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

VI.5 VI. VĀS.-PAUL. SIDDHĀNTA – LUNAR ECLIPSE 161 the Moon and the Shadow is 21 minutes of arc. Therefore, when the difference between their centres is 21′, the total obscuration begins or ends, as at M₁ or M₂ in Fig. 4, below. Fig. VI. 4 Here M is the Moon at new-moon, and S is the centre of the Shadow. M₁S = M₂S = 21′ = the difference of the semi-diameters, constant according to this Siddhānta. SM is the latitude at full moon. Therefore, Minutes of obscuration = M₁M₂ = 2(M₁M or MM₂) = 2 √(SM₁² – SM²) = 2 √(21² – latitude²) = 2 √(21² – {21 × (moon ∼ Rahu)/5}²) = 2 × 21 √(5² – (moon ∼ Rāhu)²)/5, = 2 × 21 × √(25 – (moon ∼ Rāhu)²), the simplified rule, from which by inverse operation, we get the original rule, 21 × √[{5 – (moon ∼ Rāhu)} × [10 – {5 – (moon ∼ Rāhu)}]] 4/5. The conversion of the minutes of arc of obscuration into time is, as already given, by the propor- tion, Minutes of daily motion of (Moon – Sun): Minutes of obscuration :: 60 nāḍikās:nāḍikās of obscuration. From the simplified rule it will be readily seen that when Moon ∼ Rāhu is 5°, the minutes of obscuration, and thence the time, is zero. Therefore only when the difference is less than 5°, there is obscuration, not when greater, i.e. if the correct latitude at new moon is greater than 21′, there is no total eclipse. Example 2. The Moon at new moon is rā. 8-13-24. Rāhu (Tail) is rā. 8-12-0. The daily motion of (Moon – Sun) = 750′. Find the minutes of obscuration and the time. The corrected Rāhu = rā. 8-12-0 – 1° 36′ = rā. 8-10-24 Moon ∼ Rāhu = rā. 8-13-24 – rā. 8-10-24 = 3°. By the simplified rule, the minutes of obscuration = 2 × √(25 – 3²) × 21/5 = 2 × 4 × 21/5, minutes. The duration of obscuration = 2 × 4 × 21 × 60/(5 × 750) = nā. 2-41.