28 PAÑCASIDDHĀNTIKĀ II.4 nothing is said, then the emendation gatikāṣṭhāṃśa is the proper one, which we have given. (TS also give this). If, on the other hand, we take it that the instruction is to subtract 2 minutes per ghana, taking the word projjhya in the previous instruction to be understood here also, then the emenda- tion gatyaṣṭāṃsa will be the proper one. In this case we would also have to keep the letter ṭa of the original as it is, without changing it into ṭha. But the addition of 2 minutes per ghana alone would agree with the correct mean motion for the period of our author which is in cycles etc. 110-11-7-32- 15 for 3031 days, the Vāsiṣṭha mean motion being 110-11-7-32. NP have emended the word as ṣaṣṭhāṃsa which would not give the correct result. *Example 2. Find the mean Moon for the end of the gati just before Days from Epoch, 3,0

II.6 II. VĀSIṢṬHA-SIDDHĀNTA 29 The two formulae can be written down thus: (i) If P is the number of plus-padas, {1094 + 5(P − 1)} P/63. (ii) If P' is the number of minus-padas, {2414 − 5(P' − 1)} P'/63. Example 3. Continue Ex.2 and compute the true Moon for the days given. The mean Moon got in Ex. 2 to the end of the gati = 4ʳ 24° 4' The padas obtained are 64, plus-padas (P) Adding degrees equal to P = 64, + 2 4 0 Using formula (i) intended for plus-padas, {1094 + 5 (64 − 1)} 64/63 = 1431 minutes + 0 23 51 —————————————————————————————————————————————————————————— The true Moon 7 21 55 Example 4. The Days from Epoch are 1219. Find the True Moon. 1219 + 1936 = 3155 (= days for computation). Dividing by 3031, ghana got is 1, remainder 124. Multiplying 124 by 9 and dividing by 248, the gatis got are 4. The remainder 124 are padas. This is just one half-gati and no pada is left over. r ° ' Ghana 1 × ¾ = 0ʳ 22° 30'. Deducting from 12 rāśis = 11 7 30 Adding minutes 1 × 2 + 0 0 2 Kṣepa + 1 7 29 Gatis 4, × 184 9/10 = 740 (minutes) + 0 12 20 For the half-gati, add + 6 0 4 —————————————————————————————————————————————————————————— True Moon 6 27 25 Example 5. Find the true Moon for Days from Epoch, 1228. 1228 + 1936 = 3164 (= days for computation). Dividing by 3031, ghanas got 1, remainder 133. Multiplying by 9 and dividing by 248, the quotient 4 are the gatis got, and the remainder 205 are padas left over. A half-gati (= 124 padas) can be taken from this, and the remaining 81 are minus- padas. r ° ' ghana 1 × ¾ʳ = 0ʳ 22° 30'. Deducting from 12 rāśis 11 7 30 Adding 1 × 2 minutes + 0 0 2 Adding kṣepa + 1 7 29 Gatis, 4, × 184 9/10 = 740 (minutes) + 0 12 20 For the half-gati + 6 0 4 Degrees equal to P' = 81' + 2 21 0 Using formula (ii) (as the left over are minus-padas = P'), 2414 − 5 (81 − 1) 81/63 = 2589 mts. + 1 13 9 —————————————————————————————————————————————————————————— True Moon 11 1 34 The following is the explanation of the processes: The true Moon at a given time t is: (i) the mean Moon at t plus (ii) the equation of the centre for t. (i) is given here in five parts. We shall call them (a), (b), (c), (d), (e) which are to be added up to get the total mean Moon.

30 PAÑCASIDDHĀNTIKĀ II.6 (a) (Usually called the Mūla-dhruva or Kṣepa) is the mean Moon at a point of time 1936 days before the Epoch, when the Moon's apogee and the mean Moon exactly coincided according to this Siddhānta. This is given as śaśi-muni-navayamāś ca rāśyādyāḥ, i.e. 1ʳ 7° 29ʹ. (b) is the mean motion during whole numbers of cycles of 3031 days from the point of time 1936 before Epoch, each cycle equal to 110 anomalistic revolutions of the Moon. This (b) is found by multiplying the mean motion per cycle (110 revolutions, 11 rāśis, 7 degrees, 32 minutes) by the number of cycles, called ghanas, obtained as quotient, by dividing the Days from Epoch plus 1936, by 3031. As full revolutions can be neglected, it is enough if we multiply the ghanas by 11 rāśis 7 degrees 32 minutes, which may be done as ghanas × 2ʹ + ghanas × 11ʳ 7° 30ʹ. Ghanas × 2 is given by dviguṇaghanāḥ kalāḥ (yojyāḥ). Because 16 ghanas × 11ʳ 7°30ʹ equals 15 full revolutions, it is enough if we divide out the ghanas by 16 and take the remainder alone for multiplication (for we shall be neglecting only full revolutions), which we are asked to do by ghanaṣoḍaśāhṛta-śeṣam. As 11ʳ 7° 30ʹ is ¾ rāśi less than a full revolution, we can multiply the remaining ghanas by ¾ rāśi and take this as subtractive, which we are instructed to do by projjhyādhas triguṇitaṁ caturbhaktam bhādi (rāśyādi.) Thus b is disposed of. (c) is the mean motion during the subsequent full anomalistic revolutions called gatis, which form the quotient got by dividing the remaining days by the anomalistic period, 248/9 days, (i.e. multi- plying the days left over by 9 and dividing by 248). For each gati the mean motion is 1 revolution and 184 9/10 minutes (which can be obtained by dividing the motion per ghana, viz. 110 rev. 11ʳ 7° 32ʹ by the number of gatis in a ghana, viz. 3031 × 9/248). Hence the rule to multiply the gatis by 185ʹ and deduct minutes equal to 1/10 of the gatis. This is given by viṣayadhṛtayo gatighnā gatīkāṣṭhām- śonitāḥ kalāḥ yojyāḥ. (d) What are now left of the days are ninths of days called padas (and these obviously would be less than 248). The mean motion per pada is 1 degree 27 209/248 minutes, and so padas × 1° 27 209ʹ/248 should be added to complete the mean motion till t. Of this, the Siddhānta asks us to add 1° per pada first, which is given by śeṣapadasamāṁśāmśāḥ (yojyāḥ). This forms d. (e) The residue 27 209/248 minutes per pada, forming (e), is combined with the equation of the centre (ii) and given by the two formulae of II.6. If the padas contain a half-gati (i.e. 124 padas) the value of (d) + (e) + (ii) for the half-gati part is combined together and given as 180° 4ʹ. This is got as follows. As the half-gati is equal to 124 padas, d = 124°. (e) + (ii) given by the first formula of II.6 is: {1094 + 5(124 − 1)} 124/63 = 3364ʹ = 56° 4ʹ; 124° + 56° 4ʹ = 180° 4ʹ = 6 rāśis 4 minutes, which is given by gatyardhe bhagaṇārdham deyam liptācatuṣkasaṁyuktam and which instruction has so much puzzled TS. But, of course, this is incorrect and the defect lies in the equation of the centre- part of the formulae in II.6, which give zero-value for the equation of the centre not at 124 padas, but at 133 padas, as we shall show presently. We shall first explain II.6 by showing how the formulae here combine the residual mean motion, viz. padas × 27 209/248 minutes (= e) with what is identifiable with the equation of the centre (= ii). The equation of the centre of the Vāsiṣṭha is peculiar. Usually in the Siddhāntas the equation of the centre varies as the sine of the anomaly, and therefore is zero at zero degree anomaly, going to a minimum at 90°, again rising to zero at 180°, then going to a maximum at 270°, and then falling to 0 to 360°, i.e. zero°. Thus it is negative in the first two quadrants and positive in the third and fourth quadrants and of the form, '−a sin θ, where 'a' is the maximum or minimum numerical

II.6 II. VĀSIṢṬHA-SIDDHĀNTA 31 value, and θ is the anomaly. Note that this is the first term of the series for the equation of the centre in modern astronomy, with its sign reversed, and the reversing is necessary because the ano- maly was reckoned by the ancients from the apogee, not perigee. But in the Vāsiṣṭha it is of the form — (665 — 5P) P/63 for the first two quadrants and + (665 — 5P') P'/63 for the last two. These are derivable from the equation for the Moon’s daily true motion given in III.4, (as we shall show there), which assumes the increase or decrease of motion as uniform. Here we shall assume them and derive the two formulae of II.6. As said before, (e) + (ii) is given by the formulae and (ii) is — (665 — 5P) P/63, for the first formula. Therefore (e) + (ii) = 27 209/248 P — (665 — 5P) P/63 = (63 × 27 209/248 — 665 + 5P) P/63 = (1754 — 665 + 5P) P/63 = (1089 + 5P) P/63 = {1094 + 5 (P — 1)} P/63, which is the first formula. For the second formula (ii) is + (665 — 5P') P'/63. ∴ (e) + (ii) = 27 209/248 P' + (665 — 5P') P'/63 = (27 209/248 P' × 63 + 655 — 5P') P'/63 = (1754 + 665 — 5P') P'/63 = (2419 — 5P') P'/63 = {2414 — 5 (P' — 1)} P'/63 which is the second formula. We have already shown how for the half-gati 6ʳ 0° 4' is got instead of the mean motion 6ʳ 1° 32½'. This means that there is in this an equation of the centre equal to — 88½', combined with it. So, when the equation of the centre given by + (665 — 5P') P'/63 = + 88½, then it is actually zero according to this Siddhānta. Solving this equation, we get P' = 9 or P' = 124. As P' is minus-pada, which is the original padas got less 124, we get that the equation of the centre actually becomes zero at original padas, P= 133, and P = 248. As P = 248, is the end of the gati, this is what we expect, as the anomaly has again become zero. Also, by computation or examination we can get from the equation of the cyclic part of the for- mula for the first half-gati, — (665 — 5P) P/63, the numerically greatest value of the negative equa- tion of the centre, which is — 351', for P = 66½. In the same way, from that the formula for the second half-gati, + (665 — 5P') P'/63 we can get the maximum + 351', for P' = 66½; but as there is a residue of — 88½' in the second half-gati, 351' — 88½' = 262½' is the actual maximum. The numerical mean is 307' which, we see, is very nearly equal to that of the other Hindu Siddhāntas. It is not that VM does not know that zero equation of the centre must occur at P = 124, and not at 133, for in his own Romaka and Saura it is so. Nor is it difficult for VM, a master in the science, to give the two formulae so as to have the equation of the centre zero at P = 124, (so that for a half- gati we get the correct 6ʳ 1° 32½'), retaining, at the same time, the equation of the centre desired by him. If he had given the two formulae in the form (1134 + 5P) P/63, and (2374 — 5P') P'/63, he could have secured this. But adherence to the Siddhānta has prevented him from doing this. So closely does he follow the original that he does not even give the two formulae in the more simplified forms, (1089 — 5P) P/63, and (2419 — 5P') P'/63. The following things are to be noted in connection with this Siddhānta. Of both the Sun and the Moon, the mean motion and the equation of the centre is mixed in a peculiar manner and thereby the true motion is given. We shall see that it is the same in the case of the Pauliśa also. The period of 3031 days called ghana here is the same as what is called kālānala in the Vāk- yakaraṇa, which gives for this period, the mean motion, 11ʳ 7° 31', neglecting full revolutions. The number 248 given here is there mentioned as devara.

32 PAÑCASIDDHĀNTIKĀ II.6 We can see that the remark in I.4 about the tithi of the Vāsiṣṭha being very incorrect is appropriate, but it can be shown that it is not due to the error in the Moon, but in the Sun, whose sidereal year is taken as 365-15-0 days. The Moon's motion for 3031 days is in cycles etc. 110-11-7-32 = 110 5063/5400 cycles. In one day the motion is 110 5063/5400 ÷ 3031 = 5,99,063/ (5400 × 3031) cycle. In one day the Sun's motion is 1/365¹/4 = 4/1461 cycle. The relative motion, i.e. their separation per day is 5,99,063/(5400 × 3031) − 4/1461 cycle. The time taken for a separation of one cycle, i.e. the synodic month, is in days, 1/{59,90,63/(5400 × 3031) − 4/1461} = 1461 × 3031 × 5400 ÷ (1461 × 5,99,063 − 4 × 5400 × 3031) = 7,97,09,23,800 ÷ 26,99,20,481 = 29 - 31 - 50 - 17 - 38. But the correct synodic month computed for the time near that of our author is 29 - 31 - 50 - 7 - 47. Therefore in successive synodic months the tithi comes later by days etc. 0-0-0-9-51, according to Vāsiṣṭha. In about 29¹/2 years this will accumulate to one nāḍikā. This is bad indeed, and merits VM's remark in I.4 taditarau dūravibhraṣṭau (i.e., 'The tithis of the other two have slipped far away from the real'.) Now, we have seen that according to this Siddhānta the mean Moon moves revs. 110-11-7-32-0, while the real motion for the period is revs. 110-11-7-32-15. Therefore in 3031 days the Vāsiṣṭha nakṣatra is delayed by a little more than one vināḍi. So a delay of one nāḍi is caused in 440 years only. So the delay of one nāḍi in the tithi per 29¹/2 years mentioned above, must be due almost wholly to the error in the Sun, the result of giving the time per cycle as 365-15-0 days. Again, in every 3031 days, the Vāsiṣṭha Moon lags behind the correct one by 15". A lagging behind by one degree will take place in 3031 × 60 × 60 ÷ 15 days, i.e. in about 2000 years, a very long period indeed. Bearing this in mind, we shall try to answer the question already raised, viz. whether the Vāsiṣṭha Sun and Moon are given for sunrise at Ujjain or sunset at Yavanapura; and, incidentally, we shall show that the kṣepa, 1ʳ 7° 29′ given by śaśi-muninavayāmāśca rāśyādyāḥ and obliterated by TS by their drastic emendation as yama-hṛtāś ca is necessary in II.3. The following is the mean Moon for Epoch (viz. Śaka 427 elapsed, i.e. in A.D. 505), sunset, at Yavanapura, beginning Monday, Caitra Sukla being about to begin. i. Computed for the period according to modern astronomy, assuming the ayanāṁśa to be practically 0 for the time 354° 48′ ii. According to Saura 355° 6′ iii. According to Siddhānta Śiromaṇi 355° 41′ iv. According to Romaka 356° 12′ v. According to Vāsiṣṭha, assuming that the mean Moon is given for Ujjain sunrise 355° 6′ -do- -do- for sunset at Yavanapura 346° 54′ -do- Ujjain sunrise, without Kṣepa 317° 37′ -do- sunset at Yavanapura without Kṣepa 309° 25′ (For use by anyone interested in making the calculations himself, the Kalidina etc. of Epoch is 13,17,122-37-20. Also, the Epoch is 5,09,432-22-40 days before mean sunrise at Ujjain of the first January 1900). An examination of the table will show that the Vāsiṣṭha Moon agrees with that of every other fairly well, taking it as being given for Ujjain sunrise, and taking that the kṣepa is given. If, on the other hand, it is assumed that it is given for sunset at Yavanapura, there is a difference of about 8°, which can happen only in 1600 years, which is very very unlikely; for this to happen the Vāsiṣṭha should

II.6 II. VĀSIṢṬHA-SIDDHĀNTA 33 have been written 1600 years earlier. If there is not the kṣepa, the difference is 37°, an impossible thing, not to speak of the assumption that it is for Yavanapura sunset and there is no kṣepa, both together which will make the difference 45° and worse. Hence the Vāsiṣṭha epoch is definitely at sunrise at Ujjain and not at sunset at Yavanapura (Alexandria). We have shown the kṣepa necessary. But TS have emended kalāḥ dviguṇaghanāḥ, śaśi-muni- navayamāś ca rāśyādyāḥ (verse 3) into phalaṁ dviguṇaghanāḥ śaśi-muni-yama-hṛtāś ca rāśyādyāḥ and spoiled the already correct reading and introduced two extra syllables in the last foot, which spoils the āryā metre too. (It must be noted that already there are 16 mātrās in the last foot, i.e. one mātrā extra, which can be explained away or corrected by reading rāśyādyāh as rāśyādi.) One another point: TS have expressed their inability either to interpret II.6 or to explain why 6ʳ 0° 4′ is to be added for a half-gati (vide Com, page 9). But still thinking II.6 gives the pure equation of the centre, the commentary goes on: arthāt vedārkālpa-padeṣu ṛṇam, adhikeṣu dhanam iti buddhimad- bhiḥ svayam eva ūhyam, i.e. “It goes without saying that when the padas are less than 124, the result is subtractive, and when more it is additive”, which is wrong for we have seen that the result of both the formulae are additive. Moreover, the failure, both by TS and NP, to realise that the expression ‘rāśyādyāḥ’ specifically instructs that the digits in śaśimuninavayama are to be taken as ‘beginning from rāśi’, i.e. as 1ʳ 7° 29′ and not as a whole number 2971 (TS) or as “2ʳ 9;7, 1⁰” (NP) have led to incorrect interpretations by them; also, the Notes of NP (vol. II, pp. 16-19) and the deductions made (p.19) have to be revised in the light of all that has been stated above. [नक्षत्र-तिथी] श (श्यर्ध)दलं त्रिकृतिघ्नमृक्षमंश(स्थि)ता मुहूर्ताः स्युः । व्यर्केन्दुदलं विषयाऽऽहतं तिथिस्तद्गदेवोक्तः ॥ ७ ॥ Nakṣatra and Tithi 7. Divide the True Moon by 4 and multiply by 9. What we get in the rāśi column is the nakṣatra. What is got in the degree column are the muhūrtas. Deduct the true Sun from the true Moon, divide the result by 2 and multiply by 5. Tithis are got in the rāśi column and thirtieths of tithis in the degree column. As the 27 nakṣatra-segments are divided into the 12 rāśi - segments, there are 2¼ = 9/4 nakṣatras per rāśi. Hence the instruction to divide the rāśis by 4 and multiply by 9 to get the nakṣatras. As degrees are thirtieths of rāśis, the resulting numbers in the degree column are thirtieths of nakṣatras, called muhūrtas by this Siddhānta. It must be noted that the word muhūrta originally meant the 30th part of a nakṣatra, but later came to be applied to the 30th part of a day as well, because both are practically the same in duration. The interval of longitude between the Sun and the Moon is the tithi, 12° forming one tithi, i.e. there are 2½ = 5/2 tithis per rāśi. Hence the instruction to divide the rāśis by 2 and multiply by 5 to get the tithis. 7a. A. शशादलं; B. शशखदमन्तिर b. A. मंशस्थिता; B. ॰मक्ष (B3. मक्षु) -मंशास्थिता d. A. तिथिस्तंद्॰

34 PAÑCASIDDHĀNTIKĀ II.7 Example 6. The true Sun is 10ʳ 18°, and the true Moon is 5ʳ 22°. Find the nakṣatra and the tithi. Nakṣatra: The Moon is 5ʳ 22°. Dividing by 4, (5ʳ 22°)/4 = 1ʳ 13°. Multiplying by 9, 9x 1ʳ 13° = 12 – 27, i.e. twelve nakṣatras have gone and in the 13th, (Hasta), 27 muhūrtas have gone. Tithi: Moon – Sun = 5ʳ 22° – 10ʳ 18° = 7ʳ 4°. Dividing by 2, (7ʳ 4°)/2 = 3ʳ 17°. Multiplying by 5, 3ʳ 17° × 5 = 17 – 25. Seventeen tithis are gone and in the eighteenth (Bahula Tṛtīyā) 25/30 parts have gone. [अहर्मानम्] मकरादौ 'गुण'युक्तो मेषादौ 'तिथि'युतो र(वि)दिवसः । कर्कटकादिषु सत्सु 'त्रयस्त्रिकाः' शर्वरीमानम् ॥ ८ ॥ Day-time 8. When the Sun is in the 3 rāśis, Makara etc., the Sun measured in rāśis plus three is the duration of day-time in muhūrtas. When it is in the 3 rāśis, Meṣa etc., the Sun plus fifteen is the duration of day-time. When in the 6 rāśis, Karkaṭaka etc., the Sun plus nine is the duration of the night-time. (To get the duration of the day-time, this should be subtracted from 30). Though no measure of time is mentioned here, we can infer that it is the muhūrta (2 nāḍīs) because by adding the shortest day, 12, and the longest, 18, we get 30 which must be equal to the whole day, i.e. 60 nāḍikās. Thus, for the Sun at the beginning of each rāśi, Meṣa etc., the day-time in muhūrtas is 15, 16, 17, 18, 17, 16, 15, 14, 13, 12, 13, 14. The longest day is 18 muhūrtas when the Sun is at the first point of Karkaṭaka (Cancer) at Summer solstice and the shortest 12 muhūrtas when at the first point of Makara (Capricorn) at Winter solstice. The day and night are equal at the first points of Meṣa (Aries) and Tūla (Libra), i.e. at the equinoxes. The daily increase or decrease in day-light is 4 vināḍīs per day. In essence, the same formula for day-light is found in the Vedāṅga Jyotiṣa and the Paitāmaha Siddhānta (PS, XII.5),with this difference that here the true Sun is used, but there, because they have no true Sun but only the mean Sun, the day which is proportionate to the mean Sun, is used. Evidently not understanding what is given here, TS have made a drastic change in the text, writing bhūsvarga-tithimito for meṣādau tithiyuto, intending to make this agree with the next verse giving the noon-day shadow. But, even within that verse, there is contradiction and this need not have been attempted, at such cost. To crown all their interpretation with their emendation is full of contradiction within itself, which has been set out in detail by me in a paper entitled 'Vāsiṣṭha Sun and Moon' in the Journal of Oriental Research, 25 (1955-56) 19 – 41. The uniform increase and decrease in day-time given here is wrong, of course, and it varies with the position of the Sun, being greatest at the equinoxes and falling to zero at the solstices. The maximum or minimum day-time itself varies with the latitude of the place (depending on tan. 8b. A. मेषादौ. C. भूर्स्वर्गातित्थिमितो रखेर्दिवसः c. B2. सत्सु; C.D. षट्सु A.B.C. रखेः. A2. दिक्कम् । d. B. मानाम्

II. 10 II. VĀSIṢṬHA-SIDDHĀNTA 35 latitude), what is given here being for some place having a North latitude of 35° 45′. (This matter is dealt with in the text in III.10 and IV.26). The rules for day-time is explained thus: From the beginning of Makara to the end of Mīna the Sun in rāśis increases from 9 to 12. The day-time also increases following it, from 12 muhūrtas at winter solstice to 15 at Equinox. As (9 to 12) + 3 = (12 to 15), the instruction to add 3 for the Sun in this quadrant follows. In the same way, from the beginning of Meṣa to the end of Mithuna, the Sun in rāśis increases from 0 to 3. The day-time increases from 15 at Equinox to 18 at summer solstice. (0 to 3) + 15 = (15 to 18), and this explains the addition of 15. As there is the maximum day of 18 muhūrtas at summer solstice, there is the minimum night there, of 12 muhūrtas. This increases to maximum night, 18 muhūrtas for Sun at the beginning of Makara, 6 months after. As a result, as the Sun's rāśi increases from 3 to 9, the night increases from 12 to 18. (3 to 9) + 9 = (12 to 18) and this explains the addition of 9 (three times three). Example 7. Give the day-time for the Sun at the beginning of: (i) Ṛṣabha, (ii) Kumbha and (iii) Kanyā. (i) The Sun is in the quadrant 0 to 3 rāśis. The beginning of Ṛṣabha is one rāśi. Therefore 1 + 15 = 16, muhūrtas, is the day-time. (ii) The Sun at the beginning of Kumbha is 10 rāśis. This is in the quadrant 9 to 12. Therefore 10 + 3 = 13, muhūrtas, is the day-time. (iii) The Sun at the beginning of Kanyā is 5 rāśis and is in the 6 rāśis 3 to 9. Therefore 5 + 9 = 14, muhūrtas, is the night-time. Deducting from 30, 30 − 14 = 16, muhūrtas, is the day-time. [शङ्कुच्छाया] कर्कटकादिषु भुक्तं द्विगुणं माध्यन्दिनी भवेच्छाया । मकरादिषु चाप्येवं, (किंत्वस्मिन्) मण्डलाच्छोध्यम् ॥ ९ ॥ मध्याह्नच्छायार्धं सत्रिभमर्कोऽयने भवेद्याम्ये । उदगयने संशोध्यं पञ्चदशभ्यो रविर्भवति ॥ १० ॥ Gnomonic Shadow 9. When the Sun is in the six rāśis beginning with Karkaṭaka, the number of rāśis traversed from the beginning of Karkaṭaka, multiplied by 2, is the mid- day shadow (of the twelve-digit gnomon) in digits. When the Sun is in the six rāśis beginning with Makara, find the distance in rāśis traversed by the Sun from the beginning of Makara, and multiply by 2. Subtract this from 12, to find the mid-day shadow. 10. When the Sun is in its southward-course, (i.e., in the six rāśis from Karkaṭaka), half the mid-day shadow plus three is the longitude of the Sun in rāśis. When in the northward course in the six rāśis from Makara, half the noon-shadow subtracted from fifteen is the Sun in rāśis. d. A.B.C.D.किं चास्मिन्. A. मंडलात्छोध्याम्; 9a. A1.कर्कटादिषु; B2॰दिवु. B. भक्तं B. मदलात्सोध्या b. A.B1.मध्यंदिनी. A. भवेछाया 10a. A1.॰ह्नछायार्द्धं c. A1.चाप्येव; A2.वाप्येवं d. B. पञ्चदशेभ्यो 5

36 PAÑCASIDDHĀNTIKĀ II.10 Example 8. (a) On a certain day the Sun’s longitude is 5 rāśis. (b) On another day it is rāśis 11-15. In both cases find the mid-day shadow. (a) The Sun is 5 rāśis and is within the six rāśis from Karkaṭaka, being 2 rāśis from the beginning of the first point of Karkaṭaka (i.e. 3 rāśis). So, 2 × 2 = 4 digits is the shadow. (b) The Sun’s longitude is rāśis 11-15. This is within the six rāśis from the first point of Makara (9 rāśis), the Sun’s position being 11ʳ 15° − 9ʳ 0° = 2ʳ 15° = 2½ rāśis from that point. 12 − (2½ × 2) = 7 digits is the noon-shadow, Example 9. (a) The Sun is in its southward course and the shadow is 4 digits. Find the longitude of the Sun. (b) The Sun is in its northward course and the noon-shadow is 7 digits. Find the Sun. (a) Half the shadow = 4/2 = 2. As the course is southward add 3 rāśis; the Sun’s longitude is 5 rāśis. (b) Half the shadow = 7/2 = 3½. As the Sun’s course is northward, deduct from 15. 15 − 3½ = 11½ rāśis. This is the longitude of the Sun. From the two sets of examples it can be seen that the two formulae are the inverse of each other. The formulae are explained thus: This Siddhānta assumes that the noon-shadow is zero when, at the end of its northward course, it reaches the first point of Cancer. Then as it moves southward, the shadow increases to 12 digits at the end of the course, i.e. after 6 months, when the Sun reaches the first point of Capricorn. Assuming the increase to be uniform, there is an increase of 2 digits per rāśi. As the shadow is zero for the first point of Cancer, the longitude in rāśis measured from this point, multiplied by 2 gives the shadow. Thus, if c is the Sun in rāśis measured from the first point of Cancer and s is the shadow in digits, s = 2c for the 6 months till the Sun reaches Capricorn, where the shadow is 6 × 2 = 12 digits. Then the shadow decreases at the same rate to zero at the first point of Cancer, in the course of 6 months. Therefore if c is the Sun measured from the first point of Capricorn, when the shadow is 12, and s the shadow, then s = 12 − 2c. Now for the longitude of the Sun from the noon-shadow. We have seen that for the six rāśis from Cancer, s = 2c. Therefore c = s/2. But c is counted from the first point of Cancer, i.e. 3 rāśis. There- fore the Sun’s longitude in rāśis is 3 + c = 3 + s/2, which is the same as the instruction to divide the shadow by 2 and add three rāśis. For the six rāśis from Capricorn, we have seen that s = 12 − 2c. Therefore c = (12 − s)/2. But c is counted from the first point of Capricorn. i.e. 9 rāśis. Therefore the Sun = 9 + c = 9 +(12 − s)/2 = 15 − s/2, which is the instruction given. It must be noted here that both the formulae are very rough. At summer solstice, when the Sun is at the first point of Cancer, its north declination is maximum and given by Hindu astronomy as 24°. At that time, the mid-day Sun is at the zenith at places on 24° north latitude (like the region of Ujjain) and so it is only in this region that the shadow is zero at this time. When the Sun reaches its southernmost point at winter solstice, i.e. the first point of Capricorn, its south declination is 24°. Therefore the zenith distance of the noon-Sun as seen from latitude 24° North at that time must be 48° towards the South, and the shadow at that time must be greater than 13 digits and not 12. (All this will be shown in Chapter IV). If the shadow is to be 12 digits, the Sun’s zenith-distance must be 45° and the region where the Sun is seen at this zenith-distance is 21° North latitude. Thus there is contradiction even here. In verse 8, we showed that the rule is intended for a region having about 36° North latitude, neither 24° nor 21°. Thus, so far as these things are concerned, the Siddhānta seems to be a hotch-potch.

II.11 II. VĀSIṢṬHA-SIDDHĀNTA 37 [छायातो लग्नं लग्नतः छाया च] द्वादशभिः सच्छायैर्मध्याह्नोनेर्भजें'द्रसहुताशम्' | अपराह्ने चक्रार्धाद्विशोध्य सार्कं भवति लग्नम् || ११ || Lagna from shadow and vice versa 11. Add 12 to the shadow (of the twelve-digit-gnomon, measured in digits) at any time of the day, and deduct from it the mid-day shadow for the day. Divide 36 by this and take the result. This result taken as rāśis, plus the Sun in rāśis is the lagna at the moment, if it is forenoon. If afternoon, deduct this from the Sun plus six rāśis and the lagna is got. The formulae (a) for the forenoon and (b) afternoon respectively can be expressed thus: (a) Lagna = Sun + 36/(12 + shadow – noon shadow). (b) Lagna = Sun + 6 – 36/(12 + shadow – noon shadow). What is called lagna is the Orient Ecliptic Point, i.e. the point of the ecliptic rising on the eastern horizon. Example 10. (a) On a certain day, the Sun is 9 rāśis and the mid-day shadow 12. At a time in the morning the gnomonic shadow is 36. Find the lagna for the moment. (b) On a certain day, the longitude of the Sun is 6 rāśis and the noon-shadow 6 digits. At a time in the evening the gnomonic shadow is 24 digits, find the lagna for that moment. (a) From formula (a), Lagna = 9 + 36/(12 + 36 – 12) = 9 + 1 = 10, rāśis. Hence the first point of Kumbha is rising in the east. (b) From formula (b), Lagna = 6 + 6 – 36/(12 + 24 – 6) = 12 – 1 6/30 = 10 24/30 rāśis. Hence the 25th degree of Kumbha is rising in the east. These rules are rough and there is no question of strictly proving them. But we can explain them thus. From sunrise to noon, as the altitude of the Sun increases, the lagna goes on increasing and the shadow decreasing, till it is shortest at noon. Therefore the increase in lagna can be roughly expressed as, a/(shadow + b), where a and b are constants to be determined. Now, if the place is supposed to be situated on the equator, and the ecliptic on which the Sun moves is supposed to coincide with the celestial equator, then at noon the shadow will be zero. At that time the increase in lagna (after sunrise) would be 3 rāśis, as the Sun has reached an altitude of 90°. Therefore 3 = a/(0 + b). Again, seven and a half nāḍīs after sunrise, the Sun would have risen to an altitude of 45°. So the increase in lagna now is 1 1/2 rāśis and the shadow is 12 tan 45° = 12 digits. Therefore, 1 1/2 = a/(12 + b). Solving these two equations for a and b we get a = 36, b = 12. Therefore the increase in lagna is 36/(shadow + 12), of course, on the given two assumptions. But the place may not be on the equator and the ecliptic does not coincide with the celestial equator and the Sun moving on it has a varying declination, with the result that generally the noon-shadow is not zero. According to the length of the noon-shadow at other times also there will be an increase in the shadow over what 11a. A1. द्वादशभिः; A2. द्वादभिः. A. सछायै A2. द्रसंजताशं b. B. मध्याह्नाने. B. हृतांशः. A1. ०द्रसजताशं; c. A.B. चक्रार्द्धाद्

38 PAÑCASIDDHĀNTIKĀ II.11 it would otherwise be, for which the rule has been formulated. As the deduction of the noon- shadow for the day of observation from the shadow would, to some extent, remove this extra length of shadow and bring about an approximation to the ideal condition, the noon-shadow is asked to be deducted from the shadow in the formula. Therefore the increase in lagna is given by 36/ (shadow − noon shadow + 12). As at sunrise the Sun is the lagna, adding the increase to the Sun we get the lagna, i.e. lagna = Sun + 36/(12 + shadow − noon-shadow). This is for the forenoon. In the afternoon, what happens in the forenoon with reference to the shadow is reversed, and therefore the rule gives the part of the lagna to increase from the time of observing the shadow to sunset. So, it is less than the lagna at sunset by what is got from the formula. But the lagna at sunset is the Sun plus six rāśis. Therefore the formula for the afternoon becomes: Sun + 6 − 36/(12 + shadow − noon-shadow). Again let it be remembered that the rules are rough. व्यर्के लग्ने लिप्ताः प्राक्पश्चाच्छोधितास्तु चक्रार्धात् । कार्यश्छेदः 'शून्याम्बराष्टलवणोदषट्कानाम्' ॥ १२ ॥ लब्धं द्वादशहीनं मध्याह्नच्छायया समायुक्तम् । सा विज्ञेया छाया वासिष्ठसमाससिद्धान्ते ॥१३ ॥ 12-13. Deduct the Sun from the lagna and convert the remainder into minutes of arc, if forenoon. If afternoon, deduct the minutes from a half circle, (i.e. from 10,800 minutes), and take these as minutes. Divide 64,800 by the minutes got. Add the result to the noon-shadow of date and deduct 12 from this. This is the shadow at the time of the given lagna. This is according to the succinct Vāsiṣṭha Siddhānta. The formula (a) for the forenoon, and (b) for the afternoon, respectively, are: (a) shadow = {64,800 ÷ (lagna − Sun, in minutes) + noon-shadow − 12}. (b) shadow = 64,800 ÷ {10,800 − (lagna − Sun, in minutes)} + noon-shadow − 12. Example 11. (a) On a certain day at a time in the forenoon, the Sun is 9 rāśis and the lagna 10 rāśis and the noon-shadow of date is 12 digits. Find the shadow for the time. (b) On another day, for a time in the after- noon, the sun is 6 rāśis and the lagna 10 rāśis 25 degrees and the noon shadow of date is 6 digits. Find the shadow. a] Using formula (a), shadow = 64,800 ÷ {(10 − 9) × 1800} − 12 + 12 = 36 digits. b] Using formula (b), shadow = 64,800 ÷ {10,800 − (10 5/6 − 6) × 1800} − 12 + 6 = 64,800 ÷ (10,800 − 8640) − 12 + 6 = 64,800 ÷ 2160 − 12 + 6 = 30 − 12 + 6 = 24 digits. These formulae (a) and (b) can be derived from the previous formulae (a) and (b) of verse 11, for these are only the inverse of the previous operations. 12a. B3.लिप्ता b. A.प्राक्पक्षा A.छोधितास्तु; B.छोधितासु 13. A1.लब्धं; A2.लब्ध. B3.हिनं B1.चक्राद्धति:; B3.चक्रार्धात् । द्वोतिः b. A.B.°ह्रच्छायया. B.समायुक्ता c. A.B.कायछेदः (B3.कायः छेदः) d. A2.वासिष्ट; B1.2.वाशिष्ट; B3.वाशिष्ठ

II.13 II. VĀSIṢṬHA-SIDDHĀNTA 39 The previous (a), is: lagna = Sun + 36/(12 + shadow − noon-shadow). Therefore, lagna − Sun = 36/ (12 + shadow − noon-shadow). Therefore 36/(lagna − Sun) = 12 + shadow − noon-shadow. There- fore, shadow = 36/(lagna − Sun) + noon-shadow − 12, where it is to be noted that (lagna − Sun) is in rāśis. If it is to be expressed in minutes, we have, shadow = 36/(lagna − Sun) × 1800 ÷ 1800

• noon-shadow − 12 = 36 × 1800/(lagna − Sun) in minutes + noon-shadow − 12 = 64800/(lagna − Sun) in minutes + noon − shadow − 12, which is (a) here. The previous (b) is: lagna = Sun + 6 − 36/(shadow + 12 − noon-shadow). Therefore 36/(shadow
• 12 − noon-shadow) = Sun + 6 − lagna = 6 − (lagna − Sun). Therefore 36/{6 − (lagna − Sun)} = shadow + 12 − noon-shadow. Therefore shadow = 36/{6 − (lagna − Sun)} + noon-shadow − 12, where (lagna − Sun) is in rāśis. If it is to be expressed in minutes, we have, shadow = 36/6 − { 6 − (lagna − Sun) × 1800 ÷ 1800} + noon-shadow − 12 = 36 × 1800/{6 × 1800 − (lagna − Sun) in minutes}
• noon-shadow − 12 = 64,800/{10,800 − (lagna − Sun) in minutes} + noon-shadow − 12, which is (b) here. The concluding words, Vāsiṣṭha-samāsa-siddhānte, though forming a part of the sentence giving the rules of verses 12-13, may be detached from it and taken to refer to the whole chapter II and mean ‘All this is given as in the succinct Vāsiṣṭha Siddhānta’. The word, nakṣatrādicchedaḥ is found at the conclusion in the manuscripts. Perhaps it is, nakṣatrā- dhicchedaḥ, meaning ‘Naksatra section’ and the title is appropriate because this is the chief thing given here and other things depend on it. Thus in this chapter the true Sun and Moon are given according to the Vāsiṣṭha and also other matters depending on them like the nakṣatra, the tithi, day-time shadow and lagna. (The star planets i.e. the regular planets, of this Siddhānta will be given in XVIII.) TS have professedly not understood verses 1, 5 and 6 and gone wrong in verses 3 (and 8), which means they have practically not understood the Siddhānta at all. Thibaut even thinks that verse 1 may be dealing with the Moon. But this professed ignorance did not prevent him from making the unwarranted remark in the Introduction (vide p. XXXVIII)..... “the methods are so crude and so completely omit to distinguish between mean and true astronomical quantities, that the Vasistha Siddhanta can hardly be included within Scientific Hindu Astronomy.” [इति पञ्चसिद्धान्तिकायां वराहमिहिरविरचितायां वासिष्ठसिद्धान्ते ग्रहादिगणितं नाम द्वितीयोऽध्यायः ॥] A.B. have as Colophon, नक्षत्रादि छेदः; C.D. नक्षत्रादिच्छेदः | Thus ends Chapter Two entitled ‘Vāsiṣṭha-Siddhānta: Planetary Computations etc.’ in the Pañcasiddhāntikā composed by Varāhamihira

Chapter Three

PAULIŚA-SIDDHĀNTA — PLANTERY COMPUTATIONS ETC. ३. तृतीयोऽध्यायः पौलिशसिद्धान्तः — ग्रहादिगणितम् Introductory This chapter is a compendium of the part of the Pauliśa Siddhānta dealing with the Sun, Moon and Rāhu. It has already been mentioned that the original Pauliśa is now lost, perhaps for ever. This and the Saura are the only Siddhāntas dealt with by the author in full, the others being scrappy. For some reason not known to us, at present the Pauliśa is mixed up with the Vāsiṣṭha, for, the mean Moon, together with its peculiar technical terms pada, gati and ghana are used here, without mentioning how they are got. The formula for the Moon’s daily true motion, which patently belongs to Vāsiṣṭha has strayed into this chapter, another evidence of their being mixed up. In this chapter the Sun, the Moon and Rāhu, the methods of computing the daily nakṣatra, tithi and karaṇa, the two yogas vyatīpāta and vaidhṛti, the day-light in any given place in India, and certain holy days necessary for religious observations are dealt with. [स्फुटरविः] ‘खार्क’घ्नेऽ ‘ग्निहुताशन’मपास्य ‘रूपाग्निवसुहुताशकृतैः’ | हृत्वा क्रमाद् दिनेशो मध्यः केन्द्रं सविंशांशम् ॥ १ ॥ ‘एकादशा’ ‘ऽष्टषट्कं’ ‘रूपोना सप्ततिः’ ‘ख’युक्ता च | ‘नवषट्क’’मु(त्कृ)ति’श्च क्षयः कलाः केन्द्रराशिसमाः ॥ २ ॥ ‘दश’ ‘षट्काष्टक’ ‘सप्तति’ ‘सप्ततिरेकाधिका’ च ‘नवषट्कम्’ | ‘पञ्चकृति’श्चोपचयो मध्यमसूर्यः स्फुटो भवति ॥ ३ ॥ True Sun

1. Multiply the days from Epoch by 120, deduct 33 and divide by 43,831. The mean Sun in revolutions, rāśis etc. is obtained. Add 20° to this mean Sun. What is called kendram is got. 2-3. For the first six rāśis of kendra there are the following six quantities: 11, 48, 69, 70, 54 and 26, all deductive and in minutes. For the next six rāśis are the following: 10, 48, 70, 71, 54 and 25, all additive and in minutes. (If these are taken one after another according to rāśis of the kendra gone and) applied to the mean Sun, it becomes true Sun.

III.3 III. PAULIŚA-SIDDHĀNTA 41 In short, these twelve quantities are intervals of the equation of the centre for every rāśi of the kendra, the word being used in a peculiar sense here, and not the usual one of mean anomaly. The faulty readings, munyakṛti, muptakṛta is corrected as mutkṛti, meaning 26, and thereby the excess of one syllable in the foot also gets corrected. TS and NP have emended it as makṣakṛti, meaning 25, which, by its form, seems to be less likely to be the original, and which keeps the defect of the excess of one syllable. Example 1. (a) Find the true Sun for days from Epoch 690. (b) Compute the true Sun at Epoch. (a) Multiplying the days 690 by 120 and deducting 33, we have, 690 × 120 − 33 = 82,767. Dividing by 43,831, the revolution got is 1 and remainder 38,936. Multiplying this by 12 and dividing by 43,831, the rāśis got is 10. The remainder is 28,922, which multiplied by 30 and divided by the same divisor, gives degrees 19. The remainder 34,871 multiplied by 60 and divided by the same divisor gives 48 minutes. Thus the mean Sun is, omitting revolutions, rā. 10-19-48. Kendra = Mean Sun + 20° = rā. 10-19-48 + 20° = rā. 11-9-48. For 11 full rāśis of kendra and 9° 48' of the 12th, the minutes to be applied are, −11, −48, −69, −70, −54, −26, +10, +48, +70, +71, +54, +25 × 9° 48'/30°, which added together is − 17. Applying this to the mean Sun, the true Sun is 10ʳ 19° 48' − 17' = 10ʳ 19° 31'. (b) At Epoch the days are zero. Therefore 0 × 120 − 33' = −33. Mean Sun = −33/43,831 revolu- tions = −33 × 12 × 30 × 60/43,831 minutes = −16' = 11ʳ 29° 44', (since cycles of 12 rāśis can be added or omitted). Kendram = 11ʳ 29° 44' + 20° = 0ʳ 19° 44'. As no full rāśi or kendram is gone and there are 19° 44' in the first rāśi, the minutes to be applied are: − 11' × 19° 44'/30° = − 7'. The true Sun = 11ʳ 29° 44' − 7' = 11ʳ 29° 37'. It is to be noted that the word kendram here is not used in its usual sense in later astronomical works, as the mean anomaly (counted from the Higher Apsis) which is obtained by deducting the longitude of the higher apsis from the mean planet. In fact, the use of the expression in modern western astronomy itself is different in the sense that the anomaly is obtained by subtracting the lower apsis or perigee from the mean planet. The Sūrya Siddhānta instructs that the kendram is to be obtained by deducting the mean planet from the higher apsis which is equal to the kendram given by others, subtracted from 12 rāśis. Thus the common characteristic of these different kendras is that it is used as the argument for the equation of the centre and in this sense its use is appropriate here also. So we can take it that the mean Sun itself is used as the argument in the table, the values being given for the intervals 340° − 10°, 10° − 40°, 40° − 70°, 70° − 100°etc. instead of 0° − 30°, 30° − 60°, 60° −90°, 90° − 120° etc. (vide Table)

3. Quoted by Utpala on BS 2, p.40. | b. B. श्ययुक्ता च 1a. A. हताशन | c. A1. मुन्यकृतिश्च; A2. मुन्यकृतीश्च; B. मुप्तकृतश्च; b. A. मथास्य. A1. वसुताश | C.D. मक्षकृतिश्च c. A. हत्वा; B1.2. हचा; B3. हवा. A. क्रमादितेशो; | d. A. समा B3. क्रमादिनेशो | 3a. A1. दसष०. A1. D. सप्ततिः d. B1. सर्विशांश; D. सविंशांशः | c. B. haplographical omission of सप्तति. 2a. B2. दशाष्ठ | A.B. तिनैका

42 PAÑCASIDDHĀNTIKĀ III.3

The procedure is explained thus: According to the Paulīśa Siddhānta there are 120 solar revolu- tions in 43,831 days. So, in any desired number of days, the Sun’s mean motion is days × 120 ÷ 43,831 revolutions, which multiplied by 12, 30 and 60 successively gives rāśis, degrees and minutes. The mean solar year begins 16½ nāḍikās after Epoch. Therefore to reckon days from the beginning of the year, 16½ nāḍikās or 33/120 days must be subtracted from the days from Epoch. As the days have already been multiplied by 120 and converted into one hundred and twentieth parts, we have to deduct 33 parts in order to deduct 33/120 days, which is the instruction. But what is called mean Sun here is not the real mean Sun. It is the real mean Sun plus the equa- tion of the centre for the beginning of the year (this makes it the true Sun) plus 7 minutes of arc. That it is so can be seen thus: At the beginning of the year the so-called mean Sun is zero, the Sun having made full revolutions, starting from the zero point 16½ nāḍikās from Epoch. The Kendram at that time is 0° + 20° = 20°. For this we have the difference or interval of equation of the centre, – 11 × 20°/30° = – 6′.67. Deducting this from rā 0-0-0, we have the true Sun rā. 11-29-53. Deducting the equation of the centre from this, the real mean Sun is got, for the true Sun is obtained by adding the equation of the centre to the real mean Sun. Thus the so-called mean Sun = the real mean Sun

• the equation of the centre + 7 minutes = the real mean Sun + 142′ (135′.8 + 6′.67), the equation of the centre at this point being 135′.8 (which we shall show presently). Thus, the so-called mean Sun is practically the true Sun at the beginning of the year (the difference being only 7′) and the beginning of the mean year is therefore practically the beginning of the true year. This, we have alluded to already in verses I.11-13. Now, as this mean Sun has the same rate of motion as the real mean Sun, everywhere the difference of 142′ between them is maintained. Now we shall verify the intervals, i.e. differences of equation of the centre. Let us assume that the longitude of the higher apsis is rā. 2-16-0, according to this Siddhānta. (As the original Siddhānta is lost, we cannot assert that it is so. But if the assumption works, i.e. explains everything to be explained, without leading to contradictions, then we have to take that it is correct.) This is very likely because according to the Romaka it is at rā. 2-15-0 (see VIII.2, where the instruction, “subtract (from the mean Sun) rā. 2-15-0 to get the kendra of the Sun” is given.) The Sūrya Siddhānta, Āryabhaṭīya etc. give rā.2-18-0 for the same; and the earlier Paulīśa may correctly give rā.2-16-0. Now, as the intervals are for mean Sun 340° to 10°, 10° to 40° etc., we can say deducting 76°, they are for anomaly 264° to 294°, 294° to 324°, 324° to 354° etc. (See table above). Assuming the maximum

III.3 III. PAULIŚA-SIDDHĀNTA 43 to be 140', if the equation of the centre is computed for these anomalies, and the intervals tabulated, we get – 12.9, – 45.6, – 67.7, – 71.6, – 56.3, – 26.0, + 12.9, + 45.6, +67.7, + 71.6, + 56.3,

• 26.0, in minutes. As these are not much different from the series – 11, – 48, – 69 etc., we see that our assumption about the apsis is correct and the series – 11, –48, are derivable from it. In the two series, the constants are the same in some places, differ by 1' in some, by 2' in some, the differ- ence being 3' only in one place. Even this small difference may be due to the constants of the given series being empirical, the values having been discovered by repeated observations only. In this connection it may be mentioned that the tabular values of the Romaka differ far more than this does from the same values obtained

44 PAÑCASIDDHĀNTIKĀ III.3 naturally differ. Secondly, even if all Siddhāntas had started reckoning from the same point origi- nally, the difference in the duration of their solar years would, in course of time, cause differences in longitude when computed for a particular moment of time, causing apparently a shift of the first point. Thus as the first point of the Paulīśa is about 2° east of that of the Saura or Romaka, the mean Sun is less, as in the case of the Siddhānta Śiromaṇi, which is one degree more because its first point is one degree west of that of the Saura or Romaka. It is also well-known that there is a difference of three degrees between the first point of the Caitra and Raivata Pakṣas. Now we shall explain why 20° is added to the mean Sun to form a peculiar type of Kendram to give the values. Why have the values not been given directly for intervals of mean Sun, 0° to 30°, 30° to 60° etc. The reason must be that the author has taken these values from the original Paulīśa, or the original Paulīśa itself from its source, where they would have been given for intervals of mean Sun, 0° to 30°, 30° to 60° etc. But the source or original Paulīśa's first point might have been 10° east of that of the Paulīśa here given, and it might have been shifted west in course of time to the present first point adopted, (there is evidence in the Vedas of this kind of shift being made, as evident, for instance, in the case of the beginning of the year from the Sun at Mṛgaśirā to the Sun at Kṛttikā and so on) with the result that what was originally 0° had become 10°, what had been 30° had become 40° and so on reckoned from the new point. So the values are as if they are given for 340° to 10° etc. and to make them full rāśis for convenience, 20° are asked to be added and the name kendram given to it. Thus everything is properly explained. TS have not understood the method given here. So far as the explanation of the part referring to the mean Sun goes, what they say is correct. But after that what they say is all wrong. There is the express instruction after giving the first series 11, 48 etc. that the values should be subtracted from the mean Sun. The second series 10, 48 etc. come after that in a separate sentence and a separate verse, with the instruction that the values should be added. But somehow TS have understood the two instructions to mean that the two series should first be added one to one and then the resulting new series, which they think is the equation of the centre itself, should be applied to the mean Sun. If this is done, the instruction where to add and where to subtract is lost, which they have not noted. They have failed to see that the Siddhānta gives differences of values for every 30° intervals of kendram. They have never considered why if the equation of the centre itself is given, it is given in two series which are almost identical. Again, the new series of theirs only appears to be the equation of the centre, which is because the differences of a sine-function-series is a cosine function series, which is again a sine-function-series with a lateral shift of 90° in the argument. In verifying the series they have formed by comparing it with the actual, which they have computed, they have found a difference of 6' and 7' in two terms, but waived them aside as negligible. But 6' or 7' are too large to be negligible. Further, they have failed to see that the word kendram is used here in a peculiar sense as we have already said. They have taken it to mean the regular anomaly. But the mean anomaly can be found only if the longitude of the higher apsis is given, which is nowhere to be found. They explain this by saying that the longitude of the apsis was well known and therefore not given! Different Siddhāntas give different values for the longitudes of the apsis; the Romaka gives 75°, the Saura gives 80°, and the Āryabhaṭīya and Sūrya Siddhānta give 78°. Which of these are we to take? Certain things and operations, we can understand from the context and nature of the work, but this is not a thing which can be learnt without being told, as also the instruction when to add or when to subtract the values, which, according to them, has also to be learnt otherwise. If they had only tried to work out some examples, as for instance, taking the mean Sun as 80°, using their interpretation of the rules, then they would have discovered their mistake.

III.4 III. PAULIŚA-SIDDHĀNTA 45 [चन्द्रगतिः] [वि]नवात् पदाद्दशघ्नात् सप्तांशः ‘(साश्विखस्वरो)’ भुक्तिः । गत्यर्थान्ताच्छोध्यो लिप्ताभ्यो ‘(नवमुनिवसु-थ्यः)’ ॥ ४ ॥ True Motion of Moon 4. If the padas obtained (by II.2) are plus-padas, (i.e., in the first half-gati) subtract 9 from the padas, multiply by 10 and divide by 7. Add the result to 702. The Moon's daily true motion in minutes is got. In the second half-gati, (i.e., if the padas are minus-padas), deduct 9, multiply by 10 and divide by 7 and subtract the result from 879. The resulting minutes are the daily true motion of the Moon. Example 2. Find the Moon's daily true motion for plus padas 9, 18, 27, 36, 63, 71, 72, 81, 117, 124 and minus-padas 9, 63, 71, 72, 117, 124. P = 9 daily motion = 10 (9 − 9)/7 + 702 = 702 minutes P = 18 " = 10 (18 − 9)/7 + 702 = 702 + 12 6/7 = 714 6/7 minutes P = 27 " = 10 (27 − 9)/7 + 702 = 702 + 2 × 12 6/7 = 727 5/7 minutes P = 36 " = 10 (36 − 9)/7 + 702 = 702 + 3 × 12 6/7 = 740 4/7 minutes P = 63 " = 10 (63 − 9)/7 + 702 = 702 + 6 × 12 6/7 = 779 1/7 minutes P = 71 " = 10 (71 − 9)/7 + 702 = 702 + 88 4/7 = 790 4/7 minutes P = 72 " = 10 (72 − 9)/7 + 702 = 702 + 7 × 12 6/7 = 792 minutes P = 81 " = 10 (81 − 9)/7 + 702 = 702 + 8 × 12 6/7 = 804 6/7 minutes P = 117 " = 10 (117 − 9)/7 + 702 = 702 + 12 × 12 6/7 = 856 2/7 minutes P = 124 " = 10 (124 − 9)/7 + 702 = 702 + 164 2/7 = 866 2/7 minutes P = 133 " = 10 (133 − 9)/7 + 702 = 702 + 177 1/7 = 879 1/7 minutes P' = 9 " = 879 − 10 (9 − 9)/7 = 879 − 0 = 879 minutes P' = 63 " = 879 − 10 (63 − 9)/7 = 879 − 6 × 12 6/7 = 801 6/7 minutes P' = 71 " = 879 − 10 (71 − 9)/7 = 879 − 88 4/7 = 790 3/7 minutes P' = 72 " = 879 − 10 (72 − 9)/7 = 879 − 7 × 12 6/7 = 789 minutes P' = 117 " = 879 − 10 (117 − 9)/7 = 879 − 12 × 12 6/7 = 724 5/7 minutes P' = 124 " = 879 − 10 (124 − 9)/7 = 879 − 164 2/7 = 714 5/7 minutes The daily motion when the pada is less than 9 cannot be found from the formula as it is, but we can frame a rule by considering the nature of the variation of motion. Minus-padas 1 to 8 are the successive padas after plus-padas 124 and previous to minus-padas 9. Therefore the motion must lie between 866 2/7 and 879, increasing from 866 2/7 (vide example). Therefore, minus-padas 1 to 8, multiplied by 10 and divided by 7, added to 866 2/7 will give the motion. In the same way, plus- padas 1 to 8 are successive padas after minus-padas 124, and before plus-padas 9. Therefore the motion lies between 714 5/7 and 702, decreasing gradually. Therefore plus-padas 1 to 8, multiplied by 10 and divided by 7 must be deducted from 714 5/7 to get the motion. 4a. A.B.C. नगात् पदात्. B 1.3. दशाघ्नात् b. A.B. साश्विखांवरो; C. साश्वि [खाचला] c. B. गत्यर्द्धान्ताच्छोध्यो B 1.3. भुव्तिक्तिः; B2. भवनोक्तिः d. A.B.C. वसुमुनिवथ्यः

40 PAÑCASIDDHĀNTIKĀ III.4 The two rules for the true daily motion can be shown to be connected with the two rules for finding the true motion in II.6, by deriving these from those, and the exact derivation itself is a proof of the correctness of the rules. The rule for plus-padas (i.e. applicable to the first half-gati) is, {1094 + 5(P − 1)} P/63. If m is the mean motion at the end of the last full gati, then the true Moon = m + P° + {1094 + 5(P − 1)} P/63 minutes. The daily motion of the current day is got by subtracting the true Moon at the end of the previous day from that at the end of the current day and if the pada is P at the end of the current day, it is P − 9 at the end of the previous day. Therefore the motion for the current day = [m + 9° + {1094 + 5(P − 1)} × P/63] − [m + (P − 9)° + {1094 + 5 (P − 9 − 1)} (P − 9)/ 63]′ = (m − m) + P° − (P − 9°) + {1089 + 5P)} P/63 minutes − {1089 + 5(P − 9)} (P − 9)/63 minutes = in minutes, 540 + 1089 P/63 + 5P²/63 − 1089 (P − 9)/63 − 5(P − 9)²/63 = 540 + 1089 × 9/63

• 5 × 9²/63 + 90 (P − 9)/63 = 540 + 1134/7 + 10 (P − 9)/7 = 540 + 162 + 10 (P − 9)/7 = 702 + 10 (P − 9)/7, which is the rule here given. In the same way, taking the rule for the true Moon in the second half-gati, i.e. for minus-padas, we have, the daily motion = [m + rā. 6-0-4 + P° + {2414 − 5 (P′ − 1)} P′/63 minutes] − [m + rā. 6-0-4 + (P′ − 9)° + {2414 − 5 (P′ − 9 − 1)} (P′ − 1)/63 minutes] = in minutes, 504 + {2419 − 5P′} P′/63 − {2419 − 5(P′ − 9)} (P′ − 9)/63 = 540 + 2419 P′/63 − 5P′²/63 − 2419 (P′ − 9)/63 + 5 (P′ − 9)²/63 = 540 + 2419/7 − 45/7 − 10(P′ − 9)/7 = 540 + 339 1/7 − 10 (P′ − 9)/7 = 879 1/7 − 10 (P′ − 9)/7, which is the rule for daily motion in the second half-gati, omitting the small fraction of minutes 1/7. (Note that in the example we actually got this 879 1/7 as the maximum). From the relationship between these two sets of rules shown here, we can understand that the rules for daily motion, though they have strayed into the chapter dealing with the Pauliśa, actually belongs to the Vāsiṣṭha. If they are to be used for the Pauliśa also, it is because they are interconnected and mixed up, as the use of the same technical terms, and the absence of the method to find the mean Moon, show, We can even say that the author does not intend this for the Pauliśa because another set of rules is given for this in III.9. Because the daily increase in padas is 9, the daily increase or decrease in the true motion is 10 × 9/7 = 12 6/7 minutes. By integrating the motions and deducting the mean motion during the days for which the integration is done, we can find the equation of the centre implied in the rules: For convenience let us take padas 9, 18, etc. and work out for plus-padas first. The total true motion in minutes for P/9 days, i.e. to the end of P padas is: 702 + 10 (9 − 9)/7 + 702 + 10 (18 − 9)/7 + 702
• 10 (27 − 9)/7 ..... + 702 + 10 (P − 9)/7 = 702 × P/9 + 9 × 10 {1 + 2 + 3 +..... (P − 9)/9}/7 = 702 P/9 + 9 × 10 {½(P − 9)/9} P/(9 × 7) = 702 × P/9 + 5P (P − 9)/63 = 702 × 7P/63 + 5P²/63 − 45P/63 = 4869 P/63 + 5P²/63. The mean motion per day can be found from 11.2–4 to be 790′ 35″ and for P/9 days, the mean motion is 790′ 35″ × P/9 = 5534 P/63 minutes. Subtracting this from the true motion found, the equation of the centre obtained is 4869 P/63 + 5 P²/63 − 5534 P/63 = 5P²/63 − 665 P/63 = (5P − 665) P/63. In the same way, we can find the equation of the centre connected with the second half-gati, i.e. minus padas. The total true motion = 879 1/7 × P′/9 − {5P′²/63 − 45 P′/63} = 6154 P′/63 + 45 P′/63 − 5P′²/63. Deducting the total mean motion, the equation of the centre = (6154 + 45 − 5534) P′/63 − 5P′²/63 = 665 P′/63 − 5P′²/63 = P(665 − 5P)/63. It is these two rules for the equation of the centre that we used in II.6, to derive the rules there. From inspection we see that P(5P − 665)/63 is negative for all values of P, as it ought to be in the first half-gati. P′ (665 − 5P′)/63 is positive, as it ought to be in the second half-gati for all the values of

III.4 III. PAULIŚA-SIDDHĀNTA 47 P'. The numerical maximum is for P or P' = 66½. For P = 124, the value is −88½ minutes which is cancelled by + 88½ minutes for P' = 124, with the result that at the end of the gati the equation of the centre is zero and the true moon is equal to the mean Moon. It is mainly because of the relationship between the rules of II.6 and those in this verse, established by using the corrected reading vinavāt for nagāt that we have made that correction. Also if we take nagāt, then there will be a deficiency of one syllable. The adoption of the other reading, nāgāt has not the fault of the defective syllable, but 9 is required in the proof and not 8. The reading sāśvisāmvaro is corrected as sāśvikhasvaro, to mean 702, which is the minimum daily motion, derivable from the rules. TS also have intended 702 minutes to be the minimum daily motion by their emendation sāśvikhācalā, but theirs is not likely to be the original reading with such changes in the letters. We have corrected vasumuninavabhyah into navamunivasubhyah because from the establishment of the relationship, which we did, 879 minutes is the maximum daily motion and we require this. Further, the maximum must be as far above the mean value 790½ minutes, as the minimum is below. The minimum 702' is 88½ below 790½. So the maximum must be 790½' + 88½' = 879'. But TS have not touched this, for, by their own words, they have not been able to interpret this verse. [मन्दफलम्] पदमेकोनं 'पञ्चाष्टक' घ्नं'मे(कर्तुं)पक्षविषयेभ्यः' । प्रोज्य पदघ्नं छिन्द्या-'न्नवयममुनि'भिः कला इन्दोः ॥ ५ ॥ Equation of the centre 5. Reduce the plus or minus padas by one and multiply by 40. Subtract this from 5261. Multiply the result by the padas and divide by 729. The resulting minutes are equation of the centre. The formula is: The equation of the centre = {5261 − 40 (pada − 1)} pada/729, where pada is any pada, plus or minus, without distinction. Example 3. (a) Pada = 63. Find the equation of the centre. (b) Pada = 9. Find the equation of the centre. (a). {5261 − 40 (63 − 1)} 63/729 = (5261 − 2480) 63/729 = 2781 × 7/81 = 240 1/3 minutes. (b). {5261 − 40 (9 − 1)} 9/729 = (5261 − 320) 9/729 = 4941 × 1/81 = 61 minutes. खार्काधिकं भवेद्यत् परिशोध्यं तत् पुनः शताद् विंशात् । शशिनि धनं पूर्वार्धे (गत्यर्थेऽन्त्ये) क्षयः कार्यः ॥ ६ ॥ न पदं त्रिषष्टिपूरतः प्रथमपदं सप्ततिं त्वतिक्रम्य । पदयुक्ताः षट्पञ्चगुणाश्च बिन्दुस्त्रिघनभक्ते ॥ ७ ॥ 5a. B. पदमेकानं c. A.प्रोह्या; B1.3. प्रोध्या; B2. प्रोझ्या b. A.B. ॰घ्नेमेकं तु पक्ष॰ (A2. ॰नु पक्ष॰) d. B2. कलां