पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)

IV. 26 IV. THREE PROBLEMS 99 Fig. IV. 10 In the stellar sphere CEC is the celestial equator, NP and SP being the north and south poles. NESP is the Unmaṇḍala or horizon of a place on the equator. Z is the zenith of the place, N and S being the north and south points and E is the east point. DsD is the day-diameter of the Sun, (s), in the northern hemisphere, making the declination sd. D₁s₁D₁ is the day-diameter of the Sun, (s₁), in the southern hemisphere, making the declination sd. D S D is the day-diameter of the Sun (s₁), in the southern hemisphere, making the declination s₁d₁. s and s₁ are the rising points of the Sun as seen from the place, NsEs₁S being its horizon. The altitude of the North Pole. N NP = angle NE NP, is the latitude, which is equal to SE SP, from which it is seen that for places in the northern hemisphere, the Unmaṇḍala is raised from the horizon by this angle in the north, and depressed by this angle in the south. As the Sun, in its diurnal circuit, takes exactly half a day to move from the eastern Unmaṇḍala to the western, the day-time is longer when the Sun’s declination is north, for it has to travel, after rising, an arc in the diurnal circle (equivalent to the great circle arc dE) to reach the Unmaṇḍala and an equal time while setting. The time is less when the declination is south, because before rising it has to travel less by an arc equiva- lent to Ed₁ to reach the horizon from the unmaṇḍala (and an equal time less while setting). dE and Ed₁ are the arcs of half-cara. Therefore when the declination is north, the time corresponding to 2 DE in the day-difference, (the day time being greater than 30 nāḍikās by this amount,) and when it is south the time equivalent of 2 Ed₁ is the day-difference, (the day-time being less than 30 nāḍikās by this amount). So we have to calculate dE, and Ed₁. In Δ dEs, right angled at d, by fundamental formula III, sin dE = Radius × sin Sd × Cos sEd ÷ (Cos sd × sin sEd). But, sd is the declination and sEd = 90° − N E NP= 90° − latitude. ∴ sin half-cara = 120′ × sin dec × cos (90°− lat.) ÷ {(Cos dec × sin (90° − lat))}