Something that I have wondered about recently is this.

Around the equator, day length does not vary a great deal over one year. The further north or south one goes, the more extreme becomes the contrast between night and day length over a six month period, before reversing the process around midsummer's day.

Is there a formula, based on one's degree of latitude which predicts by how many minutes each day the hours of daylight will increase or decrease over six months?

I hope this post is in the right place, as geography, maths or physics all appear valid places for its location...

Well, could you speak one at a time, please, I will not understand if you all shout at once...

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10-13-06, 06:25 AM
Sherasi
Discussion of Day Rotation for all the Planets

quote:
To calculate how long it takes for the Earth to rotate through an angle of one degree, we divide the length of a day, 24 hours, or 1440 minutes, by the 360 degrees that it turns through during that rotation, obtaining a rotational speed of 4 minutes per degree. Since the Sun's motion differs from the stars' motion by one degree, and it takes 4 minutes for the Earth to turn through one degree, it takes the Sun 4 minutes longer to go around the sky than it takes for the stars to do so, and the rotation period of the Earth is 4 minutes less than the length of its day. Since we define a day as having exactly 24 hours, the rotation period is 23 hours 56 minutes.



A More Accurate Calculation
The method of calculating the day length described above is only an approximation, because it does not directly take into account the fact that the planet continues to move around the Sun while it turns through the angle that the Sun appeared to move relative to the stars during one rotation. For the Moon, a secondary correction is required, and if we wanted very precise results, additional corrections would be required. And for Mercury and Venus, the large motion during one rotation (240 degrees for Mercury, and more than a whole circle for Venus) removes any hope of using a method of calculation based on a single rotation. Instead, we have to think in terms of the number of rotations that the planet makes in one year.
In one year, the planet rotates a certain number of times, and the stars go around the sky that number of times. If the planet were stationary relative to the Sun, so that the Sun was fixed in the sky relative to the stars, it would rise and set the same number of times as the stars, but during one year, the planet moves once around the Sun, and as a result, the Sun moves once to the east among the stars. If the planet has direct rotation, as most of them do, so that the stars move westwards across the sky, the Sun's eastward motion relative to the stars is backwards, and so it goes around the sky one less time, and

The number of days in a year = the number of rotations - 1.
If the planet has a retrograde rotation, the stars will move across the sky to the east, instead of to the west, and the Sun's eastward motion will result in its crossing the sky one time MORE than the stars. However, since we treat this kind of rotation as being negative, this simply means that we will get a larger negative number, and the equation is still correct.
As you will see, for the inner planets, the number and days and rotations in a year is fairly small, and this equation provides an easy way to calculate precise day lengths. However, for planets which have a lot of days in a year, using this equation requires very precise calculations, and it is more convenient to express this formula in a different way. To do so, we define ND as the number of days in a year, NR as the number of rotations in a year, d as the length of one day, r as the rotation period, and y as the length of the year. Then, the equation becomes

ND = NR - 1.

But ND = y / d,

and NR = y / r,

so y / d = y / r - 1

Multiplying by d and dividing by y, we obtain

1 = d / r - d / y,

or, rearranging terms, d / r - 1 = d / y.

Multiplying by r, we obtain

d - r = d (r / y) = d / NR.
In other words, the difference between the day length and the rotation period is about the same as the length of the day divided by the number of rotations in a year. Since we use this method only when the number of days and rotation periods is very large, the day length and the rotation period are almost the same, and we can use the rotation period as an approximation to the day length, giving the following result:

d - r = d / NR,
and, when d » r, d - r » r / NR.

10-13-06, 09:58 AM
Jenny Roberts
Ritz, I have been trying since you posted to find something.

Some people are so impatient! Roll Eyes

10-13-06, 02:19 PM
Professor
Although it's not exactly what you're looking for, does this help?

10-22-06, 11:52 AM
Ritzmar
First, my apologies, I have been very busy for several days, and only now have time to give this thread my full attention. Thanks to the three of you who have tried to help.

Sherasi, Many thanks for your detailed explanation which is much appreciated. And thanks for trying, Jenny, I could not find a website at all which gave any help in this matter. However Professor is offering the closest thing to what I am searching for, albeit it is not quite what I am after.

I want a formula which predicts the hours and minutes of daylight depending on the degree of latitude and the time of year. For instance, if I am at the equator the hours will not alter much over 365 days. At my latitude (c. 53°) it can be light at 10.00 pm in mid June if the sky is clear, but dark by 6.00 pm in mid December. There is a decrease by so many minutes a month from June 21st until December 21st, which then reverses at the same rate, I believe. Go to Helsinki and the degree of daylight loss per month from June till December is much more dramatic. Go to the north pole and it is absolutely dark all day 0n 21st December and never goes dark on 21st June. So there is a correlation between the passage of time from the 21st June until 21st December, and a correlation of increasing loss of daylight depending on the line of latitude.

There simply must be a standard formula which combines a) line of latitude, and b) degree/speed of daylight loss/gain in the northern/southern hemisphere.

I hope that I have been clearer this time than I was at first! Even in a tiny country like ours, there is a dramatic difference between southern England and the extreme north of Scotland as regards the speed of daylight loss and gain as we go through the year. It simply must exist as a standard formula somewhere...any ideas?

Big Grin Wink

11-03-06, 03:10 PM
Georgia85
Start with page 20 of this file and see if any of the formulas mentioned are useful. There are formulas for sunrise and sunset determination and time converters. It is way over my head but maybe it will help you?

11-05-06, 11:51 AM
Ritzmar
Well! It looks right, but I am going to have to sit down for a couple of hours to study this, as from the word go the formulae are way beyond my basic grasp of physics! But thanks very much. My son, Steven will be home later, and is scientifically-oriented. I shall ask him to have a look at your site and then see if he can explain it to me in words that I may understand (i.e. dumb it down sufficiently for my feeble brain to comprehend!)

I thought that I had heard the last of this and that no-one was interested/informed enough to reply, so many thanks once more for taking the trouble. It really does seem to encapsulate entirely the question which I have been thinking about.

Cheers, Georgia! Wink ...(x)...

11-06-06, 09:25 AM
Georgia85
Certainly not informed enough to reply but always interested in helping out Smile Was you son able to explain things to you Ritzmar?

11-07-06, 02:41 AM
Ritzmar
Not had time to ask him yet! My other son & his lovely wife arrived for dinner last night with our latest grandson, so there was no time. I shall attempt to pin him down this evening...thanks again Georgia (x)
Wink

11-22-06, 08:22 AM
Adi
Yes. And there are lots of sites which give you the sunrise/sunset times automatically or allow you to enter your Lat and Long.

Take your pick.

11-23-06, 04:22 AM
Ritzmar
Sherasi, Jenny, Professor, Georgia & Adi; many thanks for all your input which is invaluable. I really do appreciate the work which you all have put in, and I am now working my way through the various systems to clarify my own reasoning and comprehension of the question. It does seem to be far more complicated than I at first thought, but all the information here will be so useful.

Once more, a sincere "thank you" to each of you. Cheers!
Ritz.

This message has been edited. Last edited by: DorianGreyed,