Guest Post by Willis Eschenbach
[SEE UPDATE AT THE END]
Due to the recent posts by Lord Monkton and Nick Stokes, I’ve been thinking about the relationship between radiation and temperature. So I turned to the CERES dataset. Here is a scatterplot of the monthly global average surface temperature versus the monthly global average downwelling total radiation absorbed by the surface. The total radiation is the sum of the net solar radiation at the surface and the downwelling longwave radiation at the surface. I’ve removed the seasonal variations from the data.
Note that 3.7 W/m2 is the increase in downwelling longwave radiation expected from a doubling of CO2 …
When I saw that, I thought well, maybe the increase is small because there’s a lag between the absorption of the radiation and the warming. To see if that was the case, I did a cross-correlation analysis of the relationship.
No lag visible.
Now, I get busted regularly for drawing what I’m told are the wrong conclusions from the data that I present. So I’m just gonna say …
Comments?
——————————————————————————————————————
Me, I’m writing this from banks of the Kenai River in Alaska, one of my favorite spots in the world. When I got off the airplane, the aroma of the air was absolutely intoxicating. Summertime is short here but the days are long, and the air is full of the heady perfume of every plant and every animal growing and going at triple speed, making the most of the brief Alaska summer. Here’s what the sun is doing today this far north …
[UPDATE] Someone asked what temperature I’m using. I used the conversion of the upwelling longwave radiation from the surface. However, the answer is only slightly different if I use, for example, the HadCRUT surface temperature. Here is that result:
As you can see, there is no significant difference when I use the other surface temperature dataset.
My very best regards to all, may your days be as full of sunshine as mine,
w.
PS—My usual request: when you comment, please quote the exact words you are responding to, so we can all be clear about who and what you are talking about.
PPS—Bonus question. What latitude on the planet gets the most hours of sunlight per year?
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Due to orbital eccentricity, the North Pole (currently) receives the most hours of sunlight. The Earth’s eccentricity causes a slightly slower orbit during Northern Hemisphere Summer.
If Earth’s orbit was perfectly circular, all latitudes would average 12 hours sunlight per day.
Since the Earth’s axis processes, this relationship will change in the future.
“If Earth’s orbit was perfectly circular, all latitudes would average 12 hours sunlight per day.”
Dead wrong. The correct statement would be: if the polar axis were perpendicular to the earth’s orbital plane (around the sun) “all latitudes would average 12 hours sunlight per day.” The word ‘equinox’ was well defined by Chaucer: “This circle is also called the Equator, that is the measurer of the day, because when the Sun is at the start of Aries and Libra, the days are same length everyplace in the world. Therefore, these two signs are called the equinoxes.” –AGF
If the polar axis were perpendicular to the earth’s orbital plane, then all latitudes would receive 12 hours of sunlight per day, every day of the year.
With the axis not being perpendicular, the amount of sunlight varies throughout the year, but if the orbit was perfectly circular, the average amount over the year would work out to 12 hours a day.
Dead right.
And yet that’s what the OP wrote.
Methinks that if the polar axis were perpendicular to the earth’s orbital plane (about the sun) then the exact north and south poles would receive continuous sunlight, assuming no blockage from surrounding higher ground elevations. True, the sun would directly on the horizon, and perhaps half of the solar disk would be blocked by the horizon, but the poles would nevertheless being receiving continuous solar radiation in such a hypothetical situation.
Uhhh . . . “higher ground elevations” in my above post obviously includes ice.
And apologies for the typos (dropped “be” and “being” instead of “be” in last sentence). Sigh . . . too early in the morning, my time.
Final note: the solar disk subtends an apparent disk angle of 0.53 degrees in the sky at 1 AU, thus insuring that sunlight would actually fall on the top of an perfectly spherical or perfectly oblate-spheriodal Earth.
The Arctic Circle has the longest total annual daytime… 4,647 hours, while the North Pole receives 4,575. About 3 days more on the Arctic Circle than the North Pole. This is annual amount of daytime…the amount of time the Sun is above the horizon. Because of elliptic nature of the Earth’s orbit, the Southern Hemisphere is not symmetrical: the Antarctic Circle, with 4,530 hours of daylight, receives five days less of sunshine than its antipodes. The Equator has a total daytime of 4,422 hours per year. https://en.wikipedia.org/wiki/Sunshine_duration
Hmm, the question was most hours of sunlight not daylight. Given the hours of daylight vary very little with latitude (4422 to 4647 according to Earthling2) the impact of cloudiness is likely to be far more significant and the least cloudy areas of earth are at the edges of the Hadley cell ie: latitudes 30 degrees north and south – which is where most of Earth’s deserts are. It is due to the already dry descending and thus warming air at the edges of the Hadley cell which results in a permanent band of high pressure at these latitudes.
Yup, looks like about 0.5 C for a CO2 doubling (3.7 w/m2).
But I forgot, it’s all about the magical-mystery water-vapor enhancement! /sarc
WV bands are saturated. CO2 (which is the same bands except for 15 um) does nothing.
This the correct answer.
http://us-climate.blogspot.com/2015/06/daylight-and-twilight.html
Remember to account for refraction — the atmosphere bends incoming light, so the average day anywhere is a bit more than 12 hours. The sun appears in the sky for a while both before sundown and after sunup when, by a true straight line, it would be below the horizon.
The effect is surely greatest where the sun’s apparent path is most oblique, as opposed to perpendicular, to the horizon — that is, at higher latitudes both north and south during their respective summers. I now live at about 58 degrees north, and I notice that at winter solstice there is about seven hours of daylight, but now at summer solstice there is about 6.5 hours without daylight — so the situation is not symmetric!
No one processes data as interestingly as Willis Eschenbach. If his first graph is correct – and, knowing him, it probably is – then the combined effect of forcing and feedback on global warming is negligible.
In that event, it is puzzling that official climatology assigns only a plus-or-minus-10% uncertainty to the directly-forced warming of 1.05 K per CO2 doubling (i.e., per 3.7 Watts per square meter of radiative forcing), a value thrice the 0.38 K equilibrium warming after feedback has acted that Willis has found.
This is an interesting result indeed, and it bears out the conclusion of the Connollys pere et fils – also no slouches at data handling – that there is no detectable influence of the water vapor feedback in the radiosonde data.
I hope that Willis will write this result up as a paper for peer review and publication.
The earth’s axis “precesses” in a wobbly circle with a cycle length of 22,000 yrs due to the effects of the sun and moon and the other planets tugging on the bulging equator. The earth’s orbit itself also slightly processes with the focal point being the sun though this orbital precession is not nearly as dramatic as that of Mercury. The earth’s axial precession causes a gradual 90 degree shift in the location of the equinoxes to a different point in the earth’s orbit as well as the axis pointing toward a different pole star approximately every 5500 years. So the appropriate pole star 6000 years ago to indicate north would have been the star at the end of the handle of the Big Dipper. Also 11,000-16,500 years ago, coinciding with the ending of the last Ice Age, the winter solstice would have occurred at a different location in the earth’s orbit, shifted 180 degrees slightly further distant from the sun due to the eccentricity of the orbit. The current location at the winter solstice is at a point closest to the sun. One presumes 11,000 yrs ago somewhat less solar radiation was received by the northern hemisphere by the the location of the earth at winter solstice. Though I’m not sure of the difference in actual received watts/m^2. The axial tilt would have been slightly different as well though I’m not sure of the change from the vertical compared the current tilt at that time. The book Ice Age (2001) by John and Mary Gribbin introduced me to the Milankovitch Cycles (beside precession of the equinoxes, the other two being orbital eccentricity with a cycle of 100,000-110,000 yrs and variation in axial tilt currently 23.5 degrees from vertical nodding up and down with a range of 3 degrees, referred to as nutation with a cycle of 41000 years) and the exciting history of the discovery of the ratios of atomic O16:O18 isotope chemistry developed as surrogate for T anomalies and magnetometer dating science in cores of ice cap and deep ocean floor that led to the acceptance of Ice Ages, their occurence every 110,000 yrs during the past 2.5 million years or so. It’s fascinating to us amateurs.
Conclusion: Minor variations in a minuscule atmospheric gas doesn’t have much impact on earth temperatures. There are too many other massive energy transfers for such a small perturbation to have a significantly discernible effect.
I remember crossing the Kenai river on the way to Homer one year, when there must have been a salmon (steelhead?) opening–the river was a forest of fishermen. I had two instant thoughts:
1. How many fights must have broken out that day due to tangled lines?–these were (probably armed) Alaskans! (I’d seen a LOT of Alaska’s highway signs by then.)
2. How on earth did any fish manage to get upstream through the forest of legs in hip waders–it was a close-spaced grate capable of diffraction!
(There was no thought of any issue of global warming though; the world was at that time still in a settled-science panic because it was going to freeze.)
Thanks, Len. On the Kenai, that’s called “Battle Fishing”. And if you want real fun, toss in a few grizzly bears …
w.
Len: Most likely was the Red Salmon run. There are millions that go through it. Fights are surprisingly few, most fishermen are decent people and capable.
Willis: As locals, we always called it ‘Combat Fishing. Now, mostly Grizzlies are fine to fish around. They prefer the shallows to the pools we want fish out of.
Oh wait–I think it was what they call a ‘King salmon’ (Chinook) up there–and I heard a short time later on the radio that a record fish had just been caught–such that from the timing someone must have been landing it just as we drove over the highway bridge. An interesting memory.
Actually, Willis, its called Combat Fishing by locals (I was a local). After Combat Fishing for awhile, most fisher-people learn how to avoid line-fouling.
Armed Alaskans that participate in Combat Fishing know most everyone else is armed; no altercations.
Since netting fish is unlawful, the vast majority of salmon get through the lines of fisher-people. After individual fish runs are determined to be sufficient to ensure salmon survival, State officials will allow limited netting.
The excitement of a 45 pound King Salmon on the line is indescribable. The group I got into with on the Kenai Peninsula used fly rods, 20 pound test lines, and treble-hooks with the barbs removed. It took a lot of skill and resulted in some epic battles. I was once dragged down a stretch of the Anchor River near Homer into the Cook Inlet where, as the waves washed over the top of my full-body waders, the line snapped when the reel ran out of line as I ran out of ground to move forward.
You should have seen the one that got away.
It was called Al Gore.
Willis – your assignment is to report back on the effect of increasing atmospheric CO2 on the number and size of the salmon. (Mepps Flying C per chance?)
Farmer
That’s easy. It’s a very large and obviously catastrophic effect that will shrink their size by half and their numbers by 97%
Tonyb
Or it will increase their size by 14%
He’s a trout, an old one, not a salmon.
I’ve always heard it called Combat Fishing. I’ll be there next week but not to fish.
Yeah, commenters are right, it’s “combat fishing”. Gol-dang memory …
w
Easy fix to “Combat Fishing”. Hook up with a local.
https://imgur.com/gallery/T3zuY4B
Thanks, Renee, nice shots. For those interested in my history on the Kenai, see Fishing The Mighty Kenai …
w.
I feel so lucky to enjoy and study the Kenai River during this Holocene interglacial period. It’s been a glacier for 70% of the time over the past the 100,000 years.
http://dggs.alaska.gov/webpubs/dggs/gb/text/gb008.pdf
Does the FAA still send out inspectors to paint tail numbers on the mosquitoes?
Mosquitoes up there are now required to have ‘black boxes’ and collision avoidance systems.
When we took a rental RV down the Alaska Highway a couple of years ago, we were told to be careful otherwise the mosquitoes, given the chance, would carry us off and come back later for the RV.
In Alaska, you can even see the no-see-ums (Culicoides).
I remember being in Fairbanks in early-April in 1967. Water was running in the creek beds, but the ground was still covered with snow — and the mosquitoes were already out and flying around!
Out on the Aleutians it was too windy for mosquitos, one of not very many benefits to being there.
Len says:
How on earth did any fish manage to get upstream through the forest of legs in hip waders–it was a close-spaced grate capable of diffraction!
Quantum effects. Some fish magically appear past the fishermen if they’re not observed. 😉
12 degrees South gets the most DWSR, so I assume that that is proportional to the amount of sunlight received.
Just get yourself a bucket of cold water and a heat gun and try putting heat through the surface of the water, You can’t, that tells the story
Get yourself a handful of air with a dewpoint higher than that of the water. That will add heat to the water as the water vapor condenses (and forms dew?) on the water. The inverse of evaporation, wet bulb temperatures, all that good stuff.
I have said it before, and I will say it again: The Sky cannot heat the Surface of the Earth, and the Sky cannot heat itself. The Sun heats both. CO2 absorbs 15 micron IR, thermalizes it low in the Atmosphere, absorbs and reradiates it high in the Atmosphere.
The IR absorbing sky does not (generally) heat the Earth, but it does radiate some energy to the Earth in a process that is effectively radiation partial insulation, which does result in a higher temperature. An analogy (not the same process but similar in effect) would be to put a clear blanket with some vent holes over the surface.
Leonard Weinstein,
Do you know what Thermalization is? It seems that you do not. CO2 radiates at 15 microns, corresponding to a temperature of -80 C. When this radiation encounters an object which is warmer than -80 C, it is simply reflected, not absorbed, no heat is transferred from the colder gas to the warmer surface. Yes, photons do that. This is the nature of the Second Law, and if humans know anything about physics it is the First and Second Laws of Thermodynamics.
Next time it’s cold outside – do not don a jacket. After all, jacket cannot heat your skin.
If you wrap a coat around a steel ball at 98.6 degrees F, its temperature does not increase. Why not?
Your body is a heat source. Efficiency of a human body is roughly 22%, and most of the expended energy is released as heat. When you are not physically active – your produce 40-45W per m² of your skin. Therefore you are in radiative equilibrium with your environment if it’s ~10K colder than your body. Human skin is on average 33°C, that’s why temperatures around 23°C are so pleasant to us. Your body radiates 450W, and is receiving ca. 400W.
Now, what happens if it’s 0°C outside? Your body still radiates 450W, but from the environment you’ll receive just 287W. You’ll lose over hundred Joule per second, and will soon be cold and dead. Now you don your jacket. Heat transfer is slowed down and after an equilibrium is established, your jacket will be on the outside the same 10K warmer than the environment. The heat flux is the same 40W, but the jacket emit just 330W, and receives 287W. The jacket will also have temperature gradient, which will be defined by the thermal conductivity of the jacket material. In our case heat flow is known – it’s 40W/m², temperature difference is 23K, and thermal conductivity for thinsulate is 0.023W/m*K. Therefore your jacket has to be 13mm thick – then you are perfectly warm.
What happens here is that the equilibrium surface is pushed outwards, away from your skin. For that you need a medium that can slow down heat transfer. It’s the rate of the slowdown, which defines the temperature gradient. No slowdown – no gradient – no jacket effect.
Exactly the same thing happens in the atmosphere. The equilibrium surface is 5500m away and CO₂ is one of the factors that define the slowdown. The slowdown happens because absorption and re-emission take time. The time is minuscule, but not instantaneous. And exactly that makes CO₂ a “jacket-gas”.
Anyway, I think the back-radiation approach is physically not correct. My approach would be to calculate thermal conductivity of the radiative transfer and how CO₂ influences it. BTW, the lapse rate of the atmosphere is the equivalent of the temperature gradient in the jacket. If CO₂ significantly influences the thermal conductivity of the atmosphere, then the lapse rate should change. Since we don’t see it – I’d suspect that CO₂ influence is rather smallish.
Well put.
It is not a matter of conductivity (which is extremely low for gases), nor of absorption and re-emission since the absorbed energy is almost always thermalized before it has time to be re-emitted. The absorbed energy is instead transported by convection. Convection moves considerably more heat away from the surface than radiation.
@tty I’ve never meant to reduce my argument to radiation only.
Because the room temperature is 75 degrees. (Assumed steady temperature.)
Your steel ball will cool to eventually approximate 75 degrees.
If the room temperature were kept at 150 degrees F, then the steel ball also would eventually approximate 150 degrees F. (Some thermal lag in both cases. Greater or more efficient insulation = greater thermal lag. Greater steel bearing mass = greater thermal lag. Too much insulation = faster cooldown in most cases. )
Stupid question…what software do you use for your charts?
Michael, I do almost all my work in the computer language “R”. It’s the best language I’ve used in my 50+ years of programming computers.
w.
Yes I’ve seen you mention R many times over the years, ditto with Steve McIntyre.
I just wonder why I don’t see more of it in the world…so many charts are clearly done in Excel.
Excel is much easier to use than writing R code. Especially if you are only looking for quick trends and simple statistics.
Renee June 8, 2019 at 9:47 pm
I love Excel, and I can make it sit up and spit pickle juice. I’ve written hundreds of macros, and even wrote hooks into it to make it run C code … but it is slow and lousy for big datasets.
For example, each of the CERES individual datasets, such as say the surface downwelling longwave, is 180 rows by 360 columns by 216 layers. That’s a total of 153,964,800 data points for each dataset … and often I’m comparing four of them. Try that in Excel.
Now, suppose I want to add an offset to every one of those individual data points, say plus 5. The surface longwave dataset mentioned above is called surf_lw_down_all.
In Excel, I’d have to create a new dataset of the same size by writing a formula like
=A1 + 5
in each of the 153,964,800 cells.
And it would take hours to run …
In most other computer languages, to add 5 to each cell of the array you’d have to write something like:
for myrow = 1 to 180
….for mycolumn = 1 to 360
……..for mylayer = 1 to 216
…………surf_lw_down_all[ myrow, mycol, mylabel] =
surf_lw_down_all[ myrow, mycolumn, mylabel] + 5
……..next mylayer
….next mycolumn
next myrow
(ignore the dots, they’re just for formatting, WordPress won’t let you indent lines)
In R, on the other hand, I just say
surf_lw_down_all = surf_lw_down_all +5
That’s it. No need to write nexted loops to address each of the hundred and fifty million individual data points.
It takes R maybe a second to run that.
Anyhow, that’s why I use R instead of Excel. To be clear, I’ll use Excel for certain quick-and-easy calculations, but R is industrial strength.
Finally, R is free, cross-platform, has free packages for everything from obscure statistics to 3D graphics, and it has a killer user interface called RStudio. RStudio has features like, click on the name of a function to jump to wherever you defined the function, even in another file. And it has function completion, spreadsheet-style inspection of data, and lots of other great features.
Steve McIntyre talked me into learning R. I’ve been programming computers since 1963, over half a century now, and it was a real eye-opener … I’ve blessed Steve many times since then for his persistence in telling me I needed to move up to the big leagues.
Anyhow, that’s why I use R …
w.
You don’t see much of R in the world because you’re in the wrong industry. In Data Science and predictive modeling, everyone uses R or a handful of other tools (and almost everyone was lol at least use R occasionally).
It’s actually a good litmus test for the competence of anyone’s Analytics or Business Intelligence department – do they ever use R? If not, it’s a very immature group.
So can we just say an increase of 1 Watt/m² causes ~0.1K warming?
Bonus Answer – All latitudes get the same total hours each year.
The shame of it is that there are places in this world where the extra daylight is wasted as you have to sleep, and you pay for it in the winter with months of miserable cold darkness – too many years living in Scotland!
We have a winner! Every spot on the earth receives the same amount of direct sunshine … curious but true.
Thanks, Alec.
w.
I believe VaughnBeethoven is correct. The North Pole currently has (very)slightly more hours of direct sunshine.
Willis, I assume you mean the same duration of sunlight, not the same “amount” (total Joules/m2 or W/m2)
Another way of expressing this surprising truth:
For every point on the earth, on average the sun is above the horizon exactly half the time.
Chris
With Earth being inclined to its orbital plane by 23.5 degrees and having a slightly elliptical orbit about the Sun (e=.0167), Kepler’s second law would say this cannot be true. Reference the second OP to this article by Earthling2.
>>
For every point on the earth, on average the sun is above the horizon exactly half the time.
<<
Dr. Spencer is right–it’s not the same amount in W/m^2. Also, due to refraction by the atmosphere, it’s not exactly half either.
Jim
Except when there has been a solar eclipse.
My guess on the latitude with most sunlight is 90N and 90S. The sunlight actually precedes the Sun poking above and below the horizon due to refraction. At 90N/S it will “hang” on the horizon as the Sun slowly drops down or rises at the Equinoxes, giving more sunlight over the total year.
I suppose you also have to take into account the speed of the Earth around the Sun but without looking it up, I’m guessing it is about the halfway point at each Equinox, so it won’t matter between 90N and 90S.
Close?
90N, since the orbit is elliptical, perihelion during the summer, and the Sun shines also a bit below the horizon. Just a guess might be more complicated though.
Perihelion is in early January, hence northern hemisphere winter, southern hemisphere summer.
mcswell
Perihelion is in early January, hence northern hemisphere winter (Nov-Dec-Jan-Feb) has a higher solar radiation level at TOA than does northern hemisphere summer, southern hemisphere summer has a higher solar radiation level at TOA (top of atmosphere) than does southern hemisphere winter (June-July-August).
However, the tilt of the earth matters more to radiation at ground level than does the radiation at top of atmosphere.
My guess is latitude 19° 28′ 46N, where it intersects with Mauna Loa volcano.
With a clear 360-degree view to the surrounding ocean, that particular location more than 4 km above sea level is exposed to a few moments of extra sunlight every day.
And if you wanna talk about the impact of feedbacks… there is some evidence that the extra insolation at Mauna Loa is enough to melt rock!
Give this man a cigar! With earth surface rotation speed (at the equator) being 40,000 km/day or 1667 km per hour, it can be calculated that the peak of Mauna Loa (if located on the equator) would receive the first rays of sunlight at the same time as a point on the ocean surface 226 km to the east. 226/1667 x 60 is about 8 minutes. Multiply that by two for the same effect in the evening. Ok, Mauna Loa is not on the equator, and the extra light is at that very low angle but your insight wins the prize IMHO.
Kilimanjaro elevation 19341 feet, -3 S latitude.
Yes, in private communication Nick said he realized the Kilimanjaro is higher and closer to the equator but the image of lava wasn’t present, so he used Mauna Loa to demonstrate the principal.
Thanks, Greg. But it’s Willis’ question and he’s the arbiter and he says sunlight’s the same everywhere.
So, sigh…, apparently there’s no bonus for “thinking outside the oblate spheroid”.
Then again, I really didn’t deserve to win, because Kilimanjaro’s summit is supposedly FIVE kilometers above its surroundings. If that can be believed, because the feedbacks at Kilimanjaro should be the same and yet it has gone dormant. Of course most WUWT readers are aware that conjecture about the sign and strength of feedbacks on a handful of watts per square meter is a bit dodgy despite having been settled for several years.
Thanks, Nick, but I’m not the arbiter of anything. I’ve been wrong far too many times for that.
I was considering a smooth surface. I would imagine that Mt. Everest gets the most hours if we count mountains.
w.
PS—I did like “thinking outside the oblate spheroid” …
Cloud cover at 90N and 90S might affect the hours of sunlight.
Similarly cloud cover at Mauna Loa.
Yuma, Arizona , apparently takes the crown as the sunniest [directly measured] place on earth rather than those locations that are potentially the sunniest.
NASA had a go at answering the question and found that the sunniest patch of ocean was somewhere south of Hawaii and the sunniest land location was in the Sahara Desert in Niger
http://www.bbc.com/earth/story/20160127-which-spot-on-earth-gets-the-most-sunlight
Ah, the Kenai. That’s where I was born and raised. Your description was spot on. Enjoy the sun and fish.
Thanks, Josh, it’s a wonderful place.
w.
I suggest that the feedback model is not the one that should be used in assessing the reduction in in the radiation of energy to space. A parallel resistance model (with clouds and clear sky as the two variable resistances) is more likely applicable. The surface temperature of water and ice is being controlled by water cycle. Evaporation is an endothermic process so the surface temperature will tend to be at the dew point (air is saturated). Condensation in clouds occurs as water vapor transfers energy to air (at the dew point). There is not much difference between the dew point at the surface and the dew point at the bottom of a thunder cloud so there will not be much net radiation. However, when there are no clouds and the humidity is low the net long wave radiation to to the top of the atmosphere is relatively high. The electrical analogy would be the difference between the dewpoint at the surface and the frost point at TOA as the driving force or voltage and clouds as the big resistance with sum of the effects of water vapor and CO2 in clear sky as the low parallel resistance. Both vary on a daily basis with both time and space, allways seeking equilibrium at any one spot on earth.
–PPS—Bonus question. What latitude on the planet gets the most hours of sunlight per year?–
I would guess the middle of the Earth’s bulge, 0 Latitude, the equator.
Interesting answer. If you go S on the Texas coast day length increases in the winter, if you go N it does in summer so the solar panel on my trailer worked longer in New York. Since there was not as much night and it was cooler, the batteries and occupants appreciated it. Saw a picture somewhere with use of panels at high latitudes in snow, but didn’t see batteries.
In Texas summer it is too hot to measure until well after the sun sets.
What atmospheric “surface” is used for the downwelling? How far out? Radiation is emitted from a surface. No surface then it must be the voices in someone’s head.
Is the upwelling “surface” per WMS at 1.5 m above the ground or the actual ground?
The Instruments & Measurements
But wait, you say, upwelling LWIR power flux is actually measured.
Well, no it’s not.
IR instruments, e.g. pyrheliometers, radiometers, etc. don’t directly measure power flux. They measure a relative temperature compared to heated/chilled/calibration/reference thermistors or thermopiles and INFER a power flux using that comparative temperature and ASSUMING an emissivity of 1.0. The Apogee instrument instruction book actually warns the owner/operator about this potential error noting that ground/surface emissivity can be less than 1.0.
That this warning went unheeded explains why SURFRAD upwelling LWIR with an assumed and uncorrected emissivity of 1.0 measures TWICE as much upwelling LWIR as incoming ISR, a rather egregious breach of energy conservation.
This also explains why USCRN data shows that the IR (SUR_TEMP) parallels the 1.5 m air temperature, (T_HR_AVG) and not the actual ground (SOIL_TEMP_5). The actual ground is warmer than the air temperature with few exceptions, contradicting the RGHE notion that the air warms the ground.
Conclusion
So, the 396 W/m^2 upwelling LWIR and net 333 W/m^2 GHG energy loop of RGHE and the K-T diagram and RGHE claim that the air warms the ground are all illusions due to misunderstood & misapplies instruments.
Nick, you ask:
“Is the upwelling “surface” per WMS at 1.5 m above the ground or the actual ground?”
————–
I’ve been waiting on an answer to that question for some time. Last summer my IR thermometer read various surface temperatures ranging from 99 F (grass) to over 160 F (roof of car) while the ambient air was about 90 F.
I would guess that the difference could be even greater if the air temp was cooler and less humidity. LWIR is coming up from the surface. I have no idea if the ‘budget accounting’ simply uses 1.5 m air temps rather than actual but I’m watching for it. A 50 – 60 F emission temp difference is significant.
““Is the upwelling “surface” per WMS at 1.5 m above the ground or the actual ground?””
Solar SW is absorbed by the Earth’s surface (not at 1.5m) – so that is the ultimate emitting surface.
However LWIR photons are intercepted and absorbed by GHGs on their exit to space. Being re-emitted in all directions.
So it is an infinitely complex process of exchange – with the biased emitted direction for some distance that of the surface and as the photon gets higher in the atmosphere the direction of re-emission gradually gets biased to space. There is a region where this transition takes place, and it is where the atmosphere is at a temp of ~ -18C. The temp which matches the S-B relation to Earth’s absorbed solar energy. That point is where satellites see’s Earth’s temperature. And it is 255k and not 288k (average surface temp). That is the GHE at work.
The question remains unanswered. I agree with you that the emitting surface temp is actual reality. To refine the question just a bit, do the cartoons showing the radiation budget use the true surface temperature or do they use the so called surface temps that were measured at 1.5 – 2.0 meters?
The question remains the same, just a little more refined.
If the monthly radiation were to oscillate back and forth between (say) 506 and 512 W/m2 I would expect the temperature to oscillate back and forth a bit relative to what you’d get from the average of 506 and 512, but that (small?) temperature oscillation would not reflect what you’d get from a sustained 506, relative to a sustained 512.
If a small oscillation didn’t give what we’d get from a longer sustained value, we’d see a lag between the two.
w.
Which way would it lag?
The response (temperature) would lag the forcing (downwelling radiation).
w.
In my hypothetical example that would surely not happen, there would be only 2 temperature values, and it would be crazy if the higher temperature came along with the lower radiation.
A complication for interpreting the correlation is that the SW part of the radiation is probably largely the EFFECT of the temperature heating the atmosphere, rather than the CAUSE of the temperature variations.
“My guess on the latitude with most sunlight is 90N and 90S. The sunlight actually precedes the Sun poking above and below the horizon due to refraction. At 90N/S it will “hang” on the horizon as the Sun slowly drops down or rises at the Equinoxes, giving more sunlight over the total year.”
Well if 10 km elevation or higher.
Otherwise the sunlight has go thru many thickness of atmospheres, before getting to 90N and 90S.
And to get any sunlight, to reach a surface, the surface must be pointed at the Sun. So if the plane was vertical to the ground and pointed at sun.
With tropics, surface if level [and most is] that angle to have the plane at.
How about another theory. Due to the characteristics of the measuring instruments, an increase in surface temperature makes it look like the surface is absorbing more energy.
On the other hand, daytime clouds can reduce the amount of energy reaching the surface. In a desert climate, the sun dramatically warms the surface rather quickly. ie. the surface warms and cools appreciably between daytime and night.
It could be all about clouds and even the alarmists admit we don’t know as much about clouds as we should. link
Does removing Seasonal Variation reduce or hide lag?
Does the maximum Solar Radiation occur in June when the Highest Temps appear to be in August?
I drew the same conclusion, but then I started to think about the spatial resolution. You get a lot of incoming somewhere, or less incoming everywhere, and the average is the same, but the average temperature apparently is not. Interesting graph though, and takes a bit more of explaining given the ‘conservative’ let alone ‘worst case’ climate sensitivity estimates. Thanks for sharing, because the incredible stability of temperature does cast some healthy doubt on high end sensitivity speculations.
We have about the same sunshine as you there, and the max today was around +28C. The thunderstorm predicted missed us by ‘that much.
For the bonus question, how are we accounting for clouds?
If clouds reduce sunlight, I’ll go with 30 degrees north.
No clouds, Duncan. Just a count of hours of sunlight.
w.
In that case I’ll go with Santa. 90 degrees north.
Climate science doesn’t understand basic math. They are unaware of Holder’s inequality. You cannot take an ave flux for a sphere and convert it to temp via the Stefan Boltzmann equation. If you do you get a 1K increase for 3.7W.m^2. You have to calculate the flux for a billion points around the globe and then integrate that and then apply the Stefan Boltzmann equation for each one of these billion fluxpoints for a spherical surface.The correct answer is closer to 0.38 K as Willis has found out.
“No lag visible” – that’s because it’s leading.
That suggests that temperature leads radiation. That in turns suggests that cloud cover is being modulated by temperature – which explains why you’re getting 0.37C per doubling than 1.05C as is commonly quoted.
Why has this simple calculation never been done before? It undoubtedly has …. but it never got published because it did not fit the doctrine of the climate cult.
You’ve just put a stake in the heart of a >$1trillion scam.
Um…$10 to $100 Trillion.
The gross world product (GWP) is the combined gross national product of all the countries in the world equals to the total global GDP.
World economy, comprising 193 economies, in 2019 is projected around of US$88.08 trillion in nominal terms against US$84.84 trillion in 2018, according to IMF. Global GDP in terms of PPP (purchasing power parity, used by the CIA) is forecasted around of Int$143.09 trillion against Int$135.24 trillion in 2018. This number is 162% of nominal. PPP to Nominal ratio was 1.28 in 1990, all time lowest of 1.20 in 1995. But due to higher growth rate of developing economies, which have generally higher GDP in ppp terms, this ratio has increased.
In constant 2010 prices, which gives better idea about expansion over years, World economy has expanded from $11.33 tn in 1960 to $80.10 in 2017, or 7.07 times on the basis of market exchange rate as estimated by world bank comprising more than 200 economies. In purchasing power parity terms, world gdp has expanded by 2.46 since 1990, data for ppp terms available since.
http://statisticstimes.com/economy/gross-world-product.php
This century…not annually duh.
Another one. See Nickolov and Zeller and Noonworld.
WE, your second cross correlation chart is IMO a key to understanding your first. It does not say there is no lag. Take, for example, correlation 0.5. It says there can be a lag of ~ minus 0.5 to plus 1.0 years. Especially since ‘no lag’ is only 0.8. Since you removed seasonality, that surprising 1.5 year negative to positive range suggests something else is driving the Ceres system—and therefore also your first chart. If so, then the first chart shows a ‘spurious correlation’. One plausible ‘something else’ hypothesis is ENSO.
Fitting a simple multivariate multiple regression with surface T =radiation flux plus ENSO (you could just simply use pos, neg, neutral as the ENSO values: plus one, zero, minus one) tests that hypothesis by looking at the resulting significance of the Enso term coefficient.
I go with the South Pole
‘Scientists estimate that the pilot whale population in the eastern North Atlantic is above 700,000 whales, with approximately 100,000 around the Faroe Islands.’
Read more: https://metro.co.uk/2019/05/30/faroe-islands-defend-slaughter-whales-sea-turns-red-blood-9743755/?ito=cbshare