time at Greenwich, and thence the right ascension and declination scarcely be necessary. The horizontal parallax of the moon being always greater than the horizontal refraction, the rising of the moon is retarded. The computation of the quantity will be shewn in a subsequent problem. Maddy Prop. VIII. To find when a star rises heliacally (art. 327.) A star of the first magnitude rises heliacally, or first becomes. visible after having been obscured by the solar rays, when the sun is about 12° below the horizon. Let F (fig. 53) be the star rising; DA the right ascension of the star, A being the first point of aries; AXL, the ecliptic; and the arc KL perp. to the horizon = 12o. Then L is the place of the sun, when the star rises heliacally. From the right angled triangle CED by circular parts. Sin CD tan lat. tan decl. Hence, from the right ascension of the star, AD, AC is known in the triangle ACX, and also the adjacent angles A and C are known. By Napier's Analogies a cos (C+A): cos (C-A): : tan (AC): tan (AX + CX). Hence AX is known, and therefore the angle X is easily found. Whence AL the sun's longitude is known, and therefore the day when the heliacal rising takes place. To investigate the heliacal rising of a star 2000 years ago, we must decrease the present longitude of the star by 2000 x 50", 1= 28° nearly. Then with the longitude and latitude of the star, and a Luby's Trig. art. 137, p. 83. a obliquity of the ecliptic, compute the right ascension and declination, and then as above find the sun's longitude at the heliacal rising; it will be sufficiently exact to find by common proportion the number of days from the equinox from the longitude of the sun thus found, taking 59.8" for the motion in 24 hours.b Prop. IX. To find the sun's declination when the twilight is 0 shortest in a given latitude. Let Rr (fig. 54. 1) represent a parallel circle 18° below the horizon HO (art. 52). KNTYL a great circle touching Rr in T and intersecting the equator EQ so that the angle Y= HCE (the colatitude). Draw any parallel of the equator MNS. When the sun is in this parallel the arch MN is described in the same time, as the arch CY when the sun is in the equator. For let the arches MW and NX be perpendicular to the equator, then the right angled triangles MWC and NXY will be equal, and WC-XY, consequently WXCY. Therefore since MN and WX are described in equal times, MN and CY will be described in equal times. Hence the portion AT of a parallel of the equator between the horizon and its parallel Rr is described in less time than the portion of any other parallel to the equator between the same circles, because the time of describing AT= the time of describing CY= the time of describing MN, less than the time of describing MS. Hence the twilight is shortest when the sun describes the parallel AT, that is, when the sun's declination is TI. Now, because KTL touches Rr, the vertical circle ZFT is at right angles to KTL. Hence in the right angled triangles CFB and BTY, the vertical angles are equal and C=Y, therefore these triangles are equal, TB BF and each 9°. Also DF at right angles to EQ=TI the declination. Hence if we conceive the cir a Converse of prop. 4. = b The latitude of the star and obliquity of the ecliptic should also be reduced to what they were 2000 years ago, but this degree of accuracy would be quite unnecessary, in regard to any use that could be made of the result. cles of the sphere projected perpendicularly on the plane of the meridian, and FG be drawn perpendicular to HO, and to meet EQ in G, FG will be the tangent of FB, and FD= the sine of the declination, and DGF the latitude. Therefore by plane trigonometry = rad: sin CGF:: FG: FD, or rad sin lat: tan 9° : sin declination, : when the twilight is shortest This Prop. may be resolved in somewhat a more general form, as follows:: To find the declination of the sun or star, when the time of change from a given altitude A to a given altitude B is the shortest possible. In fig. 54. 2, L and M are the places of the object when the angle LPM is the least possible, the given zenith distances being LZ and ZM. Considering the triangles ZLP and ZMP, and the differential triangles that may be formed at L and M, it readily appears that the angles ZLP and ZMP are equal, and thence that MZP and LZP are the supplements of each other, as their sines are equal. Their sines are equal because the sides LP and MP are equal, and the side ZP common to the two triangles. The same may also be readily deduced as follows, but not so simply as by the differential triangles. Indeed this is an example, among many others, of the extreme facility of obtaining results by differential triangles as compared with the manner of obtaining the same by trigonometrical formulæ. . By trigonometry cos ZM = cos ZPM. sin ZP. sin PM + cos ZP. cos PM. And since ZM and ZP are constant, the differential equation is 0=sin ZPM. sin ZP. sin PM.d. ZPM + cos ZPM. sin ZP. cos PM.d. PM- cos ZP. sin PM.d. PM. therefore d. ZPM cos ZPM. sin ZP. cos PM-cos ZP. sin PM Now LPM is a minimum, and therefore d. LPM d. ZPM-d. ZPL = 0, = a This immediately appears from the differential triangles, but even the consideration of this expression is unnecessary, as the equality of the differential triangles at L and M is at once seen. Note by the Editor.-The prop. in its general state may be solved very simply as follows. Let the object cross on the parallel bm, fig. 54. 3. then the angle mPb is to be a min. Make z Ps on the sphere's surface equal to m Pb, and PSP Zand draw the arc of a great circle Zs, and join sm. Then since ZP is constant, and the angle ZPs a min. the arc Zs must be a min., but the spherical triangles bPZ, mPs being obviously in every respect equal, we have sm=Zb, Hence in the spherical triangle sm Z, the two sides sm, mZ are given, and the third side to be a min. This will obviously be the case when sm, mZ coincide. The object must consequently describe such a parallel nv, that and therefore given. Zn, sn may coincide. We may from this readily compute as follows:-let fall the perpendicular Px, then an is obviously half the sum, and xZ half the difference of the given zenith distances, which we shall call ≈ and z'. From the right angled triangles x PZ, For shortest twilight z=108° and '90°, hence cos(x+2)=—sin 9o, and cos(x-2)=cos 9o, therefore, &c. This mode of solution exemplifies the advantage of discussing by spherical geometry, previous to converting a question into formulæ. S or d. ZPMd. ZPL, also PM PL, and d. PM=d. PL; conse = quently, tan ZMP tan ZLP, and these angles are equal. Therefore one of these angles must be the supplement to the other, and consequently LZC = CZM. Conceive the circles of the sphere orthographically projected on the plane of the meridian. Then LK sin LZ. sin LZK, and IM sin MZ. sin IZM. Also o K= IK.LK =(since LZK= IZM.) cos LZ- sin LZ tan (ZM-LZ). sin LZ. And OC = KC—OK = cos LZ tan (ZM—LZ). sin LZ, and then OG (the sine of declination) =OC. sin lat. For the shortest twilight LZ= 90° and ZM = 108°, hence OC=—tan 9°, and therefore sin decl south = sin lat tan 9o. Prop. X. To find when Venus is brightest. Let S, T and V, (fig. 55,) be the Sun, the Earth, and Venus when brightest. Let TV also meet the orbit of Venus, supposed circular, in H, and join S, H, and S, V. The brightness of Venus varies as the versed sine of the exterior angle directly, (art. 110,) 2and as the square of the distance from the earth inversely. For the density of light decreases as the square of the distance from the radiating body increases. v sin SVH Hence when Venus is brightest TV2 is a maximum, and therefore v sin SHV × TH2 is a maximum, because TH × TV is constant, and therefore TH varies inversely as TV. Let TH, ST1 and SH=m. Then by plane trigono metry,a m2 + x2 - 2 mx.cos SHT=1. a Luby's Trig. p. 32. |