Chapter 5

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