Misplaced Pages

This is the current revision of this page, as edited by Gnomingstuff (talk | contribs) at 01:30, 20 November 2024 (rv 2006 vandalism). The present address (URL) is a permanent link to this version.

This redirect does not require a rating on Misplaced Pages's content assessment scale. It is of interest to the following WikiProjects:

Astronomy This redirect is within the scope of WikiProject Astronomy, which collaborates on articles related to Astronomy on Misplaced Pages.AstronomyWikipedia:WikiProject AstronomyTemplate:WikiProject AstronomyAstronomy Weather This redirect is within the scope of WikiProject Weather, which collaborates on weather and related subjects on Misplaced Pages. To participate, help improve this article or visit the project page for details. Articles Requested Project Resources Become a Member Project Talk Assessment A-Class Alerts WeatherWikipedia:WikiProject WeatherTemplate:WikiProject WeatherWeather Energy This redirect is within the scope of WikiProject Energy, a collaborative effort to improve the coverage of Energy on Misplaced Pages. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.EnergyWikipedia:WikiProject EnergyTemplate:WikiProject Energyenergy

etymology

I thought the name came from INcoming SOLar radiATION; can someone check and add that if it is true? -Rotiro

—The preceding unsigned comment was added by 24.69.197.108 (talk) 03:33, 14 December 2006 (UTC).

No, the etymology of insolation is derived from the Latin insolare - to expose to the sun. Jonathan (talk) 00:40, 13 June 2008 (UTC)

I was told at university by my professor that it comes from the phrase Intercepted Solar Radation. —Preceding unsigned comment added by 143.239.70.118 (talk) 11:53, 20 April 2010 (UTC)

Earth's Insolation

The whole section labeled Earth's insolation is improperly called that, as the section is referring to irradiance. this was mentioned earlier and will be modified. perhaps an article can be made for solar irradiance and this section can be moved there, as the figures are correct for the irradiance. Kopasa 14:00, 20 June 2007 (UTC)

I disagree that the figures are correct, I've read various different quotation of the value.
A little accuracy wouldn't hurt:
R= 6.96e8 Km (Sun's radius)
T= 5780 °K (Sun's photosphere or Effective temperature)
a= 5.6704e-8 (Stefan-Boltzmann Constant)
d= 149597876600 meters (Earth's average distance, Mariner 10), 1 AU
f= flux or Insolation.
L= 4pi·RaT = 4pi·df
Therefore, f=(RaT) / d
Then ((6.96e8 Km) (5.6704e-8) (5780°K)) / (149597876600) = 1369.912 W/m
This is the average. If you factor in the Earths's eccentricity, then the range is 1325.278 W/m to 1416.839 W/m

GabrielVelasquez (talk) 05:25, 28 December 2007 (UTC)

I'm not sure which figures' accuracy you are questioning. It seems you have made remarkably good predictions that track the observed value within 0.3%. The measured average, found here, is 1366 W/m, with an annual range of 1321 to 1412. Considering the number of significant figures in R, this looks essentially like a perfect match. My congratulations. Hertz1888 (talk) 16:37, 28 December 2007 (UTC)

1366 Watts per meter squared. I have looked at this Solar Constant Satellite mesured value in many internet and text sources, and I have found "Satellite mesured" values of anywere from 1365 to 1372 for the so called average/constant, and I don't know why you are using this particular number instead of another and would like to see a reference for that specific figure, average, or so-called constant. And as a clarification, even with the Solar Radius value, you rounded up and I'm only off this quoted value by 0.2883%, and that's if you accept this quoted value. If I recalculate using more accurate figures, using ((695950000)^2*(0.000000056704)*(5778^4) )/(149597876600^2), then I get 1367.8204 W/m, which is only off by 0.1333% GabrielVelasquez (talk) 01:49, 1 January 2008 (UTC)

The reference is right there in the article (ref. 1), and pertains to measurements made during the last three solar cycles. An average value of 1366 units (to the nearest whole number) is clearly visible in the concluding graphs (Figures 4 and 5). I think you are preserving more decimal places in your results, and possibly attaching more importance to those extra digits, than is justified by the accuracy of the observational data. The value for R, for example, has only three significant figures (without taking into account the uncertainty of the measurement, which may degrade the accuracy further).

In the solar constant article, it says roughly 1366 W/m, and I am going to add the same word to this article. I hope this response helps. Hertz1888 (talk) 05:06, 2 January 2008 (UTC)

My-god-man, I almost needed a magnifying glass for that. And you had to actually point it out. Couldn't you place that "figure 4 and 5" below with/next-to the link for that article. And incidentally, why are you rounding up? - Just kidding.
I did shift the over to the number itself.
thanks, GabrielVelasquez (talk) 07:01, 5 January 2008 (UTC)

image

I'm questioning the image Insolation.png. It seems inconsistent with the numbers here, on Misplaced Pages's solar constant article and on Misplaced Pages's Sunlight article. The image appears to show approx. 400 W/m at the top of the atmosphere whereas others have around 1360 W/m. Whilst it could be argued that the key shows yellow to be 400+ W/m the lower part of the image shows much lower figures (within the upper bound of the key) for the irradiance at the surface; Misplaced Pages's Sunlight article suggests 1050 W/m whereas the image seems to suggest a maximum of 32 W/. That is on top of the question as to whether W/m is irradiance or insolation. If irradiation then the image is incorrectly labelled. I don't know much about this topic but it seems inconsistent to me. Selfool (talk) 14:06, 6 November 2014 (UTC)

OK, having thought about a little I guess the point I was missing is in the title given to the image on the article page, as in "Annual mean insolation at the top of Earth's atmosphere (top) and at the planet's surface". So the numbers extracted from the image would be the average irradiance/insolation across an entire year, including hours when sunlight is off perpendicular, hours of darkness, etc.. I wonder if it would be useful to include the title in the image? To get a decent view of the image I'd clicked on it which took me to a view where the original title was not visible. This is related to the point raised by agr below under dubious sentence.Selfool (talk) 14:14, 7 November 2014 (UTC)

reverted

I just reverted several months of minor edits: the article has been quite garbled since mid-February when some particularly bad editing occurred. In particular, the lead sentence has been composed of the first half of one sentence and the second half of another, rendering it devoid of meaning. zowie 15:41, 30 May 2006 (UTC)

Insolation is solar energy incident on the earth's surface. A distinction should be made to clearly distinguish between the rate of insolation and the total insolation received in a given time period. The rate of insolation is energy received per unit time per unit area; expressed in watts per square meter. The total energy received by a unit area in a given time would be expressed in, perhaps, megajoules per square meter in a day; as the Australian map shows or kilowatthours per square meter per year; as the world map shows. The distinction is between power and energy. The word "insolation" is sometimes used to mean either. When it is used the distinction should be made clear.

alexselkirk1704@hotmail.com

I disagree with "The word "insolation" is sometimes used to mean either." irradiance is the word for power over area (watts or Kilowatts per square meter) and insolation is the word for energy over area (watt-hours or Kilowatt-hours per square meter). Kgrr 10:35, 19 March 2007 (UTC)

It seems in accurate to suggest that the inoslation at 1AU is equal to that at the surface of the Earth. This would completely discount atmospheric effect.

The amount of energy striking the edge of the atmosphere (1AU) is called the Solar constant. Insolation is solar radiation striking Earth or another planet. Kgrr 10:35, 19 March 2007 (UTC)

Visible light?

How much of the 1366 watts is visible light? Infrared? UV? And how does the 1000 watts at the surface split up? And, it would be good to add a graph of power vs freq spectrum... -69.87.199.151 20:43, 7 May 2007 (UTC)

See Solar radiation Solar irradiance spectrum above atmosphere and at surface -69.87.199.249 13:04, 10 May 2007 (UTC)

Sunburn maximum?

There are some interesting suntan/sunburn practical questions, and it is not clear where to look for answers. If you lie down, clearly the greatest UV etc will be at high noon (local sun time). (Because the light intensity is maximum, and it also hits the body most directly.) But if you are standing up, walking around etc, and wearing a hat, things get very complex. The sunlight itself is more intense at noon, but the exposed skin surfaces are not pointed at the sun so directly. So, perhaps in some situations the sun exposure at high noon would actually be less than some time before or after, when the sun angles in to hit the face and body more directly?-69.87.199.249 13:12, 10 May 2007 (UTC)

I gave a little thought to sunburn after my last comment. I believe that the amount of sunlight absorbed would depend on the amount of air if has to penetrate to reach us. So sunlight would be a maximum when the sun is vertically overhead - only possible between the tropics - and would be a minimum at dawn and dusk, or strictly speaking, at night! The earth also moves closer and further from the sun during the year. By far the biggest factor would be the angle of the sun in the sky. So I think what we are missing is some sort of formula for how sunlight is absorbed at different angles. One approximation might be that if the light travels through twice as much air then twice as much is absorbed. Another approximation might be exponential decay, eg if 90% of light falling vertically made it to the surface, then 90% of 90% ie 81% might reach us when the light travels through twice as much air. Above a square inch of horizontal ground there is 14 pounds of air, ie atmospheric pressure is 14 lbs per square inch or about 101325 Pascals. How much air is there facing a vertical square inch going out towards the horizon I wonder?

And of the light reaching the earth, how much would make it through to the surface if the sun were overhead. Presumably that varies with wavelength, though 'insolation', this made up word, presumably aggregates that.

There must be university professors that do this stuff for a living. It is a shame that nobody with the right knowledge is maintaining the insolation page. (No disrepect intended to those doing the work - better to have something than nothing!)

My gut feeling is that we have most of the air above us in the first 10km and looking at the horizon we might be looking through 1000km of air - at a complete guess - ie a hundred times as much. If we assumed 100 times as much sunlight got absorbed, I bet we wouldn't even see the sun at dawn and dusk. I'd guess absorbtion needs to be modelled as exponential decay. So trigonometry on latitude and time of day to give a sun's angle to the observer; further trigonometry to calculate the amount of air the sunlight goes through; exponential decay on the sunlight intensity reaching the earth to know how much reaches the observer, and then if the observer is reclined sunbathing at an angle to the sun, further trigonometry should give the sunlight per unit area of skin.

That all seems very complicated and some figures various times of day, latitudes, seasons, etc would be a very positive contribution to these pages I think. Crysta1c1ear (talk) 17:25, 19 February 2008 (UTC)

You are ignoring the ozone layer and cloud cover. Worst case for sunburn (medically) is south of the tropics. The effect of the sun's azimuth on the insolation at the Earth's surface (ie the attenuation effect) is well described in the literature. Practically speaking for a horizontal receiving surface on a cloudless day you might as well use a sine wave starting at dawn and ending at dusk. If the receiving surface is aimed at the sun at all times then it varies as sin^0.6, practically. The concept you are looking for is called the Air Mass Index. Cheers Greg Locock (talk) 22:41, 19 February 2008 (UTC)

Spectrum and other thoughts

I came here looking to find the amount of sunlight reaching the surface. I have seen someone quoting 1000 W/m2, and wanted to refute this as being too much. Diagrams of an earth energy budget give 55% reaching the ground and this article give 1366 W/m2 reaching the atmosphere, so I should have been able to find here a figure approximating to 55% of 1366 W/m2. Unfortunately the whole page seems to be about what the suns transmits and the atmosphere receives, rather than about the quantity that makes it though the atmosphere.

Furthermore, my discussion was about magnifying glasses used to burn wood, as so direct and indirect insolation figures would be useful.

Since the Misplaced Pages Insolation article is currently about what the sun transmits, I read the following with surprise.

The radiant power is distributed across the entire electromagnetic spectrum, although most of the power is in the visible light portion of the spectrum.

Since most sunlight is not in the visible spectrum as far as I know, I took that to be incorrect. However after absorbtion of sunlight by N2 and O2 in the atmosphere, most of the power making it down to the earth's surface might be visible. Does the sentence mean more within the visible spectrum than without? Or just more inside than within another smilarly sized band of the rest of the spectrum, and in the latter case, are these frequency bands or wavelength bands?

Since the radiation reaching the earth's surface will depend on the angle of inclination of the surface, and that will depend on latitude and season, maybe a table would be useful so someone could estimate for example heating at N 50 degrees in March.

People have argued whether figures should be energy or power. Those that argue for energy have nevertheless ended up saying energy 'per day', and that is a power figure.

So somewhere we should mention the peak power, the average power of a sunny day, average power all together, etc, and how they relate to each other. My figure for a magnifying glass burning paper for example assumes a plane perpendicular to the sun and somebody working out heating would need to account for latitude. Somebody working out sunburn would be interested in atmospheric scattering, as sunburn is worse at midday even though you can be perpendiular to the sun at twilight. The previous posters remarks demonstate my point.

In my opinion this page omits a lot and answers very little.

I don't even think insolation is a real word. Isn't it just somebody's made up word from incoming solar radiation.

Crysta1c1ear (talk) 14:35, 9 February 2008 (UTC)

O2 and N2 absobtion

Can anybody help me with this? I read in places solar radiation reaching the earth is around 1350 W/m2 and that most radiation is infrared absorbed by the oxygen and nitrogen in the air. And yet we also read figures like 1000 W/m2 making it down to the ground.

It appears to me that the arithmetic doesn't add up.

What sort of figures are we talking about for high energy photons absorbed by O2 and N2 heating the thermosphere to 2500? Are the high figures that I read for incoming solar radiation really figures that include a lot of reradiation from the thermosphere etc? If there is reradiation, how do the reradiated frequencies compare with those of the original radiation? Clearly there are temperature difference, 6000K for the sun and 2500K for the thermosphere. But the quantum jumps in the atmospheric atoms might be the same for receiving energy as for emitting it. So are the retransmission frequencies the same or different. Also, is this reradiation lumped together with scattering? Logically I regard it as different.

Any comments, references, help would be welcome.

Crysta1c1ear (talk) 13:07, 22 February 2008 (UTC)

There appears to be an interesting discrepancy between: 1) the 1366:1000 ratio of sunlight hitting earth's atmosphere to sunlight reaching through the atmosphere to the earth's surface. and 2) the 51% stated at NASA's graphic: http://asd-www.larc.nasa.gov/erbe/components2.gif, or at: http://en.wikipedia.org/Earth%27s_energy_budget Briancady413 (talk) 15:42, 24 August 2012 (UTC)

dubious sentence

I removed from the intro: "Sometimes, as in the text below, a long-term average intensity of incoming solar radiation will be given in units such as watts per square meter (W/m or W·m) and called insolation, with the duration (such as daily, annual, or historical) stated or only implied." It's not clear what this sentence is intended for. There is no implied duration in W/m. A Watt is one Joule per second. If the issue is over what time period the average is taken, that should be made clearer. --agr (talk) 12:08, 19 August 2008 (UTC)

units

The article states that units for insolation are energy per unit area (W.hr/m2), but the text accompanying the graphics to the right of the article gives units of power per unit area (W/m2).

Most folks, like me, looking at this page want to know how much solar energy can be captured per unit area in some region. The units kWh/kWp•y don't mean much unless you know what kWp means. The fact that experts use it all the time does not help. Fairandbalanced (talk) 20:22, 12 October 2008 (UTC)

The article says kWp means kilowatt peak rating. But I agree more explanation is needed. Presumably kWp is based on some assumed peak value for solar insolation Anyone know what that is?--agr (talk) 19:54, 13 October 2008 (UTC)

Energy is power over time, so the average power per unit area over the course of a multiple year time is still power per unit area, and works out to energy per unit area over any specific period of time, such as an hour. KWp is measured under standard conditions, at such and such temperature, and such and such insolation (1000 W/m2). I think one of the other solar articles explains that. Perhaps a link could be made to it. I see that the photovoltaics article has a link to kWp. Found it. Will make a link. 199.125.109.80 (talk) 06:04, 3 November 2008 (UTC)

The units in typical use are so mixed up, and as far as actual application to Solar Energy development, are near useless. For Solar PV, I'm thinking that the standard should be kW/m2, or W/m2, as is used in the accompanying nrel US Solar Map. If your data reflects the average over the year, as presumedly this map does, then you can ignore the "/day". Then if you divide the actual insolation by the 1000W/m2 Peak Insolation, then you'll get a percentage, which when multiplied by the actual peak Power rating of the installation, will give you the actual average Energy produced by the Installation over a year. —Preceding unsigned comment added by 24.17.155.1 (talk) 21:40, 11 January 2009 (UTC)

Equations

The equation for an ellipse is

${\displaystyle r={\frac {a(1-\varepsilon ^{2})}{1-\varepsilon \cos(\theta -\phi )}}.}$

but the article uses

${\displaystyle R_{E}={\frac {R_{o}}{1+e\cos(\theta -\varpi )}}}$

omitting

${\displaystyle (1-e^{2})}$

The following line is completely wrong

Nevertheless, θ = 0° is exactly the time of the vernal equinox, θ = 90° is exactly the time of the summer solstice, θ = 180° is exactly the time of the autumnal equinox and θ = 270° is exactly the time of the winter solstice.

By definition, the vernal equinox defines the zero angle in such a way that, for the given equation, the Earth is located at 180° at the vernal equinox. Not 0°. I know this is counter intuitive, but that really is the way it is done. Q Science (talk) 06:52, 17 December 2010 (UTC)

Average is 250 W/m^2 ?

where is the proof? I've been trying to find a global average of irradiance that hits the earth and the closest I could find was this: http://home.iprimus.com.au/nielsens/solrad.html . Which states (through calculation) that the average irradiance is 198 W/m^2

Can whoever put that up (without citation, I might add) please provide calculation of how they came up with that number? Thanks.

Ktonlai (talk) 01:44, 17 August 2011 (UTC)

Barometer.

A barometer is something that earth scientists use to measure air pressure. — Preceding unsigned comment added by 174.130.0.72 (talk) 01:08, 21 September 2011 (UTC)

OOPS

"As the angle increases between the direction at a right angle to the surface and the direction of the rays of sunlight, the insolation is reduced in proportion to the cosine of the angle; " True. But the angle shown (illustrated) is between the SURFACE and direction of the sun's rays... where the dependency is on sine, not cosine..

Suggest showing the correct angle for illustration. — Preceding unsigned comment added by 97.101.162.114 (talk) 00:42, 2 February 2012 (UTC)

ensolation

Is insolation distinct from ensolation, meaning the solar energy received or absorbed?
The well used 'Ensolation' has the advantage of being less easily confused with the almost antonym 'insulation'. (Almost as bad as we did with flammable & inflammable. It took decades to get most professionals to use non-flammable.)
Wikidity (talk) 21:45, 16 December 2013 (UTC)

Variation over the year

It is requested that a meteorological diagram or diagrams be included in this article to improve its quality. Specific illustrations, plots or diagrams can be requested at the Graphic Lab. For more information, refer to discussion on this page and/or the listing at Misplaced Pages:Requested images.

It would be interesting to show the theoretical (or actual) curves of insolation over the days of the year, like . -- Beland (talk) 16:45, 2 October 2014 (UTC)

Insolation v solar irradiance

Just finished a ce. Are these the same? The text seems to say so. If so, then merge the two articles? Lfstevens (talk) 22:07, 2 August 2015 (UTC)

Insolation

The only source of energy for the earth's atmosphere comes from the sun which has a surface temperature of more than 10,800°F. This energy travels through space for a distance of 93 million miles and reaches us as solar or radiant energy in the process called insulation. This radiation from the sun is made up of three parts of rays ▪︎White light(visible rays) ▪︎Ultra violet ▪︎Infra-red rays 139.5.255.140 (talk) 14:45, 23 April 2022 (UTC)

• Redirect-Class Astronomy pages
• NA-importance Astronomy pages
• Redirect-Class Astronomy articles of NA-importance
• Redirect-Class Weather pages
• NA-importance Weather pages
• WikiProject Weather articles
• NA-Class energy pages
• NA-importance energy pages
• Misplaced Pages requested meteorological diagrams