Expert: Jayendra Upadhye - 7/1/2005

Question
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Followup To
Question -
On the summer solstice day(June21/22), it is generally stated that the solar declination is at its maximum 23.5 deg.  On this day the sun shines directly overhead at noon at the tropic of cancer.  But the tropic of cancer is is a big parallel strectching over 360 deg of longitude. At which longitude is the sun directly overhead at noon a the tropic of cancer on solstice day?.  Theorectically, there is only one point on the tropic of cancer in given year at which the sun shines directly overhead at noon of solstice day.  Is this correct?  If it is correct, then it is wrong to say the sun shines overhead at tropic of cancer at noon on solstice day, which may imply this is so at all the logitudes of the tropic of cancer.  In most books it says so.  Are they really wrong to say so?  Please clarify.
Answer -
Hi,
Imagine yourself positioned over the plane of the solar syatem, with either solstices in the updown axis and the equinoxes each to your right hand and left hand respectively.

As the earth revolves almost a degree/day round the sun in its orbit, (365.24 days/360 deg = 1.0145 deg/day), you will notice that the sun's declination (-23.5 deg to +23.5 deg) to a man on the equator, changes at a different rate every day.

This is a sinusoidal rate. Its is the fastest at the equinoxes, and slowest at the solstices.

Meaning, at the peaks and troughs of the sine-wave spread +23.5 deg and -23.5 from the "zero line" of the equinoxes, [that is your x-axis time line, displacement from the equinox position being the y axis], represent areas where the sun slows down its rate of change in declination. At the solstice, it completely stops its declination change, and then actually reverses it after a "small" apparent pause", to go back in the other direction.

This happens with all sine wave motions.

Now the point is, Though you are mathematically right, that only one particular meridian actually experiences "true maximum" and "true minimum", as they are but instances of time, and endure for a fleeting moment, the sun physically ends up spending many days "on the block" to say..by being "almost on the tropic"!

To puny humans, "oh but the sun is 1/2 a degree down already" said at high noon, makes no difference, when the temperature is soaring close to 50 degrees centigrade already! [trust me that is on the record in India and Pakistan].

My calculations show me that if you accept 1/2 degree band as "on the tropic", the sun stays "on the block" for a full 24 days!
It spends within 1/2 degree of "true maximum" for a full 24 days!
That should convince MOST that it is "on the tropic" at the said time all round the globe at places at 23.5 latitude N or S as the case may be.

The calculation is as under.
Amplitude of the sinusoid = (+/-) 23.5 degrees (declination angle).
band of interest = 0.5 degree at either peak (+ve or -ve).
This amplitude band lies above [1 - (0.5/23.5)] 0.9787 units on the y axis that represents the declination.

The normalised unit circle represents the sun's relative circular path around the earth, as the earth itself "physically" revolves around the sun.

In the upper "hemisphere" where the sun moves from one equinox to the morthern "peak" and down to another equinox, a time of half a year passes.
The angle = sin-1(0.9787) = 78.16 degrees of travel round the orbit, is spent below this band.
On the other quater of the upper hemispgere too, the sun spends 78.16 degrees below our band of interest (0.5 deg near the peak).
This leaves only 180 - (2 * 78.16) = 23.68 degrees of orbital travel within the band.

This period corresponds to 365.25 days/360 deg * 23.68 deg= 24.0 days!

If you reduce the band further, this period will reduce too
for a 0.1 deg band, by above logic, the sun will reside for a cool 16.5 days! (8.25 up and 8.25 down from the nadir!)

It is a killing business...this heat in the summer..imagine the sun directly over head at noon each day, and not budging for a cool 16 days! We of the troppics should know.
Jayen

Thank you very kindly for your calculation of the number of days in the neighborhood of the solstice for a 0.5 deg band and  0.1 deg band.  I winder if you could give me a formula for calculating the solar declination(delta).  In some books there is geiven an empirical formula somehting like this"

         delta = 23.45 sin(350(284+n)/365) deg,

where n is number of days from Jan. 1st.  I don't care for empirical formulas.  I would like to have a rigorous mathematical formula.  I think a high powered mathematician versed in solid geometry can do it.  Do you have a formula for calculating solar declination (delta).  I also saw in the literature a figure plotting the annual delta versus time.  that figure does show a curvelike sine curve.  Is it theoretically or mathematically a sine curve?   How is it obtained?

Best wishes and high regards.

James P. Hsu  

Answer
Hi James..
I am NOT a real astronomer, so my "formula" may be correct or it might be wrong..(but i doubt that).
Actually to me one need not be a rocket scientist to calculate what i have done. If there are errors, most likely these are typos only.

The underlying concept has to be same whether i do it or a real astronomer does it.
[That can always be independently verified by a REAL table of delta against time, clearly demarcating the solstices and the equinoxes].

Coming to your question, lets look at your given formula..
delta = 23.45 sin(350(284+n)/365) deg..
I presume 23.45 is used because actual maximum declination may be closer to that than the oft used 23.5! (Can be checked easily out on the web).
Let us simply keep a variable "DecMax" as Maximum Declination, in the formula.
then:- Declination Delta = Decmax*sin(theta) where theta represents the angle travelled in deg over days numbered from the equinox.(whichever doesent matter, as in sines, we get same values as long as we start from an equinoctical value)
An equinox is always (almost if we ignore seconds) 1 quarter year or 365.25/4=91.3125 days away from the solstice.
Thus the solstice before 21st june is 22nd march (91 days before 21st june)
January 1st represents 171 days before the solstice in june.

eqn 1 :  Declination Delta = Decmax*sin(theta)
eqn 2 : Theta = [n days * (360 deg/365.25 days)] ..deg units

If you start your "day" clock exactly at the solstice on 22nd march and ignore "which meridian is facing high noon" and continue counting, then exactly 91.3125 days later..again irrespective of which meridian is having high noon..you will be at the solstice.
And to hell with the month! This is because days counted using the month bucket will cost you dear due to the stupid leap year!
However if you ignore the months and deal in days and day fractions only, then you are immunised to that problem. One can arrive at the month and "day of month by back calculation based on julian date.

Meridans are to be ignored because, if at the start meridian x was having highnoon, then at the solstice, it would have advanced 1/4th of a quarter day beyond high noon! Thanks to the quarter day the earth advances on itself during the whole year! At the end of the year, you would find the starting meridian ends up one full quarter of day beyond high noon, as the earth has completed 365 "full days' + a quarter turn extra!

So forget the thumb rule, stick to this absolute formula..
eqn 3: Delta = DecMax * Sin[n days * (360 deg/365.25 days)]
check  n=0     ; Delta = 0 (starting equinox)
check: n=91.3125; Delta = +DecMax
check: n=182.625; Delta = 0 (other equinox)
check: n=273.94;  Delta = -DecMax (other solstice)

I suspect the 350 in your formula is actually 360.
I have not spent time on finding what 284 represents, though it must represent the number of days to true equinox, from jan 1.
(80 days) and 360 - 80 = 280, which is closer to 284 that you mention.
The problem of calculating days betn 1st jan and the march equinox is complicated by the leap year! So just rely on the absolute days approach.
In that case starting from jan 1stm the formula would become
eqn 4: Delta = DecMax * Sin[(n - 80) days * (360 deg/365.25 days)] or  
5: Delta = DecMax * Sin[(n + 280) days * (360 deg/365.25 days)]

The factor we use is 3% better than the one used in the formula!
we use 360/365.25 they use 350/365!
Hope this suffices!
Pls rate the answer if you find it helpful.
Also can we close this topic and move onto other things? if you dont mind that is? ..:)

Regarding your last questions, It is a straightforward derivation from the law of simple harmonic motions. The sinewave is a natural consequence of such motion.
It is obtained by the assumptions inherent in all SHM calculations.
This ofcourse has strong direct observational and theoretical basis.

Jayen