Chapter III: Part 3

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. + (.).R.A. + (+).C.P. = G.S.T. (R.A.M.G.) If the result is
more than 24 hours, subtract 24 hours.
G.S.T. - (.).R.A. - (+).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.)

There is one term I have used which does not appear in the illustration. It is the Earth's Central Progress ((+).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. + (.).R.A. + (+).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 + (.).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. + (.).R.A. + (+).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
(.).R.A. 5 --11 --10
(+).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. + (.).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 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.

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 deg. 1.8' or 23 deg.--1'--48", 13 hours would be 23 deg. 1.6' or 23 deg.--1'--36", 12-1/2 hours would be 23 deg. 1.7' or 23 deg.--1'--42", and 13-1/2 hours would be 23 deg. 1.5' or 23 deg.--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 passing 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 passed 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 deg. N latitude. 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 deg. N. In other words, your angular distance from the earth's equator (which is another way of expressing your latitude) 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 deg.. Now if the axis of the earth were always perpendicular to the plane of the sun's orbit, then the sun would always be immediately over the equator and the sun's declination would always be 0 deg.. But you know that the axis of the earth is inclined to the plane of the sun's orbit. 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 deg.. 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 latitude on Dec. 22nd, when it starts to come North again. This explains easily the length of days. When the sun is in North latitude, 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 latitude--June 21st--i.e., when it is farthest North from the equator and nearest our zenith (which is in 40 deg. N latitude) 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 latitude, 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 deg. 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 deg. 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 class 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 deg. 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 deg. 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 deg. 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

The last fact to know at this time about the Almanac is found on pages 94-95. Here is given a list of the brighter stars with their positions respectively in the heavens, i.e., their celestial longitude or R.A. on page 94 and their celestial latitude or declination on page 95. These stars have very little apparent motion. They are practically fixed. Hence, their position in the heavens is almost the same from January to December though, of course, their position with relation to you is constantly changing, since you on the earth are constantly moving.

The relationship between these various kinds of time is clearly expressed by the following diagram, which put in your Note Book:

Assign for reading in Bowditch, Articles 294-295-296-297-299-300-301-302-303-304-305-306-307.

If any time is left, have the class work out such examples as these:

1. G.M.T. June 20th, 1919, 5h--14m--39s. In Lo. 68 deg. 49' W. Required L.S.T., G.S.T., L.M.T., L.A.T.

2. L.M.T. Oct. 15th, 1919, 6h--30m--20s A.M. In Lo. 49 deg. 35' 16" E. Required L.S.T.

3. L.M.T. May 14th, 1919, 10h--15m.--20s A.M. Lo. 56 deg. 21' 39" W. Required L.A.T.

4. W.T. April 20th, 1919, 11h--30m--14s C-W 2h--14m--59s CC 4m--30s slow. In Lo. 89 deg. 48' 30" W. Required G.M.T., G.A.T., L.M.T., L.A.T., G.S.T., L.S.T.

5. What is Declination and R.A. on May 15th, 1919, of Polaris, Arcturus, Capella, Regulus, Altair, Deneb, Vega, Aldebaran?

6. What is the sun's declination and R.A., Time at Greenwich, July 30th:

7h--14m--39s A.M.
4h--29m--14s A.M.
3h--04m--06s
11h--49m--59s
2h--14m--30s A.M.?

SATURDAY LECTURE

CORRECTION OF OBSERVED ALTITUDES

The true altitude of a heavenly body is the angular distance of its center as measured from the center of the earth. The observed altitude of a heavenly body as seen at sea by the sextant may be converted to the true altitude by the application of the following four corrections: Dip, Refraction, Parallax and Semi-diameter.

Dip of the horizon means an increase in the altitude caused by the elevation of the eye above the level of the sea. The following diagram illustrates this clearly:

If the eye is on the level of the sea at A, it is in the plane of the horizon CD, and the angles EAC and EAD are right angles or 90 deg. each. If the eye is elevated above A, say to B, it is plain that the angles EBC and EBD are greater than right angles, or in other words, that the observer sees more than a semi-circle of sky. Hence all measurements made by the sextant are too large. In other words, the elevation of the eye makes the angle too great and therefore the correction for dip is always subtracted.

Refraction is a curving of the rays of light caused by their entering the earth's atmosphere, which is a denser medium than the very light ether of the outer sky. The effect of refraction is seen when an oar is thrust into the water and looks as if it were bent. Refraction always causes a celestial object to appear higher than it really is. This refraction is greatest at the horizon and diminishes toward the zenith, where it disappears. Table 20A in Bowditch gives the correction for mean refraction. It is always subtracted from the altitude. In the higher altitudes, select the correction for the nearest degree.

You should avoid taking low altitudes (15 deg. or less) when the atmosphere is not perfectly clear. Haziness increases refraction.

Parallax is simply the difference in angular altitude of a heavenly body as measured from the center of the earth and as measured from the corresponding point on the surface of the earth. Parallax is greatest when the body is in the horizon, and disappears when it is at the zenith.

When the angular altitude of the sun in this diagram is 0, the parallax ABC is greatest. When the altitude is highest there is no parallax. The sun is so far away that its parallax never exceeds 9". The stars have practically none at all from the earth's surface. Parallax is always to be added in the case of the sun.

The semi-diameter of a heavenly body is half the angle subtended by the diameter of the visible disk at the eye of the observer. For the same body, the SD varies with the distance. Thus, the difference of the sun's SD at different times of the year is due to the change of the earth's distance from the sun.

The SD is to be added to the observed altitude in case the lower limb is brought in contact with the horizon, and subtracted if the upper limb is used. Probably most of the sights you take will be of the sun's lower limb, i.e., when the lower limb is brought in contact with the horizon, so all you need to remember is that in that event the SD is additive.

Now at first we will correct altitudes by applying each correction separately, but as soon as you get the idea, there is a short way to apply all four corrections at once. This is done in Table 46. However, disregard that for the moment. Put this in your Note-Book:

Dip is -. Table 14 Bowditch
Refraction is -. " 20 A Bowditch
Parallax is +. " 16 Bowditch
S.D. is +. Nautical Almanac

Observed altitude of Sun's lower limb is expressed (_).

True altitude is expressed -(-)-.

Remember that before an observation is at all accurate, it must be corrected to make it a true altitude. Remember also that the IE must be applied, in addition to these other corrections, in order to make the observed altitude a -(-)- altitude. So there are really five corrections to make instead of four, providing, of course, your sextant has an IE.

Examples:

1. June 20th, 1919, observed altitude of (_) 69 deg. 25' 30". IE + 2' 30". HE 16 ft. Required -(-)-.

2. April 15th, 1919, observed altitude of (_) 58 deg. 29' 40". IE - 2' 30". HE 18 ft. Required -(-)-.

3. March 4th, 1919, observed altitude of (_) 44 deg. 44' 10". IE - 4' 20". HE 20 ft. Required -(-)-.

Etc.

WEEK IV--NAVIGATION

TUESDAY LECTURE

THE LINE OF POSITION

It is practically impossible to fix your position exactly by one observation of any celestial body. The most you can expect from one sight is to fix your line of position, i.e., the line somewhere along which you are. If, for instance, you can get a sight by sextant of the sun, you may be able to work out from this sight a very accurate calculation of what your latitude is. Say it is 50 deg. N. You are practically certain, then, that you are somewhere in latitude 50 deg. N, but just where you are you cannot tell until you get another sight for your longitude. Similarly, you may be able to fix your longitude, but not be able to fix your latitude until another sight is made. Celestial Navigation, then, reduces itself to securing lines of position and by manipulating these lines of position in a way to be described later, so that they intersect. If, for instance, you know you are on one line running North and South and on another line running East and West, the only spot where you _can_ be on _both_ lines is where they intersect. This diagram will make that clear:

Just what a line of position is will now be explained. Wherever the sun is, it must be perpendicularly above the same spot on the surface of the earth marked in the accompanying diagram by S and suppose a circle be drawn around this spot as ABCDE. Then if a man at A takes an altitude, he will get precisely the same one as men at B, C, D, and E, because they are all at equal distances from the sun, and hence on the circumference of a circle whose center is S. Conversely, if several observers situated at different parts of the earth's surface take simultaneous altitudes, and these altitudes are all the same, then the observers must all be on the circumference of a circle and _only one_ circle. If they are not on that circle, the altitude they take will be greater or less than the one in question.

Now such a circle on the surface of the earth would be very large--so large that a small arc of its circumference, say 25 or 30 miles, would be practically a straight line.

Suppose S to be the point over which the sun is vertical and GF part of the circumference of a circle drawn around the point. Suppose you were at B and from an altitude of the sun, taken by sextant, you worked out your position. You would find yourself on a little arc ABC which, for all purposes in Navigation, is a straight line at right angles to the true bearing of the sun from the point S. You can readily see this from the above diagram. Suppose your observer is at H. His line is GHI, which is again a straight line at right angles to the true bearing of the sun. He is not certain he is at H. He may be at G or I. He knows, however, he is somewhere on the line GHI, though where he is on that line he cannot tell exactly. That line GHI or ABC or DEF is the line of position and such a line is called a Sumner Line, after Capt. Thomas Sumner, who explained the theory some 45 years ago. Put in your Note-Book:

Any person taking an altitude of a celestial body must be, for all practical purposes, on a straight line which is at right angles to the true bearing of the body observed.

It should be perfectly clear now that if the sun bears due North or South of the observer, i.e., if the sun is on the observer's meridian, the resulting line of position _must_ run due East and West. In other words it is a parallel of latitude. And that explains why a noon observation is the best of the day for getting your latitude accurately. Again, if the sun bears due East or West the line of position must bear due North and South. And that explains why a morning or afternoon sight--about 8-9 A.M. or 3-4 P.M., if the sun bears either East or West respectively, is the best time for determining your North and South line, or longitude.

Now suppose you take an observation at 8 A.M. and you are not sure of your D.R. latitude. Your 8 A.M. position when the sun was nearly due East, will give, you an almost accurate North and South line and longitude. Suppose that from 8 A.M. to noon you sailed NE 60 miles. Suppose at noon you get another observation. That will give you an East and West line, for then the sun bears true North and South. An East and West line is your correct latitude. Now you have an 8 A.M. observation which is nearly correct for longitude and a noon position which is correct for latitude. How can you combine the two so as to get accurately both your latitude and longitude? Put in your Note-Book:

Through the 8 A.M. position, draw a line on the chart at right angles to the sun's true bearing. Suppose the sun bore true E 1/2 S. Then your line of position would run N 1/2 E. Mark it 1st Position Line.

Now draw a line running due East and West at right angles to the N-S noon bearing of the sun and mark this line Second Position Line. Advance your First Position Line the true course and distance sailed from 8 A.M. to noon, and through the extremity draw a third line exactly parallel to the first line of position. Where a third line (the First Position Line advanced) intersects the Second Position Line, will be your position at noon. It cannot be any other if your calculations are correct. You knew you were somewhere on your 8 A.M. line, you know you are somewhere on your noon line, and the only spot where you can be on both at once is the point where they intersect. You don't necessarily have to wait until noon to work two lines. You can do it at any time if a sufficient interval of time between sights is allowed. The whole matter simply resolves itself into getting your two lines of position, having them intersect and taking the point of intersection as the position of your ship.

There is one other way to get two lines to intersect and it is one of the best of all for fixing your position accurately. It is by getting lines of position by observation of two stars. If, for instance, you can get two stars, one East and the other West of you, you can take observations of both so closely together as to be practically simultaneous. Then your Easterly star would give you a line like AA' and the westerly star the line BB' and you would be at the intersection S.

Assign for reading: Articles in Bowditch 321-322-323-324. Spend the rest of the period in getting times from the N. A., getting true altitudes from observed altitudes, working examples in Mercator sailing, etc.

WEDNESDAY LECTURE

LATITUDE BY MERIDIAN ALTITUDE

A meridian altitude is an altitude taken when the sun or other celestial body observed bears true South or North of the observer or directly overhead. In other words, when the celestial body is on your meridian and you take an altitude of the body by sextant at that instant, the altitude you get is called a meridian altitude. In the case of the sun, such a meridian altitude is at apparent noon. Now latitude is always secured most accurately at noon by means of your meridian altitude. The reason for this was explained in yesterday's lecture. The general formula for latitude by meridian altitude is (Put in your Note-Book):

Latitude by meridian altitude = Zenith Distance (ZD) +- Declination (Dec).

Zenith distance is the distance in degrees, minutes and seconds from your zenith to the center of the observed body. For simplicity's sake, we will consider the sun only as the observed body. Then the zenith distance is the distance from your zenith to the center of the sun. Now suppose that you and the sun are both North of the equator and you are North of the sun. If you can determine exactly how far North you are of the sun and how far North the sun is of the equator, you will, by adding these two measurements together, know how far North of the equator you are, i.e., your latitude. As already explained, the declination of the sun is its distance in degrees, minutes and seconds from the equator and the exact amount of declination is, of course, corrected to the proper G.M.T. Your zenith distance is the distance in the celestial sphere you are from the sun. You know that it is 90 deg. from your zenith to the horizon. Your zenith distance, therefore, is the difference between the true meridian altitude of the sun, obtained by your sextant, and 90 deg.. Hence, having secured the true meridian altitude of the sun, you have only to subtract it from 90 deg. to find your zenith distance, i.e., how far you are from the sun. This diagram will make the whole matter clear:

A = Zenith, B = Sun, C = Horizon.]

The arc ABC measures 90 deg.. That is the distance from your zenith to the horizon. Now if BC is the true meridian altitude of the sun at noon, 90 deg.-BC or AB is your zenith distance. If BC measures by sextant 60 deg., AB measures 90 deg.-60 deg. or 30 deg.. This 30 deg. is your Zenith Distance. Now suppose that from the Nautical Almanac we find that the G.M.T. corresponding to the time at which we measured the meridian altitude of the sun shows the sun's declination to be 10 deg. N. Well, if you are 30 deg. North of the sun, and the sun is 10 deg. North of the equator, you must be 40 deg. North of the equator or in latitude 40 deg. N. For that is all latitude is, namely, the distance in degrees, minutes and seconds you are due North or South of the equator. That is the first and simplest case.

Another case is when you are somewhere in North latitude and the sun's declination is South. Then the situation would, roughly, look like this:

In this case, your distance North of the equator AD would be your zenith distance AB minus the sun's declination DB. This diagram is not strictly correct, for the observer's position on the earth 0 appears to be South of the equator instead of North of the equator. That is because the diagram is on a flat piece of paper instead of on a globe. So far as illustrating the Zenith Distance minus the Declination, however, the diagram is correct. The last case is where you are, say, 10 deg. N of the sun (your zenith Distance is 10 deg.) and the sun is in 20 deg. S declination. In that case you would have to subtract your zenith distance from the sun's declination to get your latitude, for the sun's latitude (its declination) is greater than yours.

Now from these three cases we deduce the following directions, which put in your Note-Book:

Begin to measure the altitude of the sun shortly before noon. By bringing its image down to the horizon, you can detect when its altitude stops increasing and starts to decrease. At that instant the sun is on your meridian, it is noon at the ship, and the angle you read from your sextant is the meridian altitude of the sun. To work out your latitude, name the meridian altitude S if the sun is south of you and N if north of you.

Correct the observed altitude to a true altitude by Table 46. If the altitude is S, the Zenith Distance is N or vice versa. (Note to Instructor: If the sun is South of you, you are North of the sun and vice versa.)

Correct the declination for the proper G.M.T. as shown by chronometer (corrected). If zenith distance and declination are both North or both South, add them and the sum will be the latitude, N or S as indicated. If one is N, and the other S, subtract the less from the greater and the result will be the latitude in, named N or S after the greater. Example:

At sea June 15th, observed altitude of (_) 71 deg. 15' S, IE--47', HE 25 ft. CT 3h--34m--15s P.M. Required latitude of ship.

(_) 71 deg. 15' 00 S IE -- 47'
Corr. -- 36 24 HE + 10 36
------------------- ---------
-(-)- 70 deg. 38' 36" S Corr. -- 36' 24"
-- 90 00 00
-------------------
ZD 19 deg. 21' 24" N
Dec. 23 17 15 N (G.M.T. June 15--3h 34m 15s)
-------------------
Lat. 42 deg. 38' 39" N
-------------------

Assign for Night Work or to be worked in class room such examples as the following:

1. June 1st, 1919. (_) 33 deg. 50' 00" S. G.M.T. 8h 55m 44s. HE 20 ft. IE + 4' 3". Required latitude in at noon.

2. April 2nd, 1919. (_) 12 deg. 44' 30" N. CT was 2d 5h 14m 39s A.M., which was 1m 40s slow on March 1st (same CT) and 4m 29s fast on March 15th (same CT). IE -- 2' 20". HE 22 ft. Required latitude in at noon.

Assign for Night Work reading also, the following Articles in Bowditch: 344 and 223.

THURSDAY LECTURE

AZIMUTHS OF THE SUN

This is a peculiar word to spell and pronounce but its definition is really very simple. Put in your Note-Book:

The azimuth of a heavenly body is the angle at the zenith of the observer formed by the observer's meridian and a line drawn to the center of the body observed. Azimuths are named from the latitude in and toward the E in the A.M. and from the latitude in and toward the W in the P.M.

All this definition means is that, no matter where you are in N latitude, for instance, if you face N, the azimuth of the sun will be the true bearing of the sun from you. The same holds true for moon, star or planet, but in this lecture we will say nothing of the star azimuths for, in some other respects, they are found somewhat differently from the sun azimuths. Put this in your Note-Book:

To find an azimuth of the sun: Note the time of taking the azimuth by chronometer. Apply chronometer correction, if any, to get the G.M.T. Convert G.M.T. into G.A.T. by applying the equation of time. Convert G.A.T. into L.A.T. by applying the longitude in time. The result is L.A.T. or S.H.A. With the correct L.A.T., latitude and declination, enter the azimuth tables to get the sun's true bearing, i.e., its azimuth. Example:

March 15th, 1919. CT 10h -- 4m -- 32s. D.R. latitude 40 deg. 10' N, longitude 74 deg. W. Find the TZ.

G.M.T. 10h--04m--32s
Eq. T. --09 --10

G.A.T. 9h--55m--22s

G.A.T. 9h--55m--22s
Lo. in T. 4 --56 --00 (W--)

L.A.T. 4h--59m--22s
Latitude and Declination opp. name.
TZ = N 101 deg. 30'W

We will take up later a further use of azimuths to find the error of your compass. Right now all you have to keep in mind is what an azimuth is and how you apply the formulas already given you to get the information necessary to enter the Azimuth Tables for the sun's true bearing at any time of the astronomical day when the sun can be seen. In consulting these tables it must be remembered that if your L.A.T. or S.H.A. is, astronomically, 20h (A.M.), you must subtract 12 hours in order to bring the time within the scope of these tables which are arranged from apparent six o'clock A.M. to noon and from apparent noon to 6 P.M. respectively.

We are taking up sun azimuths today in order to get a thorough understanding of them before beginning a discussion of the Marc St. Hilaire Method which we will have tomorrow. You must get clearly in your minds just what a line of position is and how it is found. Yesterday I tried to explain what a line of position was, i.e., a line at right angles to the sun's or other celestial body's true bearing--in other words, a line at right angles to the sun's or other celestial body's azimuth. Today I tried to show you how to find your azimuth from the azimuth tables for any hour of the day. Tomorrow we will start to use azimuths in working out sights for lines of position by the Marc St. Hilaire Method.

Note to Instructor: Spend the rest of the time in finding sun azimuths in the tables by working out such examples as these:

1. April 29th, 1919. D.R. latitude 40 deg. 40' N, Longitude 74 deg. 55' 14" W. CT 10h--14m--24s. CC 4m--30s slow. Find TZ.

2. May 15th, 1919. D.R. latitude 19 deg. 20' S, Longitude 40 deg. 15' 44" E. CT 10h--44m--55s A.M. CC 3m--10s fast. Find TZ.

Note to Instructor:

If possible, give more examples to find TZ and also some examples on latitude by meridian altitude.

Assign for Night Work reading the following Articles in Bowditch: 371-372-373-374-375. Also, examples to find TZ.

FRIDAY LECTURE

MARC ST. HILAIRE METHOD BY A SUN SIGHT

You have learned how to get your latitude by an observation at noon. By the Marc St. Hilaire Method, which we are to take up today, you will learn how to get a line of position, at any hour of the day. By having this line of position intersect your parallel of latitude, you will be able to establish the position of your ship, both as to its latitude and longitude.

Now you have already learned that in order to get your latitude accurately, you must wait until the sun is on your meridian, i.e., bears due North or South of you, and then you apply a certain formula to get your latitude. When the sun is on or near the prime vertical (i.e., due East or West) you might apply another set of rules, which you have not yet learned, to get your longitude. By the Marc St. Hilaire method, the same set of rules apply for getting a line of position at any time of the day, no matter what the position of the observed body in the heavens may be. Just one condition is necessary, and this condition is necessary in all calculations of this character, i.e., an accurate measurement of the observed body's altitude is essential.

What we do in working out the Marc St. Hilaire method, is to assume our Dead Reckoning position to be correct. With this D. R. position as a basis, we compute an altitude of the body observed. Now this altitude would be correct if our D. R. position were correct and vice versa. At the same time we measure by sextant the altitude of the celestial body observed, say, the sun. If the computed altitude and the actual observed altitude coincide, the D. R. position is correct. If they do not, the computed altitude must be corrected and the D. R. position corrected to coincide with the observed altitude. Just how this is done will be explained in a moment. Put in your Note-Book:

_Formula for obtaining Line of Position by M. St. H. Method._

I. Three quantities must be known either from observation or from Dead Reckoning.

1. The S. H. A., marked "t."

Note: The method for finding S. H. A. (t) differs when the sun or
star is used as follows:

(a) For the Sun:
Get G.M.T. from the corrected chronometer time. Apply
the equation of time to find the G.A.T. Apply the D.R. Lo.
(-W)
(+E) and the result is L.A.T. or S.H.A. as required.

(b) For a Star:
(Note to pupils: Leave this blank to be filled in when we
take up stars in more detail.)

2. The Latitude, marked "L."

3. The Declination of the observed body, marked "D."

II. Add together the log haversine of the S.H.A. (Table 45), the log cosine of the Lat. (Table 44), and the log cosine of the Dec. (Table 44) and call the sum S. S is a log haversine and must always be less than 10. If greater than 10, subtract 10 or 20 to bring it less than 10.

III. With the log haversine S enter table 45 in the adjacent parallel column, take out the corresponding Natural Haversine, which mark N_{S}.

IV. Find the algebraic difference of the Latitude and Declination, and from Table 45 take out the Natural Haversine of this algebraic difference angle. Mark it N_{D+-L}

V. Add the N_{S} to the N_{D+-L}, and the result will be the Natural Haversine of the calculated zenith distance. Formula N_{ZD} = N_{S} + N_{D+-L}

VI. Subtract this calculated zenith distance from 90 deg. to get the calculated altitude.

VII. Find the difference between the calculated altitude and the true altitude and call it the altitude difference.

VIII. In your Azimuth Table, find the azimuth for the proper "t," L and D.

IX. Lay off the altitude difference along the azimuth either away from or toward the body observed, according as to whether the true altitude, observed by sextant, is less or greater than the calculated altitude.

X. Through the point thus reached, draw a line at right angles to the azimuth. This line will be your Line of Position, and the point thus reached, which may be read from the chart or obtained by use of Table 2 from the D. R. Position, is the nearest to the actual position of the observer which you can obtain by the use of any method from one sight only.

Example:

At sea, May 18th, 1919, A.M. (_) 29 deg. 41' 00". D.R. Latitude 41 deg. 30' N, Longitude 33 deg. 38' 45" W. WT 7h 20m 45s A.M. C-W 2h 17m 06s CC + 4m 59s. IE--30". HE 23 ft. Required Line of Position and most probable position of ship.

WT 18d -- 7h -- 20m -- 45s A.M.
-- 12
------------------------
WT 17d -- 19h -- 20m -- 45s
C-W 2 -- 17 -- 06 Corr. + 9' 34"
------------------------ IE -- 30
CT 17d -- 21h -- 37m -- 51s ------------
CC + 4 -- 59 + 9' 04"
------------------------
G.M.T. 17d -- 21h -- 42m -- 50s (_) 29 deg. 41' 00"
Eq. T. + 3 -- 47 + 9 04
------------------------ ------------
G.A.T. 17d -- 21h -- 46m -- 37s -(-)- 29 deg. 50' 04"
Lo. in T 2 -- 14 -- 35 (W--)
------------------------ log hav 9.48368
L.A.T.(t) 17d -- 19h -- 32m -- 02s log cos 9.87446
Lat. 41 deg. 30' N log cos 9.97473
Dec. 19 deg. 21' 25" N --------
log hav S 9.33287
N s .21521
L - D 22 deg. 08' 35" N D +- L .03687
--------
Calc. ZD 60 deg. 16' 30" N ZD .25208
-- 90 deg. 00 00
------------- TZ found from table to be
Cal. Alt. 29 deg. 43' 30" N 90 deg. E.
-(-)- 29 deg. 50' 04"
-------------
Alt. Diff 6' 34" Toward.

_ Course. Dist. Diff. Lat. Dep. Diff. Lo._
90 deg. 6' 34" 0 6.5 8.6

D.R. Lat. 41 deg. 30' N D.R. Lo. 33 deg. 38' 45" W
Diff. Lat. -- Diff. Lo. 8 36 E
---------- -------------
Most probable fix Lat. 41 deg. 30' N Lo. 33 deg. 30' 09" W

As azimuth is N 90 deg. E, Line of Position runs due N & S (360 deg.) through Lat. 41 deg. 30' N. Lo. 33 deg. 30' 09" W.

Assign for work in class and for Night Work examples such as the following:

1. July 11th, 1919. (_) 45 deg. 35' 30", Lat. by D. R. 50 deg. 00' N, Lo. 40 deg. 04' W. HE 15 ft. IE--4'. CT (corrected) 5h. 38m 00s P.M. Required Line of Position by Marc St. Hilaire Method and most probable fix of ship.

2. May 16th, 1919, A.M. (_) 64 deg. 01' 15", D. R. Lat. 39 deg. 45' N, Lo. 60 deg. 29' W. HE 36 ft. IE + 2' 30". CT 2h 44m 19s. Required Line of Position by Marc St. Hilaire Method and most probable fix of ship.

Etc.

SATURDAY LECTURE

EXAMPLES ON MARC ST. HILAIRE METHOD BY A SUN SIGHT

1. Nov. 1st, 1919. A.M. at ship. WT 9h 40m 15s. C--W 4h 54m 00s. D. R. Lat. 40 deg. 50' N, Lo. 73 deg. 50' W. (_) 27 deg. 59'. HE 14 ft. Required Line of Position by Marc St. Hilaire Method and most probable position of ship.

2. May 30th, 1919. P.M. at ship. D. R. Lat. 38º 14' 33" N, Lo. 15 deg. 38' 49' W. The mean of a series of observations of (_) was 39 deg. 05' 40 deg.. IE--01' 00". HE 27 ft. WT 3h 4m 49s. C--W 1h 39m 55s. C.C. fast, 01m 52s. Required Line of Position by Marc St. Hilaire Method, and most probable position of ship.

3. Oct. 21st, 1919, A.M. D. R. Lat. 40 deg. 12' 38" N, Lo. 69 deg. 48' 54" W. The mean of a series of observations of (_) was 19 deg. 21' 20". IE + 02' 10". HE 26 ft. WT 7h 58m 49s. C--W 4h 51m 45s. C. slow, 03m 03s. Required Line of Position by Marc St. Hilaire Method and most probable position of ship.

4. June 1st, 1919, P.M. at ship. Lat. D. R. 35 deg. 26' 15" S, Lo. 10 deg. 19' 50" W. W.T. 3h 30m 00s. C--W 0h 20m 38s. CC 1m 16s slow. (_) 16 deg. 15' 40". IE + 2' 10". HE 26 ft. Required Line of Position and most probable fix of ship.

5. Jan. 5th, 1919. A.M. D. R. Lat. 36 deg. 29' 38" N, Lo. 51 deg. 07' 44" W. The mean of a series of observations of (_) was 23 deg. 17' 20". IE + 01' 50". HE 19 ft. WT 7h 11m 37s. C--W 5h 59m 49s. C. slow 58s. Required Line of Position and most probable fix of ship.

WEEK V--NAVIGATION

TUESDAY LECTURE

A SHORT TALK ON THE PLANETS AND STARS IDENTIFICATION OF STARS

_1. The Planets_

You should acquaint yourself with the names of the planets and their symbols. These can be found opposite Page 1 in the Nautical Almanac. All the planets differ greatly in size and in physical condition. Three of them--Mercury, Venus and Mars--are somewhat like the earth in size and in general characteristics. So far as we know, they are solid, cool bodies similar to the earth and like the earth, surrounded by atmospheres of cool vapors. The outer planets on the other hand, i.e., Jupiter, Saturn, Uranus, and Neptune, are tremendously large--many times the size of the earth, and resemble the sun more than the earth in their physical appearance and condition. They are globes of gases and vapors so hot as to be practically self luminous. They probably contain a small solid nucleus, but the greater part of them is nothing but an immense gaseous atmosphere filled with minute liquid particles and heated to an almost unbelievably high temperature.

Of the actual surface conditions on Venus and Mercury, little is definitely known. Mercury is a very difficult object to observe on account of its proximity to the sun. It is never visible at night; it must be examined in the twilight just before sunrise or just after sunset, or in the full daylight. In either case the glare of the sun renders the planet indistinct, and the heat of the sun disturbs our atmosphere so as to make accurate visibility almost impossible. The surface of Mercury is probably rough and irregular and much like the moon. Like the moon, too, it has practically no atmosphere. Mercury rotates on its axis once in 88 days. Its day and year are of the same length. Thus the planet always presents the same face toward the sun and on that side there is perpetual day while on the other side is night--unbroken and cold beyond all imagination.

Venus resembles the earth more nearly than any other heavenly body. Its diameter is within 120 miles of the earth's diameter. The exasperating fact about Venus, however, is that it is shrouded in deep banks of clouds and vapors which make it impossible for us to secure any definite facts about it. The atmosphere about Venus is so dense that sunlight is reflected from the upper surface of the clouds around the planet and so reaches our telescopes without having penetrated to the surface at all. From time to time markings have been discovered that at first seemed real but whether they are just clouds or tops of mountains has never really been established.

Of all the planets, we know more about Mars than any other. And yet practically nothing is actually known in regard to conditions on the surface of this planet. We do know, however, that Mars more nearly resembles a miniature of our earth than any other celestial body. The diameter of Mars is 4,210 miles--almost exactly half the earth's diameter. The surface area of Mars is just about equal to the total area of dry land on the earth. Like the earth, Mars rotates about an axis inclined to the plane of its orbit, and the length of a Martian day is very nearly equal to our own. The latest determinations give the length of a Martian solar day as 24h 39m 35s. Fortunately for us, Mars is surrounded by a very light and transparent atmosphere through which we are able to discover with our telescopes, many permanent facts.

The most noticeable of these are the dazzling white "polar caps" first identified by Sir William Herschel in 1784. During the long winter in the northern hemisphere, the cap at the North pole steadily increases in size, only to diminish during the next summer under the hot rays of the sun. These discoveries establish without doubt the presence of vapors in the Martian atmosphere which precipitate with cold and evaporate with heat. The polar caps, then, are some form of snow and ice or possible hoar frost. Outside the polar caps the surface of Mars is rough, uneven and of different colors. Some of the darker markings appear to be long, straight hollows. They are the so-called "canals" discovered by Schiaparelli in 1877. The term "canal" is an unfortunate one. The word implies the existence of water and the presence of beings of sufficient intelligence and mechanical ability to construct elaborate works. Flammarion in France and Lowell in the United States claim the word is correctly used, i.e., that these markings are really canals and that Mars is actually inhabited. The consensus of opinion among the most celebrated astronomers is contrary to this view. Most astronomers agree that these canals may not exist as drawn--that they are to great extent due to defective vision. There is no conclusive proof of man-made work on Mars, nor of the existence of conscious life of any kind. It may be there but conclusive proof of it is still lacking.

_2. The Stars_