Cor. 2.-Every straight line in the plane of the primitive, and produced indefinitely, is the projection of some circle on the sphere passing through the projecting point. Cor. 3.-The stereographic projection of any point on the surface of the sphere, is distant from the centre of the primitive by the semitangent of the distance of that point from the pole opposite the projecting point. THEOREM II. Every circle of the sphere, which does not pass through the projecting point, is projected into a circle. Cor. 1.-The centres and poles of all circles parallel to the primitive have their projections in its centre. Cor. 2.-The centre and poles of every circle, inclined to the primitive, have their projections in the line of measures. Cor. 3.-All projected circles cut the primitive in two points diametrically opposite. THEOREM III. The centre of the projection of a great circle is distant from the centre of the primitive by the tangent of the inclination of the great circle to the primitive, and its radius is the secant of the same. THEOREM IV. The centre of projection of a small circle, perpendicular to the primitive, is distant from the centre of the primitive by the secant of the distance of the circle from its nearest pole, and the radius of projection is the tangent of the same. THEOREM V. The projections of the poles of any circle inclined to the primitive, are in the line of measures distant from the centre of the primitive by the tangent and cotangent of half its inclination. THEOREM VI. Any two circles upon the sphere, passing through the poles of two great circles, intercept equal arcs upon them. THEOREM VII. If, from either pole of a projected great circle, two straight lines be drawn to meet the primitive and the projection, they will intercept corresponding arcs of these circles. Solution of Right-angled Spherical Triangles. The solution of right-angled spherical triangles may be accomplished by formulæ investigated expressly for that purpose. We are indebted to Napier, however, for a comprehensive rule of great advantage to the memory, by reducing all the theorems employed in the solution of right-angled triangles to two. This is called the rule of the circular parts, and is perhaps one of the happiest examples of artificial memory that is known. DEFINITIONS. 1. If in a right-angled spherical triangle the right angle be set aside, and the five remaining parts of the triangle alone be considered, consisting of the three sides, and the two oblique angles, then the two sides containing the right-angle, and the complements of the other three, namely, of the two angles, and of the hypotenuse, are called the circular parts. II. When, of the five circular parts, any one is taken for the middle part, then, of the remaining four, the two which are immediately adjacent to it on the right and left are called adjacent parts ; and the other two, each of which is separated from the middle part by an adjacent part, are called opposite parts. This arrangement being made, the solution is obtained by the following THEOREM. In any right-angled spherical triangle, the rectangle under the radius, and the sine of the middle part is equal to the rectangle under the tangents of the adjacent parts; or to the rectangle under the COSINES of the OPPOSITE parts. This theorem, or rule, may be easily remembered, by remarking, that the first vowels in sine, tangent, cosine, are respectively the same as the first in middle, adjacent, opposite, or, Rx sin mid=rect tan adj=rect cos op.* It is usual to convert the equation under consideration into an analogy having the unknown quantity for the last term, though to those acquainted with algebra it would be more convenient to make it alone the first term of an equation, and the remaining terms, combined properly according to the rules of algebra, the last. PROBLEM I. Given three of the six parts, as, for example, the hypotenuse and one of the angles of a right-angled spherical triangle, to find the sides and the remaining angle. On the first of May, 1826, the sun's longitude was 1° 10° 32′ 12′′, and the obliquity of the ecliptic 23° 27′ 40′′; required the right ascension and declination ? t B O B Ans.-R. A. 2h 32m 27.3; dec. 14° 59′ 47′′ N. Construction.-With the chord of 60° describe the primitive circle EPQP' on the plane of the solstitial colure, and draw the diameters EQ and PP' at right angles to one another, then n will EQ represent the equator, and PP' the E polar axis. Lay off from the same line of chords E e=23° 27′ 40", the obliquity of the ecliptic, and draw the diameter el representing the ecliptic, at right angles to which draw p p', and p, p' are the poles of the ecliptic. From the line of semitan r B t * Should either of the oblique angles, or hypotenuse, be one of the parts, then, instead of the word in the formula, use that derived from its complement, that is, for sine read cosine, for cosine read sine, and so on. For the explanation of these terms, the usual treatises on astronomy may be consulted. To those acquainted with the use of the globes correct ideas relative to these problems may be readily obtained. It may be added, that the sun's longitude and the obliquity of the ecliptic are computed from astronomical tables. L gents, (Theorem I.), lay off the sun's longitude 1' 10° 32′ 12", or 40° 32'.2 on the ecliptic, from A to C, then C will be the place of the sun, and by laying off from E to n, and from Q to m from the line of chords, extents equal to the sun's declination, then will the circle n c m passing through these points be a parallel of declination. Through the points PCP draw a circle of right ascension, cutting the equator EQ at right angles in B, then will AB be the right ascension, BC the declination, and BCA the remaining angle or angle of position, as it is sometimes called, which, in astronomy, is seldom of much use. Calculation. In the triangle ABC there are given AC = 40° 32′ 12′′, and the angle BAC = : 23° 27′ 40′′, to find BC, the distance of the sun from the equator EQ, or the declination, as it is usually called. Now, since in spherical trigonometry the sines of the sides are proportional to the sines of their opposite angles, Therefore, To find AB we may employ the method of the circular parts. In the triangle ABC are given AC and the angle BAC, to find AB the right ascension. Now, since the side CA, the angle CAB, and the side AB are all connected, that which stands in the middle or the angle A is called the middle part, and the sides AC and AB adjacent to it on each side are called the adjacent parts. Consequently Rx cos A cot_AC× tan AB; and resolving this into an analogy, as is frequently done in this country, we have, As cot AC * 40° 32′ 12′′ 10.067939 To tan AB the same as before. To those acquainted with algebra, it is better, after the manner of foreign mathematicians, still to retain the form of an equation thus: cos Ax tan AC, the radius being repre tan AB= Rx cos A sented by unity; in which case ten must be rejected in the index. To log cos A 23° 27′ 40′′ Add log tan AC 40 32 12 9.962526 9.932061 9.894587 Sum tan AB 2h 32m 27.3 It may be remarked, that if the parts are all connected, that which stands in the middle is called the middle part, and the other two are called the adjacent parts. If two only are connected, and one stands by itself, then this is called the middle part, and the other two are called the opposite parts. To find the angle ACB, since the parts under consideration are still all connected, AC standing in the middle is assumed as the middle part, and the angles A and C are the adjacent parts, whence Rx cos AC cot Ax cot C, and cot C = cos AC cos ACX tan 9.880808 9.637496 9.518304 Sumcot C 71 44 42 .2 Or the comp 18 15 17 .8, is called properly the angle of position, sometimes useful in computing the parallaxes in solar eclipses and occultations of the fixed stars and planets by the moon. By assuming different parts of the triangle ABC for the middle part, may be resolved the following Examples for Exercise. 1. On the first of June, 1827, at noon on the meridian of Greenwich, the sun's longitude will be 2s 10° 9′ 45′′, the obliquity of the ecliptic 23° 27′ 36′′; required the right ascension and declination? Ans.-R. A. 4h 34m 75.6; Dec. 21° 59′ 34′′ N. 2. August 12th, 1827, the obliquity of the ecliptic being 23° 27 36", the sun's right ascension will be 9h 25m 29.3; required his longitude and declination ? Ans.-Longitude 4° 18° 56′ 28′′, Dec. 15° 9' 32" S. 3. On the 10th November, 1828, on the meridian of Greenwich, the sun's right ascension will be 15h 2m 325.7, and declination 17° 14' 12" S.; required the sun's longitude and the obliquity of the ecliptic? Ans.-Longitude 7° 18° 6' 7", and obliquity of the ecliptic 23° 27' 34". 4. On the 2d of March, 1828, when the sun's declination was 7° 5' 18" S., and the obliquity of the ecliptic 23° 27′ 35′′; required his longitude and right ascension? Ans. Longitude 11° 11° 56′ 34′′; R. A. 22h 53m 24s. PROBLEM II. When the celestial object is not upon the ecliptic, as the moon, or the planets, and some of the fixed stars, the right ascension and declination are found by the solution of two right-angled triangles. 1. On the 17th of January, 1826, at noon, on the meridian of Greenwich, the moon's longitude was 1° 11° 5′ 14′′, and her latitude 2° 34′ 3′′ N.; required her right ascension and declination, the obliquity of the ecliptic being 23° 27′ 40′′? To resolve this example it is necessary to employ two right-angled spherical triangles. In the foregoing figure, the longitude of the moon or any star S, is AD, the latitude DS, the obliquity of the ecliptic BAC, the right ascension AB and the declination BS. Now, supposing a line drawn from A to S, there would be formed the right-angled spherical triangle ADS, right-angled at D, of which AD and DS are given to find the angle DAS and the side AS. If the position S of the star is without the ecliptic, then to the obliquity of the ecliptic BAC, add the angle DAS, the sum will be the angle BAS; but if S is within the ecliptic, that is, between it and the equator, subtract the angle DAS from the obliquity BAC, and the remainder will be the angle BAS. Since the side AS, and the angle BAS, are now known, AB the right ascension, and BS the declination, may be found. Calculation. By the rule of the circular parts, first AD and DS are given to find AS, and since the last is separated from the two first by the oblique angles, it will be the middle part, and AD and DS are the opposite parts; therefore, Rx cos AS cos DS x cos AD, or cos AS cos DS x cos AD to radius unity. To log cos DS 2° 34' 3" Add log cos AD 41 5 14 = 9.999564 9.877204 9.876768 Again, to find DAS, since the right angle does not separate the parts, DA standing in the middle is called the middle part, and the side DS and the angle DAS are the adjacent parts, hence Rx sin DA = tan DS x cot DAS, and therefore, cot DAS= sin DA cot DS 11.348322 9.817702 11.166024 Hence AS and BAS are now known, to find AB and BS. First to find AB. In this case the parts are connected; therefore BAS is the middle part, and AB and AS are the adjacent parts, whence Rx cos BAStan AB x cot AS, or tan AB= tan AB=cos BAS x tan AS, hence To log cos BAS 27° 21′ 52 Add log tan AS 41 9 11 Sum=log tan AB 37° 49' 5."5 To find BS, the angle BAS and side AS are connected, and BS is disjoined, whence Rx sin BS=sin ASX sin BAS, or since the sines of the sides are proportional to the sines of their opposite angles. As sine ABS or radius Is to sine AS 41° 9'11" So is sine BAS 27 21 52 To sine Dec. BS 17 36 24."4 N 10.000000 9.818274 9.662426 9.480700 The foregoing method is general and applicable to any part of the ecliptic, provided proper attention be paid to the situation of the celestial object with respect to the ecliptic and equator. As this problem and its converse is of frequent occurrence in practical astronomy, rules and formulæ, and even tables, have been formed for the purpose of facilitating the computations. The following rules, given |