Bridging the Gap! Greenhouse Gas Theory Fail
In this article I am going to help push forward everyone’s understanding of the problem we true scientists and engineers face, when pushing back against the climate crisis lies being imposed upon us and help us all to WIN.
It is important that we do win, because we ARE being lied to on a monumental world wide scale and we need to arm ourselves with real and true knowledge in a manner that is not only right, but convincing and above all, can be shown to be right.
Now, it has quite often been stated by others that using an averaging approach to determine planetary surface temperatures is wrong and that we should instead use temperatures which are calculated using the intensity of the radiation which is incoming as using an average gives us temperatures which are just, “too low.”
For more on this you can read the book. https://www.amazon.co.uk/Slaying-Sky-Dragon-Greenhouse-Theory/dp/1901546373.
A lot of alarmists, when told this, just scoff and think it a bit ridiculous to make such suggestions, they should not do this and I am going to explain to you all why that is and show the quick and easy reason as to why that is, which anyone can do which shows exactly the mistake that averaging can lead to. You will see it can lead to some massive errors.
In my book (cover, top) I made the statement below.
“To think that the cold glass is making the ground warmer because of radiation is like saying standing beside a fire, and then saying that you’re making the fire warmer, because your skin radiation is going back to the fire. That’s just plain ridiculous.
When I have made this point to “Frizzlers” in the past, they tell me, that I don’t understand thermal dynamics of radiation, unfortunately for them I do. I know it precisely which is how I know they are wrong, I can do Mechanical Thermodynamics of radiation maths.”
And I am going to elaborate some more on how it is I PRECISELY KNOW that they are wrong.
In my book, I showed the 1st stage approximation approach to quickly determine steady state temperatures of objects in a vacuum, in space, in order to help educate people so they can be armed and fight back against climate alarmists.
0 = ά E A(a) – έ A (e) σ T4
I then also explain that.
“This is the equation which initially explains steady state temperatures of objects in space, before we start looking at specifics of how precisely energy moves from one part to the next. It is the first stage of determining temperatures of objects in space and it is an approximate approach. This shows that the energy in, must balance with the energy out, the principle of energy conservation.”
So Energy in must match energy out.
ά E A(a) = έ A (e) σ T4
I have underlined and put in bold the important bit you need to know. So let’s now, move a bit more onto some specifics and more precise corrections which need to be performed, to get a more accurate picture of what is occurring. This is what I call.
“BRIDGING THE GAP!”
Why? Because there can in many cases be a massive gap, in fact in some instances a gigantic disparity, between what an averaging approximation calculation shows and reality. Reality can be arrived at with more precise and properly performed calculations. And I promise, I will do it without using the dreaded word, “Flux.” 😀 ooooh a challenge.
In my book I showed some examples such a flat plate, a greenhouse and a cube using the average approximation approach. I showed, that even using an averaging approach, we can see that a plate with a greenhouse attached would be cooler than just a plate so how can people say “radiation greenhouse effect induces warming.” I was thinking about including some more examples and going into deeper detail but I didn’t want my book to be too overwhelming, so at that point decided against it.
Learning the basics is a necessary first step before you can move on to bigger and harder things. No one expects a baby to learn to walk by being whipped into a sprint.
So here I will elaborate a bit more, so we can all “fight back” in a better and more well equipped manner, making the gigantic assumption of course that you have all read and understood my book. If not, you should all rush out and grab one, it’s a great help in fighting off the alarmists I promise. 😉
If we “average approximate” the temperature of a flat plate in space with both sides exposed to space, we get the temperature of 331k. Because we assume that the plate is “emitting” (a more accurate word would be “losing”) energy on both sides. I.e There is incoming 1367 watts and therefore to balance this there is outgoing on both sides of 683.5 watts.
Such as in this picture here:
Diagram 1 – Planar Plate Calculation Averaged Approximation Approach
Now, this is where averaging starts to fall down and this falling down, replicates itself across all the “global warming models.” Which are by the way, quite clearly wrong.
This is because you can-not consider the side which is absorbing the energy from the sun, to be a “losing” surface. This is the side where energy is being “gained” by the object in space, not lost.
Yes, the object, the flat plate, is “emitting” on both sides and these emissions can be recorded, seen and measured using a IR meter, but that is not how you should look at it. You need to change your viewpoint. If you placed yourself in front of the plate in order to take a reading with your IR camera assuming you were on some sort of space walk, you would start to “Shield” the plate from the sun and it would have a reduction in energy gain and start to exhibit lower steady state temperatures.
The plate is gaining 1367 watts on the sunny side and losing 1367 watts on the dark side. It is not losing 1367 on both sides, that is wrong. How can the plate be losing energy to a great big ball of fire which is millions of degrees Celsius in temperature?
What you need to do, is split the sides between “gaining” and “losing.”
When all you have is an object in space and the sun, this is an easy task to perform.
So we modify the equation.
ά E A(a) = έ [A (e)- A(a)] σ T4
You see what I did there, I highlighted the important bit in red and bold. You deduct the area of absorption, (the gaining side) from your total area of losses. You don’t consider the sunny side to be losing heat, because it isn’t.
So let’s do that with our straightforward example of the plate in space.
This is the “Planar Plate Calculation”
The plate, floating in space, perfectly perpendicular to the suns rays, has 2 sides exposed to space. But the side of the plate which is receiving the suns rays, you can not count as part of your calculation of heat lost for determining steady state temperatures. So for a plate, what this means is your area for emission for losses, is cut in half. What does this mean? It means only the reverse side of the plate is used for determining heat lost to space, which means you end up with a much higher steady state temperature.
So for a plate in space, the steady state temperature is not 331k, but it is in fact, 394K.
This exactly matches the temperature you would expect if you had a flat plate and then put a block of insulation behind it, when using the average approximation approach. In fact, if we performed the calculation even more accurately, with the block of insulation it would be cooler, because there would be thermal conductance, transferring energy from the plate to the block, due to physical contact.
Diagram 2 – Corrected Planar Plate Calculation.
So what we actually get using a corrected equation for a flat plate is 394k, that’s a disparity of 63 degrees. That’s quite a lot.
And for those that want proof and complain about lack of using “peer reviewed reputable links, blah, blah, baby baby why did you link “I love my CO2 I read desmog blog”, cry whine baby, blah blah.” Read this and start to realise the truth, that you are being lied to by the alarmist loonies.
https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm
“The plate is in sunlight. Sunlight warms the plate, and the plate radiates thermal energy back into space.” & “Thus, the nominal temperature of an object, in space and in sunlight, is 394 °K.” are the important take home points you need.
This article produced by Nasa isn’t elaborating on the mistake alarmists make but it is pointing out that at a height of 300km a flat plate in space is not warmer due to friction caused by atmospheric particles and as such friction can be ignored.
However, it does show that 394K is the temperature you should arrive at for the plate in space, before considering friction. 331K, the averaging approach is just too low.
Are you as an alarmist, really going to embarrass yourself and argue against this? I’m sorry, did Nasa get their planar plate calc wrong, or are you a clueless nitwit? You truly that hooked on the corrosive climate fraud crack that you just can’t cut the habit?
Yet it is this averaging approach, which EVERY SINGLE ALARMIST, is using and relying on. All of them are. It is this error, which leads them on to make all sorts of daft proclamations, such as “tipping point”, “reductions in rate of cooling.” “Green-plate Effect” (something I addressed in my book and defeated whilst still adhering to the purely average approximate approach with all its errors) “Climate Crisis”, “Global Warming” “Radiation Greenhouse Effect” etc etc.
They are plucking random daft explanations out of a magic rabbit hat, in order to “BRIDGE THE GAP “ between reality and their mis-understanding of it.
They are just plain ignorant, it really is that bad. Or of course they know and are over-acting in order to conceal the lie.
This correction can be applied to any object, a cube or even a sphere, or yes indeed, our planet.
It is this correction, to the average approximation approach, which all global warmist’s have neglected to mention or even consider when explaining why the Earth is at the temperature it is at. They are blatantly lying to you when they say it is because of “Radiation Greenhouse Effect.” The disparity, the one caused by considering the “Gaining “ side as part of the “losing side” when performing average steady state equations is the explanation as to why there is difference between the calculated temperature of 255k which the false, lying, deceiving temperature alarmists harp on about and the reality we see an feel everyday of 288k or there-abouts.
Using an averaging approach for a sphere we would get a temperature of 279K. So lets do some quick math using the more accurate corrected approach, 1367per meter gained, 1367 per meter lost, but with only a hemi-sphere rather than a sphere for determining losing areas, making the wild assumption of isothermic conditions (which is clearly not a precise method) our sphere in space and ignoring all thermal mass, we would arrive at a steady state temperature, interestingly enough of 300k. A disparity of 21 degrees. Well, that’s your global warming greenhouse effect right there, with no gases. If we started to allow for albedo to reduce energy in & reflection, we can easily reduce to 288K, depending on what assumptions we made.
NOTHING TO DO WITH RADIATION GREENHOUSE EFFECT OR WITHOUT EVEN CONSIDERING IT.
The entirety of the Climate Claptrap industry is all based around hiding this fact and not revealing what is really happening. This is their dirty secret.
Radiation greenhouse effect, is a lie and you all need to understand this. The authorities have become fake, those we rely on to teach and look after us have turned against us and are filling us with lies on a daily basis. It is time for the deceit, fraud and corrosive climate corruption to end.
The problem, is one of lack of understanding and one of lack of capability (or in the case of alarmists –WILLINGNESS) to perform the calculations in a more accurate and correct manner. This is something the elite lying authorities long ago realised and hatched their schemes against us. Abusing the people on mass, brainwashing the mass populace to believe in fantasy, taking advantage of our lack of capability in figuring out their lies.
A capability which we all now have and exposing the truth is now something we can all do, when repelling alarmist liars.
This omission, by the alarmists in all their models and falsified research explains the world we see around us. Not Climate Crisis. Climate Crisis is not the real issue, Climate Crisis Fraud however, most certainly is. And this is best tackled by taking direct action against those whom perpetrate it, the law of every country in the world, needs to reflect this.
PRINCIPIA SCIENTIFIC INTERNATIONAL, legally registered in the UK as a company 
incorporated for charitable purposes. Head Office: 27 Old Gloucester Street, London WC1N 3AX.
Please DONATE TODAY To Help Our Non-Profit Mission To Defend The Scientific Method.
Trackback from your site.
Terence Bolger
| #
I’m sorry, this was too difficult to follow. This should have been proof-read by someone who is fluent in English.
Reply
geran
| #
The plate would be at 331 K, emitting 683.5 from both sides.
Bodies emit based on their surface temperatures.
Reply
Zoe Phin
| #
Hi Geraint!
Sorry to be a bitch, but …
‘If we “average approximate” the temperature of a flat plate in space with both sides exposed to space, we get the temperature of 331k.’
The insolation reaching a hemisphere is TSI/2 for ~12 hours only. Earth will emit TSI/4 for 12 hours of day, and store TSI/4 for the 12 hours of night
“You don’t consider the sunny side to be losing heat, because it isn’t.”
Well, actually, that’s not true. The loss is in the infrared, and the gain is in UV/Light. Some UV/Light is obviously reflected by albedo.
https://smd-prod.s3.amazonaws.com/science-pink/s3fs-public/thumbnails/image/radiation-budget-5.jpg
See? This occurs simultaneously. If the sunny side of earth did not emit infrared, the ball on the right would be all black.
Reply
Geraint Hughes
| #
You need to read the article again, you didnt understand it. There is no loss on the sunny side. If you do the calcuation as you say you get t=331k for a flat plate. If you do it how Nasa says, you get 394k. See their website. https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm
Reply
Zoe Phin
| #
Geraint,
1367 W/m^2 is normal radiation only. The hemisphere is not a perpendicular surface. You need to divide TSI by 2 for a hemisohere.
I work out the math here:
https://phzoe.wordpress.com/2019/10/31/toa-insolation/
Reply
Geraint Hughes
| #
Zoe, you say there isnt enough heat to make the average temp 300k for a sphere in space. I refer you to this link https://pdfs.semanticscholar.org/13ea/3d02e5f0e0f7e2732012ae37119d4b2e3f11.pdf If what you say is true, are you able to explain why the spherical ballon, has a iso thermic temperature of 307k, with the a/e ratio material 1.0, the grey paint? What is it you know that industry just does not? You telling me the table is wrong and everything on the link is rubbish?
Reply
Zoe Phin
| #
Geraint,
Did you read the article you linked?
“The balloon is assumed to be at a temperature of 300 K when deployed”
“Objects in daylight…”
“Equilibrium
Temperature of
Sphere in
Sunlight (K)”
“Objects at night”
“Thus, any spherical shell in low
Earth orbit in the Earth’s shadow
will have an equilibrium tempera-
ture of about 180 K. The cylindri-
cal object discussed above (with a
length and base diameter of
3 meters) will have an equilibrium
temperature of about 170 K.”
Reply
geraint hughes
| #
Zoe,
Learn to read. A / E ratio, in sunlight on the table the sphere in sunlight (in space) 307k. Explain that please. You just cut and pasted random bits and pieces, which on their own make no sense. Explain the table baby brains, go on.
Reply
Zoe Phin
| #
Geraint,
You need to explain why Earth emits on average ~390 W/m^2, not why a 3 kg ball can go from 300K to 307K for 12 hours.
JaKo
| #
Well, a very thin flat plate, perpendicular to the Sun rays and having some average emissivity/absorbance on both sides may not radiate that much into the side of the incoming radiation, but there would still be some heat loss — the outgoing radiation doesn’t have to be perpendicular to the plate and anything outside of the Sun’s apparent perimeter is still a black body of very low temperature (open space).
BTW, why doesn’t the NASA link work?
https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm
Reply
Geraint Hughes
| #
JaKo.
There is only heat gain on the sunny side. Like I say, it is emitting, but not LOSING.
Reply
Geraint Hughes
| #
This link works, i tried it. https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm
Reply
geran
| #
Link still doesn’t work.
Reply
Geraint Hughes
| #
I see it no problem, i have no idea what your issue is. Try the contents page https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/index.htm then click on the link for estimating the temperature.
Reply
geran
| #
I still can’t get it to work. It may be an older off-site NASA link that is not compatible with Safari. It appears to be some kind of indoctrination effort (k-12). Often such NASA efforts are filled with pseudoscience. I was wondering where you got the “gaining/losing” idea. I’ve never seen that before.
Zoe Phin
| #
I made an even prettier version of the insolation calculator:
https://phzoe.wordpress.com/2019/11/05/graphical-t-o-a-insolation/
One can easily add an albedo factor to the program.
Reply
Pierre D. Bernier
| #
Well, there is the hard way and the easy way to calculate it. For the hard way go to
https://climateofsophistry.com page… WATCH GREENHOUSE EFFECT REAL?
Search… The Sun’s Flux On Earth For Geothermalist Dummies
The easy way is …The Earth’s disk intercepts the Sun’s flux. That disk has a surface area of πR^2 but the flux will be spread over the hemisphere of surface 2πR^2. Hence, 1370 * πR^2 / 2πR^2 = 1370 / 2. Subtracting 30% albedo leaves 480 falling on 1 hemisphere at a time with 240 emitted by the whole sphere for equilibrium. We then have a potential of 30C coming in and a -18C going out as measured by satellites.
Reply
Zoe Phin
| #
Pierre,
“We then have a potential of 30C coming in and a -18C going out as measured by satellites.”
A new breed of climate scammers (possibly unaware of their actions) want to give the impression that you can use the same energy on the way in AND on the way out.
The SB law is for emission ONLY. It’s true that you get ~480 W/m^2 for a hemisphere as input – but only for 12 hours. Now what do you want to do with that input? It doesn’t have a temperature yet. If you want 30C then you will have to EMIT 480 W/m^2. After emitting 480 for 12 hours you have absolutely nothing left for the 12 hours of night! If you emit 240 W/m^2 for 12 day hours, then you will have 240 for the 12 hours of night.
You act like you have 480 for the day AND 240 for the night, but actually you have 0 left over for the night. You’re 24h average works out to 152K – brutally cold.
You and Postma are giving the impression that you have 30C coming in and -18C going out as a daily high to low. In reality your progression goes from 30C (12 day hours) to -273C (12 night hours).
You are subtly bumping the available energy with rhetoric. Wake up!
It’s as if you drank a liter of water, and then releived yourself of a liter of water, and now believe you processed 2 liters of water.
Reply
philf
| #
IPCC Intergovernmental Panel On Climate Change
GCM General Circulation Model (many, based on IPCC CO2 assertions)
These six links from five authors are all you really need to understand global warming.
My speculation: As the temperature went down into the Little Ice Age, limestone was deposited around the edges of bodies of water. As the temperature has recovered since, the limestone dissolved and added CO2 to the oceans, with a delay of 300-400 years. It was just an accident that this added CO2 coincided with our industrial revolution. Temperature creates CO2, not the other way around. There is proof of that. Read on.
Pangburn
Shows that temperature change over the last 170 years is due to 3 things: 1) cycling of the ocean temperature, 2) sun variations and 3) moisture in the air. There is no significant dependence of temperature on CO2.
https://globalclimatedrivers2.blogspot.com/
Connolly father & son
Shows the vertical temperature profile follows the ideal gas laws and is not caused by CO2. Millions of weather balloon scans and trillions of data points have been analyzed to come to these conclusions. One important conclusion is that there is no green house gas effect.
https://globalwarmingsolved.com/2013/11/summary-the-physics-of-the-earths-atmosphere-papers-1-3/
utube:
https://www.youtube.com/watch?v=XfRBr7PEawY
Pat Frank
Shows that GCM results cannot be extrapolated a few years, let alone 50 or 100.
https://www.frontiersin.org/articles/10.3389/feart.2019.00223/full
and
https://wattsupwiththat.com/2019/10/15/why-roy-spencers-criticism-is-wrong/
Joe Postma
Shows that the “flat earth model”of the IPCC is too simple. Their real models are built into the GCMs which don’t fit the real data.
https://climateofsophistry.com/2019/10/19/the-thing-without-the-thing/
Reply
Norman
| #
Zoe Phin
Even though your knowledge of physics if very poor (don’t lie you never got A’s in an actual physics class) this dude that wrote the article seems to have zero knowledge of any physics but is thinking he can make up his own versions by force of an egotistical personality.
Geraint Hughes if you did any actual physics like exposing one side of a plate to radiant energy you would see it does not get as hot as you seem to foolishly think. This is probably one of the most unscientific pseudoscience blogs I have ever seen. Most authors make up these totally bogus ideas and post them as if they were fact and they pretend they know physics by putting in a few established equations.
Reply
Geraint Hughes
| #
Norman, please tell NASA. They say the plate is 394K, what is your smart arse answer to that? https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm
Reply
Pierre D. Bernier
| #
zoe
Calculate for the surface -23.5 to 23.5 E-W by -23.5 to 23.5 N-S under the Sun’s zenith and you get a 1296 W/m^2 average. After albedo of 30%, gives us a heating potential of 907 W/m^2. SB Law says a mean radiative temperature of 355 °K or 82 °C. That’s very hot yet proving that the Sun alone can create the climate !!!
The lower atmospheric temperatures on Earth never go anywhere near that supposed 82 °C at any time. We know the energy to do it does come in. So where does it go ? The waters ! Stored for redistribution and our friend El-Nino. NO MORE NEED FOR GHGE OR GEO HEATING EFFECT.
Reply
Herb Rose
| #
Hi Pierre,
The energy from the sun is stored in the water in the atmosphere.
The heat in El-Nino is the same heat that is in the Gulf Stream. The rotation of the Earth and inertia of water causes the equatorial currents where the water travels east at a slower rate the rest of the globe, giving the appearance of a westerly flow at the surface. This slower easterly flow exposes the surface water in the currents to the energy of the sun for a longer period increasing the energy they absorb.
The Atlantic current strikes the coast of Brazil redirecting it north creating the Gulf Stream.
The Pacific current strikes the islands of the Philippines, Malaysia, and Indonesia blocking the flow and creating a pool of heated water. When the surface waster gets hot enough the Pacific current flows under the hotter surface water and causes that hot water to flow westerly on either side of the current creating El-Nino.
Herb
Reply
Zoe Phin
| #
Pierre,
You have a lot of trouble understanding the concept of time.
“Calculate for the surface -23.5 to 23.5 E-W by -23.5 to 23.5 N-S under the Sun’s zenith and you get a 1296 W/m^2 average.”
Yes, at that location you will get a T.O.A. insolation of 1296 W/m² for 47/360*1440=188 minutes. Did you know that a day has 1440 minutes?
“SB Law says a mean radiative temperature of 355 °K or 82 °C.”
SB Law is for EMISSION ONLY. If you want 82 °C, you will have to emit 907 W/m² for 188 minutes, leaving you with nothing for the remaining 1252 minutes. Of course we’re just talking about the zenith, so you have more energy. You can use my program to figure out 24hr avg insolation at your chosen location.
You’ve decided to stray from Postma, the half-earther, and become a 47/360th-earther. What a silly goof.
Reply
Pierre D. Bernier
| #
Zoe
/ It’s true that you get ~480 W/m^2 for a hemisphere as input – but only for 12 hours. /
Yet another show of your poor understanding.
What we have here is a snapshot in time where the Earth receives 480 W/m^2 on that hemisphere at that precise moment and radiates 240 W/m^2 from the whole sphere at that same precise moment. One second later, or 1 minute later, or 1 hour later, it is another hemisphere which receives those 480 W/m^2 while the whole sphere still radiates 240 W/m^2. We go round and round and round every second, day after day.
All energy is accounted for, in and out, and there is no need for a Greenhouse Gas Effect or a Geothermal Heating Effect.
Reply
Herb Rose,
| #
Hi Pierre,
You will learn what everybody else who has tries to teach Zoe something that it is an impossibility. “Mrs Wonderful” is so impressed with herself that she cannot accept that she is not infallible. It doesn’t matter who tells her that gravity does not preferentially attract denser objects she knows better.
Her whole geothermal nonsense is because she doesn’t understand the difference between an object absorbing energy and radiating energy. When the energy of the sun strikes the surface of the ocean it not only heats the surface but penetrates the water transferring energy to a volume of water, heating it. The longer red wavelengths are absorbed by the surface layer with shorter wavelengths penetrating deeper. The lower water is not only being heated by conduction from the water at the surface but also energy from the sun.
When the water radiates energy into the atmosphere it is a function of the surface area. The top layer must cool before the lower layers are able to lose energy. The absorption and radiating of energy by the Earth occur at different rates. There is stored solar energy in the Earth’s crust and before geothermal energy (excepting volcanos, bents, and hot spots where the added heat is radiated into space) can radiate energy into space the stored solar energy must dissipate. The greatest amount of energy from the sun strikes the equatorial region with less being absorbed in the polar regions, the greatest amount and depth of stored solar energy is in this equatorial zone. The geothermal energy radiates from the inner core in all directions to the Earth’s crust The areas of permafrost in the polar regions where the surface is thawed by the energy from the sun while at lower depths the ground is frozen shows that the entire Earth contains stored solar energy and is in equilibrium with the sun. Before geothermal can influence the surface temperature the energy from the sun must decrease enough to allow the stored solar energy to dissipate through radiation.
Herb.
Reply
Zoe Phin
| #
Herb,
I’ve debunked/explained all these points before. Why can’t you learn?
24hr avg @ surface, assuming standard NASA budget:
geothermal – 234.9 W/m² (close to blackbody)
solar – 163.3 W/m² (not even remotely close to a blackbody)
Total – 398.2 W/m²
The so-called “stored solar energy” is used up within the day (or year if you want to think about it that way).
Using Bernier’s math: 480 comes in for 12 hours. 240 is emitted and 240 is STORED for 12 hours. At night the 240 is used up. We need 398 emitted 24 hours a day – on average.
Reply
Zoe Phin
| #
Pierre,
“receives those 480 W/m^2 while the whole sphere still radiates 240 W/m^2.”
SB Law is for EMISSION ONLY. If the whole radiates 240 W/m², then it’s at -18°C.
“One second later, or 1 minute later, or 1 hour later, it is another hemisphere which”
The radiation received for 1 second, or 1 minute later, or 1 hour later, still has to last 24 hours.
“All energy is accounted for, in and out, and there is no need for a Greenhouse Gas Effect or a Geothermal Heating Effect.”
Non-sequitur. You didn’y prove anything, only showed you misunderstand time.
Even T.O.A insolation is ~340 W/m², forget albedo. Yet the surface is 390 W/m². There’s 50 W/m² missing, and we didn’t even use albedo.
Reply
Pierre D. Bernier
| #
Zoe
Why are you so hell bent on proving that you are from the alarmist’s side ?
All your doing is proving that you dont understand basic physics.
Reply
Zoe Phin
| #
Pierre,
“proving that you dont understand basic physics.”
Pure projection from a silly 47/360th-earther.
“Why are you so hell bent on proving that you are from the alarmist’s side ?”
Hilarious. Alarmists are geothermal deniers just like you. When it comes to geothermal denial, Postma and Alarmists are in bed together.
The math and physics shows what it shows. If it’s not obvious on Earth, it’s sure as heck obvious on Venus.
You’re sitting on a frying pan with the stove turned on. There’s a UV lamp also pointed down at the frying pan. You are shown that the UV lamp can’t generate the temperatures observed. You deny this fact and you deny the stove’s thermal power. It couldn’t possibly be the stove! lol
Reply
Pierre D. Bernier
| #
Zoe
I’ve asked you that question twice before on Climate of sophistry with no aswer. I’ll ask again… Who are you ?
1) An alarmist plant ?
2) A brat who can’t acknowledge anyone but herself ?
3) Someone who refuses to take her prescribed meditation ?
I’ll go for all of them !
When who ever employes you has finished with you I suggest you go get a job at Barnum and Bailey in the freak show depatment.
Reply
Zoe Phin
| #
Pierre,
I’m exactly who you see: someone smarter than you..
There’s now 3 positions:
1) Sun is not enough + GHGs
2) Sun is enough (no evidence)
3) Sun is not enough + Geothermal
Any conflation between positions #1 and #3 is SOPHISTRY.
If you try to claim I have something in common with alarmists, I will do the same for you:
Alarmists deny geothermal, just like you and Postma.
Get wise and learn how time works.
Reply
Pierre D. Bernier
| #
🙂
Zoe Phin
| #
Pierre,
I see you talking behind my back. and misrepresenting my arguments.
“The Earth receives 480 W/m2 of energy from the Sun on the exposed hemisphere. But since that the exposed side is hotter then the night side it will radiate back to space about 255 W/m2 while the night side will radiate only 225 W/m2. 480 W/m2 creates a potential for average temperatures of 30C thus creating the climate while the Earth is in energy equilibrium.”
SB is only for emission. You just said the emission never exceeds 255, ergo the sun is not enough.
The 30C only exists in your imagination.
You want to use the same energy on the way in AND on the way out. You can’t do that. Only out counts.
If you want 30C, you have 0K left over for the night … just reread all the comments here!
“I don’t know if the person I’m talking about is a sociopath/psychopath or if she’s just being paid by someone to act stupid and do that kind of shit but it would be nice if when such things are as clear as day that these persons would be banned.”
The only psychopaths I see are those who double count energy, deny it, and think the sun is enough.
Postma had me fooled for 2 years. Wake up!
Reply
tom0mason
| #
I. So the sun is not enough eh?
What are all the effects of our sun on this planet?
Explain not only the TSI figures but how ALL solar radiation, ALL solar particle emissions, ALL solar magnetic and electric field effect this planet. Show the timing of these effects, and the timescales of the natural lags this planet’s put on them.
II. So the sun is not enough eh?
How exactly do the oceans warm up?
How long do the oceans take to dissipate the heat they have?
As always with today’s version of ‘climate science™’ no due notice is taken of the timescale the real climate takes to environment variations that affect the planet. Thermohaline circulation, also called Global Ocean Conveyor or Great Ocean Conveyor Belt, the component of general oceanic circulation controlled by differences in temperature and salinity. How long does that thermohaline circulation take to do it’s round trip?
There is little in the way of research showing how, where, or when the oceans release their heat, all we have is SST measurements and estimations — we do NOT know from when and where that SST heat originated.
So can anyone show that they really know, quoting some research, that observes (not modeled nonsense) how and when the heat arrived in the oceans (last week, month, year, last 100 or 1000 years?), where it has traveled, and where the bulk of it is now. Also show (within 6 months) when it will affect the atmosphere.
Reply
Zoe Phin
| #
Tom0mason,
Unless you can show the sun creating an average 15°C at the surface … the sun is not enough, despite its enormous abilities.
Reply
tom0mason
| #
My point is and always has been we are operating from a point of ignorance.
How can we say that the sun is not supplying the 15°C you ‘believe’ it should when so much is unknown. Maybe that 15°C is just a mirage caused by science’s lack of knowledge/understanding and it’s antipathy towards investigating anything that is not of a human generated origin (like CO2, methane, etc.). Science has become bogged down in mindless reciting of what is not observable.
Just out of interest what do you (Zoe Phin) think the solar variations in particles and gravitational, magnetic, and electric effects have on the earth’s temperature and climate?
And what evidence is there for the ocean heat influencing the current SST is not a product from the Medieval warm period?
So as I have said to you before ‘go ahead and do your math’ it does not prove anything if the observations are not made.
Reply
Geraint Hughes
| #
The math as I explained is sufficient. At some point I will build a rig, which demonstrates this also.
Reply