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′

XI.5 XI. ECLIPSE DIAGRAM 239 5. Where this line cuts the eclipsed circle (f in the fig.) is the point of first contact. To get the point of last contact also, similar work should be done, using he Moon’s latitude at the time of last contact, marking it on N" – S" line, and drawing the line to the sum-circle in the opposite direction, (i.e. not west ward but earst-ward). (The point of last contact got is l in the fig.) Note: Since the directions are asked to be marked reversed in the case of the Moon eclipsed, we understand that they have to be marked as they are when the Sun is the eclipsed body. The first part of the instructions can be understood to be intended for first contact, since the second part is expressly stated to be for last contact. Since the Moon’s latitude of the moment of last contact is asked to be used to find that point, we infer that the latitude of the moment of first contact is to be used to find the first point. The drawing of lines to the westward rim and eastward rim of the sum- circle can be inferred from the known directions of first and last contacts, keeping in mind that they are marked reversed on the Moon-circle. Thus, the following is the work to be done: (i) In the case of the lunar eclipse alone, mark the directions reversed, on its rim. Take the Moon’s latitude at the time of that particular contact whose point is to be found. Measure it along N" – S" according to its own direction, and mark the point. (In the solar eclipse the latitude is to be parallax-corrected.) (In the fig. these points are L, A, for first and last contacts, respectively.) (ii) From this point, draw a line parallel to E" – W", westward or eastward, respectively, for first or last contact, to meet the rim of the sum-circle, (as in fig, Lf' or Al'.) (iii) Draw f'O or l'O to intersect the rim of the eclipsed at f or l. These are the points of first and last contacts. For example, in the fig, we shall find these: (i) The points of contat for the lunar eclipse is wanted. Therefore N W S E are reversed as s e n w. The latitudes are 19'.35 N, and 27'.65 N. Converted into digits, these are 8.3 and 11.9. According to scale, in the fig., these are 0".8 and 1".2. Measuring along ON", the points marked are L, and A, respectively. (ii) The westward parallel for first contact drawn is Lf', and the eastward parallel for last contact drawn is Al'. (iii) Joining f' and l' with O, the point of first contact got is f, and the point of last contact got is l. Thus, we find from the figure that the first contact is very near the southeast point of the Moon’s rim, and the last contact is a little to the south of its west point. 4b. A.B.विक्षेपान्तद्; C.विक्षेपं तत्; D.विक्षेपस्तत् A.C.D.दिगन्तकं; B.दिगतकं C.०दन्यं स्पृशेन्मध्यम्; D.०दन्यं [लि]खेन्मधात् c. A.B.स्पृश; C.स्पृशद्; D.स्पृशेद् 5b. A.मोक्ष्येण्वेवं B.विपर्ययशोध्यः C.द्वितीये वृत्ते; D.द्वितीयं वृत्तं c. A.खकुध्या; B.खकृतवृध्या D.खकृत्या d. A.०दन्यस्येन्मध्यं || B.०दन्यच्चेन्मध्यान्तसंपाते ||४|| d. A.B.मोक्षत्वा दिक्; C.मोक्षत्वाद् दिक्; D.मोक्षादिक् Really त संपाते belongs to verse 5. A2.विघालव्या

240 PAÑCASIDDHĀNTIKĀ XI.6 The explanation of all this has already been given. The author does not go beyond this in his graphical representation. The corrections of the text to agree with the ideas intended to be conveyed, are easily understood. [कलानामङ्गुलीकरणम्] लिप्ताद्वयेन हरिजे त्रयेण (मेषूरणे) ऽङ्गुलं भवति । अनुपातोऽन्तर [सं]स्थे कर्तव्यो दृष्टियुक्तार्थम् ॥ ६ ॥ Conversion of minutes into angles 6. So that the graphical representation may appear as the eclipse is seen actually, the minutes of arc are to be converted into digits, at 2' per digit when the Moon is near the horizon, and at 3' per digit when it is on the tenth sign, i.e. meridian, and proportionately in between. The proportion is as follows: In the fifteen nāḍikās (roughly) when the Moon rises from the horizon to the meridian, or falls from the meridian to the horizon, there is an increase of one minute of arc from 2' to 3', or decrease of one minute of arc from 3' to 2'; what is it at a given time? The number of minutes thus obtained is to be represented by one digit in the graphical represen- tation. At this rate the minutes of latitude etc. are to be converted into digits. In the example, for the hour angle of ten nāḍis, we have the rate per digit, (3 − 10/15) = 2 1/3 minutes, i.e. three digits per seven minutes. The following is the explanation of the conversion formula: The Sun and the Moon appear to be larger at the horizon, and to become smaller and smaller as they proceed to the meridian and near the zenith. This is an optical illusion, and really the size is practically the same, as can be proved by measurement, taking photographs etc. We shall not explain the phenomenon here as it is outside the pale of astronomy proper. But we must mention that the explanation given in some works like the Siddhāntaśekhara in wrong. Our author has taken that the magnification at the horizon is one and a half times that at the zenith, (practically the meridian in our latitudes) and the increase in size is proportionate to the hour angle, though this may not be strictly true. He also thinks that the orbs, which are nearly 32', appear to be about 11 digits near the zenith, and about 16 digits at the horizon. Hence his rule, two minutes per digit at the horizon, and three on the meridian. With the possibility of different people giving the actual estimate of size differently, what the author says must be taken as only approxi- mate. Therefore no harm will ensue by taking the meridian for the zenith, or by taking the mean value of the maximum hour angle, viz., 15 nāḍikas, in finding the proportion, especially in our latitudes. 6b. A. सेषूरणंगुल 1. Col. A.B. अवर्णनात्येकादशोध्यायः | c. A. तरःस्थे; B. तरस्थे [इति पञ्चसिद्धान्तिकायां वराहमिहिरविरचितायां C. इत्यनुवर्णनं नामैकादशोध्यायः | d. B. भुक्तार्थिः अनुवर्णनं नाम एकादशोऽध्यायः ]¹ D. अ [नु] वर्णनमेकादशोध्यायः | Thus ends Chapter Eleven entitled ‘Eclipse Diagram’ in the Pañcasiddhāntikā composed by Varāhamihira

Chapter Twelve

PAITĀMAHA SIDDHANTA

द्वादशोऽध्यायः

पैतामहसिद्धान्तः

Introduction In this chapter Varāhamihira deals with the Paitāmaha siddhānta, the other four having already been dealt with. As an astronomical work the Paitāmaha is of very little value, as the author has remarked in his Introduction, “the tithis of the other two, (meaning the Vāsiṣṭha and the Paitāmaha), are far from correct.” Since this siddhānta gives only the mean Sun and Moon, and therefrom the mean tithis, it cannot satisfy the requirements of the Dharmaśāstras. But it is historically important as a system that immediately followed the Vedāṅga Jyotiṣa. We shall explain at the end of the chapter how this could have subserved religious purposes in ancient times, and what merits it possesses as the basis of a civil calendar.

[अहर्गणः] रविशशिनोः पञ्च युगं वर्षाणि पितामहोपदिष्टानि | अधिमासास्त्रिंशद्भिर्मासैरवमो द्विषष्ट्याऽह्नाम् || १ || (द्व्यू) नं शकेन्द्रकालं पञ्चभिरुद्धृत्य शेषवर्षाणाम् | द्युगणं माघसिताद्यं कुर्याद्द्यु[गभानि] वह्न्युदयात् || २ ||

Days from Epoch

1. The Siddhānta of Paitāmaha teaches that the luni-solar yuga is five years. After every thirty synodic months there is an intercalary month, and there is an omitted day for every 62 lunar days or tithis.
2. Subtract two from the years of the elapsed Śaka era, and divide out the remaining year by five. The ‘days from epoch’ are to be calculated for the remaining years etc., the first day being the śuklapratipad of the month of Māgha. The nakṣastras of the Sun and the Moon, calculated by using the days, are for sunrise.

1-2 Quoted by Utpala on BS 8.22 1c. D. मासः 1d. A. मासैरवमस्त्रिपद्याप्तुं |; B. मासिरवमस्त्रिषष्ट्यार्का (B2.3. षष्ट्यंर्का) C. द्विषष्ट्या तु. 2a. A.B. द्युनं; U. द्यूनं 2b. A. पञ्चविगुधृत्य; B. पञ्चविगुहृत्य A1. वर्षाणं 2c. A.B. ॰माघसिताद्यं 2d. A.B.C.D. कुर्याद्द्यु (B. द्युध्यु; C.D. द् यु) गणं A.C.D. तदह्न्युदयात्; B. तदद्वैद्यात्; U. तद्वह्न्युदयात्

242 PAÑCASIDDHĀNTIKĀ XII.2 The period after which the Sun, Moon, and the planets all meet again at the first point of the zodiac is commonly called the yuga, meaning "the period of union." Here this is meant for the Sun and Moon alone, and this Siddhānta takes it to be five years, approximately, with a view to convenience. In the same way, the statements that there is an intercalary month after every thirty months, and an omitted day for every 62 lunar days, are approximate, and rounded off for convenience. The first tithi of the light fortnight of Māgha begins the yuga, and the year and the day begins with sun- rise. The author has not mentioned the number of intercalary months or days in the yuga, nor has he given how to get the 'days from epoch', expecting the readers to be experienced enough, by now, to know it for themselves. He has indicated it in I.16, and we have explained it under I.14-17. The only thing that is necessary is to know the years of the beginnings of the yuga, and this has been given here as two years after the śaka epoch, and every five years thereafter. We shall first compute the number of the intercalary months etc in the Yuga. The number of years in the yuga is 5, given. The solar months are 5 × 12 = 60. The intercalary months are, 60/30 = 2. The lunar or synodic months are, 60 + 2 = 62. The lunar days or tithis are 62 × 30 = 1860. The omitted lunar days are, 1860 ÷ 62 = 30. The (civil) days are, 1860 - 30 = 1830. The Moon's revolutions are, the Sun's revolutions + the synodic months = 5 + 62 = 67. The Vyatīpātas are, solar revolutions + lunar revolutions = 5 + 67 = 72. The number of days in the solar year is, 1830 ÷ 5 = 366. The days per ayana are 366 ÷ 2 = 183. This is enough for our purpose. To get the 'days':- (i) (Elapsed śaka years — 2) ÷ 5. Take the remainder alone. (ii) The solar months gone = (i) × 12 + elapsed months from Māgha. (iii) The intercalary months = (ii) ÷ 30. Take the quotient alone. (iv) The lunar months gone = (ii) + (iii). (v) The lunar days gone = (iv) × 30 + tithis gone in current month. (vi) Omitted days = (v) ÷ 62. Take the quotient alone. (vii) The days from epoch are, (v) - (vi). The tithis gone, used in (v) are actual elapsed tithis, and not those increased by one (according to verse 4. of this chapter,) for calendrical purposes. If the latter is used, subtract the calendrical elapsed tithi from the remainder got in (vi). If this is greater than 46, lessen the days from epoch by one to get the correct days. The reason for this will be explained while dealing with verse 4, following. Further, since there can be no fraction of intercalary month or avama at the beginning of a yuga carried over from a previous yuga, there is no kṣepa for these, in the computation rules. As for the explanation of these rules it has already been given in chap. I, when dealing with the rules for days from epoch according to the Romaka. The names of the five years, (not given in the text,) are: Saṁvatsara, Parivatsara, Idāvatsara, Anuvatsara, and Idvatsara. In certain Vedic śākhās, there is a slight variation in some names. As for the reading of the text, we have corrected dyūnam, into dvyūnam, since the former is meaningless in the context.

XII.3 XII. PAITĀMAHA SIDDHĀNTA 243 Example 1. Compute the ‘days from epoch’ according in the Paitāmaha, for the sunrise of calendrical date eleventh of the light fortnight of Kārttika, in the elapsed Śaka year 426. (i) (426 − 2) ÷ 5 = 424 ÷ 5. Here the remainder is 4, the years gone, in the current yuga. The fifth, Idvatsara is current. (ii) Counting from Māgha, 9 months have elapsed before (Kārttika), in the current year. ∴ the solar months gone = 4 × 12 + 9 = 57. (iii) Intercalary months = 57/30, = 1 27/30 (quotient = 1) (iv) Lunar months gone = 57 + 1 = 58. (v) The calendrical tithis gone in the month are 10. ∴ the total tithis gone = 58 × 30 + 10 = 1750. (vi) Omitted tithis = 1750 ÷ 62 = 28 14/62. (The quotient, 28 are the omitted tithis). Since the remainder, 14, minus 10, leaves 4, which is not greater than 46, the calendrical tithi itself is the tithi. (vii) Days from epoch = 1750 − 28 = 1722. [तिथिनक्षत्रादिः] सैक (त्वं) शे द्युगुणे तिथिर्भमार्कं नवाहते 'ऽक्ष्यर्कैः' । ‘दिग्रस’भागैः सप्तभिरूनं शशिभं धनिष्ठाद्यम् ॥ ३ ॥ Tithi, Nakṣatra etc. 3. Add to the ‘days’ a sixty-first part of itself. The total tithis are got, (which, divided out by thirty, leaves the tithis in the month). Multiply the ‘days’ by 9, and divide by 122. The total nakṣatras are got, (which, divided out by 27, gives its actual nakṣatra, reckoned from Śraviṣṭhā). Multiply the ‘days’ by 7 and divide by 610. Subtract this from the ‘days’. The remainder are the total nakṣatras of the Moon, (which, divided out by 27 and the remainder counted from Śraviṣṭhā, is the Moon's nakṣatra). The following is to be done:– (i) Tithi = ‘days’ + ‘days’ ÷ 61. (This, divided out by 30, and the remainder counted from śukla-pratipad, is the tithi proper). (ii) Sun’s nakṣatra = ‘days’ × 9 ÷ 122. (This, divided out by 27 and the remainder counted from Śraviṣṭhā, is the Sun’s nakṣatra). (iii) Moon’s nakṣatra = ‘days’ − ‘days’ × 7 ÷ 610. (This, divided out by 27 and the remainder counted from Śraviṣṭhā, is the Moon’s nakṣatra). Example 2. For the date of Ex. 1, find the tithi etc. The ‘days’ got there are 1722. (i) Tithi = 1722 + 1722 ÷ 61 = 1722 + 28 14/61 = 1750 14/61. Divided out by 30, the remainder is 10 14/61. ∴ the 10th Tithi of the light fortnight is gone, and 14/61 of Ekādaśī has gone at sunrise. 3. Quoted by Utpala on BS 8.22. 3a. A.B. सैकत्र्यंशत्वं (B.न्वं) चेद्युगणे; C. सैकषष्ट्यंशे द्युगुणे; D. सैकषडंशे द्युगणे; U. सैकत्रिशे b. A.B. भमार्कनचा (B. वा) हस्तेऽर्कैः (B. हस्तेष्टकैः); U. ॰नवाहतो॰ c. A.B. दिग्रह; D. भक्तैः d. A. नूनं A. धनिष्ठाद्यं; B2. धनीष्ठाधं

244 PAÑCASIDDHĀNTIKĀ XII.4 (ii) Sun’s nakṣatra = 1722 × 9 ÷ 122 = 127 4/122. Dividing out by 27, the remainder is 194/122. Counting from Dhaniṣṭhā, Citrā is gone, and the Sun is at 4/122 of Svāti. (iii) Moon’s nakṣatra = 1722 - 1722 × 7 ÷ 610 = 1722 - 19464/610 = 1702 146/610. Divided out by 27, the remainder is 1 146/610. Śraviṣṭhā is gone, and the Moon is at 146/610 of Śatabhiṣaj. It is to be noted, that two of these being known, the third can be counted from them. The following is the explanation of the rules: Under verses 1–2, the number of days in the yuga etc. have been got. Using them, the total tithis are got by the proportion, 1830: ‘days’ :: 1860: Tithis. ∴ Tithis = 1860 × ‘days’ ÷ 1830 = ‘days’ × 62 ÷ 61 = ‘days’ (1 + 1/61) = ‘days’ + ‘days’ ÷ 61. Next, since there are five solar years in the yuga, there are 5 × 27 = 135 solar nakṣatras. We have the proportion, 1830: ‘days’ :: 135: total Sun’s nakṣatra. ∴ Sun’s nakṣatra = 135 × ‘days’ ÷ 1830 =‘days’ × 9 ÷ 122. Next, since there are 67 revolutions of the Moon in the yuga, there are 67 × 27 = 1809 nakṣatras. So, we have the proportion, 1830: ‘days’ :: 1809: Moon’s nakṣatras. ∴ Total Moon’s nakṣatras. = 1809 × ‘days’ ÷ 1830 = ‘days’ × 603 ÷ 610 = days (1 - 7/610) = ‘days’ - ‘days’ × 7 ÷ 610. While the mss. readings tryaṁśatvaṁce etc. are corrupt, Bhaṭṭotpala’s reading saikatrīṁśe itself is wrong, since it contradicts facts, and we have emended it into saikartvaṁśe. TS have corrected is as saikaṣaṣṭyaṁśe gaṇe which is not proper since the correction does not follow the letters of the text. NP emends it as saikaṣaḍaṁśe for the number 61 that is required, but generally the said number is not found to be formed thus. [तिथि: व्यतिपातश्च] प्रागर्धे पर्व यदा तदो(त्त)राऽतोऽन्यथा तिथि: पूर्वा । ‘अर्क’घ्ने (व्यति)पाता द्युगणे ‘पञ्चाम्बरहुताशै’: ॥४ ॥ Vyatīpāta 4. If the moment of full or new moon falls before noon, the second of the two tithis connected with the day is the (civil) tithi for the day, otherwise the first. Multiply the ‘days’ by 12 and divide by 305. The Vyatīpātas are got. The tithi of this Siddhānta is mean tithi. Since this is less than the day, each day has parts of two tithis connected with it, and we have to fix one of them as the date of the particular day. It is this that is done by the first half of the stanza, it seems. If others like the Śrāddha-tithi are meant to be fixed here, they would be mentioned by name. The mere word tithi without an attribute must mean only the date. Agreeing that the date is meant to be fixed here, is it the date of the full or new Moon alone that is fixed here, or that of any day? From the word parva used, one may think it is only the former that is sought to be fixed. But this cannot be, since fixing the date of one particular day among so many is practically useless. If it is argued that the fixing of the day as parva or pratipad is useful to 4b. A. तदोत्तस्तोन्त्यथा; B1.3. तदा तए सोन्यथा; B2. तदा तएं तोन्यथा c. AB. व्यापिपाता; D. व्यतिपातो d. A. द्यगणे A. पंचावरहु..ाशैः; B. पञ्चाम्बरं हूताशैः (B2. दूतां शैः)

XII.4 XII. PAITĀMAHA SIDDHĀNTA 245 determine whether the anvādhāna or the iṣṭi is to be performed on that day, then the general term, tithi, need not have been used, and it would have been easier to mention the thing. Further this is a matter for the Dharmaśāstras to deal with, not for an astronomical work. Therefore, by fixing the date of the parva, the author means to fix all the subsequent dates following, upto the next parva, and make them convenient for civil use. If the dates are consecutive, one for each day, it will be con- venient for civil reckoning. If there is a jump, omitting one date in the middle (tithikṣaya), it is plainly inconvenient. The instruction in the verse secures that the dates follow without omission in the middle of the fortnight. For, if the moment of full or new moon is after noon, there will be no omitted tithi in the fortnight following, and the corresponding dates will follow one after another each day, beginning from prathamā, next day. But, if the moment of full or new moon is before noon, then there will be an omitted tithi in the fortnight following, since the remainder in getting the omitted days will be greater than 46, increasing by one each day. Therefore, if now the day of full or new moon itself is reckoned as the first date of the fortnight, then the reckoning can be continued to the end of the fortnight without omission. It is bearing in mind this idea implied by the text, that we made a distinction between the astronomical and the civil date, in giving the computation of the 'days from epoch'. The Siddhānta is only repeating here the idea of the Vedāṅga Jyotiṣa in: dyu heyaṃ parva cet pāde pādas triṃśattu saikikā | the term pāde (meaning 'quarter day') corresponding to the term ardhe in our text. The thirty-one parts mentioned form the measure of the pāda, in units of 1/124 parts of a day. We have explained the vyatīpāta in detail, in our commentary on III.20. We must remember here two things that we said there: (1) Vyatīpāta occurs when the Sun and the Moon have the same declination, (both north or both south), and when one is moving northward while the other is moving southward. (2) If the Sun, Moon, and declination are all mean, as here, and if the first point is at the solstice, as here, mid-vyatīpāta-yoga must fall when the Sun + Moon equals 12 rāśis. In the yuga, (of 1830 days) there are 5 + 67 = 72 such yogas, i.e., vyatīpātas. Therefore, for given 'days', there are, 72 × 'days' ÷ 1830 = 12 × 'days' ÷ 305, vyatīpātas, as given here. The quotient obtained are the vyatīpātas gone. But this knowledge is practically useless, and we must take it that the Siddhānta intends here to give when the vyatīpāta occurs, a time extremely propitious for gifts, japa, homa, etc. This can be found easily from the remainder. Divide this by 12. The result is days etc. gone from the last vyatīpāta. Subtracting the remainder from 305 and dividing by 12, we get the days to the middle of the next vyatīpāta. If the remainder is zero or nearly so, it is clear that the vyatīpāta is on. Example 3(a). Given the 'days' 1697, what is the tithi for that day, and the subsequent days, upto the end of the fortnight? The tithis gone = 1697 + 1697 ÷ 61 = 1697 + 27 50/61 = 1724 50/61. Dividing out by 30, the remainder is 14 50/61, i.e. 50/61 part of pūrṇimā has gone, and 11/61 part remains. Since the tithi is equal to 61/62 day, 11/61 tithi equals, 11/61 × 61/62 = 11/62 day. The full Moon ends at 11/62 day, i.e. before noon. Therefore that day itself is Prathamā, (not Pūrṇimā). (The same conclusion follows from the remainder of the omitted day, 51, in this case, (minus zero, for the civil day gone,) being greater than 46. After this, for ten days, the tithis are from the second to the eleventh, for civil purposes, though at sunrise the tithis are from the first to the tenth, the eleventh being the omitted tithi. On the next day the tithi at sunrise is the twelfth, as also the civil tithi, and so on.

246 PAÑCASIDDHĀNTIKĀ XII.5 Example 3(b). Given the 'days' 1722, find when the vyatīpāta falls. 1722 × 12 ÷ 305 = 67 229/305, i.e. 67 vyatīpātas have gone, from the beginning of the yuga. The remainder is 229. Dividing by 12, we get 19-5, days etc. gone from last vyatīpāta. Subtracting from 305, and dividing by 12, we get days etc. 6-20, to go for the middle of the next vyatīpāta, i.e, it will be falling at 20 nāḍīs on Kārttika Kṛṣṇa-dvitīyā. The correction of the reading here is easy to understand. Of TS, the latter says some farfetched thing as the meaning of the first part of this verse, which seems meaningless to me, and the former without saying anything himself, refers us to the Sanskrit commentary, evidently not understanding it himself. In explaining the Vyatīpāta, their use of the one forming the seventeenth of the series Viṣkambha etc., can serve no purpose in the present case. The same confusion is seen also in NP (see Pt II.p.82). We have already shown that the series itself did not exist at the period of Varāhamihira. [अहर्मानम्] (गतमयनादुत्तरतो) (द्यूनां) (गन्तव्य)मपि च याम्यस्य । (द्विघ्नं) 'शशिरस'भक्तं द्वादश(सहितं) दिवसमानम् ॥ ५ ॥ Duration of a day 5. Take the days gone in the Uttarāyaṇa, i.e. the northward course of the Sun, and the days to go in the Dakṣiṇāyana, i.e. the southward course. Multiply by two, divide by 61, and add 12. The duration of the day time (in muhūrtas) is got. Though the text is very corrupt here, since we know that the duration of daytime is given here, and that as mentioned in the Vedāṅga Jyotiṣa, many words occurring in that work being recognisable here, we have succeeded in reconstructing the verse using the letters found in the text, though not to our entire satisfaction. But there is no doubt about the idea intended to be conveyed, viz. that of VJ (verse 22 of the Ṛgveda version and verse 40 of the Yajurveda version). But TS and NP, in order to retain the expression dvādaśāhinam, read into the text many impossible things and make several emendations not caring even for the metre. The rule given is thus explained: According to this Siddhānta the shortest day is 12 muhūrtas, at the end of the southward course and the beginning of the northward at Winter solstice, and the longest is 18 muhūrtas at the end of the northward course and the beginning of the southward, at Summer solstice. Each course has 183 days, and the increase or decrease of daytime is considered uniform. Therefore, since there is an increase of 6 muhūrtas from 12 in the 183 days of the north- ward course, the increase for a desired number of days (counted from the beginning) is: days gone × 6 ÷ 183 = days gone × 2 ÷ 61. So, the duration is 12 + days gone × 2 ÷ 61. Taking the south- ward course, it is 12 muhūrtas at the end, proportionately greater, the earlier is the day, the increase being 6 in 183 days, at the beginning of the course. Therefore, during the southward course, the duration is 12 + the days to go in the course × 2 ÷ 61. It is to be noted that the author has not said 5a. A.B. धृतिरनयाद्युत्तरयो (B1.3. ॰त्तरयो); C. खमितमेष्य दिनमपि याम्यायनस्य C. द्व्यग्निगोष्ट्ररतः; D. [सत्रि] धृति [च] रणध्यु [मु] त्तरगे D. त्वगतद्युमपि च याम्यस्थे b. A. खमृणं तद्यमपि च याम्यस्य; c. A. द्विग्नं; B. द्विग्रं B. भक्तं B. सू (B3. ख) मृणं गतद्यमपि च याम्यास्य; d. A.B.C.D. द्वादशहीनं

XII.5 XII. PAITĀMAHA SIDDHĀNTA 247 when the northward course begins, and when the southward one. (Perhaps he has said that in this verse, and the corruption of the text has masked it.) When the Sun enters Śraviṣṭhā its northward course begins, and when it is at the middle of Āśleṣā, its southward course begins, or the first half of the year is the northward course, and the second half, the southward one. It can also be inferred from the rule for vyatīpāta. Example 4. Given the 'days' 1722, find the daytime for the day following. 1722 ÷ 366 = 4 258/366. Four years have gone, and 258 days in the fifth. The 183 days of the northward course have gone, and 75 days have elapsed in the southward course. The days to go = 183 − 75 = 108. The duration of daytime in muhūrtas = 12 + 108 × 2 ÷ 61 = 12 + 3 33/61 = 15 33/61. To conclude, it has been said in the introduction to this chapter that the Sun, Moon etc. of this Siddhānta are only mean, not true. Some think it is even worse than this, which we must investigate now. We have seen that, according to this Siddhānta, there are 366 days in the year, 5 years or 1830 days constitute the yuga, and there are 62 synodic months in it. But actually there are about 365 ¹/⁴ days in the year, and also 62 synodic months take 1830.8965 days. Therefore, instead of being at the zero point of Śraviṣṭhā at the beginning of each yuga, the Sun will be advancing by about three degrees per yuga. The synodic month being not completed, it will be really only Caturdaśī then, and at the end of every yuga, the tithi will be preceding by about one per yuga. This is common to both this Siddhānta and the Vedāṅga Jyotiṣa. But accumulation of this error is prevented by tying the Yuga to the correct Śaka year, by the instruction to use the Śaka year for finding the year of the yuga. (This is of the same nature as tying the lunar year to the solar.) It may be said, in passing, that at the period when the VJ was followed, actual observation was used to find out when a correction was wanted and the same made by simply omitting an intercalation, for which, it is possible, they could even have discovered a formula in the long run. We have made this clear in our Notes to the VJ, (Indian National Sc. Ae., New Delhi, 1985) and the during the Seminar on Indian Calendar held by the Institute of Traditional Cultures, University of Madras. (Bulletin of the Institute of Traditional Cultures, Madras, 1968, Part I, page 60.) When thus the accumulation of error is taken care of, giving 366 days to the year is extremely convenient for civil, calendrical, and even religious purposes, since it makes calculations easy in the same way as we, even now, take the year to have 365 days, and the months 31 etc. days. We get 183 days for each ayana, 122 days for each cātarmāsya, and 61 days for each ṛtu, all in whole numbers. There are 61 sāvana months in the yuga. Even the saura month has 30 days and a half, a fraction easy to work with. Further, the fortnight or pakṣa was the unit used then in the place of the modern week. We have shown, in our Notes on verses 2 and 4 above, how it was secured that the dates follow consecutively in the pakṣa, a factor so essential for a civil calendar. The approximateness of the tithis etc. being mean, is of course there, but this is only an advantage in civil reckoning. As for religious purposes, the true tithis etc. required for rites like darśa-pūrṇa-māsa were guessed from earlier observations, as we have said, and there are several indications in the Vedas that such was the case. The abhyudayeṣṭi, enjoined to be performed if the Moon is observable in the east between the anvādhāna on the previous day and the iṣṭi on the next day, is one such indication, for, if the visibility had been properly calculated from the true Sun and Moon, using dṛkkarma etc. there would be no need for abhyudayeṣṭi at all. [इति पञ्चसिद्धान्तिकायां वराहमिहिरविरचितायां पैतामहसिद्धान्तो नाम द्वादशोऽध्यायः ] Thus ends Chapter Twelve entitled 'Paitāmaha Siddhānta' in the Pañcasiddhāntikā composed by Varāhamihira Col. A.D. इति (A om इति) पैतामहसिद्धान्ते द्वादशोध्यायः B. इति पितामहसिद्धान्ते द्वादशोऽध्यायः C. इति पैतामहसिद्धान्तो नाम द्वादशोऽध्यायः

)? Wait, look at the top: there is an i-matra! Wait, look at the top of that conjunct: there is an i-kar (ि)! Wait, and the consonant: it has 'श्' on the left, and below it/connected to it: Is it 'च'? Yes, 'श्च'! And after it: 'त'? Wait, does it have 'ि' over 'श्च'? Yes, 'श्चि'! And after it: 'त'? With an anusvara or visarga? Wait, look at: "भिश्चिनुतः"? No, look at the glyph: It is: श् + च + ि = श्चि, then त + ः = तः! Wait, that is "भिश्चितः"! Wait, in line 8: "तरुनगनगरारामसरित्समुद्रादिभिश्चितः सर्वः |" WAIT! Look at line 8 again! Is that "भिश्चितः" in the main text?! Let's look at line 8: दि - भि - श्चि - तः ! Wait, where is 'श्रितः'? Look at the footnote 2b: b. B2. ०भिश्रितः ! WAIT! Look at footnote 2b: Is it ०भिश्रितः?! LET'S LOOK AT FOOTNOTE 2b: b. B2. ० then `भिश्र