Snow on a relatively warm day Outside my window it's snowing.
This isn't so strange, even for the South of England in mid-April. What is surprising is that the ambient air temperatures (according to the Met-Office) are about 7 or 8 degrees. Normally one would expect any snowflakes that formed in the clouds to have melted long before they reached the ground. Normally if the air temperature is more than 1 or 2 degrees, we get rain.
So what conditions can combine to allow for snow at temperatures above 5 degrees?
Conversely what conditions normally allow for rain at temperatures a little over freezing.
I note that there has been a steady northerly airflow for several days, and temperatures are well below normal (we would expect about 14 degrees at this time of year).  The clouds are Cumulus mediocris, and the snow is falling in very short showers of light intensity. The cloud base appears relatively high (but hard to estimate) The humidity is low (40%) and pressure is high at about 1020 mBa (Met Office values), and the snow melts rapidly when it reaches the ground.
I'm aware that a temperature inversion can result in freezing rain.  Is snow on relatively warm days caused by a very high lapse rate, and so extreme cold temperatures at cloud level?

Not being a winter precipitation expert by any stretch (we avoid such silliness in Florida!), I think a significant key may be the "the humidity is low (40%)" part.
When winter precipitation falls through warm air, it certainly gets heated by the air.  But if the air is dry, some of the outer fringes of the precipitation will melt\sublimate, taking away energy and thereby cooling the flake, helping it to survive longer.  http://www.sciencebits.com/SnowAboveFreezing looks to have a pretty great detailed page about the nitty gritty of it.
Certainly precipitation starting at colder temperatures gives the particle a little reserve.  The low conductivity and high specific heat capacity of water\ice are also notable.  But the latent heat of sublimation is over 10 times as great as the specific heat capacity, showing that losing a little ice on the surface can really counteract quite a bit of warming.

But it's important to note that at the values you gave, the ScienceBits calculator doesn't give any hopes of snow.  So what's going on?
First, it's important to impart that the temp\humidity values aren't far off (try RH=30% and T=7°C or RH=40% and T=6°C and you'll start to get values).
So I am indeed convinced that your local values were off by around that amount.
You can reasonably expect that the temperature that an app or website gives typically has error of a few degrees (with similar errors in most other variables) unless you happen to be right at the location of a reporting site.
This is due to variations over space and time.
Even small elevation changes  cause significant impact on temperatures, and the current data for a relatively flat place like New Jersey shows how other microscale effects (like surface color, roughness, wind blocking, and water\plant moisture) typically lead to at least a degree or two variation between even well-sited representative locations (not even broaching the bigger changes things like concrete, buildings, trees, rock formations, etc cause).
In addition, most weather sites only give full reports once per hour... and we certainly know how things can change in an hour, especially in the morning\evening or near fronts.
In addition, there are three ways your app\website will give you the temperature and other variables:

They just give the reading at the closest official weather reporting station.  For instance weather.gov, the US Weather Service page, will do this.  Full ASOS reporting stations are about 60 miles apart on average in the contiguous US.  You can peruse this site to see the stations in the UK or other parts of the world, but in general they're reasonably far apart.  Mesonets and private weather stations offer more options, but are usually not available to apps\weather service use, and often inconsistently calibrated\sited.

Some websites give a weighted average of nearby values... but this still won't generally include elevation effects at all.

And some use weather models to fill in the gaps and estimate to a higher degree for your location, properly incorporating some elevation data along with the fundamental meteorological equations.  However the resolution (and initial data) is often still poor enough that it doesn't match reality.  I believe the highest resolution forecast\analysis models right now is only 3km, and those are only in limited spots like the US (HRRR, NAM)).  Most models only offer values three hours apart as well, and are only run every 6-12 hours, so the data is far from precise.  Here is the Global Forecast System model forecast for 4 PM BDT (1500 UTC):

You can see some hints of the terrain in a few spots, but nearly the fine variation reality has.
This model only has a 10 km resolution, so many of the smaller hills around your area aren't included at all.
Here's a topography image for a similar scale model, the European ICON model (13 km res):

You can see the North Downs fairly well, and certainly some hint of the hills and valleys around your area.  But they likely aren't well enough represented to truly incorporate the differences in weather in such complex areas.
Likewise it looks like your closest site on Met Office's observation map is Charlwood (though an international list suggests siting at Gatwick... perhaps they're really the same site?)
Remarkably the Gatwick observations from that day actually indicated 9°C at 1600 BST.  Still it's a good 6+ miles away.  And there's also about 70 meters elevation difference between East Grinstead and Gatwick\Charlwood.  It's worth noting that in confirmation at Gatwick there were only rain reports until it got down to freezing after midnight.

Finally as you asked about the vertical profile, here is the model forecast sounding from about that time:

Additionally you can get the SkewT or a table of the actual UK data from the Camborne launch site from 12Z (1 PM BST) at this website to compare. [Here's a site with a bit of info on reading such diagrams].
I originally opined that I didn't think a high lapse rate should be likely since those tend more towards deep convection\thunderstorms.  However the data shows I'm sufficiently wrong... looks like over the lowest km, there's a remarkable 11.2°C decrease, the rare superadiabatic lapse rate.
Still, even that profile only leads to an LCL temperature of around -5°C (calculated as -4.5°C on the Camford sounding, -6.2°C on the GFS sounding using Gatwick surface pressure), which represents the cloud base of a lifted surface-based parcel at those locations (obs at Gatwick are similar).  However with such a lapse rate, by the time the parcel tops out near equilibrium level (which is a rough idea of the maximum cloud top height for a surface parcel)... it's nearer -20°C.  Certainly colder, though perhaps not what I'd think of as extreme.
But you were clearly right to think that the lapse right could be high... in retrospect it makes sense as for surface air to rise air enough for precipitation to fall despite a stout T\Td spread it'd need to rise a fair bit, likely needing decent lapse rates.  My mistake was that the instability (a rough proxy of the rate air would rise at freely [so not including the impacts of the dynamical lift of the storm system]) such great lapse rates produce is tiny (only a few J/kg on this sounding, compared to hundreds or thousands in severe thunderstorms typically)... this is because such cold air doesn't hold nearly as much moisture.
So it's quite possible to indeed get big lapse rates in cold weather and get the types of clouds you saw.  That just may not mean cloud temperatures are particularly extreme, in part because cold weather clouds tend to be more shallow due to the reduction in moist adiabat temperature gains.

In the end, I really believe the temperature was likely a few degrees lower in East Grinstead than in those surrounding areas.  Not only because of the resolution of data... but also in particular because of the weather: a persistent decent northwest wind most of the day, which flows roughly into the sharpest larger-scale terrain gradient in your area.  That would tend to force a lot of ascent, likely developing initial clouds which gradually precipitated down but not reaching to the surface... evaporating fully, but at the same time reducing the air temperature and increasing the dew point below (a process called wet bulbing), until the low-level values became more reasonable for precipitation around the hills.  I'd expect you know this type of pattern very well from the many foggy rainy days that are likely near moors like yours.  That localized process, coupled with the fairly cool temperatures deeper within the cloud, likely were enough to get the sporadic snow showers.


Was the last glacier too hard for the Neanderthal? I know several theories are proposed to explain Neanderthal extinction:

Consanguinity.
Homo Sapiens competence.
Climate change.

But Finlayson et al (2008) quote the last ones lived south of the Iberian Peninsula. This make me think Europe's climate was too cold for them, so they migrated to South.

This surprises me as I have allways considered Neanderthal a cold specie adaptated to glaciar.

The oldest potential Neanderthal bones are dated to 430 kya (Neanderthal, Wikipedia); they had faced to a previous glaciation and they survived.

I am not asking if the climate related extinction theory is the correct one. I am asking if there are evidences the climate on Europe was too cold for this specie. Also I would like to know if there are evidences they also migrated to South in the previous glaciar peak, around 150 kya ago.

Clive Finlayson, Darren A. Fa, Francisco Jiménez Espejo, Jóse S. Carrión, Geraldine Finlayson, Francisco Giles Pacheco, Joaquín Rodríguez Vidal, Chris Stringer, Francisca Martínez Ruiz (2008): "Gorham's Cave, Gibraltar—The persistence of a Neanderthal population". Quaternary International, Volume 181, Issue 1, Pages 64-71, ISSN 1040-6182.
https://doi.org/10.1016/j.quaint.2007.11.016.

First things first, the reasons for the Neandertal(N) extinction are not known definitely. There is though no evidence of violence (*).

There is an overlap in Europe of N and anatomically modern humans (AMH) of several thousand years, even if N find sites are limited to the Mousterien industry alone. This is a conservative limit because N may have used more advanced technology like blade production as well in their last millennia, which would extend N presence to ~32,000BP. Sites like Gorham's Cave fit into that picture, but this long chronology of N is debated and counterdebated :-)). Anyway, long before the last glacial maximum (~23 to 20ky BP) they were gone.

Somewhat outdated is the view that N is a cold adaptation of the human lineage. The appearance of tools for clothing (bone needles) drastically rise with the Aurignacien, the first stone tool industry linked to AMH in Europe. That of course does not mean N didn't use clothes, but that they probably had not the same proficiency as Cro Magnon & consorts in making them. Correlation of find sites and climate confirm the picture that N avoided cold climates. N populations were smaller than AMH, though not really endangered by genetic bottlenecks (there is newer work concerning Denisovans, stating they never really split from N, but i can't find it).

By now it pretty clear that N and AMH interbred on many occasions.

So why did they become extinct ? Together with other ice age megafauna in a highly stressed environment, loosing their way of subsistence ? Competition of top predators ? Inbreading nevertheless ? Dwindling numbers and simple merge with AMH ? Bad luck ? Unclear ...

(*) I want to add a remark here to the Shanidar 3 guy mentioned below. First of all, the authors do not exclude an accident. Second, the work has been criticized by Trinkaus et al. for methodological errors, as is cited here. Similar arguments apply to St. Cesaire 1 (healed skull fracture) is likely an accident, the argument for an intentional blow is too weak. I do not categorically exclude violence in the Paleolithic, the accusation is unfair, I only say there is no (undisputed) evidence. The Shanidar burials, as well e.g. La Chapelle aux Saints 1 are seen as examples for compassion for those in need of care among N. Btw.: we haven't been at the meat of that yet, but i am not handing out fuel against my argumentation ;-)

It was stated that

  "Our ancestors killed the males of other tribes and raped the few remaining females. That our ancestors killed Neandertals who looked rather different is not a far stretch from killing other anatomically modern humans who looked only slightly different."

Both claims are not backed by science, Cro Magnon and Neandertals were not that different (reconstructions Neandertal Museum, Mettmann, Germany) and normal people don't just attack others. Hunting strategies of late N and AMH in the early Aurignacien don't differ by much.

That men were killed and woman stolen happened in very isolated cases in the Neolithic and contemporary Mesolithic, and violence also becomes evident in the Epipaleolithic/Neolithic precursors ~10kyBP , in sedentary or semi-sedentary groups, but that's not the point here. Rather not did it happen in the cold steppe of oxygen isotope stage 3 in Paleolithic Europe for obvious reasons: when the loss of a hunter could endanger the whole group; and women - having one child at a time to keep up the pace of mobility - moved around with the group. At the time 40kyBP population was low, it must rather be assumed that any accidental meeting was a cause of celebration and interchange, as is reflected in the genome.


barotropic component definition In this paper the author calculated the barotropic component $u_{bt}$ of the current $u$ as:

  the vertical mean velocity weigthed with the depth differences

If I am not wrong, in a continuous case, this translates to:

$$ u_{bt} = \frac{1}{d}\int_{-d}^{0}u(z)~dz$$

However, supposing the current to be geostrophic, the following expression can be found for the barotropic component:

$$ u_{bt} = -\frac{g}{f}\frac{\partial \eta}{\partial y} $$

where $g$ is the gravity acceleration, $f$ the Coriolis parameter and $\eta$ the SSH anomaly.

Are these two expressions equivalent? If yes, why? If not, which are the differences between the hypothesis at the base of the two expressions?

The top expression is basically the depth-averaged velocity. Meanwhile the second expression only applies to geostrophic flows. The depth-averaged velocity can be caused by many mechanisms, with the geostrophic balance being one of them. The most common ones are the wind and the tides (but also differences in atmospheric pressure, sloping bottoms, ...). Both tides and wind create a barotropic pressure gradient that has to be compensated by other forces in the Navier-Stokes equations, with friction being the most common one. The geostrophic balance is a specific solution of the NS equations that neglects the effects of friction and any other forces. John Wilkin (Rutgers) has some good notes on how to estimate these currents from observations.

The barotropic pressure gradient is also the integral result for the entire water column. A good illustration is the thermal wind example:  You need a difference in slope between isobaric (pressure) and isopycnal (density) surfaces to give you a baroclinic gradient. In theory in a barotropic fluid the flow is the same at all depths: $u(z)=u_{bt}$. Clearly, in practice, any fluid will have boundary layers in which that is not the case.


Rain without clouds? Is it possible to have rain without visible clouds?

As far as I know, clouds are liquid water and eventually the liquid water in them falls down, producing rain. Therefore one would expect clouds before rain.

But is it theoretically possible to have rain when there aren't any visible clouds? Maybe if the concentration of liquid water in the air isn't too high it is possible.

I actually witnessed this in Ashland, OR several years ago. A strong winter storm had just passed when there was a sudden, brief downpour. When I looked up, there was not a cloud in the sky. What I believe happened is that the air was supersaturated with moisture when a brief breeze caused water to condensate around the few nuclei available. This seems to be confirmed by the second scenario in this explanation: Consider the very rare situation in which still air is supersaturated with moisture (its relative humidity is above 100 percent) and clouds have not formed because condensation nuclei are practically nonexistent. Water vapor will quickly condense onto the few nuclei that are present and the droplets will rapidly grow into raindrops.This is the one and only time I have ever experienced this.


How do I prepare for the International Earth Science Olympiad? Recently I have been selected for the National Camp of INTERNATIONAL EARTH SCIENCE OLYMPIAD of my country. How do I prepare for it? Resources, books, etc.

I know there are 4 parts- geology, hydrosphere, atmosphere and astronomy. Astronomy isn't much of a problem for me (I can manage it). I need the most help in geology. Please see that complex details are not asked in it.

For the syllabus, please see https://www.sites.google.com/site/ineso555/syllabus---entrance-test.

Here are a few, completely random advices I wished someone would have given me before I started to study geology. I'm aware of that this is over simplified, but I believe that it's good to start easy, gain interest and later, with experience, understand that you have been wrong.

A good book to start with:
Earth: Portrait of a Planet by Stephen Marshak. It is used in many universities for undergraduate students. It has a lot of good illustrations and goes breathy through most aspects of earth science. I'd check that and memorize the diagrams and pictures.

http://www.usgs.gov/ also have plenty of easy accessible material.

Make sure that you understand the concept of tectonics well and use that as a starting point for almost any geological discussion.

Don't forget the most basic physical properties. What works in a laboratory also works in larger scale. (E.g. Density will make light material float and heavy material sink. That's why rivers erodes the paths to the oceans and volcanoes erupt volatile rich lava. Oceanic crust is heavy and continental crust is light.)

Take one hour of your life to learn the geological periods (everyone should!). Use the periods as imaginary folders for fossils, orogenies, mass extensions and whatever you later come across. (E.g. 'Permian' will be something distinctive, just like 'Elephant', 'Qawwali' or 'Andhra'. You'll recognize it when you see/hear/feel it, even if it's difficult to describe in words!).

When you read articles, always start with the latest, and work your way back. In this way you'll start with the modern concept and not get biased by earlier models.

Keep an interest in the present. You should know some biology to understand paleontology, some oceanography to understand deposition and hydrology to understand erosion etc. When you add time, we call it geology.

And, if you start to study and research in geology or anything else, learn LaTeX and learn some basic programming, before you need it.

Good luck with the Olympiad!


Geological feature - what is it? Hello,

We were walking in Bundoran in Donegal County, Ireland today and we encountered the geological feature shown in the picture (from above from cliffs and from very close up at beach level). We have three questions about it?

How would you call it? It looks to me like the tall rocks on the sides are sea stacks but isn’t the eroded bit in the middle more like a sea stump? Or are they something different altogether?
What type of rock is it (in the close-up pics) Our guess is that it is shale, but it could also be limestone. Which one do you think it is?
Please notice the ubiquitous fossils in the close-ups. Are we correct to think they are crinoids?

I have just realised a good view from top of this feature is available on Google, so I am sharing it here as well:
https://www.google.com/maps/@54.4852881,-8.2823865,3a,75y,49.15h,69.55t/data=!3m8!1e1!3m6!1sAF1QipNS26PrjyPwTfKnxdxVE29LHtp7A8TctpDVSjl1!2e10!3e11!6shttps:%2F%2Flh5.googleusercontent.com%2Fp%2FAF1QipNS26PrjyPwTfKnxdxVE29LHtp7A8TctpDVSjl1%3Dw203-h100-k-no-pi-0-ya103.88327-ro-0-fo100!7i8192!8i4096

Than you in advance for answering all our questions.

1) I would call the formation a sea stack.  Effectively you have a band of rock which is either more resistant to the forces of erosion and weathering from the sea, or is located on top of a more resistant strata.  The more eroded portion is just that, a more eroded portion, and by its absence tend to identify the rest as a sea stack.
2) The matrix, the rock with the fossils, looks like a limey shale, or a limestone with a lot of shale or clay in it.  It appears to have a component of lime or CaCO3 in it.  This could be confirmed by placing a small amount of dilute HCL or Muriatic Acid on the rock and seeing if it fizzes.
3) There are definitely crinoid stem sections in this rock, as mentioned by JulPal.  If you look carefully at this rock you will likely find a number of other fossils.

To get a tighter description you will have to give more information about the rocks

Treatment of this rock with a dilute acid bath might show a number of fossils.  See the article here https://museum.wales/articles/2011-06-07/Fossils-in-stone-acid-preparation-of-fossils/


Were there times during human inhabitated Earth were Earth average temperature was hotter than today? This article,

Earth's temperature in the past 18,000 years

seems to show Earth temperature during 5000-3000 years ago was approximately a degree hotter than today. Is this true, were there times during human inhabited Earth were Earth average temperature was hotter than today? If so, were there other moments not listed in that graphic from former times?

Answers would be "no" and "no."

That graphic isn't sourced, but it appears to refer to the Mid-Holocene Warm Period, roughly 7,000 to 5,000 years ago.

According to NOAA and other sources, the warming of the Mid-Holocene Warm Period was limited to part of the Northern Hemisphere; some regions would have been warmer than today, but not the globe:

  It appears clear that changes in Earth's orbit have operated slowly
  over thousands and millions of years to change the amount of solar
  radiation reaching each latitudinal band of Earth during each month.
  These orbital changes can be easily calculated and predict that the
  Northern Hemisphere should have been warmer than today during the
  mid-Holocene in the summer and colder in the winter.

This graphic from the NOAA page cited above shows the estimated mid-Holocene temperature anomalies (variation from mean).
 Though humans had migrated widely the period, relatively few would have been living outside that cooler blue band.

Were there earlier warmer periods for modern-ish humanity before that? No, since earth was emerging from a glacial period starting about 11,500 years ago, and global average temperatures were gradually rebounding from about 6 degree Celsius cooler than today.

The next previous warm era would have been the Eemian interglacial period about 130,000 years ago, which again was associated with orbital variations.

You could quibble about the era of "human-inhabited" earth; homo sapiens arose about 300,000 years ago, but remained limited to Africa until about 70,000-50,000 years ago.


Error trying to download ERA5 data: Exception: Missing/incomplete configuration file: C:\Users\username/.cdsapirc I am new with Python so please be kind.
I want to download data from ERA5 and so far I have installed the CDS API package and followed all the instructions from https://cds.climate.copernicus.eu/api-how-to

I have my main file ready to use but I get an error about the configuration file. I know that this file should contain the following:

url: https://cds.climate.copernicus.eu/api/v2
key: {UID}:{key}
verify:0

I have created a text file and a configuration file with this info and I have placed them in the following address (as required by the error I get in python):

C:\Users\username\.cdsapirc

However, I still get the same error. What I am doing wrong? Do I need to create a different type of file? Do I need to write the user info in a different way?

Exception: Missing/incomplete configuration file: C:\Users\username/.cdsapirc

By the way, I'm using a windows computer.

I had the same problem but have just found the solution after an hour or so of trawling through forums! The problem was that the code 'api.py' could not find the '.cdsapirc' file. The file type has to actually be 'CDSAPIRC' and not a 'txt' file. The icon for the file will then be displayed in your folder as blank and the name will be '.cdsapirc'. I found the solution on here: https://www.wikihow.com/Create-and-Delete-Files-and-Directories-from-Windows-Command-Prompt. I created the file using the following steps:

Navigate to 'C:\Users\username' (or you can just move it later).
In Command Prompt, write 'type nul > .cdsapirc' and press Enter.
Right-click the file and press 'Edit with Notepad++' (probably works with other programs).
Paste the text that you already mentioned (key, etc).
Save and close the document.

Hope that works :)


How do they determine the "feels like" temperatures for weather data? I'm pulling air temperature and other weather stats from various APIs. There's this concept called "feels like temperature", separate from "actual temperature". Apparently, this is the temperature that it feels like to a human, depending on the wind speed and other factors.

Is this determined, as the name suggests, by a human who "feels" it? Or is it mathematically calculated based on the other data such as the wind speed, so that it isn't actually "felt" by a human at all, but rather is a scientific estimation of how a human probably "will feel" the temperature?

Personally, I have no idea how anyone can "feel" the temperature in specific terms; to me, it's either "freezing", "cold", "chilly", "OK", "warm", "hot" and "unbearable" -- any guess I would make about what temperature it "feels like" would vary with each time you asked me and not be accurate whatsoever.

The "feels like" value is calculated based on temperature, but altered depending on wind (cold temperatures) or humidity (hot temperatures).  Think of it this way:

1) if it's freezing cold outside, the wind will make if seem even colder because any body heat you have will be blown away quickly.
2) if it's hot outside, high humidity will make it seem even hotter because your sweat won't easily evaporate.

From wunderground:

  The Feels Like Temperatures map show what the outdoor temperature will feel like for the current day. Feels Like Index is a factored mixture of the Wind Chill Factor and the Heat Index. The combination of the heat index and the wind chill factor are denoted collectively by the single terms apparent temperature or relative outdoor temperature or simply Feels Like.

  Wind Chill

  Wind chill is the apparent temperature felt on exposed skin, which is a function of the air temperature and wind speed. The wind chill temperature (often popularly called the wind chill factor) is always lower than the air temperature, except at higher temperatures where wind chill is considered less important. In cases where the apparent temperature is higher than the air temperature, the heat index is used instead.

  Heat Index

  The heat index (HI) is an index that combines air temperature and dew point in an attempt to determine the human-perceived equivalent temperature — how hot it feels, termed the felt air temperature. The human body normally cools itself by perspiration, or sweating, which evaporates and carries heat away from the body. However, when the relative humidity is high the evaporation rate is reduced, so heat is removed from the body at a lower rate causing it to retain more heat than it would in dry air. Measurements have been taken based on subjective descriptions of how hot subjects feel for a given temperature and humidity, allowing an index to be made which corresponds a temperature and humidity combination to a higher temperature in dry air.

  The heat index is derived from work carried out by R. G. Steadman. Like the wind chill index, the heat index contains assumptions about the human body mass and height, clothing, and the wind speed. Significant deviations from these will result in heat index values which do not accurately reflect the perceived temperature.


Polar melting and the equator As the polar ice caps disappear, the equatorial regions must take the bulk of extra water due to the earth's spin.

My question is will the tilt of the earth be affected and what would be the consequences?

An interesting question! I suggest you do some 'slightly complex' calculations. Look up the angular momentum of the whole Earth. Then calculate pM.dF where pM is the additional angular momentum from the melted ice, redistributed as a function of latitudinal angular momentum. dF is differential redistribution caused by continent-ocean constraints. I'm confident that you will find that the additional differential off-axis angular momentum will be miniscule compared to the total. There will of course be some effect upon the Earth's spin axis, but this will be just another of the small catalogue of mass distribution effects which contributes to a somewhat chaotic wobble in the Earth's spin axis, which is typically in the order of a meter or two. When global warming gets worse and huge masses of ice /water are redistributed, then I'm guessing that the effect could be orders of magnitude more - tens or hundreds of metres maybe? Also consider that the Earth behaves like a gyroscope, so much of the additional angular momentum will exhibit as rotation around other axes, much as hurricanes rotate in the atmosphere.


How much does the gravitational and magnetic fields of a planet effect the chemistry in it? It is known that the earth has a magnetic field and gravitational field around it, I wish to ask to what degree is the chemistry on the planet is influenced by these fields? How would the chemical reactions seen everyday on earth be different if the strengths of these fields were different?

How would the chemical reactions seen everyday on earth be different if the strengths of these fields were different?

Depends. If you're just talking about higher $g$ (i.e $g>9.8\ \rm{m/s}^2$) but the same atmospheric pressure, than not much will change on the surface of the earth.
If the atmospheric pressure changes, there could be changes to chemical reactions. There are several factors controlling chemical reactions. One being, obviously, the identity of the reactants. The other one is temperature. But also important is pressure. Because pressure on the surface of the earth is always 1 atm, no one really thinks or cares about pressure. No one, until it changes. That said, for any substantial change in chemistry, pressure has to change quite a lot, which would make earth uninhabitable for humans, so there would be no one to see that chemistry changed.


Forecasting oceanic rogue waves A rogue wave is, according to NOAA, "a large, unexpected and dangerous" oceanic wave.  These are also known as 'killer' or 'freak' waves and have been a part of folklore of the past and observation of the present - such as in the rather terrifying example below (from gcaptain.com:

  Caption: A wall of water approaches the Stolt Surf in Oct. 1977 Photo:
  Karsten Petersen, www.global-mariner.com

There are a number of theories as to how these form, according to the NOAA page linked above, it can be a result of constructive interference or focusing of wave energy.  Another theory reported in the Scientific American Blog entry from 2009 The Real Sea Monsters: On the Hunt for Rogue Waves is that

  "If a wave propagates from east to west, and the current moves west to east, then a wave starts to build up," says Wu, who studies wave–current interactions in a 15-meter pool. The wave basically climbs the current's wall, rising out of what appears to be nowhere.

This question is not so much based on the actual mechanism behind rogue waves, but how we can use these models to forecast their occurrence, going from the theoretical models discussed in the Scientific American blog to develop actual proven forecast models.

What is the status of research in forecasting oceanic rogue waves?

There are few known mechanisms that lead to the generation of Rogue waves, such as the ones you mentioned, but essentially all Rogue waves are the due to the nonlinear wave dispersion characteristics of large groups of waves.

I can imagine one approach to predicting the emergence of such waves is to simulate the evolution of initial wave states with Navier-Stokes based wave models that capture the nonlinear dynamics involved in real time. However, with the limited computational power and numerical techniques we have today, that is currently impossible.

So I think the question is: Do we know of specific signatures in the ocean wave climate, which ultimately give rise to these so called "walls of water"? Presumably these wave patterns could serve as early warning indicators of an upcoming rogue wave event.

There is indication that the emergence of solitons from modulational instability or dispersion of wave packets occurs in consonance with nonlinear effects and nonlocal coupling giving rise to Roge waves. Thus, an approach based on the spectral analysis of the sea-state as a whole or of particular envelopes of propagating waves may hint at particular wave-field patterns that originate Rogue waves. Please see the sources below for more information on the current state of the art. However, at the moment the answer is that forecasting Rogue waves is still much of an unknown.

  Akhmediev, Nail, et al. "Rogue Waves: Early Warning vs Forecast."
  (2011).

  Akhmediev, N., et al. "Early detection of rogue waves in a chaotic
  wave field." Physics Letters A 375.33 (2011): 2999-3001.

  Onorato, Miguel, et al. "Rogue waves and their generating mechanisms
  in different physical contexts." Physics Reports 528.2 (2013): 47-89.


Why is almost all the landmass on one side of the globe? One side of the Earth is almost completely covered with water (the Pacific), while landmass is almost all on the other side. Is this coincidence or plays there something else? This difference is not as pronounced as the difference between North and South.

NVZ is correct. Get a palaeo-geographical atlas and see how the the distribution of the continents has changed over time. There is much more ocean than land, and it has ever been so, but there is no particular pattern to the way the continents have drifted. It's not quite random because the heat distribution in the mantle isn't random, but given the conveyor belt system of plate tectonics there will always be a tendency for crustal plates to clump together before breaking up and then clumping together somewhere else.


Why is reverse osmosis desalination the dominant method used and not heat-based methods? Also why is brine being dumped back into the ocean? So I was recently watching this video regarding desalination efforts around the world and the viability of desalination as a means to provide humans with freshwater. I had a few questions:

1 - They mention that the brine that is released as a byproduct of the process is potentially harmful to marine life, so why not just dump it in the desert? Why do we need to release back in the ocean? I'm sure the effects of this won't be that big given that desalination is a tiny percentage of the world's water production. And if anything, given that global sea levels are rising, this might help things, right?

2 - The video states that the vast majority of desalination plants these days use reverse osmosis desalination and not heat-based methods. Why is that so? How is a process that requires energy better than using solar energy to evaporate the water?

Heat-based methods require energy from the sun, yes. But the way I imagine things, we can passively let in water from the ocean, put it under some glass panel that magnifies sun energy, and then let the water evaporate out through some piping to condense somewhere else. How is powering a pump through a membrane (that requires cleaning after) better?

A lot of these thoughts are just me thinking out loud with no data to back me up because honestly, I wouldn't know what to look up regarding things like this. I'd love some science-based replies to this! I'm definitely missing something big.

Starting with question 2): you can use heat to evaporate the water but you need some sort of energy or thermal exchange to cool and condense it back to liquid. My understanding is that reverse osmosis (RO) is more energy efficient overall.

Question 1) Brine is only put into the oceans from system that are near the sea, which usually use seawater as a water source. The brine has an insignificant effect on the ocean salinity once it is mixed in but can have a local effect because of poor mixing and because differences in density between the brine and seawater cause it to sink to the bottom. Systems can be engineered to minimise the impact.

Inland, the brine can degrade groundwater resources if it is disposed to seep into the ground. Even in hot, dry climates, it may be difficult to completely evaporate it to dryness in lined basins so it doesn't seep in. Seepage basins are a viable option in some places where the groundwater is already to salty to use as a resource (for example where the salty water is extracted, put through an RO system and then reinjected to the same aquifer).

RO systems are designed so that the salty concentrate is removed during operation so the system doesn't need to be cleaned to remove the salt. However, impurities may still require cleaning of the membranes. There are ways to minimize the amount of cleaning needed.


Can someone help me identify this rock please? I recently found this at a garage sale and I would love to know rock this is. It weights 6 lbs.

-Thank you

It is quartz. The whiter parts are larger crystals that have grown together and the tan is agate. It formed like a completely filled geode.


Is the atmosphere a part of Earth? My question is quite simple, but I can't seem to find any answer to it: how do we define the limit of the object "Earth"? More specifically, can we consider the atmosphere a part of it? Or even the oceans, for that matter? I was going for a no at first glance, but that would mean that you are actually leaving the Earth when you fly on a plane. That feels weird. How do scientists define the limit of the Earth?

It's all in how one chooses to define it M.Connor.

Often in physical science courses, we define the Earth system as the lithosphere (rocks), atmosphere (air), biosphere (life), and hydrosphere (water) [also sometimes separately defining a 5th aspect, the cryosphere, which is ice]

In the end, we often just think of the Earth as everything confined by its gravity (not including in orbit) as that's everything still controlled by the Earth.

But in a very true sense, you're still leaving the Earth when you take off on a plane.
Just you're still in a state/region of the Earth's system where what goes up must come down.  So you won't be gone long!


What are the band names of the NCEP grib files? I'm downloading .grib files from NCEP: ftp://nomads.ncdc.noaa.gov/GFS/analysis_only/

Each file has 315 bands. What are their names?

Reproducibility with R:

library(stars)
library(cptcity)
met <- read_stars("ftp://nomads.ncdc.noaa.gov/GFS/analysis_only/201501/20150103/gfsanl_3_20150103_0000_000.grb")
met2 <- filter(met, band == 2)
plot(met2, col = cpt())

This file is in GRIB-1 format. Each band corresponds to a GRIB message. The band names can be found in an INV file on the same server.

For your example check out the corresponding inv file ftp://nomads.ncdc.noaa.gov/GFS/analysis_only/201501/20150103/gfsanl_3_20150103_0000_000.inv.

Another option is to use wgrib -v with -v  option.

The band names and offsets for all 315 bands can be obtained this way, e.g. for band 2

2:81534:D=2015010300:VGRD:planetary boundary layer (from Richardson no.)
 :kpds=34,220,0:anl:winds are N/S:"v wind [m/s]

That is bands 1 and 2 store u and v components of the wind speed at the boundary layer.

kpds parameters make up the product definition section identifiers. The first kpds number can be mapped to mnemonic GRIB identifiers, i.e. number 34 maps to VGRD.

The unit is m/s and the data representation is 64-bit floating point. The latter piece of metadata can be extracted from the GRIB file itself e.g. using gdalinfo.

Band 2 Block=360x1 Type=Float64, ColorInterp=Undefined
  Description = 0[-] var220 (reserved)
  Metadata:
    GRIB_COMMENT=v-component of wind [m/s]
    GRIB_ELEMENT=VGRD
    GRIB_FORECAST_SECONDS=0 sec
    GRIB_REF_TIME=  1420243200 sec UTC
    GRIB_SHORT_NAME=0-var220
    GRIB_UNIT=[m/s]
    GRIB_VALID_TIME=  1420243200 sec UTC

With the -V option wgrid utility produces a more detailed output:

rec 2:81534:date 2015010300 VGRD kpds5=34 kpds6=220 kpds7=0
    levels=(0,0) grid=3 planetary boundary layer (from Richardson no.) anl:
  VGRD=v wind [m/s]
  timerange 10 P1 0 P2 0 TimeU 1  nx 360 ny 181 GDS grid 0 num_in_ave 0 missing 0
  center 7 subcenter 0 process 81 Table 2 scan: WE:NS winds(N/S)
  latlon: lat  90.000000 to -90.000000 by 1.000000  nxny 65160
          long 0.000000 to -1.000000 by 1.000000,
          (360 x 181) scan 0 mode 128 bdsgrid 1
  min/max data -38.8 45.4  num bits 10  BDS_Ref -388  DecScale 1 BinScale 0


Is the net amount of CO2 sequestered by trees from the atmosphere simply related to the mass of new growth? My simplistic model for the carbon cycle was that plants (or trees in particular) photosynthesise and transpire. The former takes up $\small\mathsf{CO_2}$ and $\small\mathsf{H_2O}$ and the latter does the reverse but obviously not netting out. The carbon for the cellular material (in particular wood) comes entirely from atmospheric $\small\mathsf{CO_2}$ (?).

The mass of trees comes mainly from cellulose which is roughly 50% carbon. Therefore the mass of sequestered $\small\mathsf{CO_2}$ is simply related to the mass of trees, with the proportion equal to the ratio of the molecular masses of cellulose and $\small\mathsf{CO_2}$ with balanced carbon content. Wiki says cellulose is some sort of sugar type thing $\small\mathsf{(C_6H_{10}O_5)n}$ and carbon dioxide is obviously $\small\mathsf{CO_2}$ so the ratio is something like: for each kilo of tree/wood grows 1.62 kg of $\small\mathsf{CO_2}$ is sequestered?

Is this completely wide of the mark?

Simple answer: No. The process "carbon sequestration" considers the whole Earth system holistically (otherwise it would not serve its purpose), and in your case, you only look at the plant-atmosphere system so nothing can be said of the amount of carbon sequestered.

What you are describing is more close to Net Primary Production, which is defined as carbon fixation from photosynthesis minus carbon released from autotrophic respiration (which is somewhat different from transpiration that you mentioned).

Carbon sequestration refers to the transfer of atmospheric carbon into a relatively stable pool of carbon (on a climate-relevant time scale, usually at least in decadal-scale). "New growth" implies all the biomass gains of the tree over time, not necessarily "stable" biomass only. One obvious case arises for deciduous trees where large portion of "gained" leaves fall and decompose relatively rapidly.


Climate on an Earth without an Axial Tilt The Premise

The Earth rotates on its own axis, which is tilted by about 23 degrees. I am imagining a situation where such an axial tilt does not exist.

From general reading, it appears the largest impact of removing an axial tilt would be on seasonal change, in the sense that there wouldn't be any changes whatsoever. Every latitudinal belt would have its own set season all year round, with some minimal changes depending on the earth's distance from the sun in its usual orbit (ranging basically from 91.4 million miles in January compared to 91.5 million miles in July - which is less than 1%).

The Question

What would be the implication on rain cycles if the Earth had no Axial Tilt? It appears to me that rain carrying clouds move to the dictates of wind blowing to an area with a different air pressure. If our planet had no axial tilt, the Equator would get extremely hot all year round, causing water to evaporate and move north or south towards cooler areas - at least that's my understanding.

What parts of earth would receive the most rainfall if such a change were to occur?

Just throwing  some thought on the subject, it is easy to me to understand that there is a big pump moving large amount of water from one pole to the opposite one along the year, that is during summer in north hemisphere the ice melts at Arctic and mix to the sea; at the same time water start precipitating more and more on the Antarctica, piling up as snow and ice.  During the other six months the pump reverts, moving water from Antarctica to the Arctic.

In this hypothetical situation this pump will behave differently.   The water will evaporate from the sea, and precipitate towards the poles.   Those rain in midway to the pole would provide vegetation life, erosion, and all sort of things rain water does.   That water falling in the north and south pole would have no reason, in principle, to melt and reinsert in the oceans.

If the rain started to fall mostly on high latitudes, possibly the region around equator would be more exposed and deserter, increasing the heat, evaporation, and the convection of the air at high speed.   Possibly very intense winds blowing everything, moving quickly the clouds and vapor to higher latitudes, where the rain would fall.   Anyway, the hotter atmosphere should increase the loss of the lighter gases to the space due to kinetic energy gain, decreasing the atmosphere pressure little by little while becoming hotter.  In this loss there could be even some water vapor.

Then I think that in the long run the water would tend to move all the way to the poles draining drastically the remaining of the planet.  If we think of an ocean as salty (3.5%) as ours, possibly there would be deposition of large layers of salt, most of it sodium chloride. I don't really know if there would be a "life zone" around some high latitude.  Certainly tropic would have no meaning in this configuration.

That is the "big picture", but not all the story.

We have different wind flow in atmosphere and water flow in the oceans due to others factors.   One of those being the tides that move water in waves half circumference of the planet.   The main component being that from the moon, but we have other of smaller intensities like the solar influence.  Other factors are the convection due difference temperatures and pressures, and Coriolis force making it go round in its trip across latitudes.

Another factor is that we have a non homogeneous distribution of mass on planet surface, the removal of large amount of water, moved to the poles, possibly would increase the amount of wobbling, then the precession ( https://en.wikipedia.org/wiki/Axial_precession) of the planet.   Perhaps this wobbling would be responsible to a certain amount of melting at the poles, keeping the pumping working longer and, who knows, preserving some life in the planet.

A hotter atmosphere could increase the amount of clouds, if it covers all the planet we could have a very long freezing of the planet.

If we think on a dry planet with water moved to poles as ice, we could perhaps include sublimation of ice in the figure, adding the loss of water to space due to  heating of the atmosphere.   I don't think we would become a Mars, our gravity is about three times Mar's gravity.

HTH


Flux density and albedo To save the Earth, the surface of the Earth has been covered by silver foil. The foil has a albedo of 0.9. What is the flux density emitted from the surface of the Earth (assume temperature of 50 deg C)?
So I thought I could use Stefan Boltzmann law and just plugg in my values and got 569, but this is not correct. What I'm I missing?

In your calculation, did you use albedo correctly? Remember, albedo is the ratio of reflected radiation to total radiation, and 1-albedo is the ratio of absorbed radiation to total radiation.
Edit: Make sure you are plugging the numbers in correctly too.


How to define the force at the base of an oceanic crust In the book of Turcotte and Schubert, Geodynamics I'm trying to solve problem 2.8.

The pressure in the water, in the oceanic crust, and in the mantle beneath the oceanic crust are (from equation 2.19, page 78) is

$$p_w = \rho_w g y , \quad  0 \le y \le h_{w}$$
$$p_{oc} = \rho_w g h_{w} + \rho_{oc} g (y - h_{w}) , \quad  h_{w} \le y \le h_{w}+h_{oc}$$

$$p_m = \rho_w g h_w + \rho_{oc} g h_{oc} + \rho_m g (y-h_w-h_{oc}), \quad  h_w+h_{oc} \le y \le h_{cc}$$
I should integrate that to find the force at the base of the oceanic crust.

we know that

$$F=∫_0^b pdy=ρg∫_0^by dy=0.5 ρgb^2$$

so

$$F_m=∫_{h_w+h_{oc}}^{h_{cc}} p_mdy$$

then I tried this

$$F_m=∫_{h_w+h_{oc}}^{h_{cc}}\rho_w g h_w + \rho_{oc} g h_{oc} + \rho_m g (y-h_w-h_{oc})dy$$

and it didn't provide the proper result, so i tried to integrate all the pressures:

$$F_w=∫_0^{h_{w}}\rho_w g ydy = 0.5 \rho_w g h_{w}^2$$
$$F_{oc} = ∫_{h_w}^{h_{w}+h_{oc}} \rho_w g h_{w} + \rho_{oc} g (y - h_{w})dy= \rho_w g h_{w}+0.5 \rho_{oc} g (h_{w}+h_{oc})^2 - 0.5 \rho_{oc} g h_{w}^2 - \rho_{oc} g h_{w}$$

I did the same for the last integral and added them all together: $$F_o = F_w+F_{oc}+F_m$$

but I still don't get the same final expression as equation 2.20, page 78.

Can someone give me a hint to solve the integral? are the limits of my integral correct? Am I adding the right forces together?

Your approach of adding all the forces together is correct. You have a mistake in in the solution of the $F_{oc}$ integral, though.
You forgot to integrate the terms $\rho_{w}gh_{w}$ and $-\rho_{oc}gh_{w}$

It should read:

$F_{oc} = ... = \rho_{w}gh_{w}(h_{w}+h_{oc}) - \rho_{w}gh_{w}h_{w} + 0.5\rho_{oc}g(h_{w}+h_{oc})^{2} - 0.5\rho_{oc}gh_{w}^{2} - \rho_{oc}gh_{w}(h_{w}+h_{oc}) + \rho_{oc}gh_{w}h_{w} $

You have to keep this in mind for $F_{M}$ as well.


How to get the exact temperature from 1880 to 2020? Wherever I look, I find "Temperature Anamolies" which I assume is an annual temperature change relative to the previous year.
If I wanted to find the exact temperatures (based on NASA's accurate forecasts and estimates), can I get them somewhere?
https://data.giss.nasa.gov/gistemp/graphs/graph_data/Global_Mean_Estimates_based_on_Land_and_Ocean_Data/graph.txt
The above link shows the temperature change but if I add them up, the current global temperature comes out somewhere near 21 degrees as opposed to 14 degrees figure I often see everywhere..

I assume is an annual temperature change relative to the previous year.

No, that would be called the rate of change. "Anomaly" means the difference from some reference temperature.


Does "atmospheric forcing" have a stated definition? I often come across the words "atmospheric forcing" or "meteorological forcing" in journals.

I presume it just means the influence/impacts of the atmosphere upon factors such as climate, wind patterns and tidal/ocean currents.

However, this is a presumption. Am I correct?

An example is located on page 11 of this link

The example that you've linked to (Larsen, 2003, "An investigation of the Faroe Shelf Front", Univ. of Bergen) is a description of various processes that contribute to the existence and variability of an ocean front around the Faroe Isles.  As an oceanography thesis, it's mainly concerned with the ocean state (e.g., temperature, salinity) and local, internal process that affect the front.  But the introduction also mentions external process that affect the local ocean state, such as the wider-scale ocean state/fluxes, topography and atmospheric forcing.

You could loosely think of these three respectively as the lateral, lower and upper boundary conditions for the local block of ocean of interest around the Faroes. In this context it could also be phrased as the forcing of the ocean by the atmosphere, so how the ocean is affected by the exchange with the atmosphere of quantities such as heat, water and momentum.  Larsen summarizes these through descriptions of air temperature and heat flux, precipitation, evaporation and runoff, and wind speed and direction.

Often when we're interested in a particular physical realm (e.g., atmosphere, ocean, land or ice-sheets) we prescribe (i.e., force) the interactions it has with the other realms to make it easier to investigate processes within the realm.


Calculate the maximum power angle of a solar panel (photovoltaics) I want to calculate the maximum power angle of photovoltaics depending of the coordinates on Earth. That means at which angle I should align my solar panel to get the maximum power over the year?
I think to get the maximum power I should point the solar panel to the South. But at which angle I should align the solar panel?
Is it always south? When I am in australia then I should align the solar panel to the north? Or I am wrong?

This question can be answered in many different ways that take into account, or ignore, the many factors that affect the total output of a fix solar panel over a year.

In general, all answers will agree that the solar panel will have to face the north in the southern hemisphere, and the south in the northern hemisphere. This is because that is the direction where the Sun will be when it reaches the maximum elevation during the day. Or if you are within the tropics ($\pm$23.5° of latitude), that's the direction where the Sun will be most of the days when it reaches the maximum elevation.

The tricky part is the inclination angle, because the solar elevation changes trough the day, and the maximum elevation changes trough the year. In general, the mean elevation at solar noon (that usually do not match the clocks noon due to standardized time zones), is given by:

$\overline{h_{max}}=90^o-\text{Latitude}$

And through the year it will vary $\pm$23.5°.

So in a first zero-order approach you can just use that mean value. Now, to find the optimum value you have to model the Sun's position trough the year and run a complex calculation to find the angle that maximizes output. Thankfully some people have done that, and they have reduced the output to simple empirical formulas. For example, in solarpaneltilt.com they provide the following formulas:

$\text{Optimim angle}=\left\{ \begin{array}{rcl} \text{Latitude} \times 0.87 & \mbox{if} & \text{Latitude}<25^o \\ \text{Latitude} \times 0.76 + 3.1 & \mbox{if} & 25^o<\text{Latitude}<50^o\end{array}\right.$

And if you don't want to put the fingers in a calculator they provide a table with some values for different latitudes:

However, this approach pinpoints to an angle with great accuracy, while neglecting what can be one of the main factors affecting the output of your solar panels: weather. Therefore, the above approach would be very accurate if you live in a desert, but misleading if you are in an area with clear weather in winter and rainy weather in summer.

To tackle this problem you can download weather statistics for your location and find the optimum angle yourself. Ideally you would do this by calculating solar output on each day, but I'll first show you also a simpler approach month by month that you can reproduce with very little computer skills and the results will be very similar.

The general workflow is as follows:

Find the typical solar radiation reaching the surface at your location each month/day of the year. Let's call this parameter $S$.
Find the typical solar elevation over the horizon at solar noon at your location each month/day of the year. Let's call this parameter $h$,
Find the maximum solar power that a solar panel could receive if directly facing the Sun, let's call this $S_{max}$ and it will be given by

$\hspace{12em} S_{max}=\frac{S}{\sin(h)}$

Find the effective power hitting a solar panel at a set angle $\theta$. Let's call this $S_{eff}$ and it will be given by

$\hspace{12em} S_{eff}=S_{max} \cos(90^o-h-\theta)$

Adding $S_{eff}$ for each month/day of the year. Let's call this $S_{total}$.
Put all these calculations in a spreadsheet and then play with different values of $\theta$ until you find the one that produces the maximum $S_{total}$.

Now, to get the typical $S$ values for each month you can go to the KNMI Climate Explorer, click the option "ERA-interim 1979-now", then select the variable "surface net solar/longwave radiation", and click "Select Filed" on the top.
Then, you will see a bunch of options, but only focus on the first box:

Fill, the coordinates, add half a degree around your location (this model resolution is ~ 80 km, so don't get picky with accurate locations), select the "max" option (mean would work too), and hit "make time series".

You will get a time series of solar radiation in your area, and the average per month from 1979 to 2019, it will look like this:

Now, you click the "raw data" link above the plot and you will get the numbers for $S$ to feed into your spread sheet, they are in the first data column:

For daily values you start from Daily fields, and select ERA-interim 1979-now rsds field. From there is the same as before. And the plot will look like this:

Now, to get the solar elevation for each month you can use this Solar elevation angle (for a year) Calculator and just take the day closer to the center of the month to feed it into your spredsheet. For getting angles for each day, you can use this Rise/Set/Transit Times for Major Solar System Bodies calculator, this is how I filled the form for my area of interest:

Note that I choose a leap year so it matches the number of days in the output from the KNMI Climate Explorer.

Whith all this, the formulas above and a bit of spreadsheet skils you can find the optimal angle for your location.

My monthly spreadsheet looks like this:

And the daily one like this:

As you can see, for my area of interest the angles differ only by 1° between the monthly and daily calculation. And about 4° with the formulas from solarpaneltilt.com. However, in other climates the differences could be much larger.

This method can be adjusted also if you want to maximize power during the winter months, that are usually the most challenging ones for a solar power supply.
If you area have large temperature variations and want to get fancy, you can also incorporate the effect of temperature in solar panels' efficiency.


Why is Thermohaline Circulation (THC) pivotal to life on Earth? Thermohaline circulation (THC) is an instance in nature in which oceanic fluctuation of amassity, temperature and sodium matter is observed on a global scale. It is compared to and sometimes called the global conveyor belt due to its high propensity to transport vital environmental ecosystems across the globe. Yet, oceanographers have stated that less than 5% of the worlds' oceans has been explored. THC is also attributed to stratification of the planet's oceanic resources.

The question: Why is THC pivotal to life on Earth?

My answer is a bit off topic to your actual question, but addresses something you said:

  oceanographers have stated that less than 5% of the worlds' oceans has been explored

I would venture that less than 5% of the ocean floor or deeper ocean depths is perhaps where less than 5% has been explored.

We've really been over the vast majority of the ocean surface at some point (and have tools like satellites that give us significant data on the rest).  Here's the current location of floating buoys:

(Argo Website)

It looks pretty comprehensive near the surface!  Still, there's often hundreds of miles between sites, so while it means we have\do cover a fair bit of the area (more than 5%!), we aren't everywhere at once.

And it's important to note that a a significant part of thermohaline circulations are near the surface, generally the red areas on this map:

[
(Wikipedia)

But the big thing to take home is that even where we don't have as much data, we can see the large thermohaline circulations well because of its scale and relative continuity.  It's quite comparable to the upper atmosphere, where we also don't get data on large parts of the upper atmosphere, but can rather well diagnose jet stream fluctuations.  Here's the upper air sounding network:

(Weatherstem.com upper air observations lesson)

Both networks are aided some by satellite and ship/plane observations, but overall, despite fairly limited observations, the circulations are quite evident because of their size and impacts.  So we have a pretty good handle on what's going on.

Hopefully someone else is able to better address your central question about why it is vital to life on Earth!


Why the "Mission to Earth’s core" proposal suggests using liquid iron instead of lead? Nature communications article "Mission to Earth's core — a modest proposal", suggests placing a large volume of liquid iron in a crack and let it sink all the way to the Earth's core, carrying along a probe that can transmit data using seismic waves.

It is an interesting idea. Although there might be many technical problems I feel no one seems insurmountable and it is worth a try.

However, I wonder why the author suggest using a liquid iron instead of a liquid lead. I would expect lead to be cheaper and also require a smaller volume to achieve the same fracture stresses (due to higher density).
In addition, a lower melting point would facilitate the initiation of the fracture propagation near the surface.

Would lead dissolve in the surrounding magma? Would it chemically interact turning into a lighter compound?

As the original article is behind a paywall I copy here the main text:

I've already commented on this before here. This will not work regardless of whether this is iron or lead or anything else. The fact that it was published in Nature does not mean it is true, always remember that.

A 108 kg ball of iron would have about 30 metres in diameter, which is not a lot, compared to Earth scales.

Here is why it will not work:

We have very massive objects on the surface. They are called buildings. They don't sink. Even though they are not that dense, buildings stand on foundations that funnel their mass to a small area, creating very large pressures. The point of the foundations is to hold the building on solid rock so they do not sink.
The author suggests to initiate the crack with explosives. Perhaps a nuclear explosion. We had many nuclear explosions in the past - they don't make cracks. They make craters.
Crack propagation is only relevant in the lithosphere. Once you get to the asthenosphere, about 200 kilometres deep, cracks don't propagate any more and the equations the author presents are meaningless. Plastic behaviour of rocks is achieved much earlier anyway.
Even if there is a crack, it will have smaller cracks radiating from it. The liquid iron, being liquid, will fill those smaller cracks and lose from the mass of the main liquid blob. With time, the blob will get smaller and smaller until it will not be able to do anything any more.
The melting point of iron is more than 1500 °C, this higher than the temperature of the ambient rocks. It will lose heat quicker than it gains by gravitational potential and solidify.
The iron will react with the rocks around it. Particularly if it is liquid, because these things are highly reactive. It will start dissolving things into it (which will counter point 5), but it will also lose iron to the surrounding rocks. It will dissolve oxygen, carbon, sulfur, possibly alkalis like sodium, maybe silicon. These are light elements that will overall lower the density of the ball of iron, bringing it to a stop.
The oxygen fugacity of the crust and upper mantle is higher than iron-wustite, which will be imposed by the equilibrium with iron. Therefore, it will leach oxygen from the surrounding rocks, essentially turning some of the liquid iron into "rock". This will also decrease the mass of the blob of iron.

You were specifically asking about lead. All points above (except 5) are equally applicable to lead.


Floating rock found in southern florida I found the rock on the beach.  It has a redish hue and floats on water.  It feels as though it is hollow because it is light for its size.  Does anyone have an idea what this is?

This isn't a rock at all, but a seed of Entada Gigas, sometimes called the sea heart or sea bean.

(CC-BY-SA-3.0)

Entada Gigas is called the "sea bean" for good reason: it's a member of the bean family Fabaceae. In fact, this species has the family's largest seed,  a flattened heart shape 6 cm in diameter and 2 cm thick.  These hollow seeds are washed into the ocean and drift on currents for up to 2 years, on the chance that one will wash up somewhere to germinate.  It seems that sea beans have evolved to propagate in this way, similar to coconuts.


Can you have a low pressure with no clouds? My understanding is that low pressure is usually associated with cloudy skies, however in the last week I observed a day that had low pressure reading but no clouds. Is this possible or did I perhaps read the barometer wrong?

Low pressure areas are simply areas where the air is less dense than the surrounding areas.  The scale of these differences can be small, size of a small farm field, or large, covering expansive areas of the earth's surface.

Clouds form around low pressure systems typically because of the way low pressure systems develop and the air moves around them.  Typically low pressure systems start forming over relatively warmer sections of the earth, this is gross over simplification, as the air warms it becomes less dense and will tend to rise.  This rising of warm often moist air leads to cloud formation because as air rises it also becomes cooler, see Lapse Rate.  Typically at some time the temperature drops below the dew point in the air moving upwards and clouds form.  But if the relative humidity of the air is not high it is possible that the air mass in the low pressure system never rises high enough for condensation of the water vapor to occur, in this case no clouds will form.

In conclusion it is possible to read a lower air pressure at a site and still not have cloud formation.

This site has a good graphic on how low and high pressure systems relate to each other. High and Low weather systems described

Please note that I am not meteorologist.  This answer does not consider all the aspects associated with cloud formation such as how airborne particulates can affect condensation


Is it right practice to validate temperature data with reanalysis data? For a project I am working on I received raw data from many meteorology stationes.
Is it right practice to test this data against air temperature from noaa's ncep/ncar data?

No. One should not attempt to validate raw in situ measurements with reanalysis data. Reanalysis data contain partial information from observations, but are mostly model fields and are provided at very coarse resolution (~1$^\circ$ longitude). In situ measurements are subject to very local, small scale effects, and have different meaning from reanalysis. One is always free to compare the two datasets, but this would not be validation in any meaningful way.


Why does Earth have abundant oxygen in the atmosphere? Because of photosynthesis, obviously.  But then it's not actually that obvious after all, because photosynthesis is mostly balanced by respiration.

We can summarise the processes of photosynthesis and respiration like this:
$$\begin{align}
\ce{CO2 + H2O + h\nu} & \rightarrow \ce{O2 + CH2O(biomass)} \\
\ce{O2 + CH2O(biomass)} & \rightarrow \ce{CO2 + H2O}
\end{align}
$$
If these two reactions were always balanced then the concentration of $\ce{O2}$ would never have changed. If I understand correctly, there are abiotic processes that deplete $\ce{O2}$ to form iron oxides, and so in order for there to be persistent $\ce{O2}$ in the Earth's atmosphere, the biosphere as a whole must continually produce an excess of it. In order for this to happen, there must be some processes that consume biomass at a non-neglible rate without removing a corresponding amount of $\ce{O2}$ from the atmosphere.

My question is, what are the processes primarily responsible for this? Is it just due to the burial of unoxidised organic matter (to form coal, oil, etc.), or is there more to it than that?

If they are different processes, I'm interested both in what caused the excess of oxygen during the "great oxidation" event, and what maintains the excess of oxygen on the modern Earth.

This is only a partial answer as it doesn't explain why the excess of O2 stayed but one thing you have to appreciate is the fact that aerobic respiration appeared almost half a billion years after photosynthesis, so we can't really say that photosynthesis and respiration have always balanced each other.

Cyanobacteria (and with them photosynthesis) are believed to have appeared somewhen between 3 and 2.5Ga (e. g. Altermann & Kazmierczak 2003, Brocks et al. 2003) while Mitochondria (and with them aerobic respiration) are thought to have evolved somewhen between 2 and 1.5Ga (see e. g. Hedges et al. 2004).

There was therefore at least half a billion year during which cyanobacteria were producing O2 with no one able to breathe it.


What does the notation "kg kg-1" for units mean for specific humidity I am using a data set from Justin Sheffield at Princeton that includes specific humidity. The description is as follows:

  "Reanalysis 6-hourly specific humidity interpolated to 1.0deg with account for elevation changes and scaled to maintain consistency with reanalysis relative humidity and CRU TS3.0 monthly air temperature";
  The units are = "kg kg-1"

What does this mean? Does it mean kg/kg so essentially dimensionless? Because I have seen around specific humidity measured also in g/kg, maybe that was wrong?
These two pages give the same definition of specific and absolute humidity, I thought they were different things:
http://www.nzifst.org.nz/unitoperations/drying3.htm
http://tigge.ecmwf.int/tigge/d/show_object/table=parameters/name=specific_humidity/levtype=pl/

Yes, the unit is kilogram per kilogram, but it means kilogram of water (moisture) per kilogram of air.


Under which conditions is this a realistic representation of a storm? Yesterday Katie Mack (an astrophysicist) tweeted this parallel between a dust storm on Mars and a storm shown in the movie "Mad Max Fury Road" (she says "Can’t help imagining that this is what that Mad Max: Fury Road scene would have looked like from above.").

This made me ask: is the movie-storm a realistic storm? If yes, under which conditions?

In particular, what caught my attention is the shape of the cross-section of the advancing storm, with dust protruding on the bottom and clouds protruding on the top at the same time. Could it happen?

Fury Road storm seems like quite a realistic representation of a huge thunderstorm advancing over a desert. The gust front would lift the sand in the bottom ahead of the storm center. Winds inside the storm would diverge at a higher altitude causing the sand to fall ahead of the actual storm.

In reality thunderstorms do produce dust storms with their gust fronts, but there isn't as visible upper part as in the screenshot of the film. This is because the storm is moving so fast that the dust hasn't had time to get lifted up and blown forwards from above. Such storm should have very slow speed advancing on average and large updrafts lifting dust efficiently upwards. And perhaps the dust is very fine so that it acts like gas and gets easily advected.

The situation is similar to an actual cumulonimbus/thunderstorm where the top and the bottom of the cloud are pronounced. See illustration of a cumulonimbus here.


Is there a standardized way to define weather icons based on quantifiable data? Weather APIs often provide weather icons which symbolize the state of the sky at a given point, or as an average.

Is there a standardized way to define what the icon should be, based on quantifiable data such as cloud coverage or rain level (the idea being that, say, with over 50% cloud coverage a "cloud" is used, then with > 1 mm rain, "drops" are added etc.) -- or is it left purely as an interpretation to the user?

The icons that are used are really just a visual way of describing the phrasing that is used in the forecast.  You can see an example of one set of icons offered by Weather Underground here.  And they have many different icon sets that you could choose from.  The phrases themselves are linked to specific ranges of probability of precipitation, cloud cover, or other factor.  This site shows some nice tables that give some of those ranges (sample shown below):


How to compute du/dx and dv/dy in moisture flux convergence? I have $u$-wind, $v$-wind and specific humidity. I would like to compute moisture flux convergence at a grid point. So I need to compute the value of $q(\frac{du}{dx} + \frac{dv}{dy})$

My question is how to compute $\frac{du}{dx}$ and $\frac{dv}{dy}$ from $u$-wind and $v$-wind at gird point $(x,y)$?

I imagine you have a horizontal grid (x,y) and wind components u(x,y) and v(x,y).

Normally we do 1-degree of latitude constant (y-dimension) and equal to 110000 m. So your dy depends on how many degrees per grid point you have (yresolution). However, dx will vary according to the latitude.

dy=110000*yesolution;

Now, we may use centered finite differences to compute what you want. (Below is a Matlab code, but I believe generic enough to be reproducible in other languages).

for y=2:length(lat)-1
     dx=abs(110000*cos(latx(y)*(2*pi/360))*xresolution);
     for x=2:length(lon)-1
          div(x,y,1)= (u(x+1,y,1)-u(x-1,y,1))/(2*dx) + (v(x,y+1,1)-v(x,y-1,1)/(2*dy);
     end
 end

Note that you will have an empty frame around your divergence field when x=1, x=max(x), y=1 and y=max(y), once they do not have two neighboors to compute the differences. The same is observed when you do this with hdivg function in Grads, for example.

By the way, this is based on and yields the same result as Grads (I have checked!) using cdiff to reproduce hdivg.

Also note that the cosine function here works with radians, so if you compute cosines in degrees directly (i.e. cosd function in Matlab), you should omit the term scaling by (2*pi/360), which is just a conversion.

Hope it helps!


Simple Climate Models that Predict Climate Change I have a physics/maths background, and to me the most convincing evidence of a scientific fact is often a very simple model that gives reasonable results. A simplified model, which you can fully understand, is often more convincing than the numerical output of a complex model.

So I'm looking for a simple climate model, which could be used to predict how increasing $\ce{CO_2}$ concentrations would lead to an increase in temperature. For instance, this model need not account for the specific geography of the earth: it might just consider the sun's radiation passing through an atmosphere and striking a flat, homogeneous surface with some constant absorptivity. Then we could just do an energy balance to find equilibrium temperature as a function of $\ce{CO_2}$ concentration.

If I could do a back-of-the-envelope calculation to see that doubling $\ce{CO_2}$ concentrations would be expected to increase the Earth's temperature by a few degrees (I expect nothing more than this order-of-magnitude evidence) then I think that would be convincing evidence that climate change is worth worrying about. You can argue with satellite data and complex models, but you can't argue with basic physics.

So far, the only source I've found that gives such a simplified model is Arrhenius himself, in his 1896 paper. I actually find his model quite convincing, and its prediction ($\sim5^oC$ increase from doubling $\ce{CO_2}$) gives a reasonable prediction too. However, his data on the absorptivity of the atmosphere was (rather quaintly) obtained by early measurements of the intensity of moonlight at different angles in the sky. While this is ingenious, I'm sure with modern technology we could get much better estimates.

So I'm basically looking for either a modernised version of Arrhenius' calculations, or alternatively another simple model which convincingly demonstrates that increasing $\ce{CO_2}$ has the power to alter the earth's temperature by as much as a few degrees.

It sounds like you are asking for a simple climate model that treats the globe as a homogeneous system.  A Zero-dimensional energy balance model may be of interest (see this Penn State site ) which uses a starting point that is also basically described in this post: Temperature as a function of luminosity and greenhouse gas concentrations

If you solve for the temperature of the Earth using a blackbody approximation and thermal equilibrium (with an albedo of about 0.3), you are only taking into account the temperature of the Earth without a greenhouse effect (about 33 °C cooler than actual temperature).  It is very difficult to calculate the influence of infrared radiative forcing components (e.g. oxygen and carbon dioxide) to an empty atmosphere, since there are many biogeochemical/physical feedbacks in the climate system which are nonlinear.

However, if you start with current temperature/concentrations, you can add terms that represent particular increases to the GHG concentrations.  Those terms are constantly being refined by the scientific community and are summarized in this Wikipedia article about Radiative Forcing (see the link to climate sensitivity at the bottom of the page).  This is a kind of "back of the envelope" modeling approach that most climate scientists use to discuss past and future influence of climate forcers, which can be very helpful when discussing the expected climate effects of various emissions/mitigation scenarios.

The Radiative Forcing article in Wikipedia says:

  Radiative forcing can be used to estimate a subsequent change in equilibrium surface temperature (ΔTs) arising from that radiative forcing via the equation:

$$\Delta T_s =~ \lambda~\Delta F$$

  where λ is the climate sensitivity, usually with units in K/(W/m2), and ΔF is the radiative forcing. A typical value of λ for CO2 is 0.8 K/(W/m2), which gives a warming of 3K for a doubling of CO2 [from pre-industrial times, assuming RF of CO2 is ~ 3.7)].

It's important to note that most climate sensitivity parameters and radiative forcing values are derived with respect to some time period (e.g. 1750 - 2011) and GHG feedbacks are set for a particular climate regime.  Revised radiative forcing and climate sensitivity parameters are needed to make calculations for other climate regimes.

The IPCC AR5 figure of Radiative Forcing components attributes 1.65 W/m2 to the actual increase in CO2 concentrations from 1750 to 2011 (or ~1.3K in temperature rise, if lambda is 0.8).

The values of Radiative Forcing that IPCC provides are likely close enough for doing realistic scenario testing.  However, something in an entirely different feedback regime would need very different parameters.  For instance, halving the CO2 in the atmosphere would have a much greater absolute effect on temperature than doubling it; temperature increases logarithmically as CO2 concentrations increase. This is because strong CO2 absorption bands are nearly saturated under present conditions, and increasing CO2 from present conditions mainly affects the flanks of the absorption bands.  However, water precipitates out of the system easily under low-temperature conditions, so CO2 has a much greater forcing value in those types of climate regimes.

The most difficult problems with testing simple climate scenarios are related to the oceans and the biosphere, which could have countless effects on the average climate and GHG regime.


Does Magnetic Deviation depend on altitude? Magnetic declination, sometimes called magnetic variation, is the
  angle between magnetic north and true north. Declination is positive
  east of true north and negative when west. Magnetic declination
  changes over time and with location. As the compass points with local
  magnetic fields, declination value is needed to obtain true north.

This is the definition provided by National Center For Environmental Information.

As you can read, magnetic declination depends on location (lat/lon) and time but I would like to know if it also depends on the altitude and why it does or does not.

Note: possibly attach some reference to the answer.

Technically yes, but practically, usually no.

The magnetic field varies in three dimensions and the variations are not parallel to the Earth's surface. However, horizontal distances varies usually on a larger magnitude than elevation and for everyday use, the declination is only based on horizontal position.

The common model for the Earth's magnetic field, IGRF-12 (International Geomagnetic Reference Field) uses the radial distance from the center of the Earth (r) in relation to the geomagnetic conventional Earth’s mean reference spherical radius (a). In other words, the elevation. (a/r) will usually be almost 1, and the difference is less than any other uncertainties in the measurements, e.g. deviation due to magnetic anomalies.

The magnetic field is often calculated by using the published model and this code. As you can see (the code is rather easy to read), the vertical component is used to estimate the magnetic field for every location.

Online calculators are often used to correct declination for navigational purposes:

NOAA's calculator doesn't include elevation and is calculated at sea level.
BGS's and GA's calculators does, but the elevation have a small impact on the declination. Try it to see!

Simpler models, e.g the dipol model also suggest how the magnetic declination varies with distance from the Earth's magnetic center. The approximation work rather well at low latitudes and at the Earth's surface. It can also be used to model Earth's magnetic field in an astronomical context.

Apart from declination, there is also an inclination to take into account. Near the magnetic poles, the compass needle will not be horizontal, but get a dip. This can make normal field compasses almost useless near the magnetic poles (trust me, I tried..).


Why do clouds sometimes form a ring around the Sun? Someone shared a video with me in which clouds were forming a ring around the Sun. I took this screen shot of that video:

What is the reason behind this?

This optical phenomenon is called a 22° halo which is a subset of other halos.  This arises from sunlight refracting through hexagonal ice crystals, which can be found in high level cirrus clouds.  Light that would otherwise not make it to your eye enters an ice crystal and then exits at an angle of approximately 22 degrees.  This produces the arc of light you see in the video.  You see (inverted) rainbow coloring of the halo because the light is not uniformly refracted but varies from 21.7 degrees for red photons to 22.5 degrees for violet photons.

Image by donalbein, Wikemedia Commons, CC-By-SA-2.5 https://commons.wikimedia.org/wiki/File:Path_of_rays_in_a_hexagonal_prism.png

There are many more optical phenomenon you can see, but all are based around sunlight and the optical properties of ice crystals and water droplets.  Here is a phenomenal picture taken in winter showing many examples:


Madden-Julian Oscillation (MJO) - How to interpret the index? How the Madden-Julian Oscillation (MJO) index can be interpret?

Let's suppose I have got an MJO index value of 0.6 in a given day, what this does mean?

Does 0.6 represents a weak MJO?

Data from https://www.esrl.noaa.gov/psd/mjo/mjoindex/
Thanks

I prefer to use the BOM MJO index and the explanation provided over there -

  When the index is within the centre circle the MJO is considered weak, meaning it is difficult to discern using the RMM methods. Outside of this circle the index is stronger and will usually move in an anti-clockwise direction as the MJO moves from west to east. For convenience, we define 8 different MJO phases in this diagram.

So in your case your signal value is 0.6 and that means it is fairly weak in amplitude as it is inside the circle. You also need to mention the phase of the MJO phase diagram. There are eight phases

Phase 1 & 8 - Western Hemisphere And Africa

Phase 2 & 3 - Indian Ocean

Phase 4 & 5 - Maritime Continent

Phase 6 & 7 - Western Pacific.

Currently the signal is a weak one as seen in this phase diagram MJO Phase diagram

When the MJO signal is strong it's amplitude will be greater than 1 and the contour line will be outside the circle. It should be noted that the MJO is an empircal index consisting of the 850 hPa winds, OLR and 200 hPa winds.

MJO passage through phase 6 and 7 is always of global interest as the impact can be of planetary scale. Usually El Ninos are preceded by Westerly Wind Bursts and the forcing factor can be a MJO passage through phase 6 and 7.

One can look at the raw data of the signal here - RMM Index text. This provides the amplitude of the signal as well as the phase of the MJO.

Another version of the same can be seen here - MJO RMM index


Climate change model What counter heating effects are taken into account in popular climate change models?

Are things such as the cooling effect of polar ice melt and the desalination from ice melt rivers taken into account?

Both are relatively small secondary effects on global temperature, but at the same time, both are worth studying.

Ice acts like a heat sponge.   It takes heat to melt ice (that's kinda obvious) and less obvious, forming ice releases heat.   Large blocks of ice cool the air when it's above freezing and warm the air when it's below.   Even as ice changes from solid to liquid, without changing it's temperature, the energy change is significant.   This is called it's latent heat of fusion or LHF.

By this article about 2% of the trapped heat goes into melting ice.   A 2% variation on heat that would otherwise, warm the air or oceans or escape into space is almost small enough to ignore, but the effect grows much larger locally near glaciers (perhaps as high as 25%, but the source I found for that isn't one I'd call credible), but certainly, the effect is significantly higher than 2% in the arctic and antarctic.

Latent heat of fusion is also important in calculations for the total energy balance and here's an article from IPCC AR4 that discusses LHF.

As pointed out in the comments, melting sea ice also changes ocean albedo and can lead to a rather quick change in absorbed heat and a reduction in the light reflected back into space.  This also happens on land, mostly snow cover, (as glaciers melt too slowly for that kind of year to year effect), but I don't think that's what the question is asking.

The 2nd part, the change in salinity is (to my knowledge, and I invite correction from anyone who studies this professionally), but from what I've read, it's much more complicated to model than LHF.   It's certainly something that researchers are aware of, both the warming arctic and changes in salinity might effect ocean currents.   Unlike air, where temperature drives density and that leads to low and high pressure and updrafts and basically, weather, oceans have two factors that can effect water density - temperature and salinity.   Warm water is lighter than cold water, but lower salinity water is also lighter than warmer, more salty water.

Small changes in salinity in ocean surface water doesn't have any real effect on temperature directly, but the effect on ocean currents is worth studying because changes in ocean currents can effect temperature significantly, certainly on a local scale, and globally too.

Modelling this, like modeling the rate of future glacial melt, may be close to impossible and when it's written about, it can sound more alarmist than science, so it's a subject that has 2 key problems, difficult to predict (even in the hard to predict climate change models the effect of salinity changes on ocean currents are especially tricky), and it's the kind of thing that might invite shouts from the other side, like Wiesław Masłowski's infamous study that suggested the arctic might be ice-free by 2013.   He was a graduate student, not a climate scientist and his was a good study and preliminary.  It should have been subject to peer review, instead, (in part because Al Gore used his study in a speech), it became a famous example to alarmism and was subject to much ridicule that persists to this day.

Scientists are certainly aware of the possible effects of salinity changes and the effect that might have on ocean currents, but it's very difficult to model and mostly written about observed effects in the present, and mostly not written regarding future forecasts.   Predictions like "the Atlantic current could shut down" may be based in science, but low probability predictions like that tend to invite criticism.

Fresh water/ice melt is discussed in Wikipedia here and here's an article where the opposite is happening.   As arctic sea ice recedes, the lighter, less salty water on top isn't present as much as it used to be and the arctic-ocean stratification is going away.   Article on that here.

It's also worth noting that glacier fed rivers will affect the ecosystem down-river, putting less or more water into the environment leading to less or more greening and less or more water evaporating off the river into the atmosphere.   It's more common for rivers to be fed by snow melt than glacial melt, but the glacial melt should be studied too.   Changes in greening down-river and in atmospheric water vapor also effect climate, so, as someone said in the comments, there are a lot of moving parts to the climate picture.

Generally speaking, scientists are aware of all of this and as much as possible, study it and include it in models.


How to Calculate the Equilibrum Function for use in Tide Calculations using Harmonic Constituents I have found a wealth of information using Harmonic Constituents to calculate tide levels.  H(t) = Amplitude * cos ( t * harmonicSpeed + phase lag).  Yet when I try to apply this to constituents from the NOAA the results never match the tables produced.  I have since found that the NOAA constituents require an additional term, the Equilibrium value with the resulting constituent function looking more like

H(t) = Amplitude * cos ( t * harmonicSpeed + EquilibriumValue + phase lag).

I am looking for any hints on how this Equilibrium value is calculated or derived.

Addendum:
The constituents are found here, this for San Diego, Ca
Constituents for San Diego.  Sorry cannot find the source of the Equilibrium value but will when I get home from work.

The equilibrium tide is a theoretical concept developed by Newton in the 17th century that considers only the gravitational pull by the Moon and the Sun and the centrifugal forces, with no inertia, no friction and no land masses. From the NOAA definitions:

  equilibrium theory — A model under which it is assumed that the waters covering the face of the Earth instantly respond to the tide-producing forces of the Moon and Sun to form a surface of equilibrium under the action of these forces. The model disregards friction, inertia, and the irregular distribution of the land masses of the Earth. The theoretical tide formed under these conditions is known as the equilibrium tide.

Lets consider as a first cut the two tide producing forces associated with just the Moon and Earth systems. The two are the centrifugal force of Earth from the center of mass of the Moon-Earth system and the gravitational attraction due to the Moon. For the point closest to the Moon:

$$ Force = \frac {GM_1M_2}{(R-r)^2}  - \frac{GM_1M_2}{R^2} $$ where R is the distance between Earth and Moon and r is Earth's radius. It gets a bit more complex for other points on the surface because instead of R you have to use $$R\pm r cos(lat) $$ for the gravitational force.

Ultimately, it comes down to a set of horizontal forces that tend to concentrate water in two theoretical points: one, the closest to the Moon and two, the farthest from the Moon. An equilibrium state would be reached, called equilibrium tide, that results in an ellipsoid with its two bulges directed toward and away from the Moon.

In practice, this equilibrium ellipsoid does not develop because of the rotation of Earth, the Coriolis acceleration and the fact that the tidal wave feels the ocean bottom and it is subject to friction. Going beyond the equilibrium theory of the tide, the dynamic theory of tides (developed by Euler, Laplace and Bernoulli) includes all these concepts (friction, inertia, Coriolis, land masses,...) and provides a better approximation to the observed tides.

My recommendation to understand tidal constituents is to use some decent tidal harmonic analysis package like t_tide for Matlab, which is likely to be the tool that NOAA is using to generate their harmonic constituents in the first place. If you really want to get into the details, then the best option is to go to the still likely best source available. That will the 1973 book "The Analysis of Tides" by G. Godin.

Pawlowicz, R., B. Beardsley, and S. Lentz, "Classical Tidal Harmonic Analysis Including Error Estimates in MATLAB using t_tide", Computers and Geosciences, 28, 929-937 (2002).

Godin, G. (1973). The analysis of tides., by Godin, G.. Liverpool (UK): Liverpool University Press, 264 p.


Can I see curvature of the Earth on a picture I've been piloting a drone yesterday and made a video hovering at around 100m, the visibility was very good. I've shown the video to a friend of mine and we were arguing whether it's the curvature of the Earth visible at horizon, or is it just camera lens disturbing the image.

I'm rather sure it's curvature of the Earth, but still don't know how to prove it. I already checked that on the lower part of the picture the buildings-roofs are lined up straight, no disturbance. Yet the horizon is slightly bent (checked by putting a piece of paper on the screen). Which I think already proves the point: should lens disturb the picture, the disturbance would be visible both on the upper and lower part of the picture symmetricly.

I've read How high must one be for the curvature of the earth to be visible to the eye?, but it doesn't help me just yet. As far as I understand, the above link help calculating the distance to the horizon depending on the altitude. But why is the horizon bent?

I would appreciate anyone helping me straighten my thinking :).

The picture:

Make your own numbers... I will take one sample, just change your figures.

I have one 4K resolution dron camera. So I can register 3840 x 2160 pixels in one shot.

Considering the Earth as an spherical body, the distance from you to the horizon depends on Earth Radius and your height by:

  Distance = (Radius + height) * Sinus {arc-cosinus [Radius / (Radius + height)]}

So depends on your height:

| Height (m) | Dist. To Hor. (Km) |
|:----------:|:------------------:|
|    1       |         3,6        |
|    10      |        11,3        |
|    100     |        35,7        |
|    1000    |        112,9       |
|    10000   |        357,1       |

My camera have a 24mm focal (on 35mm relative), so in the photograph, I can see 84º degrees. So the distance on between the edges of my horizon is:

| Height (m) | Dist. To Hor. (Km) | Dist. Edge<>Edge (km) |
|:----------:|:------------------:|:---------------------:|
| 1          |         3,6        |          4,8          |
| 10         |        11,3        |          15,1         |
| 100        |        35,7        |          47,8         |
| 1000       |        112,9       |         151,1         |
| 10000      |        357,1       |         477,9         |

Once that you have this data, you just need to calculate the arrow of the arc expected:

  Arrow Circ. Arc = Radius * Cosinus [ Arc-Sinus (Dist/2/Radius)]

So, with this data and the initial data of the camera:

| Height (m) | Dist. To Hor. (Km) | Dist. Edge<>Edge (km) | Arrow (km) | Curvature (pixel) |
|:----------:|:------------------:|:---------------------:|------------|-------------------|
| 1          |         3,6        |          4,8          |    0,000   |         0         |
| 10         |        11,3        |          15,1         |    0,004   |         1         |
| 100        |        35,7        |          47,8         |    0,045   |         4         |
| 1000       |        112,9       |         151,1         |    0,448   |         11        |
| 10000      |        357,1       |         477,9         |    4,482   |         36        |

So finally... If you have a perfect visibility condition, your camera is completely leveled, there is no fish-eye distortion... On a 4K camera at 10 km height, the Earth curvature it will be 2% -> 36 pixels on 3840 pixels wide.

If you are interested on it, that is not the way. Human eye can not perceive clearly the Earth Sphericity at a commercial flight altitude. So forget your dron, yo should go for a balloon.

Hope it helps!


Is there less precipitation (rain/snow) in cold weather than in warm weather? To me it seems that there is a lot less precipitation when temperatures go below -5°C (23°F).

Is there any evidence for that? Could this be a local phenomenon (I live in Germany)? What could be the cause for this?

I tried searching for charts that show precipitation over temperature, but all I could find were charts that showed precipitation over months of year which is not quite the same.

In most of Western Europe, precipitation comes from westerly winds formed over the warm waters of the Gulf Stream current that flows from the Caribbean to Western Europe. That is a warm current, so these winds would be warmer and make your area warmer. Besides, as they come from a warm ocean they are moist and carry water vapor and make clouds. Low temperatures are usually caused by easterly/northerly winds originating in Siberia, these are cold and come from land. Cold winds cannot carry moisture and on top of that they travel over land rather than the ocean and are dry. So in most cases in European winter, relative warmth and precipitation go hand-in-hand.