Thursday Lecture
Sidereal Time—Right Ascension
Our last lecture was devoted to a discussion of sun time. Today we are going to talk about star time, or, using the more common words, sidereal time.
Now, just one word of review. You remember that we have learned that astronomical time is reckoned from noon of one day to noon of the next and hence the astronomical day corresponds to the 24 hours of a ship's run. The hours are counted from 0 to 24, so that 10 o'clock in the morning of October 25th is astronomically October 24th, 22 hours or 22 o'clock of October 24th.
Now Right Ascension is different from both astronomical and civil time. Right Ascension is practically celestial longitude. For instance, the position of a place on the earth is fixed by its latitude and longitude; the position of a heavenly body is fixed by its declination and right ascension. But Right Ascension is not measured in degrees and minutes nor is it measured East and West. It is reckoned in hours and minutes all the way around the sky, eastward from a certain point, through the approximate 24 hours. The point from which this celestial longitude begins is not at Greenwich, but the point where the celestial equator intersects the ecliptic in the spring of the year, i.e., the point where the sun, coming North in the Spring, crosses the celestial equator. This point is called the First Point of Aries. You will frequently hear me speak of a star having, for instance, a Right Ascension of 5h 16m 32s. I mean by that, that starting at the celestial meridian, i.e., the meridian passing through the First Point of Aries, it will take a spot on the earth 5h 16m 32s to travel until it reaches the meridian of the star in question.
Roughly speaking then, just as Greenwich Apparent Time means the distance East or West the Greenwich meridian is from the sun and Local Apparent Time means the distance East or West your ship is from the sun, so R.A.M.G. means the distance in time the Meridian of Greenwich is from the First Point of Aries, measured eastward in a circle. And this distance is the same as Greenwich Sidereal Time, i.e., Sidereal Time at Greenwich or the distance in time the meridian of Greenwich is from the First Point of Aries.
Now, what is the star time that corresponds to local time? It is called the Right Ascension of the Meridian, which means the R. A. of the meridian which intersects your zenith. Just as L.A.T. is the distance in time your meridian is from the sun, so Local Sidereal Time is the R. A. of your meridian, i.e., the distance in time your meridian is from the First Point of Aries. Put in your Note-Book:
G.S.T. and R.A.M.G. are one and the same thing.
L.S.T. and R.A.M. are one and the same thing.
G.M.T. + Circle enclosing dot.R.A. + Circle enclosing a Cross.C.P. = G.S.T. (R.A.M.G.) If the result is more than 24 hours, subtract 24 hours.
G.S.T. - Circle enclosing dot.R.A. - Circle enclosing a Cross.C.P. = G.M.T.
G.S.T. - W.Lo. = L.S.T.
+ E.Lo.
L.S.T. + W.Lo. = G.S.T.
- E.Lo.
I can explain all these formulas very easily by the following illustration which put in your Note-Book: (Note to Instructor: If possible have copies of this illustration mimeographed and distributed to each student.)
First Point of Aries
There is one term I have used which does not appear in the illustration. It is the Earth's Central Progress (Circle enclosing a Cross.C.P.). The astronomical day based on the sun, is 24 hours long, as said before. The sidereal day, however, is only 23h 56m 04s long. This is due to the fact that whereas the earth is moving in its ecliptic track around the sun while revolving on its own axis, the First Point of Aries is a fixed point and hence never moves. The correction, then, for the difference in the length of time between a sidereal day and a mean solar day is called the Earth's Central Progress and, of course, has to be figured for all amounts of time after mean noon at Greenwich, since the Sun's Right Ascension tables in the Nautical Almanac are based on time at mean noon at Greenwich.
Now you have a formula for practically all kinds of conversion except for converting L.M.T. into L.S.T. You could do it by the formula
L.M.T. + W.Lo. = G.M.T. + Circle enclosing dot.R.A. + Circle enclosing a Cross.C.P. = G.S.T. - W.Lo. = L.S.T.
- E.Lo. + E.Lo.
But that involves too many operations.
A shorter way, though not so simple perhaps, is as follows: L.M.T. + Reduction page 2 N.A. for time after local mean noon + Circle enclosing dot.R.A. of Greenwich mean noon ± Reduction page 2 N.A. for Lo. in T. (W+, E-) = L.S.T.
Note to Instructor:
Explain this formula by turning to page 107 N.A. and work it out by the formula L.M.T. + Lo. in T (W) = G.M.T. + Circle enclosing dot.R.A. + Circle enclosing a Cross.C.P. = G.S.T. - Lo. in T (W) = L.S.T. Example:
| L.M.T. | 10h 40m 30s |
| Lo. in T | 4 56 W + |
| —————— | |
| G.M.T. | 15 36 30 |
| Circle enclosing dot.R.A. | 5 11 10 |
| Circle enclosing a Cross.C.P. | 2 34 |
| —————— | |
| G.S.T. | 20 50 14 |
| Lo. W | - 4 56 |
| —————— | |
| L.S.T. | 15h 54m 14s |
Now Bowditch gets this L.S.T. in still another way. Turn to page 110, Article 290. There the formula used is L.M.T. + Circle enclosing dot.R.A. + Circle enclosing a Cross.C.P. = L.S.T, and in order to get the correct Circle enclosing dot.R.A. and Circle enclosing a Cross.C.P. the G.M.T. has to be secured by the formula L.M.T. + W.Lo. = G.M.T.
- E.Lo.
Let us work this same example in Bowditch by the other two methods. First by the formula
L.M.T. + W.Lo. = G.M.T. + Circle enclosing dot.R.A. + Circle enclosing a Cross.C.P.= G.S.T. - W.Lo. = L.S.T.
- E.Lo. + E.Lo.
L.M.T. 22d 2h 00m 00s
+ W. Lo. 5 25
————————
G.M.T. 22d 7h 25m 00s
Circle enclosing dot.R.A. 1 57 59
Circle enclosing a Cross.C.P. 1 13
————————
G.S.T. 22d 9h 24m 12s
- W. Lo. 5 25
————————
L.S.T. 22d 3h 59m 12s
The small difference between this answer and that of Bowditch's is that the Circle enclosing dot.R.A. for 1916 is slightly different from that of 1919. Bowditch used the 1916 Almanac, whereas we are working from the 1919 Almanac. Now turn to page 107 of the N.A. and let us work the same example in Bowditch by the method used here:
| L.M.T. | 2h - 00m - 00s |
| Red. for 2h | 0 - 20 |
| Circle enclosing dot.R.A. 0h | 1 - 57 - 59 |
| Red. Lo. 5h - 25m | 0 - 53 |
| ——————— | |
| L.S.T. | 3h - 59m - 12s |
The reason I am going so much into detail in explaining methods of finding L.S.T. is because, by a very simple calculation which will be explained later, we can get our latitude at night if we know the altitude of Polaris (The North Star) and if we know the L.S.T. at the time of observation. Some of you may think that the N.A. way is the simplest. It is given in the N.A., and in an examination it would be permissible for you to use the N.A. as a guide because, in an examination, I propose to let you have at hand the same books you would have in the chart house of a ship. On the other hand, the method given in the N.A. is not as clear to my mind as the method which starts with L.M.T., then finds with the Longitude the G.M.T. That gives you, roughly speaking, the distance in time Greenwich is from the sun. Add to that the sun's R.A. or the distance in time the sun is from the First Point of Aries at Greenwich Mean Noon. Add to that the correction for the time past noon. The result is G.S.T. Now all you have to do is to apply the longitude correctly to find the L.S.T., just as when you have G.M.T. and apply the longitude correctly you get L.M.T. That is a method which does not seem easy to forget, for it depends more upon simple reasoning where the others, for a beginner, depend more upon memory. However, any of the three methods is correct and can be used by you. Perhaps the best way is to work a problem by two of the three that seem easiest. In this way you can check your figures. When I give you a problem that involves finding the L.S.T. I do not care how you get the L.S.T. providing it is correct when you get it.
Assign for Night Reading in Bowditch the following Arts.: 282-283-284-285. Also the following questions:
1. Given the G.M.T. and the longitude in T which is W, what is the formula for L.S.T.?
2. Given the L.A.T. and longitude in T which is E, what is the formula for G.S.T.?
3. Given the L.S.T. and longitude in T which is W. Required G.M.T.
Etc.