Function for minutes of sunlight throughout the year?

Question

I'm tracking the changes in the amount of sunlight in San Francisco throughout a year....and I need to write a function relating to sine/cosine for the table/graph that i've created. But I have no idea how! Help!

Answer 1

As an approximation, try the equation :
Hours of Sunlight, S = M - A * cos[2 * pi * (t ± k) / T]

Your data should show a maximum range for sunlight over
a year - for San Francisco, about 9 hrs in January up to
about 15 hrs in June. Find the exact values, say x and X,
then calculate the mean to be M = (x + X) / 2, which will be
approximately 12. Put this value for M into the equation.

Next, A is the amplitude.
This is calculated to be : A = (X - x) / 2, or approximately 3.

The minus sign between M and A is for the northern
hemisphere, because the minimum amount of sunlight is
near the beginning of a year. For the southern hemisphere,
a plus sign would be required.

Next, if you want to calculate sunshine on a daily basis,
let T = 365 or 366 (the number of days in a year).
Alternatively, you may want to calculate average sunshine
on a monthly basis. In this case, let T = 12.

If calculating on a daily basis, then "t" represents days
from the start of the year. If on a monthly basis, "t"
represents months from the start of the year.

Next, "k" is an offset value, which you could probably
set to zero, but you may get a more accurate value from
your data. You have to find the time of the year when
sunlight is at a minimum. If this is, say, 10 days before
Jan 1st, then use (+ k), where k = 10. If it's 10 days
after Jan 1st, then use (- k).

We have everything now, so let's say for argument's sake
that the minimum is on Dec 25th. Thus, k = 7 and you
have to use a plus sign. The equation will then be :

Hours of Sunlight, S = 12 - 3 * cos[2 * pi * (t + 7) / 365]

January 1st is at t = 0 days.
April 22nd is at t = 111 days.

So, on Jan 1st, S = 12 - 3 * cos[2 * pi * (0 + 7) / 365]
= 9.0 hours

On Apr 22nd, S = 12 - 3 * cos[2 * pi * (111 + 7) / 365]
= 13.3 hours

Remember that the cos expression is in radians.

EDIT : I've just seen that you wanted minutes of sunshine,
so just multiply the value of 'S' by 60.

Answer 2

You need to know the sunlight on the longest day of the year. Subtract 12 minutes from each (this is a quirk of how sunlight is reckoned vs a "purely" mathematical definition). Say that is 14-1/2 hours. Now subtract 12 hours, and the remainder, say 2-1/2 hours is your amplitiude. You can use either a sin or cosine function, but you will need a lag time because the miniumum length of sun is offset from the start of the year. So try
2.5 sin (2pi)(t -100)/365.25 where t is in Julian Days, and the function is the deviation from a 12 hour day.
NOTE: This is an approximation since the earth has an elliptical orbit around the sun, and because the sin tends to be somewhat "flatter" as it approaches a minimum or maximum than what actually happens.

Answer 3

try this page:
http://herbert.gandraxa.com/herbert/lod.asp