Thus, it appears, as has been observed by Ozanam, Emerson, and Delambre, that in all plane dials, one half of the dial being traced, determines the other by the several distances from the hour line of 6, measured on two vertical lines equidistant from the meridian. Thus these distances are equal between the lines from 6 to 7 in the morning and from 5 to 6 in the evening, 6 to 8 ..... morning . . . . . . . . 4 to 6 . . . . . evening, &c. &c. &c. The position of the substyle, in this instance between HM and HA, is determined by the preceding equation for cot s, and the elevation of the style by the preceding theorem for sin E. 26. But this kind of dial, as well as horizontal and direct dials, may be constructed independent of computation. Thus: On the proposed plane ABCD assume H for the centre from which the hour lines shall diverge, and draw the vertical line HM for the hour line of 12. Produce HA to G, so that HA shall be to AG, in the ratio of sin 90° to sin D; draw G6 to make the angle AGG = L, the latitude of the place, and join H6, which will be the 6 o'clock hour line. From H draw HE parallel to G6 to meet a line or parallel to AB in E. From o the intersection of oe and HM, draw of to make the angle Eor = complement of the given declination; make of = or, draw Fs perpendicular to oe, make os = os, and join Hs which will be the substylar line. Perpendicular to Hs draw the right line 5s.1, intersecting the meridian in 12, and the 6 o'clock hour line in 6. On 6-12 as a diameter describe a semicircle to cut Hs produced in P. With centre P and radius Ps describe a semicircle, which divide into arcs of 15° each, both ways from the line P6, or, which amounts to the same, set off angles 6r'7, 7P8, 8P9, 9P10, &c. each = 15°. Through the points 5, 6, 7, 8, &c. where the lines bounding these angles cut the lines 5s.1, draw from H the lines H44, H55, H66, HT7, &c. they will be the hour lines required. For, first, since the expression for the 6 o'clock hour line, when the terms are rendered homogeneous, is - sin 90° tan AH6 = sin D tan L, we shall have tan L to the radius sin D equal to tan AH6 to the radius sin 90°; which is obviously the case with regard to the triangles GA6, HA6. Consequently, the 6 o'clock hour line is rightly determined. Again, since EH is parallel to 6G, the angle of H is equal to the latitude, and EHo the co-latitude; therefore or = or, is the cotangent of the latitude to the radius oh ; and since Eof is equal to the co-declination, os = os is the sine of the declination to the radius of = or; or os = os = sin D cot L. But os = tan ohs = cots (substylar angle with horizon) to radius Ho; therefore cot s = sin D cot L, as it ought to be; and the substyle is rightly determined. * But the substylar line is evidently a portion of the right line between the centre of the dial and the apparent pole; therefore, the apparent pole, P, lies in the }. of Hs. At the apparent pole, P, also the horary angle between 6 o'clock and 12, is 6 x 15°, or a right angle; and the angle in a semicircle is a right angle: therefore, the semicircle described upon 6s12 as a diameter, will intersect the prolongation of Hs in P, the apparent pole. The truth of the remainder of the construction is manifest”. 27. For astronomical methods of determining the meridian, and the declination of any vertical plane, the student may turn to prob. 2 of the next chapter, examples 3, 4, and 5. o 1. SINCE the science of astronomy has given birth to spherical trigonometry, it is to be expected that at least some elementary problems in astronomy may be admitted into an introduction like the present. To determine the position of points in the apparent heavens, astronomers first referred to two planes, the horizon and , the meridian (see chap. viii. § 1), which are fixed in reference to any one place on the surface of the earth. But the necessity of comparing observations at different places, has led to the introduction of other planes and circles into the science, independent of the position of the observers, and even of the figure of the earth. Thus, when the situation, of the celestial equator, and the manner of valuing the angles between meridians by the measure of time, became known, they were employed to determine the position of the heavenly bodies by means of their right ascension and their observed declination. Afterwards, as it was found that a considerable portion of the celestial phenomena relative to the planetary system, occur in the plane of the ecliptis, or in planes but little inclined to it, it was found expedient to refer the position of the stars to the same plane, that is, to determine their latitude and longitude (chap. viii. art. 17, 18). * The problems given in this chapter will suffice to show the application of the principles of dialling to the most useful cases, They who wish to pursue either the theory of the mechanical part of dialling, through all its modifications, may consult Leadbetter's Mechanic Dialling, Emerson's Dialling, the treatise on dialling in the 5th vol. of Ozanam's Course of Mathematics, and that in the 3d vol. of Dr. Hutton’s edition of Ozanam's Recreations. There is also an elegant essay on dialling by M. de Parcieux, at the end of his Trigonometrie Rectiligne et Spherique; and a neat and sinple deduction of the practice of dialling from the principles of perspective, at the end of S’Gravesando's Essay on Perspective. These, and many other branches of astronomical inquiry, which we shall not here be able to touch, depending upon the mutual relations and intersections of different circles of the sphere, fall necessarily within the department of trigonometry. A few only will here be selected. PROBLEM I. 2. Given the obliquity of the ecliptic, and either the right ascension and declination of a star, or its latitude and longitude, to find the other two, and the angle of position. Let ec in the annexed figure be a portion of the ecliptic, Eq a portion of the equa- tor, the two circles intersecting in r: the first point of Aries, in an angle, i, of 23°27'49". Let r be the elevated pole of the ecliptic, r that of the equator, PP' a portion of the solstitial colure. #. P’ and P let quadrants P'sL, PSR, of E great circles, be drawn through s, the place of the star. Then ER will be the right ascension, a, of the star, so its declination, d, or Ps, its co-declination, EL its longitude, l, . sL its latitude, L, or Ps its co-latitude, PsP’ = p, the angle of position; and PP' is given = 23° 27'49", the present measure of the obliquity of the ecliptic. It is farther evident that sp? is the complement of the longitude, and P'Ps = P'PE + EPR = 90°,+ right ascension; as indicated at the poles of their respective circles. Now, if they are the right ascension and declination which are supposed known, in addition to the obliquity, we shall have, from the triangle spp., (see chap. vi. equa. 2 and 4), cos P's = sin PP'sin SP cos P'Ps + cos Pop cos SP, Adopting the preceding literal representatives of these sides and angles, and remembering that cos P'Ps = cos (90° + a) = — sin a, these become sin i = - sin icos d sin a + cos i sind.... (1.) These two formulae may be accommodated to logarithmic computation, by taking a subsidiary angle % such that sin a tand " for then, exterminating sin a from the former, and tan, d from the latter, we shall have 3. If, on the contrary, the latitude and longitude of the star are given, we shall have the declination and |