Why Do Some Months Have 30 or 31 Days While February Has 28?

[mreq]

I want to know how that point on the orbit it's calculated.
Because i set the time to june 10th 1564 (location randomly) and the sun it's in taurus constelation.
Thanks!

 

[Janus]

I want to know how that point on the orbit it's calculated.
Because i set the time to june 10th 1564 (location randomly) and the sun it's in taurus constelation.
Thanks!

The point is simply calculated like this: You start from some date, say Jan 1, 2000, where you have made a precise observation of the Earth-Sun position relative to the Stars. Then taking the length of the Mean sidereal year, you calculate how many Sidereal years have passed since June 10th 1564. You'll get some whole number and a fraction. Multiply that fraction by 360° and you have how many degrees difference between the present position of the Earth and the one you want to calculate . (to be more accurate, you factor in the fact that the Earth travels at differ speeds at different points of its orbit.) Its just a matter of starting from a known position and date and crunching the numbers.

That being said, I do not believe that WWT does this. It is more designed for showing real time positions than calculating ancient ones. The reason I say this is that it shows the Sun in the same position in the zodiac for 6/10/1564 as it does for 6/10/2010, and it shouldn't. The Sun should be at the borderline between Cancer and Taurus in 1564.

For that, you might try a program called Skyglobe. It will calculate positions going far back or forward in time. It also compensates for the switch between Julian and Gregorian calendars in 1582.

Skyglobe has a nice feature where you can jump forward or backward by centuries, allowing you to watch the Sun drift through the zodiac due to the precession of the equinoxes.(there will be a jump due to the switch of calendars in 1582.)

 

[mreq]

Skyglobe seems to be old!
Isn't there any software to be truly profesional ? What's so hard ? (i mean like there is movie editor - which do everything ...it should be a astronomy software...)

 

[Janus]

Skyglobe seems to be old!
Isn't there any software to be truly profesional ? What's so hard ? (i mean like there is movie editor - which do everything ...it should be a astronomy software...)

So what if its old, as long as it does the job? It not like the methodology for determining the relative positions has changed. The only real difference is the interface. Sure with WWT you can zoom in and see those pretty images of the object, but that is just eye candy if all you need to know is its position.

At least Skyglobe allows you to backtrack the positions over long periods of time. I don't know how many astronomical program you are going to find that will do that; It's not as if it would be a high priority.

 

[mreq]

Another thing. If i take let's say 19 february in a bisec year, then on the next year 19 february it's not on the same position on the orbit as it was last year.
Where is that on the orbit?

 

[Janus]

We already covered that back in post #30. The Sidereal year is 365 days, 6 hr, 9 min, 9.7676 sec long. Thus in a non-leap year. the Earth will be 6 hr, 9 min, 9.7676 sec short of a complete orbit the next Feb 19. This works out to just about 1/4 of a degree or half the width of the Moon.

However, during a leap year, which is 366 days long, the year is longer than it takes for the Earth to complete an orbit by some 3/4 of a degree. So what you would get is the Earth falling behind by 1/4 of a degree for each of 3 years and then making that up in the fourth year. There will still be a minor drift caused by the difference between Tropical and sidereal year (again, as noted in post 30), but this would take years to notice.

 

[Integral]

We already covered that back in post #30. The Sidereal year is 365 days, 6 hr, 9 min, 9.7676 sec long. Thus in a non-leap year. the Earth will be 6 hr, 9 min, 9.7676 sec short of a complete orbit the next Feb 19. This works out to just about 1/4 of a degree or half the width of the Moon.

However, during a leap year, which is 366 days long, the year is longer than it takes for the Earth to complete an orbit by some 3/4 of a degree. So what you would get is the Earth falling behind by 1/4 of a degree for each of 3 years and then making that up in the fourth year. There will still be a minor drift caused by the difference between Tropical and sidereal year (again, as noted in post 30), but this would take years to notice.

This last bit of drift is compensated for with the century leap years, that is a century year (1800, 1900 etc. ) is NOT a leap year even though it is obvioulsy a multiple of 4, unless it is still divisible by 4 after dividing by 100. Thus the year 2000 was a leap year making it part of a 400 year correction cycle. I somehow feel cheated because the rare event that occurred in 2000 meant that we maintained the leap year cycle that we have all become familiar with. Since the current calendar system was established in the 1750's this is the first century which was a leap year.

 

[Janus]

This last bit of drift is compensated for with the century leap years, that is a century year (1800, 1900 etc. ) is NOT a leap year even though it is obvioulsy a multiple of 4, unless it is still divisible by 4 after dividing by 100. Thus the year 2000 was a leap year making it part of a 400 year correction cycle. I somehow feel cheated because the rare event that occurred in 2000 meant that we maintained the leap year cycle that we have all become familiar with. Since the current calendar system was established in the 1750's this is the first century which was a leap year.

Actually, I was referring to the difference between Tropical and Sidereal years. This drift is what is left over after all the calendar manipulations, which are designed to keep the calendar in step with the Tropical(seasonal) year. In other words, the drift due to the precession of the equinoxes. It works out to be about 1.4° per century.

 

[mreq]

So, Janus, if I want to calculate the sideral years from the point of year 0 how to do it ?
Thanks!

 

[Janus]

Our calendar has no year zero. But if you want to figure out how many sidereal years occur between the same date in different calendar years, you would:

Multiply the number of years between the dates by 365.
Add in the number of leap days that occurred during the period.
Divide this by 365.256363051

 

[mreq]

Janus please if you know another software, because Skyglobe it's not running on my pc.
Thanks!

 

[Chronos]

In ancient times, people relied on lunar cycles. 13 lunar cycles [28 days] equals 364 days. That was accurate enough unless your civilization persisted for centuries. After a few hundred or so years, you realized this clock was just a hair off [assuming you trusted your ancestors]. The Romans had this thing about the number 13, so they 'fixed' the calendar - probably just to annoy the encroaching barbarians.