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

X.6 X. SAURA-SIDDHĀNTA — LUNAR ECLIPSE 233 (ii) Distance in minutes between the centres of the Moon and Shadow = √((i)² + (the Moon’s latitude at the given time)²). (iii) The amount eclipsed in minutes = half-sum of angular diameters of the Moon and Shadow - (ii) B. To find the amount eclipsed in the case of the Sun. (i) “Corresponding minutes of arc” = The minutes obtained as by A (i) × the half duration not corrected for parallax ÷ the half duration corrected for parallax. (This will be a little approximate, but has been given for case of computation, since the two times are known.) (ii) Distance in minutes between the centres of the Sun and the Moon = √((i)² + (Parallax-corrected lat. of time)²). (iii) The amount eclipsed in minutes = half sum of angular diameters of the Sun and the Moon

• (ii). Example 3. Continuing Ex. 2, find the amount of the moon eclipsed 3 nāḍīs after T. A. (i) Corresponding minutes of arc = (780' - 60') × 3/60 = 36' (ii) Distance between centres = √(36² + 26.6²) = 44'.76 (having found that the Moon’s lat. at the moment is 26'.6). (iii) Amount eclipsed = 54'.34 - 44.76 = 9'.6. Example 4. At a certain solar eclipse the difference of Sun and Moon’s motions is found to be 720', the parallax-corrected latitude, 2 nāḍīs before the parallax-corrected new moon, is found to be 15', the sum of the semi-diameters is 31'.9, the un-corrected half duration is nā. 2-30, and the corrected half duration is nā. 3. Find the amount of the Sun eclipsed, at 2 nāḍīs before the parallax corrected new moon. (i) Corresponding minutes of arc = (720 × 2 ÷ 60) × nā.2 1/2 ÷ nā.3 = 24 × 5 ÷ 6 = 20' (nearly). (ii) Distance between centres = √(20² + 15²) = 25'. (iii) The amount eclipsed = 31'.9 - 25' = 6'.9. The following is the explanation of the method: Let us first take the case of the lunar eclipse. At full moon, the Moon and the Shadow are in conjunction, i.e. they have the same true longitude. Since the Shadow has the same motion as the Sun, the interval between them for any interval of time before or after full moon is the same as the interval in tithi proportionate to the time interval. Therefore there is the proportion, if for 60 nāḍīs there is the difference of the daily motion, how much for the interval in time. So the difference in motion is multiplied by the given time and divided by 60. Since the motions are measured along the ecliptic, the interval in minutes along the ecliptic is got, corresponding to the time interval. The distance between the centres is got thus: In fig.2, S is the centre of the Shadow and M is that of the Moon. SM' is the ‘corresponding minutes’ got for the interval in time. MM' is the Moon’s latitude at the given moment. Since MM' is directed towards the pole of the ecliptic, the triangle SM'M is right-angled at M'. Since the triangle, being small, can be treated as a plane triangle, we have, by the Pythagoras Theorem, the distance between the centres, SM = √(SM'² + MM'²) = √(corres. minutes² + latitude²), as given. The amount eclipsed in minutes = Rr = SR - Sr = SR - (SM - Mr) = SR + Mr - SM = sum of semi- diameters of the Shadow and the Moon, minus the distance between their centres.