Chapter IV: Part 4

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.

3. From the course to noon and speed of the ship, figure the change in longitude per hour in terms of seconds of time. For instance, suppose a ship were steaming a course of 275 deg. at the rate of 11 knots per hour in approximately 38 deg. North latitude. The change of longitude per hour for this speed would be 14' of arc or 56s of time.

4. Now the sun travels at the rate of 60 minutes or 3600 seconds per hour. To this hourly speed of the sun must be added or subtracted the hourly speed of the ship according as to whether the ship is going in an easterly or westerly direction. If, as mentioned above, the ship is steaming a course of 275 deg. (W 1/2 N) and hence changing its longitude at the rate of 56s per hour, then the net rate of approach of the sun per hour would be 3600s - 56s, or 3544s per hour.

5. Divide the total time to noon from the L.A.T. of the morning sight (expressed in seconds of time) by the net rate of approach of the sun per hour. The result will be the corrected time to noon--i.e. the time at which the sun will be on the ship's meridian when the ship is changing its longitude to the westward at the rate of 56s per hour.

6. One more step is necessary. To the watch time of the morning sight, add the corrected time to noon. The result will be the watch time of Local Apparent Noon. Thirty minutes before will be the watch time of 11:30 A.M. and at 11:30 A.M. all deck clocks should be set to the local apparent time of the place the ship will be at local apparent noon.

The following example illustrates the explanation just given and should be put in your Note Book:--

Example:--At sea, August 7th, 1919. About 7:30 A.M. by ship's time, position by observation just found to be Latitude 30 deg. 05' N, Longitude 58 deg. 08' W. WT of morning sight 6h-53m-13s A.M. C-W 4h-37m-21s. CC + 3m-38s. Course 275 deg.. Speed 11 knots. TZ N 90 deg. E. What will be the Watch Time of Local Apparent Noon?

WT 6h -- 53m -- 13s A.M.
+ 12
-------------------
18 -- 53 -- 13
C-W 4 -- 37 -- 21
-------------------
CT 23 -- 30 -- 34
CC + 3 -- 38
-------------------
G.M.T. 23 -- 34 -- 12
Eq. T. -- 5 -- 42
-------------------
G.A.T. 23 -- 28 -- 30
Lo. in T. 3 -- 52 -- 32
-------------------
L.A.T. 19 -- 35 -- 58
24 -- 00 -- 00
-------------------
Total time to Noon 4h 24m 02s

Course -- 275 deg.
Change in Lo. per hr.-- 14', 56s.

3600s
-56
-----
3544s, Net rate of approach of sun

4h
60
----
240m
+ 24
-----
264m x 60 = 15840s

15840s
+ 02
-----
15842s, Total time to noon.
3544) 15842 (4.47 hours
14176
-----
16660
14176
-----
24840
24808
-----

Corrected time to Noon 4h -- 28m -- 12s
WT of A.M. sight 6h -- 53m -- 13s
-----------------
WT of L.A.N. 11h -- 21m -- 25s
WT of 11:30 A.M. 10h -- 51m -- 25s

When, therefore, the watch reads 10h--51m--25s, the deck clocks should be set to 11.30 A.M. and thirty minutes later it will be apparent noon at the ship.

In all these calculations it is taken for granted that the speed of the ship and hence the change in longitude can be gauged accurately. A check on this can be made by comparing the longitude of the A.M. sight with the D.R. longitude of the same time. Any appreciable difference between the two can be ascribed to current. Now, if a proportionate amount of current is allowed for in reckoning the speed of the ship from the time of the A.M. sight to noon, then a proper correction can be made in the net rate of approach of the sun and the corrected time to noon will be very close to the exact time of noon. Of course there will be an error in this calculation but it will be small and the result gained will be accurate enough for ordinary work.

So much for finding the watch time of Local Apparent Noon. Careful navigators carry the process further and get the watch times of 15, 10 and 5 minutes before noon, so that by the use of constants for each one of these times, an accurate check on the noon latitude can be quickly and easily secured. We have not time in this course to explain how these constants are worked out but it is well worth knowing. The information regarding it is in Bowditch Art. 325, p. 128, and Art. 405, p. 181.

A word about the watch used by the navigator should be included here. This watch should be a good one and receive as much care, in its way, as the chronometer. It should be wound at the same time every day, carefully handled and, in other respects, treated like the fine time-piece that it is.

While authorities differ on this point, the best practice seems to be not to change the navigator's watch to correspond with the apparent time of each day's noon position. The reason for this is two-fold. First, because constant moving of the hands will have an injurious effect on the works of the watch, and second, because, by not changing the watch, the C-W remains approximately the same, and thus a good check can be kept on both the watch and the chronometer as well as on the navigator's figures in reckoning the times of his various sights.

Assign for night reading the following Arts. in Bowditch: 323, 324, 333. Also problems similar to the following:

1. At sea, July 28, 1919. Position by observation just found to be Latitude 44 deg. 58' N, Longitude 22 deg. 06' W. WT of morning sight 6h-02m-20s. CC 3m 34s slow. Course S 24 deg. W. TZ N 90 deg. E. Speed 9 knots. What will be the watch time of Local Apparent Noon?

2. At sea, August 9th, 1919. Position by observation just found to be Latitude 38 deg. 48' N, Longitude 70 deg. 46' W. WT of morning sight 8h-15m-01s A.M. C-W 3h-56m-32s. CC 3m-43s slow. Course 272 deg.. Speed 12 knots. TZ N 90 deg. E. What will be the watch time of Local Apparent Noon?

WEEK VII--NAVIGATION

TUESDAY LECTURE

COMPASS ERROR BY AN AZIMUTH

The easiest and most accurate way to find the error of your compass is, first, to find the bearing of the sun by your pelorus. If you set your pelorus, so that it will exactly coincide with the course you are steaming as shown by the compass in your chart house and then get a bearing of the sun by noting where the shadow from the pelorus vane cuts the circumference, this bearing will be the bearing of the sun by compass. At the same time, get your true bearing of the sun from the Azimuth Tables. The difference between the two will be the compass error, marked East or West according to the following rule which put in your Note-Book:

1. Express your Compass Bearing and your True Bearing by NEW compass reading.

2. If TZ is to the right of CZ, C.E. is East. Formula: True--Right--East.

3. If TZ is to the left of CZ, C.E. is West. Formula: True--Left--West.

You must now remember that what you have is a Compass Error, consisting of both Variation and Deviation. To find the Deviation, the Variation and C.E. being given, is merely to apply the rules already given you under Dead Reckoning. For instance, if you had a C.E. of 10 deg. W and a Variation of 4 deg. E, the Deviation would be 14 deg. W.

Put this example in your Note-Book:

LAT 20h 59m 57s
Lat. 4 deg. 55' N
Dec. 10 deg. 39' 30" N

Ship heading N 11 deg. W. CB of (.) S 88 deg. E. Variation 10 deg. W. What was the ship's true course and Deviation of Compass on direction ship was heading?

CZ 92 deg. (New compass reading)
TZ 80 deg. (New compass reading)
---
CE 12 deg.

CE = 12 deg. W
Variation 10 deg. W
-----
Deviation 2 deg. W

True course being sailed N 23 deg. W or 337 deg..

Let us now work out some of the following examples:

1. L.A.T. 22h--14m--18s
Lat. 30 deg. 29' S
Dec. 17 deg. 28' 44" N
Ship heading S 84 deg. W
Compass Bearing 44 deg.
Variation 10 deg. W.

Required T.C. and Deviation on ship's loading.

2. August 29th, 1919. CT 2h 29m 18s A.M. Longitude 120 deg. 19' 46" E.
Latitude 44 deg. 14' N. Ship heading 98 deg..
Compass Bearing S 42 deg. E.
Variation 4 deg. E.

Required T.C. and Deviation on ship's heading.

3. June 17th, 1919. CT 4h 18m 44s A.M. Longitude 60 deg. 14' 59" E.
Latitude 38 deg. 48' 00" S. Ship heading SW x S.
Compass Bearing 40 deg.
Variation 12 deg. W.

Required T.C. and Deviation on ship's heading.

Etc.

WEDNESDAY LECTURE

CORRECTING LONGITUDE BY A FACTOR

We are now almost ready to begin the discussion of a day's work at sea. The only method we have not taken up is the one which is the subject of today's lecture. It is a method to correct your longitude to correspond with the difference between your latitude by Dead Reckoning and your latitude by observation.

Suppose you take a sight in the morning for longitude. The only latitude you can use is a D. R. latitude, advanced from your last known position. Now suppose you run until noon and at that time take a sight for latitude. In comparing your D. R. latitude, advanced the true course and distance steamed to noon, and your latitude by observation taken at noon, suppose there is a difference of several minutes. The question is--How can we correct our longitude to correspond with this error discovered in the latitude? This is the method which put in your Note-Book: