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

280 PAÑCASIDDHĀNTIKĀ XIV.41 A is Canopus at the time of its heliacal rising and S the Sun at that time. PGAQ is the hour circle of Canopus, and G the point where it intersects the ecliptic. Assuming that the celestial longitude of Canopus is 90° and the celestial latitude 75° 20' S¹, we have ♈G = 90°, AG = 75° 20' and AA' = 75° 20' − 24° = 51° 20'. Therefore Rsin A' E = Rsin (asc. diff. of Canopus A) = Rtan δ tan θ, vide formula (2), where δ is the declination AA' of Canopus and θ the latitude of the place, = (sin δ . sin θ / cos θ) × (R / cos δ) = (sin 51°20' . palabhā / 12) × (120 / cos 51°20') because, according to Varāhamihira, R = 120' = 25 × (palabhā / 2), approx. ∴ A'E = arc (in terms of degrees) corresponding to Rsine equal to 25 × (palabhā / 2), approx. EG" = asc. diff. of G, approx. = 21 × palabhā, vināḍīs, approx. because for unit palabhā, the ascensional difference for G (the first point of Cancer) is 21 vināḍīs. Also, assuming 15 to be the time-degrees for the visibility of Canopus, G'S = G"S' approx. = 15 degrees, approx. ∴ A'S' = A'E + EG" + G"S' degrees = [10 (A'E + 15) + 21 palabhā] vināḍīs, where A'E + 15 is in degrees and palabhā in digits. Degrees multiplied by 10 are vināḍīs. Now GS is the arc of the ecliptic which rises above the horizon (of Laṅkā) in the time given by the arc A'S' of the equator. Hence it is obvious that Canopus A will rise heliacally when the Sun is at S, i.e., when Sun’s longitude = longitude of G + arc GS = longitude of G (i.e., 90°) + arc of the ecliptic which rises (at Laṅkā) in the time given by the arc A'S' of the equator.

1. Actually, the longitude of Canopus is 85°4' and the latitude of Canopus is 75°50'S.