Saturday, 30 May 2020

Another little mystery solved!

Now that I have finally stumbled across the basic physics (which the Consensus keeps well hidden) that tells you 99% of what you need to know about why there has to be a lapse rate (and hence why the sea level temperature is warmer than you'd expect from solar radiation alone) and how to calculate it (between 8K/km and 10K/km, depending on which method you use - water vapour and water droplets moderate that to the observed 6.5K/km lapse rate), lots of other things just fall into place.

The infamous IPCC 'world energy budget' diagram (scroll down here) shows that the troposphere is emitting twice as much IR radiation back down to the hard surface as it is radiating to space. So with a straight face, the Consensus segues into the claims that 'greenhouse gases' are trapping heat and/or actually reflecting it, in other words the IR bounces back and forth between hard surface and troposphere, the 'greenhouse gases' act like a layer of insulation etc.

This all seems very implausible to me. In any sane world, a hot object emits IR evenly in all directions. The Consensus are also confusing cause and effect - the lower layers aren't hotter because they are absorbing and hence emitting more radiation; they are hotter anyway because of the inevitable lapse rate, so of course they emit more radiation!

The effect is actually measurable though, so it's a question of finding the actual explanation. There is no point picking holes in a flawed explanation (I wasted far too much time and mental effort on that), just start again with the basic physics (and get cause and effect the right way round!).

The actual explanation is - as ever - quite simple.

We've already established that the actual surface of Earth is not the hard surface, it is the troposphere (and the top few inches of land or ocean that are part of the troposphere for calculation purposes). The average overall temperature of the troposphere is - unsurprisingly - the same as the 'effective temperature' (the temperature you can calculate on the basis of the amount of solar radiation coming in adjusted for albedo, on the assumption that there is no atmosphere). The temperatures lower down are higher and the temperatures higher up are lower than the effective temperature, because of the lapse rate.

So the troposphere at sea level is 288K. This is - according to the Consensus, and I am taking their word for it - emitting twice as much radiation (half of which hits the ground) as the layer of the troposphere that is high enough to be emitting radiation directly to space (half of which actually goes to space).

You can easily work out the temperature of the layer higher up that is emitting radiation to space, and then you can work out its likely altitude.

Just divide sea level temperature (= 288K) by the fourth root of 'double' (assuming the "twice as much" figure to be reliable) (= 1.19) = 242K.

How high is that layer?

Deduct 242K from 288K (= 46K) and divide that by the expected/calculated lapse rate (between 8K/km and 10K/km, call it 9K/km for sake of argument) = 5.1 km.

The Consensus figure is "about 5 km", so we are not far off!

Thursday, 28 May 2020

Fun with numbers - where does the sunshine go?

To try and put the numbers into perspective, I did some workings.

Step 1. Multiply up the amount of incoming solar radiation per second (in Watts/m2, i.e. Joules/second/m2) to find out the total number of Joules each m2 gets in a 12-hour day (= 20.6 million of 'em).

Step 2. Look up mass/kg and specific heat capacity (Joules/kg/1 K) for air, wet soil, and water.

Step 3. Adjust/tweak the main variables until you get 'sensible answers' in the last column. The main variables are:

a) how those 20.6 million are split between air/soil and air/water. (answer 80/20 and 20/80 respectively, partly due to albedo and partly to do with how good soil and water are at moving heat downwards or back up into the air again)

b) the height of the column of air which noticeably warms during the day (answer 800 metres*)

c) how far down from the surface the soil warms up (answer 4.5 inches)

d) how far down from the surface the oceans warm up (answer 39 inches).

The 'sensible answers' are that the soil surface/the air above it warms by 16K during the day; the ocean surface/the air above it warms by 4K during the day - which is why in the day time you tend to get onshore breezes and at night you get offshore breezes. The Earth is two-thirds covered in oceans and that averages out to 8K.

* Clearly, there's not a clear cut-off of 800 metres altitude. So for example, maybe the lowest layer above the land warms by 16K, 800 metres up it warms by 8K etc, and at 1.6 km (about 1 mile up), the air barely changes temperature from day to night. Same applies to soil and water, going downwards. If you are building a sand castle, you don't have to dig very far down before the sand is noticeably colder than at the surface. Go for a swim in the afternoon, the top few inches are pleasantly warm; stand chest deep further out and your feet get cold.

At night, the reverse happens, and the lapse rate flattens again. In extreme cases, the land cools so far and so fast that it drags down the temperature above it so far and so fast that you get a temperature inversion, i.e. warmer air over colder air, that's like a negative lapse rate.

Which is all good fun, but what is the relevance?

Firstly, that you don't need to worry about quite how or why energy/heat is absorbed, transferred or distributed (conduction, convection/down drafts, mixing or wind/currents, radiation, latent heat of evaporation/condensation). The sun sends us a certain amount of total energy and it warms stuff up, and we can reconcile/estimate how much stuff is warmed up by how many degrees K. Sometimes the obvious answers are the correct ones and need little further investigation.

Secondly, what it reminds us that is the daily variation, based on incoming solar radiation, is relatively small compared to the absolute temperature. At its coldest (just before dawn), the surface is (say) 284K and at its warmest (mid/late afternoon) it's 292K.

Which, as ever, makes me question the Consensus obsession with this chart. That particular one is gloriously mislabelled as "Earth's annual and global average energy budget". It's not! That's the global average energy budget per second! They don't even understand their own propaganda.

What the Consensus is trying to do is explain that you can and should work out how many people are in a shop by looking at how many go in or come out every second (or in their terms, the average difference between the number people going in and coming out, which must be zero anyway, hence meaningless). Sure, it gives you a guide, but you'd also have to know roughly how long each person remains in there. If ten customers enter and exit a corner shop every hour, there will only be one or two customers in there at any one time. If ten customers enter and exit a large car show room every hour, there might be about ten customers in the show room at any one time. (Ignoring the lock down rules).

The "customers entering and exiting" are like the sunshine that arrives every day, which is sufficient to warm the soil/ocean surface and the air above it by 8K on an average day; it cools down again by 8K on an average night. The infamous chart gives you no clue whatsoever as to what the baseline average temperature is ("the number of customers actually in the shop").

So why not just count the actual number of people in the shop (the baseline average minimum temperature)?

The answer to this is not particularly difficult: stuff warms up and then it cools down again. Basic physics. The smaller the surface area relative to the volume/mass, the slower it is to warm up or cool down. Warmth from the sun can't get very deep into the soil or the ocean, so for each 1 m2 of surface, there's a column of troposphere with a volume of +/- 11,000 m3, which can only lose heat to space (counting the stratosphere and higher layers as 'space') via the 1 m2 at the top.

Most of the energy (kinetic energy, potential energy or latent heat of evaporation/condensation) in the air and top bit of land and oceans (which is effectively part of the atmosphere for these purposes) is left over from the previous day; and most of what was left over from the previous day was left over from the day before that ad infinitum. Mathematically, this energy has a half life of about 24 days.
These ramblings have now reached full circle. My guess is that the Earth's atmosphere is set up to radiate a certain % of its energy every 24 hours. The equilibrium temperature is therefore where the extra sunlight that comes in during the 12-hour day is equal to the amount lost during the subsequent 12-hour night. If Earth's surface is radiating 2.78% (relative) every 24-hours, and is receives 8K's worth (absolute) when the sun is shining on it, then the equilibrium is 8K ÷ 0.0278 = 288K. Something like that.

Wednesday, 27 May 2020

Water, water everywhere

From the Arizona Daily Independent:

Water vapor is a powerful greenhouse gas, but its net effect in the atmosphere is to lower temperatures party because of convective heat transfer. Proponents of anthropogenic global warming (AGW) and most IPCC climate models assume the opposite (and that’s why climate model predictions diverge from reality).

AGW hypothesis: Carbon dioxide, a weak greenhouse gas, begins warming the planet. This warming evaporates water and so puts water vapor into the atmosphere which amplifies the warming effect. This is called a positive feedback.

At first look, this proposition seems logical and reasonable. But other properties of water vapor reduce temperatures and the net effect is a strong negative feedback. A positive feedback tends to destabilize a system, whereas, a negative feedback tends to keep a system in check. Just think for a minute, if water vapor had a net positive feedback effect, this planet would have had run-away global warming long ago. That alone should falsify the positive feedback hypothesis. But let’s look at some observational evidence for a negative feedback.

The graphic below compares four pairs of cities, each at about the same latitude so that each pair receives about the same amount of sunlight, and the cities are inland, away from possible tempering by sea breezes. The data is from the National Weather Service (the temperatures have been corrected for elevation differences). The difference between the pairs is that one city is in an arid climate, the other is in a humid climate. We see that the more humid city in each pair has a lower average annual temperature. The addition of water vapor to the atmosphere has a cooling effect in spite of water vapor being a greenhouse gas much more powerful than carbon dioxide.

Monday, 25 May 2020

Climate science - as easy as A, B, C.

We all accept that the actual average sea-level temperature of the Earth (288K) is about 33K higher than its 'effective temperature' i.e. what it would be if Earth had no atmosphere (255K). That's basic physics - average incoming solar radiation in W/m2 minus amount reflected as light (known figures) raises temperature of the 'surface' (as defined - see below) to whatever it needs to be to radiate the same amount of W/m2 back out to space.

The Consensus is that the entire 33K discrepancy is due to the presence of minor trace gases in the atmosphere, such as the +/- 2% water vapour or 0.04% CO2 ('greenhouse gases'). The surface converts short wave visible radiation from the Sun to longer wave infra red radiation, and this is trapped/absorbed or reflected by the 'greenhouse gases', which in turn warms up the surface even more, so it emits even more infra red in a vicious circle. Here is a typical article explaining this, which includes the energy budget diagram which doesn't add up as a bonus. Parts of that explanation are correct, parts are guesswork and assumptions, it is riddled with contradictions, leaves a lot of loose ends and unanswered questions, and overlooks some important basic physics.

Actually, if you sift out the basic physics from the guesswork and just apply the basic physics, it is quite easy to work out that the hard surface should be 45K - 50K warmer than the effective temperature. The formulas and calculations are the same whether or not there are any 'greenhouse gases', which suggest that they do not increase temperatures. As a matter of fact, there is only a 33K discrepancy, because water vapour and water in the troposphere have a moderating influence and reduce surface warming and temperature fluctuations (which accords with everyday experience and like-for-like comparisons of temperatures in humid and dry areas).

They could and should teach this as part of GCSE level physics, it wouldn't take more than two or three lessons.

A. Barry

You can easily work out that there are about 10,000 kg of air (101,325 Pascal ÷ 9.807 m/s/s) for every m2 surface. We know that 1 m2 of air at sea-level has a mass of 1.293 kg. So if atmosphere were same pressure and density all the way up, with a hard edge, it would be 7.7 km high.

You can guess intuitively that the atmosphere gets thinner as you go up and gradually tapers off into space and that there is no hard edge. So let's assume actual average density is half that at sea-level, a reasonable guess is that most of the mass of the atmosphere is up to an altitude of 15 km or so, and atmospheric pressure falls by about 7% for each km you go up, at least for the first five or ten km (which is not far off actual measurements).

If you want to calculate this properly, you use the Barometric Formula (or 'Barry', as I affectionately call it) which is based on actual ideal gas laws and gives you reasonably accurate predictions for pressures at different altitude, at least for the troposphere (which is all we really care about, i.e. is the bottom 11 km, others say bottom 13 km, it's thicker at Equator and thinner at the Poles). The formula is very clever. I can just about understand how they work it out, but I would struggle to reverse engineer it or explain how to derive it.

B. The lapse rate

In an intuitive way, you can also guess that the temperature at the top of the atmosphere is close to the temperature of the nearly empty vacuum of space, which is either close to 0K or has no measurable temperature at all, depending on your point of view. So temperature falls the higher you go, as anybody who has been up a mountain knows.

Remember that 'energy cannot be created or destroyed, it merely changes from one form to another'. Air at sea level as thermal energy (aka kinetic energy) and no potential energy (it can't fall any further down). Air higher up has the same amount of total energy - less kinetic energy and some potential energy. It is reasonable to expect the total amount of energy to be the same at different altitudes.

Once you accept this, you can work out the lapse rate. I'll show you how, just for fun and because it is important:

Potential energy in Joules = mass x gravity x height.
So J = m x g x h

Joules required to increase temperature of 1 kg of a substance by 1K = specific heat capacity ('cp' ) of that substance.
So T = J ÷ (m x cp)

We can simplify/merge those two equations as follows: T = (m x g x h)/(m x cp); cancel 'm' top and bottom; T = g x h/cp; divide both sides by h; T/h = g/cp.

(Thanks to Tallbloke for this short-cut - Wiki gives an explanation which is almost impenetrable to the layman, although it ends with exactly the same formula)

1 kg of air which is 1,000 metres higher up has got 9,807 more Joules of potential energy than 1 kg 1,000 metres lower down, so the air lower down must have 9,807 more Joules of kinetic energy (and vice versa).

How much warmer is the air 1,000 metres lower down?
Specific heat capacity of air = 1,006 J used/needed to increase 1 kg of material by 1K
T/1,000 = 9.807/1,006
T/km = 9.75K
Hence the predicted lapse rate = 9.75 K/km altitude.

[I think it makes more sense to a) use the specific heat capacity for constant volume rather than constant pressure, and b) to calculate J/m3 rather than J/kg. That means first using Barry to find pressure, at different altitudes and then finding the temperatures which balances Joules of kinetic energy and Joules of potential energy (basing calculations on density at different altitudes, not on pressure, so you need to know pressure and temperature to work out density, and the temperature is the thing you are looking for!), which is why the lapse rate I worked out was 8K/km. I'm not sure if my logic on a) and b) is 100% sound, but my method gives an answer which is closer to the real world observed typical rate of 6.5 K/km, so I'm happy with my method for now.]

C. What is the surface of the Earth?

The Consensus give a nod to Barry and the main reason for the lapse rate. They don't deny they exist, they give the formulas but then downplay them as irrelevances and draw no conclusions from them. It's like the road sign saying turn left and the sat nav saying turn left, but turning right anyway. It's all about radiation from the surface being reflected back down by those dastardly 'greenhouse gases'! Any other explanation is heresy!

The Consensus' most heinous and borderline criminal obfuscation is in their definition of the 'surface'. They define 'the surface' as the hard surface at sea level, or the surface of the oceans (two-thirds of Earth's surface is ocean).

Here are a few reasons why that is wrong:

1. If you approach Earth at speed from space, you feel it when you hit the atmosphere. That is the real surface.
2. If you scale down the earth to the size of a football, the troposphere is only 0.2 mm thick. You think it's thick when you look up at an airliner, but it's only 11 km away, you could drive that far in a few minutes.
3. There are about 10,000 kg of air (mass) for every m2 surface (see above)
4. The sun only warms the top few inches of the hard surface (or water), that's a few kg of mass per m2. So if the atmosphere is 0.2 mm thick, the hard surface or sea-level is barely a couple of molecules thick.
5. When we talk about the surface of the ocean, we mean the top of the water, not the hard surface at the bottom.
5. As far as heat distribution goes, we might as well treat the top few inches of the hard surface as the bottom part of the troposphere. It's usually the same temperature, and must be the same pressure.

Therefore, the real surface is the whole troposphere, not the hard surface or the surface of the oceans.


The troposphere is the surface, and as a whole and on average, is the temperature you would expect from incoming solar radiation = +/- 255K. This is what you would expect, and this is what you get. I don't see why anybody taking this view should be on the defensive in a discussion. It is those trying to say otherwise who are struggling.

The troposphere itself, it is not a constant 255K. There has to be - and there is - a lapse rate. There are different ways of calculating/predicting it, and it can be observed/measured, so it's about 33K warmer than 255K at the bottom (the hard surface); it's about 255K half-way up (as defined); and it's about 33K cooler than 255K at the top of the troposphere.


This A-B-C easy GCSE-level approach also explains a lot of things which the Consensus explanation can't and doesn't (and by and large, just glosses over to save embarrassment), for example:

1. Why the top of the troposphere (or the peak of a very high mountain) is colder than the 'effective' temperature, i.e. colder than it would be if Earth had no atmosphere.

2. Why the day/night temperature range on Earth (+/- 15K) is so much smaller than the day/night temperature range on the Moon (+/- 300K).

3. Why, despite the 'greenhouse effect', Earth's day-time temperatures at the surface are lower than what they would be without an atmosphere.

4. Why the 'greenhouse effect' is much stronger at night (i.e. actual temperature minus expected temperature of the night side of Earth if it had no atmosphere) than in the day time (when the Sun is blazing down on us and there is plenty of radiation sloshing about).

5.  Why 'heat rises' is a truism only observed in enclosed spaces kept above the temperature of their surroundings (central heating in buildings; actual greenhouses/polytunnels) and why convection doesn't actually transfer heat upwards. For sure, there are thermals above hot surfaces (like square miles of dark tarmac at airports, which can make landing trickier than it need be on hot days), but for every molecule that goes up, one has to come down. One molecule converts kinetic energy to potential energy and the one coming down does exactly the opposite.

6. Why there is no need to get tied in knots over which methods energy (in its various forms - visible and infra red, kinetic energy, potential energy, latent heat of evaporation/condensation etc) is distributed in the troposphere (conduction, convection or radiation). All you need to know is that energy tries to distribute itself as evenly as possible (governed by the physical laws discussed here, or by winds/weather, if you want an everyday term for a complicated process).

7. Why the troposphere emits twice as much radiation towards the ground than it does out into space. This cannot be satisfactorily explained by 'greenhouse gases trapping and/or reflecting heat', it is because the troposphere at sea level is warmer and hence emits more radiation that the layer higher up, which is colder and so emits less radiation, see here.

8a. Why it is irrelevant that N2 or O2 are transparent to, and cannot absorb or emit infra red (even if this were true, which is questionable). They can certainly warm up, and they in turn keep the hard surface at the same temperature. So even if N2 and O2 aren't emitting infra red themselves, the hard surface converts that warmth back to radiation anyway. The total infra red leaving the hard surface is the same whether it is bouncing back and forth as infra red between hard surface and troposphere (the Consensus), or whether the hard surface has to convert kinetic energy from N2 and O2 back into infra red first (the actual explanation).

8b. Why, even if the troposphere were indeed completely transparent to and unaffected by infra red radiation, incapable of absorbing or emitting it, the temperature at the hard surface would be the same as it is now. The hard surface would quickly reach 255K (accepted by the Consensus) and it would warm up the troposphere by conduction and convection until the whole troposphere (the effective surface) were 255K on average (it being incapable of radiating heat to space, that's the Consensus). That 255K would be the average, there would still be a lapse rate and so the temperature of the hard surface would increase to 288K (and the top of the troposphere would be about +/- 222K). The hard surface would then be warm enough to emit the required amount of infra red straight through the troposphere and back into space (not being able to lose any more energy to the troposphere by conduction or convection).

9. Why there is a lapse rate on all planets with an atmosphere, even Gas Giants (Jupiter, Saturn), which have no hard surface (and if they do, radiation from the Sun never gets there) and which consist mainly of 'non-greenhouse gases' (mainly hydrogen and helium); why they are insanely hot at their centres; and why those Gas Giants are actually emitting more radiation to space than they get from the Sun.

10. Why the 'greenhouse effect' on Mars is barely measurable (max 5K), even though there is forty times as much CO2/m2 surface area than there is on Earth. And why, if you go to Venus and hover at the altitude where atmospheric pressure is the same as Earth's sea-level atmospheric pressure, the temperature is the same as at Earth's sea-level (having adjusted for the fact it is closer to the Sun). This is even though there is still an enormous amount of CO2 above you!

Sunday, 24 May 2020

Weekly deaths - all causes - E&W - 2020 - up to week 19

Data from the ONS.

My assumption is that at some stage later in the year, the red line will be undershooting the other two. But we shall see.

Saturday, 23 May 2020


This whole topic is fascinating.

Donald Trump just plucked it apparently out of nowhere. Outlets like The Guardian poured scorn on the idea (this article is a joy to read). Some doctors say it helps or might help; others say it's nonsense. A lot of people say they took it with no side effects (but we'll never hear from those took it and died). Many small scale trials have had mixed or inconclusive results.

Ever the contrarian, I'd be delighted if it does help. Even a stopped clock is right twice a day (or in Trump's case, a stopped calendar, he's right once a year). As a human being, I'd be delighted it they find anything that helps, I don't know or care what that might be.

However, it appears that some researchers are taking it seriously enough to do a very large scale trial. Possibly for the satisfaction of proving that Trump is an idiot, but hey. No experiment has ever 'failed'. You do them to find out whether anything happens and if so, what happens. So even if nothing happens or the opposite of what you expect happens, that experiment has not 'failed' on its own terms. It has advanced human knowledge, however slightly.

But... are these trials really necessary?

We've been running a large scale real-world trial for decades - the drug is regularly taken against malaria, and is taken by people suffering from lupus or rheumatoid arthritis. Why don't we just go back and find out what the COVID-19 infection and death rates are among the hundreds or thousands (millions?) of people who are taking it anyway for other reasons?

Friday, 22 May 2020

Questions, questions...

Missing figures - how to make the IPCC's figures add up

Their 'energy budget' was the topic of my two previous posts. If you summarise it, you get the following table:

That is the average for any 24-hour period in total. Problems arise if you try to dis-aggregate it into 'day' and 'night' (see previous post).

1. I think they didn't understand latent heat of evaporation (they show that the surface gives 80 W/m2 to the atmosphere) and the heat/potential energy transfers (they refer to this as 'thermals' and show that the surface gives 17 W/m2 to the troposphere).

Latent heat and potential energy can't be expressed in W/m2, they are expressed in Joules! They can't be measured with light sensitive equipment or with a thermometer! Some heat at the surface is converted to other forms of stored energy entirely.

Like trees - they capture a lot of light and some heat from the sun and convert it to stored energy i.e. chemical energy. This type of energy cannot be measured with light sensitive equipment or a thermometer! You can however convert this stored chemical energy back to a lot of heat and some light by burning it.

So while the surface is losing some its W/m2, those W/m2 are not being added to the troposphere, and they just drop out of the equation. When water condenses or there is a down draft (to match a rising thermal elsewhere), the energy is converted back to heat, which warms the surface, enabling it to radiate more Wm/2 again. There is no one-way transfer from surface to atmosphere, the W/m2 disappear from the surface and re-appear at the surface when and where the condensation and down drafts happen.

2. There are at least three more or less mutually exclusive theories as to what happens when infra red radiation hits a 'greenhouse gas' molecule and the diagram mixes and matches. It can't do all three things at once, and none of them actually increases the amount of energy:

A. Absorption
Molecule briefly absorbs the photon which increases the molecule's kinetic energy; before it has time to re-emit the photon, it collides with an N2 or O2 molecule, and increases the kinetic energy of those. We are told that all warm objects emit infra red except N2 and O2, so they are stuck with the higher kinetic energy*. This leads to warming. Fine, we know you can warm things up by shining infra red at them, so this in isolation makes sense.

(* Which leads me to a different train of thought, if N2 and O2 can only warm up - conduction and convection from the surface - but never radiate infra red, how would they ever cool down, even at the top of the atmosphere? If you think this through, the atmosphere would actually be a lot warmer than it is if it had no 'greenhouse gases'. Which is almost certainly nonsense, but shows up the theory to be nonsense as well.)

But, as with the conversion from actual heat to latent heat or potential energy, conversion from infra red radiation to heat takes those W/m2 out of the equation and parks the energy sideways in a different equation (same as latent heat, potential energy or trees). Temperature is measured in K, not in W/m2.

The diagram does not take this into account, so it actually shows no warming effect.

B. Scattering or re-radiation
The molecule absorbs and re-emits the infra red photon, so it scatters it or re-radiates it at random in all directions. Fine, this also makes sense in isolation.

If you start with a random photon somewhere in the troposphere moving in a random direction, you know that there are slightly more molecules per unit volume beneath it than above it. Therefore - you would reasonably assume - it is ever so slightly more likely that it will be bounced upwards by a molecule below it than being bounced downwards by a molecule above it.

Analogy - in the pub, people crowd round the bar. You are standing quietly drinking your pint. If you move away from anybody who jostles you, you will end up at the other side of the room. But I don't know how to adjust for this - the atmosphere is radiating a total of 199 W/m2 to space, so let's assume it is also radiating 199 W/m2 back to the surface.

C. Reflection
The theory appears to be that the molecule reflects the infra red photon directly back down again, the same way that is gets noticeably warmer (and slightly brighter, if you are in a back-lit urban area) on a clear night when a large thick cloud passes you at the right altitude.

This seems the least plausible theory.

But the diagram relies heavily on the theory of reflection. Out of the 332 W/m2 'back radiation', 199 W/m2 can be explained by theory B. above, meaning the average extra 133 W/m2 must be reflected - it is the only way to reconcile their figures.

It's worse when you dis-aggregate it into day/night: by day, the atmosphere is reflecting 13% straight back down, which seems just about plausible. But at night, the atmosphere is reflecting 45% straight back down, which is clearly nonsense.

Put the two together, and that's why they show an ugly grey band of 'greenhouse gases' at cloud level which appears to bounce 85% of the radiation from the surface straight back down again.

3. But never mind, I'm assuming that enough of their numbers are honestly and accurately measured or calculated for my version to be plausible - at least on their terms. I have used their figures as far as possible, but I had to change thermals from 17 W/m2 to 31 W/m2.

If the table looks a bit blurry, click it to see it clearly.

Thursday, 21 May 2020

Missing figures round - how do you make the IPCC's figures add up?

Here's our baseline energy budget, averaged over 24 hours. I have been battling with this for the past 48 hours (hence no post yesterday) and have pinned down its fatal flaw:

We see that the Earth's surface receives, on average, 493 W/m2, consisting of 161 incoming solar radiation (ignore the one-eighth that is reflected straight back) plus 333 back radiation. The average temperature of the surface is 288K. That's our starting point.

If we had no atmosphere, the Earth's surface would receive 298 W/m2 incoming solar radiation, i.e. the 341 minus the one-eighth that would be reflected straight back, and its temperature would be about 250K - 255K (opinions differ). That's another fixed point.

So as a check, to see if we are on the right track, let's see if we can what calculate Earth's average surface temperature would be without an atmosphere.

You convert W/m2 to temperature in K as follows. You need to have a starting position (493 W/m2 and 288K). If W/m2 change, then the change in temperature to the fourth power is proportional to the change in W/m2.

298/493 = 0.605
0.605 x 288K ^ 4 (6.88 billion) = 4.16 billion
4.16 billion ^ 0.25 = 254K.

Excellent! It is widely agreed that the average temperature of the Earth, if it had no atmosphere, would be around 254K, so it stacks up so far.

Now, let's do the same exercise for night-time and day-time (working backward from temperature to find W/m2).


Typical average Earth night-time temperature is (say) 7C = 280K.
280K^4/288K^4 = 0.893. 0.893 x 493 W/m2= 441 W/m2.
Therefore, Earth's surface must be receiving 441 W/m2 at night.
At night, the earth is clearly getting no incoming solar radiation.
Let's assume the back radiation is the same (333 W/m2).
That gives us a missing figure of 108 W/m2.

Does anybody know where this missing 108 W/m2 comes from?


To have an average day-time temperature of (say) 22C = 295K, total incoming radiation must be 541 W/m2 (work it out yourself).
It's seems fair to assume that during the day, the Earth's surface is getting twice the average incoming solar radiation (that is how it is calculated) = 161 x 2 = 322.
The Earth's surface must be getting at least as much back radiation by day as it does on average = 333 W/m2.
That's a total of 655 W/m2 incoming. This would give an average day-time temperature of 309K = 36C, which is clearly nonsense.
To get the 541 W/m2, the amount of back radiation must be 114 W/m2 less than the average, which is also clearly nonsense.
UPDATE - I think I have cracked it, see next post.
IMHO, the whole thing is a load of nonsense, it just crumbles under closer inspection, so using it to try and reconcile know and sensible figures is like trying to prove that the square root of the colour blue equals a banana.

The correct way to explain/reconcile Earth's average surface temperature with incoming solar radiation - which works by day or by night, with or without an atmosphere - is intuitive, relatively easy to calculate (once you know how), and completely different to their model.

So my next challenge is to re-state all their figures to reconcile W/m2 and temperature for the overall average position; the position if we had no atmosphere; night-time and day-time. I think I know what they missed, possibly deliberately (i.e. what they missed is the latent heat of evaporation/condensation and the temperature/potential energy trade offs when air rises or falls). But maybe I'm wrong :-)

Tuesday, 19 May 2020

Sunny side up!

From, a diagram showing how the total amount of radiation received at the Earth's surface is more than incoming solar radiation:

Fair enough, this implies that day time temperatures are higher because of The Greenhouse Effect. Seems plausible. If you don't think about it for more than a few seconds.

Problem is, it ignores facts.

Our Moon is the same distance from The Sun as we are. The max daytime temperature is 127C and the night-time low is -173C (from here). That's a day-night range of 300C.  Why is this? It is because our Moon has no atmosphere.

There's no official adjusted average for day-night range on Earth (wildly different for dry deserts and humid cities) but let's go daytime average 20C and average night time 5C? A range of only 15C.

So the actual warming effect of the atmosphere is far stronger at night, when the Earth's surface is 178C* warmer than it would be without an atmosphere (i.e. same as Moon's night-time temperature).

The flip side of this is that the Earth's day-time temperature is 107C* cooler than it would be without an atmosphere (i.e. same as Moon's daytime temperature)- exactly when these  diagrams tell us that the surface should be warmer than it would be without an atmosphere/without The Greenhouse Effect.

Back to the drawing board, lads! Maybe redo those diagrams to explain why The Greenhouse Effect cools the surface of the Earth during the day? (You can't, it's as logically impossible as what the above chart is trying - and clearly failing - to explain).

* Sure, on average the Earth's surface is about 35C warmer than our Moon's surface. That's because we have an atmosphere, full stop, regardless of its constituent gases.