It is practically impossible to fix your position exactly by one observation of any celestial body. The most you can expect from one sight is to fix your line of position, i.e., the line somewhere along which you are. If, for instance, you can get a sight by sextant of the sun, you may be able to work out from this sight a very accurate calculation of what your latitude is. Say it is 50° N. You are practically certain, then, that you are somewhere in latitude 50° N, but just where you are you cannot tell until you get another sight for your longitude. Similarly, you may be able to fix your longitude, but not be able to fix your latitude until another sight is made. Celestial Navigation, then, reduces itself to securing lines of position and by manipulating these lines of position in a way to be described later, so that they intersect. If, for instance, you know you are on one line running North and South and on another line running East and West, the only spot where youcanbe onbothlines is where they intersect. This diagram will make that clear:
Intersection of NS and EW lines
Just what a line of position is will now be explained. Wherever the sun is, it must be perpendicularly above the same spot on the surface of the earth marked in the accompanying diagram by S
Line of Position
and suppose a circle be drawn around this spot as ABCDE. Then if a man at A takes an altitude, he will get precisely the same one as men at B, C, D, and E, because they are all at equal distances from the sun, and hence on the circumference of a circle whose center is S. Conversely, if several observers situated at different parts of the earth's surface take simultaneous altitudes, and these altitudes are all the same, then the observers must all be on the circumference of a circle andonly onecircle. If they are not on that circle, the altitude they take will be greater or less than the one in question.
Observers on circumference of one circle
Now such a circle on the surface of the earth would be very large - so large that a small arc of its circumference, say 25 or 30 miles, would be practically a straight line.
Suppose S to be the point over which the sun is vertical and GF part of the circumference of a circle drawn around the point. Suppose you were at B and from an altitude of the sun, taken by sextant, you worked out your position. You would find yourself on a little arc ABC which, for all purposes in Navigation, is a straight line at right angles to the true bearing of the sun from the point S. You can readily see this from the above diagram. Suppose your observer is at H. His line is GHI, which is again a straight line at right angles to the true bearing of the sun. He is not certain he is at H. He may be at G or I. He knows, however, he is somewhere on the line GHI, though where he is on that line he cannot tell exactly. That line GHI or ABC or DEF is the line of position and such a line is called a Sumner Line, after Capt. Thomas Sumner, who explained the theory some 45 years ago. Put in your Note-Book:
Any person taking an altitude of a celestial body must be, for all practical purposes, on a straight line which is at right angles to the true bearing of the body observed.
It should be perfectly clear now that if the sun bears due North or South of the observer, i.e., if the sun is on the observer's meridian, the resulting line of positionmustrun due East and West. In other words it is a parallel of latitude.And that explains why a noon observation is the best of the day for getting your latitude accurately. Again, if the sun bears due East or West the line of position must bear due North and South. And that explains why a morning or afternoon sight - about 8-9 A.M. or 3-4 P.M., if the sun bears either East or West respectively, is the best time for determining your North and South line, or longitude.
Now suppose you take an observation at 8 A.M. and you are not sure of your D.R. latitude. Your 8 A.M. position when the sun was nearly due East, will give, you an almost accurate North and South line and longitude. Suppose that from 8 A.M. to noon you sailed NE 60 miles. Suppose at noon you get another observation. That will give you an East and West line, for then the sun bears true North and South. An East and West line is your correct latitude. Now you have an 8 A.M. observation which is nearly correct for longitude and a noon position which is correct for latitude. How can you combine the two so as to get accurately both your latitude and longitude? Put in your Note-Book:
Through the 8 A.M. position, draw a line on the chart at right angles to the sun's true bearing. Suppose the sun bore true E ½ S. Then your line of position would run N ½ E. Mark it 1st Position Line.
Position Lines
Now draw a line running due East and West at right angles to the N-S noon bearing of the sun and mark this line Second Position Line. Advance your First Position Line the true course and distance sailed from 8 A.M. to noon, and through the extremity draw a third line exactly parallel to the first line of position. Where a third line (the First Position Line advanced) intersects the Second Position Line, will be your position at noon. It cannot be any other if your calculations are correct. You knew you were somewhere on your 8 A.M. line, you know you are somewhere on your noon line, and the only spot where you can be on both at once is the point where they intersect. You don't necessarily have to wait until noon to work two lines. You can do it at any time if a sufficient interval of time between sights is allowed. The whole matter simply resolves itself into getting your two lines of position, having them intersect and taking the point of intersection as the position of your ship.
There is one other way to get two lines to intersect and it is one of the best of all for fixing your position accurately. It is by getting lines of position by observation of two stars. If, for instance, you can get two stars, one East and the other West of you, you can take observations of both so closely together as to be practically simultaneous. Then your Easterly star would give you a line like AA' and the westerly star the line BB' and you would be at the intersection S.
Intersection of star observations
Assign for reading: Articles in Bowditch 321-322-323-324. Spend the rest of the period in getting times from the N. A., getting true altitudes from observed altitudes, working examples in Mercator sailing, etc.
A meridian altitude is an altitude taken when the sun or other celestial body observed bears true South or North of the observer or directly overhead. In other words, when the celestial body is on your meridian and you take an altitude of the body by sextant at that instant, the altitude you get is called a meridian altitude. In the case of the sun, such a meridian altitude is at apparent noon. Nowlatitude is always secured most accurately at noon by means of your meridian altitude. The reason for this was explained in yesterday's lecture. The general formula for latitude by meridian altitude is (Put in your Note-Book):
Latitude by meridian altitude = Zenith Distance (ZD) ± Declination (Dec).
Zenith distance is the distance in degrees, minutes and seconds from your zenith to the center of the observed body. For simplicity's sake, we will consider the sun only as the observed body. Then the zenith distance is the distance from your zenith to the center of the sun. Now suppose that you and the sun are both North of the equator and you are North of the sun. If you can determine exactly how far North you are of the sun and how far North the sun is of the equator, you will, by adding these two measurements together, know how far North of the equator you are, i.e., your latitude. As already explained, the declination of the sun is its distance in degrees, minutes and seconds from the equator and the exact amount of declination is, of course, corrected to the proper G.M.T. Your zenith distance is the distance in the celestial sphere you are from the sun. You know that it is 90° from your zenith to the horizon. Your zenith distance, therefore, is the difference between the true meridian altitude of the sun, obtained by your sextant, and 90°. Hence, having secured the true meridian altitude of the sun, you have only to subtract it from 90° to find your zenith distance, i.e., how far you are from the sun. This diagram will make the whole matter clear:
Find Zenith Distance from Meridian Altitude
A = Zenith, B = Sun, C = Horizon.
The arc ABC measures 90°. That is the distance from your zenith to the horizon. Now if BC is the true meridian altitude of the sun at noon, 90°-BC or AB is your zenith distance. If BC measures by sextant 60°, AB measures 90°-60° or 30°. This 30° is your Zenith Distance. Now suppose that from the Nautical Almanac we find that the G.M.T. corresponding to the time at which we measured the meridian altitude of the sun shows the sun's declination to be 10° N. Well, if you are 30° North of the sun, and the sun is 10° North of the equator, you must be 40° North of the equator or in latitude 40° N. For that is all latitude is, namely, the distance in degrees, minutes and seconds you are due North or South of the equator. That is the first and simplest case.
Another case is when you are somewhere in North latitude and the sun's declination is South. Then the situation would, roughly, look like this:
Observer in N latitude, Sun's declination is S
BC = Altitude of the sun, AB = Zenith Distance and DB = Sun's Declination.
In this case, your distance North of the equator AD would be your zenith distance AB minus the sun's declination DB. This diagram is not strictly correct, for the observer's position on the earth 0 appears to be South of the equator instead of North of the equator. That is because the diagram is on a flat piece of paper instead of on a globe. So far as illustrating the Zenith Distance minus the Declination, however, the diagram is correct. The last case is where you are, say, 10° N of the sun (your zenith Distance is 10°) and the sun is in 20° S declination. In that case you would have to subtract your zenith distance from the sun's declination to get your latitude, for the sun's latitude (its declination) is greater than yours.
Now from these three cases we deduce the following directions, which put in your Note-Book:
Begin to measure the altitude of the sun shortly before noon. By bringing its image down to the horizon, you can detect when its altitude stops increasing and starts to decrease. At that instant the sun is on your meridian, it is noon at the ship, and the angle you read from your sextant is the meridian altitude of the sun. To work out your latitude, name the meridian altitude S if the sun is south of you and N if north of you.
Correct the observed altitude to a true altitude by Table 46. If the altitude is S, the Zenith Distance is N or vice versa. (Note to Instructor: If the sun is South of you, you are North of the sun and vice versa.)
Correct the declination for the proper G.M.T. as shown by chronometer (corrected). If zenith distance and declination are both North or both South, add them and the sum will be the latitude, N or S as indicated. If one is N, and the other S, subtract the less from the greater and the result will be the latitude in, named N or S after the greater. Example:
At sea June 15th, observed altitude ofCircle with line under71° 15´ S, IE - 47´, HE 25 ft. CT 3h 34m 15s P.M. Required latitude of ship.
Assign for Night Work or to be worked in class room such examples as the following:
1. June 1st, 1919.Circle with line under33° 50' 00" S. G.M.T. 8h 55m 44s. HE 20 ft. IE + 4' 3". Required latitude in at noon.
2. April 2nd, 1919.Circle with line under12° 44' 30" N. CT was 2d 5h 14m 39s A.M., which was 1m 40s slow on March 1st (same CT) and 4m 29s fast on March 15th (same CT). IE - 2' 20". HE 22 ft. Required latitude in at noon.
Assign for Night Work reading also, the following Articles in Bowditch: 344 and 223.
This is a peculiar word to spell and pronounce but its definition is really very simple. Put in your Note-Book:
The azimuth of a heavenly body is the angle at the zenith of the observer formed by the observer's meridian and a line drawn to the center of the body observed. Azimuths are named from the latitude in and toward the E in the A.M. and from the latitude in and toward the W in the P.M.
All this definition means is that, no matter where you are in N latitude, for instance, if you face N, the azimuth of the sun will be the true bearing of the sun from you. The same holds true for moon, star or planet, but in this lecture we will say nothing of the star azimuths for, in some other respects, they are found somewhat differently from the sun azimuths. Put this in your Note-Book:
To find an azimuth of the sun: Note the time of taking the azimuth by chronometer. Apply chronometer correction, if any, to get the G.M.T. Convert G.M.T. into G.A.T. by applying the equation of time. Convert G.A.T. into L.A.T. by applying the longitude in time. The result is L.A.T. or S.H.A. With the correct L.A.T., latitude and declination, enter the azimuth tables to get the sun's true bearing, i.e., its azimuth. Example:
March 15th, 1919. CT 10h 4m 32s. D.R. latitude 40° 10' N, longitude 74° W. Find the TZ.
We will take up later a further use of azimuths to find the error of your compass. Right now all you have to keep in mind is what an azimuth is and how you apply the formulas already given you to get the information necessary to enter the Azimuth Tables for the sun's true bearing at any time of the astronomical day when the sun can be seen. In consulting these tables it must be remembered that if your L.A.T. or S.H.A. is, astronomically, 20h (A.M.), you must subtract 12 hours in order to bring the time within the scope of these tables which are arranged from apparent six o'clock A.M. to noon and from apparent noon to 6 P.M. respectively.
We are taking up sun azimuths today in order to get a thorough understanding of them before beginning a discussion of the Marc St. Hilaire Method which we will have tomorrow. You must get clearly in your minds just what a line of position is and how it is found. Yesterday I tried to explain what a line of position was, i.e., a line at right angles to the sun's or other celestial body's true bearing - in other words, a line at right angles to the sun's or other celestial body's azimuth. Today I tried to show you how to find your azimuth from the azimuth tables for any hour of the day. Tomorrow we will start to use azimuths in working out sights for lines of position by the Marc St. Hilaire Method.
Note to Instructor: Spend the rest of the time in finding sun azimuths in the tables by working out such examples as these:
1. April 29th, 1919. D.R. latitude 40° 40' N, Longitude 74° 55' 14" W. CT 10h 14m 24s. CC 4m 30s slow. Find TZ.
2. May 15th, 1919. D.R. latitude 19° 20' S, Longitude 40° 15' 44" E. CT 10h 44m 55s A.M. CC 3m 10s fast. Find TZ.
Note to Instructor:
If possible, give more examples to find TZ and also some examples on latitude by meridian altitude.
Assign for Night Work reading the following Articles in Bowditch: 371-372-373-374-375. Also, examples to find TZ.
You have learned how to get your latitude by an observation at noon. By the Marc St. Hilaire Method, which we are to take up today, you will learn how to get a line of position, at any hour of the day. By having this line of position intersect your parallel of latitude, you will be able to establish the position of your ship, both as to its latitude and longitude.
Now you have already learned that in order to get your latitude accurately, you must wait until the sun is on your meridian, i.e., bears due North or South of you, and then you apply a certain formula to get your latitude. When the sun is on or near the prime vertical (i.e., due East or West) you might apply another set of rules, which you have not yet learned, to get your longitude. By the Marc St. Hilaire method, the same set of rules apply for getting a line of position at any time of the day, no matter what the position of the observed body in the heavens may be. Just one condition is necessary, and this condition is necessary in all calculations of this character, i.e., an accurate measurement of the observed body's altitude is essential.
What we do in working out the Marc St. Hilaire method, is to assume our Dead Reckoning position to be correct. With this D. R. position as a basis, we compute an altitude of the body observed. Now this altitude would be correct if our D. R. position were correct and vice versa. At the same time we measure by sextant the altitude of the celestial body observed, say, the sun. If the computed altitude and the actual observed altitude coincide, the D. R. position is correct. If they do not, the computed altitude must be corrected and the D. R. position corrected to coincide with the observed altitude. Just how this is done will be explained in a moment. Put in your Note-Book:
Formula for obtaining Line of Position by M. St. H. Method.
I. Three quantities must be known either from observation or from Dead Reckoning.
II. Add together the log haversine of the S.H.A. (Table 45), the log cosine of the Lat. (Table 44), and the log cosine of the Dec. (Table 44) and call the sum S. S is a log haversine and must always be less than 10. If greater than 10, subtract 10 or 20 to bring it less than 10.
III. With the log haversine S enter table 45 in the adjacent parallel column, take out the corresponding Natural Haversine, which mark NS.
IV. Find the algebraic difference of the Latitude and Declination, and from Table 45 take out the Natural Haversine of this algebraic difference angle. Mark it ND±L
V. Add the NSto the ND±L, and the result will be the Natural Haversine of the calculated zenith distance. Formula NZD= NS+ ND±L
VI. Subtract this calculated zenith distance from 90° to get the calculated altitude.
VII. Find the difference between the calculated altitude and the true altitude and call it the altitude difference.
VIII. In your Azimuth Table, find the azimuth for the proper "t," L and D.
IX. Lay off the altitude difference along the azimuth either away from or toward the body observed, according as to whether the true altitude, observed by sextant, is less or greater than the calculated altitude.
Altitude Difference along the Azimuth
X. Through the point thus reached, draw a line at right angles to the azimuth. This line will be your Line of Position, and the point thus reached, which may be read from the chart or obtained by use of Table 2 from the D. R. Position, is the nearest to the actual position of the observer which you can obtain by the use of any method from one sight only.
Example:
At sea, May 18th, 1919, A.M.Circle with line under29° 41' 00". D.R. Latitude 41° 30' N, Longitude 33° 38' 45" W. WT 7h 20m 45s A.M. C-W 2h 17m 06s CC + 4m 59s. IE - 30". HE 23 ft. Required Line of Position and most probable position of ship.
As azimuth is N 90° E, Line of Position runs due N & S (360°) through Lat. 41° 30' N. Lo. 33° 30' 09" W.
Assign for work in class and for Night Work examples such as the following:
1. July 11th, 1919.Circle with line under45° 35' 30", Lat. by D. R. 50° 00' N, Lo. 40° 04' W. HE 15 ft. IE - 4'. CT (corrected) 5h. 38m 00s P.M. Required Line of Position by Marc St. Hilaire Method and most probable fix of ship.
2. May 16th, 1919, A.M.Circle with line under64° 01' 15", D. R. Lat. 39° 45' N, Lo. 60° 29' W. HE 36 ft. IE + 2' 30". CT 2h 44m 19s. Required Line of Position by Marc St. Hilaire Method and most probable fix of ship.
Etc.
1. Nov. 1st, 1919. A.M. at ship. WT 9h 40m 15s. C-W 4h 54m 00s. D. R. Lat. 40° 50' N, Lo. 73° 50' W.Circle with line under27° 59'. HE 14 ft. Required Line of Position by Marc St. Hilaire Method and most probable position of ship.
2. May 30th, 1919. P.M. at ship. D. R. Lat. 38º 14' 33" N, Lo. 15° 38' 49' W. The mean of a series of observations ofCircle with line underwas 39° 05' 40°. IE - 01' 00". HE 27 ft. WT 3h 4m 49s. C-W 1h 39m 55s. C.C. fast, 01m 52s. Required Line of Position by Marc St. Hilaire Method, and most probable position of ship.
3. Oct. 21st, 1919, A.M. D. R. Lat. 40° 12' 38" N, Lo. 69° 48' 54" W. The mean of a series of observations ofCircle with line underwas 19° 21' 20". IE + 02' 10". HE 26 ft. WT 7h 58m 49s. C-W 4h 51m 45s. C. slow, 03m 03s. Required Line of Position by Marc St. Hilaire Method and most probable position of ship.
4. June 1st, 1919, P.M. at ship. Lat. D. R. 35° 26' 15" S, Lo. 10° 19' 50" W. W.T. 3h 30m 00s. C-W 0h 20m 38s. CC 1m 16s slow.Circle with line under16° 15' 40". IE + 2' 10". HE 26 ft. Required Line of Position and most probable fix of ship.
5. Jan. 5th, 1919. A.M. D. R. Lat. 36° 29' 38" N, Lo. 51° 07' 44" W. The mean of a series of observations ofCircle with line underwas 23° 17' 20". IE + 01' 50". HE 19 ft. WT 7h 11m 37s. C-W 5h 59m 49s. C. slow 58s. Required Line of Position and most probable fix of ship.
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 intheir 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 wordimplies 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 starmay 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?
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.
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.?
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.
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 polestar. 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.