Can the previous weather be computed from the current situation?

The underlying equations for fluid-dynamic models are hyperbolic partial differential equations. They can generally be written in the form
$$
  \frac{\partial}{\partial t} u(t) = D(u(t))
$$
where $D$ in some way evaluates the current state of the system and its spatial derivatives.

A numerical simulation then integrates this differential equation, to extrapolate from a start state $u(t_0) = u_0$ the time-dependent $u(t)|_{t>t_0}$.

Well, if we can do that, then surely we could also solve in the inverse time direction, by considering the equation
$$
  \frac{\partial}{\partial t} u(-t) = -D(u(t))
$$
and running the integrator with $\tilde t = -t$, $\tilde D = -D$?

Actually, you quickly run into problems when you try that. The operator $D$ can be characterised by its Jacobian, which basically tells you how pertubations in the state influence the derivative. Specifically, the complex eigenvalues of the Jacobian can tell you whether a small deviation will a) amplify over time (positive real part), or b) decay (negative real part), or c) just oscillate (purely imaginary).

For physical systems the eigenvalues tend to be mostly c) or b): you get a lot of wave-like solutions which propagate / oscillate over the system, and tend to decay over time. a) however is more tricky: if you start with a small deviation from the start state, the system will over time deviate ever more and and more. Now, this kind of thing is by no means unheard of especially in meteorology; it's the essence of a chaotic system. Storms can emerge and grow stronger over time, but only by scooping up energy that's already stored in the system. At some point they'll stop.

OTOH, you always have a lot of consistently negative real-part eigenvalues. These correspond to dissipative effects: small-scale pertubations generally are smoothed out to zero by the physical effects, e.g. winds have friction, mixing of air of different temperature averages out the differences, etc.. If you now run the simulation backwards, you turn those negative real parts into positive real parts, and that means the system is suddenly massively chaotic on all length scales. Small pertubation arise out of numerical uncertainties, and grow over all bounds. You would not only end up with states different from the actual weather a week ago, but with states that are completely unlike anything the weather has ever been like – huge, erratic temperature fluctuations and small vortices with crazy wind speeds.


Yes, it's called polar motion.

The rotational pole moves continuously, as you can see from the right-hand side of this figure (below) by the Earth Orientation Centre (EOC) and the International Earth Rotation and Reference Systems Service (IERS). The figure shows about 650 days of time; mjd is modified Julian day and time goes along the locus in the polar motion diagram.

The left-hand side of the figure shows Length of Day, and I think that's chiefly why they are making these refined polar measurements — to keep track of leap seconds, etc.

It's been established, by Fong et al. (19996), among others, that earthquakes could change the earth's rotational axis, by redistributing mass on a large scale. Their table 1 shows [the sort of effect that individual earthquakes theoretically have — edited after comment from David Hammen] on the length of day; $\Psi$ is the excitation vector:

The EOC's website lists other geophysical excitations, among them:

Atmospheric angular momentum
Oceanic angular momentum
Hydrological excitation function
Axial angular momentum of the core

References

Chao, Benjamin Fong, Richard S. Gross, Yan-Ben Han (1996). Seismic excitation of the polar motion, 1977–1993. Pure and Applied Geophysics, September 1996, Volume 146, Issue 3, pp 407-419. DOI 10.1007/BF00874727

Update: This article on some new research in this field is worth a read.


Intuitive definition of the Monin-Obukhov length?

The way I understand it is the height above the surface at which buoyant (heat) production of turbulence first equals the mechanical (shear) production of turbulence.

A more useful definition is given by the American Meteorological Society:

  A parameter with dimension of length that gives a relation between parameters characterizing dynamic, thermal, and buoyant processes.

  At altitudes below this length scale, shear production of turbulence kinetic energy dominates over buoyant production of turbulence. It is defined by

  where k is von Kármán's constant, u* is the friction velocity (a measure of turbulent surface stress), g is gravitational acceleration, Tv is virtual temperature, and Qv0 is a kinematic virtual temperature flux at the surface. The parameter was first described by Obukhov in 1946, and therefore should not be called the Monin–Obukhov length, even though there is a Monin–Obukhov similarity theory that uses it. The Obukhov length, of order one to tens of meters, is the characteristic height scale of the dynamic sublayer. The Obukhov length is zero for neutral stratification, and positive (negative) for stable (unstable) stratifications. The dimensionless Obukhov length z/ L (where z is height above the surface) is used as a stability parameter, with z/L = 0 for statically neutral stability, and is positive (negative) in a typical range of 1 to 5 (-5 to -1) for stable (unstable) stratification.


Why does the annual temperature range get bigger where latitude is higher?

Welcome to the Earth Science Stack Exchange!
The bigger temperature range is because of the wide difference in the length of daylight seen in the higher latitudes. Depending on how high you are talking about, there can times where there is no sun for a month and vice versa with constant sun. Because of this there will therefore be more time for warmth and cooling as the Sun (or lack thereof) is what drives the diurnal temperature change the most.


sulfur hexafluroide as a greenhouse gas and its atmospheric journey

Carbon dioxide is heavier than air but is found in nearly equal proportions throughout the well-mixed portion of the atmosphere. Xenon is much heavier than air, almost as heavy as sulfur hexaflouride, but it too is found in nearly equal proportions throughout the well-mixed portion of the atmosphere.
Turbulence is what makes the well-mixed portion of the atmosphere be well-mixed. It is only in the uppermost layers of the atmosphere, where the air is very thin, that turbulence fails to keep things well-mixed. The turbopause, which marks the border between the well-mixed lower portions of the atmosphere and heterogenous upper portions of the atmosphere, is about 100 km above the surface. The stratosphere is only 10 to 50 km above the surface, so it is well within the well-mixed portion of the atmosphere.


Is there a minimum snow depth to be considered as presence of snow?

A measureable amount of precipitation is considered to be at least 0.01 inch of water. Since on the average 10 inches of snow melt to 1 inch of water, this corresponds on the average to 0.1 inch of snow. Anything less is considered a trace (too little to measure).


How to calculate hydrostatic equilibrium?

I am assuming you are asking for the case of sea level being 200m higher and in isostatic equilibrium. In that case we can make use of Airy's isostasy model: https://en.wikipedia.org/wiki/Isostasy#Airy

Applied to a water column over the mantle, you have to replace $\rho_c$ with your $\rho_w$. The total increase in ocean depth is $x = b_1 + h_1$, where $h_1$ are the 200 m extra sea level and $b_1$ is the depth increase to isostasy. Using the equation given in the link:
$$
x =  \frac{h_1 \rho_w}{\rho_m - \rho_w} + h_1,
$$

giving $x =$ 287 m


Have there ever been significant events of ecological or geological disasters occurring simultaneously alongside extreme meterological disasters?

Here's an amazing one:

In June 1991 category 3 Typhoon Yunya made landfall in central Luzon in the Philippines and gradually weakened as it moved across the island... however as it was near exiting the island... Mount Pinatubo sustained the largest volcanic eruption in our lifetimes -- almost directly in the center of the storm!

Here is a remarkable satellite loop showing the event unfold.

Screen cap shortly before the event unfolder showing the locations, from UW CIMSS

The CIMSS Satellite Blog has a nice post writing up the event, along with a collection of additional satellite data.


The density of a water parcel evolves as it moves in the ocean. A parcel of bottom water (e.g., Antarctic Bottom Water (AABW) the densest water mass) is much denser than the surrounding water when it is formed near the surface (in the case of the AABW in polynyas and below the ice shelf in the Weddell and Ross Seas).

As the parcel of AABW sinks, it flows down the Antarctic continental shelf and slope and moves north along the bottom at speeds of 2-8 cm/s. The issue is that the water parcel of AABW will be interacting and mixing with the surrounding water.

Thus, a parcel of AABW that when it was formed had a temperature of around -0.8°C and a salinity of 34.7, will get mixed with warmer water masses (e.g., Antarctic Intermediate Water, North Atlantic Deep Water) and as it becomes less dense it will be more easily mixed with the surroundings.

If you have surrounding waters that are more saline and with comparable temperatures, then the less saline water parcel (modified AABW for instance compared to NADW) will rise. The interior circulation of the ocean is predominantly driven by thermohaline processes (differences in density) and the thermal wind equation can be used as a simplification of the dynamics, especially away from the edges (surface, bottom).


Where is the Arctic Circle for civil and astronomical twilight?

As mentioned by Barry in the comments, you just have to substract 6 degrees of latitude to get the ring around areas were it never gets DARKER than civil twilight or darker for 24 or more hours (or 12° for nautical twilight and 18° for astronomical twilight). That would be:

"Never darker than civil twilight circle": : 60° 33′ 46.7″ (orange in the figures below)
"Never darker than nautical twilight circle": : 54° 33′ 46.7″ (green in the figures below)
"Never darker than astronomical twilight circle": : 48° 33′ 46.7″ (light blue in the figures below)

To be clear, those circles surround the places where on summer solstice there is no complete nightfall (i.e. it doesn't get completely dark according to the different twilight standards).

In turn, you can think about other set of equivalent circles where it never get BRIGHTER than a given type of twilight for at least 24 hours, those would be

"Never brighter than civil twilight circle": : 72° 33′ 46.7″ (orange dotted)
"Never brighter than nautical twilight circle": : 78° 33′ 46.7″ (green dotted)
"Never brighter than  astronomical twilight circle": : 84° 33′ 46.7″ (light blue dotted)

To be clear again, those circles surround the places where on winter solstice there is no tace of daylight (i.e. they doesn't get any daylight according to the different twilight standards).

In a map, these 6 circles, plus the Polar circle (at latitude 66° 33′ 46.7″ and in red) would look like this around the North pole

And like this around the South Pole

And to have a world view, in geographic projection, they would look like this

The twilights are defined like that (i.e. Sun 6, 12, 18° below the horizon), so no corrections have to be made due to refraction or the size of the solar disk. Such corrections would only apply if you want to find the "effective" polar circle where the Sun would/would not be visible for 24 hours even with an horizon perfectly free of obstacles.
Due to those corrections, the polar night (at the Winter Solstice) only really happens some 80-100 km poleward of the polar circle. In contrast, at the polar circle itself, the Sun can be seen even at the winter solstice.


Since rocks can have magnetic signatures, as is the case with paleomagnetism, can they store past pressures also?

Yes, but not directly, and it is very hard to know the exact pressure (in contrast with paleomagnetism where you can actually get a nice number).

One way to do it is to simply look at the minerals you have in the rock. Some minerals exist only in a certain range of pressures. The best and most known example is the two polymorphs of carbon: graphite and diamond. If you find a rock with diamonds in it then you know that (at least the diamonds) formed at great depth. There are many other mineral transformations that depend on pressure. Andalusite is low pressure whereas kyanite is high pressure. Plagioclase is low pressure, spinel is intermediate pressure and garnet is high pressure. You can look up images of those minerals for some eye candy! Note that I'm using the words "low" and "high" quite vaguely, as everything is relative and the exact numbers are not important for this case.

Another way of finding the pressure of a rock formation is by using the chemical compositions of the minerals in it. This is basically an improved version of the previous method. Instead of having a certain mineral assemblage transform to a second assemblage, we take two coexisting minerals. If we know how a certain element (for example magnesium or aluminum or whatever) behaves in a certain fashion relative to the pressure, we can measure that element in both minerals and find out the pressure.

A third way is to look at a thing called fluid inclusions. These are tiny bubbles of liquid or gas (or something in between, the distinction may be fuzzy at high pressures and temperatures), trapped in a mineral when it was forming. This is essentially a tiny pressure vessel, so using all kinds of methods we can figure out what was the pressure during that time. Here's a picture of how it looks under the microscope:

source: http://rohmin.unileoben.ac.at/de/3388/

  But is there any way they can look at rock morphology to see how high sea levels may have been?

Not using these methods. The problem is sensitivity. I was talking to you about diamond and graphite - this is extremely deep in the Earth. The pressure of adding a few metres of ocean on top of it are definitely not even going to tickle that transformation. Furthermore, for this to actually be effective, the rocks have to be directly on the seafloor. Once you start burying them in the sediment, you have additional pressure on top of them. A general rule of thumb is 100 metres of additional rock pressure equal 300 metres of water pressure. How would you know how much rock was on top of it? And as the rock gets buried deeper, how would you know what time frame your measurement represents?

So in short, no. We can't do it. There are, however, other methods of figuring out sea levels without using pressures.


Can an earthquake occurring within two plates trigger another earthquake in a far away plate boundary?

I guess that it is possible.

As mentioned by others, on top of static stress triggering, which is the effect of an earthquake in its vicinity (typically up to 2-3 rupture lengths away), another way by which an earthquake can trigger others is by dynamic stress triggering, which may have effects farther away but are hard to identify and measure.

So far, the only earthquake that seems to have demonstrated how dynamic triggering over long distances can take place is the 2012 Indian Ocean earthquakes. This earthquake had an unusual combination of characteristics: it was a very large strike-slip earthquake, it took place in oceanic crust (which allows surface waves to travel longer) and had a very large concentrated burst of Mw 8.4 - 8.5 with large slip over limited distance (I do not post sources since there are at least 3 apparently good studies with very different results, for example one study says that slip exceeded 80m locally (!)). These characteristics suggest that this earthquake was very efficient at radiating energy through the crust rather than towards the interior or through very fragmented crust.

So, what this earthquake showed, is that it is possible to speed up significant earthquakes (earthquakes that were almost ready to happen, but they were sped up up to 90 days). Still, it is only one case, the mechanism is not well understood, we are not sure why this earthquake seemed to have caused this phenomenon while other earthquakes with higher magnitudes did not show the same effects.

Concluding, while static stress triggering has only a fairly local effect, dynamic stress triggering may have a wider effect but we only have detected one such case, which could very well have been a coincidence or due to an unusual mechanism. So the answer to your question seems to be yes, but with doubts and in any case for very special earthquakes.


How to interpret precipitation probability?

Given only this information and nothing else, there isn't really a definitively correct answer.

In reality, it depends on the nature of the precipitation. 61% could mean that there is a 61 percent chance of a thunderstorm between 4 and 5 PM, but since thunderstorms are short-lived, it might only be raining for 10 out of 60 minutes.

61% could also mean that an area of precipitation has a 61% chance of passing over Dallas. If that's the case, then it might be a 61% chance of rain for the entirety of the hour and a 39% chance of no rain at all.


Difference between pyrite oxidation and dissolution

Oxidation and dissolution are not the same processes, but they are related when discussing the breakdown of pyrite in near surface environments.

Dissolution is a process(*1) in which the chemical components of a solid become part of a liquid, in the liquid state. For example, take fresh water and salt. Now dissolve the salt in the fresh water. Instead of having NaCl and H2O, you now have H2O with Na+ and Cl- ions floating in it, not solid anymore. This is not something happens easily with pyrite. You can't just dump FeS2 in water and expect to have Fe2+ and S(2)- in the water. This is because the sulfur is in the reduced state, as sulfide. The more soluble form of sulfur is sulfate, the oxidised form.

Oxidation is a process in which a compound or element bonds with (more) oxygen(*2). In order to dissolve pyrite, you have to convert the insoluble sulfide to soluble sulfate. In your previous question you noted the oxidation reaction for pyrite:

$\ce{FeS2(s) + 15/4 O2(aq) + 7/2 H2O → Fe(OH)3(s) + 4H+(aq) + 2SO4^2-(aq)}$

You start with pyrite in the solid form, and you end up with sulfate (sulfur bonded to four oxygens). The leftover iron is deposited as insoluble iron oxides and hydroxides (basically - rust), and you're also generating a bunch of protons that make everything acidic (i.e. acid mine drainage).

*1 - strictly speaking, dissolution doesn't have to be a solid in a liquid. It can be anything in anything.

*2 - actually oxidation is an electron transfer reaction and can occur with no oxygen involved at all.


How effective is seeding rainclouds? What are the indications for initiating seeding?

Cloud seeding has been attempted for well over a century, with mixed success. There is no doubt at all that, locally, rain can be increased by seeding the cloud base with such nucleating agents as dry ice or silver iodide smoke. There is a great deal of doubt as to whether there is any overall gain in rainfall - or is it just an expensive way of 'robbing Peter to pay Paul', of increasing rain in one location at the expense of rainfall further downwind. Is there any nett gain in regional rainfall? Probably not. Of course, there are many cases where locally increased rainfall may be worth the effort.

Not all cases of cloud seeding work. There are several requirements. First, the process involves vigorous convection within the cloud towers in order to loft the nucleating agent into a sufficiently super-cooled part of the cloud. If the cloud is too warm it won't work - which rules out many tropical applications. Second, the 'seed' has to be injected into the cloud base at about -10 to -20 deg C, which allows 'graupel' to form, i.e. a kind of granular ice that flash freezes onto snow or ice crystals direct from super-cooled water vapour. Third, the timing has to be right, just as the cumulus towers are beginning to form, usually in the early afternoon. Fourthy, the economics are very moot. Wikipedia ('Cloud Seeding') claims that the subject is no longer 'fringe science', which is true in that the physics have been studied seriously, but mostly untrue in the sense that it is now considered economically attractive. The great majority of studies have shown that cloud seeding is unreliable, expensive, and of dubious value, with other ways of delivering water (such as desalination) being a better bet.


Atmospheric conditions needed that blocks all bright light during sunsets and allows you to stare directly at the sun

I am familiar with this phenomenon. In London, in winter 1947, I remember being able to see sunspots with the naked eye. The sun was very orange, like the one in your photo. The reason I was able to see sunspots was,of course, that the London sky was polluted with fumes and the winter sun was low on the horizon, so the rays had a greater thickness of atmosphere to penetrate. Much of the short wavelengths were filtered out, while the longer, redder wavelengths were able to get through. Much the same thing seems to have happened when you took this photograph of Seoul. It is common knowledge that red wavelengths penetrate foggy or hazy atmospheric conditions better than shorter wavelengths.


Sediment boundarys produced by largest meteors? Vredefort, Sudbury and Chicxulub

Vredefort

No extraterrestrial iridium anomaly: http://adsabs.harvard.edu/full/1989LPSC...19..733F

I'm not sure I fully agree with them, because there is quite a lot of Ir there and their threshold of 200 pg/g seems somewhat arbitrary to me. They claim the Ir was derived from local rocks, but that's a hell lot of iridium!

Sudbury

There is an an iridium anomaly. Read all about it here (paywalled) and here (open access).

  but the KT boundary is the only one commonly mentioned…why?

Because it's the first one found, a nice story of serendipity, because it's the last really big one, and the one that killed the dinosaurs. Everything you need to make a good story - even though it's not unique.

Iridium (and other PGE) anomalies and pretty much common feature of impacts. This paper shows attempts to link mass extinction events with impacts. This is a controversial issue, because you can also link massive volcanic events with mass extinctions, not only impact. But that's not for now - they do show iridium anomalies for all impacts. That's another paper showing iridium anomalies for impacts, much older this time.


How much reduction of insolation (solar energy) would be required to stop global warming?

It's not that hard to make an estimate.

CO2 traps about 2 watts per square meter.   Direct sunlight at 1 Astronomical Unit is about 1,360 watts per square meter, but spread out over the earth, average night and day it's about 1/4th of that about 340 watts per square meter.  (note, Casey's point is valid, it's probably better to use the number of watts that hits the Earth, so, my estimate is likely a bit low)

There may be more detailed and specific ways to look at it, but this is probably pretty close.   If you shade enough sun to reduce 2 watts out of 340, or 1/170th, that should balance out climate change, at least at current levels.

From space, you can view the earth as a disk in terms of area needed for shade, so the 2D area of the earth is about 127 trillion square meters, and 1/170th of that, about 750 billion square meters, or 750,000 square KM, slightly larger than Texas and as you get closer to the sun, the area needed is reduced by the square of the relative distance.

Still, that's no easy task, to put up a shade the size of Texas into space and it's no easy task to keep it there as it would in effect by a huge solar sale and likely enormously heavy.   It's far easier to simply build mirrors on earth than try to build something in space and I've seen that discussed as a possible solution to climate change.   of-course, they'd need to be kept clean to be effective, it would require maintenance, but it's probably doable.

Source (1,360 watts):  http://earthobservatory.nasa.gov/Features/EnergyBalance/page2.php

sources, 1.5-2 watts per square meter, CO2  (I went with the 2 watts estimate cause that includes CH4).   http://www.easterbrook.ca/steve/2012/01/how-much-extra-energy-are-we-adding-to-the-earth-system/ and https://en.wikipedia.org/wiki/Radiative_forcing


How are they formed? Erosion.

Long version: The rock/cliff the cave is set in possesses different hard/softness. Some parts of it are more prone to erosion than others.

Erosion comes in the form of waves and tides, washing and battering at the base of the cliff, nibbling away the soft parts of the rock quicker. Some chunks break off, rolling in the turf, bashing away at the cliff even more. Cracks form, get washed out and bigger and finally the cave forms. As is gets deeper and larger, rainwater might drip through the roof, adding to the erosion effects. Chunks to the roof start to fall down, and finally a part of the roof collapses. Your cave now has a hole to the sky.


How much earth has to be moved to get one tonne of copper

Your questions regarding stripping ratio only apply to open pit mines. They produce much more waste rock than underground mines. Typically they mine more waste rock than ore.
Underground mines are more surgical and depending on circumstances some of the waste rock they produce is used as backfill in mined out stopes.

the stripping ratio (which is usually larger than 3:1 for copper, i.e. 3 tonnes of waste for one tonne of copper ore) and the ore grade (which is <5% for copper, i.e. out of a tonne of copper, you can extract only 50kg copper). Is my interpretation correct?

You are partially correct. In this scenario the amount of copper that would be delivered to the processing plant (also known as a mill or concentrator) in mined ore is 50 kg, i.e the amount of metal in the mill feed is 50 kg.
To get usable copper metal from the ore, after mining, it needs to treated in a processing plant to concentrate the copper bearing minerals and to remove the gangue minerals. The copper concentrate is then sent to a smelter to produce copper metal. The waste from the concentrator is sent to the tailings dam/pond/storage-facility.
If you account for the recoveries of these two processes, typically 80% for concentrating and 90% for smelting, the amount of copper metal produced from 50 kg of copper in mill feed would be 36 kg (50 x 0.8 x 0.9).

How do they stack up in the mining process? Using a 1% ore grade and 3:1 stripping ratio, does that mean that to get 1 tonne of copper, we need to move 400 tonnes of earth, as we need 100 tonnes copper ore that comes along with 300 tonnes of waste?

Yes. Such a scenario would give you 1 tonne of copper in mill feed.
Unfortunately, the paper you refer to was most likely written by primary metallurgists, or possibly chemical engineers, and not mining engineers. The emphasis of the authors is on the amount of metal recovered from mining and the amount of energy required to produce usable metal from various mining projects, not the overall quantities of material mined.
Regarding your question about the Bayan Obo mining district in China your calculations are not correct. Also, from the paper you refer to you cannot obtain the total amount of earth mined, ore and waste, because there is no reference to stripping ratio.
As you state, from Table I of your reference, the proportion of neodymium in REM (REM being rare earth metal)is 18.5%. Table II states Tonnes mined/Tonnes REM is 59. This usually refers to the tonnes of ore mined and supplied to the mill, not the total tonnes of earth mined (ore and waste). Additionally, as stated in Table II and reiterated in Table IX, the recovery from the primary processing plant is 50%.
Now, 1 tonne of REM supplied to the mill contains 0.185 t (185 kg) of neodymium. With 59 tonnes of ore mined per tonne of REM, the amount of ore mined and supplied to the mill to get 1 tonne of neodymium is 59/0.185 = 318.9 tonnes.
The mill only recovers 50% of the metal supplied to it. Therefore, for the mill to recover 1 tonne of neodymium 637.8 (318.9/0.5) tonnes of ore must be supplied to the mill.


Why are inversion layers so important in meteorology?

Farrenthope is correct, but I would like to add some detail.

An inversion occurs when the air is warmer than the ground. It is called an inversion because it is the opposite of what occurs during the day. During the day, the ground is heated by the sun and the air near the ground rises.

However, during clear and calm nights, the ground will cool faster than the air. This means that at a certain time, the air will be warmer than the ground. With nothing to move pollutants upward, the air just stagnates. People will still keep pumping pollutants out, with the pollution just accumulating. During the day, however, the pollutants are lifted upwards, and do not stagnate.

For those mathematically affluent, the equation for a concentration of a pollutant can be written as $$\frac{\partial c}{\partial t}=-\vec{v}\cdot\nabla_h c-w\frac{\partial c}{\partial z}+\nabla \cdot (K\nabla c) +S_c(t)+C$$

where $c$ is the concentration of the pollutant, $\vec{v}$ is the wind vector, $w$ is the vertical velocity, $K$ is the Fick's law constant, and $C$ is the chemistry term.

If we consider the following assumptions which are common, regardless of an inversion:

Diffusion is a small process ($K\approx0$)
The pollutant undergoes no chemical change ($C\approx0$)
The equation boils down to
$$\frac{\partial c}{\partial t}=-\vec{v}\cdot\nabla_h c-w\frac{\partial c}{\partial z}+S_c(t)$$

During an inversion, because air at the surface is not going up, there is no air going down. This means $w\approx0$. With no vertical velocity, the air near the ground is slowed by the drag the ground has on the air, making $\vec{v}\approx0$. This boils the equation down to $$\frac{\partial c}{\partial t}=S_c$$.
Since $S_c \ge0$, $c$ will just keep increasing.

A good example of a dangerous inversion is the Donora event.


Circular rain cloud observed in rain radar

A way to check if those clouds and precipitation are real is to check the same area and time with another radar.

This is the image loop of Sidney (Terrey Hills) radar for the same time (the same you used). It shows the same pattern than your image. In the animated images that arch centered in the radar looks even weirder.

This is the image loop of Newcastle radar for January 17th, which in great part overlaps the Terrey Hills radar image. It shows an arched pattern, too, and with similar radius, but this one is centered in Terrey Hills radar instead of the Newcastle radar.

Therefore, since those patterns are different for each radar, I would say that they are artefacts, and that both radars suffered from the same kind of artefacts.

Interestingly, that kind of artifact seems consistent with the situation explained in dplmmr's comment: if both radar were just scanning an individual elevation angle (the same angle for both radars) and precipitation were just in a range of heights, we could expect that kind of pattern.


How does rotating water affect turbulence?

The dye diffuses vertically because it has a higher density. So when you put the dye in from the top, it wants to go to the bottom, albeit slowly. If it were more buoyant, it would float. It does diffuse horizontally, though much slower. This can be shown that, after a long time, the water will turn the color of the dye.

As far as rotation affecting the dye, that is complicated. If you start the rotation before you put the dye in, you can get Taylor Columns. Throw in different variables/change different things about the tank, and you can find different phenomenon (see the Weather in a tank experiments).
What you see in your experiment is called "shear-induced diffusion." The movement of your pen makes shear, and that shear increases the diffusion rate. I would also think that the eddies generated help diffuse the dye.


What behaviour does an ice sheet exhibit in crossing a cliff?

This depends on a few things.

If the cliff is small but widespread or the glacier is a huge, the glacier flows uphill until the cliff is gone.

If the cliff is large the glacier flows around it, is diverted by it, or piles up behind it depending on the land around it.

Glaciers are pushed along by back pressure, like toothpaste out of a tube. As long as there is enough accumulation the toe of a glacier can be driven uphill, but like anything else glaciers take the path of least resistance so if it can be diverted by the cliff it will.

Note that the forces on the cliff will be huge, and exactly the forces rock is weak against, so the cliff is over topped it will be stripped away very quickly.


An arch is a poor analogy for tectonic plates.  The lithosphere is  supported beneath by the mantle, unlike an arch which is unsupported beneath.  If the lithospheric plates were only held together only by lateral stress, friction with surrounding plates and gravity, how would they move relative to each other?  How would forces at the plate interfaces vary?  How would subduction work without the whole thing falling apart?  I am not a geologist but I cannot see how this arrangement would work.

The mantle "fluid" that sits underneath the lithosphere convects and drives plate motion.  It accepts subducting plates and emits new oceanic crust along the mid-oceanic ridges.  The support everywhere underneath the lithosphere by the asthenosphere makes an arch a poor representation for the physics at work.

Yes, you are correct that at the tectonic plate edges there is lateral stress and friction and that the whole plate is subject to gravity.  What you are missing is  that the crust is everywhere supported underneath.  An arch relies on the lateral stress to stay standing while tectonic plates do not.


How much total heat is contained in the upper layers of the atmosphere?

Mass at 21.9.1998 for Height 140 km -1000 km was $6.16\cdot10^{10} \mathrm{kg}$.
Avg. Temperature; $725 \mathrm{K}$

Roughly the mass is divided as follows;
$\mathrm{O}$, $\mathrm{O}_2$; $252 \cdot10^9 \mathrm{kg}$, “16” Molecule amount $N = 9.5\cdot10^{35}$,
Heat capacity; $920 \mathrm{J}/(\mathrm{kg} \mathrm{K})$

$\mathrm{N}$, $\mathrm{N}_2$; $364 \cdot10^9 \mathrm{kg}$; “14” Molecule amount $N = 1.56\cdot10^{36}$,
Heat capacity; $1040 \mathrm{J}/(\mathrm{kg}\mathrm{K})$

$\mathrm{He}$; $35.2\cdot10^6 \mathrm{kg}$, “4” Molecule amount $N = 5.3\cdot10^{33}$,
 Heat capacity; $5193 \mathrm{J}/(\mathrm{kg} \mathrm{K})$

$\mathrm{H}$; $1.34\cdot10^6 \mathrm{kg}$, “1” Molecule amount $N = 0.8\cdot10^{33}$,
Heat capacity; $14304 \mathrm{J}/(\mathrm{kg}\mathrm{K})$

Total Heat capacity; $61.2\cdot10^{12} \mathrm{J}/\mathrm{K}$
Quelle; My own work.

From here you can calculate $725 \mathrm{K} \cdot 61.2 \cdot10^{12} \mathrm{J}/\mathrm{K} =44.4 \cdot10^{15} \mathrm{J}$

Please note that most of the heat is at the lower atmosphere plain because >99% of the mass is there.


How Fast Could a Cyclone at the North Pole Achieve?

There's a few problems with this question.  The first is, the Coriolis effect isn't a force in and of itself.  It's a velocity dependent effect.   If the wind speed is zero over the North or South Pole, the Coriolis effect would be zero.    If the wind speed is 100 kph over the pole, the Coriolis force would be (if this site is trustworthy) $0.000146 \times V \times \sin(latitude)$, in other words, pretty small, but over long distances, a small force can create a significant change in direction.   But there-in lies the problem to answering your question.   It would depend on the size of the cyclone and the initial wind speed of the cyclone.

The other problem is, the Coriolis force doesn't form neat little circles.  Around the polls it's more like this, so estimating a presumably circular cycolone entirely due to Coriolis is doubly problematic, because a cyclone wouldn't form.  You'd get some wind direction change but not a cyclone.

The cyclone effect is the most efficient way to move air, as lower mass warm air wants to rise to higher altitudes and higher mass cold air wants to fall to lower altitudes.  The cyclone may be initiated by the Coriolis effect but once begun, it's driven by warm water, evaporation, and variation in air mass.  If you notice, Cyclones lose strength both on land and if they turn north over the ocean.  The colder ocean reduces the wind strength because it evaporates less easily and the air over the ocean cools, so with real cyclones, as the Coriolis effect grows stronger as the cyclone moves north, the cyclone grows weaker.  Cyclones don't represent the Coriolis effect.  It only plays a role in getting them started.

As you probably know, the Coriolis effect doesn't actually move anything.  Objects move by their momentum and the Earth rotates by it's momentum and it's the earth that moves under the object that gives the impression of a Coriolis effect.   If you stand on the North Pole, the Earth rotates about about 1/2 the speed of the hour hand on your watch.   It's hard to imagine that has much effect on anything and it doesn't unless the distance is significantly large.

I can't do the precise math on this one, though I find it quite interesting, but the above is a general answer.   If you try the physics board, you might get a more mathematical answer, but I think the question might work better with a bowling ball and a friction-less earth rather than the wind and the earth and a cyclone of unspecified size.   I got curious, so I asked the question with that in mind.


How far a andesitic lava can flow and what is the speed?

Andesitic lava is viscous, compared to basaltic lava. Andesitic lava flows at a rate of a few kilometers per hour, whereas basaltic lava can flow up to 10 km/h. Because of their viscosity, andesitic lavas "rarely extend beyound 8 km from their vents".

Viscosity of andesite lava and its implications for possible drain-back processes in the 2011 eruption of the Shinmoedake volcano, Japan

Pulsatory andesite lava flow at Bagana Volcano

Volcanoes, Magma, and Volcanic Eruptions - Tulane University


Why does more sunlight hits the equator than the ecliptic path?

The ecliptic path, is a well defined trajectory when displayed on the background of the fix stars like in the following figure

(taken from physics.csbsju.edu)

However, there is not such thing as an ecliptic path on the surface of the Earth. If you thought of it as the "Ground track" of a satellite but applied for the Sun, you have to consider that the rotation of the Earth would make that "ground track" of the Sun look very different than the one on the figure above. It would be a spiral going around the Earth ~365 times at latitudes between the tropics of Cancer and Capricorn (remember that the Sun takes a full year to go trough the ecliptic path).

When the the Sun is at its Northernmost declination (the top point of the red curve above), which is ~23° all the points on Earth at latitude ~23° North (the tropic of Cancer) will receive the maximum possible solar irradiation at the solar noon of that day. Once the Sun moves down to declination 22°, it will be the turn of the points at 22° North and so on.

The reason because the Equator is the line that receive more solar radiation is because it is the one that is in average closer to the Sun's ground track (in other  words, is the place with the maximum mean solar elevation).

An equivalent visualisation would be to take the mean of the ecliptic, and by looking at the above red line is quite straightforward to see that it would be a flat line on the celestial equator, that translates to the Earth's equator.

Does that makes sense to you?


Why do clouds sometimes form a ring around the Sun?

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:


Numerical weather forecasting - resources for a programmer?

The resources you want will vary on what level of understanding you are seeking.  For example, knowing the math means you can solve equations or transform them, but it doesn't mean you understand the physics involved.

Some topics you'll want an understanding of to understand the process:

Physics

At the heart of any weather model are our primitive equations.  Navier-stokes, thermodynamic energy, the state equation, and mass continuity.
Differential equations, linear algebra

The math for solving the above equations.
Discretization, numerical stability

How to turn the equations into something than be numerically solved on a 3D grid of points discrete in space and time.  This also involves choices in grid types, grid resolution, boundary conditions and time steps.  Any numeric noise must be accounted for and some artificial diffusion is sometimes used to accomplish this.
Sub-gridscale physics

A consequence of a grid means that processes smaller than about 4x your grid spacing and certainly those entirely contained within a grid volume are not resolvable by the physics in your equations.  These physics must be parameterized.  These process can include convection, radiation, turbulence, moisture, boundary layer processes and more.

Some books that may assist you include:

Numerical Methods for Fluid Dynamics (Durran, 2010) (finite difference methods, time stepping schemes, numerical methods)
Numerical Recipes (implementations of numerical methods)
Paramaterization schemes (Stenstrud 2009) (how to represent sub-gridscale physics in your model)
Modern Fortran Explained (Metcalf, Cohen and Reid, 2011) (in the world of models you will be seeing a lot of Fortran and FORTRAN).
A good text on MPI (not necessarily needed to understand the model, but if you are interested in how it runs across nodes you'll want to understand MPI).

The next step is digging into the source code.  From this perspective I suggest taking a look at CM1, which may be easier to follow than WRF from a programming perspective and will give you a foundation for how the math is applied in code to solve our equations and produce the output you see.


Could further oil drilling remove an insulation layer of the Earth cores heat?

I think your basic assumption is incorrect. Removing oil and the 80 to 90% water associated with it would reduce the thermal conductivity of the rock formations. It removes these liquids which aid thermal conductivity across the myriad of very tiny pores that contained the liquids. In the US regulations have the affect of the water being pumped back into the ground , not necessarily into the same formation. Outside the US that is not generally the case. Without doing any arithmetic, this appears to be a minuscule affect on earth temperature.....Thinking about it a little , you could likely make a better case on the basis  of the heat in the produced fluids ( water ,oil, gas): I know difficult wells go into formations that are 400 F+ , overall average the average temperature of production is likely between 200 and 300 F. No doubt SPE has real data but I am no longer a member.


Have there been any attempts to measure profiles through deep convective systems using dropsondes?

NOAA flies routine dropsonde missions into tropical convection if there is anticipated threat to land.  You can find information about these missions in the routine NHC updates for each storm.  This data is used in estimation of hurricane intensity and some of the models assimilate this data to improve storm track and intensity forecasts.

You can learn more about NOAA's hurricane hunters here.

Dropsonde missions over the continental US are problematic.  The mesoscale predictability experiment (MPEX) use dropsondes to determine if better upper air observations could improve predictability in the mesoscale.  One big issue they had is restrictions in where they could drop from the FAA.  They had to provide notice well in advance of the flights and could only conduct missions during certain time windows.  This provided big restrictions in dropsonde missions and would probably make it unsuitable for drops into deep convection (though one might argue with air traffic routing around such convection, that would make drops feasible from an FAA standpoint).

There were some recent talks about MPEX at the 27th Conference on Severe Local Storms in a session about MPEX.  In particular, a talk by Morris Weisman (video) talks a bit about the dropsonde missions.

From another perspective was the DC3 project, which I provided nowcasting in support of aircraft operations.  This project didn't use dropsondes, but instead used aircraft in vertical flight profiles to collect observations from near-surface to tropopause levels.  These profiles were in the vicinity of deep convection, but not within.  This project was primarily interested in how the chemical composition of near-surface air was modified by convection and then transported in upper air outflow.

These aircraft were not dropping dropsondes but I'm not confident a UAV would have been able to fly too much closer to convection to drop into the storms.

In summary, maritime dropsondes are common for getting in-situ measurements of tropical convection, but for continental applications FAA restrictions on dropping things from airplanes makes routine dropsonde missions unfeasible (and I'm assuming these would be done by UAVs as getting NSF aircraft assets to do this would be restricted to short term experiments and not routine operational missions).


Why isn't $\ce{H2O}$ evenly distributed in Earth's atmosphere?

The brief answer is 'the water cycle' I think.

A longer answer is that water can exist in many states on and above the surface: at least snow, ice and liquid water on and below the surface and water vapour (gas) and aerosols of liquid water or ice (clouds) in the atmosphere. On the other hand $\mathrm{CO_2}$ can only exist in one: gas.

What that means is that $\mathrm{CO_2}$ is as well-mixed as any other long-lived gaseous component of the atmosphere over time, which is 'pretty well'.  Components of the atmosphere, such as $\mathrm{SO_2}$, which are not long-lived don't get mixed so well because they don't have time to be. For water the situation is even more complicated because there are complicated phase transitions going on within the atmosphere, and because the distribution of liquid water is extremely dependent on topography, and in particular it can move large distances from where it originates.

The phase transitions of water in the atmosphere are driven by temperature and pressure conditions in the atmosphere, and in turn alter those conditions, and the whole system is chaotic of course.  The result of some of these phase transitions is water leaving the atmosphere as precipitation.

How much water gets into the atmosphere depends on conditions on the surface and in the atmosphere, and also on where water is on the surface, which is extremely non-uniform (see above).

All of these transitions are driven by energy from the Sun which varies over the planet.

So the end result of all that is that

water vapour is not long-lived in the atmosphere as a result of all this, so it doesn't have time to become well-mixed;
water is not evenly-distributed on and below the surface because it's mostly liquid (and because snow & ice is only stable in some parts of the surface);
the behaviour of water entering, moving within, and leaving the atmosphere is extremely complex and chaotic (or 'there is weather'), resulting in large variations over a range of scales in distribution of the various states of water which can live in the atmosphere, and large variations in how much water leaves the atmosphere over various scales.


There are two "normal" mechanisms for this effect:

Clay bonding, the beach sand contains a small fraction of clay particles which bond to the grain surfaces and hold the sand together when dry, this is normally the mechanism where the beach is backed by mudstone cliffs or quite close to river mouths supplying fine sediment to the beach.

Evaporite Cementation this is the more widespread and common process then what you are seeing is a small scale, and often abortive/short-lived formation of "Beachrock" this is the most likely explanation and the cause will probably be a rainwater/seawater interaction on the surface of the beach where it is still wet with seawater but no longer wave-washed as the tide recedes.


Why are deserts hotter than other places on Earth

We all know that deserts are the hottest places on Earth.

That is not necessarily true. What is true is that the hottest places on Earth are in deserts. Note the reversal. Much of the Arctic and Antarctic are technically deserts due to extremely low precipitation. Excluding these polar deserts, some deserts are quite cool, even if they are close to the equator. For example, Arica, Chile (the driest inhabited locale in the world), has a rather mild climate temperature-wise, despite being at 18°29′ south latitude. This is the case for much of the Atacama desert.

What makes them different from other parts of the world?

I'm assuming you are asking about those deserts that are hot in the summertime rather than deserts in general. A key ingredient for extremely high temperatures is a desert, but not a high latitude desert, high altitude desert, or coastal desert.
Very low precipitation is what distinguishes desert from non-desert biomes. Lack of precipitation is not sufficient for very high temperatures, as exhibited by the very cold polar deserts and the rather cool deserts that are well outside polar areas. Much of the Atacama is at a high altitude, as are the Gobi in China and Mongolia, the Taklamakan in China, and the high desert in the Mexico and the US. While some of these deserts can be hot during summer, they aren't ridiculously hot. Areas along the coast of the previously mentioned Atacama desert and the Namib in Africa are kept somewhat cool by ocean breezes. These coastal deserts can be uniformly mild year-round.
What is needed to create the possibility for extremely high temperatures is a desert that is not at extreme latitudes, that is well removed from coastal cooling, and that is at low altitudes. A desert is needed so as to receive the full brunt of solar radiation. Equatorial regions don't work because they tend to be cloudy and rainy. High latitude areas don't work because they don't receive very much insolation. High altitude areas don't work because temperature tends to decrease with increased altitude.
A low altitude desert in the horse latitudes is exactly what is needed, and this is why Furnace Creek in Death Valley, California (supposedly the hottest place in the world) and Tirat Zvi, Isreal (supposedly the hottest place in Asia) can be very, very hot. The horse latitudes are where the Hadley cell and mid latitude cell converge, creating long-lived high pressure areas where rain is highly unlikely. Most of the world's non-polar deserts are in or near the horse latitudes.
Even better than Furnace Creek are those areas that are so ridiculously hot that nobody sane would live there and hence there are no weather stations. We don't know, for example, how hot it gets in the Lut desert in Iran. It's too hot there for people to fathom, and is almost certainly hotter than is Furnace Creek.


Geological Differentiation: variables and size requirement

Not really.

Assuming even minor igneous bodies differentiate (stocks), and that differentiation is basically a process of decantation of the solid magmatic phase (a process known as fractional crystalization) and fractionation of the fluid magmatic phase, virtually any solid mixture can differentiate. If you're not convinced, here's another argument.

Let's take an analogous situation to differentiation: exsolution. They're not the same. Exsolution involves ionic migration in a solid structure. Differentiation involves ionic migration in a fluid (magma), which lacks structure. But both involve an outcome of different chemical phases. And exsolution can happen in microscopic sizes of milimetric crystals. So its fluid-ambient counterpart would, as a logical conclusion, be also able to happen on such minute conditions, and perhaps on smaller even.

And I'm not sure what you mean that differentiation will never end? On a long term geological scale - I'm talking about hundreds of millions to billions of years - that idea is perhaps correct, assuming the upper mantle and crust are constantly changing their compositions. But if you're saying that differentiation is a process that happens widespread (the entire crust, or the entire planet), rather than on specific locations (where partial melting occurs), under a small timescale (millions of years), then you're not correct, because differentiation only occurs in mixtured fluids (and I'm not sure whether it's used for any other fluid than magma).


There are micro-tectites galore, strewn over thousands of square kilometres. But remains of the actual bolide? Hard to say, because the impact crater is now buried under 600 metres of sediment. Earlier this year the joint IODP-ICDP drilling program drilled to a depth of 1300 metres. As far as I know, their findings are not published yet.


Question Concerning Meteorite Ejecta Travel

The research doesn't provide enough information to specify the displacement process of a 0.1mm spherule over 12,000 kilometers from North Greenland to Chile.
If the suggested impact was from Greenland, the space rock would have hit 800 to 1200 meters of ice at perhaps 10km/s which is the average for that kind of event.
Mount St Helens was about 0.25 km3 of ejecta, and the Greenland impact was a lot higher, i.e. 20km3 of vaporized rock.
Volcanoes send the energy out over many hours, whereas a YD meteorite would convert the energy relatively fast, perhaps mostly in the first 2 seconds. The thermal energy would have been very high, and the ejecta would have reached every level of the stratosphere, and the earthquakes would have emptied some lakes out in North America and left some major destruction fields in North America.
The same atmospheric displacement effects for all kinds of recent volcanoes would have been out measured considerably by a YT event.
Perhaps further studies will change the impactor site from the Greenland one which is perhaps 200k years old, to one in the Southern Hemisphere. At the moment, the evidence is so fleeting that the Nature article you gave has only generated a few lines of text on Wiki.


One way to estimate the upper limit of the magnitude is through observations and measurements.  According to the article Seismic Wave Interactions Between
the Atmosphere - Ocean - Cryosphere
System and the Geosphere in Polar Regions, a couple of difficulties with determining how large ice-quakes can get, asides from the regions not monitored, are due to the sources of the recorded motion and delineating cryoseismic and tectonic observations. From the article:

  Such cryoseismic sources include the movements of ice sheets, sea-ice,
  oceanic tide-cracks, oceanic gravity waves, icebergs and the calving fronts of ice caps. At
  times, it can be hard to distinguish between the waveforms generated by local tectonic
  earthquakes and those of ice-related phenomena.

However, in the Scientific American article Long-Overlooked "Ice Quakes" Data Provides Insights into Calving Glaciers, they can be identified by their location and unique waveforms identified relatively recently.

According to the first article, unlike tectonic events, cryoseismic (= ice-quake) events are seasonal and heavily climate-dependent, with the article detailing that observations have shown that in polar regions

  microseism amplitude is attenuated during local winter for both primary and
  secondary microseisms

Observations of body-wave magnitudes (Mb) between 1987-2007 in an Antarctic station were up to 6.5-7.0

A recent observation are the related 'glacial quakes', which generated log period surface waves (greater than 25s) which, according to the article, is

  equivalent in strength to those radiated by
  standard magnitude five earthquakes, and were observable worldwide. The glacial
  earthquakes radiated little high-frequency energy, which explains why they were not
  detected or located by traditional earthquake-monitoring systems. These events are two
  magnitude units larger than previously reported seismic phenomena associated with
  glaciers, a size difference corresponding to a factor of 1,000 in a seismic moment.

Similar observations have been made in the Greenland ice sheet.

NOAA has some sample cryoseismic traces on their PMEL Acoustics page Icequakes.


Your last question was easy to answer, this one is more difficult. You could look at correlation and dependence, but you'd be better off looking at variography. That's the easier part of this question dealt with.  Most resource geologists do not deal with correlation & dependence, they do variography studies.

If you are dealing with a small data set to familiarize yourself with the concept of spatial continuity then writing code in any suitable computer language is okay. If however, you are analyzing a large or complex data set use a package which has been created to analyze such data.

To analyze a large or complex data set, the data will first have to be declustered so that data points are fairly uniformly distributed throughout the data set. The other thing that will then need to be done is to establish a three dimension search ellipsoid so that only data within the 3D ellipsoid is considered with each calculation of the variance - the 3D ellipsoid will be similar in shape to a rugby football or an Australian Rules football. The 3D ellipsoid will also have to be able to orientated in a specified search direction. It will then have to be positioned on each declustered data point and geostatistical variances determined for each point - note, the mathematical expression for geostatistical variance is slightly different to the one used by mathematicians - you can blame George Matheron for that.


What causes ringing effect when using corner filter?

Filtering is always a convolution of the signal. What you are doing in the frequency domain will also be manifested in the time domain. A sharp edge in frequency domain will create a long convolution in time domain.

Ormsby (or trapezoidal) filters cause strong ringing if the slope is steep due to the Gibbs effect. To reduce the ringing, make the slopes more gentle (eg 6, 12, 45, 60) or try to taper the filter slopes.

However, it appears that you are using a strong AGC. Check if the ringing is actually a problem before smoothening the filter.

Some of the ringings will also disappear if you intend to F/K filter the direct waves.


How fast does the Earth's inner-core spin?

Roughly once per day, but a tiny bit faster. As the website for Columbia University's Lamont-Doherty Earth Observatory says:

  The inner core rotates in the same direction as the Earth and slightly faster, completing its once-a-day rotation about two-thirds of a second faster than the entire Earth. Over the past 100 years that extra speed has gained the core a quarter-turn on the planet as a whole...

Off the top of my head, I would have doubted that the core had a significantly different rotation rate than the mantle and lithosphere, because it would create different mixing layers in the liquid outer core, resulting in tangles in the Earth's magnetic field, similar to what happens in the Sun.


How much of a mountain is below the surface?

Is there some kind of math rule for how much does a mountain extends (depth) below the surface?

Definitely! It is called isostasy. When I was a student, the lecture about isostasy started with a slide asking "why don't the mountains fall over?" (it may sound better in my native language..)

First, some background:

Mountains, and in general the Earth's crust is part of the lithosphere. This lithosphere is the outermost layer of the Earth from a mechanical point of view. It is a relatively non-dense, solid and rigid piece of rock. The lithosphere lies on top of the asthenosphere, which is a more ductile layer composed of the upper mantle. In a sense, the lithosphere "floats" in the asthenosphere. You can think of ice cubes floating on water, or a better comparison would be solid cold chocolate floating on warm (but still solid) chocolate. There are two isostasy models: the Airy model and the Pratt model. The Airy model deals only with topography variations, so this is what I will consider here.

Let's take the following example:

(source: http://www.globalchange.umich.edu/globalchange1/current/lectures/topography/topography.html)

The two different settings, ocean and mountain, would be in equilibrium if the pressure at the bottom is equal: $$ P_1 = P_2 $$ Translating pressures to depth and density: $$\rho_1 g h_1 = \rho_2 g h_2 $$. You can either solve for the above example (hint: it is not exactly in equilibrium), or you can use this relationship to derive expressions for how much root would a mountain need in order to "not fall over". The Wikipedia article on isostasy has an excellent illustrated example that you can follow and understand. It also describes the Pratt model that I did not touch in this explanation.

You can also see a nice simulation of isostasy in the source link given below the drawing, where you can play with various settings and see how the cube (i.e. crust) responds.

What happens in the Himalaya?

  I found an article like this
  http://www.livescience.com/6595-depth-himalayan-mountain-roots-revealed.html
  that says that subduction is rare, does that mean that mountains don't
  usually extend below surface?

Not exactly. Mountains extend below the surface in any case. Subduction is also something that is very common. For example, the Pacific plate subducts under South America, precisely because it is oceanic crust denser than the continental crust of the Andes mountain range.

The article does not state that subduction is rare, it states that deep subduction of continental crust under continental crust is rare. This makes sense - in contrast to the Pacific and Andes example, here you have the continental crust of India colliding with the continental crust of Asia. Both are not dense and one is not supposed to subduct under the other. What you expect to see is that they both fold together. The article you linked to talks about a discovery in which the Indian plate is subducting below the Asian plate. This is not something you can easily describe with the Airy isostasy model since there are more forces involved - horizontal and not only vertical. Actually, this is why they need the computer simulations to figure it out.

This question might also be of interest:
What is the current theory about the formation of Tibet?

Also related is this article:
Researchers rule out a “tectonic aneurysm” in the Himalayas


Are there any general-purpose climate-model coupling projects?

Coupler software (e.g. OASIS, MCT, C-Coupler) is frequently used to combine these different components of earth system models. Of course interfacing models with each other using these couplers or other methods always needs some technical work and sometimes also some algorithmic work. Therefore the number of model combinations will always be limited.


Why do Volcanoes give out so much Sulphur Dioxide and Carbon Dioxide?

the mantle is in a highly reduced state

This is not entirely correct. The core is in a highly reduced state, but the mantle is not necessarily reduced, and is quite oxidised in some places. The mantle is heterogeneous.

2.

  however I justify it's emission because carbon dioxide is continually being fed into the mantle by the subduction of carbonate sediments, is this correct?

The carbon dioxide in the mantle does not necessarily derive from subduction of carbonates.

3.

  Or is this all wrong, and the gases are being emitted because they are still being degassed from the solid mantle in the process if heterogenous seperation which separated the atmosphere from the mantle. If this is the case, why have all other gases been seperated but not the two I have mentioned?

Not entirely accurate. Although it is true that some of the gases in the atmosphere derived from the differentiation of the core-mantle-crust system, a popular theory suggests that many of the gases came after the differentiation through other planetary bodies such as comets and meteorites. You are also implying that carbon dioxide and sulphur dioxide are the only two gases that are emitted from volcanoes. No. Water vapour is also emitted (commonly much more than the other two) and also other gases. It's just that less people care about the water and it gets less publicity, because, well, it's just water. There are other gases that are emitted as well: nitrogen, the noble gases. But they are much less common than the other two. They are not the only gases.

The authors of Fluxes and sources of volatiles discharged from Kudryavy, a subduction zone volcano, Kurile Islands write:

  Several potential sources for CO2 can be identified: (1) hotspot-type mantle, (2) MORB-type mantle, (3) subducted oceanic crust (MORB), (4) marine carbonate, (5) organic C from crustal and subducted sediments

They also give a table which states that in island arc volcanoes, 12% of carbon is sourced from the mantle, 67% is sourced from carbonates and 21% is organic carbon. Note that this only gives the amount degassed from volcanoes in subduction zone arcs. This does not mean that all carbonate-hosted carbon is released from the rocks to be degassed. I read somewhere that only 50% is degassed, and the rest is lost to the deep mantle. I would also expect that MORB magmas will have significantly less carbonate-sourced carbon.

Also, The deep carbon cycle and melting in Earth's interior
may be of interest.


How are lightning and thunderstorm intensity linked?

Is the ratio of IC to CG lightning relevant?

Yes. Primarily, severe storms tend to have very few cloud-to-ground (CG) strikes. Thus, the ratio of IC:CG is likely going to be very high.1 2 However, some severe storms do produce a significant amount of positive CG strikes. They would still have a high IC:CG ratio, but they would distinguish themselves by having more positive CG strikes. However, both of these types -- low-CG and some-positive-CG -- are potentially severe.2
It is important to note that the amount of CG strikes can be dependent both on where in the lifecycle a storm is, and the physical location of a storm (CG characteristics appear to be region-dependent). In tornadic supercells, the total amount of lightning seems to decrease during tornadogenesis, and the CG rate appears to increase after the tornadic stage of that supercell ends.3
There doesn't appear to be any existing literature on amplitude as a measure of severity. Perhaps, through this work, you might be able to identify whether this is an area which should be studied further.
References
1 Donald R. MacGorman, Donald W. Burgess, Vladislav Mazur, W. David Rust, William L. Taylor, and Brenda C. Johnson, 1989: Lightning Rates Relative to Tornadic Storm Evolution on 22 May 1981. J. Atmos. Sci., 46, 221–251, doi: 10.1175/1520-0469(1989)046<0221:LRRTTS>2.0.CO;2.
2 Timothy J. Lang and Steven A. Rutledge, 2002: Relationships between Convective Storm Kinematics, Precipitation, and Lightning. Mon. Wea. Rev., 130, 2492–2506, doi: 10.1175/1520-0493(2002)130<2492:RBCSKP>2.0.CO;2.
3 Scott M. Steiger, Richard E. Orville, and Lawrence D. Carey, 2007: Total Lightning Signatures of Thunderstorm Intensity over North Texas. Part I: Supercells. Mon. Wea. Rev., 135, 3281–3302, doi: 10.1175/MWR3472.1.
Further Reading

Lawrence D. Carey, Steven A. Rutledge, and Walter A. Petersen, 2003: The Relationship between Severe Storm Reports and Cloud-to-Ground Lightning Polarity in the Contiguous United States from 1989 to 1998. Mon. Wea. Rev., 131, 1211–1228, doi: 10.1175/1520-0493(2003)131<1211:TRBSSR>2.0.CO;2.

Yeung, Linus HY, E. S. T. Lai, and S. K. S. Chiu. "Lightning Initiation and Intensity Nowcasting Based on Isothermal Radar Reflectivity-A Conceptual Model." The 33rd International Conference on Radar Meteorology. 2007.

C. Emersic, P. L. Heinselman, D. R. MacGorman, and E. C. Bruning, 2011: Lightning Activity in a Hail-Producing Storm Observed with Phased-Array Radar. Mon. Wea. Rev., 139, 1809–1825, doi: 10.1175/2010MWR3574.1.


Why is runaway global warming a possibility despite all current fossil fuel deposits once having been in the atmosphere?

The largest reservoir of carbon is the rocks (sedimentary rocks) which reach into the order of the million gigatons (Gt). Over millions of years, part of that carbon have been released to the atmosphere by volcanoes. However, the amount of carbon in the atmosphere remains fairly stable because chemical erosion of rocks and plants capture that carbon excess and put it back in the geological reservoir.
There is about 4.700 Gt of fossil fuels, and the Atmosphere contain only 760 Gt of carbon.
So, if we burn all fossil fuels slowly, lets say over many millenia, there is not a problem, plants and erosion will keep the balance.
However, if we burn it quickly, let's say over few centuries, the greenhouse effect can be so strong that can kill many plant ecosystems and rise sea level, therefore reducing the surface area of rocks exposed to chemical weathering. Both effects would reduce the capacity of the system to restore the equilibrium, therefore it could take many millenia before all that carbon is returned to the geological reservoir so that temperatures can return to normal.
An extreme runaway case, like what happened in Venus, is unlikely to happen on Earth, but we don't really know. If temperatures ever reach to the point that oceans boil, then, there is a possibility that Earth might become like Venus forever.


Understanding the color coding of a weather map

Unfortunately it's common for satellite data services to show these types of images qualitatively, omitting the units or even a colorbar.
This is an infrared image from the 10.8 micron channel (TIR1) of INSAT-3D.  That website you link to is a bit misleading because (for me at least) following the link puts the text describing the 0.65 micron visible channel under the image.  If you click "Infrared" in the sidebar of that page the text is replaced with the correct text about the 10.8 micron channel.
Given that TIR1 records are stored as 10-bit values (i.e., in the range 0 to 1023) and the color scale maxes out at 939, I suspect that they are just plotting the raw count data.  This is normally transformed into radiances by applying a linear scale and offset to put the data in physical units.
I had a quick look at the equivalent 10.8 micron image from Meteosat over the Indian Ocean (see below) and I get a similar count range and a similar image when I apply their color scale.  I also suspect that they've inverted their data, e.g., plotted count_inv = 1024 - count, to fit with the viewers' expectation that cold clouds appear white.


Why are deserts mostly located on the western side of continents?

Some of the driest deserts on Earth occur in the western side of continents and they are called Coastal Deserts. Examples of such deserts are the Atacama desert (Chile, the driest desert on Earth), the Baja California desert (USA/Mexico), the Namibia desert (southwestern Africa), and the Atlantic coastal desert of Morocco/western Sahara/Mauritania. In mid-latitudes, the western side of continents is most frequently occupied by deserts, while the eastern side tends to be covered by forests (e.g., Amazon rainforest).

. Distribution of non-polar arid regions. Source: USGS.

Coastal deserts occur east of the strongest coastal upwelling regions in the world. The upwelling is associated with eastern boundary currents flowing equatorward (due to Coriolis): the Canary Current (off Northwest Africa), the Benguela Current (off Southwest Africa), the Humboldt Current (off Chile and Peru), and the California Current (western North America). The prevailing winds in these regions are the trade winds that flow parallel to the coast and generate upwelling dynamics: surface Ekman balance is setup (in deep enough waters) with water transport being to the right (left) of the wind in the northern (southern) hemisphere; this causes a deficit along the coast that needs a compensating flow in the deeper part of the water column bringing cold, nutrient rich waters (some of the most productive regions in the world) to the surface.

 Source Commons Wikipedia.

The cold waters near the ocean surface results in a cool, stable coastal atmosphere. In this region, evaporation from the ocean is reduced and produces extremely low rainfall over land. Precipitation is limited to morning fog and produces some of the driest ecosystems on Earth. The Atacama desert is the best example of such environment with average rainfalls of 15 mm/year (the driest non-polar region). In some areas, they are trying to take advantage of the little moisture the fog (Camanchaca) brings to establish some agricultural zones. The fog droplets are too small (1-40 micrometers) to form water drops and precipitate, so they use fog-catchers to collect moisture from the fog.

 Source: newatlas.com


As mentioned in the comments, the 1km difference in average elevation of Wyoming in comparison to Montana plays a major role, as from the website Wyoming State Climate Office,

  Because of its elevation, Wyoming has a relatively cool climate.

However, there is a bit more to it than that, the presence of numerous valleys

  provide ideal pockets for the collection of cold air drainage at night. Protecting mountain ranges prevent the wind from stirring the air, and the colder heavier air settles into the valleys often sending readings well below zero.


Can the shape and orientation of a large glacial rock deposit indicate glacial flow direction

Yes, it can. There are large, elongated mounds of till carried by the glacier as it advanced and left behind as it retreated. These mounds are called drumlins. Their long axis points along the track of the now vanished glacier. They are wider, higher and steeper at the end which points toward the glacier's source and taper away gradually at the opposite end, which points away from the source. They vary in size, but are sometimes a few hundred yards in length and about 50 or 60 feet high. Most are smaller. Sometimes there is a group of them, called a drumlin field. Glaciers also leave behind mounds of debris transverse to their track called terminal moraines. A terminal moraine is formed where the end of the glacier was stationary for a while when it was retreating, and there may be several of them at intervals along the retreating glacier's track.


Frost in trees in the middle of summer (of 536)

Frost leaves a record by damaging the cell structure, creating a "frost ring". Hadd et al discuss data on frost rings in N. American trees.
There is specific discussion of the Fimbul event (in AD 536) in this paper by Helama et al. 2019 (unfortunately with a pay wall). There is also discussion that there may have been a similar event in 1627 BC.
If the data is good enough, it may be possible to get an estimate of the time of year at which the damage occurred by looking at the structure of the damaged cells and seeing how it fits with what is known about the annual cycle of cell growth. This requires more detailed analysis than the widely used approach of measuring ring thickness to estimate annual mean (or growing season mean) temperature.
There is more recent information on ultra-cold summers in the Northern hemisphere, the "year without winter".


Long-term Wind Speed Forecasting: reality or wishful thinking?

Warning: "long-term wind-speed forecasting for generation" has (at least) two very different meanings. One refers to forecasting a distribution of wind speeds; the other refers to hour-by-hour (or half-hour by half-hour) forecasting of wind speeds.

Generally, when we talk about long-term forecasting of wind speeds for wind generation, we're talking about estimating the distribution of wind speeds over the next few years. That is to say, we're not talking about what the wind speed will be exactly two years from today, at 15.00 UTC. We're talking about what the shape of the distribution of the wind speeds will be - maybe in terms of the two parameters of a Weibull distribution, together with some indication of how the distribution varies by time of day, and by season of the year.

There is some skill in hour-by-hour wind speed forecasts up to about two weeks away. But for longer than that, there's very little skill in the models at all. In that regard, if we are talking about hour-by-hour wind speed forecasts, then short term is the next 0.5 - 4 hours, and long-term is 1-2 weeks.

In both cases, I'd look at recent publications from Reading University - David Brayshaw and others - and specifically on the accuracy of wind forecasts for wind generation, I'd also recommend work by Henrik Madsen and Pierre Pinson.


What type of fossil or stone is this found at beach of the Black Sea?

Its not a fossil, it is just a septarian concretion. They come in a wide array of shapes and sizes but the crack pattern and crack fill is a dead giveaway.


Transfer precipitation forecast in mm to expected snowfall in cm

The snow density, which is the "conversion function" you are looking for, of fresh fallen snow varies with atmospheric conditions, vapour saturation and temperature since these parameters determine the snow grain morphology (See this figure from SnowCrystals.com; I leave it as a link for copyright reasons). Densities can vary from as low as < 100 kg m-3 for cold conditions to maybe > 400 kg m-3 for wet snow. This essentially cover the spectrum for a normal snow pack. If you, by your question, are thinking about weather services reporting how much snow will come from a snow fall it is by no means a an easy thing to predict since you may "know" the amount of precipitation from a model, but you will have more difficulties knowing the density of the fallen snow unless you device a classification based on the atmospheric conditions during snow fall (based on snow crystal structure as in the figure). Hence for most purposes, the snow pack thickness is of little value and most will want the water equivalent of that snow, i.e. how much liquid precipitation it corresponds to.

If snow is falling during drifting conditions, the density may become higher than the falling snow itself would have resulted in. This is because with wind drifting, snow crystals are reduced in size by crushing and can hence be more densely packed.

With time the snow pack will densify due to metamorphic processes. This will cause the snow to decrease in thickness with time. But that is perhaps a different question since yours were about fresh snow.


How does water still exist on the Earth?

Lava doesn't destroy water it evaporates it into steam.  This steam rises high into the atmosphere, cools and condenses into rain which it falls back onto the Earth's surface where it gets hot again, evaporates and rises.   This isn't too different than how rain is made today, except that the evaporation process is slower and mostly driven by sunlight, wind, running rivers and by plant transpiration, not as much by a scorching hot surface.

Rain is a very efficient form of heat transfer.  Water evaporating very efficiently cools the surface.   I read or heard somewhere, don't remember where, so take with a grain of salt, but I heard that the Earth rained constantly for millions of years when the planet was young and the surface was hot.

To thermally split water into Hydrogen and Oxygen, the temperature needs to be about 2,000 degrees C.  Lava isn't that hot, though it's possible that the very young earth might have been hot enough to actually destroy some water and turn it into it's elements, Oxygen and Hydrogen, some of which would have recombined, and some of the hydrogen could have been lost as Earth's gravity is too low to retain all of it's hydrogen, but for the most part, Lava doesn't destroy water it simply evaporates it.  It's also possible that some other chemical interactions would absorb and change the chemical property of some of the water, but for the most part, Lava just turns water it into steam, which is still water, just in a different state, like how Ice is still water, just not liquid.

Update:

On the comments below, that's on the Earth Science/Astronomy border, but there are a fair bit of unknowns on how much atmosphere and water the Earth has lost to the Solar Wind.   One estimate says the Earth is losing about 90,000 tons of hydrogen every year.   Source.  Since that hydrogen effectively comes from split water molecules by UV rays high in the atmosphere and by mass, Water is about 8 parts Oxygen to 1 part hydrogen, then 810,000 tons of water is converted to 720,000 tons of Oxygen every year.    If we multiply that by 4 billion, just cause, you know, why not run the numbers, assuming consistency, over that time, the Earth has lost 3,240 trillion tons of water and added 2,880 trillion tons of Oxygen.  Consistency is a big if, but these numbers seem reasonable enough to work with.

The oceans contain about 1.4 million trillion tons of water so the estimated 3,240 trillion tons of water lost is irrelevant, less than 1/400th of the Ocean water at the current rate of loss.

Oxygen is different.   2,880 trillion tons of Oxygen is more Oxygen than is currently in the Earth's atmosphere, but much of the Earth Oxygen formed in the upper atmosphere combined with other elements like Iron in the oceans or Methane gas, or helped form the planet's granite, so even though the solar radiation should have produced lots of Oxygen from water over time and Hydrogen that was lost, the amount of water is negligible and the Oxygen has been useful in many ways.

But I digress as those are current estimates not historical ones.   The solar wind, while it does blow hydrogen off the Earth and in effect removing water from the Earth, the effect is slow enough that it's a minimal change.

But the Earth's magnetic field wasn't always there and it isn't as old as the early and late heavy bombardment periods.  Our magnetic field is thought to be about 3.5 billion years old.  Source.  So it's entirely possible that the Young Earth lost a significant amount of water prior to the magnetic field.   Venus has no magnetic field to speak of but it's retained it's atmosphere's heavy elements but very little of it's water.   Mars is much smaller, so it's less good at retaining it's atmosphere than Venus.   Nobody knows how much atmosphere and water the young Earth lost to space.

It's entirely possible Earth lost a pretty good share of it's water to the Sun's UV rays prior to it's magnetic field, and in part due to it's high temperature, but I don't think anyone had a good guess on the precise numbers.  To return to the question above, Magma on the surface isn't a big factor unless it's hot enough to split water into individual molecules.   It's the temperature of the upper atmosphere, the atmospheric escape velocity, the solar wind and magnetic field are the larger factors.


The 1954 recorded height of 8,848 m remains the accepted figure for height. As for the elevation of Everest's base, it is 5,380 m (at 17,600 ft).


What is the most cost-effective solution to climate change based on scientific evidence?

In researching my question to provide what is known so far I found the following information. I won't accept my own answer unless it gets the most upvotes.

I found this ranked list from 'Project Drawdown' (PD) that states Managing Refrigerants is the number one solution by GT CO2 reduction by total, but with an unknown cost. Offshore wind turbines (OFTs) seems to be the best solution by total reduction. Electric vehicles (EVs) has the highest savings, but with also the highest cost.

Dividing the reduction amount by the savings leads to OVTs at 11.4 GT CO2 reduction per trillion USD a year (GCR/TUY), then Solar Farms (SFs) at 7.35 and EVs at 1.11 for the top three by savings. However, if all are calculated and ranked, the most reduction by savings is Tree Intercropping at 778 and methane digesters small at 137, the most reduction by cost is Nuclear Energy at 18254 and then Solar Water at 2033, but including the costs with negative numbers (not sure what that means) it's Landfill Methane at -1374 and Solar Farms at -458 GCR/TUY.

I'm not sure if they're including the cost of nuclear accidents or waste storage, since waste storage at least in the US has not been reconciled, and may not be fast enough, so using Solar Water at 2.033 GCR/BillionUY, it would cost 16.3 billion USD per year to reduce the CO2 we emit for energy. Based on a single strategy, that's 2.11 dollars per person in the world (7.7 billion) and $12.50 for every person in the developed world (1.3 billion). Is this correct?

In the PD FAQs they state, "Carbon pricing is a policy mechanism to implement solutions and not in itself a solution to global warming. It is the single most impactful policy proposal that would accelerate the adoption of every solution enlisted."

Adding a few other solutions I've found:

CO2 recapture in concrete can reduce it's emissions by 40% (it is 8% of global) and also the energy used. https://www.solidiatech.com/
Red mud as a carbon sink, using the hydroxides from aluminum smelting because of the high solubility of CO2 in them.
Relying on rock weathering from carbonic acid in rain and because "Coccolithophores account for half of the calcium carbonate creation in the ocean", some kind of phytoplankton farms harnessing their "potential for carbon sequestration but also for the production of high value products, such as fatty acids and pigments."


Why is a layer of the subsoil so dark on this construction site?

Regarding your question:

  So now I'm wondering if that might be the "fire layer" where the excessive heat hast left its mark in the ground.

The short answer is no. The layer seems to be about 1 meter thick. No bombardment related fire can cause blackening 1 meter deep.

Now, why is it "black"? First of all, it's not black. It's dark brown. I've seen black soils, and trust me - they are pitch black. There are several reasons for it. My best guess is that it's simply moist. You know how things appear darker when wet? Same here. My hint is the floor of the site. I'm guessing it's the same material, but it's drier so it looks lighter.

There are more reasons, some concerning the organic content of the soil (for example, peat - a highly organic black soil) or the mineral content of the soil (soil composed of high iron contents will also appear black).


Does Earth's air leak into space over time?

Earth's atmosphere does escape over time, albeit very slowly.

The distribution of kinetic energies of molecules in a gas obeys (more or less) a Maxwell-Boltzmann distribution. Notice that the graph is asymptotic, so in a suitably large population of gas molecules, there is a non-zero probability that some of those molecules will have a an arbitrarily large kinetic energy. This implies that in the population of gas molecules constituting Earth's atmosphere, some of them will have kinetic energy such that their velocity exceeds the Earth's escape velocity and provided that their paths are oriented away from any obstacles (including the Earth's surface itself) and they don't collide with anything, those molecules can escape. On average, only a small number of molecules will actually achieve all these conditions. The end result is that atmosphere does indeed escape, but the effect is tiny - only on the order of grams per second due to the process described above according to this article I found.

This is not the only process responsible however, phenomena like the solar wind also have a role in liberating gas molecules from the Earth's atmosphere.

The effect is much more pronounced on other rocky planets in the Solar system, for example Mercury has very little atmosphere due to it i) being extremely hot and ii) being bombarded by highly energetic particles from the sun much more than we are. Similarly, Mars has a predominantly carbon dioxide atmosphere that is thought to have been much thicker in the past than it is now, and therefore must have escaped somehow (assuming you don't believe stories like Total Recall where the atmosphere is intentionally held captive ;) - although in fact the Martian atmosphere does condense at the poles leading to a significant amount of atmospheric agitation with the changing of the seasons)


You may want to investigate the Delivery package available from

http://www-old.dpr.csiro.au/StochasticSeismicInversion/

'Delivery is an open-source, trace-based Bayesian seismic inversion code for use in oil reservoir characterisation at the early development and appraisal stage. See the Computers and Geosciences paper for most of the details, and the Delivery4D preliminary writeup for the time-lapse modelling.'


Just taking a guess from the context, I think it could be a "view shed", the term used in GIS (Geographic Information Systems) to refer to a visibility analysis. In other words, it is a calculation that determines which areas of a given terrain would be visible from a specific point.

It is used to know, for example, what would be the coverage of a radio transmitter on a given location, or from which areas would a lighthouse not be visible, etc. To calculate such diagram you need a DEM (Digital Elevation Model) of the area.

To give you an example I just loaded a DEM and run a calculation (using Global Mapper) of the areas visible from a 20 meters tall tower in the top of a specific hill, and within a 40 km radius. And this is how the results looks like:

Where the red shading show the areas that are visible from the tower. The rest of the terrain would be hidden behind topographic obstacles.


Is there any mineral identification techniques that works for all minerals?

Yes, it is called X-ray diffraction (XRD). It is done by machines that look like this:

I'm not sure this is in the scope of your electronic engineering project.


What causes waves to form the characteristic "breaking" shape as they approach the shoreline?

The physical process you describe is known as wave shoaling.

At the basic level, waves propagating into shallow water become shorter and higher, and consequently, steeper. In shallow water, the water particles near the crest move forward faster than those below them. Similarly, the particles near the trough move backward faster than those above them. This causes strong shearing of the near-surface body of water, eventually forming a plunging breaker, or a surf wave. For small-slope (linear) and inviscid (no friction) waves, the above is a consequence of the bottom boundary condition for the water velocity to be zero at the sea floor.

There are two fundamental and related properties of water waves that contribute to shoaling. One is the wave frequency $\omega$ remaining constant as the depth $d$ decreases. Think of this as the conservation of wave crests at any fixed point. However, the wavenumber $k$ (wavelength $\lambda$) must increase (decrease) with decreasing depth, as per the dispersion relationship of water waves (neglecting the effects of viscosity):

$$\omega^{2} = gk\tanh{(kd)}$$

where $g$ is gravitational acceleration. In shallow water, the dispersion relationship reduces to:

$$\omega = \sqrt{gd}k $$

and phase speed $C_p$ and group velocity $C_g$ are both proportional to the square root of the water depth:

$$ C_p = \dfrac{\omega}{k} = \sqrt{gd} $$

$$ C_g = \dfrac{\partial\omega}{\partial k} = \sqrt{gd} $$

Individual wave crests propagate with phase speed $C_p$. Wave groups and wave energy propagate with group velocity $C_g$. Thus, waves entering shallow water become shorter and slower. The second important property leading to shoaling is the conservation of wave energy flux, which is proportional to the group velocity and wave energy:

$$\dfrac{\partial(C_{g}E)}{\partial x}=0$$

Because the group velocity decreases, the wave energy (read: height) must increase (locally). This causes the waves to grow in height as they enter shallow water. As pointed out by @IsopycnalOscillation in a comment above, the separation of water particles from the wave crests happen because the individual orbital velocities at the top of the crest exceed by far the phase speed of the wave.

Although bottom friction is significant and non-negligible in shallow water, it does not cause shoaling. Mathematically, wave shoaling can occur for completely inviscid and small-slope (linear) waves that propagate into water of decreasing depth.


What would be the impact on tides if the earth had no tilt?

A simple way to estimate the implications is to simplify the problem. If the orbits of Earth, Sun and Moon were circular and in the same plane, and the Earth had no tilt, the only remaining tidal constituents would be $M_2$ and $S_2$ (also the overtides and combination tides: $M_4$, $MS_4$...). The rest of the tidal constituents can be expressed as linear combinations of their rates of change based on the remaining forcings: orbital eccentricity and inclination and axial tilt.

The modulation of the amplitude $A_M$ of the main constituent wave, $\eta_M = A_M cos (\omega_M t)$, by another constituent of amplitude $A_c$ is: $\eta_t= (A_M+A_ccos(\omega_c t))cos(\omega_Mt)$. Expressing it as a linear combinations of the rates of change gives $\eta_t= A_Mcos(\omega_Mt)+{A_c\over2}cos(\omega_M-\omega_c)t+{A_c\over2}cos(\omega_M+\omega_c)t$, such that the original amplitude remains unchanged, while two extra components with slightly different frequencies appear.

In the case of the effect of the axial tilt, it modulates the main frequencies based on the solar parameters (the inclination of the Moon orbit is not affected by the axial tilt). The period of the solar declination is a sidereal year (365.2564 days). In the case of the modulation of the $M_2$, the resulting periodicities are 1 cycle/solar day ± 1 cycle/sidereal year. The two resulting constituents are $P_1$ and $K_1$. $P_1$ is known as the solar declinational diurnal constituent. The case of $K_1$ is slightly more complex as the inclination of the Moon also results in a contribution at this frequency. The rest of the tidal constituents are also modified accordingly, but their amplitudes are much smaller.

In short, the removal of the axial tilt will result in the removal of the $P_1$ tidal constituent (and some other minor ones) and an amplitude decrease in the $K_1$ constituent.

$K_1$ amplitude and phase from AVISO. The maximum $K_1$ amplitudes of around 0.4m are found south of Alaska and west of the Kamchatka Peninsula in the Sea of Okhotsk. The amplitude of the $P_1$ constituent is about half of $K_1$ with a maximum of around 0.2m in the Sea of Okhotsk. Overall, the effect of the tilt on tides is pretty minor relative to the rest of the effects.


Interpreting some hydropower data. What does DSF mean in this context?

I think the DSF releases are both measures of releases. Neither is a measure of power. They are both a measure of the amount of water released, in day-second-feet, as you say.

The total column is the measure of the total amount of water released (in DSF). The power column is the amount of water released in order to generate power (in DSF). The difference between those two numbers is the amount of water (in DSF) released down the spillways (rather than to generate power), as part of overflow prevention, maintenance, or other activity.

On the realtime feed, the column labelled power looks like mean power to me in MW: the dam rating is 70MW, and the power column maxes out at 70.267, so that does indeed look like MW. So that's the average over the quarter-hour. Multiply by 0.25 to get MWh.

footnote rant:
Feet? Seriously? Feet? Who uses feet? What's wrong with these people?


SEG-D is a specialized format, while SEG-Y is a general-purpose format.

In general, SEG-D is intended for field recordings of seismic data, and SEG-Y is intended for 'seismic data exchange'. Having said this, SEG-Y is so general-purpose, and so ubiquitous, that I'm not surprised to hear (anecdotally) that people are using it for data acquisition. Certainly, the recent release of Revision 2 — which allows for high sample rates, variable trace lengths, long records, and large files — facilitates this.

As an interpreter, I use SEG-Y all the time, whereas I've never come into contact with SEG-D. Thus, I don't know enough about it to go into any detail, but a quick look over the SEG-D specification suggests it has better support for describing receivers, channels, sources, and including things like observer logs. Which is what you'd expect in an acquisition format.

SEG's formats are notoriously, er, flexible. So converting directly between the two formats may or may not be possible, depending on which fields have been used, and the lengths to which you're prepared to go to 'adapt' the target format to your use case. The usual way to convert, as it were, would be to process (i.e. demux, sort, image, and maybe stack) the field data in the SEG-D file(s), and write out one or more SEG-Y files.

Good luck!

References

SEG-D Rev 3.1 description, SEG Technical Standards Committee, 2015.
SEG-Y Rev 2.0 description, SEG Technical Standards Committee, 2017.


Why would moving the prime meridian hasten the time when it would be necessary to drop a day from the calendar?

There is no basis in logic to have to subtract a day due to a change in the prime meridian.

Both the Julian & Gregorian calendars require days to be added, not subtracted. It's why we have leap years.

Also, both the Julian & Gregorian calendars were established centuries before the prime meridian was set in the 19th century. The formula for adding days & creating leap years was set long before the prime meridian was even contemplated.