Now it would be very easy for the mariner if he could measure apparent time directly so that his clock or other instrument would always tell him just what the sun time was. It is impossible, however, to do this because the earth does not revolve at a uniform rate of speed. Consequently the sun is sometimes a little ahead and sometimes a little behind any average time. You cannot manufacture a clock which will run that way because the hours of a clock must be all of exactly the same length and it must make noon at precisely 12 o'clock every day. Hence we distinguish clock time from sun time by calling clock time, mean (or average) time and sun time, apparent or solar time. From this explanation you are ready to understand such expressions as Local Mean Time, which, in untechnical language, signifies clock time at the place where you are; Greenwich Mean Time which signifies clock time at Greenwich; Local Apparent Time, which signifies sun time at the place where you are; Greenwich Apparent Time, which signifies sun time at Greenwich.

Now the difference between apparent time and mean time can be found for any minute of the day by reference to the Nautical Almanac which we will take up later in more detail. This difference is called the Equation of Time.

There is one more fact to remember in regard to apparent and mean time. It is the relation of the sun's hour angle to apparent time. In the first place, what is a definition of the sun's HA? It is the angle at the celestial pole between the meridian intersecting any given point and the meridian intersecting the center of the sun. It is measured by the arc of the celestial equator intersected between the meridian of any point and the meridian intersecting the center of the sun.



For instance, in the above diagram, suppose PG is the meridian at Greenwich, and PS the meridian intersecting the sun. Then the angle at the pole GPS, measured by the arc GS would be the Hour Angle of Greenwich, or the Greenwich Hour Angle. And now you notice that this angular measure is exactly the same as apparent time at Greenwich or Greenwich Apparent Time, for Greenwich Apparent Time is nothing more than the distance in time Greenwich, England, or the meridian at Greenwich is from the sun, i.e., the time it takes the earth to revolve from Greenwich to the sun; and that distance is exactly measured by the Greenwich Hour Angle or the arc on the celestial equator, GS.

The same is correspondingly true of Local Apparent Time and the ship's Hour Angle. Suppose, for instance, PL is the meridian intersecting the place where your ship is. Then your ship's hour angle would be the angle at the pole intersecting the meridian of your ship and the meridian of the sun or LPS and measured by the arc LS. And you will note that this distance is exactly the same as apparent time at the ship, for Apparent Time at ship is nothing more than the distance in time which the ship is from the sun. We can sum up all this information in a few simple rules, which put in your Note-Book:

Mean Time = Clock Time.

G.M.T. = Greenwich Mean Time.

L.M.T. = Local Mean Time.

Apparent Time = Actual or Sun Time.

G.A.T. (G.H.A.) = Greenwich Apparent Time or Greenwich Hour Angle.

L.A.T. (S.H.A.) = Local Apparent Time or Ship's Hour Angle.

Difference between apparent and mean time or mean and apparent time—Equation of Time.

Right under this in your Note-Book put the following diagram, which I will explain:



You will see from this diagram that civil time commences at midnight and runs through 12 hours to noon. It then commences again and runs through 12 hours to midnight. The Civil Day, then, is from midnight to midnight, divided into two periods of 12 hours each.

The astronomical day commences at noon of the civil day of the same date. It comprises 24 hours, reckoned from O to 24, from noon of one day to noon of the next. Astronomical time, either apparent or mean, is the hour angle of the true or mean sun respectively, measured to the westward throughout its entire daily circuit.

Since the civil day begins 12 hours before the astronomical day and ends 12 hours before it, A.M. of a new civil day is P.M. of the astronomical day preceding. For instance, 6 hours A.M., April 15th civil time is equivalent to 18 hours April 14th, astronomical time.

Now, all astronomical calculations in which time is a necessary fact to be known, must be expressed in astronomical time. As chronometers have their face marked only from 0 to 12 as in the case of an ordinary watch, it is necessary to transpose this watch or chronometer time into astronomical time. No transposing is necessary if the time is P.M., as you can see from the diagram that both civil and astronomical times up to 12 P.M. are the same. But in A.M. time, such transposing is necessary. Put in your Note-Book:

Whenever local or chronometer time is A.M., deduct 12 hours from such time to get the correct astronomical time:

CT 15d— 9h—10m—30s A.M. —12 ———————————— CT 14d—21h—10m—30s ———

L.M.T. 10d— 4h—40m—16s A.M. —12 ———————————— L.M.T. 9d—16h—40m—16s

Now we come to a very important application of time. You will remember that in one of the former lectures we stated that to find our latitude, we had to find how far North or South of the equator we were, and to find our longitude, we had to find how far East or West of the meridian at Greenwich we were. Never mind about latitude for the present. We can find our longitude exactly if we know our Greenwich time and our time at ship. For instance, in the accompanying diagram:



Suppose PG is the meridian at Greenwich, then anything to the west of PG is West longitude and anything to the East of PG is East longitude. Now suppose GPS is the H.A. of G. or G.A.T.—i.e., the distance in time G. is from the sun. And L P S is the H.A. of the ship or L.A.T.—i.e., the distance in time the ship is from the sun. Then the difference between G P S and L P S is G P L, measured by the arc L G, and that is the difference that the ship, represented by its meridian PL, is from the Greenwich meridian PG. In other words, that is the ship's longitude for, as mentioned before, longitude is the distance East or West of Greenwich that any point is, measured on the arc of the celestial equator. The longitude is West, for you can see LPG or the arc LG is west of the meridian PG.

Likewise if P E is the meridian of your ship, the Longitude in time is the S.H.A. or L.A.T., E P S (the distance your ship is from the sun) less the G.H.A. or G.A.T., G P S (the distance Greenwich is from the sun) which is the angle G P E measured by the arc G E. And this Longitude is East for you can see G P E, measured by G E, is east of the Greenwich meridian, P G.

In both these cases, however, the longitude is expressed in time, i.e., so many hours, minutes and seconds from the Greenwich meridian and we wish to express this distance in degrees, minutes and seconds of arc. The earth describes a circle of 360 deg. every 24 hours. Then if you are 1 hour from Greenwich, you are 1/24 of 360 deg. or 15 deg. from Greenwich and if you are 12 hours from Greenwich, you are 1/2 of 360 deg. or 180 deg. from Greenwich. By keeping this in mind, you should be able to transpose time into degrees, minutes and seconds of arc for any fraction of time. It is, however, all worked out in Table 7 of Bowditch which turn to. (Note to Instructor: Explain this table carefully). Put in your Note-Book:

89 deg. 24' 26" = (89 deg.) 5h—56m (24') 1m—36s (26") 1—44/60s —————————— 5h—57m—37s 44/60s = 38s

4h—42m—26s 4h—40m = 70 deg. 2m—24s = 36' 2s = 30" —————- 70 deg. 36' 30"

Also put in your Note-Book this diagram and these formulas: (For diagram use illustration on p. 40.)

L.M.T. + West Lo. = G.M.T. L.A.T. + West Lo. = G.A.T. L.M.T. - East Lo. = G.M.T. L.A.T. - East Lo. = G.A.T.

G.M.T. - West Lo. = L.M.T. G.A.T. - West Lo. = L.A.T. G.M.T. + East Lo. = L.M.T. G.A.T. + East Lo. = L.A.T.

If G.M.T. or G.A.T. is greater than L.M.T. or L.A.T. respectively, Lo. is West.

If G.M.T. or G.A.T. is less than L.M.T. or L.A.T. respectively, Lo. is East.

Example:

In longitude 81 deg. 15' W, L.M.T. is April 15d—10h—17m—30s A.M. What is G.M.T.?

L.M.T. 15d—10h—17m—30s A.M. —12 ————————— L.M.T. 14d—22h—17m—30s 5 —25 W + ————————— G.M.T. 15d— 3h—42m—30s ————

G.M.T. April 15d— 3h—42m—30s L.M.T. April 15d—10h—17m—30s A.M.

In what Lo. is ship?

G.M.T. 15d 3h—42m—30s L.M.T. 14d 22h—17m—30s ————————— Lo. in T 5h—25m—00s W

Lo. = 81 deg. 15'W

Assign also for Night Work reading the following articles in Bowditch: 276-278-279-226-228-286-287-288-290-291-294 (omitting everything on page 114.)



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. + (.).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:



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. 38 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

The planets are often called wanderers in the sky because of their ever changing position. Sharply distinguished from them, therefore, are the "fixed" stars. These appear as mere points of light and always maintain the same relative positions in the heavens. Thousands of years ago the "Great Dipper" hung in the northern sky just as it will hang tonight and as it will hang for thousands of years to come. Yet these bodies are not actually fixed in space. In reality they are all in rapid motion, some moving one way and some another. It is their tremendous distance from us that makes this motion inappreciable. The sun seems far away from us, but the nearest star is 200,000 times as far away from us as is the sun. Expressed in miles, the figure is so huge as to be incomprehensible. A special unit has, therefore, been invented—a unit represented by the distance traversed by light in one year. In one second, light travels over 186,000 miles. In 8-1/3 minutes, light reaches us from the sun and, in doing so, covers the distance that would take the Vaterland over four centuries to travel. Yet the nearest star is over four "light years" distant—it is so far away that it requires over four years for its light to reach us. When you look at the stars tonight you see them, not as they are, but as they were, even centuries ago. Polaris, for instance, is distant some sixty "light years." Had it disappeared from the heavens at the time Lee surrendered to Grant, we should still be seeing it and entirely unaware of its disappearance.

Now each star in the heavens is in reality a sun, i.e., a vast globe of gas and vapor, intensely hot and in a continuous state of violent agitation, radiating forth heat and light, every pulsation of which is felt throughout the universe. So closely indeed do many of the stars resemble the sun, that the light which they emit cannot be distinguished from sunlight. Some of them are larger and hotter than the sun—some smaller and cooler. Yet the sun we see can be regarded as a typical star and from our knowledge of it we can form a fairly correct idea of the nature and characteristics of these other stars.

Anyone knows that the stars vary in brightness. Some of this variation is due partly to actual differences among the stars themselves and partly to varying distances. If all the stars were alike, then those which were farthest away would be faintest and we could judge a star's distance by its brilliancy. This is not the case, however. Some of the more brilliant stars are far more distant than some of the fainter ones. There are stars near and remote and an apparently faint star may in reality be larger and more brilliant than a star of the first magnitude. Vega, for instance, is infinitely farther away from us than the sun, yet its brightness is more than 50 times that of the sun. Polaris, still farther away, has 100 times the light and heat of the sun. In fact the sun, considered as a star, is relatively small and feeble.

3. Identification of Stars

Only the brighter stars can be used in navigation. So much light is lost in the double reflection in the mirrors of the sextant, that stars fainter than the third magnitude can seldom be observed. This reduces the number of stars available for navigation to within very narrow limits, for there are only 142 stars all told which are of the third magnitude or brighter. The Nautical Almanac gives a list of some 150 stars which may be used, but as a matter of fact, the list might be reduced to some 50 or 60 without serious detriment to the practical navigator. About 30 of these are of the second magnitude or greater and hence easily found. It is not difficult to learn to know 30 or 40 of the brighter stars, so that they can be recognized at any time. To aid in locating the stars, many different star charts and atlases have been published, but most of them are so elaborate that they confuse as much as they help. The simpler the chart, the fewer stars it pretends to locate, the better for practical purposes. Also, all charts are of necessity printed on a flat surface and such a surface can never represent in their true values, all parts of a sphere. A chart, therefore, which covers a large part of the heavens, is bound to give a distorted idea of distances or directions in some part of the sky and must be used with caution.

There are a few stars which form striking figures of one kind or another. These can always be easily located and form a starting point, so to speak, from which to begin a search for other stars. Of these groups the Great Dipper is the most prominent in the northern sky and beginning with this the other constellations can be located one by one.

When the groups or constellations are not known, then any individual star can be readily found by means of its Right Ascension, and Declination. As you have already learned, Declination is equivalent to latitude on the earth and Right Ascension practically equivalent to longitude on the earth, except that whereas longitude on the earth is measured E. and W. from Greenwich, Right Ascension is measured to the east all the way around the sky, from the First Point of Aries. With this in mind, you can easily see that if a star's R.A. is less than yours, i.e., less than L.S.T. or the R.A. of your Meridian, the star is not as far eastward in the heavens, as is your zenith. In other words it is to the west of you. And vice versa, if the Star's R.A. is greater than yours, the Star is more to the eastward than you and hence to the east of you. Moreover, as R.A. is reckoned all around a circle and in hours, each hour's difference between the Star's R.A. and yours is 1/24 of 360 deg. or 15 deg.. Hence if a star's R.A. is, for instance, 2 hours greater than yours, the star will be found to the east of your meridian and approximately 30 deg. from your meridian, providing the star is in approximately the same vertical east and west plane as is your zenith.

When the general east or west direction of any star has been determined, its north or south position can at once be found from its declination. If you are in Latitude 40 deg. N. your celestial horizon to the South will be 90 deg. from 40 deg. N. or 50 deg. S. and to the North it will be 90 deg. + 40 deg. N. = 130 deg. or 40 deg. N. (below the N. pole). The general position of the equator in the sky is always readily found according to the latitude you are in. If you are in 40 deg. N. latitude, the celestial equator would intersect the celestial sphere at a point 40 deg. South of you. Knowing this, the angular distance of a star North or South of the equator (which is its declination) should be easily found. Remember, however, that the equator in the sky is a curved line and hence a star in the East or West which looks to be slightly North of you may actually be South of you.

Put in your Note-Book:

If the star is west of you its R.A. is less than yours. If east of you, its R.A. is greater than yours. Star will be found approximately 15 deg. to east or west of you for each hourly difference between the star's R.A. and your R.A. (L.S.T.).

Having established the star's general east and west direction, its north and south position can be found from its declination.

4. Time of Meridian Passage of a Star

It is often invaluable to know first, when a certain star will be on your meridian or second, what star will be on your meridian at a certain specified time. Here is the formula for each case, which put in your Note Book:

1. To find when a certain star will cross your meridian, take from the Nautical Almanac, the R.A. of the Mean Sun for Greenwich Mean Noon of the proper astronomical day. Apply to it the correction for longitude in time (West +, East -) as per Table at bottom of page 2, Nautical Almanac, and the result will be the R.A. of the Mean Sun at local mean noon, i.e., the distance in sidereal time the mean sun is from the First Point of Aries when it is on your meridian. Subtract this from the star's R.A., i.e., the distance in sidereal time the star is from the First Point of Aries (adding 24 hours to the star's R.A., if necessary to make the subtraction possible). The result will be the distance in sidereal time the star is from your meridian i.e., the time interval from local mean noon expressed in units of sidereal time. Convert this sidereal time interval into a mean time interval by always subtracting the reduction for the proper number of hours, minutes and seconds as given in Table 8, Bowditch. The result will be the local mean time of the star's meridian passage.

Example:

April 22nd, 1919, A.M. at ship. In Lo. 75 deg. E. What is the local mean time of the Star Etamin's meridian passage?

R.A.M.S. Gr. 21d—0h 1h—54m—02s Red. for 75 deg. E (—5h) —49.3 ___

R.A.M.S. local mean noon 1h—53m—12.7s

Star's R.A. 17h—54m—44s — 1 —53 —12.7

Sidereal interval from L.M. Noon 16h—01m—31.3s Red. for Sid. Int. (Table 8) — 2 —37.3 L.M.T. 21 d 15h—58m—54s

Hence, star will cross meridian at 3h—58m—54s A.M. April 22nd.

2. To find at any hour desired what star will cross your meridian, take the R.A. of the Mean Sun for Greenwich Mean Noon of the proper astronomical day. Apply to it the correction for longitude in time (West +, East -) as per table at bottom of page 2, Nautical Almanac, and the result will be the R.A. of the Mean Sun at local mean noon; i.e., the distance in sidereal time the mean sun is from the First Point of Aries when it is on your meridian. Suppose you wish to find the star at 10 P.M. Add 10 sidereal hours to the sun's R.A. just found. The result will be the R.A. of your meridian at approximately 10 P.M.

Select in the table on p. 94 the R.A. of the star nearest in time to your R.A. just secured. Subtract the R.A. of the Mean Sun at local mean noon from the star's R.A. just found on p. 94 of the N.A. and the result will be the exact distance in sidereal time the star you have just identified is from your meridian, i.e., the time interval from local mean noon expressed in units of sidereal time. Convert this sidereal time interval into a mean time interval by always subtracting for the proper number of hours, minutes and seconds as per Table 8, Bowditch. You will then have secured the name of the star desired and the exact local mean time of the star's meridian passage.

Example No. 2: At sea Dec. 14, 1919. Desired to get a star on my meridian at 11 P.M. Lo. by D.R. 74 deg. W.

(.).R.A.G.M.N. 17h—28m—26s Corr. 74 deg. W. (4th - 56m W + ) + 0 —48.6 (.).R.A. your M. 17h—29m—14.6s + 11

R.A.M. 28h—29m—14.6s

— 24 4h—29m—14.6s

R.A. of Star Aldebaran 4h—31m—18.5s Star R.A. 28h—31m—18.5 (.).R.A. your M. 17 — 29—14.6 ———————— Sid. Int. from L.M. Noon 11h—02m—03.9s Red for Sid. Int. (Table 8) — 1 —48 ———————— L.M.T. 11h—00m—15.9s

Aldebaran, then, is the star and the exact L.M.T. of its meridian passage will be 11h 00m 15.9s

Note: If the R.A.M. is more than 24 hours, deduct 24 hours. You will know whether the star is North or South of you by its declination. If you are in North latitude, the star will be south of you if its declination is South or if its declination is North and less than your latitude. If its declination and your latitude are both North and its declination is greater, the star will be north of you. The same principle applies if you are in South latitude.

Assign any of the following to be worked in the class room or at night:

1. At sea, November 1st, 1919. In Latitude 40 deg. N., Longitude 74 deg. W. WT 8h 30m P.M. Observed unknown star about 80 deg. east of my meridian and 25 deg. south of me. What was the star?

2. At sea, December 1st, 1919. CT 10h 45m 01s. CC 20m 16s slow. In D.R. Latitude 30 deg. N., Longitude 60 deg. 30' W. Observed unknown star about 60 deg. west of meridian and about 22 deg. S. What was the star?

3. March 15th, 1919. In D.R. Latitude 10 deg. 42' N, Longitude 150 deg. 14' 28" W. CT 5h 14m 28s. CC—2m 10s. Observed unknown star almost on my meridian and about 28 deg. north of me. What was the star?

4. Aug. 3, 1919, P.M. at ship. In D.R. Latitude 37 deg. 37' N. Longitude 38 deg. 37' W. At what local mean time will the Star Antares be on the meridian?

5. What star will transit at about 4:10 A.M. on Aug. 3rd, 1919? In D.R. position Latitude 38 deg. 10' N, Longitude 34 deg. 38' W.

6. At what local mean time will the Star Arcturus transit on July 17th, 1919, in Latitude 45 deg. 35' N., Longitude 28 deg. 06' W.?



WEDNESDAY LECTURE

LATITUDE BY MERIDIAN ALTITUDE OF A STAR—LATITUDE BY POLARIS (POLE OR NORTH STAR)

To find your latitude by taking an altitude of a star when it is on your meridian, is one of the quickest and easiest of calculations in all Navigation. The formula is exactly the same as for latitude by meridian altitude of the sun. In using a star, however, you do not have to consult your Nautical Almanac to get the G.M.T. and from that the declination. All you have to do is to turn to page 95 of the Nautical Almanac, on which is given the declination for every month of the year, of any star you desire. The rest of the computation is, as said before, the same as for latitude by the sun and follows the formula Lat. = Dec. +- Z.D. (90 deg. - true altitude). As when working latitude by the sun, you subtract the Z.D. and Dec. when of opposite name and add them when of the same name. Put in your Note-Book:

Formula: Lat. = Dec. +- Z.D. (90 deg. - true altitude).

At sea, Dec. 24th, 1919. Meridian altitude Star Aldebaran 52 deg. 36' S. HE 20 ft. Required latitude of ship.

Obs. Alt. 52 deg. 36' S Corr. - 5 08 ————— True Alt. 52 deg. 30' 52" S - 90 00 00 ————— Z.D. 37 deg. 29' 08" N Dec. 16 21 00 N ————— Lat. 53 deg. 50' 08" N

Note to Instructor:

Have class work examples such as the following before taking up Latitude by Pole Star:

1. At sea, May 5th, 1919. Meridian altitude Star Capella, 70 deg. 29' S. HE 32 ft. Required latitude of ship.

2. At sea, August 14th, 1919. Meridian altitude Star Vega, 60 deg. 15' 45" N. HE 28 ft. Required latitude of ship.

Etc.

Latitude by Polaris (Pole or North Star)

You remember we examined the formula in the N.A. for Lat. by the pole star when we were discussing sidereal time some weeks ago. We will now take up a practical case of securing your latitude by this method. Before doing so, however, it may be of benefit to understand how we can get our latitude by the pole star. In the first place, imagine that the Pole Star is directly over the N pole of the earth and is fixed. If that were so, and imagine for a minute that it is so, then it would be exactly 90 deg. from the Pole Star to the celestial equator. Now, no matter where you stand, it is 90 deg. from your zenith to your true horizon. Hence if you stood at the equator, your zenith would be in the celestial equator and your true horizon would exactly cut the Pole Star. Now, supposing you went 10 deg. N of the equator. Then your northerly horizon would drop by 10 deg. and the Pole Star would have an altitude of 10 deg.. In other words, when you were in 10 deg. N latitude, the pole star would measure 10 deg. high by sextant. And so on up to 90 deg., where the Pole Star would be directly over you and you would be at the North Pole. Now all this is based upon the Pole Star being in the celestial sphere exactly over the North Pole of the earth. It is not, however. Owing to the revolution of the earth, the star appears to move in an orbit of a maximum of 1 deg. 08'. Just what part of that 1 deg. 08' is to be applied to the true altitude of the star for any time of the sidereal day, has been figured out in the table on page 107 of the Nautical Almanac. What you have to get first is the L.S.T. Find from the table the correction corresponding to the L.S.T. and apply this correction with the proper sign to the true altitude of Polaris. The result is the latitude in. Put in your Note-Book:

To get latitude by pole star, first get L.S.T. This can be secured by using any one of the three formulas given you in Week III—Thursday's Lecture on Sidereal Time and Right Ascension. Then proceed as per formula in N.A.

* * * * *

Note to Instructor:

Spend rest of time in solving examples similar to the following:

1. At sea, Feb. 14th, 1919. CT 13d 21h 52m 33s. CC 1m 14s fast. In Lo. 72 deg. 49' 00" W. IE + 1' 10". HE 15 ft. Observed altitude Polaris 42 deg. 21' 30" N. Required latitude in.

2. At sea, March 31st, 1919. In Lo. 160 deg. 15' E. CT 7h 15m 19s. Observed altitude Polaris 38 deg. 18' N. IE + 3' 00". HE 17 ft. Required latitude in.

Etc.



THURSDAY LECTURE

MARC ST. HILAIRE METHOD BY A STAR SIGHT

You have already been given instructions for finding a Line of Position by the Marc St. Hilaire Method, using a sight of the sun. Today we will work out the same method by using a sight of a star. Put this in your Note-Book here and also under I(b) of the formula given you in Week IV—Friday's Lecture:

Get G.M.T. from corrected chronometer time. With your G.M.T. find the corresponding G.S.T. according to the formula already given you. With your G.S.T. apply the D.R. longitude

(- W. Lo.) ————— (+ L. Lo.)

to get the L.S.T. With the L.S.T. and the star's R.A. subtract the less from the greater and the result is the star's H.A. at the ship or "t." In using Sun Azimuth tables always take "t" from the P.M. column. Mark Azimuth N or S according to the lat. in and E or W, according as to whether the Star is East or West of your meridian. Then proceed as in the case of a sun sight. Formula:

(-W. Lo.) G.M.T. + (.).R.A. + ()CP = G.S.T. ————- = L.S.T.—Star's R.A. (E. Lo.)

(or vice versa if Star's R.A. is greater) = Star's H.A. at ship (t). Then proceed as in case of sun sight.

Example:

On May 31st, 1919, in D.R. Lat. 50 deg. N, Lo. 45 deg. W, G.M.T. 31d 14h 33m 30s. What was Star's H.A. at ship?

G.M.T. 14h — 33m — 30s (.).R.A. 4 — 31 — 44.2 (+).C.P. 2 — 23 —————————— G.S.T. 19h — 07m — 37.2s W Lo.— 3 — 00 — 00 —————————— L.S.T. 16h — 07m — 37.2s Star's R.A.(Spica) 13 — 20 — 59 —————————— Star's H.A. (t) 2h — 46m — 38.2s

Now let us work out some examples by this method:

1. Nov. 29th, 1919. CT 30d 2h 14m 39s A.M. CC 3m 14s fast. D.R. position Lat. 41 deg. 14' N, Lo. 68 deg. 46' W. Observed altitude Star Aldebaran East of meridian 50 deg. 29' 40". HE 29 ft. Required Line of Position by Marc St. Hilaire Method and most probable position of ship.

2. Jan. 23rd, 1919. P.M. at ship. CT 3h 45m 40s. Lat. by D.R. 38 deg. 44' 19" N. Lo. 121 deg. 16' 14" E. Observed altitude Star Rigel 28 deg. 59' 20" West of meridian. IE + 4' 30". HE 42 ft. Required Line of Position by Marc St. Hilaire Method and most probable position of ship.

Assign for Night Work one or two examples similar to the above.



FRIDAY LECTURE

EXAMPLES: LATITUDE BY MERIDIAN ALTITUDE OF A STAR, LATITUDE BY POLARIS, MARC ST. HILAIRE METHOD BY A STAR SIGHT

1. At sea, Dec. 5th, 1919. Observed meridian altitude Star Aldebaran 69 deg. 28' 40" S. No IE. HE 26 ft. Required latitude in.

2. At sea, Jan. 20th, 1919. CT 21d 2h 16m 48s A.M. In longitude 56 deg. 29' 46" W. Observed altitude of Star Polaris 48 deg. 44' 30" N. IE + 10' 20". HE 37 ft. Required latitude in.

3. At sea, June 4th, 1919. A.M. at ship. CT 10h 16m 32s. CC 5m 45s fast. Lat. by D.R. 42 deg. 44' N, Longitude 53 deg. 13' 44" E. Observed altitude of Star Altair East of meridian, 52 deg. 19' 30". IE—14' 00". HE 56 ft. Required line of position by Marc St. Hilaire Method and most probable position of ship.

Etc.

* * * * *

Assign for Night Work the following Articles in Bowditch: 336 through 341, disregarding the formulas.



SATURDAY LECTURE

LONGITUDE BY CHRONOMETER SIGHT OF THE SUN (TIME SIGHT)

You have now learned, first, how to get your latitude by a meridian altitude of the sun or a star and second, how to get your Line of Position and most probable fix, including both latitude and longitude, by the Marc St. Hilaire Method, using for your calculations either the sun or a star. We are now going to take up a method of getting your longitude only. This method requires as much, if not more, calculation than the Marc St. Hilaire Method. Its results, on the other hand, are far less complete, for while the Marc St. Hilaire Method will give you a fairly accurate idea of both your latitude and longitude, this method will, at best, only give you your longitude. Moreover, you can use it for accurate results only when the sun bears almost due East or West of you, for that is the best time, as you have already learned, to get a line of position running due North and South, which is nothing more than a meridian of longitude. The only reason we explain this method at all is because it is in common practice among merchantmen and may, therefore, be of assistance to you, if you go on a merchant ship. Remember, however, that it belongs to Old Navigation as distinguished from New Navigation, exemplified by the Marc St. Hilaire Method. It is undoubtedly being used less and less among progressive, up-to-date navigators, and will continue to be used less as time goes on. The fact remains, however, that at present many merchantmen practice it, and so it will do you no harm to become familiar with the method, too.

This method is based on securing your longitude by a time sight or longitude by chronometer sight, meaning that at the time the sun bears as near due East or West as possible, you take a sight of it by sextant and at the same instant note the time by chronometer. With this information you proceed to work out your problem and secure your longitude according to the following formula. Put in your Note-Book:

To find your longitude by chronometer (or time) sight.

1. Take sight by sextant only when the sun bears as near as possible due East or West. At exact time of taking sight, note chronometer time.

2. Get G.M.T. from corrected chronometer time. Apply Equation of Time to get the corresponding G.A.T.

3. Correct observed altitude to get T.C.A. Also have at hand Lat. by D. R. and Polar Distance. (Note: Secure P. D. by subtracting Dec. from 90 deg., if Lat. and Dec. are of same name. If Lat. and Dec. are of opposite name, secure P. D. by adding Dec. to 90 deg..)

4. Add together the T.C.A. the Lat. by D.R. and the P.D. Divide the sum by 2 and call the quotient Half Sum. From the Half Sum subtract the T.C.A. and call the answer the Difference.

5. Add together the secant of the Latitude, the cosecant of the P.D., the cosine of the Half Sum and the sine of the Difference (Table 44). The result will be the log haversine of the S.H.A. or L.A.T. It must always be less than 10. If greater than 10, subtract 10 or 20 to bring it less than 10.

6. From Table 45, take out the corresponding S.H.A. (L.A.T.), reading from the top of the page if P.M. at ship, or from bottom of page if A.M. at the ship.

7. Find the difference between L.A.T. and G.A.T. This difference is Lo. in Time which turns into degrees, minutes and seconds by Table 7. If G.A.T. is greater than L.A.T. longitude is West; if G.A.T. is less than L.A.T. longitude is East. Example:

August 26th, 1919, A.M. CT 26d 2h 29m 03s A.M. CC 16m 08s slow. (_) 44 deg. 57' 00". IE—1' 30". HE 32 ft. D.R. Lat. 4 deg. 55' 32" N. Required longitude in at time of observation.

26d—2h—29m—03s A.M. - 12 ————————— CT 25d—14h—29m—03s —90 deg. 00' 00" CC+ +16 —08 Dec. 10 49 48 ————————— ——————— G.M.T. 25d 14h 45m—11s P.D. 79 deg. 10' 12" Eq. T. -2 —05 ————————— G.A.T. 25d 14h—43m—06s - 1' 30" + 9 27 (_) 44 deg. 57' 00" ————- Corr. + 7 57 Corr. + 7' 57" —————- -(-)- 45 deg. 04' 57" Lat. 4 55 32 N sec. .00160 P.D. 79 10 12 cosec. .00781 —————— 2) 129 deg. 10' 41" —————— 1/2 S 64 deg. 35' 20" cos. 9.63266 - 9 -(-)- 45 04 57 —————— Diff. 19 deg. 30' 23" sin. 9.5235O—14 ——————- 9.16557 + 5 + 5 ——————- log. hav. S.H.A. (L.A.T.) 9.16562

S.H.A. (L.A.T.) 25d—21h—00m—01s G.A.T. 25 —14 —43 —06 ————————- Lo. in T. 6h—16m—55s E

Lo. (Table 7) 94 deg. 13' 45" E

I wish to caution you about confusing this method with the one Bowditch uses, and still another which Henderson uses in his book "Elements of Navigation." It is not exactly like either one. It requires one operation less than either, however, and it also requires the use of fewer parts of the various tables involved. For that reason it is given you.

Assign for work in class room and also for work at night examples similar to the following:

1. Oct. 1st, 1919. A.M. (_) 17 deg. 15' 00". G.M.T. 1d 11h 30m 00s A.M. D.R. Lat. 40 deg. 30' N. IE—2' 20". HE 25 ft. Required longitude in.

2. Oct. 10th, 1919. P.M. (_) 25 deg. 14' 30". CT 1h 15m 20s. CC 4m 39s slow. IE—3' 10". HE 26 ft. D.R. Lat. 41 deg. 29' 00" S. Required longitude in.

3. May 27, 1919. P.M. Lat. by D.R. 40 deg. 55' N. (_) 34 deg. 4' 00". IE + 1' 10". HE 10 ft. CT 8h 55m 42s. CC 2m 02s fast. Required longitude in.

4. May 18th, 1919. A.M. (_) 29 deg. 41' 15". WT 7h 20m 45s. C-W 2h 17m 06s. CC 4m 59s slow. Latitude by D.R. 41 deg. 33' N. IE—1' 30". HE 23 ft. Required longitude in.

5. August 24th, 1919. A.M. (_) 23 deg. 32' 10". IE—2'00". HE 16 ft. In latitude 39 deg. 04' N. CT 24d 2h 47m 28s A.M. CC + 4m 28s. Required longitude in.

6. June 26th, 1919. P.M. (_) 44 deg. 08' 20". IE—2' 20". HE 37 ft. CT 8h 18m 45s. CC 3m 20s fast. Latitude by D.R. 6 deg. 43' S. Required longitude in.

7. July 29th, 1919. A.M. CT 29d 11h 14m 39s A.M. CC 2m 18s slow. (_) 28 deg. 08' 30". IE + 0' 30". HE 38 ft. Latitude by D.R. 39 deg. 48' N. Required longitude in.

8. May 22nd, 1919. P.M. CT 9h 14m 38s. CC 5m 28s slow. (_) 21 deg. 07' 40". In latitude 41 deg. 26' N. IE + 3' 10". HE 40 ft. Required longitude in.



WEEK VI—NAVIGATION



TUESDAY LECTURE

LONGITUDE BY CHRONOMETER SIGHT OF A STAR

In getting your longitude by a time sight of a star, you proceed somewhat differently from the method used when observing the sun. What you wish to get first is G.S.T., i.e., the distance in time Greenwich is from the First Point of Aries. If you can then get the distance the ship is from the First Point of Aries, the difference between the two will be the longitude in, marked East or West according as to which is greater. By looking at the diagram furnished you when we were talking of Sidereal Time, all this becomes perfectly clear. The full rule for finding longitude by a star is as follows, which put in your Note-Book:

Correct your CT to get your G.M.T. From the G.M.T. get the G.S.T. From the observed altitude of the star, obtain the star's H.A. at the ship in the same way L.A.T. is secured in case of the sun. To or from the R.A. of the star add, if West of your meridian, subtract if East of your meridian, the star's H.A. at the ship, just obtained. The result is the R.A. of the ship's meridian or L.S.T.

Find the difference between G.S.T. and L.S.T. and the result is the longitude, marked East or West according as to whether G.S.T. is less or greater than L.S.T. Note: Always take the star's H.A. from the top of the page of Table 45.

Dec. 2, 1919. A.M. Observed altitude Star Sirius 2O deg. 05' 20", West of meridian. CT 11h—45m—29s P.M. CC 1m—28s slow. IE—1' 20". HE 21 ft. Latitude by D. R. 38 deg. 57' N. Required longitude in.

CT 11h—45m—29s CC + 1 —28 —————————- G.M.T. 11h—46m—57s (.)RA 16 —37 —10.3 (+)CP 1 —56.1 —————————- G.S.T. 28h—26m—03.4s IE -1' 20" —24 HE -7 08 —————————- ———- G.S.T. 4h—26m—03.4s Corr. -8' 28"

Obs. Alt. 20 deg. 05' 20 Corr. -8 28 ————— T.C.A. 19 deg. 56' 52" Lat. 38 57 sec. .10919 P.D. 106 36 24 cosec. .01849 + 1

2 ) 165 deg. 30' 16" ——————- 1/2 S 82 deg. 45' 08" cos. 9.10106 - 13 T.C.A. 19 56 52 ——————- Diff. 62 deg. 48' 16" sin. 9.94911 + 2 ————- 9.17785 - 11 ————- log. hav. Star's H.A. at ship 9.17774

Star's H.A. 3h—02m—40s Star's R.A. 6 —41 —39 ——————— L.S.T. 9h—44m—19s G.S.T. 4 —27 —01 ——————— Lo. in T. 5h—17m—18s E

Longitude in 79 deg. 19' 30" E

Assign for Night Work or work in the class room examples similar to the following:

1. April 16, 1919, in Latitude 11 deg. 47' S. Observed altitude of the Star Aldebaran, West of the meridian 23 deg. 13' 20". CT 6h 58m 29s. CC 2m 27s fast. IE—2' 00". HE 26 ft. Required longitude in.

2. Dec. 10th, 1919. Observed altitude of Star Sirius 20 deg. 05' 40" West of meridian. CT 11h 45m 29s. CC 1m 28s slow. IE—1' 20". HE 21 ft. D.R. latitude 38 deg. 57' N. Required longitude in.

Note to Instructor: If any time in the period is left or for Night Work assign examples to be worked by Marc St. Hilaire Method, changing slightly the D.R. Lat. and Longitude just obtained by the Time Sight Method.



WEDNESDAY LECTURE

EXAMPLES ON LONGITUDE BY CHRONOMETER SIGHT OF A STAR

1. Dec. 9th, 1919. In latitude 36 deg. 48' N. Observed altitude Star Capella, East of meridian 46 deg. 18' 30". IE 2' 50" off arc. HE 33 ft. CT 10d 3h 05m 05s A.M. CC 1m 18s slow. Declination of star is 45" 55' N. Required longitude in.

2. October 26th, 1919. In latitude 39 deg. 54' S. Observed altitude Star Rigel, West of meridian 42 deg. 18' 40". CT 27d 10h 32m 55s A.M. CC 2m 18s fast. IE 4' 20" off arc. HE 42 ft. Required longitude in.

3. April 11th, 1919. P.M. at ship. In latitude 43 deg. 16' 48" S. Observed altitude Star Spica 33 deg. 18' 20", East of meridian. CT 11h 08m 44s P.M. IE 3' 20" on arc. CC 4m 18s slow. HE 39 ft. Required longitude in.

4. September 15th, 1919. P.M. at ship. In latitude 49 deg. 38'N. Observed altitude Star Deneb, East of meridian, 36 deg. 16' 50". IE 3' 40" off arc. HE 40 ft. CC 6m 18s slow. CT 10h 00m 13s P.M. Declination of star is 44 deg. 59' 36" N. Required longitude in.

If any time is left, work same examples by Marc St. Hilaire Method assuming a position near the one found by Time Sight.

Assign for Night Work any of the above examples, to be worked either as Time Sights or by the Marc St. Hilaire Method, and also the following Arts. in Bowditch: 326-327-328-329.



THURSDAY LECTURE

LATITUDE BY EX-MERIDIAN ALTITUDE OF THE SUN

You have learned that when you calculate your latitude from a meridian altitude of the sun, one of the necessary requisites is to have the sun exactly on your meridian. In fact, that is just another way of expressing meridian altitude, i.e., an altitude taken when the sun is on your meridian. Now suppose that 10 or 15 minutes before noon you fear that the sun will be clouded over at noon so that a meridian altitude cannot be secured. There is a way to calculate your latitude, even though the altitude you secure is taken by sextant some minutes before or after noon. This is called latitude by an ex-meridian altitude. It must be kept in mind that this method can be used accurately only within 26 minutes of noon, either before or after, and only then when you know your longitude accurately. Put in your Note-Book:

1. Get your L.A.T. (S.H.A.).

2. Subtract it from 24h 00m 00s, or vice versa, according as to whether L.A.T. is just before or just after local apparent noon. Call the result "Time Interval from Meridian Passage."

3. With your D.R. latitude, declination and Time Interval from Meridian Passage, enter Table 26 to get the proper amount of Variation of Altitude in one minute from meridian passage.

4. With the Time Interval from Meridian Passage and the Variation, enter Table 27 to get the total amount of Variation of Altitude.

5. Add this total amount of Variation to the true observed altitude taken before or after noon, and the result is the corrected altitude.

6. Then proceed to get your latitude according to the rules already given you for latitude by meridian altitude.

Example: At sea, Jan. 23rd, 1919. CT 4h 22m 14s. CC 1m 10s fast. Longitude 66 deg. 04' W. Latitude by D.R. 19 deg. 16' 00" N. (_) 50 deg. 51' 00" S. HE 49 ft. IE—1' 30". Required latitude in.

CT 4h - 22m - 14s CC - 1 - 10 ——————————- G.M.T. 4h - 21m - 04s Eq. T. - 11 - 50 ——————————- G.A.T. 4h - 09m - 14s Lo. in T 4 - 24 - 16 (W-) ——————————- L.A.T. 22d - 23h - 44m - 58s

24h - 00m - 00s - 23 - 44 - 58 ————————- 15m - 02s = Time Interval from Meridian Passage.

Dec. 19 deg. 34' 48" S Table 26 = 2.8 Variation Lat. 19 deg. 16' 00" N For 1 min. 0 altitude.

* * * * *

Time Interval from Meridian Passage 15m 02s - 2.8" Variation for 1 minute (Table 27) 2" = 7' 30" .8 = 3 00 ——————- 10' 30" +

IE - 1' 30" (_) 50 deg. 51' 00" HE + 8 42 + 7 12 ————- —————- Corr. + 7' 12" -(-)- 50 deg. 58' 12" + 10 30 —————- 51 deg. 08' 42" - 90 00 00 —————- ZD 38 deg. 51' 18" N Dec. 19 34 48 S —————- Lat. in 19 deg. 16' 30" N

Assign for work in class room and Night Work, examples similar to the following:

1. At sea, July 11th, 1919. Latitude by D.R. 50 deg. 01' 00" N. Longitude 40 deg. 05' 16" W. Observed ex-meridian altitude (_) 61 deg. 45' 30" S. HE 15 ft. IE—4' 10". CT (corrected) 2h 38m 00s. Required latitude in.

2. At sea, June 6th, 1919. Latitude by D. R. 49 deg. 21' N, Longitude 18 deg. 18' W. Observed ex-meridian altitude (_) 61 deg. 30' 22" S. HE 42 ft. CT 1h 06m 18s. CC—1m 14s. IE 0' 30" off the arc. Required latitude in of ship.

If any time is left, work similar examples by Marc St. Hilaire Method.



FRIDAY LECTURE

EXAMPLES: LATITUDE BY EX-MERIDIAN ALTITUDE OF THE SUN

1. Jan. 1st, 1919. WT 11h 53m 18s A.M. C-W 5h 56m 16s. Latitude by D. R. 58 deg. 05' S. Longitude 89 deg. 00' 48" W. (_) ex-meridian 55 deg. 16' 30" N. IE 2' 00" off the arc. CC 1m 28s fast. HE 36 ft. Required latitude in.

2. March 11th, 1919. CT 11d 9h 14m 39s A.M. Latitude by D. R. 39 deg. 20' N, Longitude 39 deg. 48' 16" E. (_) ex-meridian 46 deg. 17' 30" S. IE 2' 00" on the arc. CC 1m 16s slow. HE 29 ft. Required latitude in.

3. April 26th, 1919. CT 26d 4h 46m 38s A.M. Latitude by D. R. 24 deg. 25' S, Longitude 107 deg. 16' 56" E. (_) ex-meridian 52 deg. 18' 50" N. IE—2' 40". CC 3m 56s slow. HE 33 ft. Required latitude in.

4. May 10, 1919. CT 2h 18m 46s A.M. Latitude by D. R. 23 deg. 54' S, Longitude 143 deg. 20' 18" E. (_) ex-meridian 48 deg. 26' 20" N. IE 3' 20" on the arc. CC 4m 18s fast. HE 41 ft. Required latitude in.

5. June 21st, 1919. CT 4h. 56m 18s. Latitude by D. R. 42 deg. 01' N, Longitude 75 deg. 00' 18" W. (_) ex-meridian 71 deg. 29' 40" S, IE—2' 30". CC 3m 04s slow. HE 28 ft. Required latitude in.

6. Dec. 18th, 1919. WT 11h 50m 18s A.M. C-W 3h 14m 18s. Latitude by D. R. 11 deg. 55' S. Longitude 48 deg. 02' 29" W. (_) ex-meridian 78 deg. 32' 30" S. IE 3' 30" on the arc. CC 2m 44s slow. HE 35 ft. Required latitude in.

If there is any time left, give examples of latitude by meridian altitude, Marc St. Hilaire Method by sun or star sight, etc.



SATURDAY LECTURE

FINDING THE WATCH TIME OF LOCAL APPARENT NOON

Noon at the ship is the pivotal point of the day's work at sea. It is then that the navigator must report to the commanding officer the latitude and longitude by dead reckoning, the latitude and longitude by observation, the course and distance made good, the deviation of the compass and the course and distance to destination. Apparent noon, then, is a most important time to calculate accurately, and to do so when the ship is under way, is not so easy at it first appears.

If the ship is stationary, and you know the longitude you are in, the problem is simple. Then it is merely a question of starting with L.A.T. of 00h-00m-00s, adding or subtracting the longitude, according as to whether it is West or East, to get G.A.T.; applying the equation of time with sign reversed to get G.M.T.; applying the C. Cor. with sign reversed to get the C.T.; and applying the C-W to get the WT. If, for instance, this WT happens to be 11h-42m-31s, when the watch reads that number of hours, minutes and seconds, the sun will be on the meridian and it will be apparent noon.

When the ship is moving, the problem is more difficult. At first thought you might imagine that all you would have to do would be to take the difference between the L.A.T. of the morning sight and 24 hours, calculate the distance the ship would run in this time and from that determine the longitude you would be in at noon. Then proceed as in the case of the ship being stationary. But such a calculation does not take into consideration the easting or westing of the ship itself. Suppose that at the morning sight the L.A.T. is found to be 20h-10m-30s. If the ship does not move, it will be 3h-49m-30s to noon. But suppose the ship is moving eastward. Then, in addition to the speed at which the sun is approaching the ship, there must be added the speed at which the ship is moving toward the sun—i.e. the change in longitude per hour which the ship is making, expressed in minutes and seconds of time. Likewise, if the ship is moving westward, an allowance must be made for the westing of the ship. And this change of longitude in minutes and seconds of time must be subtracted from the speed of the sun's approach since the ship, in going west, is traveling away from the sun.

There are various ways to calculate this allowance for the ship's speed, among the best of which is given in Bowditch, Art. 403, p. 179. Another, and even easier way, is the following, which was explained to the writer by Lieutenant Commander R.P. Strough, formerly head of the Seamanship Department of this School:—

1. Take the morning sight for longitude when the sun is on or as near as possible to the prime vertical.

2. Subtract the L.A.T. of the morning sight from 24 hours. This will give the total time from the morning sight to noon if the ship were stationary.