when the latitude sa declination are both north or both south; but when one is north, and the other south, the addition gives the time of sunrise, and the subtraction the time of sunset. 2. At what time will the sun set when its declination is 23° 12′ N., and the latitude of the place is 42° 40′ N.? Ans. 7h 33m 4, apparent time. 3. What will be the time of sunset for places whose latitude is 42° 40′ N., when the sun's declination is 15° 21' south? Ans. 5h 1m 23s, apparent time. 4. What will be the time of sunrise and sunset for places whose latitude is 52° 30' N., when the sun's declination is 18° 42′ south? Ans. { Rises 7h 44m 423, Sets 4h 15m 18 apparent time. 5. What will be the time of sunset and of sunrise at St. Petersburgh, in lat. 59° 56', north, when the sun's declination is 23° 24', north? What will be its amplitude at these instants? Also, at what hours will it be due east and west, and what will be its altitude at such times? Sun sets at 9' 13m 30' P.M. apparent Sun rises N. of east Ans. Sun sets N. of west } 52° 26' 18" Sun is east at 6" 58m 2° A.M. Sun is west at 5 1m 58 P.M. Alt. when east and west is 27° 18' 57". ON THE APPLICATION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. One of the most important problems in navigation and astronomy, is the determination of the formula for time at the rate of 15° to one hour, will give the time from apparent noon, when any particular altitude, as TS, may have been observed. PS is known, by the sun's declination at about the time; and PZ is known, if the observer knows his latitude. Having these three sides, we can always find the sought angle at the pole, by the equations already given in formulæ (T, or U, Prop. 7, Sec. III); but these formulæ require the use of the co.latitude and the co.altitude, and the practical navigator is very averse to taking the trouble of finding the complements of arcs, when he is quite certain that formulæ can be made, comprising but the arcs themselves. The practical man, also, very properly demands the most concise practical results. No matter how much labor is spent in theorizing, provided we arrive at practical brevity; and for the especial accommodation of seamen, the following formula for finding time has been deduced. From the symmetrical formulæ (S) Prop. 7, Sec. III, we have, Now, in place of cos.ZS, we take sin.ST, which is, in fact, the same thing; and in place of cos.PZ, we take sin.lat., which is so the same. In short, let A= the altitude of the sun, L = the latitude of the observer, and D the sun's polar distance. sin.A-sin.L cos.D Then, cos.P= cos.L sin.D But, 2sin.' P1- cos.P. (See Eq. 32, Prop. 2, Sec. I, Plane Trig.) Considering (L+ D) as a single arc, and (applying Equation 16, Sec. I, Plane Trig.), we have, after dividing (L+D+ 4) sin. (L+D−4), by 2, COS. sin.' P= = 2 2 L+D+A 2 cos.S sin.(S-A) cos.S sin.(S-A) cos.L sin.D This is the final result, when the radius is unity; when the radius is R times greater, then the sin.P will be R times greater; and, therefore, the value of this sine, corresponding to our tables, is, R sin.&P = √( Ꭱ cos.L cos.S sin.(SA). sin.D. PRACTICAL PROBLEMS. 1. In lat. 39° 6' 20" North, when the sun's declination was 12° 3' 10" North, the true altitude* of the sun's center was observed to be 30° 10' 40", rising. What was This angle, converted into time at the rate of 15° to one hour, or 4 minutes to 1°, gives 4* 2′′ 56′ from apparent noon; and as the sun was rising, it was before noon or 7h57m 4' A. M. If to this the equation of time were applied, we should have the mean time; and if such time were compared with that of a clock or watch, we could determine its error. A good observer, with a good instrument, can, in this manner, determine the local time within 4 or 5 seconds. 2. In lat. 40° 21' North, the true altitude of the sun, in the forenoon, was found to be 36° 12', when the declina * The instrument used, the manner of taking the altitude, its correction for refraction, semi-diameter, and other practical or circumstantial details, do not belong to a work of this kind, but to a work on Practical Astronomy or Navigation. tion of the sun was 3° 20' South. What was the apparent time? Ans. 9 42m 40 A. M. 3. In latitude 21° 2' South, when the sun's declination was 18° 32′ North, the true altitude, in the afternoon, was found to be 40° 8'. What was the apparent time Ans. 2h 3m 57' P. M. of day? SPHERICAL TRIGONOMETRY APPLIED TO GEOGRAPHY. If we wish to find the shortest distance between two places over the surface of the earth, when the distance is considerable, we must employ Spherical Trigonometry. Suppose the least distance between Rome and New Orleans is required; we would first find the distance in degrees and parts of a degree, and then multiply that distance by the number of miles in one degree. In the solution of this problem, it is supposed that we have the latitude and longitude of both places. Then the distances, in degrees, from the north pole of the earth to Rome and to New Orleans are the two sides of a spherical triangle, the difference of longitude of the two places is the angle at the pole included between these sides, and the problem is, to determine the third side of a spherical triangle, when we have two sides and the included angle given. Let P be the north pole, R the position of Rome, and N that of New Orleans. |