The correct siderial time being ascertained by the transit of a star of known right ascension, the correct mean solar time may be found from this, as follows: The Nautical Almanac gives on p. XXII. of each month the mean time of transit of the first point of Aries, which is the zero of siderial time, and may be called from analogy the siderial noon. * By means of this and the table of time equivalents, explained at p. 147, the mean solar time corresponding to any given siderial time, may be obtained by the following rule. Mean solar time required mean time at preceding siderial noon + the equivalent to the given siderial time. EXAMPLE. To convert 7 11 1020 siderial time (the true time of meridian transit recorded above) March 8th, 1850, into mean solar time for the meriIdian of New York. Mean time at preceding siderial noon, viz. March 8th, 04 56m 3074 6 58 51 19 10 58 20 9.97 0.20 *To obtain the mean time of siderial noon at any place having a different longitude from Greenwich, it is necessary to subtract 98565 multiplied by the hours and fractions of an hour, by which the place differs in long. from Greenwich if the place be west, and to add this product if the place be east of Greenwich. As the daily gain of siderial time on solar is about 3m 56, this divided by 24 or 9.8565 will be the hourly gain or the hourly motion of the sun backward from west to east. + This is obtained by multiplying 9856 by 44.94 the difference of longitude between New York and Greenwich. This correction may be applied here instead The mean solar time may be obtained by direct observation of a meridian transit of the sun, which is made by taking the transit of each limb of the sun, that is to say observing the times when the sun's disc is tangent to the wires, both upon its western and eastern side, and taking a mean of the times as the time of transit of the sun's centre.* Or more accurately by applying as a correction additively or subtractively to the observed time of transit of the limb the time occupied by the sun's semidiameter in passing the meridian, which is given for every day in the year in the Nautical Almanac, p. I. of each month. To convert mean solar time into siderial the rule is as follows. Siderial time required = siderial time at preceding mean moon + the equivalents to the given mean time. EXAMPLE. To convert 8 5 4162 mean time at New York, into siderial time. Sider. time at preced. mean noon, Gr. viz., March 8th, 234 3 19**98 Correction for long. N. Y. + 48 68 Sum sid. time required 7' 11" 10"06+ The reasons for the above rules are sufficiently evident. PROBLEM. 93. Given the latitude of the place, and the declination of a heavenly body, to determine its altitude and azimuth when on the six o'clock hour circle. of to the mean time of siderial noon. Strictly the equivalent of 4.863 in solar intervals should be applied, or 4868 may be applied to the given siderial time additively before taking out the solar equivalents. * A colored glass over the eye piece is necessary in observing the sun. The sum amounting to more than 24, of course 24' must be rejected from it, as after reaching 24h the siderial time begins at zero again. 1. Required the altitude and azimuth of Arcturus when upon the six o'clock hour circle of New York, lat. 40° 43′ N., on the 1st of Jan. 1850; its declination on that day being 19° 57′ 56′′ N. 94. There remains one case in the solution of oblique angled spherical triangles, which we have deferred to this place because we wished to employ in it the rules for the solution of right angled triangles. This is where two sides and the angle oppositet to one of them, or two angles and the side opposite to one of them, are given; or, as it is sometimes expressed, where two of the given parts are a side and its opposite angle. In such a case we may proceed as follows: By means of the proportion, the sines of the angles are as the sines of the opposite sides (Art. 81), the unknown part opposite one of the given parts may be found. Four parts of the triangle will then be known, and two will remain unknown; these two will be a side and its opposite angle, to find which, from the vertex of the unknown angle let fall an arc perpen * The horizon and equator intersect at A, 90° from the meridian. + The term opposite is used in a more exact sense here than in Napier's rules, of which we have just been speaking. dicular upon the unknown side opposite, and the given triangle will be divided into two partial triangles, which will be right angled, and in each of which two parts will be known. Applying Napier's rules to the solution of these, the partial angles which compose the unknown angle. may be found, and their sum will be the value of the unknown angle; then the unknown side opposite may be found by the proportion, the sines of the angles are as the opposite sides; or this last side may be found by calculating the two parts of which it is composed from the right angled triangles, and adding them together, which is the better method, since it avoids ambiguity. If the perpendicular are drawn from the vertex of the unknown angle to the unknown side falls without the triangle, of course the difference, instead of the sum of the angles and sides found in the right angled triangles, is to be taken. Thus in the annexed diagram triangle let the side and angle opposite given be c and c, and the other given angle B. Then first sin c sin c: sin B: sin b : B a by means of which proportion b may be calculated and will be legitimately ambiguous. Then there will be left unknown A and a. A From A let fall a perpendicular AD upon a, b which we have not drawn, lest it should confuse the diagram, but which the student can imagine; then in the right angled triangle B A D we know two parts в and c, and also in the right angled triangle c AD we know two parts c and b, the latter having been found by the proportion above. To calculate the partial angles at A, calling that in the first right-angled triangle above mentioned w, and that in the second w', we have by Napier's rules Thus all the parts of the triangle are determined. For an application of this case of solution take the following EXAMPLE. Given the zenith distance, azimuth and polar distance, to find the hour angle and colatitude of the station. Or in the diagram given, zs, z, and Ps, to find P and pz. Supposing the declination of a star, as given by catalogue, to be 160 11' N., and its observed altitude and azimuth to be 39° 10' and 75° 10' Ν P from the south, required the hour angle of the star and latitude of the station. We have now demonstrated formulas for the solution of every possible case of plane and spherical triangles, including the more simple formulas which apply exclusively to the right angled triangles. The examination questions which follow call attention to the most important results of the investigations in the preceding pages. After these, in Appendix II. will be found many useful matters connected with Spherical Trigonometry, for which it was thought not best to interrupt the general train by which the solutions of triangles are deduced. Note. In assuming hypothetical cases, care must be taken not to suppose such as are impossible. The following are the governing principles to be observed. In plane triangles, 1. One side must be less than the sum of the other two (Geom. Ax. 13 Cor.) 2. The greater side of a triangle is opposite |