Refraction 48" - paral. 5" cor. Take the true altitude from 90°, and the remainder will be the zenith distance, to be called True altitude north when the object is south, and south when the object is north of the zenith. Reduce the declination to the time of observation. Then, if the declination and zenith distance are both north or both south, their sum is the latitude; but, if one is north and the other south, their difference is the latitude; and it is always of the same denomination as the greater. Zenith distance Latitude 50° 12' 40" S. 4 50 8 36 15 52 50 24 28 43 In observations for the latitude at sea it is customary to consider the sun's semi-diameter as always 16', and to take all the corrections to the nearest minute. When it is not stated that the altitude is taken by a back observation, a fore observation is always understood. dicates the sun's lower, and the moon's lower, and in his upper limb; her upper limb; To explain this rule, let A C (plate II. fig. 8) be the line of intersection of the planes of the ineridian and rational horizon; and FD the intersection of the planes of the meridian and equator; let Z be the zenith, P the north, and E the south pole; and let S be the true place of a celestial object on the meridian; then SA is its meridian altitude, SZ its meridian zenith dis--+ indicates that the number to which it is aftance, and SF its north declination. If S' be fixed is increasing, and that it is decreasing. the place of the object, then S' A is its meridian Example 2.-If on February 5th, 1828, at 2h. altitude, S'Z its zenith distance, and SF its 49m. apparent time, as deduced from a chronosouth declination. If S" be the place of the meter, regulated for Greenwich time, the meobject, then S" C is its meridian altitude, S" Z ridian altitude of be observed to be 49° 35′ S., its zenith distance, and S" F its north declination. height of the eye twelve feet; required the latiNow Z F, the latitude = ZS+SFZ S'. S'FFS" Z S"; which is the rule. Example 1.-If the meridian altitude of the sun's lower limb be 50° 12′ 40′′ S. on May 5th, 1828, in long. 23° W., height of the eye seventeen feet; required the latitude? >'s declin. S. 2° 3′ 57′′ S. + 2° 2′ 15 28 40 tude? In this example the Greenwich time is 2h. 49m. past midnight of February 4th. Now by the Nautical Almanac, the moon's declination, semi-diameter, and parallax, at midnight of that day,with their variations in the succeeding twelve hours, stand as under: : Cor. 2 32 37 Observed altitude) 15 49° 35' 10 Dip 0" S. 3 25 The moon's semidiameter as given in the Nautical Almanac, is that which she would appear to have if seen from the centre of the earth; if she be seen upon the horizon she will of course be nearer to the zenith of the observer, and will Moon's apparent altitude. Augmentation of >'s semidiameter. therefore appear under a greater angle. This augmentation of her semidiameter may be taken with sufficient exactness for practical purposes at sea from the following table: 0° 3° 6° 12° 15° 18° 21° 27° 30° 36° 39° 45° 48° 54° 60° 72° 90° To altitude 49° we find from this table that the augmentation of the moon's semidiameter is 11"; hence, in the above example, the semidiameter ought to have been 15′ 21′′; but it is only in finding the longitude by lunar observation, that such nicety respecting the moon's semidiameter becomes of any practical importance in the business of navigation. Semidiameter 0′′ 1′′ 2′′ 3′′ 4′′ 5" 6" 7" 8′′ 9" 10" 11" 12" 13" 14" 15" 16" Observed altitude 10° 50′ 0′′ Dip 3 56 Example 3.-If the meridian altitude of Aldebaran be 28° 5′ 10′′ N. on February 18th, 1828, height of the eye fourteen feet; required the latitude? Observed alt. 28° 5' 10" N. Refr. 4′ 47′′-par. 9′′ Latitude Example 2.-If the altitude of Capella, below the pole, be 4° 14′ 10′′, on August 10th, 1828, height of the eye thirteen feet; required the latitude? To find the latitude by the meridian altitude of an object below the pole. To the altitude add the complement of the declination, and the sum will be the latitude, of the same name with the declination. For (fig. 8, plate II.) FP and Z C are equal, being quadrants, and, if the common part Z P be omitted from each, there remains ZF the latitude =PC. Now if S" be the object below the pole, then SC is its altitude, S" D its declination, and SP the complement of its declination; and S""C+S"P PC' the latitude. In taking the sun's declination, it must be noticed that he is on the meridian twelve hours after noon. Example 1.-If the altitude of on the meridian below the pole be 10° 50' on May 20th, 1828, in long. 14° E., the height of the eye sixteen feet; required the latitude? To find the latitude from the observed altitude of a celestial object, near the meridian, the time of. observation being known. If the object is the sun, the apparent time is the meridian distance; for any other object, to the apparent time add the sun's right ascension, and from the sum subtract the right ascension of the object, and the remainder is its meridian distance. If the meridian distance is more than twelve hours, subtract it from twenty-four hours. Then add together the cosine of the meridian distance, and the tangent of the polar distance and the sum, rejecting the tens from the index, will be the tangent of arc first, which will be obtuse or acute according as the polar distance is obtuse or acute. Again, add together the secant of the polar distance, the sine of the altitude, and the cosine of arc first, and the sum, rejecting the tens from the index, will be the cosine of arc second. The sum or the difference of arcs first and second will be the co-latitude; and that latitude must be taken which agrees most nearly with the latitude by account, which will always be known nearly enough to determine which of the imputed latitudes is the true one. Let PA (fig. 9, plate II.) be the meridian on which the first point of Aries is, PS that on which the sun is, Pr that on which the star is, at the moment of observation. Then if P M be the meridian of the place of observation, S P M is the apparent time, APS the same right as cension, AP the stars, and P M the star's meridian distance. Now APS+S P M—AP x= PM, which is the rule for finding the meridian distance of a star; and the apparent line SPM is evidently the meridian distance of the sun. Next, let A B(fig. 10 plate II.) be the object's altitude, AZ its zenith distance, AP its polar distance, PZ the colatitude, and on P Z, or PZ produced, let fall from A the perpendicular A C. Then rad. cos. P cot. A P tan. PC; or rad, cos. P tan. A P. cos. P R tan. PC= = cot. A P Time of observation Greenw. time by acc. R. A. March 9th, whence, O's R. A. 24h. : 3m. 40s. Pis the object's meridian distance, and PC is arc I in the rule. Again cos AP : cos A Z (= sin. cos. PC sin. A B A B): cos. PC: cos. CZ = cos. A P :: 9h. 23m.: 1m. 26s, corr. of O's R. A.; 's R. A. 4h. 26m. 5s. *'s po!. dist. 73° 50′ 46′′. II in the rule. And when Z and P are on the same side of C the difference is Z P, otherwise their sine is Z P the colatitude. Example 1.-If the altitude of be 35° 2′, at Oh. 48m, 12s. P. M. October 7th, 1828, in lat. by account 47° 50′ N., long. 20° W., height of the eye fifteen feet, required the latitude? h. m. S. Greenwich time by acc. 24h. : 22′ 58′′ : : 2h. 8m. : 2′ 3′′ corr. of declin. 48° 12′ 12° 3', meridian dist. = 's declin. Oct. 7th. 5° 35' 39'S × 22′ 58′′ Correction. 2 3 24° 49′ 1 Co-lat. 46 46 55 90 Arc II. cos. 9.957927 Diff. arcs I and II. 35 13 C Mer. dist. 12° 3' cos. 9.990324 Pol. dist. 95 38 tan. 11:005955 sect. 11.008057 Given the time of observation and the altitude of the pole star to find the latitude. Find the meridian distance of the star as in the preceding method of finding the latitude. If the meridian distance is less than 90° consider it as a course, if it is between 90° and 270° consider the difference between it and 180° as a course; if it is more than 270° consider the difference between it and 360° as a course. Then with this course, and the polar distance of the star in minutes, enter a traverse table, and the difference 10-996279 cos I. 9:002069 of latitude will be the correction to be applied to the true altitude to obtain the latitude; addiAlt. 35° 13' sin. 9-760927 tive when the meridian distance is more than 90° and less than 270°, otherwise subtractive. Arc II. cos. 9:771053 diff. arcs I and II 483 latitude N. Example 2.-If the altitude of Aldebaran be 53° 12′ at 7h. 6m. 10s. apparent time, March 9th, 1824, in long. by account 35° W., lat. 44° N., height of the eye thirteen feet, required the latitude? For, let P (fig. 11 plate II.) represent the pole, MNA the horizon, at BCD, the circle in which the pole star moves round the pole, a circle which from its smallness may without any important error in navigation be considered as a plane one. Then PN is the latitude, and, when the star is at O, its polar distance OP taken from its altitude ON leaves P N the latitude. When it is at 180, it is directly below, as at 0 it was above the pole, and the polar distance 180 P added to its altitude 180 N gives P N the latitude. When the star is at 90 or 270 its altitude is the same as that of the pole. When it is at A or D, EP or HP taken from the altitudes A M or DQ leaves P N the latitude; when at B or C, PF or PG added to BM or CQ gives PN the latitude. Now A O is the star's meridian distance at A, O A B, O BC, and O BCD, its meridian distance at B, C, and D, respectively. The difference between O B and 180° is B 180, the measure of the angle BPG and similarly APE, G PC, and D P H, are the angles which taken as a course with AP, BP, CP, or DP, as a distance, give the corrections P E, PF, PG, and P H; the corrections to be applied to the altitudes AM, BM, CQ, and DQ, respectively to obtain the latitude PN. Example.-The altitude of Polaris was 52° 27′ 35′′; at 7h. 24m. apparent time February 2d, 1828, the sun's right ascension at the time being 21h. 2m., height of the eye fifteen feet; required the latitude? To find the latitude from two altitudes of the sun observed on the same day with the time elapsed between the observations. We take the following direct rule for solving this important problem from Mr. Riddle's Treatise on Navigation and Nautical Astronomy. Take the sun's declination for the Greenwich time which corresponds to the middle time between the two observations, and if it is of the same name with the latitude subtract it from 90°, if of a different name add it to 90°, and the sum or the remainder will be the sun's polar distance. Deduct the true altitudes from the observed ones, and, if the ship has been sailing in the interval between the observations, the first altitude must be reduced to what it would have been if taken at the same place with the second. This correction may be obtained by the following proportion. As radius is to the distance sailed, so is the cosine of the angle included between the course in the interval and the bearing of the sun at the first observation to the correction of the first altitude; to be added to it if the ship sails within less than eight points of the sun's bearing, but subtracted from it if the course is more than eight points from the sun's bearing. Or the correction may be more readily made from a traverse table, thus. Take the angle intended between the ship's course and the sun's bearing at the first observation as a course, and with it and the distance run on the interval as a distance the correction of the first altitude for the ship's change of place will be found in the column of latitude. Having now got the polar distance, the true altitudes, both for the place at which the second was observed, take half the interval between the two times of observation, and, reducing it into degrees, call it the half elapsed time, and proceed to compute the latitude at the place of the second observation by the following rule: 1. Add the sine of the half elapsed time to the sine of the polar distance, and the sum, rejecting ten from the index, will be the sine of arc first. 2. Add the secant of arc first to the cosine of the polar distance, and the sum, rejecting ten from the index, will be the cosine of arc second, which will be acute or obtuse, like the polar dis tance. 3. Add together the cosecant of arc first, the cosine of half the sum of the true altitudes, and the sine of half their difference, and the sum, rejecting the tens from the index, will be the sine. of arc third. 4. Add together the secant of arc first, the sine of half the sum of the true altitudes, the cosine of half their difference, and the secant of arc third, and the sum, rejecting the tens from the index, will be the cosine of arc fourth. 5. The differences between arc second and arc fourth is arc fifth when the zenith and the elevated pole are on the same side of the great circle passing through the apparent places of the sun at the two times of observation, otherwise their sum is arc fifth; but arc fifth can never exceed 90°. 6. Add the cosine of arc third to the cosine of arc fifth, and the sum rejecting ten from the index will be the sine of the latitude. Note 1.-When there is any doubt whether the zenith and elevated pole are on the same or different sides of the great circle passing through the places of the sun, the latitude may be computed on both suppositions, that being considered as the true latitude which agrees most nearly with the latitude by account, which will always be known nearly enough for that purpose. This additional computation will give very little trouble, as it is only arc fifth and its cosine that will require alteration. Note 2.-By this method the latitude may be found from two altitudes of the same fixed star; but, if the interval be in solar, it must be reduced into sidereal time, which may be done with sufficient exactness by adding to the elapsed time ten seconds for every hour or one second for every six minutes. Note 3.-It will expedite the calculation if all the logarithms that occur at the same opening of the book are taken out at the same time, and any little mistake in the observations will produce the less error in the result, the nearer the greater altitude is to the meridian. Note 4.-It is a curious circumstance in the history of this problem that almost all who have written upon it have introduced a correction in the wrong angle for the change of longitude be tween the observations; a correction which can nave no existence if the times are marked as they are presumed to be by the same watch, and the altitudes are reduced to what they would have been if taken at the same place. This error is found in Mackay's Navigation, and in his Treatise on the Longitude, and it was in the earlier editions of Norie's Navigation, but has been omitted since It is found in the article Navigation in the Encyclopædia Britannica; and even in the Encyclopædia Metropolitana, a work now in the course of publication. It has probably been owing to this remarkable circumstance that seamen have been so little disposed to rely on the results of double altitudes. 1821. Demonstration of the rule. Let Q, represent the sun at the place of the ship at the first observation, NAS the meridian, and B the place of the ship at the second observation. On A Q drop the perpendicular BC; then the ship in running from A to B has advanced AC directly towards the point on which the sun bore when the ship was at A; therefore AC added to the altitude observed at A will give what that altitude would have been if observed at B. Now NAQ is the bearing of the sun and NA B the ship's course, and B A C the difference of those angles; and rad. : A B :: cos. BAC: AC, the correction of the first altitude. If the ship run from A to B', from the sun, then AC' computed in the same way is the correction of the first altitude, subtractive, to reduce it to what it would have been if taken at the same place as the second. Having now obtained the altitudes, both at the place of the second observation, let Z (fig. 13 plate II.) be the zenith; P the pole ; A and B the places of the sun at the two times of observation; AC, BD, the altitudes, AZ, BZ, the zenith distances; AP, BP, the equal polar distances; ZPA, and Z P B, the times of observation, or the meridian distances of the sun at the times of observation. Then APB is the elapsel time, and, if PE be a perpendicular on AB, then A B and AP B will both be bisected by it; whence APE or EPB will be the half elapsed time. Join E Z, and from Z on EP draw the perpendicular Z F. Then A E is arc first, E Pare second, ZF are third, E F are fourth, and E. P are fifth; which as the figure is drawn is the difference of arcs second and fourth; but, if Z and P had been on different sides of AB, FP would have been the sum of EP and EF. Lastly, rad. cos. Z P, or rad. sin. .at. cos. ZF cos. F P, whence Example 1.-On April 4th, 1828, in lat. 35° N., long. 28° W., by account, at 8h. 10m. 40s. A. M. the alt. of was 38° 12', bearing by compass E. S. E; and at 11h. 22m. 14s. A. M., the alt. of was 60° 18′+, the ship having run in the interval S. W. by S. four miles per hour, height of the eve sixteen feet, required the latitude? Here the ship is running from the sun, within seven points of that opposite to his bearing; and the distance run between the observations is about thirteen miles; whence the correction of the first altitude is about 3'. |