Is this a valid method for determining the angle of the moon at a given time and place?

Question

Moon On April 11th 1979, longitude 86oW 12h58pm

1 angle from greenwich at 0h
april 11th, midnight 02LI41
april 12th, midnight 15LI05

A - Get the difference between these two days
(15-2)o + (5-41)'' = 13o - 36''
= 12o24
= 12o + 24'' *1o/60''
= 12.4o

B - Get the offset at 12h58
12h58 = 12h + 58min * 1h/60min
= 12.96h
12.96h * 12.4o / 24h = 6.696o --> difference added to value at midnight
War time not in effect.

C - Add the offset to basic value of the day
Libra's spawn is between 270o and 300o
And starts at 270o
2Li41 = 270o + 2o41
= 272o41
272o + 41'' * 1o/60'' = 272.683o
+ 6.696
= 279.379o

2. adjustment for longitude
+86o --> earth spins from east to west
would be -86o if the longitude was 86o East
= 365.379o

3 adjustment of moon's position during that offset of time
365.379o * 1 lunar cycle / 27.32 terrestral cycles = 13,374048316251830161054172767204o
365.379o + 13.374o = 378.753o

4. Delay of light from moon to earth
1.2 seconds approx.
1.2s * 1h/3600s = 0.0002777....h
0.00027h * 15o/h on earth = 0.004166...o
378.753o + 0.004o = 378.757o
Nearly negligible

5 angle translated to sign
378.753o > 360o
Therefore the cycle is completed once
378.757o - 1cycle * 360o/cycle = 18.757o

6 Spawn of capricorn 0o to 30o

Hence 18.753o means 18.757o capricorn
Or
18o + 0.757'' * 1o/60'' = 18o45 capricorn

The Moon hits earth with a 18CP45 angle type
at longitude 86o, 12h58pm

Brant: lol. Of course i do that for astrology! These guys need help. And about latitude my answer goes like this, the orbit of the moon is horizontally FIXED. : P Otherwise we would not get a perfect half moon.

Ok that was only a hack. The appeareance of the moon changes horizontally, that's what i meant. It indicates that the aspect between the sun and the moon and changes only in one dimension. If the light/shadow was growing vertically and horizontally, I guess i would go after latitude too.

Third the classic view of longitude already includes a difference of physical lenght, shrinking towards the poles and growing towards the equator. Not considering that difference and using latitude instead to get it again, appears to scrap the calculus.

Of course i am not sure about what i advance but it seems to bear fruit and that is why i am seeking your advice.

I appreciate the answers and insight you gave me, thanks (in spite of any difference remaining in our respective views of the problem and its solution)

Answer 1

First, the earth rotates west to east, not east to west.

Since the right ascension and declination of the moon at any given time is easily determined from ephemerides and is not affected by latitude or longitude, one must presume that you want the altitude/azimuth.

What possible reason would there be to figure the amount of time it takes for light to get from the moon to the earth and then call it negligible? Of course it's negligible, and the time it takes doesn't change enough for it to ever be anything BUT negligible. And why would you express any number to 30 decimal places, and then round off to two? This looks like a lot of made up numerage, just for looks.

All of this calculation is useless since you have not included the latitude. This will have an effect of the altitude of the moon, and thus its angle in the sky. Now it seems that this labor is for astrological purposes. Why bother? Astrology is bunk, no matter how many numbers you use. Explain why there would be any difference for a person who is born ten minutes earlier or 500 miles to the west or north. Tell me how you know these "effects" and the differences that supposedly result from such small changes in the angle of the moon.

If you can't do it and show logical reasoning, based on statistics or direct observation, then this is all fantasy and folly.

Answer 2

Your notation is so non-standard and the problem so ill-defined that I can't give a precise answer. Also, all of this stuff about Capricorn and Libra indicates that your purpose is astrology, and you should realize that astrology has no scientific basis.

But I'll point out some of the problems here:

1) You specify the time (12:58 PM) but don't say whether this is UT or CST or some other time zone.

2) The position of the moon in the sky is often given in one of two ways -- geocentric (as seen from the center of the earth) and topocentric (as seen from a location on the surface of the earth). If you want the geocentric position, you don't need your longitude, but only the time (given in UT). If you want the topocentric position, you need both your longitude and your latitude. Since you specify the longitude but not the latitude, it's not clear what you're after.

If you want to do accurate astronomical calculations (for astronomy or whatever else you have in mind), you should be armed with the right information. An excellent source is Willman-Bell, which publishes the following two books by the famous celestial mechanic Jean Meeus:

Astronomical Algorithms
Astronomical Formulae for Calculators

web site: http://www.willbell.com/

To use these, you need a pocket calculator or else you have to create software to implement the algorithms. Willman-Bell also sells software that does calculations like this, but I'm not personally familiar with it; take a look at their web site.

To get you started, I'll give you the results of some software I have on my computer (based on Meeus' algorithms).

ephemeris for moon
1979 April 11
12:58 PM CST = 1858 UT (I assumed you meant 1258 CST.)
longitude = 86 W
latitude = 40 N (I specified an arbitrary latitude.)
height = 0 meters

true ecliptical coordinates:
lambda = +192 deg 27 min 41.9 sec
beta = +2 deg 13 min 23 sec

true equatorial coordinates (geocentric):
alpha = 12 h 49 m 19.0 s
delta = -2 deg 52 min 36 sec

topocentric coordinates (from location on earth specified above):
alpha = 12 h 50 m 31.7 s
delta = -3 deg 25 min 41 sec

horizontal coordinates (i.e., azimuth and altitude):
A = -141 deg 26 min 34 sec
h = - 46 deg 53 min 4 sec
(Meeus likes to use azimuth based on due south; this azimuth corresponds to a position 38.6 degrees east of north.)

If you have a method that agrees with the above numbers, you'll know that you're on the right track.

-- edit:
After reading your comment, I have to reassert that the topocentric position of the moon depends on latitude. The moon's orbit is neither in the plane of the ecliptic (it's about 5 degrees off) nor the plane of the equator (the inclination varies over the years between about 18 and 29 degrees). If you ignore the latitude, you're guaranteed to get an incorrect answer in *both* coordinates, not just beta or delta; whether the error is large enough to be of concern, only you can say.

Regarding a perfect half moon -- that comes about when the earth-moon-sun angle is 90 degrees. It doesn't have to do with anything being horizontally fixed.

-- final thought:
On thinking about your comments further, it seems that you are trying to calculate the position of some sort of fictitious astrological moon that orbits in the plane of the ecliptic or the equator or some such thing. If that's the case, I can't help you. My algorithms are for the real moon, not a fictitious one. The good news for you is that a horoscope based on bad calculations is just as accurate as one based on good ones.

Answer 3

trie it practically