# LST formula

1. May 6, 2013

### mishima

Hi, I've been messing with the found-on-web formula

LST = 100.46 + 0.985647 * days past J2000 + degrees longitude + degrees UT

to calculate LST using the other 3 variables. I understand where all the terms come from except for the 100.46. What is that?

2. May 7, 2013

### Mordred

Its really hard to tell what formula your relating to. Are you talking about the local standard time formula?

3. May 7, 2013

### mishima

Hi, its the local sidereal time. I got the formula from the website below, which says it is a good approximation for a range of 100 years since J2000.0. I've also looked through the book it references, but am having trouble finding the exact source of the "100.46" figure.

http://www.stargazing.net/kepler/altaz.html

You can find the formula under the heading "local sidereal time". Thanks for taking a look.

4. May 7, 2013

5. May 7, 2013

### mishima

I think it is a general formula because I have compared with some online calculators.

I'm not sure what you mean that LST requires RA and DEC. I was under the impression LST was how many degrees the local meridian has rotated away from the first point of Aries on celestial sphere. It has a definition independent of the equatorial coordinate system, it seems. Its true that the first point of Aries is also 0 RA. And then there is the relation between LST, hour angle, and RA (also on that website) that

hour angle = lst - right ascension

But I don't think RA and DEC are built into the 100.46, thanks anyways though. :)

6. May 7, 2013

### Mordred

Last edited: May 7, 2013
7. May 7, 2013

### Mordred

I'm wondering if the 100.46 is the acceleration correction.

http://www.trans4mind.com/personal_development/astrology/Calculations/calcLocalSidTimeMidnight.htm

Because sidereal time is faster than regular time, we need to correct this figure by allowing 10 seconds for every hour of the local mean time. A sidereal day is 4 minutes short of a regular day, so we need to accelerate regular time to accord with sidereal time.

10 hours 15 minutes 40 seconds

The acceleration is 100 +2.5 seconds (or 103 seconds). We add this to our sidereal time so far:

1 hour 13 minutes 42 seconds + 1 minute 43 seconds (103 seconds) =

1 hour 15 minutes 25 seconds

8. May 7, 2013

### mishima

Well as I understand the 0.985647 figure comes from 360 degrees in 365.2421 days (tropical year, ie solar year). It converts an amount of Julian days in this epoch (J2000) into an amount of degrees.

It comes from the geometry of the definition of sidereal time. Take 0.985647 degrees per day and divide by 15 degrees per hour you get 0.0657098 hr per day. When this amount of hours is converted to minutes/seconds, you get 3 minutes and 56 seconds which is exactly the difference between the solar and sidereal days.

Like the days being converted to degrees, longitude is also expressed in degrees. Finally, the 15 times UT converts UT into degrees (there are 15 degrees per hour). So I'm confident the 100.46 is also a degree measure.

9. May 7, 2013

### Mordred

fair enough I searched all over the first link for a corresponding calculation for the 100.4. Finally gave up on the article itself so looked at other articles. That was when I noticed the correlation in the last post.
Sidereal isn't something I'm too familiar with so trying to help you out was informative for myself as well.

I looked over numerous articles the one I posted is the only one I found a similar value under the acceleration section.

10. May 7, 2013

### mishima

Heh, well I appreciate the help, its good to put some thoughts into writing when dealing with something. It somehow offsets the calculation by a certain fixed degree. Maybe it has to do with the conditions during J2000 at Greenwich...will check later with software.

11. May 7, 2013

### Mordred

Glad to help. I know there are several others that may be able to help. Now that its clearer what your after.
Some of them are not as active on PF as I though lol.
I cannot shake the feeling that the first term is a correction factor. The second term is 100% as you described it. Almost every article describes it.

12. May 7, 2013

### mishima

How about this reasoning: on J2000 the second term would vanish because "days since J2000" = 0. Similarly if longitude were 0 degrees (like at Greenwich), the 3rd term would vanish. Finally, if UT was 0 the last term would also vanish. Thus the 100.46 must be the LST of Greenwich on J2000 at 0:00 UT (all variables in equation = 0). In other words, it is the amount of degrees away from the meridian that the first point of Aries is at that date/time.

13. May 7, 2013

### mishima

Ah yep, that must be it since this online LST calculator is showing 6.6 hrs as the LST for Greenwich at 0 UT on J2000, which is equal to about 100.46 degrees.

14. May 7, 2013

### Mordred

That does follow, I think you nailed it.