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. + (.).R.A. + (+).C.P. = L.S.T, and in order to get the correct (.).R.A. and (+).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. + (.).R.A. + (+).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 (.).R.A. 1 --57 --59 (+).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 (.).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 (.).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 lat.i.tude at night if we know the alt.i.tude 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.
a.s.sign 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.
FRIDAY LECTURE
THE NAUTICAL ALMANAC
For the last two days we have been discussing Time--sun time or solar time and star time or sidereal time. Now let us examine the Nautical Almanac to see how that time is registered and how we read the various kinds of time for any instant of the day or night. Before starting in, put a large cross on pages 4 and 5. For any calculations you are going to make, these pages are unnecessary and they are liable to lead to confusion.
Sun time of the mean sun at Greenwich is given for every minute of the day in the year 1919 in the pages from 6 to 30. This is indicated by the column to the left headed G.M.T. Turn to page 6 under Wednesday, Jan.
1st. You can see that the even hours are given from 0 to 24. Remember that these are expressed in astronomical time, so that if you had Jan.
2nd--10 hours A.M., you would not look in the column under Jan. 2nd but under the column for Jan. 1st, 22 hours, since 10 A.M. Jan. 2nd is 22 o"clock Jan. 1st, and no reading is used in this Almanac except a reading expressed in astronomical time. Now at the bottom of the column under Jan. 1st you see the letters H.D. That stands for "hourly difference" and represents the amount to be added or subtracted for an odd hour from the nearest even hour. In this instance it is .2. You note that even hours 2, 4, 6, etc., are given. To find an odd hour during this astronomical day, subtract .2 from the preceding even hour. For any fraction of an hour you simply take the corresponding fraction of the H.D. and subtract it from the preceding even hour. For instance, the declination for Jan. 1st--12 hours would be 23 1.8" or 23--1"--48", 13 hours would be 23 1.6" or 23--1"--36", 12-1/2 hours would be 23 1.7"
or 23--1"--42", and 13-1/2 hours would be 23 1.5" or 23--1"--30".
Now to the right of the hours you note there is given the corresponding amount of Declination and the Equation of Time. Before going further, let us review a few facts about Declination. The declination of a celestial body is its angular distance N or S of the celestial equator or equinoctial. Now get clearly in your mind how we measure the angular distance from the celestial equator of any heavenly body. It is measured by the angle one of whose sides is an imaginary line drawn to the center of the earth and the other of whose sides is an imaginary line pa.s.sing from the center of the earth into the celestial sphere through the center of the heavenly body whose declination you desire. Now as you stand on any part of the earth, you are standing at right angles to the earth itself. Hence if this imaginary line pa.s.sed through you it would intersect the celestial sphere at your zenith, i.e., the point in the celestial sphere which is directly above you. Now suppose you happen to be standing at a certain point on the earth and suppose that point was in 15 N lat.i.tude. And suppose at noon the center of the sun was directly over you, i.e., the center of the sun and your zenith were one and the same point. Then the declination of the sun at that moment would be 15 N. In other words, your angular distance from the earth"s equator (which is another way of expressing your lat.i.tude) would be precisely the same as the angular distance of the center of the sun from the celestial equator. Suppose you were standing directly on the equator and the center of the sun was directly over you, then the declination of the sun would be 0. Now if the axis of the earth were always perpendicular to the plane of the sun"s...o...b..t, then the sun would always be immediately over the equator and the sun"s declination would always be 0. But you know that the axis of the earth is inclined to the plane of the sun"s...o...b..t. As the earth, then, revolves around the sun, the amount of the declination increases and then decreases according to the location of the earth at any one time with relation to the sun. On March 21st and Sept. 23rd, 1919, the sun is directly over the equator and the declination is 0. From March 21st to June 21st the sun is coming North and the declination is increasing until on June 21st--12 hours--it reaches its highest declination. From then on the sun starts to travel South, crosses the equator on Sept. 23d and reaches its highest declination in South lat.i.tude on Dec. 22nd, when it starts to come North again. This explains easily the length of days. When the sun is in North lat.i.tude, it is nearer our zenith, i.e., higher in the heavens. It can, therefore, be seen for a longer time during the 24 hours that it takes the earth to revolve on its axis. Hence, when the sun reaches its highest declination in North lat.i.tude--June 21st--i.e., when it is farthest North from the equator and nearest our zenith (which is in 40 N lat.i.tude) it can be seen for the longest length of time. In other words, that day is the longest of the year. For the same reason, Dec.
22nd, when the sun reaches its highest declination in South lat.i.tude, i.e., when it is farthest away to the South, is the shortest day in the year for us; for on that day, the sun being farthest away from our zenith and hence lowest down toward the horizon, can be seen for the shortest length of time.
Put in your Note-Book:
North Declination is expressed +.
South Declination is expressed ?-.
Now turn to page 6 of the Nautical Almanac. You will see opposite Jan.
1st 0h, a declination of ?23 4.2". Every calculation in this Almanac is based on time at Greenwich, i.e., G.M.T. So at 0h Jan. 1st at Greenwich--that is at noon--the Sun"s declination is S 23 4.2".
You learned in the lecture the other day on solar time, that the difference between mean time and apparent time was called the equation of time. This equation of time, with the sign showing in which way it is to be applied, is given for any minute of any day in the column marked "Equation of Time." You will also notice that there is an H.D. for equations of time just as there is for each declination, and this H.D.
should be used when finding the equation of time for an odd hour.
Put in your Note-Book:
1. The equation of time is to be applied as given in the Nautical Almanac when changing Mean Time into Apparent Time.
2. When changing Apparent Time into Mean Time, reverse the sign as given in the Nautical Almanac.
That is all there is to finding sun time, either mean or apparent, for any instant of any day in the year 1919. Do not forget, however, that all this data is based upon Greenwich Mean Time. To find Local Mean Time you must apply the Longitude you are in. To find Local Apparent Time you must first secure G.A.T. from G.M.T. and then apply the Longitude.
(Note to Instructor: Make the cla.s.s work out conversions here if you have time to do so and can finish the rest of the lecture by the end of the period.)
So much for time by the sun. Now let us examine time by the stars--sidereal time. Turn to pages 2-3. There you find the Right Ascension of the Mean Sun at Greenwich Mean Noon for every day in the year. You remember that, roughly speaking, the Sun"s Right Ascension was the distance in time the sun was from the First Point of Aries. So these tables give that distance (expressed in time) for noon at Greenwich of every day. For the correction to be applied for all time after noon at Greenwich (i.e., (+).C.P.), use the table at the bottom of the page. For instance, the (.).R.A. at Greenwich 9h 24m on Jan. 1st would be
(.).R.A. 18h--40m--21s (+).C.P. 1 33 -------------- 18h--41m--54s
Now we must go back to some of the formulas we learned when discussing star time and apply them with the information we now have from the Nautical Almanac. If the G.M.T. on April 20th is 4h--16m--30s, what is the G.S.T. for the same moment? That is, when Greenwich is 4h--16m--30s from the sun, how far is Greenwich from the First Point of Aries? You remember the formula was G.S.T. = G.M.T. + (.).R.A. + (+).C.P.
G.M.T. 4h--16m--30s (.).R.A. 1 --50 -- 6 (+).C.P. 0--42 -------------- G.S.T. 6h--07m--18s
Suppose you were in Lo. 74 W. What would be the R.A.M. (L.S.T.)? You remember the formula for L.S.T. from G.S.T. was the same relatively as L.M.T. from G.M.T., i.e.,
L.S.T. = G.S.T. - W. Lo.
+ E. Lo,
Here it would be
G.S.T. 6h--07m--18s (74 W) - 4 --56 --00 --------------- L.S.T. 1h--11m--18s
Now these are not a collection of abstruse formulas that you are learning just for the sake of practice. They are used every clear night on board ship, or should be, and are just as vital to know as time by the sun.
Suppose you are at sea in Lo. 70 W and your CT is October 20th 6h--4m--30s A.M., CC 2m--30s fast. You wish to get the R.A. of your M, i.e., the L.S.T. How would you go about it? The first thing to do would be to get your G.M.T. It is CT--CC.
20d--06h--04m--30s A.M.
--12 ------------------ CT 19d--18h--04m--30s CC --02 --30 ------------------ G.M.T. 19d--18h--02m--00s
Then get your G.S.T.
Oct. 19d--18h--02m--00s (.).R.A. 13 --47 --38.5 (+).C.P 2 --57.7 -------------------- 19d--31h--52m--36.2s --24 -------------------- G.S.T. 19d-- 7h--52m--36.2s
Then get your L.S.T.
G.S.T. 7h--52m--36.2s W.Lo (--) 4 --40 -------------- L.S.T. 3h--12m--36.2s