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

140 PAÑCASIDDHĀNTIKĀ V. 3 Moon, the time for a segment equal to 12° 56′.6 to rise is, 228 × 12° 56′.6 ÷ 30 = 99 vināḍis. This is less than 2 nāḍis and so the Moon will not be visible that day. As the time got is far less than the requirement, repetition of the work is unnecessary. The steps are explained thus: [चित्र : Fig. V. 1 - Cel. Eq., D, M, M1 (M'), Ecliptic, Diurnal Circle, South, W, S, L, North] Fig. V. 1 Here, WD is the celestial equator. SM'C is the ecliptic and LM' is the diurnal circle of the Moon projected on the ecliptic. S is the Sun, M is the Moon and M' is the same projected on the ecliptic. MM' is the Moon’s latitude. SM' is the difference in longitude which is found in step (i). WS is the Sun’s declination, and DM' is the Moon’s mean declination. ∴ SL is the difference of the declinations, found in step (ii). Assuming the triangles as plane triangles, in the right angled triangle SLM', LM'² = SM'² − SL² = (SM' + SL) (SM' − SL) ∴ LM' = √(SM' + SL) (SM' − SL), and LM' being the square root, it is equal to √(diff. in long. + diff. in dec.) (diff. in long − diff. in dec.).... (step iii) M'C is the result and it is found thus: As MM' is perpendicular to SC, triangle MM'C is right angled at M'. ∴ angle SM'L = angle CMM' Therefore the two triangles are similar. ∴ M'C/MM' = SL/LM .