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

IX.20 | IX. SAURA-SIDDHĀNTA — SOLAR ECLIPSE | 219 मध्यज्याकृतिविश्लेषितां पृथक् स्थाप्य मूलमेकस्याः । सवितुर्दृक्शेपाख्यं संस्मृत्यर्थं पृथक् स्थाप्यम् ॥ २० ॥ Dṛkṣepa of the Sun 19. Find the sine of the longitude of the Orient Ecliptic Point (o.e.p.) at new moon, multiply by the sine of maximum declination (of the Sun, 48′ 48″) and divide by the sine of the colatitude. (This is sine amplitude of o.e.p., called Udayajyā.) Multiply this by the sine of the zenith distance (z.d.) of m.e.p. already found, and divide by 120′. Square the result, and subtract from the square of the sine z.d. of the m.e.p. 20. Set the remainder in two places. In one place, find its square root. This is the sine of the zenith distance of the nonagesimal (z.d. of n.) called the Sun's dṛk-kṣepa. Keep this safe aside for future work. The following is to be done: (i) Using the vināḍīs of ascensional differences of the place, the o.e.p. at new moon is to be found. (ii) Sine amplitude of o.e.p. = sine z.d. of m.e.p. × 48′ 48″ ÷sine colatitude. (iii) Sine (m.e.p. ~ nonagesimal) = sin z.d. of m.e.p. (ii) ÷ 120. (iv) Square of sine (z.d. of n) = (sine z.d. of m.e.p.)² – (iii)². (v) Sine z.d. of n = √(iv). Example 13. Continue example 10, given already lat = 10° 24′ and new moon is at nā. 20-40. (i) Let us take it that using the ascensional differences of the place (given by chap. IV), the o.e.p. found is rā. 5-27-56. (ii) Sine amplitude of o.e.p. = sine rā. 5-27-56 × 48′ 48″ ÷ sin (90° – 10° 24′) = sin 2° 4″ × 48′ 48″ ÷ sin 79° 36′ = 4′ 20″ × 48′ 48″ ÷ 118′,0 = 1′ 48″. (iii) Sine (m.e.p. ~ ṇ) = 28′ 9″ × 1′ 48″ ÷ 120 = 25″. (iv) Sin² (z.d. of n) = (28′ 9″)² – (25″)² = 792′ .25. (v) Sin (z.d. of n) = √792.25 = 28′ 9″. 19a. A.B. विलग्ना ज्या | b. B. ०षितां; C.D. ०षिता. A.B.D. स्थाप्या; C. स्थाऽपि. b. A. काष्टांत; B3. कापांत० | A.B. ०मेकस्या c. B. मध्यमज्याघ्नी | c. A. ०दृक्षेपाख्यं; B. ०दृक्क्षेपाख्यं 20a. B1. ०तति०; B2.3. ०तति०. D. ०विशे० | d. B2. पृथक्थो य ||

220 PAÑCASIDDHĀNTIKĀ IX.21 The following is the explanation of the rule: See fig. 2. There, n is the nonagesimal, i.e. o.e.p. minus three rāśis. Z is the zenith. n.z is the zenith distance of n, and sine nz, called the 'Sun's dṛk-kṣepa', is wanted here, to get the Moon's parallax in latitude. This Siddhānta takes the spherical triangle Z n M, right angled at n, to be approximately equal to a plane right-angled triangle, with the sines of the arcs as straight lines and finds the dṛk-kṣepa by, (sine zn)² = (sine MZ)² – (sine nM)² = sin² z.d. of m.e.p. – sin² (m.e.p. ~n). (The correct method has already been expounded by us in Chap. VI, when dealing with the solar eclipse according to the Pauliśa.) Sin (z.d. of M.e.p.) has been got already. The other quantity required, viz. sin (M.e.p. ~ n), is got by the well-known formulae relating to spherical right angled triangles, (already given by us), sin (M.e.p. ~ n) = sin (z.d. of M.e.p.) ×sin MZn ÷ 120'. But, MZn = O'ZW = OZE, which is the amplitude of the o.e.p. Its sine, Udayā, given here = the declination of the o.e.p. × 120' ÷ sine colatitude = sin longitude of o.e.p. × 48' 48" ÷ sin colatitude, as given in chap. IV. [शङ्कु:] दृ[क्]क्षेपकृतिं जह्यात् त्रिज्यावर्गात् ततोऽस्य यन्मूलम् । लग्नाऽर्कविवरमौर्व्या गुणितं त्रिज्योद्धृतं शङ्कुः ॥ २१ ॥ Gnomon 21. Subtract from 14,400, the square of sin z.d. of n, (kept unused in the other place in the previous work), and find its square-root. Multiply this by the sine of the distance between the Sun and the o.e.p., and divide by 120'. The result, which is the sine of the Sun's altitude, is called Śaṅku, i.e. the Sun's Śaṅku. ∴ Śaṅku = √ 14,400 – sin² (z.d. of n) × sin (o.e.p. ~ sun) ÷ 120'. [sin² (z.d. of n) has already been got, and kept apart] Example 14. To complete example 10. From Ex. 13, by (i), o.e.p. = rā. 5-27-56, and by (iii), sin² (z.d. of n) = 792.25, from Ex. 10, Sun = rā. 2-0-0. Śaṅku = √ 14,400 – 792.25 × sin (rā. 5-27-26 – rā. 2-0-0) ÷ 120' = 116' 39" × sin (rā. 3-27-26) ÷ 120' = 116' 39" × 106' 7" ÷ 120' = 103' 10". The rule is derived as follows: From fig. 2 it can be seen that the Sun's altitude is 90° – Zs. ∴ Śaṅṅku = Sine Sun's altitude = Cos Zs = Cos ZW × Cos ns ÷ 120' (by the well-known formula, already given) = √ radius² – sin² Zn × sin Os ÷ 120', (∵ ns is os – 90°). sin Zn is sine z.d. of n already found, and its square has already been got and kept apart for use here. ∴ Śaṅku = √ 14,400 – sin² z.d. of n × sin (o.e. P ~ sun) ÷ 120' as given by the author. 21a. A. दृक्षेप; B. दक्षेप. B. कृति A. जह्या c. B.विवरे b. A. वर्गात्रितोस्प. A. °वन्मूलं; B. °पन्मूलं d. B. गुणित. A. त्रिज्योदृवृतं; B. त्रिज्योधृत

IX.23 IX. SAURA-SIDDHĀNTA — SOLAR ECLIPSE 221 [लम्बितपर्वान्तः] शङ्क्वङ्गुलाख्यविंशतिशतकृ(त्योर)न्तरेण विश्लेषात् | स्थि(त)वर्गान्मूलं द्विनवकाहतं (त)द्वि(भ)ज्य कक्षाभ्याम् || २२ || भागविशेषा(त्ति)थिवत्तिथ्य(न्तनाम) पुनः पुनस्तत् स्यात् | एवं मृग्यः कालस्तूत्पन्नो यावदविशेषः || २३ || Parallax-corrected New Moon 22-23. Subtract the square of the Sun's śaṅku got above from 14,400. From the remainder subtract the square of the Sun's dṛk-kṣepa kept apart in the previous work and find its square root, (technically called Dṛggati). Multiply this by 18 and divide by each of the kakṣās of the Sun and the Moon. Find the respective arcs (in minutes) and get their difference. Treat this as the minutes of tithi and find the tithi-nāḍīkās for this. Subtract the nāḍīkās from the time of new moon if forenoon, and add, if afternoon. The parallax-corrected new moon (p.c.n.) is got. Repeat the oper- ation of finding the p.c.n., till there is no difference (in time) in two successive opera- tions. This is the p.c.n. (to be used in the subsequent work). Though there is no doubt about the idea here, it is difficult to get the idea from the words used, on account of several corrupt readings. In verse 22, a word is broken at the end of the third foot, and there are 18 mātrās in the fourth, sinning against the Ārya metre. In the same verse, in the sec- ond foot NP has emended viśleṣāt into viśeṣitāt against the manuscript readings, an emendation that is not needed. In the 23rd verse, evam mṛgyaḥ kālaḥ is a repetition. TS have succumbed to this diffi- culty and give the wrong interpretation that the difference between 14,400 and the square of the śaṅku, should be subtracted from the square of the dṛk-kṣepa, unaware that this is impossible since the latter would always be less than the former. We shall show this in the explanation. As for calling the Sun's śaṅku as 'digits of śaṅku' we have seen it being technically called so in chap. IV. The method enunciated here is as follows: (i) Dṛggati = √(14,400 − śaṅku² − dṛk-kṣepa²). (ii) (a) Sin Sun's parallax in long. = 18 × (i) ÷ Sun's kakṣā (b) Sin Moon's parallax in long. = 18 × (i) ÷ Moon's kakṣā. From the two sines, the arcs should be obtained in minutes. The Moon's minus the Sun's parallax is the (effective) parallax in longitude. 22a. B. ॰ख्यं विंशति b. A. शतकृशोनंतरेण; B. शततशोनन्तरेण (B2. त्तरेण; B3. त्तरेण) D. विशेषि[त]त् c. A.B1.2. स्थिति d. A1.B. हतं सद्भिभाज्य A1. कक्ष्याभ्यां 23a. A. विशेषस्तिथि; B.C. विशेषास्तिथि b. A. तिथ्यर्द्धान्तामतः; D. तिथ्यर्द्धातामनः; C. तिथ्यन्तान्नामतः; D. तिथ्यन्तोऽतः पुनः d. A. ऽक्षूत्पन्नो; B. तत्पन्नो B. यावदवशेषः

222 PAÑCASIDDHĀNTIKĀ IX.23 (iii) Nāḍis of parallax = the parallax in longitude found in (ii) × 60 ÷ the motion of the tithi per day. Subtracting the nāḍis from new moon in the forenoon, and adding in the afternoon, the p.c.n. is got. Finding the m.e.p. etc. of the p.c.n., the work should be repeated upto getting the nāḍis in (iii). These nāḍis are to be subtracted or added to the original new noon. A better p.c.n. is got. Using this time the work may be further repeated for a still better approximation. Example 15. To continue Ex. 10 In the last example the śaṅku got is 103′ 10″. In Ex. 10, the motion per day of the Sun and the Moon found are 57′ and 810″, the Sun's kakṣā found is 16,641 and the Moon's 1171.2. The square of the dṛk-kṣepa kept apart, is 792.25. From these: (i) Dṛggati = √(14,400 − (103 1/6)² − 792.25) = 54′ 27″. (ii) (a) Sine Sun's par. in long. = 18 × 54′ 27″ ÷ 16,641 = 3″.6. (b) Sine Moon's par. in long. = 18 × 54′ 27″ ÷ 1171.2 = 50″.2. The Sun's parallax is arc of 3″.6 = 1′.7. The Moon's parallax is arc of 50″.2 = 24′.0. The parallax in longitude = 24′.0 − 1′.7 = 22′.3. (iii) Nāḍis of par. = 22′.3 × 60 ÷ (810′ − 57′) = 1-47. Since new moon is afternoon, adding to the time of new moon, the p.c.n. = nā. 20-40 + nā. 1-47 = nā. 22-27. We shall repeat the operation for a better approximation. (The motions of the Sun and Moon per day, and their kakṣās need not be done again.) The nāḍis of the p.c.n. after midday = 22-27 − 15-40 = 6-47 = 407 vināḍis. The right ascension, corresponding to the interval of 407 vināḍis, using the ascensional differences of zero latitude = 10°

• 10° + 10° + 10° × 85.2 ÷ 108.8 (for vināḍis 105.4 + 107.6 + 108.8 + 85.2) = 37° 50′, after the Sun ( = rā. 2-0-2, 2′ more for the 2 nāḍis later). ∴ The m.e.p. = rā. 2-0-2 + rā. 1-7-50 = rā. 3-7-52. The declination of m.e.p. = 23° 46′ north. The zenith distance of the point = 23° 46′ − 10° 24′ = 13° 22′, north. Sine z.d. of m.e.p. = 27′ 44″. The o.e.p. at p.c.n. = rā. 6-8-48 (using as before the ascensional differences of 10° 24′). Since amplitude of the point = 7′ 34″. Sine (m.e.p. ~ n) = 27′ 44″ × 7′ 34″ ÷ 120 = 1′ 45″. Sine ² (z.d. of n) = (27′ 44″)² − (1′ 45″ )² = 765.89. Sine (z.d. of n) = 27′ 40″. Śaṅku = √(14,400 − 765.89) × sine (rā. 4-8-46) ÷ 120′ = 91′ 1″. Dṛggati = √(14,400 − (91′ 1″)² − 765.89) = 73′ 9″. Sine Sun's par. in long. = 18 × 73′ 9″ ÷ 16,641 = 4″.8. Parallax = 2′.2. Sine Moon's par. in long = 18 × 73′ 9″ ÷ 1171.2 = 1′ 7″.5. Parallax = 32′.2. Relative parallax = 30′.0. Nāḍis of parallax = 30′ × 60 ÷ 753′ = nā. 2-23. Adding to time of new moon, the closer p.c.n. = nā. 20-40 + nā. 2-23 = nā. 23-3.

IX.24 IX. SAURA-SIDDHĀNTA — SOLAR ECLIPSE 223 Repeating the work, the p.c.n. got will be about nā.23-20. The rule is thus explained: It has been shown in the context of the Paulīśa solar eclipse that the relative total parallax is obtained by multiplying the relative horizontal parallax (π) by sine zenith distance of the Sun (dṛgjyā) and dividing by the radius. In this Siddhānta, the horizontal parallaxes of the Sun and the Moon are got separately by dividing by their distances for the sake of exactness. But the Sun's dṛgjyā is used for the Moon too, since the difference is very small in the neighbour- hood of new moon, with the solar eclipse occurring. In fig. 2, dṛgjyā = sin Zs, and the relative parallax = ss'. Its projection on the ecliptic, sl, is the relative parallax in longitude, by which (Moon — Sun) has got to be increased or decreased to get their apparent difference in longitude. In the figure, since s is west of n, it is subtractive, and p.c.n. is later, and therefore the nāḍīs of parallax are additive. (When the Sun is east of n and parallax is additive, clearly the nāḍīs are subtractive.) Since (Moon — Sun) is tithi element, the relative parallax is treated like tithi, and multiplied by 60 and divided by the daily motion to get the nāḍīs of parallax. Now, sl is found thus in this Siddhānta: sl² = ss'² - s'l². (∵ the triangle ss'l is right-angled at l, and so small that it may be considered plane.) = ss'² - ss'².sin² l ss' = ss'² - ss'².sin²Zn ÷ sin² Zs (∵ triangle Zns is right angled at n) = π² dṛgjyā² - π² sin² (z.d. of n) (∵ ss' = π × dṛgjyā = sin zs, and Zn is the z.d. of n) = π² (radius² - Śaṅku² - sin² z.d. of n). (dṛgjyā² = radius² - Śaṅku²) ∴ sl² = π √(120² - śaṅku² - sin² z.d. of n) = π × dṛggati, as given But, π = the Moon's horizontal parallax - the Sun's horizontal parallax. ∴ the dṛggati is multiplied by each and then subtracted. It has been said already, in previous two solar eclipse contexts, that the sine of the horizontal parallax is obtained by dividing the earth's radius by the respective distance. Since the author uses as the divisor not the actual distance but the respective distance divided by 43, the earth's radius also has to be taken divided by 43. The author takes 788 yojanas as the earth's radius, adopting the value of the Āryabhaṭīya and multiplying it by 3/2 to express it in the yojana measure of the Ārdharātrika etc. systems. (These systems give the earth's radius as 800 yojanas.) Dividing it by 43 we get 18.3, and the author gives it as 18, corrected to the nearest unit's place. Further, since Zn is perpendicular to the ecliptic, Zs, the zenith distance of the Sun at any position on the ecliptic, is always greater than Zn, and, accordingly, their sines also, since the arcs are all less than 90°, i.e. (radius² - śaṅku²) is always greater than sin² z.d. of n. Therefore, the interpretation of TS that the former is to be subtracted from the latter is wrong, as mentioned already. The need for successive approximation by repetition of work is plain. [नति:] अविशेषाद् (दृक्क्षे)पं 'वस्वेक'घ्नं विभज्य कक्षाभ्याम् । लब्धान्तरचापांशा मध्यज्यादिग्वशेन नतिः ॥ २४ ॥

224 PAÑCASIDDHĀNTIKĀ IX.25 Parallax in latitude 24. Take the sine z.d. of n last got in the successive approximation, multiply by 18, and divide by the respective kakṣās. The respective sine parallax in latitude is got. The arc of their difference is the relative parallax in latitude and its direction is that of sine z.d. of m.e.p. (i.e. of M from Z.) Since the sines are very small, it is immaterial whether the arcs are found first and their difference is taken, or whether the arc of the difference of the sines is taken, both being the same practically. But the latter will entail less work. a) Sine parallax lat. of the Sun = 18 × sin z.d. of n ÷ Sun's kakṣā. b) Sin parallax in lat. of the Moon = 18 × sin z.d. of n ÷ Moon's kakṣā. (b) − (a) is the sine of the relative parallax in lat. whose arc is to be found, and its direction is that of M from Z. Example 16. To continue example 10. The sine z.d. of n, last got in the successive approximation in the last example is 27′ 26″ say. (a) Sun's sine par. in lat. = 18 × 27′ 26″ ÷ 16,641 = 1″.8. (b) Moon's par. in lat. = 18 × 27′ 26″ ÷ 1171.2 = 25″.3. Sin relative par. in lat. = 25″.3. − 1″.8 = 23″.5. Rel. par. in lat. = arc of 23″.5 = 11′.2, north, since M is north. The rule is thus derived: The zenith distance of the nonagesimal, Zn, is the Sun's dṛkkṣepa. In the context of the solar eclipse, according to the Paulīśa, it was shown how the parallax in latitude (p.c.lat) is to be got by multiplying this dṛk-kṣepa by the relative horizontal parallax and using it as a correction to the Moon's latitude to obtain the corrected latitude. Here, the parallax is derived separately for each of the Sun and the Moon, for the sake of greater accuracy. Now, the direction of the nonagesimal and the sine of its zenith distance is the same as that of sine z.d. of M, and the direction of the apparent shifting of the Sun and the Moon by parallax, as resolved on the line of latitude, is the same as that of sine z.d. of n, (as ls' in fig. 2). Therefore, the direction of the parallax correction is the direction of sine z.d. of M, as mentioned by the author. (In the fig. it is north.) Thus everything is explained. ज्याविधिना विक्षेपं तत्कालं प्राप्य तेन सहितोना । स्पष्ट[T] नतिः प्रमाणैः स्वैस्स्वैर्ग्रासं स्थितं च वदेत् ॥ २५ ॥ 25. The Moon's latitude at the time taken is to be got by using the sine (of Moon ~ Rāhu), and this is to be added to or subtracted from the parallax correction in latitude (according to their direction). This is the parallax- corrected latitude (p.c. lat.). This is to be got separately for each of the times separately and from them the times of total obscuration and total duration are to be got. 24a. A. दृक्षेपं; B. दृक्षेप 25a. B. विक्षेप d. A. स्वैस्वैग्रसं; B. स्वैस्वैग्रांमं b. B. चस्वेकग्रं (B2.3.°ग्रं) d. B. ज्याद्विख b. A. प्राथ c. A.B. स्पष्टनति D. स्थिति

IX.26 IX. SAURA-SIDDHĀNTA — SOLAR ECLIPSE 225 The maximum latitude is given in verse 6 to be 270'. The use of sine (Moon − Rāhu) has been already indicated in the context of the Romaka, and therefore only indicated here. In correcting, like directions are additive, and unlike directions subtractive, the resulting direction being that of the greater. The parallax-corrected latitude is to be got for each of the time of first contact, last con- tact, and middle, the last serving for immersion and emergence too, as these times are near enough to the middle. The separate computation of the corrected latitude suggests that the nāḍīs of parallax also are to be computed separately for the different times. Thus, the following is to be done: (i) The uncorrected latitude = 270' × sine (Moon − Rāhu) ÷ 120 = sine (Moon − Rāhu) × 9 ÷ 4. (If Moon − Rāhu) is less than 6 rāśis, the latitude is north, otherwise south. Rāhu here means the Head of Rāhu). (ii) Parallax-corrected latitude = latitude ± relative parallax in latitude (+ if of the same direc- tion, and ~ if of different directions, the resulting direction being that of the greater). Example 17. (To continue Ex.10,) find the parallax-corrected latitude at the final parallax corrected new moon, i.e. at nāḍī 23-20. This time is nāḍīs 2-40 later than new moon. So, from the data given in Ex. 10, Rāhu-head = rā. 7-29-24, and Moon = rā. 2-0-0 + 810' × 2 2/3 ÷ 60 = rā. 2-0-36. Moon − Rāhu = rā. 6-1-12. From this, Moon's latitude = 9 × sine (rā. 6-1-12) ÷ 4 = 5'.7, south. (∵ Moon − Rāhu-head) > rā. 6-0-0.) (ii) Parallax corrected lat. = 5'.7 ~ 11'.2 = 5'.5, north (∵ of different directions, north being greater.) These rules have been explained before. [विमर्दकालः] अवनतिवर्गं जह्याद् रवीन्दुपरिमाणयोगदलवर्गात् । तन्मूला(त्तु) द्विगुणात्(ति)थिभुक्तवदादिशेत् कालम् ॥ २६ ॥ Duration of the eclipse 26. Subtract the square of the parallax - corrected latitude from the square of the sum of the semi-diameters of the Sun and the Moon and find the square root. Double this, and find the time for it, treating it as the motion of tithi. (The duration of the eclipse it got.) This verse has already occurred as verse 16 of chap. VIII and fully explained there. The only difference is two mis-readings here. Example 18. To continue Ex. 10. In Ex. 11, the angular diameter of the Sun has been found to be 31', and of the Moon, 33' and the daily motion of the tithi 753'. The parallax-corrected lat. has been found to be 5'.5. From these, Duration in nāḍikās = 2 × 60 × √(31 + 33)/2 − 5.5² ÷ 753 = 2 × 60 × 31.52 ÷ 753 = *nā.*5-1. 26b. B3. ॰न्दुः. B. परिपरिमाणग्दल (B3. ॰योगदल) c. A.B. मूलात् d. A.B1.2.तिथिभुक्ति. B. ॰वदादिकेत्कालं

226 | PAÑCASIDDHĀNTIKĀ | IX.27 It has already been mentioned that half the duration subtracted from the final parallax-corrected new moon is the beginning and added to it is the end of the eclipse. Further, if the difference of the semi-diameters is used in the work, instead of the sum, the duration of total eclipse is got. Here, if the Moon's is greater, there is actual total obscuration. If the Sun's is greater, there is annular eclipse. The author expects us to be conversant with these things. तिथ्यवना(मो) ग्रहणादिना(म)विश्लेषि[तो यु]तः स्थित्याम् । गोलाऽन्यत्वे देयस्त्ववनामो[मौ]क्षि(क)स्यैवम् ॥ २७ ॥ 27. Find the nāḍis of parallax for the time of the beginning. If the time of beginning and the new moon are both in the forenoon or both in the after- noon, find the difference of the nāḍis of parallax and add it to the half duration to get the correct half duration to be subtracted from the time of the corrected new moon. If one is before noon and the other afternoon, add the nāḍis of parallax, and add it to the half-duration to get the correct half-duration (to be subtracted from the time of parallax-corrected new moon). Do the same for the time of the end of the eclipse, (to find the correct half duration to be added to the parallax-corrected new moon, to get the correct last contact). The following example will make the meaning clear. Example 19. To continue Ex. 10. We have already obtained, par.c. new moon = nā. 23-20, parallax-correction for new moon = nā. 2-40, and the total duration nā. 5-1. Applying the half duration on both sides of p.c.n, the approx. time of first contact = nā. 20-50, last contact = nā. 25-51. The parallax correction in time for first contact using verses (22-23) is nā. 1-50. As both the new moon and time of first contact are in the same part of the day, i.e. afternoon, the difference between their parallax correction = nā. 2-40 − nā. 1-50 = nā. 0-50. This is to be added to the half duration to get the first half duration. Adding, nā. 2-30 + nā. 0-50 = nā. 3-20. Subtracting this from the parallax-corrected new moon, the correct time of first contact = nā. 23-20 − nā. 3-20 = nā. 20-0, after sunrise. Next, the parallax-correction for time of approx. last contact is nā. 3-30. As both new moon and last contact are in the afternoon, subtracting the corrections, for both from each other, we have nā. 3-30 − nā. 2-40 = nā. 0-50. Adding this to the half-duration we have, nā. 2-31 + nā. 0-50 = nā. 3-21, for the correct second half duration. Adding this to the p.c.n, the correct time of last contact = nā. 23-20 + nā. 3-21 = nā. 26-41 after sunrise. The following is the explanation of the rules for the correction given here and the justification for our interpretation. At first the duration is given neglecting the effect of parallax on the time. If the parallax is taken into account, the duration will always be longer than otherwise, as we have said. This can be seen from the following consideration. Let us take the case when the end of new moon 27a. A.B. नाम b. A.B.C.D. ॰दिना च वि॰. A.B. विश्लेषित; c. B. ॰न्य चेदेय C.D. विश्लेषितः d. A. स्त्वनामो A.B. Hapl. om of मौ

IX.27 IX. SAURA-SIDDHĀNTA — SOLAR ECLIPSE 227 is before noon. The first contact being earlier still, its interval from noon is greater, and therefore its nāḍīs of parallax too is greater than those of the new moon. Since both are subtractive, the first half duration is lengthened, the first contact happening earlier. Therefore the difference is to be added to the duration. In the case taken, the last contact may happen before noon or after noon. If before noon, the interval from noon upto the last contact is less than that upto new moon. There- fore the nāḍīs of parallax of the last contact is less than those of the new moon, and both are sub- tractive. Therefore the second half duration also is lengthened, the last contact happening later. So the difference is, here too, additive to the duration. If the last contact is afternoon, the naḍīs of parallax are clearly additive to the time of last contact, and the last contact happens later. But the parallax-corrected new moon occurs earlier, and so the second half-duration is lengthened both ways, and so the sum of the parallaxes is added to the duration. Thus in all three possibilities of the first case, there is only additive correction. Let us now take the second case, viz. that the new moon occurs after noon. Clearly what is said for the first contact in the first case applies to the last contact in the second case, and vice versa, but the additiveness and subtractiveness alone have to be interchanged. Therefore, here too, in all three possibilities the differences or sums, have to be added to the duration, as we have said in giving the meaning of the verse. Not understanding the above, TS have interpreted the verse in such a way that the instruction will result in lessening the duration, which is contrary to facts. Now, for the readings. In the second foot of the verse three mātrās are missing, and to restore them we have read viśleṣita as viśleṣito yutaḥ in accordance with the meaning. The emendation, by TS and NP, of the manuscript reading viśleṣitasthityām into viśleṣitaḥsthityā does not express the intended idea fully. In the fourth foot two mātrās are missing, and to restore them we have read the meaningless nāmokṣi as nāmo maukṣī. To conclude: In the introduction to this chapter we said that the Sun, Moon, and Rāhu, together with the methods of computing them are better in the Saura than in the Romaka. Now, we have seen that in the computation of the solar eclipse also, the Saura excels. For instance, the mean angular diameters of the Sun and the Moon are 30' and 34' according to the Romakas, while they are 32' and 32' according to the Saura, very near the correct 32' and 31', respectively. Computing true diameters and the parallax using the distance of the instant of eclipse, and using the true motion of the time of eclipse for getting the duration etc. are commendable in the Saura. Getting the sine z.d.n. by using the sine of the zenith distance of m.e.p. is a better method than that used by the Romaka, as also the method of successive approximation for various things like parallax in time of new moon etc. The abandoning of the Romaka's faulty correction of the Moon's position in its own orbit, is itself praiseworthy. With such good features, the Saura is easily the best of the five Siddhāntas. [इति पञ्चसिद्धान्तिकायां वराहमिहिरविरचितायां सूर्यसिद्धान्तेऽर्कग्रहणं नाम नवमोऽध्यायः ]¹

1. Col. A.B.D. इति (B. om इति) सूर्यसिद्धान्तेऽर्कग्रहणं (B. णनाम) नवमोध्यायः | C. इति सूर्यसिद्धान्ते सूर्यग्रहणं नाम नवमोऽध्यायः | Thus ends Chapter Nine entitled ‘Saura-Siddhānta: Solar Eclipse’ in the Pañcasiddhāntikā composed by Varāhamihira

Chapter Ten SAURA-SIDDHĀNTA — LUNAR ECLIPSE १०. दशमोऽध्यायः सौरसिद्धान्तः — चन्द्रग्रहणम् Introduction In this chapter the method of computing the lunar eclipse according to the Saura Siddhānta is given. Since the true Sun and the Moon and Rāhu, the true distances of the Sun and the Moon, and the Moon’s angular diameter and latitude, have already been given in chap. IX, the angular diameter of the Shadow alone is given here, as also the computation of the times of contacts etc. The last three stanzas give the amount of eclipse at a desired time, as also the beginning and end of total phase of the eclipses, both of the Sun and the Moon. [तमोबिम्बमानम्] रविकक्षा नवतिगुणा ‘षडष्टदस्रो’द्धृतेन्दुकक्षायाः । छेदः ‘षट्त्रि’घ्नाया ल(ब्धे) नोनश्च षड्वर्गः ॥ १ ॥ ‘वियदर्क’गुणे शशिक(क्ष्य)या हृते कार्मुकं त(मो)व्यासः ॥ २ a ॥ Diameter of the Shadow 1-2a. Multiply the Moon’s true distances in its orbit by 36, and divide by the Sun’s true distance multiplied by 90 and divided by 286. Subtract this result from 36, multiply by 120, divide by the Moon’s true distance and get the arc of the resulting sine. This is the angular diameter of the Shadow. The following is asked to be done: (i) ‘Result’ = 36 × Moon’s true dist. ÷ (90 × Sun’s true dist. ÷ 286) = 36 × Moon’s true dist. × 286 ÷ (90 × Sun’s true distance). (ii) Sine angular diameter of Shadow = (36 − ‘result’) 120 ÷ Moon’s true distance. Or, simplifying, this is equal to: {36/Moon’s true distance − (36/Sun’s true distance) × 286/90} × 120 = 4320/Moon’s true distance − 13,728/Sun’s true distance. 1a. A. कक्ष्या. B. नवतीगुणा b. A. द्रुतेन्दु. A. B 1. 3. कक्ष्यायाः C. षडश्च c. B 1. 3. षद्रिघ्नाया d. A. लघोनो; B 1. 3. लधोनातश्च 2a. B 1. 3. वियदर्वगुणे b. A. B. कक्ष्याया; C. D. कक्ष्या. A. तमोर्व्यांसः; B 1. 3. तयोव्याघ्रः

X.2 X. SAURA-SIDDHĀNTA — LUNAR ECLIPSE 229 Or, (since the arc is small, multiplying this by 3438 and dividing by 120), the angular diameter of the Shadow in minutes = 1,23,768 ÷ Moon's true distance − 3,93,307 ÷ Sun's true distance. Example 1. On a certain day at the time of full moon (T) the true Sun is rā. 10-0-0, the true Moon is rā. 4-0-0, Rāhu Head is rā. 3-25-0, the Sun's motion per day for the time is 60′, and the Moon's 780′. Compute the lunar eclipse. The Sun's true dist. = 9,48,558 ÷ 60 = 15,809 (by IX.15). The Moon's true dist. = 9,48,680 ÷ 780 = 1216.3 (by IX.15). The Moon's angular dia. = 38,640 ÷ 1216.3 = 31′.77 (by IX.16). The Moon's lat. at T = 9 × sine (rā. 4-0-0 − rā. 3-25-0) ÷ 4 = 23′.5, north. From the true distances got above, the angular diameter of the Shadow = 1,23,768 ÷ 1216.3 − 3,93,307 ÷ 15,809 = 101′.76 − 24′.88 = 76′.9. The following is the explanation of the method for finding the Moon's angular diameter: The actual diameter of the Shadow is the diameter of the circular section of the Shadow-cone (formed by the earth intercepting the Sun's light,) at the Moon's orbit, at the time of full moon. This is represented in fig. 1, below by U′U″. Fig. X. 1 The angle subtended by this at the centre of the earth, E, is the angular diameter desired to be computed here. In the figure, S is the centre of the Sun, E, that of the Earth, and U that of the Shadow section. SS′ is the Sun's radius, EE′ is the Earth's, and UU′ is that of the Shadow. SE is the true distance of the earth from the Sun, and EU, that of the Moon from the earth. S′E′U′ is the direct common tangent to the orbs of the Sun and the Moon. As the distance are very great when compared with the radii, EE′, SS′, and UU′ are practically parallel. Draw ES″ parallel to E′S′, and U′E″ parallel to U′E′. The triangles, ES″S and UE″E, are similar. Therefore, E″E/EU = S″S/SE.

230 PAÑCASIDDHĀNTIKĀ X.4 ∴ E"E = S"S × EU/SE = EU × (S'S - S' S")/SE = EU × (S'S - E'E)/SE. In order to get the angular diameter of the Shadow, we require the radius of the Shadow, UU', = EE' - EE" = EE' - {EU (SS' - EE')/SE} = 18 - Moon's true dist. × (Sun's radius - 18) ÷ Sun's true dist. = 18 - Moon's true dist. × {18 × 5,14,787 ÷ (2 × 18 × 3438) - 18} ÷ Sun's true dist. (Here, 18 is the number obtained by reducing the earth's radius by 43, and given by the author in giving the parallax, see IX.23.) 18 × 5,14,787 ÷ (2 × 18 × 3438) is the Sun's radius, since, of two orbs, the parallax as viewed from one is the angular semi-diameter as viewed from the other. Therefore: Sun's minutes of parallax: Sun's angular semi-diameter in minutes :: 18: Sun's reduced radius. But the Sun's minutes of parallax = 18 × 3438 ÷ the Sun's true distance, and the Sun's semi- diameter in minutes = 5,14,787 ÷ (2 ×the Sun's true distance) = 18 - Moon's true distance × 18 × 284.4 ÷ (90 × Sun's true dist.) Since, sine angular diameter of the Shadow is got by multiplying the radius by 2, and the max. tabular sine and dividing by the Moon's true distance, sine angular diameter of Shadow = {36 - Moon's true dist. × 36 ÷ (90 × Sun's true dist. ÷ 284.4)} × 120 ÷ Moon's true distance, almost the same as the author has given, but with 284.4 instead of 286. If the number had been 17.9 instead of 18 taken as a whole number for convenience, then we shall get 286 itself, as given by the author. We have already shown that the formula can be simplified. The author must have given it in the involved form for indicating the geometrical construction by way of proof. When the numbers occurring are seen to be correct in the way shown by us, it is quite improper for TS to read ṣaḍaṣṭadasra (278) as ṣaḍaśvadasra (276) and to agree with this, making the Sun's reduced diameter as 146, in their proof, instead of the correct 149.73, got by dividing 5,14,787 by 3438. [विमर्दकालः] चन्द्रतमोव्यासयु(तिं) द्वाभ्यां हृत्वा ततो वर्गात् ॥ २ b ॥ विक्षेपवर्गहीनादासन्नपदे 'वियद्विद्विचन्द्र'घ्ने | सूर्येन्दुभुक्तिविवरो(द्धृ)ते स्थिते(र्ना)डिका लब्धाः || ३ || प्रग्रहणे(न्दोः) कृत्वा विक्षेप [म] तोजनया स्थि(ति) र्भवति । एवं भूयो भूयः स्थित्य(वि) शेषः कृतो यावत् ॥ ४ ॥ Duration of the Eclipse 2b-3. Add the angular diameters of the Moon and the Shadow, divide by two, and square it. Subtract the square of the Moon's latitude from this, and find

X.4 X. SAURA-SIDDHĀNTA — LUNAR ECLIPSE 231 the square root. Multiply this by 120 and divide by the difference of the motions per day of the Sun and the Moon pertaining to the time of eclipse. The duration of the eclipse is got in nāḍikās. 4. Find the Moon’s latitude at first contact and using this find a more correct duration. Repeat this till there is no difference between the previous and the next durations. Note: Though the total duration alone is given here, we are expected to know how to find the first and last contacts from this, from previous contexts. In the successive approximation, what is said for the first contact must be taken for the last contact also. Therefore the following is asked to be done: (i) Rough duration = 120 √(half-sum of angular diameters)² – lat.² ÷ difference of instantane- ous daily motions. (ii) T ± half (i), are the rough first and last contacts. (iii) Using the latitude of the rough first contact and repeating (i) gives successively better first contacts. (iv) Using the latitude of the rough last contact, and repeating (i), gives successively better last contacts. (It should be noted that the shorter the duration the greater are the number of repetitions required.) Example 2. Continue Ex.1. (i) Rough duration in nāḍis 120 √{ (76.9 + 31.77)/2 }² – 23.5² ÷ (780 – 60) = 120 √ 54.34² – 23.5² ÷ 720 = 120 × 49 ÷ 720 = nāḍis 8-10. (ii) Rough times first and last contacts = T – nā. 4-5: T + nā. 4-5. (iii) The Moon at rough first contact = rā. 3-29-6.9, Rāhu then = rā. 3-25-0.2. Moon – Rāhu = 4° 6′.7. From this the Moon’s latitude is 19′.35, north. Using this, a more correct duration for first contact = √54.34² – 19.35² × 120 ÷ 720 = nā. 8-28. Subtracting half this from the time of full moon, the first contact is at T – nā. 4-14. There is no need to repeat, since the duration is long. (iv) The Moon at rough last contact is rā. 4-0-53.1, and Rāhu then, rā. 3-24-59.8. From this the Moon’s lat. is 27′.65, north. Using this, a more correct duration for last contact = √54.34² – 27.65² × 120 ÷ 720 = nā. 7-48. Adding half this to full moon time, the last contact is at T + nā. 3-54. There is no need to repeat. The method has been explained several times before, which need not be repeated here. As for understanding that the successive approximation is for getting the last contact also, though men- 2c. A.B. युतिः. B1.3. हत्या; B3. हत्या 4a. A.B.C. प्रग्रहणेन्दुः 3b. A.B. वियद्वि b. A.B. om. म. A.B. स्थितेः d. A.B. धृते. A.B.C.D. स्थिते for स्थिते; A. लब्धा d. A.B. स्थित्यवशेषः

232 PAÑCASIDDHĀNTIKĀ X.6 tioned only for the first contact, the two common statements anāyā sthitir bhavati and sthityaviśeṣaḥ kṛto yāvat, indicate this. [इष्टकालग्रासः] अर्केन्दुभुक्तिविवरं वाञ्छितनाडीहतं तु षष्टिहृतम् । स्थितिलिप्तास्ताभ्यस्त(त्त)कालेन्दोश्च वि(क्षे)पात् ॥ ५ ॥ कृतियोगपदं शोध्यं शशिराहुकला(प्र)माणयोगदलात् । यच्छेषं तद् ग्रस्तं ज्ञेयं तत्कालमर्केन्द्बोः ॥ ६ ॥ Obscuration at any desired moment 5-6 Take the nāḍīs before or after full or new moon upto the times for which the amount eclipsed is wanted. Multiply this by the difference of the Sun's and Moon's daily motions, (mentioned above), and divide by 60. The 'corres- ponding minutes of arc' are got. Square this, square the Moon's latitude for the moment, add them, and get the square root. Subtract this from the half- sum of the diameters of the eclipsing and the eclipsed bodies. The remainder is the minutes of arc eclipsed, at the moment taken, of the Moon in the case of the lunar eclipse, and of the Sun in the case of the solar eclipse. It is clear that by 'corresponding minutes of arc' is meant here, the distance in minutes between the Moon and the shadow, measured along the ecliptic. From the instruction it is clear that the nāḍīs taken is the interval between full or new moon and the moment for which the amount of eclipse is wanted. It is clear from the context that the Shadow is meant by the word Rāhu. Though from the mention of the Shadow, and the Moon's latitude without any mention of parallax, this seems to be given for the lunar eclipse only, the expression arkendvoḥ at the end shows that this is meant for the solar eclipse also. The author thinks that the reader has acquired sufficient knowledge, by now, to make the necessary changes when applying the rule to the solar eclipse. Therefore, in the case of the solar eclipse, the amount eclipsed is got by using in the rule, the parallax-corrected latitude for latitude, the Sun's and the Moon's angular diameters for those of the Moon and the Shadow, and the parallax-corrected difference of daily motions for the mere difference of daily motions. Thus, the following is instructed to be done: A. To find the amount eclipsed in the case of the Moon (i) "Corresponding minutes of arc" = difference of instantaneous daily motions of Sun and Moon × interval in nāḍīs from full moon ÷ 60. 5c. C.तत्स्थितिलिप्ताविवरत्. A.B.ताभ्यस्ता; D.ताभ्यस्तु d. A.तान्तकालेन्दोश्च (A2.तातत्का). AB.विशेषात् 6a. B.ततियोग० b. B.शशिराङ्ग (B2.B.ब्ज) कलां. A.C.D.कलाद्यमान; B.कलाघमाण c. A1.यछेपं; A2.यछे षं d. B1.3.मर्केन्दो:

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.

234 PAÑCASIDDHĀNTIKĀ X.6 Moon's orbit M R r Ecliptic S M' Fig. X. 2 What has been proved for the lunar eclipse can be taken for the solar eclipse also, with the neces- sary changes. The difference in motions should be here corrected for parallax and used. That this corrected difference is always less than the uncorrected will be clear, when we consider that always the parallax-corrected half duration is always greater than the uncorrected, which we have already proved. Therefore it is clear that by multiplying the “corresponding minutes” by the uncorrected half duration, and dividing by the parallax-corrected half duration, will give the parallax-corrected “corresponding minutes”. It is also clear that in the case of the solar eclipse we must use in the proof the parallax-corrected latitude in the place of the uncorrected latitude, the Sun for the Moon, and the Moon for the Shadow. When this is done, the proof is exactly similar to that for the lunar eclipse. Another thing is to be noted. The Hindu astronomers took the amount of eclipse at full moon or corrected new moon as the maximum and called it the magnitude, (grāsa-pramāṇa), though actually this is only very nearly the maximum and the actual maximum occurs a little earlier or later. If we take this full or new moon itself for doing the present work, since the time interval is zero, and thereby the ‘corresponding minutes’ are also zero, the latitude itself becomes the distance between the centres. Therefore we got that the amount eclipsed in this case (i.e. the magnitude) is to be got by subtracting the latitude of full or corrected new moon, from the half sum of the angular diameters, as already given. Now for the readings: vāñcchitanāḍī (‘desired time’) is meant here the interval in time from the full or new moon, either before or after. But TS have taken the expression to mean ‘the desired point of time’, and in order to get the meaning of ‘interval’ that is wanted for use, have emended the already correct sthitiliptāstābhyas tat into tatsthitiliptāvivarāt, which is unnecessary. Another thing must be said here. If the reading had been tithiliptāḥ instead of sthitiliptāḥ given by the manuscripts and accepted by us, it would have been better; for this would mean the minutes of tithi, as indeed these are, being part of a tithi by nature.

X.7 X. SAURA-SIDDHĀNTA — LUNAR ECLIPSE 235 [पूर्णग्रासकालः] अन्त्याद्ययोर्विशेषा (द) वनतिविक्षेपवर्गविवरपदम् | द्विगुणं तिथिवत् कृत्वा विमर्दकालोऽर्कचन्द्रमसोः || ७ || Time of total obscuration 7. Take the difference of the angular semi-diameters, instead of their sum. Square it, subtract the square of the parallax-corrected latitude (in the case of the solar eclipse) or of the latitude (in the case of the lunar,) find the square root, double it, and treat it as tithi, (i.e. multiply by 60, and divide by the difference of the parallax-corrected daily motions for the solar eclipse, or of the mere daily motions in the case of the lunar). The time of total obscuration is got. In short, everything done for the duration, using the difference of the semi-diameters instead of the sum, is to be done for this. Halving this time and subtracting from or adding to the corrected new moon or full moon gives the first approximate times of immersion and emergence. In the case of the lunar eclipse, successive approximation should be done. In the solar eclipse this is not necessary, because the times of immersion and emergence are very close to the corrected new moon. The parallax-correction for the motion alone need be taken into account and that once for all. Another point to be noted is that, in the solar eclipse, if the Sun’s angular diameter is greater than the Moon’s, instead of a total eclipse there will be an annular (ring-like) eclipse, since the Moon will not be big enough to hide the Sun. The times got, in this case, give the beginning and end of the annular phase. Also, the given examples cannot be continued to illustrate this section, because under the conditions got there will be no total phase, since the latitudes are greater than the differ- ence of the semi-diameters. The proof of the rules given here has already been given in connection with the eclipses according to the Vāsiṣṭha and Pauliśa with graphical illustrations. Now for the text, and readings: The text does not instruct that the difference of the semi-diameters should be squared before adding to the square of the latitude. But mathematical principles indicate it, since the addition of an unsquared quantity with a squared one is unwarranted. We have corrected viśeṣāvavanati into viśeṣādavanati while TS have corrected it into viśeṣāddalanati and NP into viśeṣārdhabhapati. It is clear that they have taken more liberty with the text than necessary, and it is also purposeless. [इति पञ्चसिद्धान्तिकायां वराहमिहिरविरचितायां सूर्यसिद्धान्ते चन्द्रग्रहणं नाम दशमोऽध्यायः |]¹

1. Col. A.D. चन्द्रग्रहणं दशमोऽध्यायः; B. चन्द्रग्रहणे.दशमोध्यायः C. इति चन्द्रग्रहणं नाम दशमोऽध्यायः Thus ends Chapter Ten entitled ‘Saura-Siddhānta: Lunar Eclipse’ in the Pāñcasiddhāntikā composed by Varāhamihira 7a.b. A. अंत्याययो विंशेषाववनति; B. अताद्ययार्विशेषावनत्ति C. विशेषाद्दलनति; D. विशेषाधर्भपति b. B1.3. विचरपदं c. B1. कृचा d. B. कालो चन्द्र

Chapter Eleven

ECLIPSE DIAGRAM

एकादशोऽध्यायः ग्रहणपरिलेषः Introduction Since the distinction among eclipse-types, and various ideas mentioned therein, will not be clear without graphical representation, the author 'follows up' the chapters on eclipses with one solely devoted to this subject. [अपमण्डलाद्यङ्कनम्] यष्ट्या वि(द्धा)ङ्गुलया वृत्तं परिलिख्य संप्रसार्य दिशम् । अ(न्त्या)द्यदलैक्येना(थ य)दपरमर्धेन चाद्यस्य ॥ १ ॥ चन्द्रा(म्ब)रान्तरांशोत्क्रमज्यया ज्यां निहत्य वैषुवतीम् । 'खार्का'शांशा(नु)दयास्तमयोदग्द्याम्यतो दद्यात् ॥ २ ॥ Marking the ecliptic etc.

1. Using the stick-instrument with notch-marks of digits, draw the circle called the 'sum-circle', having for its radius the half sum of the diameters con- verted into digits. Mark the east-west and north-south lines. (E-W, and N-S, in fig. 1). Similarly, using the semi-diameter of the eclipsed body, converted into digits as radius, draw the 'eclipsed body circle', concentric with the sum-circle. (See fig.)
2. Find the versine of the hour-angle (of the Moon at mid-eclipse) and multiply this by the tabular sine of the latitude of the observer and divide by 120. Find the arc of degrees of the resulting sine. If the hour-angle is east, lay the degrees north of the east-point, if west, south of the east-point. The east-point with reference to the equator is thus got. (E', in the figure. E'-W' is the cor- responding east-west.) 1a. B. षष्ठया A.B. विध्यंगुलयाः; C.D. विध्यङ्गुलया b. A1. वृत्तं; B1.3. वृत्तं B3. दिशां; C.D. दिशः c. A. अंताद्यदलैक्योना; B1.3. अन्ताद्युदलैक्योनात् c-d C.D. ०दलैक्येनाद्यमपर d. A.यदपर; B1.3. पदपर B2.चापस्य 2a. A. चंद्रावरात्तरांशो; B. चन्द्रावतरांशो; D. न्तरेशात् b. B. तक्रमज्याथाज्यां A.B. विहृत्य B. वैषुवती c. A. खार्काशादुदया; B.खा (B1.3.ख) र्कीशांशाम्बुरदया; d. A. दम्न मयोतुदग्द्याम्यतो: (A2. gap for मयो तु दग); B. स्तमयोनुदग्द्याम्यतो; D. स्मयात्तुदग्द्याम्यतो

XI.3 XI. ECLIPSE DIAGRAM 237 सत्रिगृहस्य हिमांशोरपक्रमांशान् यथादिशं कुर्यात् । प्रागपरसिद्धिरवेवं (चक्रा)द् याम्योत्तरे ज्ञेये ॥ ३ ॥ 3. Add three rāśis to the Moon’s longitude and find the degrees of declina- tion of this point. If the declination is north, lay the degrees north of Eʹ, if south, south of Eʹ. This is the east-point with respect to the ecliptic (E″ in the figure.) Draw the straight line through the centre, E″ OW″ . E″ – W″ is the ecliptic east-west. By means of circles, (i.e. by drawing the perpendicular bisector), get the ecliptic north-south, viz. N″ – S″. [Diagram: Fig. XI.1 showing celestial/ecliptic circles with points N, N″, S, S″, E, Eʹ, E″, W, Wʹ, W″, O, A, b, f, fʹ, l, lʹ, s, w, "Eclipsed Circle", and "Sum-Circle"] Fig. XI.1 For illustration, we shall represent in the figure the lunar eclipse worked out in the examples of chap. X. The angular diameters of the Moon and the Shadow got there are 31ʹ.8 and 76ʹ.9. The Moon’s lat. at first and last contacts are, respectively, 19ʹ.35 and 27ʹ.65, both north. The first and second half durations are nā. 4-14 and nā. 3-54, respectively. Let us assume that at T, the hour angle, is 10 nāḍīs, i.e. 60°, west and the latitude of the observer is 10° 24ʹ (N). The Moon’s longitude has already been given as rā.4-0-0. (i) The half sum of the angular diameters = 108ʹ. 7 ÷ 2 = 54ʹ.35. This is to be converted into digits using the formula of verse 6, below, and used as the radius of the sum-circle. It is, 54.35 ÷ (3 – 10/15) = 23.3 digits. According to the scale in the figure, 1 unit = 10 digits, this is 2″.33. 3b. B. ॰मांशात् द्यथा c. A. सिधिरेवं. C. मत्स्याद्; D. बकाद् d. A.B. वकाद् A. याम्यान्तरे; B. याम्योतर D. ज्ञेये [च]

238 PAÑCASIDDHĀNTIKĀ XI.5 (ii) The semi-diameter of the eclipsed body, (here the Moon), is 31′.8 ÷ 2 = 15′.9 = 16 ÷ (3 – 10/15) in digits, = 6.8. According to the scale used, this is represented as, "0.7, radius of eclipsed circle. (iii) For the hour angle of 60°, the tabular versine = 60′, and tabular sine latitude of observer is 21′ 40″. From these, the tabular sine of the angle of deflection caused by lat. = 60′ × 21′ 40″ ÷ 120′ = 10′ 50″. The angle of deflection = 5° 11′, south of the east point, since the hour angle is west. This is angle, EOE′, and E′ – W′ is the equatorial east-west. (iv) Longitude of Moon + rā. 3-0-0 = rā. 7-0-0. The declination of this point is 11° 44′, south, which is the southward defection from E′, represented as angle E′