TIME-ALTITUDE-AZIMUTH, EX-MERIDIAN AND PRIME-VERTICAL DIAGRAMS.
For Identifying Unknown Stars, Finding the Hour Angle, Altitude, and Azimuth of a Known Body, Hour Angles and Times of Rising and Setting and Crossing the Prime-Vertical; the Compass Error, Latitude and Longitude, and also to facilitate the Plotting of Sumner's Lines on the Chart, and for finding the Great Circle Course and Distance.
1. To identify an unknown star in cloudy weather we have given, or may obtain:
(1) The Greenwich mean time.
(2) The ship's position by D. R.
(3) The star's observed altitude.
(4) The star's observed azimuth.
Find the true altitude and the true azimuth, and the local sidereal time.
In the spherical triangle ABC, if we have given the two sides a, b and the angle C, we may find A and B from formulas (1) and (2) and then c from (3) and (4).
2. Captain Lecky's "General Utility—A, B, C, and D Tables" may be used for finding the hour angle of a star from its altitude and azimuth, but not its declination.
F. B. S. Lawlor, R. N. R., has shown how certain tables in None's Epitome may be used for the identification of an unknown star by its altitude and azimuth.
3. Realizing the importance of this problem to the practical navigator, the writer suggested that the time-azimuth tables issued by the Navy Department be modified as shown in the enclosed table, in which for each degree of latitude and declination is given the altitude as well as the azimuth corresponding to the given hour angles. This can be done by means of the formula sin b = cos (L—d)-2 cos L cos d sin 1/2 t. Referring to the example on page I—Entering the table with the given altitude 29°30' and azimuth N. 740 W., the hour angle 4 hrs. 50 min and declination 28° 30' N. are taken out at once.
4. Among the valuable works on Navigation published by A. C. Johnson, R. N., is "Combined Time and Altitude-Azimuth Tables," which from the name would seem to be just such tables as those proposed, but upon examination they prove quite different, namely, for finding the azimuth of a known body in high latitudes.
5. The graphical method by Captain C. D. Sigsbee, U. S. Navy, is a good one, but it is believed that for the solution of this problem with the greatest speed and accuracy, the Time-Altitude- Azimuth Diagrams will be found more convenient than any other way. These are based on the graphical solution of the spherical triangle by what may be called Tri-rectangular Coordinates.
6. In the astronomical triangle PMZ, suppose L and d constant while t and h vary, and if for the different values of t we compute the corresponding values of h and Z, then we can construct two curves, one having hour angles as ordinates and altitudes as abscissas, while the other has hour angles as ordinates and azimuths as abscissas, the common ordinate forming the connecting link between the two.
7. Referring to Sheet I, the ordinates represent hour angles and the horizontal index lines are drawn at intervals of 10 min. of hour angle. The time-altitude curves, shown by the full lines in the left hand lower corner of the diagram, are laid off with hour angles as ordinates and the corresponding altitudes as abscissas. One of these curves is drawn for each degree of declination used. The time-azimuth curves shown by the dotted lines in the center of the diagram are plotted with the hour angles as ordinates and the corresponding azimuths as abscissas. One of these curves is also drawn for each degree of declination used.
8. The curves shown by the full lines extending from the right hand lower corner of Sheet I to the top of the diagram are the Portions of the time-altitude curves near the meridian expanded on a large scale for declinations 0° and 23°, and the abscissas represent the difference in minutes to be added to the observed ex-meridian altitude to obtain the observed meridian altitude, at the place of observation, which altitude or rather the latitude found from it must be corrected for the run of the ship to noon.
9. From an examination of the diagrams on Sheet I it will be seen that the following problems may be readily and accurately solved:
Problem 1.—To identify an unknown star.
Any horizontal index line cuts all of the time-altitude and time-azimuth curves, and perpendiculars from the points of its intersection with these curves (corresponding to the same declination, and these are all numbered so as to be readily identified) to the base line will intercept on the base line the altitude and azimuth of the body having the given declination and the assumed hour angle, hence in order to identify an unknown star: At the same time that the altitude is observed take a compass bearing of the star and note the time by watch. Reduce the observed altitude to true, find the true azimuth and the local sidereal time.
10. On the base line lay off the true altitude and azimuth, and at these points erect perpendiculars to the base line, or follow up the sheet along the vertical lines with the forefinger of each hand, until an index line is reached on which each perpendicular intersects its corresponding curve of the same degree of declination. We thus find the hour angle and declination corresponding to the given altitude and azimuth. Apply the hour angle to the local sidereal time for the star's right ascension and with this and the declination taken from the curves used, take the star from the list in the Nautical Almanac.
11. Example: At sea, Feb. 26, 1901, about 5.30 p. m.; G. M. T. 9 hrs. 36 m. 00 s., Lat. 35° 00 N., Long., 6o° 00 W. by D. R. Observed the altitude of a star (2nd mag.) through a rift in the clouds 200 (true) bearing N. 86° 30' W. (true). Find the star.
Lay off the altitude and azimuth on the base line and follow up the vertical lines at these points and we find that the altitude and azimuth perpendiculars intersect on the index line corresponding to 5 hrs. oo min. on the altitude and azimuth curves of 14° declination, hence
* H. A.. 5 hrs. oo min.
* dec. = 14° oo'
Applying the hour angle to the L. S. T. we obtain—
* R. A. 22-59-08 and dec. 14° N., hence from the star list we find the star must be a Pegasi (Markab).
In the same way the example given on page I can be solved and we find—
* H. A.= 4 hrs. 50 min., * dec. 28° N. and * R. A.-=o hrs. 04 min.
These results are almost exactly the same as those obtained by solving the problem by trigonometry; by increasing the scale of the diagrams even greater accuracy may be expected.
12. PROBLEM II.—To find the hour angle, altitude or azimuth of a known body at any given time.
If any one of these is known or can be obtained the other two may be found directly from the diagram. Take an altitude or azimuth of the star and its azimuth or altitude and hour angle may be found front Sheet I. Or if it be desired to locate some particular star at a certain time, find its hour angle at this time, and from Sheet I find its altitude and azimuth, which placed on the sextant and pelorus will enable the observer to pick up the star at once with the minimum amount of calculation.
13. PROBLEM III.—To find the time of rising and setting. In practice the altitudes will be laid off from o° to 9o° and the azimuths from o° to 18o°. This was not done on the sample Sheet I for lack of space. The points where the altitude curves cut the ordinate at o° will give the hour angle of rising and setting, and this applied to the time of meridian passage will give the times of rising and setting. The azimuths at rising and tin may be obtained from the intersection of the index lines through these points with the azimuth curves.
14. PROBLEM IV.—To find the approximate altitude and hour angle of a body on the prime vertical.
This problem is useful in constructing the prime vertical diagram for finding the longitude. Erect a perpendicular to the base line at 9o°. The ordinate of the intersection of this perpendicular with the azimuth curve corresponding to the declination of the body will give its approximate hour angle and the intersection of the index line through this point with the altitude curve will give its altitude. Thus the altitude of a Andromeda when on the prime vertical is about 540 and its hour angle about 2 hours 44 min.
15. PROBLEM V.—To find the compass error.
Sheet I may be used as either a time-azimuth diagram for the sun or the stars or an altitude-azimuth diagram as is most convenient.
Example : Feb. 26, 19o1. At sea 7.00 p. m. observed altitude of a Andromeda 24° (true) bearing by compass N. 76° W. Find the compass error.
16. PROBLEM VI.—To find the latitude—(1) by ex-meridian altitudes.
Example: At sea March 21, 1902, Lat. 350 oo' by D. R., Long. 150 00, W.
Observed altitude of 54° 30', L. A. T. ii hrs. 40 min., 30 sec. from A. M. time sight, corrected for run. Entering the right hand margin Sheet I with the sun's hour angle 19 min. 30 sec. from the curve for declination o° (sun's Dec. March 21) we find the difference o° 18' to be added to the observed altitude to obtain the meridian altitude at the place of observation, and the latitude found from it must be corrected for the run of the ship to noon.
17. PROBLEM VII.—To find the latitude (2) by star double altitudes.
This method is described by A. C. Johnson, R. N., as follows: If two altitudes be taken, and the hour angles worked out with the latitude by D. R., their sum, if on opposite sides of the meridian, or their difference, if on the same side, should, if the latitude is correct, agree with the difference of their right ascensions; if there is any discrepancy it shows that the latitude used is incorrect. (For complete description of this method see "Handbook for Star Double Altitudes," by A. C. Johnson, R. N., published by J. D. Potter, II king Street, Tower Hill, E., London.)
18. PROBLEM VIII.—To find the latitude (3) given the altitude and hour angle of a known star.
19. PROBLEM IL—To find the exact hour angle and the longitude.
The hour angle and hence the longitude may be accurately found with the aid of the hour angle or prime vertical diagrams shown on Sheet II.
Here the portion of the time-altitude curve near the prime vertical is expanded on a large scale, as was done with the e:c' meridian altitude curves. As near the prime vertical the altitude varies uniformly with the hour angle, this part of the curve is a straight line, and it is only necessary to establish two points within the limits of the diagram, though a third should be established as a check. Of course the hour angle diagram may be plotted at a distance from the prime vertical, though more points will have to be found to establish the curve. This diagram will be found very convenient for navigators making long passages on nearly east and west courses, as the same curves may then be used from day to day, and the hour angle taken directly from the diagram without any calculation whatever, and with the salve degree of accuracy as though it were calculated in the usual way. This will be demonstrated by some examples given later. This diagram is particularly convenient for use with stars as the declination is practically constant, but it may be used also with the sun at the sacrifice of a very little additional time and labor than is necessary with a star.
Sheet I will be found very convenient for checking the hour angles when working time sights in the usual way.
20. About 4.00 p. m., May 20, 1902. In Lat. 350 10' N., find the hour angle of the sun west of the meridian, true altitude of .35° 43' and corrected Dec. i90 47' N.
Referring to Sheet I, it is seen that at this altitude the sun is near the prime vertical and to make a diagram that may be used from day to day when desired, with Lat. 350 N. and Dec. 19 00' N., compute the hour angles for altitudes 330 oo' and 36 00’ N these are 4 hrs. o9 min. 55 sec. and 3 hrs. 55 min. 16 sec., respectively. Plot the points A and B and draw a straight line between them, Sheet II. If another point be established with any intermediate altitude this point will be found to fall exactly on the line AB. With altitudes 34° 00’ and 36° oo' and Dec. 20° 00' N In the same way establish the points C and D and draw the line CD. The distance EF between AB and CD is due to the difference in declination of one degree. Take EF in the dividers and apply it to the scale of minutes at the bottom.
21. The fact that the hour angles taken from the diagrams do not in all cases agree with the computed values of the same more closely is due to inaccuracies in laying off the diagrams and the smallness of the scale, as the principle on which the diagrams are constructed is correct. The blank sheets for the prime vertical or hour angle diagrams should be constructed and engraved with the utmost care and accuracy on a scale sufficiently large to enable the hour angle to be taken from it to the second, the sheet being of sufficient size to include at least half an hour of hour angle. A Mercator chart on a large scale for low latitudes makes all excellent sheet for plotting the hour angle diagram and the writer has obtained results in this way which agreed with the computed angles to a second. In practice a number of these blank diagrams (Sheet II) engraved on good hard paper should be furnished. On these the time-altitude curves would ordinarily be drawn in pencil, and erased when of no further use, to make room for new ones, though it will sometimes be found convenient to ink these in for the stars. The lines indicating the hour angles, altitudes and declinations should be marked as shown on Sheet II. The hour angle and declination are marked near the time-altitude curves and the altitudes written over their corresponding declinations, thus 11. i mean alt. 350 for dec. 190, and alt. 13° for dec. 50, etc. In this way a number of curves may be drawn on a single sheet without any risk of confusion. From an examination of Sheet I it will be seen that the time-altitude curve is in many eases a straight line for some distance on either side of the prime vertical, and is thus available for plotting on a large scale for use ill finding the hour angle.
22. PROBLEM X.—To find the great circle course and distance between two given points.
It is evident that the problems of great circle sailing may be readily solved by means of the time-altitude-azimuth diagrams by simply renaming the parts of the triangle and diagrams.
23. PROBLEM XI.—To find the latitude of the vertex.
From P draw PV perpendicular to the Great Circle Track, then V is the vertex, and in the right triangle APV we have given:
PVA = 9o°; AP =90°— L; and PAV = course from A found by Problem X. Then we have sin PV = sin A cos L, Where L = latitude of A.
24. PROBLEM XII.—To find the meridian of the vertex and its distance from the point of departure.
Enter the latitude of the place of departure with the initial course as an azimuth, then the intersection of the ordinate through this point with the azimuth curve of the same declination as the latitude of the vertex will give the index line whose corresponding hour angle will be the longitude of the vertex from the place of departure. The abscissa of the intersection of this index line with the altitude curve of the same declination as the latitude of the vertex will give the distance to the meridian of the vertex from the place of departure.
25. PROBLEM XIII.—To find the latitudes and longitudes of any number of points in order to plot the track on a chart, having given the latitude and longitude of the vertex.
26. PROBLEM XIV.—When the point of departure and the destination are on opposite sides of the equator, to find the course and distance and the longitude of the intersection of the great circle with the equator.
Find the latitude of the vertex by Problem XI. This latitude is equal to the inclination of the plane of the great circle to the equator or course from the intersection towards B = 9o°— L (v). Then in latitude o° with 9o°— L(v) as an azimuth, the intersection of the ordinate at this point with the curve of the same declination as the latitude of the destination (B), will determine the index line of the hour angle which is the longitude of the intersection from the destination. By using the latitude of the departure (A) and 9o° ± L(v) as an azimuth, the longitude from A can be found, entering in this case the sheet of same latitude as A—declination contrary name—and using curve for o° declination.
27. From the above it will be seen that the Time Altitude Azimuth and Ex-meridian and Prime Vertical or Hour Angle Diagrams represent graphically and with mathematical accuracy the paths of the different heavenly bodies whose declinations are used, both in altitude and azimuth, from rising to meridian passage or from meridian passage to setting. In addition to identifying an unknown star with ease and accuracy, and solving all the ordinary problems of astronomical navigation, the most important problems of Great Circle sailing may be quickly solved with the aid of these curves.
28. The diagrams are constructed on sheets 15" x 20", one for each degree of latitude from 0° to 6o° for latitude and declination same and contrary name, with declinations from o° to 62°, including those of the principal stars of the first and second magnitude.