The operation, which gives 119° 53′ 8′′ for the azimuth from the north, and 2h 55m for the time from noon, is left for the learner's exercise. Note. From such an example as this it will be easy to find the meridian, by simply drawing a line to make with the direction in which the sun is at the moment of observation, an angle equal to the azimuth found by this process. And this method, provided the sun's altitude be corrected for refraction and parallax, will be found very commodious in the determination of a meridian, as the changes in declination, which prevail in the method by corresponding altitudes, are in this avoided entirely. The same method serves evidently to determine the variation of the compass. Example V. In the latitude of 52° 12′ 35′′ N. when the sun's declination is 15° 54′ 25′′ N. what is the sun's azimuth at 9h 5m in the morning? The azimuth, which is 119° 53′ 8′′ from the north, may be readily found, as well as the angle at the sun and the zenith distance, by the equations marked 14 in the preceding investigation. The work is left for the pupil's exercise. Note. An operation, like that required in this example, will serve to find the declination of any vertical plane, by knowing when the sun begins to shine upon it, or quits it. Example VI. At Peterborough, in latitude 52° 32′ N., on February 21, 1801, at what time from that star's being on the meridian, was the apparent motion of Dubhe or a Ursa majoris vertical; and what were the altitude and azimuth of the star at that time? On the year and day specified, the declination, d, of the star was 62° 50′. The logarithmic opperation cor responding to equa. 9, viz. cos P cot d tan l will give 47° 57′ 46′′, equivalent to 3h 11m 51, in time, for the hour angle from the meridian. Also, equa. 10, or sin z = cos d sec 1, will give 48° 38′ 28′′ azimuth from the north: And equa. 11, or sin a sin / cosec d, will give 63° 8′ 29′′, for the altitude of the star. The operation is again left for the learner's exercise. Note. Allied to the principle of this example is the method of finding, not merely the meridian, but the latitude, simply by observing two azimuths of the same known fixed star with the azimuth compass. If the azimuth of a circumpolar star be taken on each side of the meridian when at the greatest from the elevated pole, that is, when the apparent motion of the star is vertical; then it is evident that half the sum of these two azimuths will be the true azimuth from the north or south, according as the latitude of the place is north or south, and that half the difference will be the variation of the needle. Also, recurring to the diagram at page 158, there are given in the triangle Pzs right angled at s, the star's codeclination Ps, and the azimuth Pzs, whence the colatitude zp becomes known by means of Case 4 of right angled triangles; from which we have At what time on the 9th of August, 1817, will the twilight begin and end at Cambridge? Here the sun's declination for the proposed time is 15° 55′ 25′′ N. and the latitude 52° 12′ 35′′ N. Putting 108° for zs, 37° 47′ 25′′ for pz, and 74° 4′ 35′′ for Ps, in equa, 12, the resulting hour angle reduced to time will give 10h 12 for the time from noon when twilight begins and ends. Im PROBLEM III. 13. Given the latitude of a place to find the time and the duration of shortest twilight. Let p be the elevated pole of the world, z the zenith of the observer, s the sun 18° below the horizon, zs = 90° +18°, or more generally zs =90° + 2a. The observer, whose zenith is z, will then see the commencement of the morning twilight. By reason of the diurnal rotation, the declination circle H Z B Q Ꭱ Ps turning about the axis will bring the sun from s to s in the horizon, if it be the heavens which turn about the earth on the contrary hypothesis of the earth turning on its axis, the zenith z will approach nearer to s by describing about r the little circle zmbq, such that the distances PZ, Pm, Pb, PQ, are all equal. The same consequences as to this problem may be drawn from both hypotheses; but the latter is a little the most convenient. When, therefore, the zenith shall have descended from z to any point m, such that ms = 90° the sun will appear at 90° from the zenith, the day will commence and the twilight terminate; and the arc zm of the little circle will be the measure of the angle zpm, and consequently of the duration of the twilight. To determine this angle draw the arc Zвm of a great circle, and to its middle B the perpendicular arc PB; then will sin & zm sin zm sin zem = sin ZPB = sin PZ COS L Now, in the spherical triangle zms, sm+mz> zs, or 90° + mz> 90° + 2a. Therefore mz > 2a, and mz > a. Let mza+x: then [because 1 = - cos x = 2 sin 23x. ch. iv. art. 23.] COS L sin a + + COS L COS L 2 sin x cos (a + 1x) COS L Now x is essentially positive, and a and 3x small angles such that a + x < 90°; therefore sin zêm > sin a COS L It is farther evident that the twilight will be longer as x is greater, and shorter as x is less; and that it will be the shortest possible if x can be nothing; because in that case the second term of the second member of the equation will vanish. But this is what occurs when the triangle zms is reduced to the arc zs, that is, when the point m falls on b, or, when the distance PS is such that the part zb of the vertical zs lying within the little circle zmo is equal to 2a, and the exterior part bs equal to 90°. It is also manifest that if Ps increases the opposite angle rzs will become enlarged, and that on the contrary the said angle will diminish if es diminishes. In these variations the point m will approach to or recede from z, the intercepted part zb will vary, and it may vary between the limited O and zq= 2PZ. Thus the intercepted part may have all values between 0 and 2(90° — L) = 180° -2L, and consequently may have the value 2a. Hence, in the case where zb= 2a, and bs= 90°, the shortest twilight will obtain; and its semiduration in degrees will be found by the equation. sin ZPB = sin a COS L duration in time = ZPB (2.) .... 14. Other theorems may be deduced from the resulting construction. On the arc zb = 2a and with the complement zp of the latitude, constitute the isosceles triangle zPb, and let fall the perpendicular rm; zrb will be the angle which measures the duration. Prolong zb till bs 90°, and draw the arc PS, which will be the sun's polar distance for the day of the shortest twilight. m therefore cos zm: cos ms :: cos rz: cos PS; or, cos a: sin a :: sin L: sin D; whence sin D = — sin a cos a sin L = tan a sin L (3.) 15. Farther, the right angled triangle rzm, gives tan zm = cos z tan PZ (chap. vi. 26.) and cos z=tan zm cot PZ = tan a tan L (4.) Now z is the sun's azimuth at the commencement of the twilight, and rzb = pbz = 180° - pbs. But rbs is the sun's azimuth at the instant when his centre is at the true horizon: therefore the sun's azimuths at the begin ning and end of the crepusculum are supplements to each other on the day of the shortest twilight. Hence, since cOS PZS tan a tan L (5.) we have cos pbs = - tan a tan L 16. zps and bps are the hour angles at the beginning and end of the twilight; let the former be denoted by P', the latter by P: then ZPS bps PP = zrb, the angle which measures the duration of twilight. Hence we have from what has been done above, |