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Answered Promptly. th / e c— (Gos ag Ui <a Fz N ‘ didi qi SSS = LATITUDE AND LONGITUDE: BY EQUAL ALTI- TUDES NEAR THE MERIDIAN AND THE 2 GREATEST ALTITUDE OF A HEAVENLY BODY. The subject has already been discussed in previous ar- ticles, but may still be enlarged upon to show the magni- tude of all the errors of the official method in use by way of substituting easy, exact and practical solutions. In the correct solution of the problem it has to be borne in mind that the sum of the hour angles of equal altitudes east and west in two different places, equals the elapsed time between observations, plus or minus the difference of longitude in time of the two places, plus when the place of the second observation is east from the place of the first ob- servation, and minus when west of it. As hour angles and the difference of altitudes with the _ meridian altitude are the co-ordinates of a parabola, the axis of which is the meridian, the curves for two different places may be represented on the same axis with the distance be- tween vertexes equal to the difference of latitude of those places. The distance of any point of the curves from _the axis ona line at right angles to the latter represents the hour angle for this point; and the distances on the same line of two points of different curves, the relation of their hour angles for equal altitudes. If the hour angles in two different places are respectively a, and a, for a certain altitude, a curve may be constructed midway between the two original ones with an hour angle on a, +a, -and a parameter corresponding to the same line — middle latitude of the two places. This curve being shifted east on the line indicating hour angles, until its end point coincides with the first curve, the other end point will coin- cide with the second curve, the shifted curve thus repre- senting the curve due to a change of position during obser- vations. The distance of the axis of the shifted curve from the original axis equals half the difference of the original hour angles, and at the same time is the hour angle of the greatest altitude for middle latitude at the middle of the time of observations. The equation of the reduction to the meridian reads:! 22512 sin, 1’ 2 (tan b ie tan c) in which H —h is expressed in arc minutes and t in time minutes; H denoting the meridian altitude, h the ex-meri- dian altitude, t its hour angle, b the latitude and c the de- clination. H—h= 225 sin. 1” Putting = lbs 2 (tan b 4 tan c) P = 30,558 (tan b dhe tan c) and expressing H —h by y. Bliss LIQUID Giri COMPASS Dyas oe eB ae, (1) which is the formula of co-ordinates for the -parabola, p be- ing its parameter. Changing the beginning of co-ordinates from the vertex to a point of the curve a minute east for which y = q form- ula (1) transforms into (a=st)? Meme bard ocas p and as pq = a? we obtain by subtraction t — (2a —t) Pp t denoting the time from the beginning of co-ordinates. Distinguishing quantities at the first and second observa- tions by the indices 1 and 2, the following two equations obtain t Vt = oe aay =f) Pi t Yo ==-> (2a,'—-t) 2 and if all quantities pertaining to a curve midway between the two preceding ones are distinguished by the index 0, we a,+a, 2 and py to correspond to middle latitude. Hence the form- ula pertaining to change of position will read: have, according to the explanation given above ae 0 which is the correction to find from the observed altitude the greatest altitude. To find the hour angles we have Pi ds = 4] and p,q, =a} Pi di — Pe Qe ai — az Bi > = Pa Gace Pads a, +a, Putting p, =p) —xandp,=p,+x i Po, a oe Is Rade da 2, f= a, — a, ih Po (di — 42) —x (qi + qe) 2 ay P2— Py As x = ————-and q, + q, = q, approx 2 Bis 88 which = —; and as q, — q, = difference of lati- Po Made in seven sizes by JOHN BLISS & CO, 128 Front Street, New York, is finely finished sensitive, accurate and durable. and is extremely steady. Is the best Liquid Com: pass ever made in this or any country. For sale by ship chandlers generally. Moves quickly | tude, which may be denoted by u, we obtain by substitution PoY P2— Py 4 a, — a, = See: Sb 2a, 2 Po a, — a, Pot (Po-= pao and —=} — —— }...... (4) 2 ao Po a; — 4, —— is the hour angle of the greatest altitude; and 2 : + a, ; ay from it and —— 2 dition and subtraction. The reduction of the greatest altitude to the meridian: =a, arefound a, and a, by simple ad- apa. \ eo r= ) Pras elim iatede; wre tatarate stenanaed (5) 2 Po Pol ee? : =— aa approximately ..... (52). 2 a) The preceding fovinuies are very nearly exact and, there- _ fore, preferable to those published in the MARINE RECORD of October rz last. te The following example illustrates the reckoning: Example: In 59° and 61° N. latitude by account true equal altitudes of the sun were found 17° 45’ 437; true course in the interval N. 30° E; distance run 138.56 miles; time be- tween observations th. 49m. 49.4s. Declination 10° S.; find latitude and ship’s time at the middle of the elapsed time. 30° course and 138.56 dist. give diff. lat. 120, depar. 69.28. Mean lat. by account 60°, and 69.28 dep. give diff. of longi- tude 138.56 = 9.24 m. B. elapsed time rh. 49.82 _ corrected elaps. time th. 59.06m. = 2 ao 99:53. = 4p Poo = 58.31, u = 120, p, — p, = 4.28 Hence by formula (4) a,—a, i {58.31 120 4.28 = -=1 Se eas ced 2 59-53 58.31 = 4 (117.54 — 4.37) 113.17 4 = 28.29 hour angle of greatest altitude. : 28.29? The reduction by formula (5) r= — £3.93 : 58.31 Correction of observed true altitude by 59.357 formula (3) y= = 60.41 58.31 Observed true altitude 17° 45.72” Sum 18° 59.867 from 90° 0,007 — Meridian Z. D, 71° 0.14/ N. Declination 10 )s True lattitude in at the middle of observations 61° 0.14’ 'N.