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Insolation curves: solar energy on earth

  1. Sep 12, 2008 #1
    I've been involved for a long time going over astronomical influences on climate. My job is to be an astronomer and I don't know about climate. Maybe you think this is impossible, but I think it would be too complex to say whether a record of some sort within the ice was the driving force of all climate there is, or just a record of yet another thing coming from the sky.

    In any case the big deal in this area is calculating common Insolation curves presented in the following way:

    1. without atmospheric extinction
    Insolation=Solar constant * sin(altitude of sun above horizon)
    at one geographic spot, at one moment during the day

    2. with extinction
    Inso=So * sin(a) * k^(1/sin(a))

    In the short term (eg. 100 years) the math behind general spherical astronomy is easy. The case 1 is integrable, the case 2 is only for numeric computation. We calculate altitude of sun every day, any time.

    In the very very long term the orbit of Earth is changing shape, and axis of Earth is precessing. Hence, there is some more complex general formula for this case which will tell us the altitude of the sun depending from geographical latitude, obliquity, eccentricity, ecliptic longitude of the sun...

    Although I demonstrate all this to some point within my powers as a student and a jerk, I live in a system that hates quoting and demands evidence.

    Given this compilation of data:
    we can look into a couple of graphs:

    Notice one odd thing: In the middle of summer these two very distant geographical latitudes, have the same local maximum of Insolation - a whole 500 W/m^2.

    Would someone care to comment on this?
    you won't be quoted I promise!
  2. jcsd
  3. Sep 12, 2008 #2
    Before going into the details, the milankovitch cycles generate more problems than they solve, generating the funniest hypotheses.

    Study for instance the publications here on the page of Lorraine Lisiecki.

    One could also wonder how the faintest radiative forcings could generate such powerful reactions in the ocean. Or perhaps that powerful reactions in the ocean triggered climate changes on land.

    It may be a completely different story, which may include the movement of the sun around the barycentre of the solar system, the generated changes in gravity and length of day and the reaction of the oceans.

    How is that for starters?
  4. Sep 12, 2008 #3
    Proposal is interesting but for later. Now the game is all set and there's no back. :-)

    I've been running the matlab code they gave. For example (press F5) if I execute: (average for thousand years - ignore, latitude 15, day number 180, type of day normal calendar)

    >> daily_insolation(1,15,180,1)

    ans =


    >> daily_insolation(1,65,180,1)

    ans =


    The difference (W/m^2) is not great!
  5. Sep 12, 2008 #4
    >> daily_insolation(0.001,15,1:1:365,1)

    for table (0.001 is ignored by program)
  6. Sep 13, 2008 #5
    Perhaps that the insolation is the average daily value. But at 65N the duration of the day in summer time is much longer than at 15N. So although the max w/m2 at noon is somewhat lower, daily insolation starts earlier and last longer. hence the daily average value may be near equal.

    The real difference obviously is in winter time when 15N gets not much less than in summer time while 65N is in a deep dip by then, low sun, short day.

    Does that help?
    Last edited: Sep 13, 2008
  7. Sep 13, 2008 #6
    You're guessing good!

    I'm looking for that eye-catching orbital configuration
    eccentricity = 0.05
    obliquity = 23.3
    argument of perihelion should be that which denotes earth in perihelion = measured from spring equinox the angle should be 90 degrees unless their program requires something special... like 90+180 yaddayadda

    >> ecc=0.05; obl=23.3; omega=90;
    >> daily_insolation([ecc,obl,omega],15,1:1:365,-1)

    the extreme is about 507 W/m^2 in the mid summer at latitude 15 degrees
    and at latitude 65 it is about 545 W/m^2.

    When I put omega at 180 degrees (I don't know why) the extremes are around 490 watts at 15 deg
    and at 65 deg 498 watts => the difference is mere 10 W/m^2

    So they do come really close!

    I don't know how to plot everything that I want as a wander around MATLAB variables...

    Sometimes I lose variables during calculation:

    >> ecc=0.05; obl=23.3; omega=90;
    >> [day lat]=meshgrid(1:5:365, -90:90);
    >> [Fsw,ecc,obl,omega]=daily_insolation(0,lat,day);
    >> [c,h]=contour(day,lat,Fsw,[0:50:500]);
    >> clabel(c,h)

    >> disp([ecc,obl,omega])
    0.0172 23.4460 281.3700 (should be 0.05, 23.3, 90)
  8. Sep 13, 2008 #7
    Excuse me, earth should be in perihelion according to the questionable graph in the first post, and there should be a summer solstice in the northern hemisphere at that time - at that day

    Omega is then +90 degrees from spring equinox. The authors of the program say that equinox is always 80th day. Solstice is on 160th day then.

    >> ecc=0.05; obl=23.3; omega=90;
    >> daily_insolation([ecc,obl,omega],65,160,-1)

    ans =


    >> daily_insolation([ecc,obl,omega],15,160,-1)

    ans =


    It doesn't look clear.
  9. Sep 14, 2008 #8
    Calculated by hand for 21st of June
    insolation with extinction
    for 15 north and 65 north latitude

    1080 W/m^2
    753 W/m^2

    Altitude over horizon changes the value a lot, but the length of day was 13 hours at 15 degrees north, and 22 hours in Sweden 65 north. Hence the integrated daily average must be the key.

    I tried this orbital configuration:

    >> ecc=0.05; obl=23.3; omega=90;
    >> [day lat]=meshgrid(1:5:365, -90:90);
    >> Fsw=daily_insolation([ecc,obl,omega],lat,day,-1);
    >> [c,h]=contour(day,lat,Fsw,[0:50:500]);
    >> clabel(c,h)

    And got a contour graph with vast area from 15 north to the north pole of over 500 W/m^2 daily average insolation over one year period.

    Some orbital configuration must be responsible or driving over the edge everything into insane daily average for one day period in far northern solstice.
  10. Nov 12, 2008 #9
    Hello everybody!

    Yes, there is a difference between 442.5 and 478.3 W/m^2!
    A factor of 1.08 which, if cooling were proportional to T^4 (Kelvin), would change T by a factor of 1.02 or 6K! This is a lot!
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