difference of a few seconds will sometimes be found, owing to the small quantities neglected. To illustrate this, the following examples are given : EXAMPLE XI. The sun's correct central altitude was 32° 25', his declination 17° N. Eight hours afterwards, by a watch, bis correct central altitude was 30° 8', and declination 16° 55′ N. Required the latitude, supposing the latitude by account to be 53° 20′ N. The tabular correction corresponding to the first altitude 32° 25', declination 17° N., and latitude by account 53° 20′ N., is 80′′. Multiplying this by the difference of the declination 17° -16° 55'5'300', the product (rejecting the two right-hand figures) is 240.004', the correction of altitude. This is to be subtracted from 32° 25', because the sun recedes from the elevated pole, while the declination changes from 17° N. to 16° 55" N.; therefore the corrected first altitude is 32° 21". Using this with the second altitude 30° 8', the second declination 16° 55', and the elapsed time 8 hours, the calculation may be thus made by the first method, as follows: COL. 1. El. time 8 [P. M.] Cosec. 10.06247 .Cosec. 10.53614 . Cosec. 10.08168 Cosine 9.74812 ..... Cosine 9.74812 Cosec. 10.28512 B 31° 18′ N. Cosec. 10.28426 sum alts. 31 14 Cosine 9.93196 [E is the sum of B, Z, if of the same name; difference, if o' a different name.] Latitude 53 25 N. Sine 9.90471 As it is entirely arbitrary which altitude is considered as the first, or the one to be corrected, it may not be amiss to repeat the operation, considering 30° 8' as the first altitude, and 16° 55′ as the first declination. The tabular number corresponding to these quantities, and the latitude by account, is 79', which, being multiplied by the change of declination 300", (rejecting the two right-hand figures,) gives 237"=3′ 57′′, or 4 nearly. This is to be added to 30° 8' to get the corrected first altitude 30° 12′, because the sun approaches the elevated pole, while his declination changes from 16° 55′ to 17°. Assuming, therefore, the corrected first altitude as 30° 12', the second altitude 32° 25′, the second declination corresponding thereto 17° N., and the elapsed time, as before, 8 hours, the calculation may be then made as follows: sum alts. 31° 184 Cosine 9.93165 Cosec. 10.28429 B 31° 27' N. Cosec. 10.28256 So that the latitude is exactly the same by both methods. If the middle time between the two observations be required, it would be obtained by adding the log. tangent of C 8.30263, to the log. secant of E 10.22493; the sum, rejecting 10 in the index, is 8.52756, which, being sought for in the log. tangents, correspond in the Col. P. M. to 0h 15m 26', whose half 0 7m 43" is the middle time between the two observations. Taking the sum and difference of this and half the elapsed time, 4", gives the times from noon when the observations were made, 4 7 43 and 3 52m 17, the one being before noon, the other afternoon. The same result is obtained whichever altitude is corrected. EXAMPLE XII. Suppose we have, at the same moment of time, the moon's correct central altitude 55° 20, the moon's declination 0° 36′ N.; the sun's correct central altitude 37° 40', his declination 0° 17' S; the hour angle, or difference of the right ascensions of the sun and moon, as given by the Nautical Almanac, 5 hours; required the true latitude, the latitude by account being 23° 20′ N. The tabular correction corresponding to the latitude by account 23° 20′ N., the sun's altitude 37° 40', (considered as the first altitude,) and the declination 0° 17′ S., is 50", and the change of the two declinations from 0° 17′ S. to 0° 36′ N. is (53′ =) 3180"; this being multiplied by 50, and the two right-hand figures rejected, gives the correction of altitude 1590′′ — 26′ 30′′; this is to be added to the altitude 37° 40', because the change of the sun's declination from 0° 17′ S. to 0° 36' N., approaches the sun to the elevated pole; therefore the sun's corrected altitude is 38° 6' 30", or simply 38° 6'′. Using this with the moon's altitude 55° 20', the moon's declination 0° 36′ N., and the hour angle 5 hours, the latitude may be found by the first method, in the following If the moon's altitude, 55° 20′, be considered as the first altitude, and corrected, the tabular number corresponding to this altitude, the moon's declination 0° 36′ N., and the latitude by account 23° 20′ N. will be 70". Multiplying this by the change of declination 3180", and neglecting the two right-hand figures, gives the correction of altitude 2226" 37' 6", or simply 37', which is to be subtracted from the moon's altitude 55° 20′ to obtain the corrected altitude 54° 43', because the change from 0° 36′ N. to 0° 17′ S. makes the moon recede from the elevated pole. Using the corrected altitude 54° 43', the sun's declination 0° 17' S., and the sun's altitude 37° 40′, with the hour angle 5h, the latitude may be found by the first method, in the following [Z less than 90°, named N. or S. like the bearing of the zenith.] Z Sec. 10.03970 Z24 7 N. [E is the sum of B, Z, if of the same name; difference if of a different name.] E23 46 N. Sine 9.60532 Latitude 23 24 N. Sine 9.59906 Which agrees with the preceding calculation. In taking the half-sum and half-difference of the altitudes, it will be convenient to prove the accuracy of the calculation by adding this half-sum to the half-difference, for the sum will be the greater altitude. The difference of the same numbers will be the least altitude. Thus, in the present example, EXAMPLES FOR EXERCISE IN THIS THIRD METHOD. 1. The sun's correct central altitude was 41° 33′ 12′′, his declination 14° N. After an interval of 1h 30m, his correct central altitude was 50° 1′ 12′′, and declination 13° 58′ 38′′; latitude by account 52° 5′ N. Required the true latitude. The tabular number corresponding to the altitude 41° 33′ 12′′ is 87", and this being taken for the first altitude, is, when corrected, 41° 32′ 0"; the second altitude is 50° 1' 12", the elapsed time 1h 30m, and the declination 13° 58′ 38′′ N. These make the latitude 52° 5' N. Or, by taking 50° 1′ 12′′ for the first altitude, and using the corresponding declination, the tabular number is 95", the corrected first altitude becomes 50° 2′ 30; using this, with the second altitude 41° 33′ 12", the declination 14° N., and the elapsed time 1h 30m, we find that the latitude becomes, as before, 52° 5' N. 2. Given the correct central altitude of the moon 53° 43', her declination 14° 16' N. After an interval, in which the hour angle was 1h 44m 15, her correct central altitude was 42° 29', and declination 13° 52′ N.; the latitude by account 48° 54′ N. Required the true latitude. With the first altitude and first declination the tabular number is 98", and the corrected first altitude 53° 19′ 28′′, the second altitude 42° 29′; with which the declination 13° 52′ N., and the corrected elapsed time or hour angle 1h 44m 15, we find that the latitude is 48° 55′ N. Or, by taking 42° 29′ for the first altitude, and 13° 52′ N. for the first declination, the tabular correction will be 83", and the corrected first altitude 42° 49′; using this, and the second altitude 53° 43', the corresponding second declination 14° 16′ N., and the hour angle 1h 44m 15, we find the latitude to be 48° 54' N., nearly; agreeing with the former calculation. 3. Given the correct central altitude of the moon, 55° 38', her declination 0° 20′ S. After an interval in which the hour angle was 5h 30m 49, her correct central altitude was 29° 57', and her declination 1° 10 N.; the latitude by account 23° 25′ S. Required the true latitude. With the first altitude 55° 38', and the first declination 0° 20′ S., the tabular correc tion is 71", and the first corrected altitude 54° 34' 6". Using this with the second altitude 29° 57', the second declination 1° 10' N., and the hour angle 5h 30m 49", the true altitude will be found 23°23′ S. Or, by taking 29° 57' for the first altitude, and 1° 10′ N. for the first declination, the tabular correction will be 45", and the first corrected altitude 30° 37'. Using this with the second altitude 55° 38', the second declination 0° 20′ S., and the hour angle 5h 30m 49, the true latitude will be found to be 23° 24′ S., nearly agreeing with the preceding calculations. In making the calculations of these three examples, the seconds were noticed, which is always best to be done, particularly when the altitudes are nearly equal; some difference might be found in the above results if the nearest minutes only were taken. Thus, Example XI., calculating to the nearest minute, gives the latitude 53° 28'. If the calculation be made as in page 191, it becomes 53° 25', differing 3'. This difference would be avoided by taking the angles to seconds, and in some extreme cases it would require the use of 6 or 7 places of decimals. FOURTH METHOD. To find the latitude by double altitudes of the same or different objects, the declinations being different. This method, like the first, requires only the use of Table XXVII.; and the words sine, cosine, &c., are written for log. sine, log. cosine, &c. The logarithms are arranged in these columns as in the first method, according to the following formula, which ought to be written down before the calculation is commenced; this will simplify the operation, and may prevent mistakes. In this formula it is said that C is of the same affection as B; the meaning of which is, that if B is less than 90°, C also is less than 90°; and if B is greater than 90°, C also is greater than 90°. Likewise A is of the same affection as the hour angle H, meaning that if the hour angle is less than 6 hours or 90°, A will be less than 90°; and if the hour angle exceed 6 hours, the angle A will exceed 90°. In some late works on navigation, no notice is taken of the cases where the hour angle exceeds 90°, or the distance of the objects exceeds 90°, and on that account the rules appear less subject to different cases than the following rule, which embraces all possible cases, and the apparent simplicity of the rules referred to, arises from their imperfections and incompleteness. RULE. 1. Find the hour angle H, and take out the corresponding secant, which put in Col. 1, and its tangent in Col. 3. 2. Take the declination d, corresponding to the greatest altitude, place its tangent in Col. 1, its sine in Col. 2. 3. The sum of the two logarithms in Col. 1 (rejecting 10 in the index) is the tangent of the angle A, which is less than 90° if the hour angle is less than 6 hours, (or 90°,) but greater than 90° if the hour angle is greater than 6 hours. This angle is to be marked north and south, with a different name from the declination d, at the greatest altitude. The cosecant of A is to be placed in Col. 2, its cosine in Col. 3. 4. Place the declination D, corresponding to the least altitude, below the angle A, and if they are of the same name, take their sum, but if of different names, take their difference, and call this sum, or difference, the angle B, making it north or south, like the greatest of the two quantities A, D. The cosine of B is to be placed in Col. 2, its cosecant in Col. 3. 5. The sum of the three logarithms in Col. 3 (rejecting 20 in the index) is the cotangent of an angle F, (less than 90°,) which is to be taken out and marked north or south, with a different name from B. 6. The sum of the three logarithms in Col. 2 (rejecting 20 in the index) is the cosine of the angle C, which is to be taken less than 90° if B is less than 90°, but greater than 90° if B is greater than 90°. The angle C, and its cosecant, are to be placed in Col. 1. 7. Place the altitudes below C, take the half-sum of these three quantities, subtract the greatest altitude from the half-sum, and note the remainder. Place the secant of the least altitude in Col. 1, its cotangent in Col. 2, its sine in Col. 3; the cosine of the half-sum in Col. 1, and the sine of the remainder in Col. 1. The sum of the four * The hour angle is the same as the elapsed time in double altitudes of the sun. This time is turned into degrees by Table XXI., but it is more simple to double the hour angle, and find it in Col. P. M Table XXVII., and take out its corresponding tangent. If this double angle exceeds 12h, reject 125, and find the remainder in Col. A. M., and take out its corresponding tangent. In the following examples this double angle is marked with the letters P. M. annexed. This rule is easily remembered in three places in which it occurs, from the circumstance that sis the first letter of sum and same, and d the first letter of difference and different. If the sum be taken to find B, and it exceed 180°, subtract it from 360°, and call the remainder B, last logarithms of Col. 1, (rejecting 20 in the index,) being divided by 2, gives the sine of an acute angle, which being found and doubled, gives the zenith angle Z, which is to be named north if the zenith and north pole are on the same side of the arc or great circle, passing through the two objects, (or the two observed places of the same object,) but south if the zenith and south pole are on the same side of that great circle.* 8. Take the sum of the angles Z and F if they are of the same name, but their difference if of different names; this sum or difference is the angle G, to be marked north or south, like the greatest of the angles Z, F.† The sine of G is to be placed in Col. 2. 9. The sum of the two lower logarithms of Col. 2 (rejecting 10 in the index) is the tangent of an angle I, which is to be taken out (less than 90°) and marked north or south like G. The secant of I is to be placed in Col. 3. 10. Write the declination D, corresponding to the least altitude below I, take their sum if of the same names, their difference if of different names. This sum or difference is the angle K, of the same name as the greater of these two quantities. The sine of K is to be placed in Col. 3. 11. The sum of the three last logarithms in Col. 3 is the sine of the required latitude, of the same name as K. EXAMPLE XIII. Given the sun's correct central altitude 41° 33', and his declination 14° N. After an interval of 1h 30m, by watch, his correct central altitude was 50°, and his declination 13° 58′ N. Required the latitude, the sun being south of the observer when on the meridian. If the latitude had been south, Z, instead of being 57° 18' north, would be 57° 18′ south; G=54° 38' S., I=42° 37' S., K=28° 37′ S., and the latitude 25° 34′ S. The labor of making this extra calculation is but little, and where any doubt exists of the name of Z, it is best to make the computation both ways; this, however, will rarely happen. The calculations of this example, and most of the following ones, are made to the nearest minute; where great accuracy is required, it will be proper to take the logarithms and angles corresponding to seconds. *This case occurs also in the first and second methods of solution, and it must be determined on the spot by the situation of the objects. In double altitudes of the sun, moon, or planets, when the elapsed time is not very great, the angle Z is generally to be marked with the bearing of the zenith from the observed object, when at its greatest altitude on the meridian, which in north latitudes, without the tropies, is in general north; in south latitudes, without the tropics, south. Sometimes, when the sun passes the meridian near the zenith, it may be doubtful whether the zenith be north or south; in which case the problem may be solved for both cases, (which increases the labor but little,) and that one of the two computed latitudes selected which agrees best with the ship's reckoning; but it is generally safest not to use observations of this kind, which are generally liable to great errors from small mistakes in the altitudes. If the sum be taken to find G, and it exceed 180°, subtract it from 360°, and call the remainder G, with a different name from Z or F |