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° or 15°. 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° 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° N. your celestial horizon to the South will be 90° from 40° N. or 50° S. and to the North it will be 90° + 40° N. = 130° or 40° 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° N. latitude, the celestial equator would intersect the celestial sphere at a point 40° 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° 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° 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° 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° W.

.R.A.G.M.N. 17h 28m 26s

Corr. 74° 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° N., Longitude 74° W. WT 8h 30m P.M. Observed unknown star about 80° east of my meridian and 25° 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° N., Longitude 60° 30' W. Observed unknown star about 60° west of meridian and about 22° S. What was the star?

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

4. Aug. 3, 1919, P.M. at ship. In D.R. Latitude 37° 37' N. Longitude 38° 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° 10' N, Longitude 34° 38' W.

6. At what local mean time will the Star Arcturus transit on July 17th, 1919, in Latitude 45° 35' N., Longitude 28° 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° - 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° - true altitude).

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

Obs. Alt. 52° 36' S

Corr. - 5 08

—————

True Alt. 52° 30' 52" S

- 90 00 00

—————

Z.D. 37° 29' 08" N

Dec. 16 21 00 N

—————

Lat. 53° 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° 29' S. HE 32 ft. Required latitude of ship.

2. At sea, August 14th, 1919. Meridian altitude Star Vega, 60° 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° from the Pole Star to the celestial equator. Now, no matter where you stand, it is 90° 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° N of the equator. Then your northerly horizon would drop by 10° and the Pole Star would have an altitude of 10°. In other words, when you were in 10° N latitude, the pole star would measure 10° high by sextant. And so on up to 90°, 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° 08'. Just what part of that 1° 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° 49' 00" W. IE + 1' 10". HE 15 ft. Observed altitude Polaris 42° 21' 30" N. Required latitude in.

2. At sea, March 31st, 1919. In Lo. 160° 15' E. CT 7h 15m 19s. Observed altitude Polaris 38° 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:

G.M.T. + .R.A. + CP = G.S.T. (-W. Lo.) = 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° N, Lo. 45° 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° 14' N, Lo. 68° 46' W. Observed altitude Star Aldebaran East of meridian 50° 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° 44' 19" N. Lo. 121° 16' 14" E. Observed altitude Star Rigel 28° 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° 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° 29' 46" W. Observed altitude of Star Polaris 48° 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° 44' N, Longitude 53° 13' 44" E. Observed altitude of Star Altair East of meridian, 52° 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°, if Lat. and Dec. are of same name. If Lat. and Dec. are of opposite name, secure P. D. by adding Dec. to 90°.)

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° 57' 00". IE - 1' 30". HE 32 ft. D.R. Lat. 4° 55' 32" N. Required longitude in at time of observation.

26d 2h 29m 03s A.M.

- 12

—————————

CT 25d 14h 29m 03s - 90° 00' 00"

CC+ +16 08 Dec. 10 49 48

————————— ———————

G. M. T. 25d 14h 45m 11s P. D. 79° 10' 12"

Eq. T. -2 05

—————————

G. A. T. 25d 14h 43m 06s - 1' 30"

+ 9 27

44° 57' 00" ————

Corr. 7 57 Corr. + 7' 57"

—————

45° 04' 57"

Lat. 4 55 32 N sec. .00160

P. D. 79 10 12 cosec. .00781

——————

2) 129° 10' 41"

——————

½ S 64° 35' 20" cos. 9.63266 -9

45 04 57

——————

Diff. 19° 30' 23" sin. 9.52350 -14

——————

9.16557 +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° 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° 15' 00". G.M.T. 1d 11h 30m 00s A.M. D.R. Lat. 40° 30' N. IE - 2' 20". HE 25 ft. Required longitude in.

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

3. May 27, 1919. P.M. Lat. by D.R. 40° 55' N. 34° 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° 41' 15". WT 7h 20m 45s. C-W 2h 17m 06s. CC 4m 59s slow. Latitude by D.R. 41° 33' N. IE - 1' 30". HE 23 ft. Required longitude in.

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

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

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

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