Talk:Insolation

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[edit] 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)

[edit] 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·R2aT4 = 4pi·d2f
Therefore, f=(R2aT4) / d2
Then ((6.96e8 Km)2 (5.6704e-8) (5780°K)4) / (149597876600)2 = 1369.912 W/m2
This is the average. If you factor in the Earths's eccentricity, then the range is 1325.278 W/m2 to 1416.839 W/m2

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/m2, 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/m2, 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/m2, 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 [1] over to the number itself.
thanks, GabrielVelasquez (talk) 07:01, 5 January 2008 (UTC)

[edit] 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)

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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)

[edit] 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).

[edit] 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 [1] -69.87.199.249 13:04, 10 May 2007 (UTC)

[edit] 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)

[edit] 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 Wikipedia 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)

[edit] 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)