How does Earth's thermal radiation into space vary by location and time of day? CHEOPS (also 1, 2, 3) is a space telescope recently put in low Earth orbit to look for exoplanets around other stars. It uses a cooled detector array and so uses a heat shield to avoid thermal radiation from the Sun as well as the Earth directly below.

Thermal radiation from the Earth is a significant problem for spacecraft thermal management in low Earth orbit, so it's important to know how much thermal radiation Earth emits into space.

If we use an average emissivity $\epsilon$ of 0.8 and assume a temperature of 0 °C, then using

$$P= \sigma \epsilon T^4$$

where the Stefan-Boltzmann constant $\sigma$ is 5.67 $\times 10^{-8}$ W m-2 K-4 we can estimate it to be about 250 W/m2.

But what is it really, and how does it vary (on average) by the type of terrain under a satellite and time of day?

From this answer to For a given 48 hour observing period, what fraction of the celestial sphere is available to CHEOPS? (originally from the CHEOPS Red Book ESA/SRE(2013)7
November 2013

The yearly average Top Of Atmosphere (TOA) Outgoing Longwave Radiation (OLR) is around 145–345 W/m², as measured by the AIRS instrument:

Source: Wikimedia Commons contributors, "File:AIRS OLR.png," Wikimedia Commons, the free media repository.

Now, this is an average that includes both day and night.  How much does it vary?  This depends on the location.  Based on GERB data, a 2007 paper estimates (please pay attention to the thick solid line only):

Source: Comer, Slingo, and Allan (2007)

Even in places with a very large diurnal temperature cycle such as the Sahara, the difference between maximum and minimum is only about 10%.  In oceanic areas it is closer to 2%.  That means that if you take the mean (shown in the top figure) and subtract around 1–5% depending on the region (the drier, the more difference) you get a reasonable estimate of the nighttime outgoing longwave radiation.

If you want a more precise estimate, you can always download satellite data and calculate it yourself.

Full source for second figure: Comer, Slingo, and Allan (2007), Observations of the diurnal cycle of outgoing longwave radiation from the Geostationary Earth Radiation Budget instrument, Geophysical Research Letters, Vol. 34, L02823, doi:10.1029/2006GL028229


Transform Boundary and its relation with Earth's rotation I have read about all 3 types of boundaries in the book : "Essentials of Geography" by Lutgens et. al.
But there is nothing mentioned about Transform Boundary and Earth's rotation effect on it.

I know the basics that since MoR are not always in straight lines, so there is a differential moment between two portions of the plate which cause this transform boundary that is perpendicular to the MoR.

But now, when I was reading one of the 'official' book of this subject by NCERT I got stuck on this paragraph :

Transform Boundaries : Where the crust is neither produced nor destroyed as
the plates slide horizontally past each other.Transform faults are the planes of
separation generally perpendicular to the midoceanic ridges. As the eruptions do
not take all along the entire crest at the same time, there is a differential
movement of a portion of the plate away from the axis of the earth. Also, the
rotation of the earth has its effect on the separated blocks of the plate portions.

[This] is where this is mentioned.
Now I'm really not able to relate Earth's rotation, axis, and its effect on transform boundary. Neither Google is giving any meaningful result.

NCERT is an Indian document written from the perspective of the prevailing non-scientific educational culture, in which (for example) astrology is considered to be bona-fide science! Therefore it is not authoritative, and certainly not written accurately from the perspective of peer-reviewed scientific sources. It has been criticized before, and in this case it is quite incorrect to write that transform faults are significantly affected by the Earth's rotation. Rather, the NCERT quotation that you cite seems to confuse the Earth's pole of rotation with the Euler pole of rotation. The latter is a key concept of the plate tectonic movement between any two plates. The Euler pole of rotation can be at any angle with respect to the Earth's rotational axis.


What is the purpose of analyzing the bed-load of a river? A group of students and I have gone to a field trip where we performed tasks like: measuring the velocity of the river flow, finding the width and depth to produce a cross section along with valley profile and gradient.

I am a bit confused though regarding bedload analysis. What would the shapes of the rocks collected tell about the erosion at the point?

Typically rivers can be divided into three areas when it comes to erosion:

far upstream there is the area where most erosion happens
in the middle mostly transport of material happens
downstream rivers mostly deposit material

Which also can be determined by the shape and size of rocks in a streambed, which correlate directly to erosional force at that point of the river.

larger, rough rocks are rather freshly broken off the bedrock, it takes a lot of force to move these rocks, ergo there is a lot of erosion happening, due to a combination of inclines, amount of water and shape of the riverbed
smooth, smaller pebbles have been in the river for a longer time. They usually have been transported some distance from where they came into the riverbed, which wore them down to their current form. It also takes considerably less force to push/roll a smallish pebble compared to rough chunks of rock. Ergo the erosion at this point of the river is smaller
sand and silt (which still are some form of rock) are typical riverbeds close to the rivers mouth. It takes ages to grind rocks to sand, and it is much easier to transport sand over long distances, so when the sand drops to the riverbed, the erosional force is close to nothing.

For some more information, look here e.g.: http://www.alevelgeography.com/the-long-profile-changing-processes-types-of-erosion-transportation-and-deposition/


Resources to determine temperature probability for a given location on a given date What tools or resources are available (free) on the web to determine the probability of a given point on the earth to be at or above (or below) a specific temperature on a specific date.

Some [extreme] examples:

At the South Pole, the probability of it being > 90-F on June 1 approaches 0% (I'm guessing).

Conversely, the probability of it being < 32-F on the equator (at or near sea level) on Dec 1 approaches 0%, as well.

Another way to phrase what I'm looking for:  On what date in the calendar year does the probability of the air temperature being above (or below) X approach 0% for a given location/city?

So, say, for Washington, D.C.-- after (or before) what date in the year is it virtually impossible for the temperature to be above, say... 80-F?

The most basic way to measure probability is to look at historical
records. There are several sources for this data, including:

http://wunderground.com
ftp://ftp3.ncdc.noaa.gov/pub/data/noaa/isd-lite/

This assumes no climate change. If you believe in climate change, you
will need to factor that in.


Is there a name for a type of algorithm or processing which uses only one source of data as input? Many algorithms in remote sensing take one or more different datasets as inputs and return some output. An example might be computing the amount of sulfur dioxide (SO2) injected into the stratosphere after a volcanic eruption, having two main input datasets: 1) the SO2 amount and layer altitude (e.g., Eumetsat IASI Level-2 SO2 data) and 2) the altitude of the tropopause (e.g. NCEP/NCAR reanalysis data). Even though each input dataset contains multiple fields which are used in the computation, there are only two sources of data.
I am working on a remote sensing tool that uses only one input dataset (containing multiple fields) which accomplishes a task that has previously required more than one dataset with quite a bit of uncertainty coming from the other datasets. I think that's kind of novel.
Consequently, I am wondering if there is an existing name for an algorithm which uses only one dataset that distinguishes it from previous algorithms using multiple input sets?
My instincts (read: "love of coffee") tell me to refer to it as a "single-origin" algorithm, but surely there must be a name for this already.
Any insight is greatly appreciated.

I think you are thinking of an injective function  or mapping.
However this work you are doing sounds sort of suspect.  Usually adding additional data (with appropriate cleaning) reduces uncertainty.  If the additional data is adding to the uncertainty that might indicate that the original data set is noisy or unreliable.


Young researcher hoping to volunteer for Antarctica fieldwork I am a postdoctoral scholar who is interested participating in Antarctica fieldwork/exploration as a volunteer. My background is in biology, but I am happy to participate in any fieldwork opportunities, because I hope to use such an experience to better understand how teams perform effective science in harsh environments. I have worked in scientific teams in remote environments and believe I can grow a lot for future endeavors by working in even more harsh environments like Antarctica.

As a side note, I am a U.S. citizen who has extensive experience overseas (especially in Japan).

For the past few years, I have tried:

1) Contacting numerous scientists who lead Antarctica teamwork. 2) Contacting Palmer LTER and McMurdo LTER.
3) Applying to the ANSMET program.

So far I have not had much luck, which is understandable given the limited slots. I wanted to inquire if anyone on this forum may know of other avenues someone in my position can apply to increase their chance of obtaining such a volunteer position? Any advice would be so very much appreciated! Thank you.

Try the long game:

Start reading up on the research that is being done.  For the interesting ones, start corresponding with the authors asking questions about the work.  They have to be good questions, showing that you know what you are talking about.
Look at who funds each researcher.  Check the papers for those 'thank-yous' that give credit to various individuals and institutions.  Check the institutions to see how they hire support personnel.  If you can't get a job as a researcher, look at getting a job just as a tech.
Most researchers are connected with one more more institutions.  Some of these specialize in polar science.  Some may supply support crews that span scientific disciplines.  E.g. They may have a pool of snow-cat operators that any of the scientists can draw upon to move equipment and personnel.  This is more likely with the permanent bases.
Can you speak well?  Do you have a large grasp of polar science in many fields?  Look at being a guide on the cruise boats that go to the arctic.  If you are quick witted, and can reduce difficult topics to digestible chunks, this is a possibility.  Train for it by writing a science blog.
Write a polar science blog.  Turn dry science into good clear interesting commentary about what's happing, how it happens, why it's important.  This becomes something to put on your resume, and if you get a decent following may itself get you there as a science journalist.
Prepare yourself. Learn how to dress and take care of yourself in Arctic regions.  Someone they don't have to train as much is cheaper for them.
Consider having your appendix removed.  At one point this was recommended as an appendectomy in Antarctica is double plus un-fun.


Was the Caribbean Sea a closed sea during the last glaciation? Taking a look at Google Earth the straits between the Caribbean Sea and the Atlantic Ocean are not very deep.

The sealevel during the last glaciation was 120 meters lower than the present level.
Was the Caribbean Sea a closed or semi-closed sea during the last glaciation or the straits were deep enough to make it a connected sea?

Thanks to @Eartworm I learned today I was totally uncorrect: there are more than 2000 meters deep in the straits.

Tool to see bathymetry, navionics.com


Why are weather kites no longer used (much) operationally? Figure 1.12 from the IPCC AR5 WG1 report (reproduced below) illustrates the different instruments that have been used for weather forecasting.

Most of the methods have been in continuous use since their inception, except one: kites were used, but later disbanded.

Why is this so?  This Monash university article from 1999 makes the point that they are superseded by radiosondes and aircraft.  However, atmospheric measurements on aircraft are very expensive and mainly used for research as opposed to operational weather forecasting; and, unlike weather kites, they do not (normally) fly in the boundary layer¹.  Radiosondes are only launched once or twice per 24 hours per location.  Passive (operational) space-borne measurements have a lower vertical resolution, in particular near the boundary layer.  I imagine that kites should provide a low-cost, high-resolution (both temporally and vertically) profile of the boundary layer.  Why is it they are not really used operationally anymore?

Figure 1.12 | Development of capabilities of observations. Top: Changes in the mix and increasing diversity of observations over time create challenges for a consistent climate record (adapted from Brönnimann et al., 2008). Bottom left: First year of temperature data in Global Historical Climatology Network (GHCN) daily database (available at http://www.ncdc.noaa.gov/oa/climate/ghcn-daily/; Menne et al., 2012). Bottom right: Number of satellite instruments from which data have been assimilated in the European Centre for Medium-Range Weather Forecasts production streams for each year from 1996 to 2010. This figure is used as an example to demonstrate the fivefold increase in the usage of satellite data over this time period.  Figure reference: Cubasch, U., D. Wuebbles, D. Chen, M.C. Facchini, D. Frame, N. Mahowald and J.-G. Winther, 2013: Introduction. In: Climate Change 2013: The Physical Science Basis. Contribution of Working Group I to the Fifth Assessment Report of the Intergovernmental Panel on Climate Change [Stocker, T.F., D. Qin, G.-K. Plattner, M. Tignor, S.K. Allen, J. Boschung, A. Nauels, Y. Xia, V. Bex and P.M. Midgley (eds.)]. Cambridge University Press, Cambridge, United Kingdom and New York, NY, USA.

¹Research aircraft sometimes do.  Flying in a full-sized jet 20 metre above the ocean surface is an interesting experience!

Most data we collect about weather is done regularly and for all altitudes of the troposphere.  Kites, though, are at the mercy of the weather and do not fly high enough.  So when there is ice or snow or other meteorological conditions that don't allow a kite to fly, you get no data.  We now use methods to measure the weather that don't break in bad weather.  The answer to your question is succinctly put in this NOAA webpage where they state these difficulties with kites:

The average altitude reached was only about 3 km.
Data could not be evaluated until after the kite was reeled in...
Observations could only be taken in good weather with winds neither too light or too strong.
There was danger of the kite breaking away and endangering lives and property.

In the 1940s, weather balloons were being released and could gather an entire profile of data in all weather conditions.  This is far superior to a kite.  That being said, some people do still use kites: High-Flying Science, with Strings Attached.


What is the physical meaning of the von Kármán constant? I know that the von Kármán constant ($k\approx0.4$) is useful in turbulence modeling. Something that bothers me is my lack of understanding where $k$ comes from or the physical meaning behind it. If $k$ is mainly used as a proportionality in the log-wind profile, why is it within the formulation of the Monin-Obukhov length? What is the physical meaning behind $k$? Is it a universal value (i.e. valid on extraterrestrial planets)? Why is $k$ considered important enough to be eponymous, while I see $\frac{1}{k}$ more often in formulations than $k$?

K is in fact just an empirical proportionality value without any significant meaning. In fact each of the tracers (such as temperature, humidity and ...) should have had their own K BUT we dump that coefficient into the (star)* version of that tracer, for example T* (near surface temperature) or Hum* (near surface humidity), so we can still use K in their respective formulation.

Monin-Obukhov length has K in it because that length is related to u*. That length comes from dividing the Shear production and buoyancy procution/dissipation term in turbulence kinetic energy (TKE) equation. As you know Shear production is a function of momentum transfer and u*. Since there is a u* there is a k in there too.

Unlike other stuff in turbulence, the story of K is unfortunately boring.


Do solar panels reduce the amount of suns rays reflected back into the atmosphere ? Therefore reducing global warming The way I understand the basics of global warming is we are putting to much Co2 into the atmosphere which increases the size / density of atmosphere shell, sun rays enter the atmosphere then there is some reflection off the earth and then the reflected rays get trapped within the atmosphere

What i was wondering if solar panels absorb the sun rays, is there then less of the suns rays reflected back into the atmosphere, or is some of the strength reduced ?

Thanks

Whether solar panels reduce the amount of suns rays reflected back into space depends on their albedo and the albedo of the surface that they cover.  Desert sand, for example, is fairly reflective and solar panels might reflect less sunlight back into space than deserts, but it's not just the reflection that matters.  You also have to account for energy production.

If enough solar panels are built that fossil fuel generated electricity (usually coal or natural gas), can be reduced, by 5% or 10%, that works out to a pretty useful drop in CO2 footprint.   Less CO2 from burning fossil fuels means less warming.   The ratio of surface temperature by absorbed sunlight vs reflected sunlight is on average, less significant than the heat trapping of CO2.   Explaining that gets kind of long and mathy and it's full of approximations, but as a ballpark estimate, 1 joule of energy produced by fossil fuels and the CO2 that's released by that process, that CO2 warms the atmosphere by several joules.   We can see this by estimating the added heat to the air and oceans as a result of climate change and comparing that to the energy use of all mankind.  I might add a crude calculation of my own later, but here's an article that says the ratio of heat captured by man made climate change to waste heat has a ratio of 100 to 1.

However you calculated it, the added heat by increases in CO2 exceeds any albedo variation by at least an order of magnitude, possibly two.   So even if Solar panels did absorb and retain more heat from sunlight than the surface area they cover, they still should lead to cooling by reducing the CO2 footprint.   The more important question is whether energy from solar panels can effectively reduce CO2 output from power-plants, because power-plants need to cover demand and solar energy can be intermittent.   Still, if there is any measurable reduction in fossil fuel generated electricity, there should be a net cooling due to how effective CO2 is in trapping heat even if the solar panels aren't efficient reflectors of sunlight.

Also,

  The way I understand the basics of global warming is we are putting to
  much Co2 into the atmosphere which increases the size / density of
  atmosphere shell.

This is incorrect.  It's not the size or density of the atmosphere but it's opacity to the long-wavelength infrared light radiating from  the Earth that drives climate change.  (feedback mechanisms like melting ice and rising albedo play a role too), but changes in the Atmosphere's density or mass are largely irrelevant.

Burning fossil fuels doesn't change the density or mass of the Atmosphere much anyway and they might even make the atmosphere a little lighter, though the overall calculation gets pretty complex with plant growth and oceanic absorption.

When fossil fuels are burned, O2 is taken from the air, combined with hydrocarbons in the fuel which becomes CO2 and H20.  The net mass change of O2 lost and CO2 gained depends on the Carbon to Hydrogen ratio of the fuel burned, but the effect is pretty small either way.   High hydrogen fuels like Methane, the primary ingredient in natural gas, takes more O2 from the air by mass than it returns in CO2.   More Carbon rich hydrocarbons like coal, the mass ratio is closer to even.   But in either case, the 30 plus billion tons of CO2 that's produced every year is negligible compared to the 5.1 million billion tons of the atmosphere.   That's less than 1 part in 100,000 by mass per year.

What drives climate change is what happens to the long-wavelength infra-red light that leaves the Earth.   The longer wavelength photons that leave the Earth sometimes hit a molecule of CO2 that temporarily wiggles, before reflecting that light-energy back.   That's the basis of the Greenhouse effect.   The reflection back to the Earth of long wavelength infrared light.   and more CO2 means more IR light gets reflected, which adds heat to the atmosphere.

In a nutshell it's the color of the sky, that is, color in wavelengths we can't see, so calling it color isn't quite right either, but that's the gist of the greenhouse effect.  Opacity not Density, though if you were to make Earth's atmosphere a lot more massive, say 10%-20%, that could have an effect, but we don't have the means to do that.

  sun rays enter the atmosphere then there is some reflection off the
  earth and then the reflected rays get trapped within the atmosphere

This is generally true.  Reflectivity is an important aspect of Earth's energy budget and Albedo of a solar panel is one aspect of solarpanel design.   That said, Solar panels cover such a small percentage of the Earth's surface, that their effect on reflection is negligible.   Also, reflected rays are less likely to get trapped in the atmosphere than infrared rays.   The Atmosphere is largely transparent to visible light and reflected rays of light are mostly in the visible spectrum.

An interesting and related question to ask is, instead of installing solar panels, what if we paint things white (well, you might not want a white roof in winter, so that might not be a perfect solution) but in someplace where it's always warm/hot all year round and white roofs are common.  White paint would reflect more sunlight back into space than a typical solarpanel would.   The math gets tricky where you have to weigh the fossil fuel use reduced against the increase in light reflected.  Here's an article that touches on that in some detail.

I also want to add that the dark color of solar panels is designed to absorb sunlight and turn it into electricity.  A reflective solar-panel kind of defeats the purpose, unless you're talking concentrated solar power which works based on reflection, but those aren't typical solar panels, though they are solar energy.


Locating Missing Airplanes using Remote-Sensing Satellites The Maylaysian jetliner that went missing has yet to be found. There are many remote sensing satellites, equipped with radars, currently in service. Together, they provide good coverage of the entire Earth. Why is it their data could not be used effectively in tracking down the path of an airplane? If the airplane lands on the ocean, shouldn't we be able to tell from satellite data?

Civilian earth observing SAR satellites do not always operate over empty swaths of oceans (specially in the middle of Indian Ocean) in order to save power. They are not designed to track a fast moving target and have repeat cycle of at least a few days. Then there is the issue of resolution. High res acquisitions modes (with satellites like TerraSAR-X) are pre-programmed in advance.

So in theory it is possible that a flying aircraft might show up in the amplitude image (a pixel or two wide) if a satellite was indeed flying over the Indian ocean and acquiring data over the area where the plane was flying. But then all you have is a single image. And once the plane is under water you wont see anything.


How to identify chemical property and structure of rocks? I am interested in identifying the properties of rocks. For example, black opal is shiny and can emit 5 different colors based on different lighting conditions. Is there a way for an amateur to be able to find out the crystal structure and understand why these different lights are being emitted?
Also is there a way to guess the chemical property of rocks based on just knowing their composition? If yes, is there something an amateur can self-learn to analyze these?
If it requires certain equipment, and I can't theoretically analyze it, I might be able to ask for permission from the chemistry department, so is there something I can do there? Although it'd be great not needing to use that instrument since getting perm is hard.
Thanks a lot!

What you are asking about is mineralogy,

a subject of geology specializing in the scientific study of the chemistry, crystal structure, and physical (including optical) properties of minerals and mineralized artifacts. Specific studies within mineralogy include the processes of mineral origin and formation, classification of minerals, their geographical distribution, as well as their utilization.

It is an essential part of all geology courses. Allied to this is study of how rocks form, petrology.
There is a lot to learn within mineralogy and mineralogists are a specilized group of geologists.


How is/was continental drift monitored? I am curious about current technology but I am particularly interested in what techniques were employed prior to the advent of satellite technology.

Scientific GPS and SLR have been used for some time now, and the measurements are rather accurate. Not only do we measure horizontal movements of tectonic plates, but also uplift as e.g. in the Tibetan and Colorado plateaus.

Before the GPS was introduced, paleomagnetism were used in some studies to estimate the velocities; The age of oceanic crust are estimated by various methods and the travel distance is estimated from the width of the magnetic polarized bands in the ocean floor. To understand paleogeography and make good models we need to incorporate all branches of geology. E.g. fossil records, paleomagnetism, radiometric age estimation of rocks.

Some active faults (E.g. San Andreas) can also be directly measured with laser ranging or, in some cases, with a measurement tape to measure deformation of trenches or railway tracks.

Still, scientists are trying to understand the exact driving forces behind plate tectonics. Modelling of mantel dynamics is an interesting field of research and many questions remains.


petrophysical crossplots I recently developed a computer program that generates some petrophysical and elastic parameters from well logs such as water saturation and hydrocarbon saturation, bulk volume, irreducible water saturation, porosity, permeability, shear impedance, acoustic impedance, and poisson ratio. I would like to know which parameters to cross-plot. For example, can you use a cross-plot of depth versus acoustic impedance for detection of hydrocarbons? My analysis is aimed at reservoir characterization.

Short answer: it depends.

The usual workflow is, very roughly, something like this:

Condition all your data. Compute any 'missing' data you need or want (e.g. porosity $\phi$, lithofacies, saturation $S_\mathrm{W}$, S-wave velocity $V_\mathrm{S}$).
Inspect crossplots of relevant properties like velocity vs porosity, acoustic impedance vs elastic impedance (or maybe Poisson's ratio or ${V_\mathrm{P}}/{V_\mathrm{S}}$), and so on, colouring points by depth, $S_\mathrm{W}$ or lithofacies, say. Look for helpful discrimination.
Use judgment, research, advice, or inspiration to decide on how to use the relationships you find... perhaps you are aiming for seismic inversion, or facies prediction, or log modeling, or something else.

Before doing much else, I recommend picking up Avseth et al. (Cambridge, 2005) (and/or perhaps  Dvorkin et al., Cambridge, 2014) and reading it from cover to cover. Here are some short reviews and more details on these books. Alessandro Amato's two recent tutorials in SEG's Leading Edge will help too, especially if you're using Python. If you're in MATLAB, a lot of the code for those books, which are both by the rock physics team at Stanford, is available online.

Here are some nice $V_\mathrm{P}/V_\mathrm{S}$ vs acoustic impedance $I_\mathrm{P}$ crossplots from Amato's second tutorial to inspire you:


How to locally identify a cold front passage? I know that the passage of a cold front affects local conditions by decreasing the temperature and shifting the winds. Some precipitation could happen as well.

How do software that analyze weather station data identify a cold front passage? Is there a certain change in temperature, wind, pressure, or other variable that allows the software to identify a front?

A cold front lies in a trough of low pressure and separates air masses.  As you mention in your question there are differences in temperature and the winds on either side of the front, though these can be subtle depending on latitude and season.

A "classic" (mid-latitude NH) cold frontal passage might be recorded as southerly or southwesterly winds, increasing in in strength, temperatures rising and pressure falling.  As FROPA occurs the pressure will bottom out and start to rise, winds shift to northwesterly or northerly, temperatures start to fall and skies clear.

A "classic" (mid-latitude NH) warm frontal passage might be seen as temperatures stable, calm or weak southerly or easterly winds, high clouds with bases lowering over time and pressure falling.  The FROPA will see pressure at its minimum and starting to rise, winds shifting to southerly and temperature warming.

The common factors the changes in pressure, wind and temperature and can be used to determine frontal passage.  Look for a pressure minimum, a change in temperature trend and a shift in wind.  A cold frontal will also generally bring drier air, and a warm front humid air, so this can also be used.

Here is a meteogram from SLC:

In this plot, cold frontal passage occurs around 7Z, marked by a minimum pressure trending upward, dropping temperatures and dew points, winds shifting to northerly and skies starting to clear once the precipitation ends.

Individual frontal passages will be different in exact details, but these general set of trends should help you out.


July/August hottest months (northern hemisphere): because of continued warming or other effects? In the northern hemisphere, July and August are the hottest months statistically1, but the days start shortening after June 21st.

Is this just because the Northern Hemisphere keeps heating up after June 21st (just like the hottest part of a day is not when the sun is highest, but some hours later), or are other effects at play?

1 To my limited experience, Europe specifically, but maybe other northern parts as well. And the questions goes for other months in the southern hemisphere, of course.

The amount of surface heating from the sun is a function of time of year and latitude.  You are aware of the dependence on the time of year with the sun over the equator at the equinoxes and ~ $\pm$23$^\circ$ N at the solstices.  The varying position of the sun overhead means the projection of a solid angle from the sun onto the surface of the earth will vary as well.  This just means that when the sun is low, the energy in a "beam" of sunlight is spread over a very large area and warming will be reduced.  This is why a bright sunny day in the winter cannot provide the same heating as the bright sunny day in the summer.

Surface energy loss is a function of the temperature of the surface, and atmospheric energy loss is a function of the local atmospheric temperature.  There are some complications here, namely clouds and greenhouse gases that prevent some radiative losses.

At the summer solstice when days are long and the sun is at its highest it is easy to understand that we are getting more energy than we are losing, and the days should be warm.  What isn't so apparent is that after the days start shortening and the sun lowering in the sky that the days continue to warm.  This is simply because we are still getting more energy than we are losing.  It isn't until the energy losses (which happen 24 hours a day, regardless of day/night) finally overcome our energy gains (which only happen when the sun is out) that the days will start to cool.

From www.atmos.washington.edu

This image is from the northern hemisphere, but it is unclear what latitude it is for.  The summer solstice is the maximum in solar insolation and the winter solstice the minimum.  Radiative losses increase with temperature, and it is the point at which temperature has risen enough such that the loss is equal to the (diminishing) solar input.  This is where the maximum temperature will occur.  The lag between the solstice and the max temperature will vary with latitude.

As mentioned in a comment in the other answer, large bodies of water exaggerate this seasonal lag in temperature.  This is for the same reasons as for land, with the added effect of the large heat capacity of water (about 3x that of land).

You see the same effect in daytime maximum and minimum temperatures where (barring anything but solar influence) the daily max is often in the late afternoon and the daily min is often just before sunrise.  This is the same reasoning as above, just on a daily scale rather than seasonal.  The sun is overhead at solar noon (this varies with local noon based on longitude, time zone (daylight saving time)) but continues to provide solar heating after solar noon.  It isn't until the radiative losses overcome the radiative gains (afternoon to late afternoon) that the temperature will drop.  Likewise when the sun goes down temperatures will cool down until the sum comes back up.

From www.vsc.edu


Waterlevel increase calculation So I tried to calculate the rise of waterlevel myself. But the results I have gotten seem to be pretty low:

Given the earths radius, I clalculated different Events:

1) If all landice melts there is an increase of the earths radius of ~54m. This number should be a bit higher because only ~70% of earths surface is water. For the ice I also didn't took the change of density into account because its relatively negligible.

2) Then I calculated the new radius given the annual loss of ice 2011-2014. This number was surprisingly low with only ~1mm each year.

3) And for the extreme increase of water temperature globally of 9°C I took the Global Water Volume into account. But the increase of 5.2m is not that big either.

So are my numbers as expected or did I do something wrong? Because these numbers don't explain the predicted massive increase of the sea level in the near future.

Beside some unnecessary simplifications (like assuming the whole Earth is cover with water) The numbers looks right. IPCC predictions are between 0.3 and 0.8 meters by year 2100 for the most likely scenarios, and I find that perfectly consistent with your numbers.

Your calculation yields about 0.98 mm per year from glaciers, which using the actual surface of the sea (361,132,000 km$^2$), gives a more accurate value of 1.39 mm per year.

Look at the IPCC observations and models summarized on table 13.1 of the Chapter "Sea level Rise" of the Assesment Report 5:

For the latest period, if you add up all land ice contributions you get that they are between 0.87 and 2.05 mm/yr, totally consistent with your estimation (of ~1 mm).

The largest single contribution is thermal expansion with 0.8 to 1.4 mm/yr.

Also note that the observed contribution don't quite add up to the observed sea level rise, so we are either sub-estimating some contributions or ignoring significant ones.

If you multiply the current 3.2 mm/yr by 80 years you get ~26 cm by year 2100, but the point is that the rate of sea level rise is increasing (see this answer). That is why all predictions are above that value.

For future predictions, thermal expansion is still the leading cause of sea level rise, as you can see on figure 3.10 of the same IPCC report:

I find these predictions consistent with your numbers if you consider the current acceleration on the sea level rise rate. Unless you are talking of other predictions, not the IPCC ones. Or maybe you think that half a meter of sea level rise over 80 years is not much. If that's the case, I suggest you to research what would be the impacts in land loss, infrastructure loss and human populations displacements of such sea level rise.


How do I convert a latitude/longitude coordinate to a NOAA GFS grid cell location? Looking at this GFS http://nomads.ncep.noaa.gov:9090/dods/gfs_0p50/gfs20170707/gfs_0p50_12z.ascii?tmax2m[0:2][124:128][279:283]

The range defined by the second and third set of brackets specify the grid coordinates, however I'm not sure how to convert between what NOAA is using and the common geographic coordinate system.

What would the NOAA equivalent be for this coordinate pair: (36.25, -78.75)?

** All of the links given with times will need to be adjusted to current dates if reading this at later times or else will appear dead **

Once upon a time I dealt with such files, but not in quite a few years... and Google wasn't the best help at first in finding documentation on what the .ascii is exactly... but thanks to Kevin Goebbert's code here, I was able to find the description page for the model in question is located by removing the .ascii and inputs (and thus here):

And so right now it lists:

Longitude:
0.00000000000°E to 359.50000000000°E
     (720 points, avg. res. 0.5°)
Latitude:
-90.00000000000°N to 90.00000000000°N
     (361 points, avg. res. 0.5°)
Altitude:
1000.00000000000 to 1.00000000000
     (47 points, avg. res. 21.717)
Time:
12Z07JUL2017 to 12Z17JUL2017
     (81 points, avg. res. 0.125 days)

The p50 indicates a grid space of only half a degree in that file, so you couldn't actually use your point in this moel.

However, use the 0.25 resolution instead (description):

So first is time first is time, the second is longitude points starting at the Prime Meridian and going east, and the third is latitude starting at the South Pole and going north.  (GFS time steps have always been a mire... but based on this full list of data files and the info page previously given, it indeed appears the data provided here is just every 3 hours for the first 10 days.)

So to pick your value for the second input, it'd be something like

  ifpositive: (longitude)*4 - 1
  ifnegative: (180+longitude)*4 + 719

And for latitude:

  360+(latitude)*4

And so if I didn't make a mistake, that'd be this point (lat gridpoint 505, lon gridpoint 1124).


Rainfall units conversion from depth to rate/intensity Given 6-hour rainfal data with units in meters(m), how can one convert the 6-hourly rainfall depth in meters(m) to 1-hourly rainfall intensity or rate in kg/m^2/s?
Below is a sample data value to convert.
rainfall depth (m)= 0.0014061

This is what I have tried but not sure it is correct:
To convert the 6h value to the hourly rate, I applied a multiplicative factor:
factor=0.046296296 kg/m2/s

based on the reasoning that the equivalent to:
m (in 6 hours) to mm/s = 1000/21600 (i.e. 0.0462963)

How do I use the multiplicative factor (0.046296296 kg/m2/s) to convert 6-hour rainfall depth to hourly?
Maybe I am thinking too much here.
Any thoughts?
Thank you.

Your 6-hour rainfall data represent the depth of water fallen in 6 hours. In order to have the average rainfall intensity in m/s corresponding to that interval of time, you have to divide the rainfall depth for the number of seconds in the time interval or for mm/s you multiply for 1000/21600. If you want the intensity in m/h you divide for 6 hours.
You have only the cumulative rainfall in 6 hours so you can estimate the intensity dividing equally the volume of rainfall over the 6 hours. The information remains the same. The real intensity was not constant over the 6 hours, obviously. You cannot, therefore, obtain the real 1-hour rainfall depth for each hours from 6-hours rainfall depth but only an average value that assumes it constant in each hour. If you have the measured 1-hour rainfall depth data in meter and you calculate the intensity in m/s from that data the results will be different.
I hope to have correctly understood your doubts and that this made things clearer.


Can we carbon-date the remains of homo floresiensis found in 2003? According to the Wikipedia article on the species Homo floresiensis, the remains discovered in 2003 consist of unfossilized bones. I would assume that means they are still composed of the original organic material left behind when the human specimen died thousands of years ago. Shoudn't that mean radiocarbon dating would be a good method to date the reamains?

Many articles on Homo floresiensis also discuss how the remains were originally dated to ~12,000 years ago, but that this estimate was later revised to 60–100,000 years ago. However, everything I can find indicates that mostly geological dating methods were used, not radiometric dating.

Why not? The Wikipedia entry on carbon dating says that it can only be used reliably to date specimens up to ~50,000 years, but could carbon dating then at least place a lower limit on the age of these remains? And why wouldn't it have been used back when they thought the specimens were only ~12,000 years old?

ScienceMag says:

  Rather than damaging the fossils by dating them directly, the team looked to the sediments in which they were found. They discovered pieces of charcoal in sediments at similar depths, and considered those to be proxies for the ages of the fossils themselves. The charcoal bits were dated to around 19,000 and 13,000 to 11,000 years before present.
  But Liang Bau boasts a devilishly complicated geological history, as layers of silt and clay interleave with layers of weathered limestone, loose gravel, and volcanic ash. In many places, these layers have been scoured by erosion, altered by seeping water, and jumbled by tectonic activity.
  ...
  Dating on the gravelly sediment layer containing the fossils suggested it was deposited between 100,000 and 60,000 years ago; just above it is a layer of volcanic ash that was dated to about 60,000 years ago. This suggests that the fossils themselves couldn’t possibly be younger than 60,000 years old, the team reports online today in Nature.
  To cap it off, the team has now directly dated the fossils themselves. Originally, researchers shied away from that analysis for fear of damaging them, Tocheri says. But this time, they felt they needed to do it. “It wasn’t until [we] were reasonably convinced that [the fossils] were probably older than 50,000 years old that we decided we needed to date them directly to be absolutely certain,” he says.
  The team dated three arm bones found at different locations within the cave during previous excavations. Using uranium-thorium dating—which is based on the radioactive decay of uranium and thorium isotopes in a material—they found dates ranging from about 66,000 to 87,000 years old.

I could not find how the "charcoal bits" were dated, but it would be in the range for carbon dating.


Stability correction factor in the Wavewatch III wave model I'm studying the Wavewatch III wave model (http://polar.ncep.noaa.gov/waves/index2.shtml). I'm confused by the stability correction factor, named ASF in the code. What is its purpose? Is it related to the physics, or is it something to do with numerical methods and the CFL condition? I could not immediately figure this out from the code.

The stability correction factor ASF is related to the effects of atmospheric stability (function of buoyancy and shear) on wave growth, and has been implemented in Wavewatch3 in the Tolman and Chalikov 1996 input source term. The code where the correction happens can be found in w3updtmd.ftn:

! 5.  Stability correction ( !/STAB2 )
!     Original settings :
!
!     SHSTAB =    1.4
!     OFSTAB =   -0.01
!     CCNG   =   -0.1
!     CCPS   =    0.1
!     FFNG   = -150.
!     FFPS   =  150.
!
!/STAB2      STAB0  = ZWIND * GRAV / 273.
!
!/STAB2      DO ISEA=1, NSEA
!/STAB2        STAB   = STAB0 * AS(ISEA) / MAX(5.,U10(ISEA))**2
!/STAB2        STAB   = MAX ( -1. , MIN ( 1. , STAB ) )
!
!/STAB2        THARG1 = MAX ( 0. , FFNG*(STAB-OFSTAB))
!/STAB2        THARG2 = MAX ( 0. , FFPS*(STAB-OFSTAB))
!/STAB2        COR1   = CCNG * TANH(THARG1)
!/STAB2        COR2   = CCPS * TANH(THARG2)
!
!/STAB2        ASF(ISEA) = SQRT ( (1.+COR1+COR2)/SHSTAB )
!/STAB2        U10(ISEA) = U10(ISEA) / ASF(ISEA)
!/STAB2        END DO

The correction essentially accounts for atmospheric stability based on air-sea temperature difference, and is applied to calculate the "effective" wind speed at 10-m height, from the input wind speed. Note that this code is active only when /STAB2 switch is enabled (optional), and works only with the /ST2 switch (Tolman and Chalikov 1996 source term package). It's implementation and test results have been described in Tolman 2002.

This is also documented in the manual for version 4.18, see equations (2.86)-(2.89) in section 2.3.7.

References:

Tolman, H. L. and D. V. Chalikov, 1996: Source terms in a third-generation wind-wave model. J. Phys. Oceanogr., 26, 2,497–2,518.
Tolman, H. L., 2002: Validation of WAVEWATCH III version 1.15 for a global domain. Tech. Note 213, NOAA/NWS/NCEP/OMB, 33 pp.


What causes 'positive' lightning to be rarer, higher charged, and longer lasting than 'negative' lightning? Why is 'positive' lightning rarer, brighter, higher charged, and longer lasting than 'negative' lightning?  Are there any other unique characteristics to positive lightning that sets it apart from more common negative lightning?

Positive Lightning vs Negative Lightning

Rarity: 1:10 - 1:20 (on average)
Charge: 10:1 (on average)

(I'd imagine that brightness and duration are correlated to the charge).

Positive lightning is indeed very interesting. According to the National Weather Service, the reason why it has a higher amperage and voltage is due to the due to the amount of air it must get through to reach the ground. Otherwise, it would never reach the ground.

  Since it originates in the upper levels of a storm, the amount of air
  it must burn through to reach the ground usually much greater.
  Therefore, its electric field typically is much stronger than a
  negative strike. Its flash duration is longer, and its peak charge and
  potential can be ten times greater than a negative strike; as much as
  300,000 amperes and one billion volts!

This would also account for the brightness of positive lightning and the extended duration, as a higher amperage & voltage strike would appear brighter and take longer to dissipate.

This page talks about why positive lightning occurs. It says:

  "Positive lightning" is a simple process of completing an electrical
  circuit from the cloud to the ground and back to the cloud.  As
  negative lightning strikes the ground, the ground becomes negatively
  charged over a wide area.  As the storm moves over this area and past
  it, the H+ in the upper reaches of the cloud feels this potential
  difference from the ground and sends out a positive streamer which,
  for all intents, looks like normal negative lightning.  The ground
  sends up a negative streamer and once these two join, electrons from
  the ground flow upwards to the H+ in the top of the cloud. The ground
  then becomes neutral once again.  The ground will hold the negative
  charge of dozens of negative strikes, building up a huge potential,
  before positive lightning will occur.   This is why positive lightning
  is so powerful.

Positive lightning is probably so rare because it takes so many negative strikes in order for enough energy to build up to cause a positive lightening strike.

There are quite a few interesting facts about positive lightning:

  Some interesting properties of positive lightning:

  Positive lightning can be the dominate type of cloud-to-ground during the winter months and in the dissipating stage of a
  thunderstorm.
  Positive lightning has been identified as a major source for the recently discovered sprites and elves. Sprites and elves are most
  likely lightning discharges but occur from 18-60 miles (30-95 km) in
  altitude, well above the parent thunderstorm.
  Positive lightning is usually composed of one stroke (negative lightning typically consists of two or more strokes)

Another NOAA site says that positive lightning can strike "near the edge of a cloud", or even 10 miles away from the cloud! This means that a positive lighting strike could occur somewhere where the sky is blue, and it's not raining.

In addition, there is also a type of lightning called "bipolar" lightning, which changes its polarity.


Are wetlands effective against climate change? In the region where I live, there is a trend that conservation organizations are advocating the conservation and re-valuation of wetlands.
Aside from boosting biodiversity and providing other ecological services, the conservationists argue that wetlands capture a lot of carbon.
I know however that wetlands are notorious for steady state methane emission [1][2], so it sounds dubious whether a single-time carbon-capture or carbon-release can compensate a significant steady-state emission of a potent greenhouse gas.
Are there cost-benefit-analyses out there regarding the effect of wetlands on climate? What is your take?

There's a misinterpretation in the "steady state" part, that's not what the linked paper says. It clearly states the figures are temperature dependent, and it says that in response to global warming wetlands are going to release more Methane and tries to quantify the amount. It concludes that

This finding highlights the importance of limiting global warming below 2°C to avoid substantial climate feedbacks driven by methane emissions from natural wetlands.

'Feedback' is the keyword here. Such a feedback is an accelerated response of a system to global warming. Grossly simplifying, in case of small shallow wetlands CH4 release is connected to the level of oxic and anoxic conditions, the latter expanding not only in the table of a local 'puddle' but also geographically from lower to higher latitudes with temperature and so releasing more CH4 in a warmer climate, which warms the climate and so on. But the ability of these ecosystems to bind carbon from the atmosphere drops with rising temperature, and the amount of released carbon rises.
That is by no means an argument against the conservationists who are advocating conservation and re-valuation to help sequester carbon.

This analysis can help inform practitioners and landscape managers [...] foster the worldwide implementation of wetland restoration, creation, and conservation projects for sustainable development and climate change mitigation and adaptation.

So, yes, for now wetlands are effective against climate change as long as anthropogenic global warming is limited.


How many trees would I have to plant to solve Global Warming? According to NASA, causes of the Earth's greenhouse effect include water vapor, carbon dioxide, methane, nitrous oxide, and CFCs.  Carbon dioxide gets the most press, and NASA's page says:

Carbon dioxide (CO2). A minor but very important component of the atmosphere, carbon dioxide is released through natural processes such as respiration and volcano eruptions and through human activities such as deforestation, land use changes, and burning fossil fuels. Humans have increased atmospheric CO2 concentration by a third since the Industrial Revolution began. This is the most important long-lived "forcing" of climate change.

An article by NC State University says that a healthy tree can store 13 pounds of carbon per year.  As I understand it, carbon dioxide is processed by the tree: the carbon is stored, and the oxygen released.
Given that, how many trees would I need to plant to solve the global climate change crisis?  Should I optimize for a specific type of tree, or would pine work as well as oak or black walnut?

In 2010 anthropogenic emissions (not including land use change) were approximately 9167 million metric tonnes.  Your data on trees holding 13 lbs (5.9 kg) of carbon per year equates to 169.6 trees per metric tonne of emissions.

So to take up all of the emissions from 2010 you would need 1,545,000,000,000 trees.  A mature forest has only about 100 trees per acre (400 per hectare), so you would need 15,545,000,000 acres of mature forest.  This equals an area of 24,290,000 mi2 (62,910,000 km2).  This is approximately the land area of Asia, Europe, and Australia combined!
The surface area of land on the planet is about 150,000,000 km2, so in principle we would need to add cover onto 42% of the current land (or we could take soil from deep ocean floors to landfill 1/5th of the oceans!) in order to plant enough trees to solve the problem.

This also assumes that the 13 pounds (5.9 kg) of carbon figure is for mature forests, rather than for growing trees (see comments for further discussion).  If this value is for young developing trees, it would indicate less be attainable in the longrun.  If the 13 pounds is for a developed forest, perhaps a greater amount could be removed in early years as a temporary quicker fix.

There is also the problem that forested land is likely to have a lower albedo than the land surface that it covers, and hence the planet will reflect less sunlight back into space which would lead to some extra warming, so we would also need to compensate for that somehow.  Apparently about 26% of Earth's land is already covered with forests.  I rather doubt over half of the uncovered land surface is suitable for new forests, the continents have large bands either side of the equator that are generally too arid, and the regions close to the poles are too cold.

In short, it isn't going to work, even with the most generous assumptions about forest CO2 exchanges (unless of course I have made an arithmetic error, which is definitely a possibility).


Why is the bedrock buried so deeply in much of northern Minnesota? Although parts of northern Minnesota have bedrock exposed at the surface (especially in the Boundary Waters area and along the short of Lake Superior), the more usual situation here is for the bedrock to be buried very deeply indeed - with, in places, more than a hundred meters of soil overlying it, according to this map.

This doesn't make sense to me; the whole area was buried under an ice sheet during the most recent glaciation, and the hundreds of meters of moving ice would have scraped everything down to bedrock (and scraped off large amounts of the bedrock itself) and prevented any new soil from forming until the ice sheet retreated.  Northern Minnesota has only been ice-sheet-free for a blink of an eye, geologically speaking - how did hundreds of meters of soil manage to form in such a short period of time?

As njuffa mentioned I don't believe you are looking at soil 100s of meters deep you are looking at unconsolidated sediments 100s of meters deep.

Depending on the geological process it is possible to deposit 100s of meters of sediments in a very short time.  The conditions for deposition of large amounts of sediments are frequently found during the period of time that the glaciers are melting.  During this time the sides of the glacier end up forming the glacial fluvial terraces and the toe of the glacier often breaks into pieces of ice that the sediment can build around.

This site from the Minnesota State University describes the majority of the lakes in Minnesota as Kettle Lakes, which suggests that the tail end of the ice age was a time of significant sediment deposition.  The North Eastern area would have likely be covered by ice while the South Western portions would have been experiencing the deposition of materials from the rapidly melting glaciers.  There is a whole wikipedia site devoted to the glaciation of Minnesota that describes the events relatively well.


Is moonlight a significant condition for making life possible on earth? We know of course that light from the sun is a fundamental component to making life possible on earth because of the energy it provides. Does the light that is reflected off of the moon make a significant impact on the earth as well?

For example - at night, would it be at lot colder on earth without moonlight? And would it basically be pitch-dark on earth without moonlight? Or do stars and other astronomical entities provide a significant amount of light compared to the moon?

One significant biological event that requires a specific "type" of moonlight is the reproduction of corals in Great Barrier Reef, in Australia.

'One week each year in spring, after a full moon, millions of corals release eggs and sperm in what Bill Leggat, a co-author of the new study, called "a slow symphony."'

'An ancient light-sensitive gene has been isolated that appears to act as a trigger for the annual mass spawning of corals across a third of a million square kilometres of Australia’s Great Barrier Reef, shortly after a full moon.'


Trees: What does "highly tapering stem" mean? Trees: What does "highly tapering stem" mean?

Taper is a function measuring steepness of change in crown diameter when height increases upwards.

But "highly tapering" (sometimes also "high-degree of taper")? Is it "a large change in taper" (large steepness) or a "fine change in taper over a long height" (low steepness, but high fineness of change)?

A "highly tapering stem" is simply that the trunk of the tree is less cylindrical and more conical in profile. It's a problem because while you may have a lot of timber volume you don't have as much timber that is usable because you'll end up with a higher proportion of sawmill waste.

I'm English and have a degree in forestry. Regardless of comments elsewhere, a "highly tapering stem" is perfectly good terminology, although I agree that it could be clearer. I would tend to say "a high degree of tapering" or a "more tapered stem" for clarity's sake.

A full description of this can be found in Forest Mensuration: a handbook for practitioners by Matthews and Mackie on page 114. (They refer to a tapering log, because they are discussing felled timber).


Epicenter location of the 900-930 A.D.,7.4 Magnitude Seattle Earthquake? I am preparing a seismic hazard map of Seattle and I was curious about the great 7.4 magnitude earthquake that occurred in the Seattle area during 900-930 A.D. On a second note, is it possible to infer the epicenters of paleoearthquakes (earthquakes that happened more than 800 years back)? And to what confidence level?

If the quake you describe was on the Seattle Fault, you can probably guess the start point and the end point of the fault movement based just on the magnitude of the quake.  (Quakes that "snap" longer portions of the fault release more energy.)  The start point will give you the epicenter location.


Searching for ground water I would like to attempt to build a well-type structure in a woodland area. This woodland is mainly inhabited with Silver Birch and Sycamores and there is a shallow stream running down off of a moorland around 80ft from where the well would be ideally placed, but this can move.

My thoughts are that due to the prevalence of water there is likely to be a body of water underneath the ground simply through the natural process of infiltration.

So my questions are the following:

1) Am I likely to come across water if I dug this well?

2) How deep will I have to dig before I reach water or impermeable ground?

I live in Colorado, USA.  All gas stations must have shallow wells drilled around the gasoline tanks to monitor for gasoline tank rupture leaks.  The tanks usually lie above the water table so any gas leaks will travel to the water table and start flowing on top of the groundwater.  One hundred percent contamination happens when a flashlight is shined down the well and a gasoline sheen is seen on the water.  Less contamination is measured by sampling the water. The depth to the water table is also measured.

The point I am getting to, is that all gas stations in Colorado have to be measured every three months and the results sent to the State.  I would suggest that you check for a gas station near where you live and get a recent published water report that you can study.  One of the maps is a water table depth map.  The map will have a contour map with the depth below surface contours and arrows perpendicular to contour lines that show the direction of groundwater flow.

I would suggest for you to see if your State or Provence has a similar contamination monitoring program. This would give you an idea how the water table behaves in your area. If you go to your nearest gas station and look for circular steel plates about 6 inches in diameter placed flush on the ground around the tanks, gas pumps, and parking lot, then you will know if the station is monitored.  The top of the plate has a filled in triangle with the triangle points touching the circle circumference. Sometimes the plate has a warning that says "DO NOT FILL".  The the steel plate and map symbol looks like Unicode symbol U1F7D5.


Hydrology - Depression cone VS transmissivity My question concerns the depression cone of a pumping well in a confined aquifer: Say the transmissivity of the aquifer is 1200 m^2/day how will the cone change if the transmissivity is 600 m^2/day? All the aquifers parameters stay constant in both scenarios and so is the pumping rate (say- 300 cm/hr).

The main interest is what happens to the radius of influence (where there will be no change in head from the aquifers regional head)? and also what happens to the head at the well it self?

Thanks, Asher

Based on the Theis equation, increasing the transmissivity will increase the radius of influence and decrease the total drawdown.  Drawdown is directly proportional to the pumping rate and inversely proportional to aquifer transmissivity and storativity.

$$s = \frac{Q}{4T\pi}\int_u^\infty \frac{e^{-y}}y dy$$
and $$u = \frac{r^{2}s}{4Tt}$$
Where s is drawdown, r is radius from well, t is time, T is transmissivity, Q is pumping rate, and u is the well function.

Applying this equation, you can produce plots like this:

for these plots, I used Q = 500, S = 0.01, and t = 4


How can scientists be confident of isotope ratios in past eras and varied locations, used for radiometric dating? Radiometric dating relies on past isotope ratio predictions being very reliable.

However, this is not necessarily so. For example, in Uranium forensics, 235U and 238U ratios are about the same for all sources, but 234U varies and can sometimes be used to identify the source region. So, how can we be so sure that the isotope ratios used for radiometric dating are as reliable as we assume. What if the ratios of 238U and 235U we observe are a signature of the earth, like U234 is a signature of geographical regions?

Also, for meteorites, scientists say that the radiometric dating shows when the rock was last melted. Well, what if it was melted more than once? Does each melting reset the isotope ratios in the rock? If a rock of 4 billion years was fully melted, what happens to the isotope ratios in the rock? In pages that explain radiometric dating, that I have read, don't even mention this scenario - or the following one.

And what about the assumption that we can predict the isotopic ratios of the solar system 4 billion years ago? how is the assumption justified?

"Doctor! It hurts when I do this! «bonk»" "So don't do that then!"

One way around the issue of an unknown initial distribution of isotopes is to not do that. One of the most reliable mechanisms for dating the Earth is uranium-lead dating of zircons. Uranium can take the place of zirconium in the zircon crystal because of the chemical similarity of uranium is and zirconium. Lead on the other hand is very dissimilar chemically from zirconium and uranium. The crystal formation process strongly rejects lead. At the time a zirconium crystal forms it will contain no lead. All the lead in zircons comes from the decay of uranium. Two decay chains lead to lead, with 235U eventually decaying to 207Pb and 238U eventually decaying to 206Pb. Those two decay chains provide two independent mechanisms to date the same crystal.

Another approach is to find that initial distribution of isotopes. This is the approach Claire Patterson used in his estimate for the age of the Earth. See Claire Patterson, "Age of meteorites and the earth", Geochim Cosmochim Ac 10 (1956). He used a non-radioactive iron meteorites to establish the primordial distribution of isotopes of lead. With this as a baseline, he then used the disequilibrium ratios in meteorites to find the age of the Earth.

With regard to varying 234U/238U ratios, that's a red herring used by some who don't like the results of radioactive dating in a failed attempt to throw suspicion at those dates. The 256 thousand year half life of 234U means that there is no primordial 234U on the Earth. All of the 234U we see now came from the decay of 238U, via 238U→234Th→234Pa→234U. Those decays weaken rock at the site of the decays. When the rock weathers and is exposed to water, that weakened rock around a 234U atom means that 234U is preferentially leached from the rock.


Why (actually) is the night sky so bright in the city? How far up is that happening? The recent news reported by the New York Times, the BBC and the NPR web sites (for example) as well as in the Iceland Review and Iceland Magazine sites, about the shutting-off of Reykjavik's streetlights in order to better enjoy the Aurora (northern lights), got me thinking.
There are many life-long city dwellers who've never really seen a relatively dark sky or more than a handful of stars. Even people who could do so sometimes have never made the trip out of the city far enough for the experience.
So I was curious if this "lights-out" technique could ever be tried in a city or somewhere else besides in Iceland.
I'd like to estimate how big of a radius of "lights-out" would an observer need for the sky to be noticeably darker. But for that I'd have to know what is actually doing the scattering, and how far up does it extend.
My question: Why (actually) is the night sky so bright in the city? How far up is that happening?
I'm also curious if it is generally stronger in some climates vs others.

What's actually happening is scattering of light, both off of aerosol particles and nitrogen and oxygen molecules.

For a review of quantitative models of light pollution as a function of distance from cities see Light Pollution Modeling . Some consider height as well as distance.

To get a rough estimate of what size of lights-out event would be needed to obtain a particular improvement in sky darkness, I would suggest looking at the light pollution map at Dark Site Finder or The night sky in the World.

  Colours correspond to ratios between the artificial sky brightness and the natural sky brightness of: <0.01 (black), 0.01-0.11 (dark-gray), 0.11-0.33 (blue), 0.33-1 (green), 1-3 (yellow), 3-9 (orange), 9-27 (red), >27 (white)

Look at a coastal city such as Miami (which is extremely light-polluted) and see how the degree of light pollution drops off with distance from the city. Or look at a very isolated city such as Bismark, North Dakota.  I would estimate 20 to 40 kilometers away from a source you see a factor of 10 reduction in artificial brightness.


Why is calcium so much more abundant in/on Earth's crust (and oceans) than it is elsewhere? (relatively speaking...) Inside the entire Earth as a whole, and throughout our Solar System and galaxy, etc., magnesium is at least 14 times (or more) as common as calcium, atom-per-atom.
But, on Earth's crust and oceans, calcium is at least as abundant, atom-per-atom, as magnesium....
Why is this?    Calcium IS denser/heavier....  They're both Group-2....

Two factors need to be considered here.  Magnesium ions are smaller, thus forming more compact formula units, and they are also often mixed with iron ions which make the formula unit weight heavier than you expect.  Magnesium-rich mafic rock, being also iron-bearing and with compact ions, literally packs on the pounds and tends to sink deeper into the crust and the mantle compared with rock (even mafic rock) containing more calcium.
A false bargain
Magnesium may form lighter atoms than calcium, but it also forms smaller atoms--especially when the metal forms cations as in the most common minerals where the metal is bonded to oxygen.  So the magnesium compounds have less formula weight but pack that mass into less volume.  If we compare densities of corresponding calcium and magnesium compounds we see that these effects almost cancel out.  Data for these examples are taken from Wikipedia:
$\text{CaO}$:  3.34 g/cm³
$\text{MgO}$:  3.6 g/cm³
$\text{Ca}_2\text{SiO}_4$:  3.28-3.33 g/cm³ (larnite)
$\text{Mg}_2\text{SiO}_4$:  3.21-3.33 g/cm³ (forsterite)
So there is only a little density difference overall -- in a world where mafic rocks do not contain iron.  Ah, but they do.
Going green, but packing heavy
Why is jade usually green?  The most common jade mineral is nephrite, which is mainly a slightly hydrated calcium-magnesium silicate -- with a little iron mixed in to get that green color, at least at low levels of iron.  The formula for this mineral is a mouthful:  $\text{Ca}_2\color{green}{\text{(Mg,Fe)}}_5\text{Si}_8\text{O}_{22}\text{(OH)}_2$.  As indicated by the text in that suitable green color, the iron is intermixed with the magnesium, not with the calcium.  The same holds true with the simpler, and also green-colored, olivine, $\text{(Mg,Fe)}_2\text{SiO}_{4}$, except that olivine can contain more iron than is usually found in nephrite.
Magnesium and calcium both lie in Group 2 and both form ions with +2 charges.  But that alone does not mean they are intermixed in mineral structures.  To fit nicely into the precise, tightly packed solid phase ions must not only have consistent charges at a lattice site, they must also have consistent sizes.  Calcium ions, with an ionic radius of 114 pm (ionic radius data from Wikipedia), are simply too big to fit where there should be magnesium ions with a radius of 86 pm.  Iron with +2 charge in its high-spin state (oxides of transition metals usually have high-spin ions) comes in at 92 pm, not perfect but a much more acceptable charge/size doppelganger for magnesium.  While nephrite in green jade has only a little iron, olivine can have a full range of iron/magnesium ratio from forsterite ($\text{Mg}_2\text{SiO}_4$) to the dark fayalite $\text{Fe}_2\text{SiO}_4$, with a wide range of densities to match -- all the way from 3.2 to 4.5 g/cm³.  This range overwhelms the small difference we saw earlier between calcium and magnesium compounds without iron noted earlier.
So the combination of a common, small ionic size for iron and magnesium with a relatively heavy atomic weight for iron makes magnesium-rich rocks sufficiently iron-rich and closely-packed to weigh them down.  Calcium adds extra bulk from its own ions and remains iron free in its lattice sites of mineral lattices, so calcium-bearing minerals are less dense and tend to float into the crust (and even continental crust).
Earth's Cream of the Crop
This answer may be compared with the one here regarding the Martian crust.  There, we find more magnesium relative to calcium.  While the calcium vs magnesium/iron differentiation described above occurs there, lack of a specific continental crust means the differentiation is less.  Accordingly, the composition of the Martian crust is closer to that of the magnesium- and iron-rich mantle than that of the terrestrial crust.


How does radiogenic argon-40 get into the atmosphere? There's an awful lot of argon in the atmosphere: about 0.93% of the atmosphere is argon[1], making it the third-most abundant gas after N2 and O2. As I understand it, most of it is produced by the following decay path:

$$
\ce{^{40}_{19}K + ^{0}_{-1}e- -> ^{40}_{18}Ar + \gamma{} + \nu_{e}}
%edit
$$

How does all this radiogenic argon make it into the atmosphere, though? Most of the decaying potassium-40 must surely be at some depth beneath the surface, thus trapping the argon-40 in the interior of the Earth, right?

Most of this 40Ar makes its way to the atmosphere in two primary ways. Firstly, since most of the K is concentrated in the continental crust, volcanic eruptions on the continents over geological time have concentrated Ar in the atmosphere. Since it is too heavy and inert to decrease in any appreciable method it has a long residence time. Secondly, volcanism at mid-oceanic ridges releases Ar into the ocean and then into the atmosphere. The current combined rate of these volcanic events is ~ 2.2×109 grams/year.


Is it normal for a cyclonic storm to form at the latitude of New York, USA? I am following climate activist Greta Thunberg's sail to N America from Europe on https://tracker.borisherrmannracing.com/. The model shows a cyclonic storm forming south of Nova Scotia, east of New York on August 26, 2019. The same thing can be seen on https://earth.nullschool.net/. I wasn't aware that this type of storm formed so far north and am wondering why this storm is forming and if this is a rare occurrence that may be related to climate change.

Edit to add: Here is the model of air temperature. It appears to me that the storm forms at the interface between cool air from the north and warm air from the south and that the core is warm, but perhaps someone with more expertise can weigh in.

A cyclone is nothing more than a low-pressure system. It can surely form in the North Atlantic; in fact, most low-pressure systems that reach Europe form in the North Atlantic.
You are probably thinking about a tropical cyclone, which indeed cannot form that far north. Among others, a tropical cyclone needs high sea temperatures (>27 C), which are usually not present at such high latitudes.

About the present cyclone south of Nova Scotia: I think it is a cold core cyclone, as 500 hPa temperature are lower than in surroundings (which is e.g. not the case for tropical storm Dorian, which is currently east from the Lesser Antilles). Therefore, it's not a tropical cyclone but an ordinary extra-tropical cyclone, which can perfectly form at those latitudes.


Does organic decay release the same CO2 as burning? Does a fallen tree left to decay release essentially all of its stored carbon as $\small\mathsf{CO_2}$ to the atmosphere? Or is the carbon uptaken by termites and bacteria, preventing its release to the atmosphere?

I can't quantify the answer, but intuitively, nearly all of a burning tree's carbon is converted to $\small\mathsf{CO_2}$ - assuming the fire is hot enough. You are correct in assuming that termites and bacteria convert some of the carbon to other organic species, as do fungi, yeasts, moulds, and slow weathering. In addition, some of the rotting tree's carbon is converted to soluble inorganic ions, humic, fulvic and tannic acids. So no, organic decay would not yield the same $\small\mathsf{CO_2}$ as burning.

This would make an interesting school or 1st-year uni science experiment. Burn some wood in a sealed environment, measure the $\small\mathsf{CO_2}$ evolved, then repeat the experiment with wood in a wet organic-rich isolated biome, and compare the results. I suspect it would make for an interesting paper with potential climate change policy implications.


What do opaque rims on biotite in a volcanic rock mean? In a volcanic region, i have examined two samples petrographically. One of them has biotites with extensive opacitic rims. The other samples biotites hasn't got this property. They have the same modal mineral contents. What makes the difference?

There are several options.

During decompression, biotite is no longer stable and the hydrogen volatilises, leaving behind fine-grained intergrowth of oxide and silicate minerals.
Oxidation of the rims during exposure of the lava to the atmosphere. Higher Fe3+ contents would make the biotite darker.

It is impossible to tell without a detailed electron microscopy analysis. However, my guess it that it is option 1.


Why is smoke considered a factor of global cooling when analyzing nuclear winter scenarios, but arent when analyzing global warming scenarios? In many articles on the internet (I'll cite a couple for the sake of the question), smoke from the fires (composed mainly of $\small\mathsf{CO_2}$ and CO) following a nuclear war would cool the atmosphere and would cause a "nuclear winter".

  Nuclear winter is a severe and prolonged global climatic cooling
  effect hypothesized1 to occur after widespread firestorms
  following a nuclear war.[3] The hypothesis is based on the fact that
  such fires can inject soot into the stratosphere, where it can block
  some direct sunlight from reaching the surface of the Earth.

Nuclear winter

  In the 1980s, using simple climate models, we discovered that global
  nuclear arsenals, if used on cities and industrial areas, could
  produce a nuclear winter and lead to global famine.

  Smoke from the fires would last for years in the upper atmosphere,
  blocking sunlight, and making it cold, dark and dry at the Earth’s
  surface.

Did bombing during second world war cool global temperature

But when we are taking about global warming, oil burning (which releases $\small\mathsf{CO_2}$) contributes towards global warming.

Why is smoke considered a factor in global cooling when analyzing nuclear winter scenarios, but one of warming when analyzing global warming scenarios?

The question contains a false assumption that climate science fails to consider the cooling effects of aerosol particles, including from smoke, that are the principle forcing expected for nuclear winter scenarios. Climate science does attempt to quantify every significant effect, both warming and cooling - not only the warming effect from increased greenhouse gases - $\small\mathsf{CO_2}$.

Whilst not as recent as I would like, the graphic shows the principal climate forcings that climate science has identified and includes the cooling affects of aerosols.

Aerosols from burning forests are short-lived, and the $\small\mathsf{CO_2}$ from burning vegetation should be drawn back down by regrowth, whether forest regrowth or grassland. The most significant, enduring climate consequence of Amazon forest being cleared and burned is not the aerosols or the $\small\mathsf{CO_2}$ from the fires, it is from land clearing turning a large existing Carbon Sink into a carbon source.

Turning the Amazon to savannah would release 200 billion metric tons of $\small\mathsf{CO_2}$. The cooling effect of the aerosols from burning will not persist long, but the raised $\small\mathsf{CO_2}$ levels from the loss of a large Carbon Sink will.


Why are clouds darker before it rains? What causes the darker color of the clouds (nimbus) when it is about to rain? Non-precipitating clouds (e.g., cumulus, stratus) tend to be white, but rain clouds become gray and almost black before it rains. What is different in those clouds?

Source: Wikipedia commons.

Fair weather cumulus clouds are white because there is a clear path for sunlight to reflect off the cloud and into your eye.  Even when there are many of them, they are not vertically developed and photons can still scatter off of them to you.

As cumulus develop into towering cumulus and cumulonimbus they become horizontally widespread and quite high (the tops of severe convection overshoot into the stratosphere and the convection in the tropics can go even higher, owing to a higher tropopause).  A developed CB cloud also features an anvil.  These clouds are also optically thick, dense with cloudwater, rain, snow and ice.

For a cloud to appear white, photons representing the full visible spectrum must be able to arrive at your eye.  For an optically thick cloud of appreciable thickness, transmission isn't an option.  Note that both the physical thickness and the density of the cloud will increase its optical thickness.  This means that if the cloud size doesn't change but more and bigger rain drops are in the cloud, it will be optically thicker and appear darker. If there isn't a path from the sun to a surface of the cloud you can see, scattering isn't an option.  Water emits in the infrared, so emission isn't an option.  If visible photons cannot arrive at your eye from the cloud, then the cloud will appear dark.  The presence of hydrometeors (rain drops, snow, hail) in a rain cloud will further decrease transmission and darken the cloud.

Putting it together, as that white cumulus grows and becomes tens of thousands of feet tall, it will still appear white if the sun is on your side of the cloud.  It will appear darker if the sun is behind the cloud (transmission through the cloud).  As the cloud becomes horizontally thick, transmission will decrease, darkening the cloud if you aren't in a position to observe scattering.  Once an anvil starts to form, if you are anywhere near the storm you are likely in shadow and there probably isn't a path for scattering sun->cloud->eye.

In the picture you posted it appears the sun is behind the storm.  You can see the anvil in the upper left part of the image is bright white on its edge where you can see scattering from the sun.  As you see the anvil toward the storm it darkens because now you are seeing photons transmitted through the anvil, and less are making it to your eye.  The updraft is colored similarly, with some light areas with scattering and the bulk of the surface quite dark due to no scattering and low transmission.  Under the storm you see some bright clouds via scattering from the sun.  You can see this spectrum doesn't appear quite white, a consequence of different wavelengths scattering differently (and this effect can give rise to the green aura people associate with hail, which is really just related to having tall storms causing long path lengths and multiple scattering necessary for photons to reach your eye).


Data for plotting tephigrams I want to plot some tephigrams.
Where can I find pressure, temperature and dewpoint data in text format?

Since OP wants data in text format to plot pressure, temperature and dew point data this site University of Wyoming soundings map does provide data in a text format as shown in this example for Essen Germany.


What would happen if Earth suddenly stopped rotating and revolving at the same time? What will happen if earth stops rotating as well revolving? Will we all be moving with its velocity? It should get displaced from its orbit and move in a tangential direction.

Two very different processes! First, if the Earth suddenly stopped rotating then the angular momentum would result in everyone and everything on the Earth's surface being flung sideways at a rate proportional to the distance from the Earth's spin axis. That is, no change at the poles, to 1670 km/hour at the equator. There would also be oceanic tsunami's which would sweep over all but the world's highest mountains. Or would they? The amount of energy required to suddenly halt the Earth's angular momentum is so great that its' dissipation would melt the Earths crust, and vaporize the oceans, so maybe there wouldn't be any oceans to form a tsunami, in which case we would all be caught up in a global cyclone of super-heated steam.
As for Earth ceasing to revolve around the sun, with no centripetal force to balance gravitational attraction we would simply start upon an accelerating drift towards the sun. However, If we assume that our moon stops orbiting the Earth-moon center of gravity, at the same time as Earth (presumably by the same hypothetical process as envisioned by the questioner), then long before we burn up with increasing proximity to the sun, the moon and Earth would collide, also by gravitational attraction.


In what sense did the Oklo reactor "trap" its own nuclear waste? A number of popular writings on the natural fission reactor at Oklo, Gabon (e.g. here) state that some of the energetic byproducts (krypton and xenon, presumably 85Kr and 133Xe) of the reactor were "trapped" in aluminophosphate.

How, exactly, does "aluminophosphate" (I assume this is some sort of framework mineral?) trap gaseous elements? Is it related to the mechanism by which clathrates trap gases?

According to Record of Cycling Operation of the Natural Nuclear Reactor in the Oklo/Okelobondo Area in Gabon, the Xe is trapped in crystalline cage-like structures formed of the aluminum phosphate that are similar to zeolites.

The reference also explains that the aluminum phosphate structures grow quickly under hydrothermal conditions of 270-300 degrees C.


What are the non-anthropogenic causes of The Netherlands' low elevation? A large area of The Netherlands is below sea level. There are two non-anthropogenic reasons that I can think of:

Compaction of delta sediments that lie below the country,
A side effect of the isostatic rebound after the deglaciation of Scandinavia.

Are there any other reasons? What are the relative proportions in magnitude of these factors?

Forming of coastline

During the last ice age, the North Sea was dry. When the ice melted sea levels slowly started to rise again and due to tides and currents
a barrier of dunes was formed along what approximately is today's coast line. This created an area of land that fell dry during ebb-tide and flooded during high tide (this can still be seen in the 'Waddenzee' in the North of the Netherlands where you can walk to some of the islands during low tide). The big rivers that flow through the Netherlands brought in more sand, slowly keeping larger parts of the land dry.

Isostatic rebound

During the ice age, the surface of Scandinavia was pushed down. After the ice melted it started to rise again and pushed the Western and Northern part of Netherlands down. Strangely enough the Southern and Eastern parts of the Netherlands are rising, so it seems the Netherlands is tilting.
I'm not sure how large the isostatic effect has been, but we know that the North of the Netherlands is still going down with about 2cm per century (source in Dutch).

Human influence

I know you've asked for non-anthropogenic causes, but I'm going to include this anyway because it seems human influence has had a much bigger effect on elevation than the isostatic rebound.

In the 11th century the Dutch started to actively shape the land by building dikes and later also by using wind mills to pump out water. The Flevopolder is an example of a large part of land that has been created by the Dutch in the 1950s and 60s. When groundwater levels became lower the moors settled and started oxidizing. Researchers think that the settling and oxidation of moors today is responsible for up to 15mm decline per year (source in Dutch).

Additionally in the 16th and 17th century a lot of peat was removed from the moors and used as fuel. Peat removal created new lakes, but some of those lakes were pumped dry later and were used as farmland. Also, the weight of dikes and houses on moors still cause subsidence today in areas in the West.

In the Northeastern part of the Netherlands gas extraction has also caused local elevation drops of up to 30cm (source in Dutch) in the last 40 years.

Other sources (all in Dutch):

http://www.geologievannederland.nl/landschap/landschapsvormen/strandwal (forming of shoreline)
http://nl.wikipedia.org/wiki/Veenpolder_(poldertype) (Dutch moors)


Why does the emission of aircrafts at higher altitudes have a greater effect on the climate? I am trying to understand the footprint of a flight.
First Wikipedia says (without quoting a source):

  The level and effects of CO2 emissions are currently believed to be broadly the same regardless of altitude.

But then there are numerous references to radiative forcing, which is supposedly almost doubling the effects of the emissions (at this point only co2, or also NOx and others?). These are never explained (it says "see below" once, but there follows no explanation below.

The article on radiative forcing says that

  Typically, radiative forcing is quantified at the tropopause in units of watts per square meter of the Earth's surface.

However this does not seem to explain why quantifying it "at the tropopause" makes sense, or what that means, since the measure given is measured on the surface. Does this imply that emissions below / above the tropopause are not considered contributing to radiative forcing? (I presume they are.)

Also, how come that emissions at greater heights force more? After all, at lower heights the same amount can cover a greater surface area at the same thickness / density. Maybe, the change in wavelength of incoming radiation is greater since the traveled distance is greater, and thus more is absorbed?

Lastly, does CO2 not ascend after being emitted at surface level, such that it will enter the stratosphere just as well as if it is emitted there? Or are there further factors partially mitigating this?

You can see I have too many questions, and find the Wikipedia articles very unrevealing / lacking.

Thank you!

Initial theories of the atmosphere assumed that it was "well mixed". You will see that term mentioned in many articles, and also the fairly made point that it is an assumption. Historical measurements of CO2 concentration had been pretty sporadic with the exception of Mauna Loa. It's instructive to look at the raw data from that observatory and the adjustments made to produce a smooth curve to view changes year by year. Note also the daily variations due to plants. A NASA satellite was launched in 2014 to measure CO2 high in the atmosphere so there is now evidence CO2 is not well mixed but we do not have enough data to look at long term trends as it has not given us even 3 years worth of data yet.
It seems to be claimed that about 40% of the CO2 produced from burning fossil fuels comes to show itself as CO2 concentration increases. The rest must go somewhere else (or we got the calculation wrong !). It is often assumed 30-40% goes into the oceans. Although I have seen claims that the earth has greened by about 18% in the past 40 years (increased intensive agriculture) I have not seen the increased CO2 from this source taken into account yet in calculating which proportion of the CO2 increase in the atmosphere is due to that. At a first examination this would seem to be quite important. If the 18% greening is responsible for a significant percentage of CO2 increase then clearly fossil fuel CO2 is less important. This looks like an important area that needs to be looked at. Maybe careful study of satellite data over green areas compared to deserts/ice will reveal something.


Hypothetically, would digging a big trench from the ocean and flooding desert areas be a bad solution to combatting sea level rise? My question is basically the title. Ignoring most of the feasibility of the enormous engineering and earthmoving requirements (I guess I mean to say, given an unlimited budget) in very hot areas, would this effectively solve the problem of sea level rise as a problem? If not, what am I missing?

I'm not sure if this question is within the scope of this board, but I like pushing the boundaries :)

Yes, it would be a bad solution.

Aside from the construction and maintenance cost, the impact to the populations affected, and the impact on the environment around the destination, the benefits would be minimal, as roughly calculated in the answer to Volume of Land Below Sea Level

  So to reduce the sea level by 1m the total area of these depressions
  needs to be around 5,000,000 km^2. This is around half the total land
  area of the USA. I don't have the figures to hand but given that these
  area would probably all fit into an area considerably smaller than
  Texas - I estimate that flooding land which is below sea level will
  reduce the sea level by exactly 'not very much at all' (give or take a
  millimeter - and very much less than 1m in any event).

There is also a rate of flow issue. You basically want to create rivers flowing inland at a rate to offset or reverse sea level rise, which means the flow rate of those rivers has to approach the volume of global sea rise. Global sea level is rising by an average of about 3.4mm per year, or 1,230 cubic kilometers per year according to NASA (and the rate is increasing). That amount is about the flow of the Congo River (the second largest river in the world) or roughly 19 percent of the flow of the Amazon River.

Fed by gravity, these are among the widest and deepest rivers of the world; if you're going to move that volume of water, you need to build channels to each basin with a net flow similar to that capacity, with that gradient, that won't leach saltwater into the groundwater of surrounding communities. (The engineering issues likely pale in comparison to the political challenges.)

And keep in mind that transferring water inland is only a temporary solution. Water evaporates and by spreading water over a larger area you're increasing its surface area; Death Valley has an evaporation rate of 140 inches or 3.5 meters per year. (In Southern California, the Salton Sea is drying out and causing a variety of salinity and dust issues due to lack of inflows.) So you're transferring ~60-degree seawater into a salt pan where it's going to heat up and evaporate and leave behind a toxic brine, then fall or flow back into the ocean, contributing in some small way to the thermal expansion of the ocean.

(Also, the net volume of CO2 released by building and maintaining these massive projects would also contribute to global warming in some non-negligible way.)


How much does COVID-19 affect weather forecasts? The World Meteorological Organization (WMO) is concerned about the impact of the COVID-19 pandemic on the quantity and quality of weather observations and forecasts, as well as atmospheric and climate monitoring.

Source: https://public.wmo.int/en/media/press-release/wmo-concerned-about-impact-of-covid-19-observing-system

Apparently, meteorological data sent by aircrafts play a big role in establishing weather forecasts. Due to SARS-CoV-2, fewer airplanes than usual are up in the sky, therefore less data is available.

The chart on that page shows a significant decrease in AMDAR data, however, is there any data available on how much this actually increases the prediction error of weather forecasts?

Is there a project that tracks and publishes data regarding weather prediction errors on a regular basis that would help with analyzing the correlation between incoming data and prediction error?

Welcome to Earth Science SE!

Given that the crisis is still ongoing, I doubt we know the full extent of the impact (forecasting the weather is hard enough, forecasting the impact of a pandemic is outside my expertise). But, at least one conference  was supposed t
o go to was cancelled. And I know the lack of airplane obs is impacting data assimilation. Beyond that, I am not sure. Some hypotheses perhaps include:

Is the change in the quality of life may be impacting forecaster accuracy?
What impact does the delay in communication of scientific research have on forecast accuracy?
How does this impact the future science budget?
How does the absence of airplane data degrade forecasts (could be evaluated using a OSSE experiment)? On a related note, how
How is the communication of weather forecasts by in-studio meteorologists compare to meteorologists that are in self-quarantine?

How this ordeal has affected weather forecasts may be a good cross-disciplinary study for the future.

To fully answer your question, you can look at Benjamin et al. (2006) to evaluate the relative impacts of different observational systems. Certainly there have been some changes since then, but this can give a cursory idea of what we might be looking at.


Regarding car exhaust and its possible carcinogenic effects Can some of the gaseous chemicals in many of the car exhaust by-products of the combustion engine be carcinogenic or cancer-causing?. Could car exhaust behave in a ways that are similar to 'second' hand smoke' from cigarettes that is known to cause many health problems? Can such pollution always be cleaned up or at least 'filtered out'?

Yes almost definitely. Its not just gas, but additives, lubricants/oils, coatings, coolant (some of which leaks into the combustion chamber) etc. The full list of known carcinogens is available here.

Also see the EPA report here which says ...

"EPA estimates that mobile (car, truck, and bus) sources of air toxics account for as much as half of all cancers attributed to outdoor sources of air toxics. This estimate is not based on actual cancer cases, but on models that predict the maximum number of cancers that could be expected from current levels of
exposure to mobile source emissions."