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[Q] Calculating the intensity of sunlight based on latitude


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#1 smuikas

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Posted 15 January 2009 - 12:01 AM

Hi there,

Is there a readily available formula for calculating the intensity of solar radiation bombarding the atmosphere based on the earth's axial tilt, the latitude, and so forth?

For example, calculating the intensity at 65° North at the two solstices.

I'm most interested in attempting to calculate the percentage difference of the two intensities when compared to equatorial intensity.

Hypothetically (since I don't have the formula), 65° North would receive 30% of equatorial intensity at winter solstice, and 50% of equatorial intensity at summer solstice.

Any help is much appreciated!

#2 freeztar

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Posted 15 January 2009 - 12:19 AM

Check out this wiki page:

The solar constant is the amount of incoming solar electromagnetic radiation per unit area, measured on the outer surface of Earth's atmosphere in a plane perpendicular to the rays. The solar constant includes all types of solar radiation, not just the visible light. It is measured by satellite to be roughly 1366 watts per square meter (W/m²),[2] though this fluctuates by about 6.9% during a year (from 1412 W/m² in early January to 1321 W/m² in early July) due to the earth's varying distance from the Sun, and typically by much less than one part per thousand from day to day. Thus, for the whole Earth (which has a cross section of 127,400,000 km²), the power is 1.740×1017 W, plus or minus 3.5%. The solar constant does not remain constant over long periods of time (see Solar variation). The approximate average value cited,[2] 1366 W/m², is equivalent to 1.96 calories per minute per square centimeter, or 1.96 langleys (Ly) per minute.

The Earth receives a total amount of radiation determined by its cross section (π·RE²), but as it rotates this energy is distributed across the entire surface area (4·π·RE²). Hence the average incoming solar radiation (sometimes called the solar irradiance), taking into account the angle at which the rays strike and that at any one moment half the planet does not receive any solar radiation, is one-fourth the solar constant (approximately 342 W/m²). At any given moment, the amount of Solar radiation received at a location on the Earth's surface depends on the state of the atmosphere and the location's latitude.


Sunlight - Wikipedia, the free encyclopedia

I think what you are after is called Insolation. BTW, there's a great formula for such a need. :valhappy_smile:

#3 smuikas

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Posted 15 January 2009 - 12:55 AM

I've read through those several times and haven't been able to find what I'm looking for there. :/

Basically this is the kind of figure I'm trying to find out how to calculate:

(from wikipedia): For example, at latitudes of 65 degrees the change in solar energy in summer & winter can vary by more than 25% as a result of the Earth's orbital variation.

Wikipedia says this, but it doesn't say how to calculate it.

#4 Jay-qu

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Posted 15 January 2009 - 04:13 AM

I have drawn a picture - attached below of how I understand the question. In December the light comes from the right and in June the light comes from the left.

Note that these values are for incident radiation through a 1m square area held parallel to the ground.

On the June solstice at 65 degrees latitude, you will be 48.4 degrees from 'up' in the plane of the solar system. This means you will receive: L*sin(48.4) where L is the incident radiation. Since in July we are furthest from the sun and get 1321W/m^2 lets take Junes incident radiation as ~1335W/m^2. This means you receive 998.3W/m^2 MAX.
At the December Solstice we are closer to the sun but you make an angle of 1.6 degrees with the normal to the plane of the solar system. In Janurary our solar radiation peaks at 1412W/m^2 so lets guess that in december its around 1400W/m^2. This gives you 39W/m^2 MIN.

The percentages"
At June solstice 65degrees latitude receives 82% of what the equator gets.
At December solstice 65degrees latitude receives 3% of what the equator gets.

Attached Thumbnails

  • earthAxis.jpg


#5 smuikas

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Posted 15 January 2009 - 05:16 PM

Perfect!! Thank you!

Hmm, are you sure that formula is correct? for:

sin((90 - 65) + 23.4) * 1335

I'm getting a result of -1277.4526216098511 ....

#6 Jay-qu

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Posted 15 January 2009 - 06:08 PM

Perfect!! Thank you!

Hmm, are you sure that formula is correct? for:

sin((90 - 65) + 23.4) * 1335

I'm getting a result of -1277.4526216098511 ....

..your calculator is in radians..

;)

#7 smuikas

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Posted 15 January 2009 - 07:31 PM

I really should get sleep before I do maths ;)