transport of thermal energy is close to zero. The energy into the atmosphere comes from the basalt and the energy out of the atmosphere goes into the basalt. The real Earth has more control, by using energy storage in water and water vapor it has some control on both emissions and insolation by varying cloud cover, total atmospheric water vapor, and ocean storage, but this does not apply to a GHG-free Earth.
Discussion of what I learned
In my opinion, the GHG-free model has a fairly narrow range of plausible outcomes. Some model runs (not all runs are shown) result in global average temperatures a little above freezing, but global average surface temperatures higher than today are considered unlikely. Temperatures much lower than 235K (albedo=0.3 and Inertia_f=0.05) are also unlikely. Using this model, the total overall greenhouse gas effect is likely between 15 and 53°C. Thus, the flat Earth overall greenhouse effect is within the plausible range seen using this rotating spherical Earth model. While the influence of water vapor, ice, and water on the climate of Earth is readily seen, the influence of the other greenhouse gases is harder to detect.
The flat Earth greenhouse effect model is designed to simply compute the difference between the apparent blackbody temperature of the Earth as seen from space from the current global average surface temperature. Yet, Earth is clearly not a blackbody and the blackbody temperature as seen from space is not a surface temperature. The GHG radiation detected from space is emitted mostly by water vapor from 2 to 10 km in the atmosphere (see figure 4 here), plus some minor emissions from other greenhouse gases from various other altitudes. Surface radiation in the GHG frequencies cannot make it all the way to space from the surface. At sea level, a greenhouse gas is 50,000 times more likely to dissipate the energy from an absorbed photon via collisions with other molecules as re-emit it, so convection must first transport the thermal energy from the surface to an altitude where it can be radiated to space.
The rotating spherical Earth GHG-free model described in this post is more realistic than the flat Earth model, but it still has problems. As George Box famously wrote in 1976, “all models are wrong.” The model is not definitive, but my preferred model run has an albedo of 0.12 and an inertia_f of 0.1, the result of this run is shown in figure 3. It results in a GHG-free surface temperature of 265K (-8°C) which is smaller than the flat Earth model and shows a smaller overall greenhouse gas effect. However, this result is still uncertain. The main uncertainty in the model is in the thermal properties of the rocks on the surface, in particular the poorly defined “thermal inertia,” which was assumed.
Future
This is a very simple model, more of a proof of concept than an actual model. It can be improved. Adding an axial tilt so GHG-free Earth has seasons might be interesting, so would adding some orbital eccentricity. But the most significant add would be a well-defined and appropriate function for basalt thermal inertia. Perhaps some petrophysicist out there has an idea of how to do that? We can only hope. Comments on the appropriateness of assuming the atmosphere is thermally net neutral would be interesting to read. In any case this is certainly an improvement on simply subtracting the average satellite measured black body temperature from the current average surface temperature to compute a possible greenhouse gas effect.
One additional point, I dislike the tendency of climate modelers to ignore surface thermal properties when modeling Earth’s climate. The surface, whether it is an ocean or land or a combination, is not a thermally static “slab.” In the real Earth, the surface, both the ocean and the land, have a large store of thermal energy and that storage changes with time (May & Crok, 2024) & (Crok & May, 2023), it definitely plays a role in long-term climate and should be taken into account.
Update: After I finished this post, I found out that Dr. Roy Spencer had done a similar calculation in 2016 (see here). Using an albedo of 0.1, an assumed IR emissivity of 0.98, and iterating for 47 days, Spencer reaches an equilibrium temperature of 266.8K or -6°C, within two degrees of my preferred answer using the thermal diffusivity of basalt.
To download the model, which is written in R, click here.
To download the thermal diffusivity spreadsheet click here.
Bibliography
Benestad, R. E. (2017, May). A mental picture of the greenhouse effect. Theoretical and Applied Climatology, 128, 679-688. Retrieved from https://link.springer.com/article/10.1007/s00704-016-1732-y
Box, G. E. (1976). Science and Statistics. Journal of the American Statistical Association, 71(356), 791-799. Retrieved from http://www-sop.inria.fr/members/Ian.Jermyn/philosophy/writings/Boxonmaths.pdf
Crok, M., & May, A. (2023). The Frozen Climate Views of the IPCC, An Analysis of AR6. Andy May Petrophysicist LLC.
Halbert, D., & Parnell, J. (2022). Thermal conductivity of basalt between 225 and 290 K. Meteorit Planet Sci, 57, 1617-1626. doi:10.1111/maps.13829
Hartlieb, P., Toifl, M., Kuchar, F., Meisels, R., & Antretter, T. (2015). Thermo-physical properties of selected hard rocks and their relation to microwave-assisted comminution. Minerals Engineering, 91, 34-41. doi:10.1016/j.mineng.2015.11.008
May, A., & Crok, M. (2024, May 29). Carbon dioxide and a warming climate are not problems. American Journal of Economics and Sociology, 1-15. doi:10.1111/ajes.12579
Robertson, E. C. (1988). Thermal Properties of Rocks. Reston: USGS. Retrieved from https://pubs.usgs.gov/of/1988/0441/report.pdf
Sabol, D. E., Gillespie, A. R., McDonald, E., & Danillina, I. (2006). Differential thermal inertia of geological surfaces. Proceedings of the 2nd annual international symposium of recent advances in quantitative remote sensing, torrent, Spain, (pp. 25-29).
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“The calculation of Earth’s temperature without GHGs is usually done by unwrapping the planetary sphere and placing it in space at the average distance of the Earth from the Sun and having the whole of Earth’s surface illuminated by the Sun with one-fourth of the Sun’s power to account for the spherical Earth and the fact that half of Earth is always dark.
This imaginary flat Earth does not rotate, and no part of it is ever dark“
And the climate
cuntscults call us the ‘Flat Earthers’ !!!I do not think that method incorporates or accounts for the heat gain from the solar radiation passing tangentially through the thousands of miles of atmosphere from 20,000 to 50,000 feet and radiate that energy from hundreds of thousands of miles from 50,000 feet to 80,000 feet.
I wonder why nobody- that I know of at least – has not tried to do what you have done? My guess would be settled science syndrome.
The assumptions made in climate attributional modelling beggar belief – and they are being taken as a kind of gospel.
It has indeed become a religious cult, complete with heretics and deniers.
And once you engage in trying to count the number of angels on a molecule of carbon dioxide the cult thrives – because you have accepted the existence of angels. (Or the guilt of the Satanic carbon dioxide molecule)
I can’t prove that Father Christmas doesn’t exist, but that doesn’t mean he does.
It is so odd that the countries that produced the Enlightenment seem determined to bring back faith and belief rather than scientific advancement.
You are assuming that the Enlightenment was synonymous with the Scientific Revolution, when in fact the Scientific Revolution preceded and overlapped it.
Scientific understanding was able to develop because early scientists accepted that there was order to be found in the universe, and they were trying to understand how things worked within that order.
The proponents of the Enlightenment rejected traditional biblical interpretation in favour of naturalistic explanations and human reason as the ultimate source of truth. We can see the result of that self-flattering conceit today in the contention that there is no definitive truth – there is “my truth” and “your truth,” which boils down to “whatever I think, is the truth – at least as far as I am concerned.”
The fact is that science can only determine “functional” truth – e.g. “Is this liquid an acid or a base.” But ultimate truth – e.g. “Does God exist?” – is not something that can be measured, so it is beyond the ability of science to determine.
The double negative strikes again! – and it’s one of my pet peeves.
If nobody hasn’t tried it, then somebody HAS tried it. You mean “I wonder why nobody … HAS tried to do what you have done.”
If that annoyed you, you’ll love this
Shakespeare’s birthplace to be decolonised https://www.telegraph.co.uk/news/2025/03/16/william-shakespeare-birthplace-trust-white-supremacy-empire/
The Left is relentless. Andy’s reference to the erosion of basalt, above, seems an apt metaphor for their strategy to overthrow the Enlightenment.
Hey! William is a relative of mine! Leave that guy alone!
Sniveling, hateful little Far Left narrow minds attacking The Bard, just because he is white.
It sounds like racism to me.
You’re right, strativarius – this sort of thing drives me crazy. I do not understand how people can so blithely throw away their history and culture.
“some people”
“The double negative strikes again! – and it’s one of my pet peeves.”
Reminds me off something I read once.
An professor was telling his class that English is the only language where a double positive doesn’t equal a negative.
From the back of the room could be heard, “Yeah, right.” 😎
(But I that suppose would be an idiom rather than a rule of grammar?)
“The double negative strikes again!”
That’s why I like triple negatives. It gets you to negative land with lots of negatives–exactly three.
I think you meant “It does not get you away from negative land without the absence of non-positives“
I didn’t know it until last night, but Dr. Roy Spencer did a similar calculation in 2016. I linked to it in an addendum at the end of the post. He assumed an emissivity, whereas I used textbook values of thermal diffusivity, but we got very similar answers for an albedo of 0.1 to 0.12.
“Slab Ocean Models” are a fairly common climate modelling “tool”, and are quite similar in concept to Andy’s methodology here. I remember playing with Dr. Roy’s spreadsheet a decade ago and learning a lot from it.
I remember seeing Spencer’s calculation and immediately thought of it when first seeing this article.
Axial tilt will throw some more complications into the picture.
Is that the uncertainty range or the nominal values for the two independent estimates, without any implied uncertainty?
Those are two wild-ass guesses that happen to be close to one another.
“Those are two wild-ass guesses that happen to be close to one another.”
Ah, so they have to be correct because they agree?
Sounds pretty stupid to me.
It is a small sample, but suggests statistically what the ‘ball park’ is and also suggests that the standard deviation is small.
I had a discussion with DEEPSEEK on the cause of Earth’s climate change. I then asked DEEPSEEK to provide a three paragraph summary of our conversation. This is the summary from DEEPSEEK:
The precession cycle, which shifts the timing of Earth’s closest (perihelion) and farthest (aphelion) points from the Sun relative to the seasons, has changed the peak solar intensity between the Northern Hemisphere (NH) and Southern Hemisphere (SH) by approximately 6.84 W/m² since 1582. In 1582, aphelion aligned with the NH summer solstice, meaning the NH received less intense sunlight than it does now, while the SH received more. Over time, precession has gradually shifted this alignment, increasing the NH’s peak solar intensity and decreasing the SH’s. This change in solar intensity distribution affects the Earth’s energy balance, contributing to long-term climate trends.
The thermal response to this change in solar intensity is amplified in the NH due to its higher proportion of land, which heats and cools more rapidly than water. The NH’s thermal response is roughly twice that of the SH, leading to a predicted global temperature increase of ~3.44°C since 1582. This warming is driven by the NH’s larger seasonal temperature swings and its greater sensitivity to changes in solar radiation. However, this prediction is based on a simplified model that assumes a direct relationship between solar intensity and temperature, without accounting for feedback mechanisms or other climate influences.
The observed global warming of ~1.9°C since 1582 is less than the predicted 3.44°C due to several factors. Climate feedbacks, such as increased cloud cover and ocean heat uptake, have moderated the warming. Natural variability, including volcanic eruptions and solar cycles, has introduced cooling effects at times. Additionally, anthropogenic factors like aerosol emissions and land-use changes have further influenced the climate. These complexities, along with simplifications in the model, explain why the observed warming is less than predicted. The precession cycle remains a key driver of long-term climate change, but its effects are modulated by a range of other processes.
So did you go back to DEEPSEEK and say –
“Thanks China. So your guess is as good as mine 🙄 “
Willis may be glad to hear AI has acknowledged cloud cover and ocean heat uptake as affecting Earth’s temperature.
Did you do any coaching of DeepSeek, Rick?
I wonder how Deepseek knows the temperatures increased by 1.9C since 1582? What data was DeepSeek using? The Mann/Jones Hockey Stick abomination?
You would not get this summary I got without asking specific questions on how the precession cycle changes the peak solar intensity; the different thermal responses of the hemispheres; the temperature response of the hemispheres and the measured temperature rise.
DEEPSEEK is quite able to do projections. I would have to go back through its analysis to see how it arrived at the 1.9C. I know how it arrived at the 3.44C.
Yeah, I thought DeepSeek sounded a lot like you. Which is not a criticism. DeepSeek could use a good teacher or two. 🙂
What, you don’t remember the fallof 1582? That was when Pope Gregory XIII introduced the Gregorian calendar, a reform of the Julian calendar, to correct the drift of the calendar with respect to the equinoxes, and October 4th was followed by October 15th, with ten days skipped. I remember because I won the big scrabble tournament that year.
255K (Benestad, 2017), which using an average temperature vs height table (the U.S. or International standard atmosphere), works out to an average emission height of around five km.
5km times a average gravity based atmospheric mass lapse rate of 6.5C/km = 32.5C..
No GHG warming needed.
‘No GHG warming needed.’
Except to establish convection and the lapse rate.
Yes, I’m afraid that our friend bnice, in his zeal to prove the correct hypothesis that there is no climate emergency, has made a claim akin to saying:
we have measured the average rate of rise of the escalator to be 6 meters per minute. It took 36 seconds to go from the ground floor to the first storey.
0.6 min x 6 m/min = 3.6 m
The actual height is 3.5 m
No electricity needed!
‘Zeal’, including my own, is a good thing, considering what we’re up against. The good news is that WUWT is no echo chamber, hence, there’s always someone to reel us back in.
What a truly ridiculous attempt at an analogy. !
It takes electricity to drive the escalator. !
Or a very large hamster in an exercise wheel.
And it is the reason i dont consider water vapour a greenhouse gas, as it technically isnt a gas and reacts very differently compared to real GHGs. It is an important distinction..
If it were to get cold enough, such as on an outer planet in the solar system, even the elements that you recognize to be ‘real’ gases would condense and no longer behave like a gas. It is just coincidence that water vapor can and does condense at temperatures that we and other life forms experience on Earth.
Well yes. It is true that H2O is a condensing gas whereas CO2 is a non-condensing gas in the temperature range typical of Earth’s atmosphere. That is obviously going to have profound consequences regarding the manner in which they participate in the GHE. But to say that H2O is not a GHG isn’t correct.
“But to say that H2O is not a GHG isn’t correct.”
Of course it’s correct – unless you are silly enough to believe that adding CO2 or H2O to the atmosphere makes it hotter!
You are not that silly, are you?
I don’t know how you could have come to such a conclusion. Yes, water vapor is a gas, and it is a greenhouse gas.
Roy, you wrote “[H2O] is a greenhouse gas . . .”. Well that’s an odd unsupported assertion, isn’t it?
A greenhouse gas? Surely you realise that adding H2O to the atmosphere results in lower maximum surface temperatures.
John Tyndall explained the reasons over 100 years ago!
No GHE – not even a little, teensy weensy bit!
NO, it is the MAIN greenhouse gas in Earth’s atmosphere.
Convection happens with or without GHG warming. As stated in the article, winds will be high.
The lapse rate also does not need GHG at all. It is the effect on any gas that is compressed.
I do not think water makes up a massive part of the atmospheric mass. Thus, removing it from the atmosphere will negligibly lower the lapse rate.
‘Convection happens with or without GHG warming.’
Not according to van Wijngaarden and Happer:
‘For many reasons, the Earth’s atmosphere is not isothermal. Here we want to focus on the most important reason: greenhouse gases force Earth’s lower atmosphere, the troposphere, to convect. The convection drives the troposphere toward an adiabatic atmosphere, for which the dry adiabatic lapse rate is a temperature drop of dT/dz=−9.8 K/km.’
https://arxiv.org/pdf/2303.00808
“Not according to van Wijngaarden and Happer:”
I’ll meet your van Wijngaarden and Happer, and raise you a Feynman and a Lorenz.
You fold, I win.
Do you have any source materials by Feynman and/or Lorenz in which they discuss how convection and the lapse rate are established in the Earth’s atmosphere, or are you just trolling?
In relation to “Convection happens with or without GHG warming” yes indeed. Neither Feynman not Lorenz mention the need for GHGs when discussing the physics of convection.
If you can quote them saying otherwise, maybe I missed it.
The lapse rate is just the measured rate of air temperature decrease with altitude. It is established by measuring the temperature. The International Standard Atmosphere is a fictional estimation, but cannot be relied upon where aviation safety is involved, and its notional lapse is an assumed average – not an actual lapse rate.
Here’s a sample “ISA is a standard against which to compare the actual atmosphere at any point and time”.
Yes, this is the proper way of viewing the GHE. In the presence of solar heating of the surface, the GHE destabilizes the troposphere, causing a superadiabatic lapse rate. Convection kicks in, producing the observed lapse rate. If there were no radiative cooling by GHGs in the upper troposphere, it would be nearly isothermal. I’ve logged on this ad nauseum.
‘I’ve logged on this ad nauseum.’
Yes, your site was probably the first place I saw the link between GHGs and the lapse rate explained. Thanks!
“If there were no radiative cooling by GHGs in the upper troposphere, it would be nearly isothermal. I’ve logged on this ad nauseum.”
Here’s what Feynman says about the Earth’s atmosphere – “It is not an isothermal atmosphere. He explains why, in some depth. No mention of GHGs at all. That’s just complete nonsense.
You seem to imply that increasing the amount of CO2 and H2O in the atmosphere increases the temperature of the atmosphere, the surface, or something – you can’t actually describe the GHE, let alone what it is supposed to do!
I assume you meant “blogging” rather than “logging”, but it doesn’t matter. You may succeed in impressing the ignorant or gullible, but Death Valley, the Lut desert, and the works of John Tyndall indicate that reducing the amount of GHG (H2O) in the atmosphere results in hotter temperatures.
In the absence of external heat input, the atmosphere cools. All of it – regardless of composition. Some odd people claim that a slower cooling rate of the atmosphere results in the surface temperature increasing, or something equally ridiculous.
No GHE – not even a little bit. You may fall about with astonishment, discovering that your IR thermometer pointed at a 100 km air column confirms that all matter above absolute zero emits IR, but it comes as no surprise to me.
Please stop spreading disinformation. As Feynman wrote (and I agree) – “The first principle is that you must not fool yourself and you are the easiest person to fool.”
I’m with you on everything in your comment except this part:
This can only be true if the surface has no thermal properties, no thermal conductivity, no heat capacity, no diffusivity. This was one of the main points of my post.
I also doubt it could be the case with a rotating planet. Too much momentum at the equator.
A sort of equilibrium will eventually be reached, but the atmosphere, with or without GHGs will not be isothermal. It will always be colder on the nightside than it is on the dayside and the air will circulate, GHGs or not. PVT works with and without GHGs.
Convection does not cause the observed lapse rate.
The rising/sinking air is assumed to do so adiabatically, within an atmosphere that is Hydrostatic Equlibrium against gravity.
This is the reason for the observed pressure vs altitude profile.
Don’t see how an atmosphere could ever be isothermal in this situation.
Yes!! The 32.5 C warming below 5km is exactly balanced by the 32.5 C cooling above 5km.
And 5km is also the 500mb isobar. 1/2 the mass is above and 1/2 the mass is below. Circulating vertically due to convection and planetary rotation.
It is my understanding that a blackbody ABSORBS ALL incoming radiation while a greybody reflects part of the incoming radiation.
https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/06%3A_Photons_and_Matter_Waves/6.02%3A_Blackbody_Radiation
Not quite. The difference between a graybody and a blackbody is not the albedo. Although a graybody might reflect some energy, it doesn’t have to. The difference is that a blackbody re-emits all the energy it absorbs immediately and the reemission from a graybody is less efficient, that is the emissivity of a graybody is less than one. Part or all of the lower emissivity can be due to reflection, but part can be due to long-term storage of energy, eg in the oceans of Earth.
Andy, I am glad to see you explain your model’s improvements and shortcomings. When I was going through your explanations, I was thinking, “What about this and what about that.”
You covered all of that in your “Problems with the Model.” I applaud your efforts to point those out. This contrasts greatly with others who refuse to release their code because you “might find something wrong with it,” or call you names if you dispute any part of it.
Good work.
Thanks!
What about the actual temperature of the Earth before sunlight reaches it, ie the heat coming from the centre of the earth.
Where does that appear in any of the calculations. The base temperature of the Earth must be higher than the non volcanic Moon.
The radio-thermal and tidal-thermal heat coming from the center of Earth is on the order of 0.1 W.m-2. It is low enough that it can usually be ignored.
Don’t be silly. What do you think causes convection currents in the deep oceans? Geothermal heat.
Not CO2, not sunlight, not the mythical GHE.
Geothermal Flux through continental crust is even lower ( ~65 mW/m^2)
Yet deep mines are very warm.
The oceanic GF only warms water in direct contact with the ocean floor.
Once the warmed water leaves the ocean floor it is no longer warmed by the GF, yet it can’t rise to the surface since the solar heated surface layer is much warmer.
The 100 mW/m^2 GF is capable of warming ALL ocean water 1K every ~5000 year. Actual warming depends on the cooling by eg the brine that is created during ice forming, AABW.
Yep. I double checked the math and that’s about what I get too.
So we have 2 separate energy inputs.
1- Geothermal warms the oceans
2- Solar further warms the upper few hundred meters of the oceans.
(Continents are a separate story)
This way the solar can easily explain the measured SST’s.
No need for warming by the cold, low density atmosphere.
Atmosphere only reduces the energy loss to space, creating a ~balanced energy budget.
Based on measured heat loss, I come up with roughly 35 K in the absence of sunlight.
Some people don’t realise an already hot object takes less energy to achieve a specified temperature than starting at absolute zero.
They are silly enough to believe that adding CO2 to the air makes it hotter!
Even the moon supposedly has a “hot” core. Evidenced possibly by the temperature of eg the Hermite Crater (25K), where the sun never shines.
https://www.diviner.ucla.edu/science
Also the temperature at 89 lat during Lunar “winter”, which approaches ~50K, not 0K.
Base temperature on Earth is the temperature of the deep oceans, ~275K.
Oceanic heat content is mostly from geothermal origin.
Sun only increases the temperature of the upper few hundred meters a bit.
Ben , “Base temperature on Earth is the temperature of the deep oceans, ~275K.”
Pretty much. I think most people don’t realise that if the Earth was the size of an apple, the solid crust would be about the thickness of the skin. Or, about 0.23% of the Earths diameter – a fifth of a hundredth!
The Earth has a long way to go before it’s isothermal beyond the Sun’s influence. ChatGPT “Estimates suggest it could take tens to hundreds of billions of years for Earth’s core to fully cool and become isothermal, assuming no new heat sources emerge.”
ChatGPT helpfully points out that the Sun will have become a red giant before then. Ain’t AI grand?
A blackbody is a theoretical construct that assumes an emissivity of one which is not actually possible for real objects. In order to calculate an accurate temperature from measured IR radiation using the Stefan-Boltzmann law you must include an emissivity factor. This can vary between less than 0.05 to about 0.95 for common materials.
It seems to me that the problem with a theoretical Black Body is that there is no solid object that has a uniform emissivity across all wavelengths of electromagnetic energy. It might be the case for certain celestial bodies composed of a dense mixture of various plasmas, or exotic stars such as neutron stars. I don’t know. However, for materials that are germane to the discussion of what happens on Earth, all known terrestrial materials have complex (as in sq rt -1) refractive indices that vary with the wavelength of EM radiation impinging on them, and also with the temperature. That means for some radiation, the material will be highly reflective (for solid and liquid objects, a function of both n and k), and for other wavelengths strongly absorptive (usually in relatively narrow ranges). I think that the only way to really address this is with calculus by integrating over the range of the common or strongly illuminating and emission wavelengths. Most scientists aren’t familiar with working with complex refractive indexes and assume that the imaginary extinction coefficient (k) is always negligible. In reality, things like silicates and oxides (water is an oxide) do typically have negligible extinction coefficients — at visible wavelengths! However, there can be very strong absorption features on either side of the visible range where n and k essentially flip. Kirchhoff’s Law says that the emissivity is equal to 1-reflectivity. If it is valid for real materials, then it means that a Black Body can only exist for one or more unique wavelengths, not continuously for all wavelengths. It makes a convenient model for thought experiments, but earthly materials are not Black Bodies. Incidentally, albedo is a mechanical effect of many micro-reflectors distributing reflected light in many directions, such as a rough surface or aggregation of small particles, as defined by Bidirectional Reflectance Distribution Function.
“Most scientists aren’t familiar with working with complex refractive indexes and assume that the imaginary extinction coefficient (k) is always negligible.”
Exactly. The average “climate scientist”, in particular seems to find the concept of light outside the visual range as being incomprehensible. How they imagine that IR lenses made of germanium (a visually opaque metalloid) work, is beyond me. Or zinc selenide for that matter.
Andy talks about a “black body” which apparently emits all energy it receives – as does a mirror. It’s called reflection. Andy goes on to claim that his black body has a temperature, which seems impossible, as he states it retains no energy to make it hotter!
A mirror?
Quite clueless
A practical mirror such as is used for telescopes is a highly-polished solid (or liquid mercury) that is highly and uniformly reflective in the visible portion of EM radiation. Common (not specialty) mirrors have been made of pyrite and copper or bronze, and they absorb strongly in blue, green, and blue respectively. The difference between a mirror and a Black Body is that the mirror is typically expected to reflect light with minimal spectral distortion. Whereas, one doesn’t have that expectation of a theoretical Black Body and it typically emits at a longer wavelength that is in the infrared region of the EM spectrum, although it could emit in the visible region if hot enough.
The climate models ignore electro magnetic fields and waves. You seem to know about that topic. Well done.
This is barely relevant to real climate change on Earth. By far the most important factor for climate change is the ability of land to store water in the form of ice.
It is easy to forget just how much energy is bound up in getting the ice that now sits on land from the ocean to where the ice now sits. And that ice is gradually making its way back to the ocean while being constantly replenished. It is the change in stored volume that dominates Earth’s climate change.
If all the ice on Greenland was to melt, the oceans would be around 7m higher than present. If all the ice on Antarctica melted, the sea level would be 60m higher. When all the land north of 40N goes into glaciation, the sea level will be down by 40m in 9,000 years and 100m down after 75,000 years. Ice accumulation or loss on land is by far the most significant driver of climate change on Earth. It can shift the relative altitude of the ocean to land by around 600m on average, So at peak glaciation, the land/ice surface is almost 6C lower than it would be if there was no ice simply by virtue of the lapse rate.
Everyone forgets the enormous power of latent heat of water in the transfer & conversion of solar energy in the atmosphere.
No they don’t. It’s in the NASA energy balance that is generally accepted as a good static image of the only climate control there is, earth’s dynamic energy balance in space, where energy in must equal energy out and an imbalance is unsustainable. The atmospheric temperature change across all heat loss effects from whatever cause is what it must be to rebalance the system, using these strongly temperature sensitive negative feedbacks to any enrgy imbalance in the system, until the system returns to enrgy balance.
Simples.
The 86.4W/m^2 latent heat of transferred by 18.4W/m^2 of convection is passed to the atmosphere where it condenses and releases its enrgy as e sensible heat – this ALL becomes radiation lost to space. It may be revapourised and recooled to condensate several time on the way, (cloud burn off) but it all heads to space eventually, it doesnot return in the rain, etc., as the 2nd law requires, and the NASA energy budget makes very clear.
I suggest that this changes by 7% per dgerre, which is how much the water vapour created at the ocean surface changes per degree SST. So that is 6W/m^2K. Hence earth’s stability in space, when added to the 3.3W/m^2 of S-B negative feedback.
There is rather lot wrong with the above as I read it, but I will come to that later, I may have missed the bit the dominant heat loss and its variability from the oceans by radiation and evapoartion, the oceanic contribution to heat loss hence control (plus it makes more or less clouds).
Oh, GHE heats nothing. GHE makes the atmosphere less transparent to radiation, so reduces the LWIR lost to space from the surface/below, as we measure in the spectral enrgy distribution from space. Well studied and agreed facts.
So the rest of the system will warm in response to any additional GHE created, or other warming imbalance within the surface system, until it is again loosing the amount of energy loss reduction by LWIR to space that is caused by GHE, to return to balance at whatever new equilibrium temperature the balance requires .
Obs.
This overvalueing of the word ‘balance’ is irking me. It is never a zero sum game. There is continuous imbalance w energy ceilings of equilibrium ie how hot and cold it potentially can be depending on the state of the parameters. And the system is open. Input and output in a never settled equilibrium.
I know, a ‘reactionary’ thought..😉
“So the rest of the system will warm in response to any additional GHE created, or other warming imbalance within the surface system . . .”
Complete nonsense. The “system” will not “warm” in response to being insulated from the Sun’s rays, any more than air will “warm” by having CO2 added.
No experimental support, and your speculation remains just that.
Were it not for the major feedback loop of clouds that reduce the incoming energy. it is not a simple binary system of just two components.
Let us add some rationality here. (Liquid-) Water has an absorptivity of 0.94 and an emissivity of 0.91. We can calculate..
((342*0.94/0.91)/5.67e-8)^0.25= 281K
Water, as it is, with given insolation, would take on a temperature not so much lower than what we have. The surface is not just water of course and the broadband absorptivities/emissivities of different land types are uncertain. Yet it should not change this result too much, and we can guess like 280K for the whole surface.
This figure does not consider any (dynamic) changes in the state of water (aka glaciation) or changes in convection/advection. But all in all it tells us the atmosphere will be adding some 7-8K to the surface temperature. A long shot from the 33K greenhouse effect. Why so?
There is something people have not realized about the GHE. It is not a reality, but just a “perspective”. A theoretical way to see things given certain circumstances. It is like saying the moon is a big as the sun, which is true, seen from Earth. But it is only circumstantially true.
Equally the GHE comes with 3 distinct constraints:
Point 1 is not true and an undue simplification. Again, water as a hemispheric spectral emissivity of only 0.91. Surface emissions will not be 390W/m2, but just 360W/m2. The delta with regard to OLR (240W/m2) is just 120W/m2, the GHE is thus being overstated.
Point 2 is evidently ambiguous. The biggest contributor to the GHE are clouds!!! People are endlessly confused over this, as they assume it was WV instead. But WV is vastly overstated because of the surface emissivity issue. Basically the error margin there (395-360 = 35W/m2) is wrongly attributed to the GH-property of WV.
On top of that people are usually unaware of the overlap issue. We need to distinguish net and gross contributions of single GH agents. A 30W/m2 of “cloud radiative effect” is just the net cloud contribution, their gross contribution is ~70W/m2. With WV these figures are only about 25W/m2 and 50W/m2 respectively.
Anyhow, clouds are the biggest sole contributor to the GHE. But they also reduce the energy budget by ~50W/m2. For the “perspective GHE” this will not matter, as the energy budget is considered given. But in reality of course it matters.
Point 3 is being totally ignored. The tropospheric lapse rate of ~6.5K/km is largely due to WV. Without WV the lapse rate would be more like an unstable-adiabat of 10.8K/km. Without WV reducing the lapse rate, the GHE would be like 66% (=10.8/6.5) larger. WV thus, just like clouds, excerts a huge cooling effect.
To put things into perspective: the two biggest GH-agents, clouds and WV, are also massively cooling the Earth. With the only theoretic perspective of the GHE, those properties are getting ignored. But in reality they matter a lot.
We can do a simple calculation over what happens if we removed both of them, or all water from the atmosphere respectively. The energy budget would grow to 290W/m2, because 50W/m2 not being reflected by clouds. The non-condensing GHGs (CO2, O3, CH4..) add up to a gross GHE of ~45W/m2. With a lapse rate 66% larger that would grow to some 70W/m2. While the surface emits 360W/m2, these emissions will get reduced by said 70W/m2 to only 290W/m2. So without all the water in the atmosphere, Earth would absorb 290W/m2 AND emit 290W/m2, with basically no change in surface temperature.
This is consistent with what I pointed out at the start. Effectively it are just the non-condensing GHGs warming Earth, with about 7-8K.
Gosh, i never realised that clouds were gasses..🙂
I guess i have never understood fluid dynamics..
That’s ignorance and gullibility in action.
No, it is sarcasm in action.
Guess there is plenty more stuff you never realized or understood..
I suppose you believe your mindreading ability is real, and not just a figment of your imagination?
“Water has an absorptivity of 0.94 and an emissivity of 0.91.”
At what wavelength(s)? An absorptivity of 0.94 implies a specular reflectance of 0.06, for normal (vertical) incidence, which is high for pure water. (It should be about 0.02 for green light.) Similarly, an emissivity of 0.91 implies a reflectance of 0.09, which is even higher than the estimate from absorptivity/reflectance. Are you sure that the numbers you provided are for the same wavelength?
Hemispheric & spectral!
Here are the data as reflectance..
What are you implying? Adding gaseous H2O to air won’t make it hotter. It results in surface cooling, with lower maximum temperatures. Read Tyndall’s works if you disbelieve me.
Nor will CO2.
No GHE “warming” at all.
Thx for sharing your incompetence
“Thx for sharing your incompetence”
Am I wounded or deeply hurt? Not at all.
Maybe I would be if you could describe this “incompetence” based on facts, rather than the contents of your imagination.
Otherwise, why would I be offended or insulted by the opinions of an obvious nutter?
Seems this is what society has come to. People are only just spamming 3 phrase NPC statements, and that is all they can do.
That graph is meaningless unless the surface condition is specified as well as the complex refractive index of the reflector. How are you defining “hemispheric reflectivity?” What is the angle of incidence illuminating the hemisphere?
I know it is “meaningless” to you, just like pearls are meaningless to swine. In the real world however this is essential. Equally your question on what is hemispheric reflectivity (or reflectance?, dont care) is just plain stupid. What angle of incidence? Every(!) – integrated over the hemisphere.
Insulting me does not answer the direct question about the properties of the material represented in the graph. It is simply deflection, an attempt to distract me with an insult. It suggests that you don’t know, that you just slapped something up to snow the lay readers.
I know what a BRDF is. I’m not sure that you do. However, again, your answer is not responsive to my question. I asked you to define how the hemispheric reflectance is specified and you again resorted to an insult.
I suggest that you educate yourself by starting with reading this:
https://en.wikipedia.org/wiki/Bidirectional_reflectance_distribution_function
Ooooooh! Pearls before swine, is it?
You wouldn’t support a GHE which you can’t even describe in unambiguous terms, would you?
In the real world, there is no GHE, adding CO2 to air does not make it hotter, the Earth has cooled over the past four and a half billion years, and continues to do so.
Do you live in a fantasy world, perhaps?
Why not use all available data about our moon?
Some TSI, slower albedo (.11 vs .30) and lower rotational speed.
Its Average Surface Temperature is ~197K, with the daytime temps being mostly radiative balance temperatures, and the nighttime temps rapidly cooling towards 25-50 K, probably the base temperature due to the geothermal flux.
https://www.diviner.ucla.edu/science
The moon doesn’t have an atmosphere so it is not a model of an atmosphere without GHG’s. It is a model of a planet without no atmosphere. Andy has calculated a surface temperature. More important is the air temperature at any given height like 1 km. The surface temperature model doesn’t incorporate a convective heat transfer function . The air above the hottest point on the chart will be heated mightily and rise. As it has no ability to cool by radiation, that energy will remain in the air until it gets so hot that it will lose the heat by convection back to the surface at night. The equilibrium temperature at which that will occur is well above the boiling point of water.
Correct, but imo it is very relevant to be able to explain why the AST on Earth is presently > 90K higher than the AST on the moon, and even more relevant why the deep ocean temperature is > 70K higher than the AST of the moon.
Read Nikolov and Zeller’s various papers and all will explained.
“Andy has calculated a surface temperature.” Instrumentation has measured surface temperature.
Whom do you choose to believe?
“As it has no ability to cool by radiation, that energy will remain in the air until it gets so hot that it will lose the heat by convection back to the surface at night.”
You are obviously joking, or particularly dim. All matter above absolute zero radiates infrared radiation. Unless you can find reproducible experimental results that show otherwise, of course.
Over to you.
Michael
Your statement that all matter above absolute zero radiates infrared photons is incorrect. It is because this is not true that lasers can cool matter to a whisker above absolute zero.
There is a reason gases are called radiative and non-radiative for atmospheric discussions. Oxygen is only vaguely radiating in the IR range – a tiny bit – but not for any meaningful discussion.
An atmosphere that cannot cool by IR emission will heat to the point that it emits in another band. A good example is the Northern Lights which are red and green mostly, because atomic and molecular oxygen radiate in the visible range at different altitudes.
The idea that the atmosphere above the ground would be frozen -18 C in the absence of GHG’s is nonsense. The entire argument about “33 degrees of warming” is based on the idea that convection heat transfer from the surface will stop in the absence of GHG’s. Mechanical engineering 101 teaches how heat is transferred. Three mechanisms for it. Read and learn.
Interested to learn where convection entered the GHE.
According to Lacis et al 2010 the GHE is a radiaitve process.
Lacis et al 2010 (and 2013) is possibly the most influential paper on the subject (Control knob etc.)
“The Sun is the source of energy that heats
Earth. Besides direct solar heating of the ground,
there is also indirect longwave (LW) warming
arising from the thermal radiation that is emitted
by the ground, then absorbed locally within the
atmosphere, from which it is re-emitted in both
upward and downward directions, further heating
the ground and maintaining the temperature gra-
dient in the atmosphere. This radiative interaction
is the greenhouse effect,”
They also specifically mention the backradiation “further heating the ground”, not the atmosphere above it.
If the surface heats the atmosphere then that makes the surface the hot body and the atmosphere the cold body.
How does a cold body, i.e., the atmosphere, “further heat the ground”?
No place I have looked at ever explained this with experimental science nor with any accepted heat transfer mathematics. It is just accepted that models show that it can be done.
From
https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html
Read this carefully. “q” can only become <0 when the cold body Tc⁴ is larger than the hot body Th⁴. This would seem to be impossible.
Don’t ask me, the whole GHE idea is total nonsense.
Our ice-cold atmosphere does not warm the surface, and most certainly not our almost 4 km deep oceans.
I’m curious where convection entered the GHE.
Afaik the GHE is purely a radiative effect, and a nonexistent one at that.
+100
Excellent work. Were you able to model the lapse rate?
This is potentially an unresolved question as all rotating planets with atmospheres show complex rotating vertical bands of air that can potentially explain a vertical temperature gradient.
One must question why the effective radiating height of the atmosphere at 5km is also the 50% atmospheric pressure isobar. This seems impossibly coincidental and suggests there is a temperature gradient resulting from convection, due to the conversion between work and temperature.
I did not try and model the GHG-free atmosphere or the lapse rate. I assumed the atmosphere was transparent to sunlight and planetary emissions, I also assumed no scattering and that the heat it received from the surface (via conduction) was equal to the heat it sent to the surface via conduction. In other words, I effectively ignored the atmosphere as if it were not there. I did figure out the thermal diffusivity of the air, but other than to note it was similar to the diffusivity of the basalt, I didn’t do anything with it.
One caveat, I did assume that the wind speed was high, and the atmosphere was effective at moving heat from the dayside to the nightside. This lessens the difference between the nightside and dayside temperatures.
Thanks Andy. I very much prefer back to basics articles like yours. And the use of models for sensitivity analysis in place of fortune telling is a welcome change.
For my part I feel the lapse rate is falsely attributed as a radiative effect, ignoring that the equation contains no allowance for radiation.
This has led to an underappreciation of the role of convection in establishing planetary temperatures.
I completely agree. The tropospheric lapse rate is mostly due to convection and radiative models must assume the lapse rate for the reason. More here:
https://andymaypetrophysicist.com/2025/02/01/energy-and-matter/
Which equation do you refer to? The adiabatic one?
The lapse rate is mostly governed by the Hydrostatic Equlibrium against gravity an atmosphere is in. At every altitude the pressure must equal the weight of the air above that level. This mostly decides the temperature, should the air warm at some level, it becomes less dense than the surrounding air and rises, when cooling it sinks.
While rising or sinking the temperature of that parcel of air changes adiabatically, either according the dry or wet adiabat depending on whether condensation/ evaporation occurs.
“I assumed the atmosphere was transparent to sunlight.”
Not the smartest assumption, seeing that some 35% of the Sun’s energy doesn’t even reach the surface, resulting in lower maximum temperatures.
Assume away. Assume there is a greenhouse effect. You can’t describe it of course, so you write “The overall greenhouse effect (GHE) is often defined as the difference between Earth’s average global temperature without greenhouse gases (GHGs) and with them.” So the GHE is a number, is it? What are the other ways the GHE is defined? A different number, perhaps?
Seems nonsensical to me.
Andy,
Not sure if I am getting this right. Are you saying in effect that the heat my hand feels when close to a simple electric stovetop element would not be felt if there were no GHG to help radiate it? Do you have time and effort to do more comparisons with observations of our moon which is free of GHG and yet radiates away its daily dose of sunlight to arrive at a constant temperature, as I think is not contested? I read your link to radiation from atmospheric nitrogen and oxygen but it has not had time to sink in. In summary, why would earth with or without atmospheric GHG not reach the same temperature after centuries, as could be the case when reading paleo temperature/CO2 estimates that are not in lockstep? I am happy to claim up if my ignorance would take too much time to redress. Geoff S
This model uses radiation and conduction only to transfer heat, convective transfer in my model is net zero between the atmosphere and the surface. Radiation to space is unimpeded by the atmosphere, and conduction to/from the atmosphere to the surface is effective. I assume that the atmosphere is somewhat effective in moving some of the dayside heat to the nightside and that the basalt interior also carries some heat to the nightside.
Thus, your hand feels the radiation from the stove in my model, GHGs or not.
The Earth does not reach the same equilibrium temperature without GHGs because of lower surface storage of thermal energy. Basalt does not provide enough storage and the atmosphere none. Thus, the nightside without GHGs is much colder and the dayside much warmer.
Exactly. GHGs reduce maximum daylight temperatures, and reduce the speed of surface cooling at night!
That wasn’t so hard, was it?
No “global warming” due to a non existent GHE!
5km is the effective radiating height. It is also the 50% atmosperic pressure iso bar. 33/2 C of warming of 1/2 the mass of the atmosphere takes place below this altitude. 33/2 C of cooling of 1/2 the mass of the atmosphere takes place above this attitude.
This is such a remarkable coincidence unexplained by GHG theory but obvious in terms of work done by convection.
Build a giant planetary size refrigerator without any doors. The warm coils on the bottom resting on the surface and the cold coils at 10 km altitude.
Plug this in to a planetary sized solar array. The warm coils will warm the surface, the energy to do this will come from the cold coils at 10 km atmosphere, all driven by solar power. Run the fridge fast enough to maintain the temperarure gradient.
This is no perpetual motion machine. It creates a greenhouse effect without any greenhouse. The working fluid on our planet is air, and the pressure orifice is gravity. The motor/engine to set this in motion is the sun and the rotation of the planet.
Is your refrigerator running? Better go catch it!
“Build a giant planetary size refrigerator without any doors.”
Are you a fantasizing GHE enthusiast by any chance?
Or just exceptionally dim? What are you going to come out with next – “Grow a set of wings and fly to the moon?”
So conventional models used to calculate the effects of greenhouse gasses are literally Flat Earth models?
Yes. The more mathematically precise description would be “one-dimensional plane parallel”. In the case of Schwarzschild’s radiative equilibrium of the solar atmosphere this was a reasonable assumption. The sun is approximately spherically symmetric in all respects, and the surface can be treated as locally flat.
In addition, these models are subject to constant insolation, the surface is at a constant temperature with zero heat capacity, and there can be neither convection nor conduction, and therefore no atmospheric motion. A radiative equilibrium cannot exist in a convecting atmosphere. Schwarzschild acknowledged this in his 1906 paper on radiative equilibrium in the solar atmosphere.
https://andymaypetrophysicist.com/2025/02/15/schwarzschild-about-the-equilibrium-of-the-solar-atmosphere/?amp=1
No, these are instructive models used in college textbooks. Climate models include full implementations of the relevant physics. The entire argument is something of a straw man.
The computer models parameterize clouds, for one major item. Anyone want to hum an old Joanie Mitchell song?
So models include sub grid-scale processes? Cool.
The global climate models do not manifest a GHE. They assume it is true, parametarize it, and use it as an input.
The models from whence the GHE sprang forth are the one-dimensional plane parallel models “a la Schwarzschild” as described above.
Models do not assume a GHE, they derive it from first principles. It arises from the modeled radiative transfer equations. One-dimensional plane-parallel models were foundational in first describing how infrared radiation interacts with the atmosphere, but are not the basis of modern GCMs.
‘Models do not assume a GHE, they derive it from first principles.’
First principles? Really?
‘Schwarzschild’s equation can not be used without first specifying the temperature, pressure, and composition of the medium through which radiation is traveling. When these parameters are first measured with a radiosonde, the observed spectrum of the downward flux of thermal infrared (DLR) agrees closely with calculations and varies dramatically with location. Where dI is negative, absorption is greater than emission, and net effect is to locally warm the atmosphere. Where dI is positive, the net effect is “radiative cooling”. By repeated approximation, Schwarzschild’s equation can be used to calculate the equilibrium temperature change caused by an increase in GHGs, but only in the upper atmosphere where heat transport by convection is unimportant.’
https://en.wikipedia.org/wiki/Schwarzschild's_equation_for_radiative_transfer
From Wikipedia, of course, which isn’t exactly critical of climate alarmism.
Bottom line, if one already knows the physical characteristics of the troposphere, one can ‘shoe horn’ a radiative model to ‘fit’ a predominantly convective lower troposphere, the goal being, I guess, to support an alarmist narrative of global doom. I think I’m beginning to see cracks forming in the foundations of the AGW consensus.
The models require initial conditions as input parameters, but the modeled state is a result of the underlying physics. And radiative transfer is just one component of climate models. Convection is explicitly modeled in GCMs using fluid dynamics equations, not ‘shoehorned’ in. The interplay of radiation and convection is what determines the overall energy balance.
Yes. Most GCMs use a radiative transfer scheme like the RRTM. There is no GHE module or explicit assumption of a GHE in a GCM. It just appears organically as part of the number crunching.
Note that the RRTM is used in both weather and climate models. For example the RRTM-G is used in the GFS, ECMWF, and other global numerical weather prediction models that are used for decision making (including those consequential to life) on a daily basis. It’s even essential in the storm scale models like the HRRR and the experimental WoF.
The RRTM is also used to design, develop, and operate space based radiometers like those aboard the GOES.
‘Most GCMs use a radiative transfer scheme like the RRTM. There is no GHE module or explicit assumption of a GHE in a GCM. It just appears organically as part of the number crunching.’
Therein lies the problem, i.e., using a radiative transfer scheme to model a highly convective lower troposphere where free emission of photons by IR active molecules is effectively quashed by collisions with other molecules.
No doubt it can be made to ‘fit’ any profile, but are the results realistic? Take another look at Figure 1. in your previously referenced paper by Trenberth et al (2009). Do you really think that convection accounts for only a paltry 17 W/m^2 leaving the Earth’s surface? Again, look at the word count from his paper:
convection – 1
convective – 0
radiation – 39
radiative – 14
‘The RRTM is also used to design, develop, and operate space based radiometers like those aboard the GOES.’
Undoubtedly. The upper troposphere, and beyond, is where radiative heat transfer is actually applicable because thermally excited IR active molecules are effectively able to emit photons before colliding with other molecules.
I’m not seeing the problem. If it is convection you are concerned with then understand that GCMs also model convection (plus a bunch of other crucial factors) simultaneously with radiative transfer.
It’s not just the upper troposphere. The GOES satellites are observing the lower troposphere and the surface skin emissions as well. if the RRTM and other radiative transfer models like it are wrong then the GOES satellites (and other space and land based radiometers) would be useless to us.
I have yet to find any evidence of how radiative path loss is handled by the satellites since they can’t measure water, water vapor, dust, etc in the path being measured. Somehow it just seems to be *assumed* that what the satellites read is somehow 100% accurate without having to know the path loss involved with each sample reading.
Responding to many incorrect beliefs about models expressed above.
To begin, the scalar radiative transfer equation which is the basis of Schwarzschild’s “radiative equilibrium” is not an energy equation. It is an equation based on the propagation of “spectral radiance” which a phenomenological, heuristic model not based on first principles. Read Curtis Mobley’s “A Short History of Radiative Transfer Theory” where this is discussed in some detail.
Schwarzschild himself in his 1906 paper acknowledged that radiative equilibrium requires the absence of conduction and convection. I suspect few here have read Schwarzschild’s paper. A link to a translation is available in one of Frank’s comments above. Schwarzschild in creating his “radiative equilibrium” model of the sun first had to posit a temperature profile based on Schuster’s “two-stream” equation of radiative transfer.
The one-dimensional plane parallel equilibrium climate models used by Harde, van Wijngaarden and Happer, and others require detailed temperature and composition profiles in order to produce the modeled spectra. The models ignore convection, conduction, diurnal variation, and a myriad of other terrestrial characteristics.
One can execute these models without oxygen and nitrogen in the atmosphere and they will produce the same spectrum except for slight variations due to pressure broadening. They are modeling a condition that cannot exist on Earth.
That these models can produce a facsimile of the spectrum observed in space can be explained by the real processes at work in the atmosphere which produce a radiation field generated by collisional excitation and de-excitation. When the real physical processes are taken into account, there is no “GHE”.
Regarding GCMS, the RRTMs (rapid radiative transfer models) are “shortcut” versions of the one-dimensional RTE applied at grid scale that are parameterizations to reduce computation time.
Likewise, convection is parameterized at grid scale. Neither the proper equations nor the computing power exist to model the fluid dynamics of the atmosphere.
The GHE, which manifests in the ECMs, is assumed in the GCMs. There are dozens of various sources of “forcing” that are parameters in the model. Even the so-called “Equilibrium Climate Sensitivity” (ECS) is a parameter.
Planck had to do the same in his seminal work on heat radiation.
MPAS is a GCM. It is open source. The source code can be found here.
Would you mind pointing out the module in the source code in which the GHE is assumed so that we can discuss it?
Not always. The general rule of thumb is if the horizontal resolution exceeds 4 km the model will usually parametrize convection otherwise it will usually solve the primitive equations explicitly in its numerical core. MPAS is an example of a model that can be configured to do it either way.
This is moot since parameterization does not imply that a GCM isn’t useful or physical. In fact, parameterizing certain physical processes can actually make the GCM more useful and more physical.
On page 21 and 22 of the MPAS User Guide v 8.2.0
https://www2.mmm.ucar.edu/projects/mpas/mpas_atmosphere_users_guide_8.2.0.pdf,
RRTMG is explicitly noted as part of the “Physics Suite”.
Incorporation of the mythical GHE is implicit in using radiative transfer to describe energy transport in a convecting atmosphere.
It sure is. So is cu_tiedtke, mp_wsm6, sf_noah, bl_ysu, sf_mynn, bl_ysu_gwdo, and a bunch of other physics modules. Notice that there is no GHE module.
The GHE isn’t explicitly programmed in the RRTM either. If you disagree show me where it exists in the source code available here.
You don’t seem to understand that it is the application of radiative transfer that manifests the GHE, which is a mathematical artifact of the models.
Claiming you can’t find a “GHE module” is a pretty weak argument.
Of course it is the application of radiative transfer by which the GHE manifests. The GHE is a radiative effect.
You said in regard to the models handling of the GHE…“They assume it is true, parametarize it, and use it as an input.”
If it is truly assumed true and used as an input then it should be easy to find that GHE input in the MPAS and/or RRTM source code.
And this is hardly a weak argument. We are testing your hypothesis by looking at the source code. It’s either going to be there or it isn’t. I’m confident, due to spending over a decade looking at GCM source code, compiling them, configuring them, and running them that it doesn’t exist. But I’m willing to give you the benefit the doubt. All you have to do is deliver.
If you can’t identify that input then I don’t really have any other choice but to accept that you’re hypothesis is false. Whether that means you are the source of the myths that you indict others of I’ll leave for others to decide.
So you are a coder? Are you saying that there is nothing in the code that takes into account the posited impact of IR active gases in the atmosphere? What do you think radiative transfer models do?
Of course the code takes into account the impact of IR on the gas species of the atmosphere. That’s the whole point of the model.
What the model doesn’t do is assume the GHE to be true and/or used as an input. No where did anybody input 4 W.m-2 (or whatever value you want to use) as the RF for 2xCO2. It is an output from the model arising organically from computations.
Similarly the GHE is not assumed to be true and/or used as input in MPAS. No where did anybody input 3 K (or whatever value you want to use) as an input for warming that arises as a result of 2xCO2. It is an output from the model that arises organically from the computations.
This isn’t to say that no climate model assumes the GHE to be true and uses a specific RF or ECS as an input. Many of them of do. It’s just not typical of the global circulation models or radiative transfer models.
A good (though not perfect) analogy here is the model of general relativity. No where in the model is it assumed or used as an input that the orbit of Mercury would have a precession component. It is an output of the model that arises organically from the computations.
https://journals.ametsoc.org/view/journals/amsm/57/1/amsmonographs-d-15-0041.1.pdf?utm_source=perplexity
Open the article and read the section on the “Validation of RRTMG”
Yes. Scientists have validated RRTM and its sister model RRTMG. Validating the RRTMG (which is a good thing) does not mean it assumes the GHE to be true and/or uses it as an input. The ~4 W.m-2 radiative forcing figure for 2xCO2 is an output of the RRTM model; not an input.
Let’s jump back to the analogy of Mercury. The observation of precession and validation of general relativity model does not mean that the precession of Mercury was explicitly stated or input into in the model.
‘The ~4 W.m-2 radiative forcing figure for 2xCO2 is an output of the RRTM model; not an input.’
No one is accusing the modelers of hiding the ‘answer’ in their codes. The issue, which we all seem to be dancing around, is whether RTTM(G), or any other RTC, is an appropriate approach to modeling the lower troposphere, where excited IR active gas molecules are dissipated / thermalized by collisions with other gases on the order of 50K:1, relative to the occurrence of the spontaneous emissions.
I have not reviewed the ametsoc journal article that Tom linked to, but was struck by the following statement in its introduction:
‘One especially critical physical process is the flow of solar and thermal radiant energy through the atmosphere, which controls planetary
heating and cooling and drives the large-scale dynamics that moves energy from the tropics toward the poles.’
So, what ‘large scale dynamics’ are these? Again, referring to Trenberth (2009) Figure 1., it’s all radiation all the time in the lower troposphere, but for a paltry 17 W/m^ due to convection and another bit for evaporation. Where is the evidence that thermal radiation from the surface that is predominantly converted to sensible heat within meters of the surface doesn’t drive these dynamics?
I’ll assume that the RTTM(G) code, and the models themselves, are modern marvels of coding. But I am also aware that for all their complexity, Pat Frank (and many, many others) have pointed out since the earliest IPCC GCM composites that the radiative effect of adding CO2 emissions can be accurately emulated by simple linear scaling of these emissions.
I’m not saying this to set anyone’s hair on fire, but to point out that for all its complexity, radiative heat transfer is a lot more ‘tractable’ than modeling convection and other mass and energy flows on a rotating planet where these flows are never in equilibrium.
100%!
You are using 1360 in order to derive a temperature. Where is the outgoing radiation during the day to offset the incoming radiation? It is the NET value that determines the temperature, not just the incoming radiation.
It’s a problem with the math!
If temperature is related to radiation then you get the following relationships:
daytime: Rin – Rout ==> NetRday
nighttime: Rin – Rout ==> NetRnight
If you use 1360 to determine daytime temps then you have ignored that there is an Rout during the day just like at night. It is the *NET* radiation that determines the temperature
Offering a small correction in that the link to the translation of Schwarzschild’s paper was actually provided in one of your comments, here:
https://wattsupwiththat.com/2025/03/16/the-earth-without-greenhouse-gases/#comment-4049904
I would, however, ask if you could take a gander at (and comment upon) the short Wikipedia article on the Schwarzschild equation I linked to earlier, specifically to what it says about the ‘surface budget fallacy’ in the section on saturation. Thanks!
https://en.wikipedia.org/wiki/Schwarzschild's_equation_for_radiative_transfer
While there is much that is good in the article, it is written from the mainstream perspective that radiation propagates from the surface to the top of the atmosphere.
This leads to concepts like “saturation”, “radiative forcing”, and the “surface budget fallacy”. All of this is because of the myopic obsession on the absorptive properties of the so-called “greenhouse gases.” In fact, because they are constantly subject to non-radiative deactivation (thermalization), they can never be saturated.
You are correct, Frank. There is so much mythology in this sub thread it is somewhat mind boggling. I will attempt to summarize in another comment below.
Solar irradiance as well as albedo vary strongly with wavelength, so absorption should be calculated as a function of wavelength. Also, it is assumed the solar spectral irradiance is a blackbody curve, but this really isn’t true — different BB temperatures for the sun can obtained depending on how the curve is fitted.
Greenhouse gases have little to do with it. They don’t just absorb thermal energy, they also emit it. The earth’s atmosphere is partially heated by the sun, thanks to those gases. But mostly, the earth surface itself is heated by the sun, and that surface heat is conducted to the air, and convected aloft. That can be seen by any of the thousands of weather balloon traces.
You walk around in heated air during a sunny day. Most of that thermal energy is in NON-GREENHOUSE gases – nitrogen, oxygen, argon. They don’t radiate, so they TRAP thermal energy. The ONLY way they lose their thermal energy is via conduction to cooler molecules (cue the radiating gases) and conduction with the earth’s surface. Again, weather balloons prove this perfectly. Air heats from the surface up when the sun is shining, and cools from the surface up when it’s not. For pete’s sake, we WALK around in this every single day, and no one seems to understand it. CO2? It is a cooling influence because it radiates thermal energy at all times. As long as it is above 0K it is radiating. And ‘cooling’ anything that bumps into it.
Close to the surface GHGs absorb thermal energy emitted by Earth’s surface, but they only emit the energy 1/50,000th of the time. The rest of the time they pass the energy on to neighboring molecules via thermalization, warming the lower atmosphere. This causes convection.
Higher in the atmosphere where the molecules are farther apart, they start to emit radiation to space and cool the atmosphere. This is the emission layer and the cause of the lapse rate. Emissions from the emission layer (~2-10km) are nearly all from water vapor. CO2 and ozone emit at higher altitudes.
https://andymaypetrophysicist.com/2025/02/01/energy-and-matter/
Not sure where that comes from. GHG’s don’t wait to absorb a photon to emit one. They are constantly emitting thermal energy. They gain thermal energy by absorbing photons, AND by conduction via collisions.
If there were no ghg’s, earth would be warmer, since all radiation would from earth’s surface. The surface would HAVE to be warmer in order to radiate more energy.
Convection is caused by ALL air molecules touching a heated surface – caused by the sun. Convection cannot be achieved by gh gases, with the exception of thermals in clouds. But it is not because water vapor is a gh gas, it is because of sun’s action on the surface. There is no ‘opposite” of conduction at night. It is pure radiation and conduction from air to surface.
The “radiative temperature” calc with or without greenhouse gases depends on whether you assume we get 340 watts from the Sun at some chosen planetary Albedo. Most demonstrative calcs assume an Albedo of .3 and thus 240 watts of sunlight. Without clouds and oceans, Earth’s Albedo would probably be something like the Moon’s. But of course it isn’t, so the discussion is hopefully used to illustrate the principles involved, but often isn’t….
“Not sure where that comes from. GHG’s don’t wait to absorb a photon to emit one.”
However, if the emitted photons are not ‘replaced,’ then the emitter cools down and the peak of the emission curve shifts to longer wavelengths.
Andy
they also absorb energy from collision at the same exact rate.
Not exactly the same rate. Thermalization adds sensible heat to the lower atmosphere which causes convection, aka thermodynamic “work,” aka “weather.” Weather moves the heat around until it gets high enough so that it can be passed on to (mostly) water vapor molecules which pick it up by reverse thermalization.
Water vapor is more easily excited to a state where it can emit IR than other GHG molecules and between 2 and 10 km it does the lion’s share of the emissions.
I don’t know how much of the original energy absorbed in the lower atmosphere is turned into work, but it is probably significant.
The kinetic energy is distributed through all the molecules in the atmosphere only a small proportion of the molecules have enough kinetic energy to excite a ground state CO2 molecule to its first vibrational excited state from which it can emit a 15µm photon.
“a small proportion of the molecules have enough kinetic energy to excite a ground state CO2 molecule to its first vibrational excited state from which it can emit a 15µm photon.”
No. At even 20 C, the molecules in the atmosphere have an average velocity of oxygen molecules is around 480 m/sec.
Now the minimum energy required to excite the 101 state molecules f a CO2 molecule is about 0.22 eV. You can work out the rest.
So you are babbling nonsense, as usual.
Actually as usual your ‘facts’ leave something to be desired.
The bending modes of CO2 which is what we are discussing here require an excitation energy of 0.082 eV (667cm-1), compared with the average kinetic energy of a N2 molecule at 20ºC 0f 0,039 eV. So in order to collisional excite a CO2 molecule a N2 molecule would need to have at least twice the average kinetic energy. Check out the Maxwell-Boltzmann distribution to find out what fraction have that energy.
https://files.mtstatic.com/site_4334/54006/0?Expires=1742251906&Signature=DcblPROX1n-vxxgnJMSK~BrrS1dt-hdT~If6ij8cbHzL~AIU7Za4HgojscXNuivTRe4yubBeF4EsATM8ahxzknmFj4hwtNA1~A8srrz~c2IXVC0~w1~-J4U3wNpaJbd2sEE~fEDiuO6Z6bkBx4o9IlQiNK9AV~GPQt7lKfQz2fA_&Key-Pair-Id=APKAJ5Y6AV4GI7A555NA
Plus, Co2 has only 1 vibrational mode compared to the 3 WV has.
Water vapor has 3 categories of transitions that can absorb photons, rotation, vibrational, and electronic. In all combinations it has more than 37,000 states. I think CO2 has three total, but only two in the spectrum of interest, 15 and 4.2 micrometers. As far as climate change is concerned, only the 15-micrometer absorption is significant. The 15-micron band is attached. The illustrations are by Markus Ott and Heinz Hug.
The bending mode has two modes at right angles to each other.
It’s easier to excite water vapor to emit a photon because it absorbs/emits across almost the whole GHG portion of the spectrum. CO2 is limited, but if there is 15-micron wave out there, it will absorb it.
Andy, you wrote –
“but if there is 15-micron wave out there, it will absorb it.”
And then? You really don’t understand any of this do you?
Any matter colder than about -80 C will absorb 15 micron waves happily, and heat as a result.
Any matter colder than -80 will not emit 15 micron waves in any appreciable quantity – not even CO2!
It is a matter of observation that smartly compressing air containing CO2 to about 20 bar or so will raise its temperature to more than 500 C. For example, in a diesel engine.
You are not going to tell me that the CO2 is at a different temperature to the air which contains it, are you?
Go on, give me some comedic relief! Follow it up by saying that adding CO2 to air makes it hotter!
“They don’t radiate so they trap”
WRONG, Wrong, O2 and N2, are transparent so they want to let the IR zip through to outer space at the speed of light. Some IR, even lots of IR at specific resonant frequencies, hits the few CO2 and H20 molecules in the mix, which vibrates them and then they lose that additional energy by bumping into the couple of thousand or so N2 and O2 molecules surrounding them. Plus the bulk mass of IR absorptive gases release and absorb photons according to Planck’s curve temperature and their emissivity while this bumping is going on..
You will notice that I said nothing about radiation heating inert molecules. I said the sun heats the surface, the surface heats ALL air molecules touching the surface, and the heated air rises through convection. There is no opposite of convection at night, so the day’s heat is lost by earth’s radiation to space, and ghg radiation to space. The heated air cools off by contact with the surface, which radiates and cools. Why do you think days are warmer than nights? (usually) It sure as hell isn’t because of gh gases.
Generally speaking, I agree with what you said except for one thing. As the ground surface cools at night, and if there is a slope to the surface the air will flow laterally to lower elevations. This air can be warmed some by the work of compression (adiabatic lapse rate), but it will mix with other air, continue to cool by contact with the ground. Where I lived long ago on the east flank of the Rockies, in summer we’d get a strong breeze by around 2 am from this on clear nights.
So this is a bit like the convection during the day. Otherwise it is difficult to see how such deep layers of cool air develop at night.
Cool air develops simply because the surface radiates at night, and air in contact with it cools also. I agree that with a slope, it will follow the slope. It will also form cold pockets, which can become even colder, as the surface is always radiating. It is the reason you can have frost on roofs, cars and lawns, even though the air measured at 2 meters will be above 32F.
“ I agree that with a slope, it will follow the slope.”
Causing katabatic winds of over 300 kmh in Antarctica, on occasion. GHE believers have other odd beliefs – they imply that climate affects weather. How stupid is that – climate is just the statistics of weather observations.
Quite divorced from reality, people like that.
“So this is a bit like the convection during the day”
Word salad. A bit like? Big bit? Little bit?
Next you’ll be saying that adding CO2 makes it hotter, or something equally silly.
From jshotsky post:”…surface heat is conducted to the air,”
He never mentioned IR. You say he is wrong about something he never said.
The heat N2 and O2 acquire from the surface can only be removed from the atmosphere by N2 or O2 passing it to CO2 or WV to emit.
Once an N2 or O2 has obtained via conduction heat, that heat is trapped until it can be relieved by CO2 or WV. If that didn’t happen over time the atmosphere would over heat.
CO2 and WV cool the atmosphere.
Do you mean that N2 and O2 can’t EMIT?
They are bad at absorbing IR, but are capable of absorbing eg UV.
See eg the ozone layer
“The heat N2 and O2 acquire from the surface can only be removed from the atmosphere by N2 or O2 passing it to CO2 or WV to emit.”
Complete and unmitigated nonsense. All matter emits radiation. As Tyndall demonstrated, air with all CO2 and H2O removed could be heated with infrared rays, and radiated infrared, resulting in cooling when the infrared source was removed.
Dream on.
mkelly,
Then our Moon cannot cool, because it has no atmosphere hence no GHG?
Yet the sun makes it hotter. If it cannot cool, we get runaway heating.
There has to be a mechanism for cooling other than by gas IR going to space. What is it?
Geoff S
The moon cools by emitting IR to outer space….no gas required! MF seems to be unaware of Earth’s atmospheric “window” in the 8-14 micron range. The moon’s “ window” has a much wider range….
“MF seems to be unaware of Earth’s atmospheric “window” in the 8-14 micron range.”
I note that you put the word “window” in quote marks.
The Earth starts cooling slightly after the solar zenith, and continues to do so until sufficient insolation occurs.
It does this by emitting IR to outer space, as it has done for four and a half billion years. That’s why the surface is no longer molten.
i hope you are not implying that adding CO2 to the atmosphere makes the Earth hotter, are you? Only someone suffering from a mental deficit – say extreme gullibility – would believe that.
“cooling” means dT/dt < 0. Adding CO2 and WV to the atmosphere causes dT/dt > 0 which is the definition of “warming”.
I see this conflation of “cooling” with “radiating” a lot here on WUWT. They are not the same thing. A body can be radiating, but not necessarily cooling. For a body subject only to radiation cooling occurs when and only when the amount of energy it is emitting is more than the amount it is absorbing. Obviously if the body is subject to conduction and/or convection you have to sum up those contributions as well, but the general rule still applies. That is cooling occurs when Ein < Eout.
Your explanation is incomplete. From Planck’s Theory of Heat Radiation.
A body radiates soley based on its temperature, not the bodies it is associated with.
Body Bi has warmed Body A to 100°C. Body A radiates based on that temperature toward both Body B and Body Bi. Body Bi cools with respect to Body A because the net radiation is toward Body A. Body A cools with respect to Body B because the net radiation is toward Body B.
“A body radiates soley based on its temperature, not the bodies it is associated with.”
Yup.
How do you know if a body is radiating more energy than it’s receiving?
The temperature is dropping – that’s called cooling.
That’s how you know the Earth has radiated more energy over the last four and a half billion years than it received. It has cooled. Maybe not in your imagination, but that’s your prerogative.
“Once an N2 or O2 has obtained via conduction heat, that heat is trapped until it can be relieved by CO2 or WV.”
According to the LLM, Copilot:
Is a molecule of nitrogen not an object?
Gas molecules are not ‘objects’ as described by your copilot.
Phil, of course they are. You probably even claim they collide with each other!
Are you now claiming that mysterious “non-objects” can bounce around like billiard balls?
All matter above absolute zero radiates infrared light. Atoms are matter.
“A blackbody is an object that absorbs all of the radiation that it receives (that is, it does not reflect any light, nor does it allow any light to pass through it and out the other side).”
Phil, you wrote “A blackbody is an object that absorbs all of the radiation that it receives . . .”, for no reason at all, that I can see.
Have you finally lost touch with reality completely?
Seems the name “blackbody radiator” is pretty odd.
If you want to assign a colour to a radiating body I would think “red hot” etc.
A body that absorbs all incoming radiation appears black.
When some incoming radiation is reflected the object becomes visible, a grey body.
https://www.pnas.org/doi/10.1073/pnas.0900155106
If O2 and N2 are transparent, that means they effectively have zero reflectivity and zero absorbance (very long path length to absorb all incoming photons). By Kirchhoff’s Law, they should an emissivity close to 1. Actually, they do have a non-negligible reflectivity at short wavelengths, which is why Rayleigh Scattering, which makes the sky blue, occurs.
Huh
As you say, “effectively zero reflectivity and zero absorbance”, yes BUT
absorptivity + reflectivity + transparency = 1
and
Absorptivity = emissivity for a Black body at fixed temp.
so by Kirchhoff then emissivity is close to zero for N2 and O2 if they are close to transparent, since emissivity is relative to a BB
https://www.epsilonengineer.com/radiation.html
Go on, make some air hotter by adding CO2 to it.
Or just babble some more nonsense to obscure the fact that you cannot even describe the mythical GHE.
Sad, isn’t it?
The sun makes the air hotter. I will prove it to you most mornings. Can also show you with an IR camera that CO2 absorbs more IR than air….so provable fact is “babble” to you ?
Sun warms the surface first and then the surface warms the air, not the other way around.
You can’t describe the GHE, so you babble about the sun emitting energy. You don’t even own an “IR camera”, do you?
You don’t seem to realise that IR, by definition, includes all radiation wavelengths longer than visible light. Yes, really.
Accept reality – less GHGs result in higher surface temperatures, as John Tyndall pointed out over 100 years ago.
Go on, describe the GHE in clear, unambiguous language. Only joking, I know you can’t.
“They don’t radiate, so they TRAP thermal energy.”
Hate to mention this in case you have a stroke, but you are talking nonsense. All matter above absolute zero radiates infrared. You can’t stop it, or store it.
Microwaves, by definition, are infrared. Go ahead and store some, if you can.
Maybe you are a wee bit gullible, or ignorant, or both.
Technically being transparent and emitting are different phenomena.
A good example is an IR lens made of the metalloid germanium.
Germanium has a melting point of over 900 C. At 800 C it’s glowing red.
And yet it is transparent enough to IR to make camera lenses from!
Or you can just put your hand against your transparent windscreen after your car has been in the hot sun for a bit.
And to clarify another point N2 and O2 at Earthly temperatures say +65 to -65 C radiate zero, zilch, nada, but being transparent they let through the IR from whatever surface is “behind” them.
If O2 comes ozone by way of UV or other radiation, then Ozone (O3) exhibits several infrared (IR) absorption and emission bands.
“And to clarify another point N2 and O2 at Earthly temperatures say +65 to -65 C radiate zero, zilch, nada, but being transparent . . .”
Can’t support this bizarre assertion with experimental data, can you? Tyndall heated dry, CO2 free air with infrared. He observed by experiment that that same air radiated infrared, and cooled as a result.
All matter above absolute zero radiates infrared light.
Adding CO2 to air does not make it hotter.
Very interesting, Andy.
“Comments on the appropriateness of assuming the atmosphere is thermally net neutral would be interesting to read.”
Even assuming heating and cooling of the fully transparent atmosphere is only by contact with a warmer or cooler surface, a circulation must develop. This means there will be advection, with a net transport from the equator to the poles. This may not be a net-neutral influence on the surface temperature, because of 4th power S-B emission.
See the reference on the first page here about “horizontal heating gradients.”
https://a.atmos.uw.edu/academics/classes/2010Q2/545/545_Ch_1.pdf
I’m replying to myself here about the “flat earth” conceptual “model” that is the subject of critique in this article.. Some commenters have asked whether the CMIP models are literally “flat.”
No. They are 3D gridded models of the earth with atmosphere, land, and ocean components. Sunlight is modelled in day-night cycles with geometry as you would expect, etc.
Example here from NCAR.
https://ncar.github.io/CAM/doc/build/html/cam5_scientific_guide/index.html
So it is important not to project the “flat earth” description onto the CMIP models themselves.
Thank you for listening.
While the CMIP models do use a 3-D grid, it is my understanding that in each grid cell they use a simplified version of the “flat earth” model in order to compute the “radiative transfer” in each cell.
As an example, this document at Section 4.10 Radiative Transfer describes how the NCAR CAM5 model uses the RRTMG radiation code for longwave and shortwave radiative transfer. No doubt all the CMIP models use similar code, or maybe even the same code.
I have no idea whether the RRTMG or similar code handles the IR absorption/heating/emission for clouds, H2O, CO2, CH4, etc. realistically. But in any case, that is the potential issue.
https://www2.cesm.ucar.edu/models/cesm2/atmosphere/docs/description/cam5_desc.pdf
FYI, there was much discussion on this topic amongst Andy, Markus Ott, and myself.
The topic came out of discussions of a scenario posited by Paul Linsay. You can watch Paul’s presentation on the Tom Nelson Channel here:
https://youtu.be/hQt_I-RvGF4?si=CGWu2rKUSU1eRXx3
where the discussion of this starts around 28:00. It might be more efficient to download Paul’s paper here:
https://substack-post-media.s3.us-east-1.amazonaws.com/post-files/121426003/02098f46-f503-46ca-9ff8-6b26e0cc206b.pdf?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Content-Sha256=UNSIGNED-PAYLOAD&X-Amz-Credential=ASIAUM3FPD6BVFP7PI64%2F20250316%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20250316T211225Z&X-Amz-Expires=3600&X-Amz-Security-Token=IQoJb3JpZ2luX2VjENj%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJGMEQCICSc7tvxI11okKRdZRveLfQUqvormEugTwxcopRBcc2cAiBi9LGYBEyIQGxquDR771bQXh2yn2ZmJI%2FHFBBtBtkIUCr5AwgxEAQaDDMwMjQ3MTEyNjkxNSIMZ5PkBVokTZ6uqTqXKtYDavnqexHtRo3pgUx8GxJm896kBI7PCB%2FdCX2%2BfCN6gMD%2BsLBMaDlupDp%2B9miYYbfLYDjvf88GweCJm6YtcS5PVEIOYEp6HLcKbrqsGZ3upas8L6QZYqHLnfnHcAnTsD68CFQLsi%2FCnRVYvkgXTWZ%2Fwzd9WNkAhoYTXZ1cYwOho1Fh9knCswpQsG%2FWxzCp8aYXEmBHqkxtcbyBMyggee6lCCtnymQ9wtF1VZopVrH9DW3pDzbgt%2BlQCyo5lPn9aO0GvxyGnvEzwGFgVFcKHZNqLqaUicO2p0KmHHaV8jSoplng87rdLazoH%2Fcx6RMW7eV5QeTILGxNtzCtboJbS5fqNBd7fj0uNY5oWbfmvALtryqlI5LeDfd%2FkRW6EIIY785YYwIqnFge0YSBbokTrWjj%2FsuvK7zxdjl4idP7lssx737f3It44ihPKCwJWR3KhHnukFaDlbAH53XpDHrUbeweOJx9xco1BzredtM6WSkkpihLCMvQxzzOXKtUbfsaATIfvqvk1reWKCnW0LowcSQ8E28O%2F%2F68Nh7yKqIXwmaiVVhUP%2B5NYRkzmgilY2f9smQ9OTAolSzYvD9n9qYAKktDWIY1tFfXYTGiefUhOFtwDbnYkK9kB5gww%2BPbvgY6pgFeDOF0g8PxlMzii8sVnovSq1beKJKubTl7rT9gBW%2FNq4%2FRMCTbVekHx4uBC6hdpvxF3R3D97PYjEGsejudWqasDyL8UY%2BjDATeiRn2QUcToMpkPQzlIVj8S%2BUhHKftSxIMgzJbh4MCrardn%2FhQb5C8mZ7TegYu8tFgnBnHuhlQ3yU9HwZEWiNH2F456JGqrJo5DW6lMC7PbI5XR7hXenMjzw1Q%2Bm1L&X-Amz-Signature=5a9f69d705564b1bda679708d2dc696e61b550280aad4297731a7d0fec06e24d&X-Amz-SignedHeaders=host&response-content-disposition=attachment%3B%20filename%3D%22An_Indepth_Analysis_Of_Climate_Model_Assumptions.pdf%22&x-id=GetObject
The basic question is, in the absence of IR active gases in the atmosphere, how can heat accumulated in the atmosphere escape to space?
While Paul’s conjecture leads to an extreme equilibrium condition at odds with Andy’s position, it has some merit. He provides an explanation of his view of the dynamics in his paper.
The exclusion of conduction and convection is a necessary condition for Schwarzschild’s “radiative equilibrium”, as I explained in an earlier comment.
https://wattsupwiththat.com/2025/03/16/the-earth-without-greenhouse-gases/#comment-4049904
Radiative transfer is not a factor in this model of a completely transparent atmosphere.
I will add only that the ratio of volumetric heat capacity for basalt vs air is ~ 3000:1. 1 mm of basalt has the same heat capacity as 3 meters of air.
It will be interesting to see how the discussion evolves.
Andy,
Nice article. I’ve got to cogitate on some of this to get a better understanding.
Thanks for doing away with the flat earth scenario and average insolation. That isn’t the way the earth exists. There is a gradient from the equator to the poles as you show. Because of the T⁴ factor with radiation, there will be a temperature gradient throughout. I need to think more about your model and how this affects it.
The division by 4 only caters for the difference in surface area between a sphere and a circle.
To account for the day/night difference simple solution is to divide incoming radiation by 2 and average the result with the temperature of the dark side, 0K (or 3K if you want to take the Cosmic Background Radiation into account)
So 1364 W/m^2 minus 30% for albedo gives 954,8 W/m^2, divided by 2 is 477,4 W/m^2
Resulting radiative balance temperature 303K.
Averaging with the dark side gives (303K + 3K)/2 = 153K
This makes the measured ~197K AST of our moon simple to explain.
Radiative balance temperature on the dayside plus less colder nightside due to a base temperature caused by the geothermal flux plus carry over heat from the dayside.
See my comment
Coupled Model Intercomparison Project: CMIP
Can I assume that most of the CMIP models are flat?
You can safely assume all of them are meaningless since the radiative impact of errors in their specification of cloud coverage far exceeds the presumed radiative impact of anthropogenic CO2 emissions.
This is complicated to answer. CMIP models are three-dimensional spherical models, so they are not flat. However, they do not properly account for convection either. They are essentially over 100,000 one dimensional columnar models placed outside a sphere. The vertical columns of the one-dimensional columns are divided into cells, or “parcels.” Within each parcel the air doesn’t move and a state of “local thermodynamic equilibrium” is assumed. The parcels are not in thermodynamic equilibrium with each other, so they exchange mass and energy between them.
The main problem is the parcels are over 12,000 sq km at the equator and can contain multiple thunderstorms, so the local thermodynamic equilibrium assumption inside each parcel is nonsense. For more see here:
https://andymaypetrophysicist.com/2025/02/01/energy-and-matter/
I am a little confused by your phrasing here. I get the idea of multiple one dimensional columnar models (and generally agree with your criticisms). But when you say the “exchange mass and energy between them”, that seems to imply that the columns aren’t one dimensional, or are you stating that these “parcels” only exchange mass and energy radially, up the column?
In a GCM they can exchange mass and energy to any neighbor, up and down and sideways. The reason they don’t emulate convection is that they are too large.
Do any of the EnviroWhacos have an explanation as to how their theory / model incorporates the Temperature with height. Appears to me that there are mor holes above 50,000 feet than below 50,000 feet, How does that heat get up there to be radiated into Space?
Convection
Andy, you may not have noticed that the atmosphere cools at night, when convection caused by solar input is least.
You don’t understand atmospheric physics at all, do you?
Educate us instead of insulting us.
What would you like to know that you can’t find out for yourself without asking me?
Ask away.
Benestad is using a spherical Earth model. See equation (1) on pg. 681.
Who is it exactly that is using a flat Earth model with 21.9e24 j of energy delivered at TOA over one orbital cycle vs the 5.5e24 j that is actually delivered for a spherical Earth model?
I’m asking because I’ve never seen even a single publication that uses flat Earth geometry. It’s always spherical Earth geometry.
For those that don’t know it is the divide by 4 that is the tell that it is a spherical Earth model.
In Benestad equation (1) S is the solar constant of 1360 W.m-2. However this is the flux where the incident angle θ = 90 deg.
Since the Earth is a sphere its surface curves away such that the solar flux is being spread out over a larger area. There is a point where the solar flux comes in at θ = 0 deg. So we have to project the flux down the hemisphere. To do this we weight S based on its incident angle θ of the ring that wraps the sphere. This is the sin(θ) component. Then the length of the ring as it wraps the sphere increases as θ decreases. This is the cos(θ) component. The area is then given by sin(θ) * cos(θ) for the infinitesimal ring dθ. We then integrate the rings down the hemisphere via ∫[sin(θ) * cos(θ)]dθ to get the weighting multiplier which converts S from a flat geometry to spherical geometry.
We now have to consider that Earth has two hemispheres. There is the lit side and the unlit side of the Earthly sphere. This means we have to apply the ∫[sin(θ) * cos(θ)]dθ multiplier twice. Ya know…because Earth has two hemispheres. This makes the final multiplier ∫[sin(θ) * cos(θ)]dθ * ∫[sin(θ) * cos(θ)]dθ. The bounds on both integrals is from 0 to 90 degrees. So each definite integral calculates out to 0.5 which makes the final multiplier 0.5 * 0.5 = 1/4.
This is why Benestad divides S by 4. Note that S is flat geometry while S/4 is spherical geometry.
It’s also important to note that the solar constant S = 1360 W.m-2 actually represents the average zenith flux over one orbital cycle. If we do the full spatial and temporal integration of the Earth’s surface and its orbit we get the double integral ∫{t:0 to 31556736} ∫{θ:-π/2 to π/2} [2r^2*s(t)*cos^2(θ)*dt*dθ] where s(t) is a function of the zenith flux at time t as result of Earth’s distance from the Sun. When you calculate this integral you’ll get 5.5e24 joules. Note that (S/4)*31556736s * 510e12 m^2 = 5.5e24 joules as well. This proves that S/4 is the mathematical shortcut for both Earth’s spherical geometry and it’s elliptical orbit.
The zenith solar flux can be estimated using s(t) = 1360 W.m-2 * cos(ω*t) where t is the time in seconds from the point where Earth is at perihelion and ω is the angular velocity in radians/second and is itself estimated to be ω = 1.99e-7 rad/s.
Below is the result of the full spatial and temporal integration compared with the S/4 shortcut both of which show 5.5e24 joules of energy delivered at TOA over one orbital cycle.
Doh. Typo. I left off the amplitude α caused by the elliptical orbit. That should be s(t) = S * α*cos(ω*t) where S ~ 1360 W.m-2 and α ~ 50 W.m-2.
I’m not sure what happened to the image. I’ll repost here.
I suppose that this is supposed to prove the impossible notion that adding CO2 to the atmosphere makes it hotter?
There seems to be a problem here. If you are using vertical slices, then it seems that you are calculating an upper-bound because a bundle of rays gets spread out because of the angle of incidence being zero at zenith and infinite at a glancing angle (theta=0). At each second, the incoming flux should the sum of all the individual foot prints which is the normal flux divided by the size of the footprint. The next step, which you don’t discuss, should be the specular reflection off the oceans, where the reflectance increases with the angle of incidence, reaching 100% at a glancing angle. All of the solid terrestrial elements have an associated BRDF that changes with the angles of incidence and again should be handled for each and every footprint.
I’m not using vertical slices.
I don’t discuss it because it is completely and entirely irrelevant. Specular reflection off the oceans has zero impact on the solar radiation flux imparted on the spherical shell of Earth at TOA. That flux is S/4 regardless of the albedo or whether the reflection back to space is 0%, 100%, or some value in between.
If you don’t think 5.5e24 joules of energy is received by the Sun over one orbital cycle then what do you think it is? Show your work.
“If you don’t think 5.5e24 joules of energy is received by the Sun over one orbital cycle then what do you think it is? Show your work.”
If you don’t think that the Earth emits all the energy it receives from the Sun over one orbital cycle (plus a little of the Earth’s internal heat), what supports your fantasy?
Why do you think that the Earth has cooled over the past four and a half billion years?
It is important because the TOA flux doesn’t tell one how much light reaches absorbers at the surface and get converted to heat. Clouds aren’t the only thing that control ‘albedo.’
That has nothing to do with Andy’s claim that S/4 is a flat Earth model which is what I’m addressing.
You seem more defensive than usual on this topic. Anyone who has taken high school geometry knows that the ratio of the area of a circle to the area of a sphere of equal radius is 1:4. (The Greeks figured this out long before the invention of the calculus). The real issue is whether or not the resulting non-diurnal ‘flat Earth’ that presumes continuous and equal isolation for every point on Earth is an appropriate model for the study of climate.
If Andy understands the 1:4 relationship he is doing everything he can convince people otherwise. And a quick google search shows that WUWT has many articles insinuating that S/4 is a flat Earth model. So no, I don’t think anyone who has taken high school geometry knows this.
Again…who is assuming the Earth is flat, non-diurnal, and with equal isolation for every point on Earth? [Benestad 2015] and [Trenberth et al. 2009] certainly didn’t assume any of those things.
This is nothing compared to my posts defending the 1LOT and 2LOT.
Thanks for the links. Let’s focus on Trenberth et al. 2009. Are you certain that all of the fluxes in Figure 1 (page 314) have been adequately adjusted for the ‘spatial and temporal’ differences of a real Earth using the process described at the bottom of page 315?
Not that this is the biggest problem with climate modeling in general or the paper in particular. I did a quick word count on the above article and got the following results:
convection – 1
convective – 0
radiation – 39
radiative – 14
Doesn’t seem all that applicable to what really happens within the Earth’s lower troposphere, does it?
No. The procedure in the spatial and temporal sampling box is relevant to radiant exitances only.
I’m not sure what the troposphere has to do with this. But, the spatial and temporal sampling box is applicable to Andy’s claim that [Trenberth et al. 2009] is using a flat Earth model since the content in that box is using spherical geometry model.
‘I’m not sure what the troposphere has to do with this.’
Quite a bit. Trenberth’s Figure 1. clearly gives the impression that radiative transfer is by far the dominant mechanism of energy transfer from the Earth’s surface to the tropopause. This clearly cannot be true given that the spontaneous emission of photons by IR active gas molecules within meters of the surface is largely prevented by collisions with non-IR active molecules.
https://andymaypetrophysicist.com/wp-content/uploads/2025/01/Shula_Ott_Collaboration_Rev_5_Multipart_For_Wuwt_16jul2024.pdf
Let me make sure I have this straight so I’m not accused of putting words in your mouth. Are you trying to argue that because you don’t agree with one of the values in the diagram that necessarily means the model is using flat geometry?
One of the values? You’re missing the forest for the trees – the issue is that Schwarzschild’s model for radiative transfer is not applicable to Earth’s lower troposphere.
The forest being discussed in this thread is that Andy thinks S/4 is a flat geometry model. I see the forest just fine. If you want to discuss a different forest that’s fine. Post a new top level comment and I’ll participate as I get time.
You tried supporting the S/4 model with some ditzy calculus and got it wrong.
The issue is distributing the insolation properly over a point on the equator in order to calculate the temperature distribution at that point. The average value to do this 0.637, not 0.5. This allows one to take the peak value of insolation, 1360, and calculate an average insolation over that single point from sunrise to sunset. In other words, 1360*(1/(b-a)∫sin(x) [0,π] = 1360*0.637 = 866 at a point on the equator. That is not 240. Then as you go from the equator to the pole, that average is reduced by a cosine average which from 0 to pi is 0.637 again. That gives you the average temperature spread around half of a sphere. Even this is a very simplistic view of the earth.
Dividing the surface area of a sphere by 4 is basically ending up with the area of the flat surface if one cuts a sphere into two equal pieces. It takes some funky assumptions to reduce the sphere surface area into a much smaller flat surface and then try to find an average insolation where equal areas all receive the same insolation. It is an attempt to convert a 3D problem into an unrealistic 2D solution.
Use the procedure you just outlined and compute the total energy Earth receives from the Sun at TOA over one orbital cycle. Present your result as a single value with units of joules. Don’t deflect. Don’t divert. This isn’t a trick question. Earth receives a finite amount of energy from the Sun during its orbit around it. It is calculable and known. Go.
Strawman. Andy’s premise has nothing to do with the energy received over a year. Here is what Andy proposed.
You keep wanting to somehow toss an S/4 as how to characterize the average insolation and consequent earth temperature into the picture. S/4 is not a correct area to use in determining the insolation arriving at each point of the Earth’s surface. It is not even a duplicate of the area of a sphere.
Why don’t you address how your analysis of sine and cosine arrived at an average value of 0.5 for each of the functions in order to support S/4, rather than 0.637. That will be a step forward in the discussion. Show us the formula that you used to calculate the average of a sine from 0 to π. Here is what my books show.
[1/(b – a)] ∫ₐᵇ sin(x)dx
Do you know how to calculate the total energy Earth receives from the Sun at TOA in one year using calculus or not?
∫sin(θ)cos(θ)dθ over {θ: 0 to π/2} = 0.5. This multiplier applied twice is 0.5 * 0.5 = 0.25 = 1/4. That’s where the divide by 4 comes from in S/4.
Also, 0.5 isn’t an average.
If I thought the annual amount insolation was pertinent I would calculate it. However, you are trying to create a strawman that somehow the annual insolation has an effect on the distribution reaching the earth. You need to explain how the annual value, regardless of its measurement effects the daily distribution of energy to the surface.
I don’t care whether the value is 100 or 10,000 it doesn’t affect the method used to calculate the distribution to the earth. I have used the value of 1360, as you did when you said:
The maximum average value reaching the lit side is 0.637*1360 = 866 at the equator as described by Andy. The maximum average to the pole is 0.637*0.637*1360 = 552.
Tequator = (866/5.67×10⁻⁸)¹/⁴ = 352K
Tlatitude = (552/5.67×10⁻⁸)¹/⁴ = 314K
This is what you’re going with? Like…you don’t want to rethink it at all eh? Just dive right in?
Do you understand that a sphere has to be analyzed using 3d calculus and not 2D plane geometry? I.e. coordinates are r, Θ, φ and not just r and Θ.
The only one that needs to rethink is you.
“∫sin(θ)cos(θ)dθ over {θ: 0 to π/2} = 0.5.”
So what?
From the perspective of a point on the equator the sun travels a path from 0 radians to Π not 0 to Π/2, i.e. from sunrise to sunset is a distance of Π and is represented by the angle (Θ).
From the perspective of a point on the equator the path to the pole from that point is Π/2 radians and is represented by the angle (φ).
Standard 3D polar coordinates –> r, Θ, φ
From the perspective of that point on the equator as the sun travels its path in the sky the area under the curve is evaluated from 0 to Π/2 and then from Π/2 to Π. The integral of sin(x) from 0 to Π/2 is cos(Π/2) – cos(0) or (0 – 1) = -1. The integral from Π/2 to Π is cos(Π/2) – cos(Π) = 0 – (-1) = 1. Forget the minus signs, the total area is 2, 1 from 0 to Π/2 and 1 from Π/2 to Π. To get the average daily insolation at that point on the equator you divide 2 by the total distance of Π radians. 2/Π = .636.
Now, that average value gets modulated by the cos(φ) as you travel from the equator to the pole. And φ goes from 0 to Π/2. So the integral is sin(φ) evaluated from 0 to Π/2 which is (0 – 1). Again, forget the sign, the area under the curve is 1. Again to get the average you divide by the distance traveled which is Π/2. So, once again you wind up with the average value being .636.
So the average insolation received daily by the quadrant in which the point being evaluated lies is (.636 * max) * .636 = .4 * max.
The issue here is that it is a 3D problem and not a 2D problem. You can’t just say the surface area of a quadrant of a sphere is Πr^2, i.e. 4Πr^2/4. The energy absorbed at any point is based on the normal (i.e. 90deg) component of the radiation received. That varies by the sine relative to longitude (Θ) and by the cos relative to the latitude (φ).
You can’t just integrate sin(Θ)cos(Θ) because 1. the evaluation terms are different (0 to Π) and (0 to Π/2) and, 2. because the value based on latitude is not Θ but is φ when using polar coordinates.
I would note that the integral of sin(x)cos(x) is sin(2x)/2. Evaluated from 0 to Π/2 this would give sin(2 * (Π/2))/2 – sin(2 * 0)/2 = sin(Π)/2 which equals 0 (zero) and not 0.5. For sin(2x)/2 to equal 0.5 “2x” would have to be Π/2 meaning “x” would have to be Π//4 which is an angle of 45deg. That doesn’t make any sense at all for the path the sun travels at any point on the globe.
Admittedly this analysis makes two unstated assumptions, 1) only that portion of the radiation that is normal to the surface is available for absorption and, 2. all normal radiation is absorbed. The physics are more complicated than this but it’s a proper starting point.
These assumptions make far more sense that assuming that the surface area of a quadrant of a sphere can be assumed to be a flat plane and that all radiation is thus normal to that flat plane.
Your bad math is telling you that the average TOA flux of the Earth is 544 W.m-2.
Given Earth’s albedo is about 0.3 that is 544 W.m-2 * (1-0.3) = 380 W.m-2 absorbed.
Are you sure you don’t want to rethink this?
When you average it over a sphere that is correct. To be sure, that doesn’t include scattering, albedo, etc. It also assumes the sun follows a path directly over the equator
That is why the +33 deg from GHG is bogus.
The average available energy for ABSORPTION from radiation is .4 * 1360. Do you have the faintest idea if what albedo is meant to convey? Albedo is related to angle of incidence. For instance, more sunlight is reflected from snow when the angle of incidence approaches horizontal. That’s why snow melts faster at noon than at sunrise or sunset. The EM wave from the sun has the same total energy all the time but the angle of incidence changes and therefore the albedo. Less energy is available in the normal component of the wavefront as it impacts the surface. Thus the use of the cos(φ) to evaluate how much energy is available to become absorbed.
How much of the available energy actually gets absorbed is based on many things. For example, black, smooth surfaces absorb light energy better than shiny, silver ones do. But that doesn’t mean that a black, smooth surface will absorb all impinging radiation regardless of the angle of incidence.
Clyde S has tried to explain this to you as wall but you seem to be unwilling to learn basic 3D geometry and basic physics. This isn’t just a “numbers is numbers” game like you and climate science seme to always employ.
Wow. So both of you think this? Really?
I just sent you a Chatgpt explanation for how to calculate energy absorption when the angle of incidence is not 90 degrees. You can confirm this in any thermodynamics textbook.
this is not esoteric stuff. Any college sophomore in engineering can confirm ut fir you.
That is correct.
No one has used the term absorbed, only you.
Don’t try to create a strawman on a different subject.
This calculation is the AVERAGE insolation on the earth. It should not be used to calculate a temperature distribution because temperature uses a T⁴ term. It does not mean EACH POINT receives that value.
It is an average value over the entire 1/2 sphere represented by a given point on the earth from sunrise to sunset.
I guess no one is going to accuse you of flip-flopping regarding the matter then.
Your brother used that term. He even made it all caps.
“What I said was the ABSORBED energy by the earth is .4 * 1360 because only the normal component can be absorbed.”
Your reading comprehension skills are as bad as bellman’s.
I have said multiple times that the value is the MAXIMUM AVAILABLE for absorption, not that it is the amount actually absorbed.
I think we’re all good with the genesis of the ‘/4’ factor. The question re. ‘flat Earth’ is whether or not one obtains the same average surface temperature (T_s) on a rotating sphere if it a) is heated internally to emit 340 W/m^2 from its surface, or b) intersects a flux of 1360 W/m^2.
As for the ‘different forest’, it’s been somewhat suppressed, but I think we’ll see more discussion of it going forward.
Andy May and Jim Gorman clearly aren’t good with it.
That is good question indeed. But again the myth that climate scientists just blindly use averages in flat geometry is simply wrong. For example, [Trenberth et al. 2009] does the full spatial and temporal sampling over the real spherical Earth with the real elliptical orbit to create the diagram of values for Earth’s energy budget. In fact, it is because climate scientists like Trenberth et al. have done the hard work that we know that Earth’s surface rectification effect is about 6 W.m-2 or 1 K. There is even a special section dedicated to discussing this.
BTW…Earth’s surface emittance is closer to 400 W.m-2.
‘BTW…Earth’s surface emittance is closer to 400 W.m-2.’
For simplicity, I assumed that the sphere has a cloud-free, optically transparent atmosphere.
Too bad you can’t show how S/4 is arrived at using 3D geometry.
Do you still confirm that the average of a sine is 0.5?*Peak Value?
I already showed it. Everybody including you can see my derivation. What none of us can see is your derivation. So lay it out for us. You either know how to do it or don’t. If you can’t come up with the derivation in your next post then I don’t really have any other choice but to accept that you don’t know how to do it.
Nobody said anything about the average of sine being 0.5 except you. What I said is that ∫sin(θ)cos(θ)dθ over {θ: 0 to π/2} = 0.5. And I stand by that whether you understand how the math works or not.
“I already showed it. Everybody including you can see my derivation.”
And your derivation is wrong. sin(θ)cos(θ) is an incorrect integral.
And the integral of sin(x)cos(x) is sin(2x)/2, if you need it I can show you the derivation, it involves letting u = sin(x) so that du = cos(x)dx. . From 0 to π/2 sin(2x)/2 does *not* evaluate to 0.5. It evaluates to sin(π)/2 = 0.
The correct integral in 3D space is sin(θ)cos(φ).
Analyzing what happens with a curved surface in space lies in the scope of vector calculus and not in 2D plane geometry.
Apparently Trenberth el al, don’t do 3D geometry very well.
Every thermodynamics textbook I have shows the use of the cosine to modulate how much radiation gets absorbed from radiation that is not normal to the surface. There is no difference between 1. assuming the angle of radiation is not normal everywhere to a flat plane and 2. assuming a parallel wave front for the radiation and a curved surface.
You *still* have to use polar coordinates r, Θ, φ in the analysis and not just r and Θ.
That is patently false. You can use either one and you’ll get the same answer either way. The answer is 5.5e24 joules delivered at TOA of Earth over a 1 year period.
Show us another coordinate system for defining a sphere and how the surface changes. Don’t just pick a point and say x,y,z.
“That is patently false”
You couldn’t be more full of *it*!
Locating a point in 3D space requires the specification of three different coordinates. It doesn’t matter if you are using cartesian, cylindrical, or polar coordinates.
And the issue is *NOT* what is delivered by a parallel wave front at the TOA, it is what is absorbed at the receiving boundary. What is received is a WAVEFRONT. What is absorbed is the normal component of that wavefront. The radiation from the sun is an EM *wave*, not a group of bullets described as *photons*. The energy absorbed doesn’t become a *photon* until it is absorbed. The amount of energy available to become a *photon* is determined by the normal component of the wavefront.
It’s why a radio EM wave can continue to propagate after the energy in the wavefront is too small to provide a *photon* to a detector. It’s why the search is always on for more sensitive detectors, ie ones that require less energy to trigger.
PS. You haven’t answered what the integral of sin(x)cos(x) actually evaluates to from zero to pi/2. That should be your first clue that something is wrong with your derivation.
Let me get this straight…because I accept that the surface area of a sphere is 4x the surface area of its largest cross section and that 340 W.m-2 is the average flux delivered by the Sun at the TOA of Earth that means I’m “full of *it*!”?
Have you considered, even for a brief moment, the rest of world is correct on these matters and that it is only you that is wrong?
I did so here, here, here, and here. And I’ll say again. The integral of sin(x)cos(x) from zero to π/2 is 0.5. No matter how many times you ask the answer will always be 0.5. It will still be 0.5 tomorrow when you ask again. It will still be 0.5 even though there are other things that also equal 0.5. 0.5 is the answer.
You are full of it because you refuse to learn basic physics.
It simply doesn’t matter what the integral if sin(x)cos(x) evaluates to. The formulation is wrong.
It should be sin(theta)cos(phi) with theta from 0 to pi and phi from 0 to pi/2. That’s because you are dealing with an EM wavefront.
Any electrical engineer or physicist can explain this to you.
from chatgpt: at an angle \theta (measured from the surface normal), the effective absorbed heat must consider the reduced projected area. The heat absorbed is given by:
Q = \alpha \cdot G \cdot A \cdot \cos(\theta) \cdot t
where:
• Q = heat absorbed (J)
• \alpha = absorptivity of the surface (dimensionless, between 0 and 1)
• G = incident radiation intensity (W/m²)
• A = total surface area exposed (m²)
• \theta = angle between the radiation direction and the surface normal (degrees or radians)
• t = time duration of exposure (s)
• \cos(\theta) accounts for the reduced effective area receiving radiation
Key Considerations
1. When \theta = 0^\circ (radiation is normal to the surface), \cos(0) = 1 and the full intensity is absorbed.
2. When \theta = 90^\circ (radiation is parallel to the surface), \cos(90) = 0, meaning no radiation is absorbed.
3. For diffuse surfaces, the effective absorptivity might need to be integrated over all incident angles if radiation comes from multiple directions.
You simply can’t assume a flat plane with all radiation perpendicular to every point on the plane. That ignores theta.
The rest of the world isn’t wrong. Only you and I’m guessing climate science.
We are not all good with ‘/4’ factor. You can’t just quarter an orange, peel a quarter, lay the peel out flat, and then say that a light shined on it has all radiation impacting the surface at 90deg. Doing so is an illegal conversion from 3D space to 2D space as far as physics is concerned.
The coordinates involved in 3D space are r, Θ, φ and not just 2D coordinates r, Θ.
Think about it. The surface area of a sphere that has been sliced horizontally at the equator is actually 4Πr^2/2 PLUS the area of the flat plane at point of slicing which would be Πr^2. So you get 4Πr^2/2 + Πr^2 as the total surface area of that half-sphere.
The assumption that the surface area of 1/2 of the curved half-sphere is the same as the area of the flat plane at the equator (the point of slicing) , i.e. Πr^2, doesn’t make logical sense.
Patently False.
The surface area of the largest cross section of a sphere is πr^2.
The surface area of the sphere itself is 4πr^2.
This is a whole new level of absurdity.
Judas Priest. Cut an orange in half. Take a spray can of paint and paint every single surface of that orange. Not just the peel but EVERY surface. How much paint did you use?
What is the surface area exposed when you slice that orange at the equator?
I have thought about it. The Sun is 93 million miles away, so we can treat it as a parallel light source. The Earth, which is a sphere, intercepts a circular profile, i.e., a disk, from that light source. On an instantaneous basis, the energy impinging on that disk is all the energy the Earth is going to receive from the Sun. As a thought experiment, the same amount of energy could also be received by the Earth by instantaneously ‘illuminating’ its entire surface with a uniform source having 1/4 the intensity of the Sun. QED.
The problem is not with the ‘1/4’ factor, it’s with invoking ‘flat Earth’ assumptions, e.g., ignoring a diurnal cycle and the very uneven heating (a crummy paint job to invoke one metaphor) of the rotating Earth’s surface, to determine its ‘average’ surface temperature in the absence of GHGs.
The problem with your thought experiment is that the radiation is intercepted by every square meter on the entire surface not just the area of a flat disk.
If any m² at TOA intercepts 1360 w/m² ray from a plane wave, then that ray travels on to intercept the Earth’s surface. The absorption of the ray is based on the angle that the interception occurs at on the surface.
Here is a graphic showing the problem.
If you placed a flat disk whose radius is similar to the earth, between the earth and the sun, then nothing really changes. All the radiation passing through the disk intercepts the earth.
The only way to change the insolation reaching the earth is to change the amount of the sun’s energy flowing through that flat disk. Far be it from me to calculate a different value than 1360 W/m². But, if that is the correct amount flowing through every m² of that flat disk, then I stand behind the calculation of the distribution of insolation at the earth.
‘If you placed a flat disk whose radius is similar to the earth, between the earth and the sun, then nothing really changes. All the radiation passing through the disk intercepts the earth.’
Exactly! In your conceptualized diagram, the parallel rays that would impinge on the disk are the same rays that would impinge on the sphere. The difference is uniform intensity on the disk vs. non-uniform intensity on the sphere, plus the added fun and games arising from the sphere’s rotation, tilt, elliptical orbit, etc.
The problem occurs when 1360 is divided by a factor of 4, basically creating a flat surface. Doing so, then using that value to compute an average temperature for the globe is less than scientific.
Imo it is the other way around. The amount of radiation Earth receives is 1360 W/m^2 * the ara of a disk with the same radius as Earth.
Since a sphere has4 times the area of a disk, dividing by 4 spreads the radiation around a full sphere. Doing so ignores the fact that only half the Earth receives solar at any given time.
So as a first approximation dividing by 2 (spreading over half the sphere)
isn’t too bad.
For the moon this works out quite good actually.
If you do the actual 3D geometry and integrate from 0 to pi and 0 to pi/2 you get .4 for the factor. A little less than 1/2 as the average insolation for each quadrant over the period of a day.
No, the area is curved. Look at this. Does it look like the area is flat?
The curved surface results in a varying angle of incidence which affects the amount absorbed.
Nope. The insolation isn’t “spread”. It is a continuous function. In this example, when the sun is directly above you (90°) you receive the full amount of 1360. Move to the West one mile and the sun lags you a little, but when zenith occurs that point will also receive 1360.
When you “spread” you are in essence adding in zero insolation that dilutes what is actually received when the sun is shining.
Here is a graph from:
https://www.globe.gov/explore-science/scientists-blog/archived-posts/sciblog/2008/02/27/how-the-temperature-varies-during-the-day-and-night/comment-page-1/index.html
This graph of surface temperature slides around the earth with a period of 24 hours. Every point on the same latitude will see this same graph of absorbed insolation as the sun passes over. No average.
I do want to point out this is a very basic ideal look. Heck, the earth isn’t even a sphere. The earth is tilted. The actual 3D geometry get very complicated with numerous sin, cos, and even cotangent terms for some of the angles. But the basics must be mastered first.
Let’s forget the disk/sphere discussion for a moment.
We’ll take a cone and point it towards a light source (sun).
Do you think the shadow it casts will change with the height of the cone?
It doesn’t matter whether you point a disk, a sphere or a cone towards the sun, as long as they have the same radius they will cast the same shadow. The shadow is a disk with the same radius as the 3 objects, and stands for the number of square meters of (sun)light that is intercepted.
If the area of the cone is eg 3 times that of the shadow disk. the average radiation is 1/3 of the incoming W/m^2
Thus for a sphere it is 1/2 of the incoming W/m^2.
This discussion is only relevant for blockbody calculations.
On the real Earth we have eg to take the spec. heat capacity of the surface into consideration plus all other factors of which you named just a few above.
There IS NO shadow. The earth sits directly in the radiation field of the sun. Every inch and micrometer is exposed to that field. The amount of absorption at any given point is dependent upon the angle of incidence to they rays.
Look at this diagram closely.
Every one of those brown rays and the unseen millions in between all contain 1360 W/m². The rays intersect with every square inch of the sphere from pole to pole.
It is important to remember that the earth is a pebble sitting a large field of radiation that emenates from the sun.
Draw a horizontal line that cuts the earth in two. That is the equator. The point on the equator that is at 90° to the sun will absorb 1360*cos(0°) = 1360*1 = 1360. Now envision a point near to the Noth Pole at the same longitude as the point on the equator. It will absorb 1360*cos(89°) = 1360*0.017 = 23.
At that single point in time, the average insolation from the equator to the pole on that longitude is calculated as:
1360 * [1 / (π/2 – 0)] ∫cos(θ)dθ [0,π/2] = 1360*0.637 = 866
Now let’s examine a point on the equator to the east of the current maximum (90°). Let’s say at π/4 (45°). This point on the equator will be well on the downward side of the sine wave of insolation. What will it be receiving? We calculate that by 1360*sin(π/4) = 1360*0.707 = 962. Now how about that point near the pole on the same longitude? It will be receiving 962*cos(89°) = 16. What is the avg at this longitude?
962 * [1 / (π/2 – 0)] ∫cos(θ)dθ [0,π/2] = 962*0.637 = 552
I hope you can see the average of the sine at the equator for sunrise to sunset is 1360*0.637. This makes the average from the equator to the pole is then 1360*0.637*0.637 from sunrise to sunset.
Looking from the light-source towards an object there will always be a shadow BEHIND the object (a cylinder of no light/radiation).
Whether the object is a disk, a sphere or a cone, as long as their radius is the same the size of the shadow will be the same.
Difference is the AREA receiving the incoming radiation, but the amount of radiation intercepted will always be the area of the shadow times the W/m^2 of the radiation.
A disk has the same area as its shadow, (half a) sphere 2x the area and a cone totally depending on the height of the cone.
So for the 1360 W/m^2 case AVERAGE radiation on a disk is 1360 W/m^2, on half a sphere 680 W/m^2 and a cone unknown.
It is useless to think about rotation, sunrise/sunset etc for a blackbody, because they don’t store energy. (and don’t exist in the real universe)
“Difference is the AREA receiving the incoming radiation, but the amount of radiation intercepted will always be the area of the shadow times the W/m^2 of the radiation.”
The radiation intercepted will be the same, The amount available for absorption will be different because the angle of incidence is not based on the shadow but on the slope of the curve at the point of incidence. The curve at the point of incidence will be different for a flat object than it will for a spherical object than it will for a cone than it will for a parabolic shaped object.
“So for the 1360 W/m^2 case AVERAGE radiation on a disk is 1360 W/m^2, on half a sphere 680 W/m^2 and a cone unknown.”
The EM wave from the sun is a PLANE WAVE, it is not a point source. It strikes every single point on the sphere with an intensity of 1360. The amount available for absorption, however, is based on the angle of incidence. Every point on the total sphere in daylight gets the same intensity of radiation as every point on the half-sphere. Again, the difference is the angle of incidence of that plane wave at the point on the sphere.
“Looking from the light-source towards an object there will always be a shadow BEHIND the object (a cylinder of no light/radiation).”
I should have pointed this out better. The shadow is a flat plane. The surface of the earth is a curve. Flat-plane surface ≠ curved surface in 3D analysis.
I agree with you on the 1/4 factor. There are other issues with your statement that I want to address though.
First…ignoring the diurnal cycle is not assuming flat geometry. It is assuming no rotation.
Second…no one is ignoring the diurnal cycle and rotation anyhow. For example [Trenberth et al. 2009]’s energy budget is based on ~1520 diurnal cycles spanning ~4 orbits of the Sun.
Third…non one is ignoring the nonhomogeneous nature of the incoming and outgoing energy fluxes. For example [Trenberth et al. 2009] do the full spatial and temporal integration.
Fourth…no one is using these energy budget’s to determine Earth’s actual average temperature. In fact, [Trenberth et al. 2009] even calculate the error that arises when someone unwisely tries to do it.
Thanks. We’ll continue to disagree on points 1 – 3. As for point 4, I wish I had a dollar for every instance where someone plugged 240 W/m^2 into the Stefan-Boltzmann equation and concluded that the Earth’s average surface temperature sans GHGs was -18C.
These aren’t opinions in which disagree can be justified. These are facts.
That’s not its actual average surface temperature. That’s its black-body temperature. See NASA’s Earth Fact Sheet.
Nasa also has a Moon Fact Sheet:
BB temp Earth 254K
BB temp moon 270K
The difference is due to the use of the different albedo of both.
The 254K (255K) is also the Global Mean Effective Temperature according Lacis et al 2010
A (very?) influential paper with Gavin Schmidt as one of the authors.
In the 2013 remake even James Hansen featured.
They describe the GHE, and assign even a number to it: 33K.
Since the moon has no GHE one might assume that the actual temperature is close to the BB temp of 270K.
Measured AST is ~197K.
So whatever the cause, we must have an explanation for the >90K higher AST on Earth vs the moon.
‘That’s its black-body temperature.’
Which absent GHGs (it’s been a few days, but please recall the title of this article), would be its average surface temperature. -18C vs. +15C from NASA’s fact sheet, yields the ‘well-known’ 33C GHE, which I also wish I had a dollar for every time its cited.
It is one more way to use simple algebraic averages for a complicated system. If you want to deal with a T⁴ term (temperature) and trig based variables such as the sun “passing” across the sky, an “average insolation” is worthless unless you may be looking at the earth from a different body in the solar system .
I think it comes down, as you often say, to the applicability of averaging functions, i.e., for many functions we know that E{f(x)} is definitely not equal to f(E{x}). For example, in the early days of interest rate derivatives, it was quite common for Wall Street sharpies to pick-off less sophisticated corporate customers who didn’t recognize the importance of making convexity adjustments in their pricing.
the simple function sin(x)cos(x) evaluates to 0 (zero) from 0 to pi/2. That should have been a clue right from the start that something is wrong with this formula.
“The real issue is whether or not the resulting non-diurnal ‘flat Earth’ that presumes continuous and equal isolation for every point on Earth is an appropriate model for the study of climate.”
I concur that this is the real issue.
The answer is no.
Tom, you wrote “I concur that this is the real issue.”
I think the real issue is that increasing GHGs in the atmosphere reduces maximum surface temperatures.
No GHE. A myth, promoted by dreamers and fantasists who refuse to accept reality.
You do realize your math is circular. The integration does nothing more than provide the average value of the portion of a sine/cosine function underneath the function. In other words, yes, dividing by 4.
Read this.
https://www.eeeguide.com/measuring-the-sine-wave/#google_vignette
The sun travels one half (0 – π) of a sine wave from sunrise to sunset. The average value is 0.637. That occurs only directly under the sun.
The average from 0 to π/2 of a cos is (north to pole) is also 0.637.
However, leave it to a statistician to look for an average. All that does is assign the same insolation to every point.
The issue is that only points directly under the sun see the maximum value and therefore, has the highest temperature based on T⁴. As the sun approaches zenith heating occurs. After zenith, cooling begins.
As Andy has tried to tell you, the surface is a heat sink. If you write a gradient which is the proper way to deal with heat, you will see increasing heat, then see decreasing heat. No average is used. College thermodynamics requires calculus to deal with gradients that change in time. Averages don’t cut it.
An average insolation ignores the T⁴ factor and really misses the mark on what occurs with heating.
My training was as an EE. It would have been nice to just convert AC signals to an average or even RMS and proceed with circuit analysis. It just doesn’t work that way.
Yes, irradiation (or the radiative flux at a control volume) is the sum of absorbtivity, reflectivity, and transmissivity. For most of what we are talking about transmissivity is zero, so we are left with absorbtivity (a) and reflectivity (r). Forgive me for not having the proper greek fonts installed, lol. The big problem with averaging for a rotating body is that both a and r will vary with incident position between the terminators. That variability becomes less or unpredictable when the rotating body has surfaces of different materials (land/water) and becomes variably opaque (clouds) so ‘averaging’ is pretty much useless excerpt for the broadest back of the envelope calculations.
“Climate scientists” are a strange lot.” They probably believe that because 240 VAC averages 0, electrical components only need to withstand 0 volts, and putting 240 VAC across your body can’t possibly hurt you!
They also manage to “average” sea levels to 0.1 mm (less than the thickness of a thick human hair), measured from satellites 300 km away, on a constantly varying rough surface (tides and surface water movements).
The basis of this odd behaviour seems to be belief that adding CO2 to air makes it hotter!
Lunacy writ large.
How much is reflected is also based on the angle of incidence.
There appears to some resistance to my post here. Specifically it is the ∫[sin(θ) * cos(θ)]dθ multiplier that I speak of that some are challenging so I thought I’d draw a diagram so that it is easier to visualize. I apologize in advance for the drawing. My artistic abilities are weak.
Note #1: Yes, I know the drawing is 2D. That doesn’t mean I think spheres are 2D objects. I’m fully aware that a sphere is a 3D object. If it was easy to use a globe and take a video I would have done that, but a flat piece of paper and a scanner is all I had. I’m trying my best to represent a 3D sphere on a 2D piece of paper which forces me into conventions like using solid lines to represent points on the sphere that are in view and dotted lines to represent points the sphere that on the other side of it. I am not going to respond to ridiculous challenges that I didn’t actually consider a sphere because my diagram is drawn on a 2D piece of paper.
Note #2. This is not the only way to skin the cat. There are multiple valid ways of integrating a flux (in W.m-2) down a sphere. This just happens to be my preferred way of doing. If someone wants to post an alternate derivation which also shows the correct result of S/4 feel free to your solution.
Some facts I assumed to be true as part of the derivation include the following. If there is any challenge to these we can discuss them individually.
Circumference = 2πrArc Length = r*θArea of Sphere = 4πr^2
BTW…after posting this I realized how absurd it is that I actually had to type an entire paragraph (Note #1) explaining how a drawing on a 2D canvas does not imply flat geometry. WUWT authors and commenters can be an unreasonably pedantic bunch sometimes.
I’m with you on the /4 debate, but please keep in mind that it’s always possible to represent in 2D a process that can’t exist in 3D:
Yeah, there are some real mind bending illusions out there. The Penrose stairs is a good example indeed.
Your formulation is wrong. It is finding the volume of a slice of cake.
2πr * r * Θ –> 2πr^2 * Θ
2πr^2 is an area of an entire circle. With the addition of Θ, you are finding the area subtended by the angle Θ from the center of the sphere to the surface of the sphere.
When you integrate an area over a portion a sphere you wind up with a VOLUME, not a surface area. In essence, the volume of a slice of a round cake!
The arc length should be used alone. r * Θ integrated over the quarter of a sphere will give you the surface area.
You have also erred in making the assumption that Θ (arc angle) =ⱷ (elevation). You have tried to convert a 3D surface integral into a 2D integral. You should be using a ∯ type integral instead of just a ∫. The integral should be dΘdⱷ.So the base integral would be ∫ r *Θ cos( ⱷ)dΘdⱷ. You can’t just arbitrarily say Θ = ⱷ without risking being confused.
In addition, finding the surface area is *not* sufficient. The angle of incidence has to be considered as well. That would be your S * cos().
I still say you have no understanding of 3D geometry or of vector calculus. We had to do a lot of 3D stuff in the EE curriculum such as finding power impinging on an antenna from a NON-spherical transmitting antenna. So you didn’t have a parallel, equal power across the wavefront, EM wave impinging on the receiving antenna.
I haven’t even done a detailed look at your formulation to see if it is correct dimensionally. I somehow doubt it since you wound up finding volume instead of surface area.
This is *very* easy using polar coordinates, r/Θ/ⱷ. It is nothing more than a surface integral: ∯sin(Θ)cos(ⱷ)dΘdⱷ with the assumption of a unit circle, i.e. r = 1, the assumption that Θ is the subtended angle, and ⱷ is the elevation, i.e. the angle of incidence
Why you are so adamant about not doing it the simple, RIGHT way is beyond me. I suspect you are still trying to come with a way to justify your assumption that the earth represents a flat plane to the incoming radiation from the sun.
I’ve been traveling, so haven’t had time to post on this thread. Now that I’m back, I am dumbfounded by the confusion some posters display.
I have to say that bdgwx is completely correct, both in concept and in detailed mathematics. His critics simply do not know what they are talking about. Let’s look at his mathematical derivation shown above:
He is integrating over a hemisphere to find the total solar flux on that hemisphere, and from that the average flux density over the full spherical surface. He is NOT, as Tim urges him to do, simply integrating to find the surface area of the hemisphere. (And then Tim confusedly says that the angle of incidence must be considered as well, which bdgwx has explicitly done.)
I think part of Tim’s confusion is that bdgwx has already implicitly done the first integration of the required two integrations for a surface integral – because it is a trivial integration. Consider the “ring” he starts with. A differential element of that ring has “width” of r*dPhi. It has a “length” of r*sin(Phi)*dTheta.
Integrating this around the ring – from Theta = 0 to Theta = 2*Pi, he gets an area for the ring of
2 * Pi * r * sin(Phi) * r * dPhi
As I said, not a very interesting integral, so he did it “by inspection”, as my professors used to say.
We seem to be agreed that the flux density impinging on a ring at angle Phi is S*cos(Phi). This is constant around the ring, so it is a simple multiplication, but it is not constant between rings, so we must formally do this integration, as shown. Since his first, implicit, integration is for a full circle of Theta, this second integration only needs to be from Phi = 0 to Pi/2. I have double checked his result and it is correct.
Some things to emphasize:
It IS a proper double integration over a spherical surface, and it does use spherical coordinates. It’s just that the first integral was done by inspection.
It does NOT assume that Theta = Phi. And it does NOT find a volume.
Since the flux density varies with Phi, integrating just to find the full surface area is not helpful or proper.
Finally, it is bizarre that he should have to go to such lengths to demonstrate the result. It should be obvious (from another boring integral done by inspection) that the total solar flux impinging on the earth is
F = S * (Pi * r^2)
Since the earth has a total surface area of 4 * Pi * r^2, the average flux density is
S’ = F / (4 * Pi * r^2) = S * (Pi * r^2) / (4 * Pi * r^2) = S / 4
I would expect any of the technical professionals who work for me, and any of the university students I teach, to be able to see this immediately.
“I have to say that bdgwx is completely correct, both in concept and in detailed mathematics.”
His integral is finding the VOLUME of a half-quadrant of a sphere, i.e. πr^2 rotated through 90deg, times 2 to in order to get a total for the whole quadrant.
The volume of the spherical quadrant has no relationship to the W/m^2 absorbed by a point on the surface.
So how can calculating the VOLUME involved be correct in concept and in detailed mathematics?
The volume of the spherical quadrant has no relationship to the W/m^2 absorbed by a point on the surface.
“He is NOT, as Tim urges him to do, simply integrating to find the surface area of the hemisphere.”
I am NOT finding the surface area. I am integrating to find the average absorbed energy over the surface of a quadrant of a sphere. That requires a surface integral over the SURFACE, not an integral finding the volume of the quadrant.
That surface integral is done over the limits of 0 – π for longitude and 0 – π/2 for latitude. It is modulated by the angle of incidence, i.e. the cosine of the latitude and by the sine of the longitude.
There seems to be some confusion even over what W/m^2 actually is. Watts are joules/sec. It is a flux. So the surface integral is for a single point in time, joules/sec-m^2. To get the *total* amount absorbed it has to be integrated over time as well.
“We seem to be agreed that the flux density impinging on a ring at angle Phi is S*cos(Phi). This is constant around the ring, so it is a simple multiplication, but it is not constant between rings, so we must formally do this integration,”
No, at any point in time the angle of incidence is both a function of the point on the ring as well as what ring is being considered.
When the sun is overhead, a point in time, the angle of incidence at a point π/2 away is *not* the same as at the point where the sun is overhead. That is why the sun’s insolation during the day at a point looks like a sine wave. It starts off at zero at sunrise, reaches maximum at noon, and goes back to zero at sunset.
It’s why the *curve* of the earth matters, assuming it is a flat plane of the same area as the surface of the quadrant is not mathematically correct.
“Integrating this around the ring – from Theta = 0 to Theta = 2*Pi, he gets an area for the ring of
2 * Pi * r * sin(Phi) * r * dPhi”
A “ring” has no area. dΦ has no width, it is a point. A point has no width or length. A derivative is the slope of a curve at a point. It is defined as the difference, h, between to points on the curve as h goes to zero. When h approaches zero what happens to the area of the ring?
Take a close look at the integral:
2 * π * r * r * sin(Φ) dΦ ==> 2 * (πr^2) * sin(Φ) dΦ
Exactly what does πr^2 represent? It is the area of a circle of radius “r”. I.e. it is the area *inside* the ring, from the center of the sphere to the circumference of the ring. It is then modulated by sin(Φ) to get the area inside an arc of the circumference instead of the entire circumference.
When that area is swept from the equator to the pole you get a VOLUME.
Now, tell me how that concept and math can be completely correct? The amount of energy absorbed has nothing to do with the volume of interest, only its surface area.
Yikes! I’m amazed at how you manage to get virtually every single point wrong!
You say: “His integral is finding the VOLUME…”
No, it is not! If it were finding volume, it would have to integrate over dR radius, from 0 to the surface. It most certainly does not do that. He is very clearly doing a SURFACE integral, as anyone competent would recognize. (Are you confused by the radial construction lines he drew for illustration?)
You say: “I am integrating to find the average absorbed energy over the surface of a quadrant of a sphere.”
If you actually understood multi-variable calculus, you would see that is exactly what he is doing (except he is integrating a full hemisphere of the surface). He starts with an infinitesimal “patch” ON THE SURFACE of width:
r * dPhi
and of length:
r * sin(Phi) * dTheta
This is standard analysis for the spherical coordinates that he is using. The solar power flux density on this patch is S * cos(Phi).
The first integration of the two necessary for a surface integral is over Theta from 0 to 2*Pi. So we need to integrate:
S * cos(Phi) * r * dPhi * r * sin(Phi) * dTheta
Since Phi is constant for this integration, the only thing that remains under the integral is dTheta. The integral of dTheta from 0 to 2 * Pi is simply (2*Pi – 0) = 2 * Pi.
So the result is:
S * cos(Phi) * r * dPhi * r * sin(Phi) * 2 * Pi.
This is simply the solar flux density times the (infinitesimal) area of the ring.
You say: “A “ring” has no area. dΦ has no width, it is a point.”
OMG! – as my kids would say! You don’t understand the most basic concept in calculus, that of the infinitesimal! No wonder you get everything wrong! If you integrate an infinitesimal, you get a finite value. If you integrate zero, you get zero. The ring has infinitesimal, not zero, area.
You say: Exactly what does πr^2 represent? It is the area of a circle of radius “r”.
No, in this case, 2 * Pi * r * sin(Phi) is the length of the ring ON THE SURFACE, and r * dPhi is the (infinitesimal) width of the ring ON THE SURFACE.
Then when you integrate this intermediate SURFACE result over the angle Phi from 0 to Pi/2 ON THE SURFACE, you get S/2 average power flux density on the SURFACE of the lit hemisphere. With the other hemisphere getting 0, the average power flux density on the SURFACE of the full sphere is S/2/2 = S/4. QED.
By the way, if the result truly represented a VOLUME, it would have R^3 in it, not R^2. I see you do not know how to apply this simple type of sanity check.
“No, it is not! If it were finding volume, it would have to integrate over dR radius,”
So πr^2 is *NOT* the area of a circle?
πr^2 is DERIVED FROM INTEGRATING BY dR!!!!
Do you truly, honestly believe that πr^2 is *not* the area of a circle?
πr^2 = ∫ 2 π r dr where 2r is the diameter and 2πr is the circumference of a circle. The integral goes from r = 0 to r.
In essence this makes the right use of integration a triple integral which is what is used to find the volume of an object. πr^2 is just a shorthand way of reducing the third integral using r.
An area, πr^2, swept through an angle *is* a volume!
“except he is integrating a full hemisphere of the surface”
More malarky. You can’t integrate a full hemisphere using π/2 as the integration limit. Do you understand what the “2” is in the integral? He is finding half of the hemisphere and then multiplying it by 2.
“an infinitesimal “patch””
An infinitesimal patch has an area of zero.
and his integral has r * φ, not r * dφ. r * φ is the arc length in the piece of longitude being analyzed.
You seem to have missed the fact that he is using the same variable, φ, for both longitude and latitude. It makes it difficult to see what the integral is doing unless you break it apart and re-combine it into a standard form.
2 * π * r * r *is* the area of a circle. The S * cos(?) is the piece encompassing the elevation or latitude. In normal polar coordinates this would be shown as cos(φ) The term sin(φ) should be sin(θ) in normal polar coordinates.
The integral *should* be given as
S ∫_θ ∫_φ sin(θ) cos(φ) dθ dφ
with θ from (0 to π) and φ from (0 to π) but with a plane wave impacting a surface the term “r” can be assumed to be the unit radius, i.e. just 1.
“The solar power flux density on this patch is S * cos(Phi).”
I agree. The normal vector is related to the cosine of the angle of incidence.
“the most basic concept in calculus, that of the infinitesimal! “
I can only quote the definition of a derivative – it is the slope of the tangent to the curve at a POINT. A point has no area. A slope has no area. The derivative of f(x^2) ==> lim (h->0) (2x + h) = 2x. 2x is *not* an area, it is a slope. When you integrate a curve you are adding up the heights derived from f(x). The sum of those heights is the area under the curve.
“No, in this case, 2 * Pi * r * sin(Phi) is the length of the ring ON THE SURFACE, and r * dPhi is the (infinitesimal) width of the ring ON THE SURFACE.”
In other words you can’t reduce the expression 2 * π * r * r to 2πr^2? Have you forgotten your basic algebra rules?
the length of the ring on the surface is actually r * θ and not rsin(θ). His graphic shows this correctly. ( note his statement: arc length = r * θ)
“By the way, if the result truly represented a VOLUME, it would have R^3 in it, not R^2. I see you do not know how to apply this simple type of sanity check. “
Why in Pete’s name would you expect to see an r^3? The derivation of the volume of a sphere is done the same way the area of a circle is done.
V = ∫ πr^2 dz where z goes from -r to +r. In other words it is the sum of the areas of the circles making up the sphere. This is *EXACTLY* what bdgwx has done when he integrates πr^2 from 0 to π/2 . He has added up the areas in the parts of the circles he has defined. You can deny that you can reduce the expression in his integral to this but you are denying that standard algebraic operations can’t be done on his expression.
“Do you truly, honestly believe that πr^2 is *not* the area of a circle?”
Ooh, fun game! Can I play too?
Do you truly, honestly believe that 2πr^2 is *not* the surface area of a hemisphere?
Why did you remove the factor of 2?
And in this case the presence of two factors of r come from the two dimensions of the infinitesimal SURFACE patch that starts the surface integral. You seem not to know your spherical coordinates.
You’ve got to come up with better arguments than that!
“You can’t integrate a full hemisphere using π/2 as the integration limit.”
Yes, you can if your first integral was from 0 to 2π, as his first (implicit) integration was. Look at the drawing. His first integration gave him a full ring going from θ = 0 to 2π. For the second integral to cover the full hemisphere, it only needs to go from φ = 0 to π/2. Very clear if you look at the drawing.
“Why in Pete’s name would you expect to see an r^3 [for a volume]?”
Ummm, because volumes are 3-dimensional? Because the volume of a sphere is 4/3 * πr^3? The units don’t even work out right if it only has r^2.
“V = ∫ πr^2 dz where z goes from -r to +r” is NOT the integral for volume of a sphere. Not even close. You are just showing that multi-variable calculus in spherical coordinates is way beyond your skill set.
“You seem to have missed the fact that he is using the same variable, φ, for both longitude and latitude.”
No, I DID NOT! I have carefully explained, multiple times now, that he first did an integration over θ to obtain the ring. The second integration was over φ, with two trig terms of φ, one for the surface area, and the other for the flux density. It seems this is too difficult for you to grasp!
“Why did you remove the factor of 2?”
The factor of 2 was used by bdgwx because he made the unstated assumption that Θ = ⱷ. Since ⱷ can only go from 0 to π/2 then Θ can only do the same. Yet since a quadrant (sunrise to sunset) goes from 0 to π you have to multiply the result by 2 in order to get the full quadrant totals. This unstated assumption let him use the same symbol for both the horizontal and vertical angle – which is unnecessary. So I didn’t use that assumption. My Θ goes from 0 to π to cover the whole quadrant. And my ⱷ goes from 0 to π/2 to cover the equator to the pole.
“And in this case the presence of two factors of r come from the two dimensions of the infinitesimal SURFACE patch that starts the surface integral.“
It doesn’t matter where it comes from. Not one iota. What matters is the result of the factors in the integration expression. And combining r * r into r^2 is a legitimate algebraic operation. Thus the integration is evaluating πr^2, the area under the subtending arc, swing through 0 to π/2 – i.e. a volume. A volume is the integral of the area and r^2 becomes r^3 when you do the integration.
“Ummm, because volumes are 3-dimensional? Because the volume of a sphere is 4/3 * πr^3? The units don’t even work out right if it only has r^2.”
I didn’t give you the full integral in my previous post. I was just trying to show that the integral of r^2 ==> r^3. It’s a little more complicated because the actual volume is 4πr^3/3. But the r^2 still winds up as r^3 when integrated.
“No, I DID NOT! I have carefully explained, multiple times now, that he first did an integration over θ to obtain the ring.”
I’ll repeat, both integrations were from 0 to π/2 and the use of the same symbol for both. Look at the last expression, sin and cos have the *exact* same symbol for both.
He is finding the volume and not the area. There is simply no other way to interpret πr^2. You can deny that r * r in the expression can’t be reduced to r^2 but that’s *is* denying that you can’t do simple algebraic operations on an expression for some reason.
As I said before it is the wrong formulation of determine the amount available for insolation at each point on the surface.
My first comment is that if the average flux at every point on earth is S / 4, then how does one explain the tropics and the poles being at vastly different temperatures?
Also, how does one obtain an accurate temperature profile of the earth that explains the poles being colder than the tropics when all points receive exactly the same insolation, “S / 4”?
The biggest misunderstanding here, and in climate science, is about the difference between a point source and a plane EM wave. You and bdgwx are describing what occurs with a point source.
With a point source, the illumination from the source illuminates a larger and larger area as you move toward the pole. This reduces the available insolation per unit of area. In essence, an example is that at zenith on the equator you would have 1 W/m² for each m², while at the pole that 1 W/m² would then cover 2 m², making the resultant 0.5 W/m².
At the earth’s distance from the sun, the insolation is no longer a point source, it is a plane EM wave. A plane EM wave is a continuous function with the energy all moving in a constant direction at a constant value. Wikipedia says it this way:
From Wikipedia, the free encyclopedia
(bold by me)
This diagram shows what this means.
Each of the brown rays have the same intensity and since it is a continuous function, every point in between also has the same intensity. That means every point on the earth receives the same insolation. The only difference is that cos(φ), which is the angle of incidence, controls the amount available for absorption.
Planck described it this way:
“varied relatively to the direction of the radiation”, that means as the sphere has a different angle to the direction of the plane EM wave, the radiation available to be absorbed is reduced by cos θ, because more is reflected away.
Let me summarize,
Looking at a point P on the equator as a constant, with the sun going from sunrise to sunset, that point experiences a sine function from 0 to π. The average value of a sine wave from 0 to π is given by:
(1/(b – a)) ∫ sin(θ)dθ [0,π] = 2/π = 0.637
Looking at the same longitude from the equator to the pole, the average from 0 to π/2 will be:
(1/(b – a)) ∫ cos(φ)dφ [0,π/2] = 2/π = 0.637
If you don’t believe these, find the cosine of angles 1, …, 90 and average all 90 entries.
The end result is that the AVERAGE insolation is:
1360 * 0.637 * 0.637 = ~552
This also gives a pretty accurate picture of average temperature.
(552 *0.7/5.67×10⁻⁸)¹/⁴ = 287 -273 = 14.3°C
That’s versus a -17°C with an average of 240.
You ask: “if the average flux at every point on earth is S / 4, then how does one explain the tropics and the poles being at vastly different temperatures?”
You completely misunderstand the analysis. We are looking at the situation at an instant in time, and computing the average OVER THE AREA OF THE EARTH at that instant. Some locations get more than the average at this instant, some less. (Funny thing about averages…)
It is NOT a statement that every point on earth gets the same amount of solar flux averaged over time. Some will get more, some will get less. A point on the equator at an equinox will get the full S power flux density at noon, S/2 at 8am and 4pm, and 0 from 6pm to 6am. A point at 60 degrees latitude at the equinox will get S/2 at noon, S/4 at 8am and 4pm, and 0 from 6pm to 6am.
You say: “The biggest misunderstanding here, and in climate science, is about the difference between a point source and a plane EM wave. You and bdgwx are describing what occurs with a point source.”
Actually, we are assuming purely parallel radiative flux, what you are calling a plane wave. If we wanted to do more precise calculations, we would have to take into account the 1/2-degree spread of solar insolation.
You say: “that means as the sphere has a different angle to the direction of the plane EM wave, the radiation available to be absorbed is reduced by cos θ, because more is reflected away.”
No, the reduction by cos(Theta) has nothing to do with reflection — we are not even talking about reflection here. Try this:
In a dark room, hold a flashlight 1 meter above the ground, pointing straight down. Observe the area illuminated on the floor. Now tilt the flashlight up 60 degrees from straight down, and observe the area now illuminated. It will be twice as big, because it is spread out more, so the flux density will be half as much. This is where the reduction by cos(Theta) comes from.
Your point about the average value of a sinusoidal function over a fourth or half cycle is completely irrelevant to the issue at hand. You have to do the full surface integral as bdgwx has done — and he really did do a surface integral, even if you cannot understand it.
Your description is of a point source. You still haven’t studied a plane wave have you?
Here are two diagrams showing different wave types.
The first diagram shows how a plane wave front has equal intensity throughout the plane. The begining of the arrow illustrates how the plane is much larger than the earth.
The second diagram has a point source on the left and a plane wave on the right. When dealing with making field strength readings of an antenna, one had to make near field readings close to the antenna to assess the pattern from a point source. Several wavelengths away the reading becomes constant as the wave front becomes planar.
You can’t properly analyze the impact of a plane wave front on the earth using a near field analysis.
I am not talking about a point source. I am talking about a highly collimated beam of significant radius from the parabolic reflector of the flashlight. My argument holds.
I have designed electro-optical sensors that are completely dependent on highly collimated light. Getting the optics right was a major challenge.
The collimated beam of definite radius, producing what you call a plane wave, yields the phenomenon I claimed. Once again, this should be a trivial point, but it is beyond you!
You are talking about a point source.
from your reply: “Now tilt the flashlight up 60 degrees from straight down, and observe the area now illuminated. It will be twice as big,”
How do you tilt a plane wave from the sun from the perspective of the earth?
It is considered to be a plane wave because the earth subtends such a small portion of the spherical wavefront from the sun. You can’t tilt the spherical wavefront from the sun! Rotate the sun and it still puts out a spherical wavefront and won’t change the radiation impinging the earth one iota.
The sun is *not* a flashlight with a collimated beam from a parabolic reflector. It is a radiating object putting out a spherical wavefront. I’m not sure how you can “tilt” an object putting out a spherical wave of radiation!
I’ve posted a graphic and text from a 1942 textbook on how to analyze the sun-earth system radiation. Here is a second one from a 1974 heat transfer textbook.
As far as the sun is concerned, Θ1 is 0 (zero) so N1 is a plane wave coming at A2. The r^2 is the inverse square law to establish the intensity of the wave travelling a distance and for our purposes we already know what the intensity is, 1360, so we can assume r=1.
The integral then becomes
I ∫ cos(Θ2) dA2 ==> I ∫_Θ ∫_ⱷ [sin(Θ)/π] [cos(ⱷ)/(π/2)] dΘ dⱷ
with Θ from 0 to π and ⱷ from 0 to π/2
You are talking about a point source. A plane wave is vastly larger than the earth and totally encopsses the earth. Your flaslight lens would need to be large enough to encompass the sphere with a highly collimated beam with parallel rays throughout. You do know that is why even highly focused tight beam flashlights lose intensity at a distance. They don’t have parallel rays. That is the reason lasers can brightly illuminate objects at long distances.
A plane wave has a constant intensity at every point on the plane. Every point on the plane is traveling in the same direction with the same intensity.
The point where the plane first intersects the sphere (0° latitude and 0° angle of incidence and cos 0=1) strikes that point with an intensity of 1360.
Move up 1° on the same longitude and that point is struck with another point on the plane wavefront at an equal intensity of 1360. Only now the angle of incidence is 1°.
And so on until 89° latitude is struck with a point on the plane at an intensity of 1360 and an angle of incident of 89° and cos 89=0.017.
These two illustrations give a good description of the difference.
I don’t have time now to respond to every point, so I want to focus on one argument.
Let’s consider the “bed of nails” analogy you (or your brother?) proposed to Phil. This is the “finite element” version so we don’t argue about infinitesimals.
To put some numbers on it, let’s say there is a grid with 1-mm spacing between nails on perpendicular axes. Each parallel nail represents a certain power flux in a collimated beam (yielding the plane wave you like to talk about). A square centimeter contains 100 of these nails. This is “S”.
Now consider a few cases:
Case 1: You press the bed of nails on a flat surface perpendicular to the nails (parallel to the wave front). Each square centimeter of this surface will contact 100 nails. So far so good.
Case 2: You press the bed of nails on a flat surface tilted 60 degrees from the perpendicular that we had in Case 1. In the direction of the tilt, the nail spacing will be 2 mm between points. In the cross direction, the nail spacing will still be 1 mm. A square centimeter of this surface will contact only 50 nails, yielding half the power flux density of the perpendicular case. (This is my case with the highly collimated flashlight.)
Case 3: You press the bed of nails on a hemispheric surface of radius R. Let’s use R=100mm. The number of nails that will make contact with the hemisphere is Pi * 100 * 100 = 31416 nails. (The bed is much larger than this, to satisfy your plane wave analogy.) At top center of the hemisphere, the nail spacing will still be 1 mm in each direction.
The further away from the top center of the hemisphere, the wider apart the point spacings on the surface are. When the surface is 60 degrees away from top center, the density of points is one half of that at top center (we know this from Case 2). Further down, the density of points is even less.
The end result is that the 31416 nails that come in contact with the hemisphere are spread over an area of 2 * Pi * R * R = 62832 mm^2, for an average density 50 nails per square centimeter of hemispheric surface. Ths is S/2. If the hemisphere is half of a total sphere, the other hemisphere makes contact with zero nails, to the average density over the full sphere is half of this, or S/4.
bdgwx showed this result with proper calculus, first integrating over Theta, then over Phi, to cover the full hemisphere. Despite your objections, he did integrate separately over Theta and Phi, and he did integrate over the entire hemisphere.
But you do need to deal with the infinitesimal since the plane wave is a continuous surface. That is why I said use a nail bed with 1 million nails on a baseball. One billion would be even better to cover the baseball. In the end, you must integrate the plane wave as containing an infinitesimal number of points.
Look at the graphic I showed you where I is shown as a constant. That means any point you choose on the plain wave has a constant value. In the limit, it is a continuous function. Otherwise, you need to come up with a function that describes the plane wave as having discrete space with no intensity values between separate rays. As a continuous function, that function will touch every point on the spherical surface with a constant value of flux. You can’t just ignore that fact.
YOU were the one who introduced the finite-element bed-of-nails analogy. For several days now, I have been defending and explaining bdgwx’s (entirely correct) surface integral.
Of course, the surface integral is the better way, if there is a closed-form solution to the integral – as there is for this (idealized) case. But several people here have not been able to comprehend his solution.
The finite-element analogy you suggested provides a good approximation of the continuous surface integral. And what it does show is that an area of Pi*r^2 of the flat bed of nails, representing the plane wave of solar insolation with flux density S, covers a hemispheric surface area of 2*Pi*R^2, yielding an average flux density on the surface of hemisphere of S/2.
I’ve shown you why bdgwx’s formulation is actually finding the volume. Yet you continue to cling to the claim that r*r is not r^2, that you cannot combine terms in an expression.
2πS ∫ cos(θ) * r * sin(θ) * r dθ *DOES REDUCE TO
2πS ∫ cos(θ) * sin(θ) * r * r dθ ==> 2πS ∫ cos(θ) * sin(θ) * r^2 dθ
WHETHER YOU LIKE IT OR NOT.
So you do wind up with an integral of r^2 which, when integrated gives you an r^3 term. And an r^3 term means you are finding a volume.
I’ve even given you the derivation of the r^3 part of the volume of a sphere and you cling to the claim that r^3 doesn’t appear in the expression knowing full well that the integral of r^2 is r^3 + C.
What the finite-element does *NOT* do is handle the vector calculus. I’ve shown you examples from two textbooks where the transfer of heat is a vector sum using cos(x) to determine the normal part of the vector yet you cling to the belief that a plane wave hitting a curved surface is somehow *not* a vector calculus problem.
The surface area is handled in the vector integration by integrating from 0 to π in the x-y plane and from 0 to π/2 in the z-direction using the cos(φ) term to find the normal to the surface.
This is why climate science is so far off the rails. It’s like no one in or defending climate science has ever taken vector calculus or thermodynamics using vector calculus at the university.
Algebra Mistake #49:
2πS ∫ cos(θ) * sin(θ) * r^2 dθ does not yield an r^3 term when computed.
2πS ∫ cos(θ) * sin(θ) * r^2 dθ does not compute a volume.
The units of this integral is m^2. The units of the constant in front of the integral is W/m^2. The units of the whole expression is W.
The integral you gave is:
∫ [ S * cos(Φ) ] * [ 2πr * sin(Φ)] * [r] dΦ
I have highlighted the two “r” terms in your integral.
With a rearrangement of the terms this becomes:
∫ [ S * cos(Φ) ] * [2]* [πr] * [r] * [sin(Φ)] dΦ
which then reduces to
∫ [ S * cos(Φ) ] * [2] * [πr^2] * [sin(Φ)] dΦ
Simple algebraic operations
See the πr^2 term?
What you have defined is the area of a circle limited by the arclength Φ. Then you have swept that area through 0 to π/2 (from the pole to the equator).
That is a VOLUME.
Are you going to be like Bo and deny that πr^2 is the area of a circle? And that it is actually the area of the cross-section of the sphere?
You defined the arc length as r * Φ but left Φ out of the integration expression. Why you tried to use Φ for both the angle between the x-y plane/z-axis and, in addition, for the angle *in* the x-y plane is beyond me. Polar coordinates usually have the angles as Θ and Φ, take your pick as to which one is in the x-y plane and which one is the angle between the x-y plane and the z-axis.
So, I’ll ask again:
Do you assert that you cannot reduce terms in your integral expression? Do you deny that πr^2 is the area of a circle?
Go here to see how to find the surface area of a sphere.
https://physicsnotebook.com/determine-the-surface-area-of-a-sphere-of-radius-r-by-using-the-spherical-polar-co-ordinate-system/
There is no π in this integration. For the unit sphere, which is all you need to find the average insolation in a quadrant, r = 1 so the r^2 = 1.
And you wind up with the exact integral I suggested.
∫ sin(Θ)dΘ from 0 to π and ∫ cos(Φ) dΦ from 0 to π/2
The cos(Φ) term is used to determine the normal vector as you move around the quadrant.
This isn’t difficult. I’ve had to do it any number of times, many of them having to do with antenna work. But it doesn’t matter, an EM wave is an EM wave no matter what the frequency is. Calculating the capture efficiency of a spherical antenna in a plane wave is no different.
(hint: why did you define the arc length rΘ and then leave out the Θ while adding in the entire circumference 2πr? If you had used one or the other (e.g. 2πr/8 or rΘ) in the expression you wouldn’t be finding a volume.)
This is getting increasingly surreal!
You tell us that we must use the “plane wave” form for insolation, when it is completely obvious from the drawing and the math used that we are doing so.
You tell us that we must use the cosine of the angle of incidence on the surface, when we explicitly have been doing so.
You continue to tell us that we must integrate separately over Theta and Phi, when we have carefully and repeatedly explained that we have done exactly this.
You introduce a finite-element approximation to make your argument, then complain that it is not valid when we use it.
You claim that because you can isolate from a surface integral expression the Pi*r*r factors, this is proof that it is really the area of a circular cross section inside the sphere, when you could just as easily isolate the 2*Pi*r*r factors to show that it is the surface area of a hemisphere.
You claim that an integral with an r^2 term (and not an r^3 term) is a 3-dimensional volume value, and not a 2-dimensional area, because…{I can’t even paraphrase an argument here. The best I can come up with is if you do a subsequent integral over R, you could get a volume. But that is irrelevant!}
YOU are the one who doesn’t under vector calculus, making the most basic of mistakes, repeatedly.
YOU are the one who doesn’t understand thermodynamics, being totally confused as to why the angle of incidence is important.
Do you realize how utterly foolish you are making yourself?
“You tell us that we must use the cosine of the angle of incidence on the surface, when we explicitly have been doing so.”
Really? You have been defending bdgwx’s integration expression. It is wrong. It doesn’t find surface area, it finds volume. So the cos(θ) term is meaningless.
“You continue to tell us that we must integrate separately over Theta and Phi, when we have carefully and repeatedly explained that we have done exactly this.”
Integrating over Theta and Phi is meaningless when the rest of the expression is wrong!
“area of a circular cross section inside the sphere,”
When you sweep an area INSIDE the sphere through space you get a VOLUME, not a surface area.
“You claim that an integral with an r^2 term (and not an r^3 term) is a 3-dimensional volume value”
Have you ever used a kid’s bubble making toy? It’s an area (a circle) containing soap. When you sweep it through space you get a VOLUME – i.e. the bubble! Sweeping an area inside a sphere through space gives you a VOLUME. If it’s just part of the area inside the sphere you will get what looks like a slice of cake from a spherical cake!
“YOU are the one who doesn’t under vector calculus, making the most basic of mistakes, repeatedly.”
You’ve shown that you don’t even know how a bubble making toy works to create a volume by sweeping an area through an arc . And I am the one that doesn’t understand vector calculus?
Seriously? You’re going to stick with that laughably STUPID argument?
You look at an expression that is very obviously valid for the surface area of a ring on a sphere. Anyone with modest competence in multi-variable calculus can realize this by simple inspection.
You notice that it has some terms in common with the expression for the area of a flat circle. Therefore, you conclude that it must actually be an expression for the area of a flat circle, and not of a ring.
Can’t you see how ridiculous that argument is? I guess it’s obvious that you can’t!
And your bubble toy analogy does not work AT ALL! Sweeping a circular surface perpendicularly through space yields a cylindrical volume (R^2 * L), not a spherical volume! You do understand the difference between a cylinder and a sphere, don’t you? (I’m not sure you do…) And your proposed math does not describe at all what happens when the bubble breaks loose from the toy.
Yes, YOU are the one who does not understand vector calculus!
“Seriously? You’re going to stick with that laughably STUPID argument?”
You didn’t even bother to go look at the web site I gave you apparently.
“Sweeping a circular surface perpendicularly through space yields a cylindrical volume (R^2 * L), not a spherical volume! ” (bolding mine, tpg)
I didn’t say the bubble toy would create a sphere. I said it would create a VOLUME, which you apparently agree with. See the operative words in your quote (which I bolded)
“And your proposed math does not describe at all what happens when the bubble breaks loose from the toy.”
It doesn’t change from a volume to an area! Which is what you must believe!
BWAHAHAHAHAHA!
Let’s look at the link that you cited approvingly to bdgwx explaining how to find the surface area of a sphere.
It’s EXACTLY the solution that bdgwx used! Let’s break it down.
First, note that it exchanges the names of Theta and Phi angles from what bdgwx used. Second, it leaves out the cosine angle-of-incidence term for the insolation.
Its first step to integrate dPhi (what bdgwx would call dTheta) from 0 to 2*Pi. This yields a value of (2*Pi – 0) = 2*Pi.
So now, its expression for the surface area A is:
A = 2 * Pi * r^2 * Integral (sin(Theta) dTheta) from [0 to Pi]
In bdgwx’s nomenclature, it’s the Integral (Sin(Phi) dPhi). And since bdgwx is only calculating the surface area of a hemisphere, he only integrates from 0 to Pi/2.
But wait! There’s the Pi * r^2 term!!! By your logic, this can’t be a surface integral! It must be that of a flat interior circle!
Your confusion runs so deep that you can’t even realize that the site YOU linked to supports bdgwx’s analysis, and refutes yours!
“First, note that it exchanges the names of Theta and Phi angles from what bdgwx used.”
So what? It is still a surface area integral.
“Second, it leaves out the cosine angle-of-incidence term for the insolation.”
That’s because it is a surface area integral, not an absorption integral. I told you this in my post!
“nd since bdgwx is only calculating the surface area of a hemisphere, he only integrates from 0 to Pi/2.”
A hemisphere goes from 0 to 2π. Sunrise to sunset is half of this or 0->π.
π/2 only covers the morning or evening. That’s where the factor “2” comes in. You double the absorption in order to get the value for sunrise to sunset.
“Your confusion runs so deep that you can’t even realize that the site YOU linked to supports bdgwx’s analysis, and refutes yours!”
The site I linked to supports *MY* integral, not bdgwx”s.
I integrate sin(θ)dθ from 0-π just like the site does. This covers the average insolation as the sun travels over a point on the equator.
I integrate cos(φ)dφ from 0 to π/2 (we are only finding a quarter of the sphere, 2π/4 and not the entire sphere) just like the site does.
The integral of sin(θ)dθ from 0-π is .67. The integral of cos(φ)dφ is .67.
Exactly what I found. So the absorption is .67 * .67 * I = 0.4
You can argue this till you a blue in the face but you have yet to actually show where my math is wrong.
I *can* show where bdgwx used both 2πr and rθ (but he dropped the θ out of the actual integration formula. Why would you include the 2πr? That’s the entire circumference! Why did he drop θ when rθ gives the arc length of interest? Was it because it would have generated a θ^2 term in the evaluation?
So far all you’ve done is employ the argumentative fallacy of Argument by Dismissal. I have shown you my math and confirmed its applicability with an actual reference. Why don’t you explain each term or deleted term in bdgwx’s formulation and show how my algebraic operation of reducing like terms is somehow not valid? Don’t just say it’s wrong, actually show how its wrong.
Algebra Mistake #50.
integral[sin(θ), dθ, 0, π] does not evaluate to 0.67.
The answer is actually is 2.
Remember, we are looking for an average value.
The average of a sine from 0 to π is
(1/(b-a)) ∫sin(x)dx [0,π], where b = π, a = 0.
(1/(π-0)) ∫sin(x)dx [0,π] = 2/π = 0.637
This is a standard function used in EE all the time.
Here is a reference .
https://www.mathway.com/popular-problems/Calculus/590732
You are correct. But the average (i.e. divide by π) is .67
Oh…gotcha. If it’s just a typo and you meant to write the divide by π and just inadvertently left it out then don’t really care. I’m certainly guilty of numerous typos myself. It’s the bona-fide mistakes that I’m most concerned with.
Wow! The errors just keep coming!
You say: “I integrate sin(θ)dθ from 0-π just like the site does. … I integrate cos(φ)dφ from 0 to π/2 … just like the site does.”
No, it is NOT just like the site does! The site first integrates dφ, not cos(φ)dφ . Big difference!
You say: “There is no π in this [the site’s] integration.”
But there is! The first integral is of dφ from 0 to 2π, which gives a result of 2π – 0 = 2π. This results in the surface area of what bdgwx calls his “ring”.
There is also an r^2 term in this equation, so there is a πr^2 term here. In bdgwx’s analysis, you claim this is PROOF that it is an equation for a circular cross section through the sphere. But here, you accept it as a part of the surface area of the sphere. Why the difference? (Hint: there is none!)
After this first integration, by the site’s analysis, we have the equation:
A = 2πr^2 ∫ sin(Θ)dΘ
The second integration sums the surface areas of all the rings on the hemisphere (in bdgwx’s case) or full sphere (in the site’s case).
Now let’s add the angle of incidence effect on power flux density, applying the cos(Θ) term. We get:
S’ = S * 2πr^2 ∫ cos(Θ) sin(Θ)dΘ
where S’ is the integrated power flux (in watts) impinging on the full hemisphere, and S is the power flux density (~1360 W/m^2).
We must integrate from 0 to π/2 to get the full hemisphere.
Here is where you make your key mistake. The cos and sin terms are in the same integral, but you treat them as two separate integrals. There is only one dΘ, so you must integrate the product of the two terms. In other words:
∫ cos(Θ) sin(Θ)dΘ does NOT equal ∫ cos(Θ)dΘ ∫ sin(Θ)dΘ
Doing the integral properly, we get:
S’ = S * 2πr^2 * [-cos(2Θ) / 4] evaluated at 0 and π/2
S’ = S * 2πr^2 * [(1 + 1) / 4] = S * πr^2
Since the earth’s full surface area is 4πr^2, the average insolation over the full sphere (that is, day and night) is:
Savg = S * πr^2 / 4πr^2 = S/4
On a side note, I think you confuse yourself by trying to track the insolation on a given point on the sphere over time. All this analysis is for an instant in time. But at any instant, these conditions are true for the sphere as a whole.
“The site first integrates dφ, not cos(φ)dφ”
I don’t think you understand what is going on here. The web site is finding the surface area. I am working with the insolation over the surface area.
I am trying to find the average insolation in the quadrant. That requires using an angle of incidence.
“But here, you accept it as a part of the surface area of the sphere. Why the difference? (Hint: there is none!)”
The integral at the web site is
A = r^2 ∫sin(θ)dθ ∫ dφ
There is no 2π in this integral. Do you see one?
The integral of θ becomes 2 and the integral of φ becomes π/2 so you wind up with 4πr^2. You aren’t sweeping an area, 2πr^2, through an angle in this formulation. The 2π comes from the integration, it isn’t given as constant in the integral to begin with.
Show me how calculating the average insolation in the quadrant is somehow wrong.
Step 1:What is the average daily insolation at a point on the equator at any point in time?
The sun travels across the sky from 0 to π and the temperature profile is a sine wave starting at 0 at 0° reaching a maximum at π/2, and then dropping to 0 again at π. The integral of that sine wave divided by the base of π is 2/π = .67.
So now we know what the average daily insolation is at a point on the equator with the sun travelling along the equator. .67I
Step 2: Now, as we move north the insolation available for absorption is based on the normal vector as determined by cos(φ), where φ is the angle of incidence.
So we integrate the average insolation at the equator from the equator to the pole. We need a function that is 1 at 0° (the equator) and 0 at the π/2 (the pole). The cos(φ) function fits the bill perfectly and matches the need for an angle of incidence function as well.
So we wind up an integral of (1/π) ∫ sin(θ) dθ = 2/π = .67 as our base multiplied by I = 1360, i.e. .67I
and the integral (.67I) (2/π) ∫ cos(φ) dφ = (.67I)2/π = .67 * .67 * I = .4I
This is how you integrate the average insolation over the quadrant. It just so happens to be similar to how you find the surface area as the web site I linked to shows.
This is getting more and more bizarre!
You say: “I don’t think you understand what is going on here. The web site is finding the surface area. I am working with the insolation over the surface area.”
Doesn’t matter! The insolation is a function of θ [cosθ], not of φ. So it is constant for this first integration, which means it is pulled out from inside the integral. So it is just ∫ dφ, not ∫cos(φ) dφ. This is Calculus 101 stuff, but you don’t get it!
Then you say: “The integral at the web site is
A = r^2 ∫sin(θ)dθ ∫ dφ
There is no 2π in this integral. Do you see one?”
Yes, I do! Look at the next line, where they evaluate the definite integral. dφ integrates into φ, and it is evaluated at 2π and 0. They just did not put the limits of the definite integral on the integration symbol itself.
Next you say: “Show me how calculating the average insolation in the quadrant is somehow wrong.”
I already did! In your Step 2, as you are integrating from the equator to the pole, you make a crucial error that I explained to you last time. (Side note: Despite my warnings, you are not consistent in your use of θ and φ – I will use θ for this second integral, as the site you recommended does.)
You have two functions of θ in this integral: sin(θ) for the reduced size of the elements as θ approaches 0, and cos(θ) for the increased insolation. You MUST integrate the PRODUCT of these two terms in a single integral. You do not!
Then you shout: “SHOW ME WHERE THE MATH IS WRONG.”
So I shout back: “I ALREADY DID!”
In my last post, I explained very carefully how the integral of the product (correct) is NOT the same as the product of the integrals (what you mistakenly did). I stated it very clearly with:
∫ cos(Θ) sin(Θ)dΘ does NOT equal ∫ cos(Θ)dΘ ∫ sin(Θ)dΘ
Then I explained how to do the integral correctly:
S’ = S * 2πr^2 * [-cos(2Θ) / 4] evaluated at 0 and π/2
S’ = S * 2πr^2 * [(1 + 1) / 4] = S * πr^2
Since the earth’s full surface area is 4πr^2, the average insolation over the full sphere (that is, day and night) is:
Savg = S * πr^2 / 4πr^2 = S/4
So I showed you very explicitly WHERE THE MATH IS WRONG!
Finally you say: “you have yet to explain when bdgwx has included both (2πr), the diameter around the equator, and “r” without the θ which would be the arc length for the area in question, not 2πr which would be the entire circumference.”
Let’s go through it step by step. I will use the nomenclature of the site you use. Step 1 is to integrate the SURFACE infinitesimal element of r*sinθ*dφ (PS in the picture on the site) from φ = 0 to 2π to get a full ring around the sphere. Of course, r and sinθ (and cosθ if you are including insolation) are constant here, so go outside the integral.
The result is a ring with a finite LENGTH of 2πr*sinθ and an infinitesimal WIDTH of r*dθ (PQ in the picture), yielding an (infinitesimal) SURFACE AREA of 2πr^2*sinθ*dθ. Note that despite the presence of πr^2 in this expression, it is NOT an expression of a flat slice through the sphere.
If you are including insolation, the result is: 2πr^2*sinθ*cosθ*dθ.
Step 2 is to integrate OVER THE SURFACE along the cross-angle θ from 0 to 2π to sum up the SURFACE AREA (or insolation on the surface area) of all of the rings ON THE SURFACE of the hemisphere facing the sun.
Note carefully that the location where θ=0 in the site’s nomenclature (φ=0 in bdgwx’s analysis) is not the earth’s pole. It is the location where the sun is directly overhead. At the equinox, this would be the point on the equator at local noon time. Points where θ=2π are where the sun is rising or setting at the horizon.
And I emphasize once again that all of these analyses of insolation on the sphere are looking at an instant of time. There is no “dt” in these to integrate over time. That is a separate step and separate issue.
“This is Calculus 101 stuff, but you don’t get it!”
A trivial, Cliffie Clavin, maybe correction. 45-50 years ago, Methods of Integration was our 2nd Calc course. Your point, that Kansas must teach alt.Calc, still stands…
Fair point! Maybe “Calculus 102 stuff” would have been more accurate.
My multi-variable calculus class at MIT 50 years ago was labeled as 18.02, even though I took it first semester of freshman year. It was a great class, taught by George Thomas, author of one the most widely used calculus textbooks ever.
I still have my autographed copy, which I used to double-check some of my arguments here. When my daughter took multi-variable calculus in college recently, they were still using Thomas’ textbook.
Knowing there is more than one way I asked my son yesterday how he would solve this problem. After giving it a couple minutes of thought he proceeded with sin(θ)*cos(θ) route. What caught me off guard was the reason though. I mean it is an intuitive approach, but his reasoning was purely technical and driven by a no CAS mindset in which he often has to do everything by hand. He was already looking ahead at how the integrals were going to play out and recognized immediately that it is a special derivative/antiderivative pair. As he proceeds to explain this means integration by parts collapses into 1/2 of the square of the u term…1/2*sin(θ)^2. And because it is well known that sin(0) = 0 and sin(pi/2) = 1 he already knew it was going to evaluate to 0.5. I was like damn I’ll just plug whatever comes up into a CAS. I did not pick up on that line of reasoning.
Prodigious for a 2nd grader. All I wanted to do was draw war pictures…
Lol. Just to be clear he is a senior in HS taking calculus III. To my knowledge he is the only senior in the district at this level. And the funny thing is…he hates math. Anyway more relevant is that this is probably at least a calculus II problem because integration by parts is required. Speaking of that I agree with Ed that this may be a crucial mistake the Gorman’s are making as it seems they think you can integrate sin and cos separately and then combine the results which is obviously false
“Speaking of that I agree with Ed that this may be a crucial mistake the Gorman’s are making as it seems they think you can integrate sin and cos separately and then combine the results which is obviously false”
I noticed that several days ago, but didn’t have the courage of my convictions to press it.
At any point on the equator what is the average daily absorbed insolation?
I see you have no answer. Not surprising. It would show that my integration is correct.
You still haven’t shown that the average insolation at a point on the equator is *not* the integral of sin(theta) from 0 to pi.
You still haven’t shown that the average insolation at any point above the equator is cos(phi) * average insolation at the equator.
Can you show that the average insolation at a point on the equator is *NOT*
I ∫ sin(θ dθ = 1360 * .67?
“You still haven’t shown that the average insolation at a point on the equator is *not* the integral of sin(theta) from 0 to pi.
You still haven’t shown that the average insolation at any point above the equator is cos(phi) * average insolation at the equator.”
Irrelevant! You are incorrectly integrating along two lines when you should be integrating over a surface. That is your fundamental error here, as I have explained multiple times with the explicit mathematics.
I’ll ask you.
What *is* the average daily insolation absorbed at a point on the equator?
Are you honest enough to answer?
“Irrelevant! You are incorrectly integrating along two lines when you should be integrating over a surface.”
You *still* don’t understand. The point is to find the total insolation absorbed at any point on the surface. The temperature at any point is *NOT* the total insolation absorbed in the quadrant. the temperature at any point is the total insolation absorbed at that point!
I’ll ask again: What *is* the average daily insolation absorbed at a point on the equator?
Then comes this question: What *is* the average daily insolation absorbed at a point with a latitude of 45° and the same longitude as the previous point on the equator?
Until you can answer those two questions you don’t have a hope of calculating the average temperature of the quadrant.
Ahh, the misconceptions just continue to come flooding in!
You say: “The point is to find the total insolation absorbed at any point on the surface.”
That’s funny, the rest of us have been doing the math for the rate of insolation at any instant. What thread have you been reading?
But your point has some interesting issues, too, so I’ll look at it.
You ask: “What is the average daily insolation absorbed a a point on the equator?”
For the case of a day at the equinox, the average daily insolation impinging on a point on the equator (how much is actually absorbed is a separate question, dependent on material properties) is basically what you calculate. (I have never disagreed, just said that it is fundamentally irrelevant to the surface integral solution.)
During the 12 hours of daylight, we have:
Savg = S * ∫ cos(θ) dθ / π , integrating from -π/2 to +π/2
Savg = S * [-sin(θ)] / π evaluated at π/2 and -π/2
Savg = S * [1 – -1] / π = 2 / π = S * 0.637
as you have calculated. But of course, for the 12 hours of night, the insolation is 0. So overall, the daily average insolation is
Savg = (1/2) * (S*0.637 + 0) = S * 0.318 = 1360 * 0.318 = 433 W/m^2
I also agree that at 45 degree latitude, the values are 0.707 times this.
But then you go completely off the rails! You say:
“The temperature at any point is the total insolation absorbed at that point!” Even if, by “is” you mean “is determined by”, this completely wrong. It is determined by a whole host of factors.
The (usually unstated) assumption behind the “33C greenhouse effect” is that the horizontal heat transfer mechanisms are so powerful that any temperature differences between day and night, or between equator and pole, are erased by this horizontal heat transfer.
At the other extreme is the case where there is effectively no horizontal heat transfer, and the thermal capacitance of the surface elements that receive insolation is so small that a point on the surface always has a temperature that is very close to that which would radiate away the same power flux as it is receiving in insolation. The moon comes very close to this extreme case.
I think that is what you mean by “The temperature at any point is [determined by] the total [average] insolation absorbed at that point!” But it is not the average insolation that determines the temperature of a point on the moon – it is the instantaneous insolation. Points on the moon go through HUGE temperature swings (>100C) each day/night cycle.
And the earth has significant horizontal heat transfer mechanisms, both in the atmosphere and the ocean. Furthermore, since outgoing radiation is proportional to the 4th order of absolute temperature, an area with the same overall insolation will have a higher average temperature if the temperature range is lower due to higher horizontal heat transfer than it would be with lower horizontal heat transfer.
“That’s funny, the rest of us have been doing the math for the rate of insolation at any instant.”
The rate of insolation is *NOT* the rate of absorption. The rate of absorption is based on the angle of incidence.
“(I have never disagreed, just said that it is fundamentally irrelevant to the surface integral solution.)”
The surface integral has no part in determining the absorbed insolation at any point on the sphere. Every point on the sphere receives 1360 W/m^2, the only difference being the angle of incidence. And the average temperature for a a quadrant is *NOT* the total absorbed insolation for all points on the quadrant and therefore the surace integral has no meaning. The average temperature of the quadrant is based on the average temperature for all points on the quadrant. That is *not* determined by total insolation absorbed in the quadrant.
If the average insolation aborbed per longitude line is .67 * (.67 *1360) then that will determine the average temperature for that longitude line.
That will be the average temperature for *all* longitude lines and, therefore, the average temperature for the entire quadrant.
Integrate the longitude lines from 0 to π and then divide by π to determine the average and you get (1) * (.67 * .67 * 1360).
That gives you an average temperature of about 287K for the quadrant.
“as you have calculated. But of course, for the 12 hours of night, the insolation is 0.”
You don’t *need* to average in the nighttime hours of insolation to determine the average insolation. The temperature profile at night is an exponential decay based on T^x being radiated beginning with the temperature at sunset. What you would need to know is the average nighttime temperature (not the minimum nighttime temperature) based on that exponential decay. What is the average value of an exponential decay curve?
Ans: f_avg = T/(λt) (1 -e^(-λt) )
where T is beginning temperature, λ is the decay rate, and “t” is the time interval
Your job will be to determine what “T” is, the beginning temperature at sunset, and what λ is.
“It is determined by a whole host of factors.”
Don’t move the goalposts. We are working with a simplified model here. The same radiative simple (and incorrect) model used to get 255K for the average temperature and 33K for the GHE contribution. We don’t need to know the other factors for this simple model.
“he (usually unstated) assumption behind the “33C greenhouse effect” is that the horizontal heat transfer mechanisms are so powerful that any temperature differences between day and night, or between equator and pole, are erased by this horizontal heat transfer.”
No, the 33C greenhouse effect is from “back radiation” by GHe’s in the atmosphere according to the simple radiative model.
“But it is not the average insolation that determines the temperature of a point on the moon – it is the instantaneous insolation.”
This isn’t true either. The “rock” of the moon represents a huge heat sink with thermal resistance. If you are going to argue “other factors” then argue them all. The heat sink of the moon means not *all* heat absorbed is immediately emitted.
As I said, stop moving the goalposts. The whole point is that the radiative impact of the plane wave from the sun is not 1360/4.
You ever wonder why no one ever shows any math refuting this.
The 24 hour average of insolation is an almost worthless number. I talked to an astronomer friend about this. He said about the only useful thing you can do is with that number is rank planets from hot to cold but even that can give errors depending on orbit, rotational speed, composition, etc.
Because of the T⁴ exponential a linear average of insolation, especially using 0 for night, is very inaccurate for determining actual physical, may I say real, planetary temperature. Using zero disregards the fact that the “other” side does have a real physical temperature even when it is receiving no Insolation.
In reading through the commentary below it turns out that the Gorman’s are, in fact, challenging some of the assumption I made. One of those assumptions I made was the formula for circumference C = 2πr.
My intention was not to derive the formula for circumference so I left out the ∫rdΦ {Φ:0 to 2π} integral and instead used the well known formula. Note that ∫rdΦ {Φ:0 to 2π} = 2πr. My rationale was that S isn’t being modulated in the Φ dimension so it served more as a distracting trivial integral. So instead of explicitly stating that trivial integral I just jumped straight to 2πr in hopes that it would make the salient points in my derivation easier to follow.
A well deserved shout-out goes to Ed Bo for quickly identifying my intent!
“My rationale was that S isn’t being modulated in the Φ “
But it *is* being modulated as you show with the S * cos(Φ) factor.
The entire circumference, 2πr, is not being illuminated both day and night. So why have the entire circumference involved? You defined the arc length as r Φ. That is the factor that should be used, not 2πr * r.
I only mention the angle θ (theta) in my diagram. The angle Φ (phi) is not mentioned. I admit this may be hard to make out because I use an diagonal line in the symbol. That combined with my poor handwriting could certainly being causing ambiguity that I never intended.
You show the same symbol for all angles. Use whatever symbol you want. The symbol isn’t the issue.
The issue is that you are using both the total circumference as well the arc length.
Right. The solar flux spread is modulated in the θ dimension. The surface area of the ring which the solar flux is being received is also modulated in the θ dimension. This is clearly depicted in my drawing.
But the surface area of a quadrant is NOT determined by the entire circumference of the sphere at the equator. It is determined by the radius multiplied by the subtending angle.As you clearly show with the formula for arc length: r * θ
The real problem is that you are only calculating what the factor is to determine a part of a sphere, in this case, 1/2 of a sphere.
I don’t think any one denies that half of a sphere has a surface area of 2π (based on a radius = 1. Half of a sphere has two quadrants with one facing the sun. That means it’s surface area is π m².
The issue becomes how to apportion the incoming insolation. There two methods.
Flat disk cross-section
Assume the sphere has a radius = 1 m.
The incoming insolation is a plane wave. That means EVERY point on the wave is perpendicular to every point on the flat disk. Consequently, the total power incident over the whole disk is:
1360 W/m² × π(1)² m² = 4263 W
The power is spread over two quadrants so the power per quadrant is:
4263 W ÷ 2 = 2136 W
One of the flat disk’s quadrants is the shadow from the matching quadrant on the spherical side. That means the spherical quadrant also has 2136 W incident upon it.
The surface area of the spherical quadrant is:
4πr² ÷ 4 = πr² = π(1)² = π m²
To find the average insolation one must divide the insolation by the surface area:
2136 W ÷ π m² = 680 W/m²
If one uses an albedo of 0.3, then the average insolation that is absorbed is:
680 W/m² × 0.7 = 476 W/m²
Convert this to temperature:
(476 ÷ (5.67 x 10⁻⁸))¹/⁴ = 303K
The other way is what I have shown.
Use peak Insolation with sine and cosine distribution over a quadrant
Start with the peak Insolation, and find the average of the resulting sine curve.
Then move up a constant longitude to the average point of a cosine wave. This gives an average value of the insolation in the quadrant:
1360 W/m² × 0.637 × 0.637 = 552 W/m²
Using an albedo of 0.3 one obtains:
552 W/m² × 0.7 = 386 W/m²
Convert this to temperature:
(386 ÷ (5.67×10⁻⁸))¹/⁴ = 287K
Funny how close this is to the accepted value of the Earth’s radiating temperature.
What’s left is to see what an oblate spheroid does.
You left out the dark side which receives no insolation.
So averaging your 386 W/m^2 with 0 W/m^2 results in 193 W/m^2.
which results in a SB temperature of 241,5K, which is in the same ballpark as the 255K the GHE uses as the Effective temperature of Earth.
No, I didn’t leave it out because zero insolation doesn’t mean 0 radiation out.
The dark side has a temperature because it does store heat from daytime insolation. So the dark side does radiate heat at a non-zero level. That radiation would need to be included to obtain a 24 hour average temperature.
To calculate nighttime radiation there are several things to know. The stored heat, the beginning temperature, the functional relationship that describes the gradient of heat loss. It is not a sine function. It is some kind of polynomial with many terms. Something like a capacitor discharging.
I have no way to know these items so I only used what I did know. The insolation during the day. I will note that this is a maximum because the earth does remove some radiation by storing heat rather than immediately radiating it.
By converting incoming radiation in a SB temperature you calculated a radiative balance temperature using a theoretical blackbody which does not store energy, so rotation is irrelevant.
In real life on Earth these things do play a role, but then you can’t calculate a temperature with only the amount of radiation as a given.
example: solar delivers max ~30 MJ/m^2 in a 24 hr period, just enough to increase the temperature of 7 m^3 water 1K.
It is totally relevant. The insolation creates a sine curve as it traverses a fixed point on the earth from sunrise to sunset.
Remember, this is an analysis using ideal conditions. It assumes the sun traverses the equator and no tilted earth. It includes an emissivity of 1. I don’t proclaim that it is the be all and end all. It is a start.
However, it does introduce the fact that the insolation is a plane wave. This is an important fact that is never discussed.
It also dismisses the fact that every point on the earth receives and absorbs exactly the same average insolation, i.e., 1360÷4 W/m². This average as currently used ignores the T⁴ factor in determining temperature. It would mean every point on the earth would be at the same temperature. That is not scientific.
If you wish to criticize, address the mathematics by showing your own calculations with the assumptions behind them.
nit pick. Each point on earth may receive the same average insolation but the absorption will be different because of the angle of incidence.
The point is that your comment brings you back to computing what the absorption distribution actually is.
He most definitely is not. Equation 1 divides by 4. The full text is attached.
That’s what makes it a spherical geometry model.
No where in the text do the authors say the Earth is flat or use math that implies it is flat. In fact, the text uses math that is consistent with spherical geometry.
No, it doesn’t make it a spherical geometry model. It makes it an incorrect evaluation.
Spherical geometry uses polar coordinates radius, theta, and phi. Not just radius and theta.
Physics then makes use of the angle of incidence to analyze what happens with radiation hitting the surface of the sphere based on the polar coordinates.
BTW, do you dispute that the ∫sin(x)cos(x)dx –> sin(2x)/2?
What does sin(2x)/2 evaluate to from 0 to π/2?
Uh, here’s my AI answer..
Solution The integral \(\int \sin (x)\cos (x)dx\) does not equal {\sin (2x)}{2}\), but rather -1/4) cos 2x + C,
And here’s the cuemath answer
https://www.cuemath.com/calculus/integration-of-sinx-cosx/
What is Integration of Sin x Cos x?
The integration of sin x cos x gives the area under the curve of the function f(x) = sin x cos x and yields different equivalent answers when evaluated using different methods of integration. The integration of sin x cos x yields (-1/4) cos 2x + C as the integral of sin x cos x using the sin 2x formula of trigonometry. Mathematically, the integral fo sin x cos x is written as ∫sin x cos x dx = (-1/4) cos 2x + C, where C is the constant of integration, ∫ denotes the sign of integration and dx shows that the integration is with respect to x. Let us go through the formulas for the integration of sin x cos x.
Correct me if I’m wrong, but for the area under the curve between zero and pi/2 I get 0.5.
And what integration limits did you use?
You must integrate the sin(x) from 0 to π and cos(x) from 0 to π/2.
The result is not sin*cos. An integral of that function doesn’t give yo what is needed.
From sunrise to sunset you have a sine function with a peak value of 1360. That means at any given point (in this case) when the sun is at zenith, the point will be receiving 1360. Since the sun is at 90° or π/2 radians, sin(π/2) = 1. As anyone with experience with EE can tell you the average of that sine is “peak*0.637.
Then as you go toward a pole, the 1360 is reduced by a cosine function such that at 90° or π/2, the value of cos(π/2) = 0. Again, the average is peak*0.637.
Since we now have averages, we can multiply the peak by 0.637² to get a total average.
1360*0.637*0.637 = ~552
Is this far out of line? Let’s see.
552 * 0.7 = 386 W/m²
T = (386/5.67×10⁻⁸)¹/⁴ = 287K
Funny how that is about what the consensus is for the current GAT of:
287 – 273 = 14°C
Too bad that is without GHG’s.
“And what integration limits did you use?”
“Correct me if I’m wrong, but for the area under the curve between zero and pi/2 I get 0.5.”
“The result is not sin*cos. An integral of that function doesn’t give yo what is needed.”
All I was responding to was your erroneous integration:
“BTW, do you dispute that the ∫sin(x)cos(x)dx –> sin(2x)/2?”
https://wattsupwiththat.com/2025/03/16/the-earth-without-greenhouse-gases/#comment-4051223
Do you still think so?
My integration is not in error.
The average of a sine is:
(1/(π – 0))∫ sin(x)dx [0,π] = 2/π = 0.637
(1/(π/2 – 0))∫ cos(x)dx [0,π/2] = 2/π = 0.637
Did you forget the average of a trig function is:
(1/(b – a))∫sin(x)dx [a,b]
You v world.
Both AI and cuemath schooled you on why you are boned up here. This is not Freddie Fulcrum, “One the one hand. On the other hand.”. It’s you v world.
We’re Gonzo here folks. Back into Dan Kahan System 2, “negative standard deviation” territory…
Folks, this is Dennis Kucinich level lack of self awareness. “When I am President.” “When I am President”. Jon Stewart wanted to grab him by both collars, get in his face, and JUST yell, “DUDE!!!”, at the top of his lungs.
You are very mistaken here. When you input incorrect information to a calculator, guess what you get, garbage.
As I said. sin(x)cos(x)dx is not a valid trig function for determining the average value of the insolation distribution. The very first mistake is to integrate that combined function over the same interval, either [0,ππ)] or [0,ππ/2]. That gives you an answer that is worthless for determining the average angle of incident across the sphere.
Think about this. The equation for a circle is x² + y² = r². How do you integrate that function when rotated 360 degrees around one of the x or y axis, not to get the volume, but to get the surface area?
Figure S4 is attached.
Do you see the divide by 4? That’s the tell that this figure is using spherical geometry.
You can be maddingly dense at times. Look at the attached energy diagram from Trenberth (2009). Notice the input is 341 W/m^2. Where is that from? The solar energy that reaches Earth’s orbit is 1361 W/m^2. Where is the nightside? Where are the tropics and the poles? This Trenberth diagram, just like Benestad’s description, is totally flat-earth and divorced from reality.
Exactly. That is 1/4 the TSI of 1364 W.m-2. This is the tell that the [Trenberth et al. 2009] energy budget is for a spherical geometry.
Like I said in the derivation above it is embodied by one of the ∫[sin(θ) * cos(θ)]dθ = 0.5 multipliers. The day (lit) side is 0.5 * 1360 W.m-2 = 680 W.m-2. The night (unlit) side is 0.5 * 0 W.m-2 = 0 W.m-2. The full pole to pole integration is thus around 340 W.m-2.
BTW…the [Trenberth et al. 2009] diagram actually includes ~1520 day/night cycles in its analysis period. How do you think the authors should have visually depicted that many day/night cycles in one diagram?
That’s a map projection Andy. That’s what people do when they are forced to represent the 3 dimensional spherical Earth onto a 2 dimensional flat canvas. Using the [Trenberth et al. 2009] diagram without consideration of the math to claim that [Trenberth et al. 2009] based their study on a flat Earth model would be no different and as absurd as me using your figure 1 to claim that you also based your article on a flat Earth model. That makes your statement here a case of the pot calling the kettle black and reduces your critique of their work (and others) to how well they draw diagrams instead of how well they perform the math.
Let me get this straight…you don’t understand the math behind the divide by 4 and the purpose of using map projections to represent the Earth on a 2D canvas and I’m the one who is dense? Really?
Nada yet, but he just needs time to man up. Historically, it’s not been easy for him to do so, but I think he’ll come thru. Even Dr. Frank finally dismissed a YUGE mess of diatribe a few months ago with an Emily Litella “Never Mind”, after he realized how badly he boned…
What was that conversation about? Do you have a link for it?
Got me. No, no link, and my hard target search didn’t yield any results. So, I can’t backup my claim that Dr. Frank actually admitted a mistake, once. I recall it being one of those million instead of billion type errors that he stuck with over many exchanges. When even he realized the futility, he literally “Never Mind”ed and ducked out.
FYI, BIL THINKS he’s a guitar player, and we went to open mike at Moonshine Blues Bar 2 Sunday’s ago, to hear him. Thankfully, he was definitely the worst. Aging demo, very good musicians, good food catered from upstairs. Have you ever been?
I was just curious because I’ve never seen Dr. Frank admit a mistake or retract a false claim. In fact, what I’ve observed is that he’ll dig in his heels and make yet more mistakes or false claims in defense of the original mistake or false claim. I’m not saying that is a universal behavioral treat for him. I’m just saying that’s what I observed.
‘In fact, what I’ve observed is that he’ll dig in his heels and make yet more mistakes or false claims in defense of the original mistake or false claim.’
Well, since bob didn’t come through, maybe you can post a few links for us, based on your observations, of course.
https://pubpeer.com/publications/391B1C150212A84C6051D7A2A7F119
And there’s even a name for Dr, Franks defensive style:
https://diagrammonkey.wordpress.com/2019/10/09/hagfishing/
“The problem comes in pointing out exactly what is wrong with the analysis, not because this is a particularly difficult thing to do, but because of Frank’s defence mechanism.
There is a species of fish known as the hagfish. It’s not a nice name really, but I suspect that the fish doesn’t much care about names. Nor are they particularly attractive to humans, though, again, I’m not sure the hagfish could give a hoot. How hagfish feel about each other is a matter of supposition. One must accept the possibility of attraction between them. After all, there are hagfish. However, what really sets hagfish aside from the other horrors of the deep is their defensive reaction. When provoked, the hagfish exudes a substance that rapidly expands generating huge volumes of fibrous mucus. Anything taking an experimental nip finds itself jammed to the gills with a choking gobful of slime. After that, most predators lose interest or pretend they’d mistaken the hagfish for an old friend.”
You and John (Kennedy) should both take John’s excellent advice:
“If you can’t be civil, be gone.”
Not “civil”? Have you spent any time in this fora?
No 47’ian elementary school names. No epiphets. Unlike a half dozen others here, just data and documentation….
And still no Andy May sighting? Maybe we can expect a whole other post, to deflect…
If Andy does actually understand the geometry behind the divide by 4 and why people use 2D map projections then he sure is going to great lengths to convince people otherwise. To what end…I have no idea.
The “criticisms” I read did not really disprove the conclusion. Lots of arguments about an average having a label describing the period of the data.
I must say that I agree with Dr. Frank. The sum of 5 temperatures divided by the value of 5 temperatures taken over a week gives a label of value/period. If you don’t include the period, you have no knowledge of what it is an average of. A week, month, year, etc.
None of the comments I saw had a clue what a measurement uncertainty is. None of them addressed the issue with any mathematical derivations. That by itself makes the “criticism” nothing more than pure speculation.
“The “criticisms” I read did not really disprove the conclusion.”
A non denial denial. Is the “criticism” valid, or no?
IMO, when you bone up this soon, especially with a process that you profess to understand, but that even I can see your problem with, the rest might as well just be fruit of the half informed tree..
Sorry, not seeing how the link you provided has any relevance to your characterization of Dr. Frank’s behavior, above.
Dr. Frank arbitrarily converts 4 W.m-2 to 4 W.m-2.year-1.
Multiple people point out the egregious mistake which is consequential to his conclusion. Dr. Wunsch (who was the reviewer) retracts his previous endorsement.
Dr. Frank then defends this egregious mistake by redefining what it means to average an intensive property and assign it units it never had effectively conflating it with a rate.
Multiple people point out the new absurdity.
Dr. Frank then insinuates the reviewers are ignorant.
AFAIK to this day he still thinks a monthly average temperature has units of C.month-1. For example he would tell you the latest UAH temperature anomaly was 0.5 C.month-1 instead of what it actually is 0.5 C.
Ah, yes, error propagation in GCMs. If you’ll recall, Pat’s point was that regardless of whatever time frame the modelers chose to dribble their presumed CO2 forcings into the GCMs, at every iteration the radiative effect of these forcings was overwhelmed by the uncertainty of the radiative effect of cloud cover due to the GCMs’ demonstrable misspecification of same.
You may not care for Pat’s nomenclature, but I can assure you that his argument carries a lot of weight among many technically trained folks outside of those within the climate community. Who knows, perhaps a cogent explanation of why GCM projections are meaningless will loom large at an upcoming review of the EPA’s Endangerment Finding?
“…but I can assure you that his argument carries a lot of weight among many technically trained folks outside of those within the climate community.”
Name one. Even the peer reviewer he finally found to recommend approval of his 2019 paper, later told us that he did so only in the spirit of spirited exchange. Upon further pondering about it’s obvious boners, he withdrew it.
I have yet to see a single cite that actually used Dr. Frank’s “methods” in their work. Chem buddies and bulking up bio’s. Per Austin Powers “That’s about it”..
“bios” s/b “biblios”. bobob (as pinned by other “civil” commenters) regrets the error…
‘I have yet to see a single cite that actually used Dr. Frank’s “methods” in their work.’
I wouldn’t expect to given that significant government funding for climate alarmism directly supports the political rationale for increased government control over the economy and society. Also works for wars, banking, pandemics, etc.
“the political rationale for increased government control over the economy and society. Also works for wars, banking, pandemics, etc.”
Look beyond the conspiratorial plots you find inside of your eyelids. Yah, an international secret squirrel conspiracy to avoid using the alt. statistical “tools” that no one else has even deigned to apply, and that ALL in superterranea have technically rebutted, from the ground up, with Khan Academy, type evaluation.
I’ll end with another memorable movie quote.
“Pretty weak, Conehead”.
‘
“divorced from reality.”
As are you. The GHE is a number?
You could always try and convince the truly ignorant and gullible that adding GHGs to air makes something hotter, and that we’re all doomed.
Do you think that makes sense?
The value of 341 W/m^2 arises because the solar energy that is incident on the Earth is 1361 W/m^2 times the cross sectional area of the Earth:
1361xπr^2,
average that over the surface area of the Earth (4πr^2)
1361πr^2/4πr^2 = 1361/4 W/m^2
That is fine if your goal just finding a average insolation.
If you want to move on to a temperature calculation, you must recognize:
Write a gradient that includes both a sine variable for insolation and a T⁴ variable for temperature.
My back of the envelope shows an average temp of 320K with an albedo of 0.3 at the equator. This turns into an average of 288K for the globe.
Don’t forget, the “average” of a sine is 0.637, NOT 0.5. Also, the average of a cosine from 0 to π/2 is also 0.637.
every point on the sphere see’s a sin wave function for the insolation from the sun.At zeneth, i.e. at the point the sun is directly over the point, the insolation is 1 at peak. The average value of that sine wave is about .64 times max or (.64)*1361.
That insolation is modulated by the height above the equator. The modulating function must be 1 at the equator and 0 at the poles. That gives a cosine function of cos(latitude), i.e. 1 at the equator (cos(0) and 0 at the pole. The average of a cosine function from 0 to pi/2 is, again, about .64.
So the total insolation received by all points in the quadrant of the sphere would be (.634)(.634)1361 = 557
This means that the insolation received at a location at the latitude 45deg would get (.64)1361 = 871 each day modulated by the cosine of 45deg = .707 or an average of 670. A station at the 60deg latitude would get 871 x .5 = 435.
Now, how do you convert that average daytime insolation per quadrant to the average quadrant temperature? And since climate is *NOT* determined by temperature how do you convert an average climate for the quadrant?
(ps. not sure about the term quadrant, if you split a sphere into two hemispheres and then split the hemispheres in half you get four “quarants”, i.e. a daytime and a nighttime for each hemisphere. seems like quadrants to me. )
But that sine function will vary both by surface material reflectivity (heavy leafing forest, versus open water) most notably as you reach the rotational terminator. Also, the opacity of the atmosphere (clouds, one of willis’ emergent phenoma) will vary. The idea that ‘averaging’ can have meaningful results is almost ridiculous.
Of course that will occur. It is why I used a 70% figure in my post.
The issue is that you can’t just “average” insolation over a flat earth. Insolation is only useful as an input to determine a temperature. As such, any error or uncertainty with insolation results in further error in subsequent calculations.
Additionally temperature is very sensitive to insolation due to the T⁴ factor.
Where bdgwx arrived at an integral result of 0.5 I have no idea. The resulting figure of the integration of both sin and cosine over the correct intervals is 2/π or 0.637. In all my EE education, only 0.637 and 0.707 (RMS for power) were common “constants”.
I see what you are saying. I agree that averaging is not the proper way to deal with sin/cos functions. I was pointing out that insolation is not the only thing modulated by rotation, but the effect of that insolation is also modulated by the rotational position between the dawn/dusk terminator. Surface reflectivity and absorbtivity are functions of angle of incidence as well as surface make up which also have great effects on effects on that T^4 parameter and just trying to average that away is not an accurate representation.
“Where bdgwx arrived at an integral result of 0.5 I have no idea.”
It doesn’t really matter. He’s just trying to avoid admitting that adding CO2 to air doesn’t make it any hotter.
The GHE he can’t or won’t describe doesn’t exist, so he’s trying to show that he is an intelligent chap anyway.
S/4 = 1360 W.m-2 / 4 = 340 W.m-2 is the average flux received at TOA.
340 W.m-2 * 1 year * 510e12 m^2 = 5.5e24 joules.
I used an average to calculate the total energy Earth receives in one year. That is meaningful and a lot simpler than doing the full integration which I show elsewhere in the comments.
Note that reflectivity is irrelevant here because this is the amount of energy received at the TOA shell.
My question for you is…do you truly think 5.5e24 joules is a ridiculous result? If so what value do you get?
The TOA shell is is the problem. S/4 is not the area receiving insolation. That is 2πr² at any point in time. There is a point on the sphere where the sun at zenith. At that point, the insolation is 1360. The earth below that point is where maximum insolation occurs and the temperature is higher than any other point on the shell.
Why else do you think temperatures decrease as you get further from the equator?
No matter how you work it, S/4 becomes πr². That is the area of a flat circular surface. That is 2D geometry. It is the area of a circle formed when you cut a sphere into 2 equal pieces, πr². It has no relationship to the area of a sphere. πr² is not half the area of a sphere.
What is your aversion to using 3D geometry and trig functions to calculate proper insolation over 1/2 of a sphere?
Interesting. I actually did the 3D geometry with trig functions to derive S/4. You, OTOH, were only able to say it is 0.637*0.637*S, which is equivalent to S/2.46, without providing any math backing it up.
If you just need the total energy Earth receives in 1 year at TOA, why not just do
1360 W/m^2 * the area of a disk with the same radius as Earth * the number of seconds in 1 year?
You can do it that way too. The only difference with your approach and my approach is the order of operations. I did (1360 W.m-2 / 4) * 510e12 * 1 year whereas you did 1360 W.m-2 * (510e12 / 4) * 1 year.
the average of 10 and 8 is 9. Congratulations you can punch numbers into a calculator.
The “shell” is only a control volume boundary that energy moves through. it goes in, it comes out. Pretending it is a suface is an assumption for simplification because you can’t model the actual effects of the atmosphere over the planetary surface and how the 1360 solar flux is actually distributed. You just guess.
Why are these averages ridiculous? Because with that method everything that happens inside the control volume is unkown. It’s essentially a black box, but people are using it to try and define process on the planet surface(inside) which is poor physics and it’s ridiculous because it has become apparent over the last 50yrs of ‘projections’ that this line of modelling does not reflect what is happening on the planet yet people still insist on doing it.
Computing an average doesn’t mean the data upon which that average was computed somehow disappears. It’s still there to be analyzed in finer detail for those who want to do that.
You can compute anything you want, but whether your computation means something is a different story. I’m sure you could develop a formula to calculate how many angels could dance on the head of a pin. Portraying a rotating sphere with variable surface characteristics subject to point source incident radiation as a flat disc with uniform flux, in an attempt to understand what is happening on the surface of the sphere, is just poor physics.
I think the total amount of energy Earth receives from the Sun at TOA calculated as (1360 W.m-2 / 4) * 510e12 * 1 year = 5.5e24 joules means something. Do you think it means something?
Who is treating the Earth as a flat disk?
YOUR STATEMENT: “I think the total amount of energy Earth receives from the Sun at TOA calculated as (1360 W.m-2 / 4) * 510e12 * 1 year = 5.5e24 joules means something. Do you think it means something?”
The 1360/4 is a wasted calculation and nothing more than a boundary condition to a bad model. One of the most simplistic things you can calculate, and frankly it’s a scalar value(magnitude) and is unimportant to the point at hand. The issue is how it is distributed across the spherical rotating surface with various properties, and how that surface interacts with it. In other words, whether it’s 1360 or 5 is immaterial.
YOUR STATEMENT: “Who is treating the Earth as a flat disk?”
well, you are….
YOUR STATEMENT: “That’s a map projection Andy. That’s what people do when they are forced to represent the 3 dimensional spherical Earth onto a 2 dimensional flat canvas.”
Yet that is a false argument, no one is forcing anyone,especially you , to represent a , in your words, “3 dimensional spherical onto a 2 dimensional flat canvas” YOUR WORDS . Both gorman and I are pointing out exactly why that is a bad assumption.
You are trying to argue out of both sides of your mouth. First, the trenberth diagram is accurate, then it is a “forced representation”. Which is it?
Presenting that chart as somehow reflective of real physics is …well.. stupid.
Don’t double down on stupid.
5.5e24 joules is the correct answer.
S/4 is a spherical geometry model. The Earth is shaped like a sphere.
I divide S by 4 precisely because the Earth is a sphere.
I also do the full spatial and temporal integration here proving that S/4 is a valid model for the spherically shaped Earth.
To do anything else would be assuming the Earth is not spherically shaped.
If you want to represent the real 3D Earth onto a 2D canvas you are absolutely forced to use some kind of map projection.
It is accurate at least in so far as it is using a spherical model and the figures in the diagram represent the energy budget of the real spherically shaped Earth.
Remember, Andy’s insinuation is that the W.m-2 figures in these energy budget models are not correct because he thinks they are generated using flat geometry (which is absurd). It is Andy’s misunderstanding of the math (particularly the divide by 4) that I’m addressing.
Excellent!
The vector calculus math used for 3 dimensional objects has been around since the last 1800’s. It revolutionized Maxwell’s EM equations and every EE that learns about radiation must deal with it.
If someone wanted to really use a flat disk to simulate the surface area of half a sphere would use a 2πr² flat disk. But, you would still need to deal with a function that evaluates to 0 at the edges. An equal average over the disk assumes a sudden break at the edge which is totally unphysical.
The Earth is losing energy at a rate of 44 TW or so.
Maybe you could calculate the yearly loss in joules?
Or not bother wasting your time – the Earth is a ball of glowing rock, sitting in space a long way from the Sun. It cools, unless you have discovered new laws of physics in the past day or so.
The πr² is not the surface being illuminated. That is what size a circle is if you cut a sphere in half. You get a spherical side of 2πr² and a flat circle of πr². The illuminated side has twice the area of the flat circle.
Assuming no other factors, the point on the spherical side that is at zenith receives the full insolation and other points various amounts of sin and cos combinations. The average of a sine is 0.637 and the average of cosine is the same. Sine would apply to longitude (equator) and cosine to latitude.
This cross sectional viewpoint is a poor way to convert a 3D object to a flat 2D object and is totally unscientific and not representative of reality.
Jim, even worse is that the insolation has to penetrate even greater depths of atmosphere as the poles are approached, so the optical density of the atmosphere increases. Maximal at the terminator where it stops directly falling on the ground anyway!
Add in a few complications like optical effects – Brewster, Fresnel and the rest, air density affecting refractive losses, the height of the atmosphere varying due to the rotation of the earth, and it is no wonder that “climate scientists” take refuge in word-salad jargon and “averages”.
Phil is certainly an enthusiastic GHE believer, but might be characterised as not the brightest bulb in the box.
It’s the cross-sectional area as I said, it defines the total amount of the sun’s energy that reaches the Top of the Atmosphere (TOA). What someone chooses to do with that number in further analysis is up to them. It’s nothing to do with a ‘flat earth’.
Good luck.
Above he just told me the surface area of a hemisphere is 4πr^2/2 + πr^2.
He also says the flux from the Sun received by Earth at TOA is 0.4 * 1360 W.m-2 = 544 W.m-2.
This is what we’re working with right now.
Cut an orange exactly in half. Turn it to look where you sliced. Do you see pulp? What is the surface area of that pulp? Does that surface area not add to the surface area of the 3D object you now hold? Or does that surface not exist in your world? Your grasp of 3D geometry is sadly lacking.
If your are going to quote me then provide the exact quote and not your misunderstanding of what I said.
What I said was the ABSORBED energy by the earth is .4 * 1360 because only the normal component can be absorbed. I’m sure that makes no sense to you because you don’t understand 3D geometry. That’s your problem not mine.
Oh…so the ABSORBED energy by the Earth is 0.4 * 1360 W.m-2 = 544 W.m-2. Got it.
That is *NOT* what I said. the 544 W/m^2 is the average available energy for one quadrant of the hemisphere that *could* be absorbed based on the geometry of the system.
Whether this is absorbed or not depends on a lot of other factors.
I’m not sure you even understand what W/m^2 means.
A watt is a joule/sec. So W/m^2 is actually joule/sec-m^2 or 1 joule/sec for each m^2.
Pick a point on the equator. The average available joule/sec available for each m^2 at that point is .64 * 1360 = 870 as the sun travels from sunrise to sunset. A point at 45deg latitude will have a different average. It’s available maximum will be .7 * 1360 = 961. So the average joule/sec available for each m^2 will only be 961 * .64 = 615.
Integrate this over the whole quadrant and you get a total average available joule/sec for each m^2 of the quadrant as .64 * .64 * 1360 = 557.
Again, this is the MAXIMUM possible based on the 3D geometry of the system. Whether that is reached or not requires applying other modulating factors.
For instance, the ΔT from the number of joules absorbed by the solid earth is based on the formula of ΔT = Joules/mass-specific heat. For a gas the formula is the same but mass becomes the number of moles.
It’s not obvious that the climate models properly handle specific heat. The specific heat of the earth’s surface across an entire quadrant can vary significantly. Same for the atmosphere. For the atmosphere the specific heat varies greatly based on pressure and humidity. Especially for pressure there is a big dependence on elevation. For humidity there is a big dependence on geography and terrain. How does climate science and the climate models handle this?
This is also where climate science goes off the rails. Trying to calculate the temperature rise in the entire quadrant using the average insolation of 557 W/m^2 actually tells you nothing about the climate anywhere in the quadrant. Temperature, even temperature change, tells you *NOTHING* about the climate. A quadrant of the sphere that is earth can have *multiple* different climates within it and the average insolation can tell you nothing about that.
Temperature is an intensive property. That means that an “average” temperature is meaningless because you can’t add temperatures as you can mass. The closest you can get is to ASSUME a temperature gradient between points and pick a midpoint value from the gradient. The problem is that the gradient should be based on ENTHALPY (i.e. heat) of the intervening media and not on Temperature. Temperature is *NOT* a valid proxy for enthalpy, especially for a gas, because of the modulating factors of pressure and humidity. An example would be “what is the temperature gradient between a point on the east slope of a mountain and a station on the west side of the mountain”? If the midpoint location is the top of the mountain, developing a temperature gradient based on the two temperatures of the measuring station is bound to fail in giving a valid result for the midpoint location. Not even using enthalpy would help entirely in this situation but it would be a much better estimate because pressure and humidity would involved.
This is the metric that should be used when discussing Earth’s wheather/climate.
Data is readily available, eg https://www.pveducation.org/pvcdrom/properties-of-sunlight/isoflux-contour-plots
As an example take 20 MJ/m^2. This is the amount of energy the sun delivers to a square meter over a 24hr period.
Just enough to increase the temperature of a 5m water column 1K.
If maximum absorption happens the average number of joules delivered per quadrant of a hemisphere on average would be about 50Mj/m^2. Since maximum absorption probably doesn’t happen anywhere, 20Mj/m^2 over 24 hours sounds about right.
Yep. That’s about what I get with a full spatial and temporal integration.
Is this any help in curing your delusion Mr Gorman?
That’s bizarre !
I know the correct vid was posted, as I saw it here before I left.
I’ll try again ….
I see both just fine. And that is a good video BTW.
Even more bizarre, I do now too.
It was showing a pic of the UAH graph and when clicked went to a post about it.
If you can’t see the pulp side of an orange cut in half and realize that the pulp side has a surface then you are *NOT* seeing just fine.
And which one of those disks has the same surface area as and topology of the curved surface behind it.
Do you think insolation is concentrated like thru a lens? Remember it takes two of those disks to equal the surface area of the curved side!
Are you familiar with vector calculus?
Are you familiar with common sense ?
And with the concept that fighting reality is useless, and worse, self defeating ?
They’re going down with the ship here, Mr. Banton. Inherent bad wiring leads to overamped amygdalae, triggering fight or flight. Since they are protected by the fact that they need not/do not have to defend themselves in any above ground forum, they choose fight.
Remember negative standard deviations? “That’s my story and I’m stickin’ with it”…
Maybe you should show how you determine the the number of m² exist on the surface of the earth being illuminated. Then show us what function you use to evaluate the value striking each of those from the equator to the poles.
We want to see some math, not opinion.
Your math should explain why the tropics are warm and the poles are cold. Maybe something to do with trig functions would be helpful.
Tell us exactly how common sense informs one that using 1/2 the surface area irradiated by the sun tells you anything.
πr² ≠ 2πr
πr² ≠ 2πr²
As I’ve said before it tells you how much energy reaches the top of the atmosphere as that diagram clearly shows!
So you can’t see the pulp side of an orange cut in half either? You live in a different reality than most people who can see the pulp and realize that the flat side is also a surface!
Tell us how many of those flat disks must be used to emulate the surface area of the spherical side?
I’ll give you a hint. πr² ≠ 2πr².
Why is it so hard for you to understand that we are NOT “emulating the surface area of the spherical side” !
The object of the exercise is to determine the solar energy absorbed by the half sphere.
It is not the same.
And in doing that we see that it is the same as that absorbed by the flat disk (pir^2) of the Earth’s shadow. (See posted vid above … and any other proof found in text books for that matter).
Which is what is universally acknowledged, except by some rabbit-hole dwelling contrarians here, who think ‘winning’ their position is achieved by driving their interlocutors crazy and finally them having to depart in frustration.
Hint: That’s not how science works.
The video you posted has nothing to do with calculating the radiation absorbed by the surface area of a sphere. The video’s purpose is to show a proof of what the surface area of sphere is without using calculus to solve it. Is that why you are so stuck in using a flat disk? I hope not.
Here is an explanation of how I learned to calculate the surface area..
Derivation of Formula for Total Surface Area of the Sphere by Integration | Derivation of Formulas Review at MATHalino
Here is another method from MIT.
Surface Area of a Sphere
Why don’t you show us a book that shows the mathematics of how the absorbed radiation impinging on a sphere is calculated by using a flat disk of πr².
I think you will find that the proof is associated with determining the total radiation impinging on the earth. In other words, by inserting a flat disk that has the same area as a cross section of the earth between the sun and the earth.
As I already have to posted to FrankN, this is an appropriate method to determine the value of the radiation impinging the earth when the EM is a plane wave. It is not an appropriate method for determining the amount of radiation absorbed by the sphere because it assumes that all radiation impinges at a normal angle (90°). And remember, you have already said the angle of incident is important.
I haven’t even mentioned that the earth is not a sphere, it is an oblate spheroid which requires even more factors to determine the surface area and how it is absorbed. And it requires even more complicated calculus to determine the angles of incidence over that surface area.
here is what actually happens
The amount absorbed is *not* the same. The area of one quadrant of a sphere (i.e. from 0 to pi on the half-sphere) is 4πr^2/4 = πr^2
The shape of that area is *NOT* the same as a flat disk.
Did any climate scientist ever take vector calculus?
Did any computer programmer doing climate models ever take vector calculus?
This is really no different than splitting the force of gravity into two components in an inclined plane experiment, one component normal to the plane and one tangential to the plane in order to evaluate friction vs acceleration down the plane.
And how does the Earth get cut in half!
Only the normal component gets absorbed?
“And how does the Earth get cut in half!”
In the Theater of the Mind. Cutting the earth in half is no different conceptually than cutting an orange in half.
“Only the normal component gets absorbed?”
I’ve attached an image with pages from a 1942 thermo textbook that shows what I’m trying to convey. When looking at the figure θ2 for the sun and the earth is 90deg. So the heat absorbed is based on the cosine of θ1.
So, yes, only the normal component gets absorbed. Your image shows refraction, not reflection. Nor are any magnitudes of the vectors given. Light hitting even the surface of a body of water results in both refraction and reflection, if you haven’t seen the glare of a reflected sun off a lake or pond then you must have spent your life in a basement. If you don’t show all vectors and their magnitudes then you aren’t showing the entire picture.
When you get down votes with no rebuttal you are over the target!
“ If you don’t show all vectors and their magnitudes then you aren’t showing the entire picture.”
And claiming that “Only the normal component gets absorbed” is certainly not showing the whole picture!
ROTFLMAO!
You don’t even understand what you have posted!
The x-axis is the angle FROM normal. The y-axis is the amount of reflection.
At 0° from normal, THERE IS NO REFLECTION. Zero, nada, nothing reflected.
At 89.99999° there is 99.99999% reflection.
The angle of incidence is the angle between normal and the actual ray. Look it up.
From Britannica:
And in case you don’t believe Britannica.
https://hep.physics.illinois.edu/home/serrede/P436/Lecture_Notes/P436_Lect_06p5.pdf
“The x-axis is the angle FROM normal. The y-axis is the amount of reflection.
At 0° from normal, THERE IS NO REFLECTION. Zero, nada, nothing reflected.
At 89.99999° there is 99.99999% reflection.”
Yes, so, for example, light at an angle of incidence of 45º is over 90% absorbed according to the graph.
“Yes, so, for example, light at an angle of incidence of 45º is over 90% absorbed according to the graph.”
The integral based on angle of incidence, when it is applied to earth, assumes the earth is an opaque object.
For radiation there are three factors to be considered.
α – absorptivity
ρ – reflectivity
τ – transmissivity
α + ρ + τ = 1
For my analysis, I have assumed τ = 0. I have assumed that α and ρ are a function of the angle of incidence. α goes down the angle of incidence goes up and ρ goes up and the angle of incidence goes up. I have assumed that the relationship is cos(Θ)
That is a common assumption in radiative heat transfer, at least as a baseline for the beginning of an analysis. In the real world α/ρ/τ have a complex relationship.
It’s why I have stated that what I have derived is an AVAILABLE amount for absorption. What actually gets absorbed is a complex function.
No. It varies with the angle θ. From Planck’s Theory of Heat Radiation.
(bold by me)
Look at your diagram closely. See where it says θᵣ = Angle of refraction? Why do you assume that physical quantity has any relationship to absorption of EM waves. Ask yourself what the value of cos(π/2) is. Then ask yourself if that has something to do with the poles being cold.
No. It varies with the angle θ. From Planck’s Theory of Heat Radiation.”
Tell Tim that, it was his comment that I was responding to.
“No. It varies with the angle θ. From Planck’s Theory of Heat Radiation.”
Tell Tim that, it was his comment that I was responding to.”
Exactly what do you think sin(θ) and cos(ⱷ) in my derivation represent if it isn’t Planck’s θ? Don’t get lost in the greek letters being used. Look at what they represent.
No you don’t understand 3D geometry, “.4 * 1360″ is more energy than reaches the top of the atmosphere!
Are you brain modified today?
Read this. It shows various values with NASA recommending 1360. Look at the chapter on “Solar Radiation”.
See the “top-of-atmosphere” in the link?
https://www.sciencedirect.com/topics/earth-and-planetary-sciences/top-of-atmosphere
I do understand 3D geometry. You can’t obtain an EE without learning about Maxwell’s Equations which define the 3D components of EM waves and studying the vector calculus used in them.
The total surface area of the 3D object is 1/2 of a sphere plus the surface area of the flat area.
If you are going to deal with a rotating object you should understand how it actually works.
No! It is about dividing 1360/4 = 340*0.7 = ~240.
Here is a diagram. Tell how the sun’s readiation only intersects πr² surface area rather than 2πr².
“Tell how the sun’s readiation only intersects πr² surface area rather than 2πr²”
Of course it “intersects” 2pir^2 !
But the object of the exercise is to calculate the solar energy absorbed.
And that is dependent on the angle of incidence upon the Earth.
Ever noticed that it’s hottest when the sun is over-head ?
Which is where the area of the shadow pir^2 comes from.
And because the Earth is a rotating once in every 24 hrs that means the whole planet shares that energy in that time.
It is therefore 2x2pir^2 which is 4pir^2
So 1/4 of that hits the Earth’s “shadow” (pir^2) every 24 hours.
Just another Gorman (one or the other) rabbit-hole that leaves peeps tearing their hair out and saving their sanity by absenting the thread.
Please find and post a link to any text that says otherwise to the above.
Or else have the integrity to admit you are away with the fairies and you do not know better than the people that wrote those texts.
BTW: I know full well the outcome of the reply, but someone has to stand up for common-sense (though I know that it is in short supply in the current Trumpian world).
I’m off now to preserve my sanity.
Ta Ta
Take a look at this image from a 1942 thermodynamics textbook. They knew back then how to calculate heat transfer from an EM wave not at 90deg to the receiving surface. My newest textbooks all show the same thing.
There is no common sense in what you post. You apparently have never even experienced glare from the sun bouncing off the surface of a body of water. It’s exactly the same concept. Not everything impinging on a surface gets absorbed when the angle of incidence is different than 90deg.
Show me with some backup math where the textbook images are wrong. Otherwise you are just indulging the argumentative fallacy of Argument by Dismissal.
You first need an accurate calculation of the insolation reaching the earth.
You just agreed with me on this. The angle of incidence varies with trig values, not some average of a flat disk. Congratulations!
Now show us those sine and cosine functions with their integrals over the proper intervals which define the sphericalabsorption.
Shadow? There is no object between the sun and the earth that casts a shadow. Have you perhaps had an hallucination?
You do understand that the sun only shines for 12 hours (in an ideal system like this). You can’t average zero in for half the time and determine a scientific value.
It is therefore 2x2pir^2 which is 4pir^2
Every point on the earth experiences the sine curve of insolation during the period of time that the sun shines. If you want to accurately define the maximum you must calculate what occurs from sunrise to sunset. That sets the boundary at which cooling begins.
Heating and cooling are two different process with different gradients. Warming has a sine function as the sun rises and sets. Cooling has an exponential function with totally different parameters.
Glad to be of help!
I have shown you a detailed mathematical solution. You have shown nothing, not even when you admit the angle of incidence is part of the solution. You just parrot the old divide by 4 because of a shadow.
Talk about believing in fairies.
Here are two solution of integrals. Tell us what is wrong with them.
BTW, these are from symbolab if you like to play with them.
I see no one has offered any actual refutation of the math of 3D objects in 3D space. Saying a sphere is a flat plane has no justification based on a mathematical transform.
The CAGW and climate science defenders can’t even write the proper form of analysis for incoming and outcoming radiation.
Radiation shown as “R”
Daytime: At any point on the sphere: Rin – Rout
Nighttime: At any point on the sphere: Rin – Rout
If you are going to average Rin over 24 hours then you must also average Rout over 24 hours – because the earth *does* radiate to space even during the day.
You can’t just say that Rin determines the daytime temperature without applying the unstated assumption that there is no Rout during the day. The temperature is related to the NET VALUE of the Rin – Rout, not just Rin.
If 1360 is the Rin value then what is the Rout value?
The last time I used symbolab to verify computation of a simple partial derivative you told me it was wrong.
No, I didn’t say symbolab was wrong, I said the formulation you arrived at for uncertainty was wrong so the answer given was correct, what you used for the partial was incorrect.
In other words garbage in, garbage out.
“You first need an accurate calculation of the insolation reaching the earth.”
Yes, which is given by the insolation times the cross sectional area of the Earth, i.e. 1360xπr^2.
During 24 hrs that is shared over the whole surface area of the Earth, 4πr^2.
That would be the value if there were no atmosphere, scattering by clouds etc. needs to be accounted for to calculate the amount reaching the surface.
The cross-sectional area of the earth is *NOT* a flat plane. It is a curved surface.
If the earth was a flat plane it’s temperature profile would look like a bandwidth limited square wave. An exponential rise based on thermal inertia, a flat top once thermal equilibrium is reached, and then an exponential decay at night.
Yes, each point on the earth is impinged by a plane wave of 1360 for about 12 hours. But the amount absorbed, AT ANY POINT IN TIME, is based on the angle of incidence to the CURVED surface of the earth, it is the amount ABSORBED that determines the temperature profile. As the sun traverses the sky that amount absorbed starts at zero at sunrise, goes to 1360 at noon, and finishes at zero at sunset.
“During 24 hrs that is shared”
The insolation is not “shared”. Every point gets 1360 impinging on it at all times during the day. It’s the amount absorbed that makes the difference. The amount absorbed is a function of the angle of incidence throughout the day and therefore an “average” absorption can be calculated based on the angle of incidence function.
And the *average* insolation absorbed for quadrant of a sphere, at any point in time, will be about 0.4 * 1360 joules/sec-m^2 =
340 joules/sec-m^2.
One *real* issue here is whether or not you can integrate that over 12 hours and come up with something REAL that will give you an average temperature over the quadrant. An average temperature for a quadrant would lead one to assume a similar climate for the entire quadrant, *NOT* a real fact in evidence.
A second real issue is that calculating a temperature for the quadrant at any point in time, based on this “average” insolation absorbed, has the same issue. First, temperature is not climate and, second temperature is a standalone function of its own because it is an intensive property.
At it’s base it is a problem with using averages as climate science likes to do. Averages would tell you that Las Vegas and Miami have similar climates based on their temperatures. Or that Bangor, MN radiates at the same level as Atlanta because they are in the same quadrant.
What you *can* do is calculate the absorbed insolation for each point in the quadrant and project its temperature consistent with ignoring all the other factors that determine temperature such as pressure and humidity.
“The cross-sectional area of the earth is *NOT* a flat plane.”
By definition it certainly is!
‘Any cross-sectional area of a sphere is a circle, and its area is calculated using the formula πr², where ‘r’ is the radius of the circle (which is also the radius of the sphere).’
The plane wave is not impinging on a cross-section. It is impinging on a curved surface. How many pictures from textbooks do you need to prove this? When I get home I’ll upload another one.
Don’t bother you didn’t understand it the first time!
The area of the solar radiation that is removed by the Earth is the Earth’s cross section, the shadow cast by the Earth is a circle with the radius of the Earth.
A plane wave is a continuous function with the same intensity at every point on the plane. In essence, it contains an infinite number of parallel rays all with the same intensity moving in the same direction.
Each point on the spherical surface earth intercepts a point on the plane wave.
Show us the math whereby 1/2 of the spherical surface escapes being impacted by a point on the plane wave.
I am reminded of the 3D artistic tool made of a plane with numerous pins that can be pressed onto a 3D shape to make a copy of the 3D shape. The finer the density of the pins, the more accurate the copy.
If I could find one with 1 million pins that would barely cover a baseball, and I pressed the baseball into it, would half the spherical surface not be represented? Would it look like a flat circle of πr²?
The projection of the earths spherical surface onto a flat plane, i.e. its shadow, doesn’t represent the absorbing surface of the sphere. A plane wave is normal to every point on a flat plane perpendicular to the plane wave. A plane wave is *NOT* normal to all points on a sphere.
Since heat transfer is based on the normal component of the vector, i.e. the angle of incidence, the heat transfer to a flat plane is not equivalent to the heat transfer to a flat plane.
This is why I keep saying you have to know VECTOR calculus to do a proper analysis. It is no different that breaking down the force of gravity into a component that is normal to an inclined plane and a component that is parallel to the inclined plane. Only the normal component contributes to the friction between a block sitting on the plane, just as only the normal component of an EM wave contributes to the heat transfer. An inclined plane placed against a sphere will have a different angle as you rotate it around the sphere vertically. The angle of that inclined plane will actually be the slope of the sphere at the contact point, i.e. the derivative of the curve at the point which is actually the slope of the tangent to that point on the curve.
Doing this with a flat plane, i.e. the shadow of the sphere, gives the same angle everywhere on the flat plane, i..e 90°.
The sphere and the flat plane are different objects in 3D vector space regardless of what their surface area is.
The question I was answering was: “You first need an accurate calculation of the insolation reaching the earth.”
Which is given by the insolation times the cross sectional area of the Earth, i.e. 1360xπr^2. How it is distributed across the Earth’s surface is determined by factors such as angle of incidence, atmospheric scattering etc.
The insolation ABSORBED by the earth is the question, not the total insolation. The plane wave hits most of the earth at an angle. Thus the amount AVAILABLE to be aborbed is I * cos(x). It has nothing to do with the cross-section. The cross-section is a flat plane and any insolation from a plane wave would be hitting it at a constant angle.
Once more. The heat transfer is like what you have to do with the “insolation” of gravity hitting an inclined plane. Only the normal part of the gravity “insolation” affects the friction between the inclined plane and an object on the inclined plane. The “shadow” produced by that inclined plane, your “cross-section”, is not useful in determining the friction force. And if that inclined plane is actually warped into a curve then its shadow is even less useful.
“The insolation ABSORBED by the earth is the question, not the total insolation.”
As I said the amount reaching the Earth is given by 1360πr^2 the amount absorbed depends on the interaction with the atmosphere and surface and will of course be less than 1360πr^2.
It depends on more than the atmosphere. It’s a vector function, the only heat available for absorption is the normal vector to the surface. So it also depends on the angle of incidence. If you lived on the equator you don’t get 1360-atmosphere all day. The average insolation during the day is .67*(1360-atmos). move away from the equator and it goes down even further.
Let’s go through what this means. We can use a unit circle where “r = 1”.
• When the incoming plane wave intersects the flat plane of the cross section, everything is perpendicular. This gives us total Watts presented to the earth as:
1360 × π(1)² = 4273 W
• The cross section has two quadrants, north and south, so each receives 1/2 the power.
4273 W × 1/2 quadrants = 2136 W / quadrant
• The spherical side also has two quadrants that coincide with the quadrants of the cross section. The power presented to each of those quadrants will be 2136 W also.
• Each quadrant on the spherical side has π(1)² m². (4πr²/4 = π(1)²)
• The average intensity offered to each quadrant on the spherical side will be:
2136 W ÷ π(1)² = 680 W/m²
• Just for grins, let’s see what that gives us for temperature with an albedo of 0.3.
680 W/m² × 0.7 = 476 W/m²
using the SB equation and an emissivity of 1 gives us:
T = (476 ÷ 5.67 ×10⁻⁸)¹÷⁴ = 303K
I have shown this below.
What does this tell you?
You can not determine a real temperature distribution for the earth since each square meter has only an average value. This shows each square meter would radiate at 303K
It is not reasonable to divide 1360 W/m² by 4 since there are only 2 quadrants receiving insolation at any given time. If you divide 1360 W/m² by two, guess what you get for an average per square meter in the two lit quadrants? And, you still can’t derive a temperature distribution.
Why do you keep rambling on about quadrants, they don’t have anything to do with anything? The light from the sun is a parallel beam of 1360W/m^2, on your 1m radius sphere that means the hemisphere facing the sun intercepts 1360πr^2 W, that’s it! Dividing 1360 W/m² by 2 is geometric nonsense.
So your flat cross section doesn’t have a north and south hemisphere (quadrant)?
The “north” quadrant on the cross section is the shadow of the top quadrant of the spherical side facing the sun.
The “south” quadrant of the cross section is the shadow of the bottom bottom quadrant of the spherical side facing the sun.
I agree and you’ll notice I did not do this. I calculated the total power absorbed by the flat cross section and used that to determine the subsequent calculations.
You and bdgwx are advocates for using averages and that is what I have tried to show and that it doesn’t work.
You should notice that 1360 W/m² is not a total. It is a rate PER m². That is each and every square meter gets 1360 W. A 1m radius results in an area of πr² which becomes π with r =1. That gives 4263 total watts absorbed.
You just can’t through the issue that a spherical area has more surface area than a flat cross section.
If the cross section absorbs 4263, then so does the spherical surface area. You should be able to see that the special surface must also absorb that amount or there will lost energy.
You’ll need to show your math to show what I have done differently.
“So your flat cross section doesn’t have a north and south hemisphere (quadrant)?”
It does but it’s not relevant.
“You should notice that 1360 W/m² is not a total. It is a rate PER m².”
Exactly which is why I multiply by πr^2 to determine the amount arriving at the ToA.
As in your example, which had no atmosphere, the total that reaches the surface is the same, to determine the amount reaching each square m of the surface involves trig. For example, at 45º from the solar zenith the amount per m^2 is 1360/(√2), so using that method you can calculate the distribution of energy across the surface.
Also on the Earth you would need to make allowance for the atmosphere and the albedo of the surface, but you still start with the amount at the top of the atmosphere.
Glad to see you have agreed with what I’ve been saying.
It’s what I’ve always said.
The issue isn’t what is intercepted. The issue is what gets absorbed. What is absorbed is based on the normal vector to the surface. And that changes as you move north or south from the equator.
It depends on the angle from the solar zenith not the equator.
Nope. The earth is so far from the sun that it subtends a very small piece of the spherical wavefront. Such a small piece that the wavefront can be considered to be a plane wave as far as the earth is considered. Therefore the angle with which it intersects the earth is dependent solely on the curvature of the earth and it follows a cos(φ) function, completely normal at the equator (cos(φ) = 1) and totally absorbed. The angle is completely tangential at the pole (cos(φ) = 0) and nothing is absorbed from the plane wave. This is a simplified baseline model. The tilt of the earth is a modulating factor for the path of the sun, i.e. the length of the daytime period. But the radiation from the sun can still be considered to be a plane wave.
No convection ? Really ? So the non ghg’s won’t heat up ? So 90% of today’s atmosphere never heats up ? Otherwise an interesting exercise … you obviously get an incorrect answer with your bad assumption …
The convection in my model is net zero with respect to heat from/to the atmosphere from the surface. I most definitely did not assume no convection, in fact the wind speed will be very high. I just assume that any heat picked up by the wind on the dayside is deposited on the nightside.
Many bad assumptions, piled on wild guesses.
He doesn’t like the scientific method, and prefers fantasy to facts.
The insult is uncalled for.
What insult? Andy referred to his “wild ass guesses”, so I was being relatively polite, I thought.
My opinion about bad assumptions is shared by others, and supported by facts.
Andy shows no appreciation of the scientific method.
Maybe your opinions differ to mine, and are equally valid. I share Winston Churchill’s opinion about freedom of speech “Everyone is in favor of free speech. Hardly a day passes without its being extolled, but some people’s idea of it is that they are free to say what they like, but if anyone else says anything back, that is an outrage.”
I just decline to accept when someone offers offense, and choose not to feel annoyed or insulted because of what someone says. It’s not that hard, is it?
If you are trying to determine what impact GHGs are having on the temperature of this Earth, you need to be using this Earth as your baseline. Trying to figure out what the Earths albedo would have been had there never been any water on this planet is a dead end.
Thank you
“Climate scientists” love what coulda, woulda, shoulda, been.
Who could possibly disprove a figment of a nutter’s imagination?
The Earth has an atmosphere. End of story.
Andy, why do you use basalt for the surface?
At least on Earth > 70% of the surface is ocean, so why not use water?
Sun delivers maximum ~30 MJ/m^2 over a 24hr period.
https://www.pveducation.org/pvcdrom/properties-of-sunlight/isoflux-contour-plots
Using eg 20 MJ/m^2, this translates to the capacity to increase the temperature of ~5 m^3 water 1K.
So all radiative balance calculations are pretty much useless on our water world
‘…so why not use water?’
From the article:
‘On GHG-free Earth there is no water, so I assumed the surface is bare basalt.’
Doesn’t make much sense to me. Maybe use a theoretical substance that has the specific heat capacity and penetration characteristics of water without the ability to evaporate.
Earths surface is mostly water, so any usable model should take this fact into account.
Ben, Basalt is the most common rock on Earth’s surface, excluding water. That is why I used it. Water vapor is, by far, the most powerful GHG. To compute a GHG-free temperature I could not include oceans, water, or clouds. That was one of my main points.
“Water vapor is, by far, the most powerful GHG.”
What are you suggesting? Surely you aren’t silly enough to believe that adding H2O to the atmosphere somehow “warms the globe?”
Complete tosh, and the hottest places on Earth (like Death Valley) are notable for their severe lack of “the most powerful GHG”!
A fellow called John Tyndall made this general observation over 100 years ago, and performed a series of meticulous experiments to find out why. You might care to read his work, or not, as you choose.
Your fantasies are not reality.
Andy
Thanks for the effort. I note a comment from you down the chat that raises the critical question I have , and which I had addressed a few times in the WUWT comments (with no response).
You wrote, “I effectively ignored the atmosphere as if it were not there.”
This is a major difference between your model and an atmosphere without GHG’s. There are three wants to get heat into the atmosphere: convection, radiation to receiver gases and conduction (which is so little I’ll ignore it).
Radiation is well handled by saying the atmosphere doesn’t receive or emit radiation due to the absence of IR capturing gases. Fine. But the convective heating remains. The hot basalt will cool by convection to the air, as well as diffuse heat into its own mass. Because hot air rises, there is far more cooling to the air than to the thermal mass of the rock. The question then becomes, what is the equilibrium temperature of the atmosphere at, say, 1 km altitude, when there are no GHG’s?
Hot air rises and it doesn’t fall just because the ground cools by radiation at night. The bottom of a chest freezer doesn’t get warm because the lid is open. This can be demonstrated at any grocery store. Each day in your 36,500 runs the air would be heated hotter and hotter by the surface, and has no way to cool other than to cool against the surface at night after the surface completed its cooling by radiation, which is not instant as you point out.
First, cooling of the atmosphere can only happen at night if there is wind, and there are strong forces working to stratify the atmosphere, hotter above and cooler below. The equilibrium temperature of a no-GHG atmosphere would have to be well above 100 degrees C, whatever the average surface temperature. At night the air temperature near the surface would rapidly drop and then a strong stratification would occur as the wind speed dropped zero. The air temperature above the ground would remain high, and go higher the next day.
As stated in the article, this wind speed would be high as winds would take thermal energy from the dayside to the nightside. The Earth rotates from west to east at a speed of over 1,000 MPH at the equator, so the winds would be easterlies at the equator and given the 170K difference in daytime versus nighttime temperature and the resulting air pressure difference the speed would be very high, hundreds of MPH.
My statement, that you quoted out of context, “I effectively ignored the atmosphere as if it were not there.” simply meant that I assumed that the heat picked up by the atmosphere on the dayside (via conduction) was deposited in the cooler polar regions or the nightside, again via conduction. In other words there is no net gain or loss in the atmosphere of thermal energy. This is logical if there are no GHGs. Most heat held in the atmosphere in the real world is held by water vapor, water, and ice.
“that I assumed that the heat picked up by the atmosphere on the dayside (via conduction) was deposited in the cooler polar regions or the nightside, again via conduction. In other words there is no net gain or loss in the atmosphere of thermal energy.”
Bad assumption? You might as well assume that adding CO2 to air makes it hotter – and call it the GHE! Save you having to waste your time writing reams of pointless word salad.
Just a thought.