Graha_Laghava

In this work, we have provided an English Exposition - with explanatory notes, derivations and worked examples - of the popu- lar Sanskrit karana text, Grahalaghavam . The author of this hand- book of Indian astronomy is Ganega Daivajfia who belonged to the early 16th century. There is a general belief that after Bhaskara II there was a decline in the development of mathematics and astronomy in India. While there may be some truth in this belief, thanks to historical reasons, it is also true that the post - Bhaskara period saw an intensely creative activity in mathematics in the regions south of the Vindhyas. In fact Kerala became the cradle of tremendous and rich mathematical output - often anticipating developments in the European mathematics. In other parts of India too there shone great luminaries, like Ganega Daivajfia, upholding the great tradition of Indian astronomy and mathematics. In fact, the astronomical works of no other astronomer are in use among the makers of traditional paficdnigas (astronomical almanacs) in most parts of India today as those of the great and popu- lar astronomer, Ganega Daivajiia. 1, Date and place of Ganega and Kegava Daivajiia Ganeéga Daivajfia’s father was the famous astronomer Kegava Daivajfia and his mother’s name was Laksmi. He was born in 1507 AD (Saka 1429) at a place called Nandigrdma on the western sea- coast. S.B. Dikshit points out that Nandigrama is at present a vil- lage called Nandagaon in the Janjeera State in the Konkan region. It lies about 40 miles to the south of Mumbai (Bombay). The family belonged to the gotra (patrilineal ancestry) of Kusika. Ganega’s grand-father (Keéava’s father) was Kamalakara, also an eminent astronomer. Ganeéa’s teacher in astronomy was his father himself while Kegava’s guru was Vaijandatha. ## Page 2 (ii) GRAHALAGHAVAM OF GAN ESA DAIVAJNA Ganeéa’s father, KeSava, composed several works and com- mentaries on astronomy and astrology among which his astronomi- cal work, Grahakautuka was highly respected. In fact, KeSava is regarded as one of the best observational astronomers of ancient and medieval India. In his Mitaksaraé auto-commentary on the Graha kautuka, KeSava observes: “The figures as calculated from the Brahma, Aryabhata and Saura siddh 4ntas exhibit a vast difference in the positions of Mer- cury (Budha) and Venus (Sukra). Saturn (Sani) has shown an ex- cess of five degrees when actually observed in the sky at the time of conjunction with stars and planets and while setting and rising... Similarly, a difference is recorded in epochal positions and in the annual rates of motion....”. “Hence the future calculators should calculate planetary po- sitions by adopting the figures of revolutions increased or decreased in conformity with the actual observed phenomena of conjunctions, rising and setting of stars and planets in their own times. The writer (Keava) has accordingly found, out the mean position of the Moon, instead of its maximum equation of centre, by reversed steps, from the observation of the lunar eclipse at the ending moment of the fullmoon, since the equation of centre is neither positive nor nega- tive. The Moon’s apogee was finally fixed by reversing the steps of calculation, after observing the eclipse at the moment of the fullmoon in the celestial globe of observation since the maximum correction is neither additive nor subtractive. The moon’s position was found to be 5 minutes (of arc) less as compared to that calculated from the Suryasiddhanta. The apogee agreed with that of the Brahmapakga Thus, the writer has calculated the positions of planets by a short method after observing their actual positions at the present time”. 2. Works of Ganesa Daivajiia Ganega composed several important works on astronomy, as- trology etc among which his astronomical treatise Grahaldghavam is the most famous. In fact, the remarkable popularity of the Graha ## Page 3

laghavam surpassed that of his father’s Grahakautuka which was truly an important text in its own right. Ganeéa’s other works are: Laghu- and Brhat-Tithi Cintamani, a commentary on Bhaskara’s Siddhanta Siromani, a commentary on Bhaskara’s Lildvati (called Buddhivilasini). Vivaha vrnddvana tikd, Muhurta tattvati ka, Srdddha nirnaya etc. Ganega himself mentions another work of his, Parvanirnaya. Among Gane§Sa’s works, the Grahalaghavam appears to have been composed first, believed to be when he was just 13 years old. The epoch of the Grahalaghavam is March 19, 1520 AD, Monday. The work, Laghu tithi cintamani was composed in §aka 1447 (1525 AD) and the Buddhivilasini commentary on Bhaskara’s Lilavati in the year 1545 AD. Another work, Pata sarani, was com- posed some time after 1538 AD. Ganeéa’s Vrndavanatika gives the date of its composition in a very interesting fashion: “Take 12 as the number for the samvatsara (hayan). Add one (lava) for ayana to it. Add 6 to the sum of these two numbers (i.e, to 12 + 1) so as to obtain (19) as the num- ber of the yoga. Add 4 to the sum which would give 23 as the number of the naksatra and one (lava) for the paksa. If a paksa is added to one more paksa (i.e, 1 + 2), it would give 3 as the number denoting the week day. Take 1 as the tithi number and 11 as the month num- ber. Multiply the sum of all these numbers by 21 and increase the product by 9 (nanda).The result is the Saka year number”. This gives us the following: Samvat Ayana Yoga Naks Paksa Vara Tithi Masa Saka (12 + 1 + 19 + 23 + 1 + 38 + 1 +11)x21+9=1500 i.e(71 x 21) + 9 = 1491+ 9 = 1500. This means that the text, Vivdha vrnddvana tika was com- posed on Magha (11th lunar month) Sukla (1st pakga) Pratipat (1st tithi) of Saka 1500 (i.e., 1578 AD), Bahudhanya (12th) samvatsara, during Uttardyana (1st ayana), Dhanigtha (23rd) nakgatra and ## Page 4 (iv) GRAHALAGHAVAM OF GAN ESA DAIVAJNA parigha (19th) yoga. The date of the said text suggests that GaneSa was about 71 years old when he composed it. However, Dikshit refers to a manu- script copy of the Vrndavanatika which gives the year of its compo- sition as Saka 1476 (i.e. 1554 AD), Ananda samvatsara, which is 24 years earlier than the above-cited date (1578 AD) In the Grahalaghavam, the positions of planets have been given for the moment of sunrise of Monday, the newmoon day of Phalguna of Saka 1441 corresponding to March 19, 1520 AD (Julian). 3. Special features of Grahalaghavam Ganeéga has simplified the method of computations of the po- sitions of planets which is otherwise laborious by the traditional method. To avoid a huge number for the ahargana (number of civil days since the epoch), GaneSa has adopted an ahargana cycle of 4016 days, approximately constituting 11 solar years. Therefore, the modified ahargana, being the remainder exceeding a completed number of cycles (of 4016 days each), never exceeds 4016 days and is hence handy. From the point of view of a paficanga-maker or a beginner who is ignorant of trigonometry, Grahaldghavam is easy to use since GaneSa has completely dispensed with the trigonometric functions. In fact, the dropping of trigonometric ratios has by no means seriously affected the accuracy of results. Even the other karana texts, using the sine tables, generally give the values of sines of angles only in intervals of 15° or 334°. Therefore, the sine-values for intermediate angles, by linear interpolation, are only approximate. ## Page 5 INTRODUCTION (v) 4. Popularity of Grahalaghavam and its commentators As mentioned earlier, the Grahalaghavam of Ganega Daivajfia is the most popular astronomical text, among the ancient and medi- eval texts, currently used in most parts of India. Further, among the karana works (hand-books) on Indian astronomy, the Grahalaghavam is considered as the most comprehensive, exhaus- tive and easy to use text. The Grahalaghavam carries with it very useful and authori- tative commentaries by reputed astronomers like Gangadhara (1586 A.D.), Mallari (1602 A.D.) and Vigvanatha (around 1612 A.D.) Gangadhara’s commentary on the Grahalaghavam is called Manorama. His father was Narayana who authored Muharta martanda. Gangadhara was a Vajasancyi Brahmin belonging to the Kausikagotra, the same as that of Ganega Daivajiia. Gan gadhara lived in a village called Tapar, lying to the north of Ghrsnesvara (Lord Siva) temple which is to the north of Devagiri (Daulatabad). Visvanatha was brother of the highly accomplished astrono- mer Visnu who composed a karana with the epochal year 1608 AD This karana text is based on the Surya siddhanta. Visnu also wrote a commentary on Ganega Daivajfia’s Brhat tithi cintam ani in which he explains the theory also. Viévanatha has written an udaharana on his brother’s karana. The two popular commentators of Grahalaghavam are Viévandtha and Mallari. They were born in illustrious Maharastrian Brahmin families of astronomers. Mallari’s father was Divakara, a pupil of Ganega. Kamalakara, author of the Siddhanta tattva viveka and Rariganatha who wrote a commentary on the Suryasiddhanta were descendants of Vigvanatha and Mallari. Nrsimha, nephew and pupil of Ganega, wrote his commentary Harsakaumudi in 1548. The other commentators are Gagadhara (1586), Narayana (Kaéi, before 1635) and Kamalakara (before 1662). ## Page 6 (vi) GRAHALAGHAVAM OF GAN ESA DAIVAJNA Viévanatha has given a large number of examples in his com- mentary to illustrate the methods of the Grahaldghavam. The Graha laghavam is extensively used by the paficdriga-makers par- ticularly in Maharashtra, Gujarat, northern parts of Karnataka, the Hyderabad Deccan region of Andhra Pradesh and by the Deccanis of Varanasi, Gwalior and Indore. It is pointed out that even the gov- ernment almanacs published at Indore and Gwalior used the Grahalaghavam and the Tithicintdmani of Ganega Daivajfia. 5. Some special results of Ganega Daivajiia (i) Phase of the Moon From the newmoon to the fullmoon, the phase of the Moon increases, given by phase = (1+ cos d)/2 where d is the elongation SME of the earth from the Sun as seen from the Moon. In fact, we have d = 180° - E, so that phase = (1- cos E)/2 where E is the elongation of the Moon from the Sun as seen from the earth. On a newmoon day, E = 0° so that the phase of the Moon is zero. On a fullmoon day, E = 180° so that the phase of the Moon is one. In Indian astronomy, the phase of the Moon is measured by the width of the illuminated part of the Moon which is called sita (or Sukla). The width of the unilluminated part, equal to the differ- ence between the Moon’s diameter and the sita is called asita. In fact, (1 -S)x(Moon's ang. diameter) 180 sita= where M and S denote the celestial longitudes of the Moon and the Sun respectively in degrees. ## Page 7 INTRODUCTION (vii)

sita -(-4}r 5 arigulas where T is the number of tithis elapsed in the bright fort- night (Sukla pakga) and the Moon’s diameter is taken as 12 arigulas. Of course, this formula is approximate and a similar formula is given by Brahmagupta. (ii) Rationale for four sides to form trapezium Bhdskara II investigates the possibility of four sides forming a trapezium. He gives the condition: “in a trapezium the sum of the other flank side and the face is smaller than the sum of the smaller flank side and the base”. (Lildvati, 185) Ganeéa Daivajfia has provided a rationale for this statement of Bhaskara II. (iii) Proof of the Sulva theorem (Pythagoras’ theorem) Ganega Daivajfia in his Buddhivildsini commentary on the Lilavati provides a fully geometrical proof for the geometrico-alge- braic rationale provided by Bhaskara II for the so-called Pythagoras’ theorem on a right-angled triangle. A

Let ABC be a triangle right-angled at A (Fig 1). Let AD be the perpendicular to BC. Then the triangles ADB, CDA and CAB (the vertices are in the order of correspondence of equal angles) are similar. Therefore, from triangles ADB and CAB, we have ## Page 8 (viii)

GRAHALAGHAVAM OF GANESA DAIVAJNA

_ AC? _ DC

From (1) and (2), we get

BC

or

(iv) Construction of rational quadrilaterals

to obtain a quadrilateral with rational sides, says that four rational right-angled traingles (jatyas) are to be formed out of two basic ra- tional right triangles as follows: If m?- n?, 2mn, m? + n? and p?- q’, 2pq, p?+ @ are the sides of two rational right triangles, then according to Ganeéa, the following are the traingles out of which the quadrilaterals are built up : @ (am? ~ n*) (p?- q?), 2mn (p? - q?), (p? = q?) (m? +n?) (ii) (m? - n®) (2pq), 4mnpgq, 2pq (m? + n*) (iii) (p? - gq?) (m? - n®), 2pq (m? - n?), (p? + g?) (m? - n*) (iv) (p? - q?) 2mn, 4pqmn, (p?+ q?)(2mn) However, T.A. Sarasvati Amma wonders whether this is the procedure intended by Brahmagupta and Bhaskara II. ## Page 9 INTRODUCTION (ix) (v) Evaluation of 7 Aryabhata (b. 476 AD) has given the value of zas T= ease 3.1416 20000 and he points out that this value is approximate (asanna).

7 = 3927 3.1416 and 2 = 25> =3,1415929 1250 113 the former value being the same as the one given by Aryabhata (obtained by removing the common factor 16 in the ratio). The mode of arriving at this value of x is by considering the perimeter of a regular polygon inscribed in a circle. The ratio of the perimeter of the polygon to the diameter of the circle approximates the constant x. Of course, the more the sides of the polygon, the bet- ter is the approximation. GaneSga Daivajfia suggests that the number of sides of the in- scribed polygon, starting from 12, is successively doubled as 24, 48 until it is 384. The diameter of the circle is taken as 100 units. Then the ratio of the perimeter of the 384-sided polygon inscribed in a circle of diameter 100 would give the approximate value of x as 3927/1250. GaneSa’s commentary, Buddhivilasini on the Lilavati, is an extremely useful text to understand the rationales for the formulae and methods used by Bhaskara II and his predecessors. 6. Grahalaghavam (GL) In our present work, for each Sloka we have given a working English translation along with mathematical explanation, deriva- tions, worked examples, tables and diagrams wherever necessary. This work is based on the printed edition of GL, with commentaries ## Page 10 (x) GRAHALAGHAVAM OF GAN ESA DAIVAJNA of Mallari and Vigvanatha, edited by Pt. Kedaranath Joshi and pub- lished by Motilal Banarsidass, Varanasi, 1981. The text of Grahalaghavam consists of 187 Slokas distributed into 14 chapters. However, the commentaries of Mallari and Vigvana tha contain a fifteenth chapter of 15 Slokas called Paficariga candragrahana. The text seems to have undergone quite a few changes in the hands of descendants and commentators. However, the popularity of the Grahalaghavam remains unchallenged. The mean positions of the Sun, the Moon and the five taragrahas viz. Kuja, Budha, Guru, Sukra and Sani are computed in Chapter 1. The epoch adopted by Ganeéa Daivajfia is March 19, 1520 AD (J), Mon- day, at the mean sunrise at Ujjayini. We have provided derivations for the formulae of the text using the basic data of the numbers of revolutions (bhaganas), of the heavenly bodies in the course of a Mahayuga of 432 x 10 years. The true positions of the Sun and the Moon using their equa- tions of centre (Mandaphalam) are worked out in the Spastadhikara (Chapter 2). The approximation for the Mandaphalam used by Ganeéa avoiding the trigonometric ratios is adequately explained. In fact, this justified device has greatly simplified computations. In Chapter 3 computations of the true positions of the five star-planets are worked out by applying the two important equa- tions namely Mandaphalam and Sighraphalam. While the first equation corresponds to the “equation of centre” due to the eccen- tricities of orbits, the second one corresponds to converting the helio- centric positions to the geocentric. We have demonstrated the procedures of the text with worked examples - one relevant to the time of the commentator Viévanatha (early 16th Cent). The second example is for a modern date. In the Tripragnadhikara (Chapter 4) the problems of dik, dega and k4la i.e., direction, place and time are discussed. While the mathematical expressions involve the trigonometric ratios jyd and kotijya, the greatness of the text lies in providing justified approxi- ## Page 11 INTRODUCTION (xi) mations by avoiding these trigonometric ratios. The prupose of this exercise is mainly to simplify the computations for the benefit of the Paficdriga makers and lay students. This intended purpose is amply served. The important param- eters of kranti (declination), aksAméa (latitude of a place), lagna (as- cendant), palabha (shadow of the gnomon, sanku) etc. are explained with examples. The Lunar and Solar eclipses are studied in Chapters 5 and 6. In these chapters computations of the circumstances of the two types of eclipses are explained. The determinations of the sparsa (beginning), madhya (middle) and mokga (end) of an ordinary eclipse and of the sammilanam and the unmilanam of a total eclipse (khagrasa) are demonstrated with traditional and modern examples. In the case of solar eclipse, the effects of the parallax on the longi- tude (lambanam) and on the latitude (nati) of the Moon are ex- plained. The speciality of GL lies in computing the posibility and cir- cumstances of eclipses without using ahargana, through masagana (the number of completed lunar months). This procedure is explained in detail in Chapters 7 and 8. We have elucidated the procedures with examples. In Chapter 9 (Udayastadhikara), an exhaustive study of the heliacal rising and setting of the planets and their retrograde mo- tion (vakragati) etc., using the sighrakendra (the angular distance of a planet from the Sun) is presented. The shadow-related problems regarding planets and stars and using them to find out directions and elapsed time are explained in Chapters 16 and 11. Finding the lagna and the time by observing a star on the meridian is also discussed. Chapter 12 (Srrigonnatyadhikara) deals with the rising of the Moon’s “horns”. Finding the ‘valanam’ (deflection) is also explained in this chapter. ## Page 12 (xii) GRAHALAGHAVAM OF GAN ESA DAIVAJNA In Chapter 13, conjunctions of planets (grahayuti) is pre- sented for which their angular diameters have to be determined. The method of determining the days elapsed (gata) or to be completed (gamya) for the occurrence of a conjunction is explained. Similarly, the conjunction of a planet with the Moon (an occultation) can be predicted by determining the Sara (latitude), which is corrected for its nati. Two important yogas between the Sun and the Moon viz., Vyatipata and called Vaidhrtipatas are discussed in Chapter 14. These occur when the declinations (kranti) of the Sun and the Moon have the same magnitude and lie either on the same side or on op- posite sides of the celestial equator. In Chapter 15, obtaining the true positions, diameters etc. of the Sun and the Moon from the count of elapsed lunar months (masagana) is discussed in detail. In concluding Chapter 16, the author Ganeéa explains how to obtain planetary positions etc for the times before his epoch. At the end of this text, Appendices and a detailed Bibliogra- phy are included. The Appendices include tables of mandaphala (equation of centre) and the sighraphala of the planets, the list of 27 yogas, Bhaskara I’s. approximate formula for sine and the udayamanas of rasis. In the course of the text, mathematical derivations, compari- son with modern results and illustrative diagrams are provided. ## Page 13 CHAPTER 1 ‘MapHYAMADHIKARA (Mean Positions of Planets) Sloka 1 : The invocatory verse has two interpretations : (a) KeSava's (i.e, Bala Mukunda's) words viz, the Vedas (i.e, Srutis) are supreme. By observing the practices enjoined in the Vedas the mind gets purified leading to the Supreme Knowledge and which (Vedas) have brief, difficult and immutable words implying many meanings which were made simple and clear through the later explanatory works (i.e, Smrtis and commentaries) by the incarnations of Lord Visnu. Victory to such words of the Lord ! (b) The author’s father’s i.e., KeSava’s words are glorious. By study- ing his works the mind becomes clear and absorbs the knowledge of stars and planets. Though the words are brief these are clear in their meanings and further flawless and which are elucidated further by the commentar- ies of his disciples. Remarks : Ganeéa Daivajfia’s preceptor and father was KeSava Daivajfia who composed a handbook of astronomy called Grahakautukam. This text was very popular before his son’s work Grahalaghavam outshone it. Sloka 2: O mathematician ! Meditate on that form of Visnu (Rama’s) which rendered Siva’s bow into three pieces and threw it to the ground, who is decorated with the garland of sattva quality, who gives life to every living being, who grants the wishes, who has taken the human incarna- tion, who runs the universe and is very handsome. (b) O mathematician ! You read this handbook (karana grantha) which is devoid of sines and cosines (the trigonometric functions), which is decorated with constant multipliers and divisors (i.e., parameters), which contains operations with angles (like anomaly etc.), which provides proper knowledge of the equation of centre etc., which also adopts the knowledge of shadow of the gnomon (sariku) and which is mind-capturing and error- free. ## Page 14

Sloka 3 : Even though very great scholars composed (astronomical) handbooks, their (mathematical) results are not achieved if sines (and arcs) are left out. Therefore, I (Ganeéa) set forth to compose a (planetary) hand- book of procedures which are very simple and clear. Now, the author describes the procedure for finding the Ahargana (the number of civil days elapsed since the chosen epoch) : Slokas 4 and 5 : The procedure described in these two lokas is explained below, for convenience, in a simple algorithmic style. The epoch chosen by Ganega Daivajfia is Salivahana Saka Varsa (year) 1442 Caitra sukla pratipat, corresponding to March 19, 1520 AD (Julian), Monday. The Ahargana for a given lunar date is determined as follows: (i) Subtract 1442 from the Salivahana Saka year (elapsed) of the given date to get the years elapsed (gatabdi). (ii) Divide the remainder by 11. The quotient is called cakra (cycles) Cc. (iii) | Multiply the remainder, obtained in step (ii), by 12 and to the product add the number of lunar months elapsed, counting Caitra as 1 etc. The sum thus obtained is called ‘mean lunar months’ (madhyama méasagana) denoted by M.

(iv) number of adhikamasas is given by the quotient of 33 (v) ‘True lunar months (spasta masagana) = Mean lunar months + M +2C +10 adhikamasas = M + Quotient of ““~"" = TM (vi) Mean ahargana (Madhyama ahargana) MAH = (1M x 80) + T+ © ## Page 15

where TI is the number of tithis elapsed in the given lunar month. 1 (vii) Ksaya dinas = Quotient of [4 Madhyama Ahargana) = KD (viii) True ahargana (savana dinas) i.e, the number of civil days, TAH = Mean ahargana — ksaya dinas

= MAH - Quotient of z (MAH) (ix) | However, since the average values of parameters are considered in the above computations, 1 day may have to be either added to or sub- tracted from the result of (viii) to get the actual true ahargana. This is done by verifying the weekday as follows : (a) | Multiply the cakras by 5 ie., find 5C, Add saévana ahargana to this ie., find (5C + TAH). (b) Divide the result of (a) by 7 and find the remainder.

pet _ If R= 0, then it is Monday; R= 1 then it is Tuesday; and so on. (c) If the calculated weekday is a day next to the actual weekday, then subtract 1 from TAH and if it is one day less than the actual weekday, then add 1 to TAH. [See Note (1) and (2) appearing later]. ## Page 16 S4

GRAHALAGHAVAM OF GANESA DAIVAJNA

Sa.Saka 1534, Vaigakha Parnima, Monday = May 16, 1612 A.D. (Gregorian) @ (ii) (iii) (iv) (v) (vi)

Gatabdi = 1534 — 1442 = 92 years (from the epoch) Divide the remainder in (i) by 11: the quotient, cakras = 8 = C and the remainder = 4 Multiplying the remainder from (ii) ie, 4 by 12 and adding the number of lunar months elapsed in the given year we get (4 x 12)+1=49 = M is the madhyama masa gana No. of Adhikamasas = M+2C +10 _49 +2(8)+10 _75 33 33 33 Quotient = 2. M + No. of adhikamasas = 49 + 2 = 51 = TM is the spasta masagana (a) Mean ahargana (Madhyama ahargana) =(TM x 30) + (No. of tithis elapsed in the given lunar month) =(51 x 30) + 14 = 1544 1 (b) Add INT rg) ie., INT (8/6) = 1 to the result of (vi) (a)

(Note : INT stands for the integer value). ## Page 17 (vii) (viii) (ix)

Ksayadinas = INT (““) = INT (8) =24 = KD 64 64 Savana ahargana (i.e., No. of civil days in the running cakra) = MAH - KD = 1545 - 24 = 1521 = TAH Week day verification : 5C + TAH = 5(8) + 1521 = 1561 1561

‘That is, the weekday comes out as Monday. Since the weekday obtained from calculation is the same as the actual weekday (known), nothing needs to be added to or subracted from TAH. Therefore, True ahargana = 1521 No. of cakras = 8 Note 1 : Sometimés when (Saka year 1442) is divided by 11, to get cakras the remainder could be 0. In that case even 2 may have to be added to or subtracted from the obtained sdvana dinas to get the true ahargana for the weekday. See the following example. Example 2: Saka 1574 Caitra Sukla Pratipat Ravivara (i.e, April 7, 1652 A.D., Sunday) @ Gi) Gii) Gatabdi = 1574 — 1442 = 182, years since the epoch. 132 132) | c=Int [-=-]=12 Remainder of | 77] is 0. (Ox 12)+0=02M. ## Page 18 S6 (iv) (wv) (vi) (vii) (viii) (ix)

GRAHALAGHAVAM OF GANESA DAIVAJNA

Number of adhikamasas = INT [ee]

33 33 ive, adhikamasas = INT [3] =1 TM = M + No. of adhikamasas = 0+ 1=1 1 Mean ahargana: MAH=(1 x 30) +0+INTIE (12)] = 30+ 2 = 32. Ksaya dinas, KD = INT MAH) er 32 =0 64 64 Savana ahargana = MAH - KD = 32 - 0 = 32 Cakras = 12, Ahargana = 32 (in the running cakra)

Weekday verification : R = Remainder of ( 7 92 i.e., remainder of (2) or R=1 ie., Tuesday. But the actual weedkay is Sunday. Therefore subtracting 2 from the savana ahargana we get Actual ahargana = 32 - 2 = 30,

## Page 19 MADHYAMADHIKARA (CHAP. 1) s7 (x) Christian date : No. of civil days since epoch = (cakras x 4016) + ahargana = 12 (4016) + 30 = 48,222 Kali ahargana of the GL epoch = 16,87,850 (fixed) Kali Ahargana of the given date : 17,36,072

1600 A.D.(G) : —_17,16,982 (under kali Ahargana) Year 52 : 18,993 April 7 : 97 Total : 17,836,072 corresponding to April 7, 1652 AD (G) Note 2: Sometimes there could be an adhikamdsa in a particular given lunar year. (i) If the given date is before the adhikamasa of that lunar year, then subtract 1 from the number of adhikamasas obtained in the calculation. Gi) Ifthe given date is after the adhikamasa of that lunar year, then add 1 to the number of adhikamasas obtained in the calculation. This is demonstrated in the following example. Example 3 : Saka 1555 Caitra Sukla Pratipat, Friday [March 11, 1633]. In this year Vaiéakha is the adhikamasa which comes after the given date. We shall find the cakra and ahargana for the given date : @ Gatabdi : 1555 - 1442 = 113 ## Page 20 S8 (ii) ii) (iv)

113

}- 10 , Remainder = 3 Mean masagana = (Remainder x 12) + No. of elapsed lunar months = (3 x 12)+ 0 = 36 = M

-INT [“ cons -INT “ist 33 =INT [$]-2 (Note : Remainder = 0) Since the given date falls before the adhika Vaisékha masa, subtract 1 from the number obtained above. Therefore, the actual adhikamasas elapsed = 2 - 1 = 1. wv) (vi) (vii) (viii) (ix) True lunar months = M + No. of adhikamasas = 36 + 1 = 37 = TM Mean Ahargana, MAH = (37 x 30) + Tithis elapsed (e 10 + INT a = 1110+ 0+ INT € =1110+1=1111. Ksayadinas, KD = INT [| =INT kal =17 64 64 Savana ahargana = MAH- KD = 1111-17 = 1094 = TAH

5C+ TAH =65 (10) + 1094 = 1144; Remainder of [4] 3 ## Page 21 (&) MADHYAMADHIKARA (CHAP. 1) so i.e., Thursday; but the actual weekday : Friday Therefore, the true ahargana = TAH + 1 = 1095 Christian date : No. of civil days since epoch = 10 (4016) + 1095 = 41,255 Kali ahargana of the GL epoch : 16,87,850 .. Kali ahargana of the date : 17,29,105 From Table 1.1 to 1.3 under kali ahargana we have 1600 AD(G) > 17,16,982 33 : 12,053 March 11 : 70 Total : 17,29,105 corresponding to March 11, 1633 A.D. (Gregorian). Example 4 : Saka 1530 (Bhadrapada is adhikamasa) Kartika Sukla Pratipat, Saturday. We have @ Gi) (ii) (iv) Gatabdi = 1530 — 1442 = 88 88 ‘ Cakra, C = INT (3) =8, Remainder = 0 Mean lunar months M = (0 x 12)+7=7. No. of Adhikamasas

=INT [ 33 33

Since Kartika month (i.e, the given month) occurs after the adhika Bhadrapada masa, add 1 to the calculated number of ## Page 22 S10

adhikamasas. Therefore, number of adhikamdasas = 1+ 1 = 2. (v) True lunar months TM = M + No. of adhikamasas = 7+ 2=9 (vi) Mean ahargana : MAH =(TM x 30) + (Tithis elapsed in the given month) (vii) (viii) (ix) @) +INT (5) = (9 x 30) +0+ INT () = 271 Keaya dinas = INT (4) 64 Savana ahargana TAH = MAH - KD = 271 - 4 = 267 Weekday verification : 5C + TAH = 40 + 267 = 307 Remainder of (@)-s ie., Sunday.

: True ahargana = TAH ~ 1 = 266

No. of civil days since epoch = 8 (4016) + 266 = 32,394 Kali ahargana of GL epoch : 16,87,850. :. Kali ahargana of the given date : 17,20,244

1600 A.D. > 17,16,982 Year 08 : 2,922 December 6 340 Total 17,20,244 corresponding to December 6,1608 AD (Gregorian). Finding the Christian date from the ahargana and vice versa In the above examples we saw how to get the cakras and the ahargana from the given Candramana (i.e, lunar) date. Now, we shall see ## Page 23

how from the thus obtained cakras and ahargana the corresponding Chris- tian date can be obtained. In Table 1.1 the Julian days and the aharganas for the epochs of the Kaliyuga and the Graha laghavam are given for the beginnings of the Christian centuries from — 3200 (Julian) to 2200 AD (Gregorian).

@ Gi) (iii) The epoch adopted for the Kali era is the mean midnight between February 18th and 19th of 3102 BC (i.e, the year -3101). For any year before Christ (BC), for mathematical convenience, the negative sign is prefixed to 1 less than the numerical value of the Christian year. For example, 46 BC is considered as —45 and 3102 BC as -3101. This convention is adopted since 1 BC is taken as the “0” year of the Christian era. The epoch of the Graha laghavam (GL) : Ganega Daivajfia in his GL has adopted the mean sunrise (at Ujjayini) of March 19, 1520 (Julian) AD, Monday as the epoch. Julian days (JD) : The reckoning of the Julian days starts from the mean noon (GMT) on January 1, 4713 BC, Monday. On that day, at the mean noon (GMT), JD = 0. The procedure for finding the Christian date from the cakras and the ahargana : (i) Multiply the number of cakras C by 4016 (i.e, the number of days in a cakra) ie., find 4016 C. To this 4016 C add the ahargana A ie, find (4016 C+ A). The Kali ahargana for the GL epoch is 16,87,850. Add this constant to (4016 C + A)i.e, find (4016 C+A + 16,87,850). This gives the Kali ahargana for the required date. From Tables 1.1 to 1.3, for the thus obtained Kali ahargana the corresponding Christian date can be obtained as shown in the follow- ing example. ## Page 24 si2 GRAHALAGHAVAM OF GANESA DAIVAJNA Finding the weekday from the GL ahargana Let C and A respectively be the cakras and the ahargana accord- ing to GL. Multiply C by 5 and to this product add A i.e., find (5C + A). Dividing (5C + A) by 7, let the remainder be R. If R = 0, then the given date falls on a Monday; if R = 1, Tuesday ete. Example : In the example considered above, C = 8, and A = 1521. There- fore, 5C + A = 5 (8) + 1521 = 1561. When 1561 is divided by 7, the re- mainder R = 0. Therefore, the given date is a Monday. Finding the ahargana from the Christian date : In Table 1.1, for the beginning of the Christian century (column 1) in which the given date lies - Kali ahargana (column 3) and the cakras and the (balance) ahargana according to GL are given in columns 4 and 5. For example, consider October 7th of the year 2001 AD. For this year the cen- tury beginning year 2000 (G). From Table 1.1 to 1.3, we have

Ca. ahar. 2000 (G) 18,63,079 43 2541 24,51,545 Year 1 365 0 365 365 October 7 280 0 280 280 18,63,724 43 3,186 24,52,190 Weekday from GL ahargana : Here C = 43 and A = 3186. . BC+ A = 215 + 3186 = 3401 Dividing (5C + A) by 7, remainder R = 6, we get Sunday Note : In Table 1.1, column 2 gives the Julian days (JD). To find the weekday from JD of the given date, divide JD by 7 and let R be the remainder. If R = 0, it is Monday; if R = 1, Tuesday etc. ## Page 25

MADHYAMADHIKARA (CHAP. 1) S13

Table 1.1: JD, Kali and Graha Laghavam Aharganas

Year Julian Days Ahargana Cakras Ahargana -3200 (J) 552258 —-36208 430 2822 -3100 (J) 588783 317 421 3203 -3000 (J) 625308 36842 412 3584 —2900 (J) 661833 73367 403 3965 —2800 (J) 698358 109892 -393 330 2700 (J) 734883 146417 -384 711 2600 (J) 771408 182942 -375 1092 2500 (J) 807933 219467 -366 1473 2400 (J) 844458 255992 -357 1854 -2300 (J) 880983 292517 348 2235 -2200(5) 917508 329042 339 2616 -2100 (J) 954033 365567 -330 2997 -2000 (J) 990558 402092 —321 3378 -1900 (J) 1027083 438617 312 3759 1800 (J) 1063608 475142 302 124 —1700 (J) 1100133 511667 -293 505 —1600 (J) 1136658 548192 284 886 -1500 (J) 1173183 584717 -275 1267 -1400 (J) 1209708 621242 266 1648 -1300 (J) 1246233 657767 -257 2029 —1200 (J) 1282758 694292 ~248 2410 -1100 (J) 1319283 730817 -239 2791 -1000 (J) 1355808 767342 230 3172 -900 (J) 1392333 803867 -221 3553 -800 (J) 1428858 840392 212 3934 -700 (J) 1465383 876917 202 299 -600 (J) 1501908 913442 193 680 500 (J) 1538433 949967 -184 1061 (Contd...) ## Page 26 S14 Chris. Year 400 (J) -300 (J) —200 (J) -100 (J) 0) 100 (J) 200 (J) 300 (J) 400 (J) 500 (J) 600 (J) 700 (J) 800 (J) 900 (J) 1000 (J) 1100 (J) 1200 (J) 1300 (J) 1400 (J) 1500 (J) 1500 (G) 1600 (G) 1700 (G) 1800 (G) 1900 (G) 2000 (G) 2100 (G) 2200 (G)

GRAHALAGHAVAM OF GANESA DAIVAJNA

Julian Days 1574958 1611483 1648008 1684533 1721058 1757583 1794108 1830633 1867158 1903683 1940208 1976733 2013258 2049783 2086308 2122833 2159358 2195883 2232408 2268933 2268923 2305448 2341972 2378496 2415020 2451545 2488069 2524593 Kali Ahargana 986492 1023017 1059542 1096067 1132592 1169117 1205642 1242167 1278692 1315217 1851742 13888267 1424792 1461317 1497842 1534367 1570892 1607417 1643942 1680467 1680457 1716982 1753506 1790030 1826554 1863079 1899603 1936127 Graha Laghavam Ca. Ahargana -175 1442 -166 1823 157 2204 -148 2585 -139 2966 —180 3347 -121 3728 -111 93 -102 474 -93 855 84 1236 75 1617 66 1998 57 2379 48 2760 -39 3141 -30 3522 -21 3903 -11 268 -2 649 -2 639 7 1020 16 1400 25 1780 34 2160 43 2541 52 2921 61 3301 ## Page 27 Days 365 730 1095 1461 1826 2191 2556 2922 3287 3652 4017 4383 4748 5113 5478 5844 6209 6574 6939 7305 7670 8035 8400 8766 9131 9496 9861

sis

Graha Ca.

WNNNNNHHE HHH HERB RR OCD OCOD OC OOO

Ahar. 0 365 730 1095 1461 1826 2191 2556 2922 3287 3652 1 367 732 1097 1462 1828 2193 2558 2923 3289 3654 368 734 1099 1464 1829 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 Days 10227 10592 10957 11322 11688 12053 12418 12783 13149 13514 18879 14244 14610 14975 15340 15705 16071 16436 16801 17166 17532 17897 18262 18627 18993 19358 19723 20088

Ahar. 2195 2560 2925 3290 3656 5 370 735 1101 1466 1831 2196 2562 2927 3292 3657 372 737 1102 1468 1833 2198 2563 2929 3294 3659

(Contd...) ## Page 28

Table 1.2 (Contd.) Year Days Graha Laghava Year Days Graha Laghava Ca. Ahar. Ca. Ahar. 56 20454 5 374 78 28489 d 377 57 20819 5 739 79 28854 iu 742 58 =. 21184 5 1104 80 29220 7 1108 59 =. 21549 5 1469 81 29585 7 1478 60 21915 5 1835 82 29950 7 1838 61 22280 5 2200 83 30315 7 2203 62 22645 5 2565 = 8480681 dU 2569 63 23010 5 2930 85 31046 tf 2934 64 23376 5 8296 86 31411 7 3299 65 23741 5 3661 87 31776 7 3664 66 24106 6 10 88 32142 8 14 67 = 24471 6 875 89 32507 8 379 68 = 24837 6 741 90 32872 8 744 69 25202 6 1106 91 33237 8 1109 70 25567 6 14719233608 8 1475 71 25932 6 1836 93 33968 8 1840 72 26298 6 2202 94 34833 8 2205 73 26663 6 2567 95 34698 8 2570 74 27028 6 2932 96 35064 8 2936 xi) 27393 6 3297 97 35429 8 3301 76 27759 6 3663 98 35794 8 3666 17 28124 7 12 99 36159 9 15 Note : (1) In Table 1.3, the first two columns are headed by C and B which stand respectively for a common (non leap) year and bissextile (leap) year. For a given date in a leap year, only for January and February, the column headed by B must be used. For other months even in a leap year and for all months in a common year the first column under C must be used. (2) In Table 1.1, the letters J and G in brackets represent respectively the Julian and the Gregorian calendars. ## Page 29

Table 1.3 : Ahargana for Days of a Year Dates Jan. Feb. Mar. Apr. May Jun. July Aug. Sep. Oct. Nov.Dec. cB 01 £40 31 - - - - a A = = a I 12 1 32 60 91 121 152 182 218 244 274 305 335 Pye je 74 33 61 92 122 153 183 214 245 275 306 336 Cth Go} 34 62 93 123 154 184 215 246 276 307 337 45 4 35, 63 94 124 155 185 216 247 277 308 338 5 6 5 36 64 95 125 156 186 217 248 278 309 339 6 7 6 37 65 96 126 157 187 218 249 279 310 340 UC ey 38 66 97 127 158 188 219 250 280 311 341 B20) 6) 30) 67) 98) 128) 159) 189) 220) 251. 281. 312) 342) 9 10 9 40 68 99 129 160 190 221 252 282 313 343 10 11 10 41 69 100 130 161 191 222 253 283 314 344 ahh ob} ahh = c}} 70 101 131 162 192 223 254 284 315 345 12 13 12 48 7 102 132 163 193 224 255 285 316 346 138 14 13 44 72 103 133 164 194 225 256 286 317 347 1415 14 45 73 104 134 165 195 226 257 287 318 348 1516 15 46 74 105 1385 166 196 227 258 288 319 349 1G ot 16) 47-7512 106 136 = 167. 197, 228) 1259) 289) 320 2350) 17:18 17 48 76 107 137 168 198 229 260 290 321 351 18 19 18 49 77 108 138 169 199 230 261 291 322 352 19 20 19 50 78 #4109 139 170 200 231 262 292 323 353 20 21 20 51 79 110 140 171 201 282 263 293 324 354 21 22 21 52 80 111 141 172 202 2838 264 294 325 355 22 23 22 53 81 112 142 173 203 234 265 295 326 356 23 24 23 54 82 118 143 174 204 235 266 296 327 357 24 25 24 55 838 114 144 175 205 286 267 297 328 358 25 26 25 56 84 115 145 176 206 237 268 298 329 359 26 27 26 57 85 116 146 177 207 288 269 299 330 360 27 28 27 58 86 117 147 178 208 239 270 300 331 361 28 29 28 59 87 118 148 179 209 240 271 301 332 362 29:30 29 - 88 119 149 180 210 241 272 302 333 363 30312 308 89 i120 150 181 211 242 273 303 334 364 31 - 31 - 90 - 151 - 212 243 - 304 - 365 ## Page 30

Sloksa 6,7,8 : The dhruvas and ksepakas of all planets Ksepaka is the mean position of a heavenly body at the time of the epoch and dhruvaka is the multiplier of the completed cakras for a given date. In fact, dhruvaka is the residual motion of a body in a cakra after removing the completed revolutions. The dhruvakas and ksepakas of the heavenly bodies are tabulated below.

Table 1.4 Dhruvakas of bodies

rocca Kendra Kendra |Rasi 0 0 9 7 1 4 0 1 7 Améa (*) 1 3 2 2 25 3 26 14 15 Kala(‘) 49 46 45 50 32 27 18 2 42 Vikala(") 11 il 0 0 0 0O 0 0 0

Ravi Candra Cand- Rahu Kuja Budha Guru Sukra Sani rocca Kendra Kendra Rasi 11 11 5 0 10 8 7 7 9 Améa (°)19 19 17 27 7 29 2 20 15 Kala(‘) 41 6 33 38 8 33 16 9 21 Sloka 9 : The use of dhruvaka and ksepaka are explained : From the motion of a body obtained from the ahargana subtract the product of the dhruvaka and the cakra and to it add the ksepaka. This gives ## Page 31

the mean position of the body for the mean sunrise (at Lanka and Ujjayini). In the case of the moon, the distance (in yojanas) between one’s place and the central meridian (rekha), chosen as the meridian passing through Lanka and Ujjayini, is divided by 6 to get the correction in kalas (minutes of arc). This is added to or subtracted from the earlier obtained position of the moon according as one’s place is to the west or to the east of the central meridian (rekha). Explanation : The above correction described for the moon is referred to as desantara samskara due to difference in the sunrise timings at the given place and at Lanka. The desantara correction for the moon =__—_The distance in yojanas __ nagon’s daily motion Circumference of earth in yojanas Consider the mean daily motion = 790°35” and paridhi (cireumfer- ence) of the earth = 4967 yojanas. We get the Desantara correction for the moon = 2035" (Distance in Yojanas ) = oe x(distance in yojanas ) 6.2827 x(distance in yojanas ) al= where the distance in yojanas is the shortest distance of the given place from the meridian passing through Lanka and Ujjayini. The modern known values of the earth’s equatorial and polar radii are respectively 6378.16 km and 6356.775 km. Considering the mean radius of the earth in miles and the circumference of earth given as 4967 yojanas in the GL commen- tary, we get 1 yojana ~ 5 miles. The circumference of the earth as 4967 yojanas is taken by Bhaskara II in his Siddhanta Siromani. ## Page 32

Sloka 10 : The method of calculating the mean longitudes of Ravi, Budha and Sukra has been given in this sloka in a systematic manner. It is as follows : (i) Divide ahargana A by 70 and then subtract the quotient from A to get the resultant in degrees. (ii) Divide A by 150. (The result will be in kalas). Divide the result by 60 in order to convert it into degrees. (iii) Subtract the result of step (ii) from that of step (i). (iv) Multiply cakra = C by dhruvaka D and subtract the product from step (iii) (v) Add ksepaka K to step (iv). The result gives the mean planet in degrees. ie. i =4/4-4-—4*_|-cxp\+Kx Mean longitude of the planet = { 7 a. } Note : In traditional Indian astronomy the mean longitudes of Budha and Sukra are taken the same as that of the sun. Example : For the given date, samvat 2036, Saka 1901

A = Ahargana = 3328, C = Cakra = 41 D = Dhruvaka=0"1°4911"— K = Ksepaka = 11" 19° 41’ =349°41" A A ‘ _ {4-2 -—4— |- Mean longitude of the sun = {( 70 150 =) cxp}ex ## Page 33

= 3328 — 3328 __3328 _ — 41x 0"1°49'11"b + 11419 041" 70 150x 60 = 3555°.1621 = 10" 15° 9° 43".6 [removing completed revolutions]. Note : Step (iii) gives aharganotpannagraha (i.e., mean longitude derived from the ahargana).

Given date : 11th August 1998. For the given date, A = 2033, C= 43. For the sun, D= 0" 1° 49’ 11" and K=349°41" | i =|[4-4-—*— |-exp|+x Moan longitude ofthe sn = || a we) . | 2033 2033 = || 2033 -——— - ===} 43 x 194911" |+349°41' «975° -[( 70 ae) x |. = 2275".1665 = 115° 9° 59°.5 [removing completed revolutions]. Another method : Instead of the method explained in sloka 10, we can as well adopt the following method in which the ahargana is multiplied by the mean daily motion. In the example considered earlier, A = 3328, C = 41, D = 1°49°11’, K = 11" 19° 41’ and the mean daily motion of the sun, m = 59’08"10". (Note : Here, 1"= 60” ie, one vikala = 60 prativikalas]. Therefore, Ahargana generated mean longitude = Ax m = 3328 x (59°08°10" ) = 3328 x 0°.9856018 = 3280°.083 = 40°04’58" = 1* 10° 04’ 58°(removing the multiples of revolutions). ## Page 34 s22 GRAHALAGHAVAM OF GANESA DAIVAJNA Now, Cx D=41° x 1°49'11" = 74°3631" = 2" 14° 36°31" (Ax m)-(Cx D) = 17 10°0458" -2" 14°36°31" = 10” 25°28°27" Adding the Kshepaka, K = 11" 19° 41’, we get 10* 25°28°27" + 11719°41° = 10°15°09°27" This is the mean position of the Sun, Budha and Sukra.

The mean position of the Moon can be determined as follows: (i) Multiply ahargana A by 14 and divide the product by 17. (ii) Subtract step (i) from 14 x A. (iii) Divide A by 140 x 60. Subtract the quotient from step (ii); the differ- ence gives the ahargana-derived Moon. (iv) Subtract the product of dhruvaka and cakra (C x D) from step (iii). (v) Add ksepaka K to step (iv) which gives the mean position of the Moon. i.e., Mean longitude of the Moon = axis 418) A_\ eyplax 17) 140 x60 Example : A = 3328, C=41, D= 3°46°11", K=11°19°6" Axl4 A i =4/ 4x14 - 5 Mean longitude of the Moon {( x if } Tone }-cxpex ## Page 35 MADHYAMADHIKARA (CHAP. 1) 823 = (2028 x14 SAA) BELLI — 41x 3°46'11" +11°19°6'0" 17 140 x 60 = 44045°.439 = 4"5°26°20° (removing completed revolutions). Modern Example : Given date : 11th August 1998 For the given date, A = 2033, C = 43 For the Moon : D = 3°46’11", K=11*19°6" ane) A ‘ =||4xl4—- - Mean longitude of the Moon = (( i7iae sraorco |-cxpex = 2033 x14 — 2033 x14 peas — 43x 3°46'11" +11°19°6" 17 140 x 60 = 26974°.525 = 117 4°31’= 334°31" removing completed revolutions. Sloka 11 : Mean positions of Candrocca (Moon's apogee) and Rahu. The method is explained as below : (i) Divide ahargana A by 9. (ii) Again divide ahargana A by 70 and it is in kalas. Divide the result by 60 to get it in degrees. (iii) Adding step (i) and step (ii), we get ahargana derived mandocca. ## Page 36

(iv) Subtract the product of cakra and dhruvaka (C x D) from step (iii) (v) Add ksepaka K to step (iv).

A A . ie, | Mean position of Mandocca = iG 0s =) aoe | a Example : A= 1521, C=8, D=9"2°45’, K=5"17°33" sti -|(4+-4_] -exp Mean position of Mandocca = || 9 * 79x60 +K =| (21, 1521) gs oraegse | 5%17°33" 9 "70x60 = 314°54°43" = 10°14°5 4°43" Modern Example : Given date : 11th August 1998 For the given date, A = 2033, C= 43

A _ 5 :. Mean position of Mandocca = lé + xa, -CxD\+K (7 2033 =|——+ y Rooas! . pRym99) 7 a) 43 x 9 oes’ + 51 P33 ## Page 37

= -11334°.327 = 185°.67294 (adding multiples of 360°) = 185°40°22" = 6 5° 40’ 29". Mean position of Moon’s Pata (Rahu) The following is the method to determine the mean position of Pata (Rahu). (i) Divide ahargana A by 19 (the result in degrees) (ii) Divide ahargana A by 45. Divide the result in kalas by 60 to get it in degrees (iii) Add step (i) with step (ii) (iv) Divide step (iii) by 30 to get the result in Rasis and subtract the quo- tient from 12. This gives cakra Suddha Rahu. (v) Subtract the product of D and C from step (iv). (vi) Add K to step (v). Then we will get A A s+

={360°-(44—4 |) -cxd+k 19 "45 x60 Example : A= 1521, C=8, D= 72°50", K= 27°38" F o(4,—4_\W_ Mean longitude of Rahu {a0 (4-xa}} CxD+K ## Page 38

= 5360°— Cae 2 —7°2°50'x 8 + 27°38" 19 45x60 = 44°.350702 = 44°21°2".5 = 1" 14°21°2.5". Modern example : Given date : 11th August 1998 For the given date, A = 2038, C= 43 For Rahu : D= 7" 2°50’, K= 27°38" ote A . Mean longitude of Rahu = {os = (4 + xa} CxD+K _ 360° (203 2033 } = 43 x 7"2°50" + 27°38" 19” 45x60 = —8871°.953 = 128°.04704 (adding multiples of 360°) = 128°2°49" = 4R g° 9" 49"

(i) Multiply ahargana A by 10 (ii) Divide step (i) by 19 (the result is in degrees) (iii) Divide step (i) by 73 (the result is in kalas) (iv) Subtract step (iii) from step (ii), the result of which will be the ahargana derived mean Kuja. (v) Take the product of cakra and dhruvaka and subtract the product. (Cx D) from step (iv). ## Page 39 MADHYAMADHIKARA (CHAP. 1) $27 (vi) Add ksepaka (K) to step (v). Then 10x A°_ 10x A" 19 73 Example : Consider the date May 15, 1612 (G), Monday. A= 1521, C= 8, D= 1" 25°32’, K=10"7°8" :. Mean longitude of Kuja 10 x 1521 °10x1521' 19 73 |- 8x 1"25°32'+10"7°8" = 800°.52632 — 208”.35616 — 8 x 55°.53333 + 307°.13333 = 659°.92038 = 9"29°55°13" [removing the completed revolutions] Modern Example: Given date : 11th August 1998 For the given date : A = 2033, C= 43 For Kuja : D = 1"25°32’, K= 10"7°8° 104° 10 9 7B -. Mean longitude of Kuja = [ |-e xD+K 10 x 2033 *_Wwa2th) _ 43 x 1"25°32'+10*7°8" 19 B =~ 1015°.4416 = 64°.558448 = 64°33°80° = 2% 4° 337 30" (by adding multiples of 360°) ## Page 40

The method of finding sighrakendra of Budha is as follows : Sighra normaly of Budha (i) Multiply ahargana by 3. (ii) Divide 3A by 28. (iii) Add step (i) and step (ii); the result will be in degrees. A (iv) Subtract 38 kalas (minutes of arc) from step (iii) the result of which will be the ahargana-derived mean Budha kendra. (v) Subtract the product of cakra and dhruvaka from step (iv) (vi) Add ksepaka (K) to (v). : 3A yay i.e, Sighrakendra of Budha = es +34) (4) | Ghia 3A 4 )] . [co +4)-(ata ~cxD+K Example : A= 1521, C=8, D= 4" 3°27' = 123° 27' and K = 88 29° 33’ = 269° 33' i.e, Mean Sighrakendra of Budha - (s+24) -[ 7 ) -CxD+K =1\ 28 38 x60 ## Page 41

38 x60 ee +3 x1s21} -( fe] ) —8 x 123° 27'+269° 33’ = 4007°- 2472 = 47° 14° 49"=18 17° 14’ 49" [removing the completed revolutions].

Given date : 11th August 1998 For the given date A = 2033, C= 43 We have for the Budha kendra: D= 4" 3° 27', K=8" 29° 33' 3434) - 2 ) -CxD+K 28 38 x60 3x 2083, 5. 9039 -(2083. | _ 43x 4B 3 a7 + 8F og ay 28 38x 60

1277 °-12976 = 197°-12976 = 197° 7' 47" (Removing the multiples of 360 °) Note : Usually the practice, in other Siddhantic texts, is first to find the Sighrocca of Budha and then to subtract the mean Budha (same as the mean Ravi) from it to get the Sighrakendra of Budha. Similar is the practice in the case of Sukra’s Sighrakendra. However, Grahalaghavam gives the method of determining the Sighrakendras of Budha and Sukra directly without finding the sighroccas. ## Page 42

Sloka 13 : Mean positions of Guru and Sukra kendra are explained. (i Mean position of Guru :

(i) Divide ahargana A by 12. The result is in degrees. (ii) Divide ahargana A by 70. This will be in kalas (minutes).

(iv) Subtract the product of cakra(C) and dhruvaka (D) from (iii). (v) Add ksepaka (K) to (iv). AY (A i.e., Mean longitude of Guru = (a) - (4) —CxD+K -|(4)- ee) | cD OK *1\12, 70 x60 Example: A=1521, C=8, D= 26° 18° and K=7% 9° 16' = 212° 16 -CxD+K AY A_y -. Mean longitude of Guru = (4) -s <=) = 1a) e(eis2t —8 x 26° 18'+ 212° 16’ “ 12 70 x60 ## Page 43

= 128 °-25452 = 128° 15'16"= 4" 8° 15’ 16" Modern example : Given date : 11th August 1998

and for Guru D= 0% 26° 18’ , K= 7% 2° 16° Ay AY Now, Mean longitude of Guru = (é) (aa) renee _ (2) + 2033 ) 43 x 26° 18'47% 2° 16! =) 12 70 x60 749° -70071 = 330 °-29929 (adding 1080°) = 330°17'57"= 11R 0° 17' 57" Mean sighrakendra of Sukra : The method to find out Sukra kendra is as follows : (i) Multiply ahargana A by 3. (ii) Divide step (i) by 5. (iii) Divide step (i) by 181. (iv) Add step (ii) and step (iii). The result is in degrees. (v) Subtract the product of cakra (C) and dhruvaka (D) from step (iv) (vi) Add ksepaka K to step (v) ## Page 44

i.e., Mean position of Sukra Kendra = Example : A = 1521, C=8, D=1"14°2' = 44°2' ,

p 34 3A] :, Mean position of Sukra Kendra = [4 + 4) -CxD+K oe 3x 1521 7 + = —8 x 44° 2' + 230°6" 5 181 = 815° 4135” = 95° 41’ 36" = 3% 5° 41’ 35” [removing the completed revolutions]. Modern example : Given date : 11th August 1998 For the given date, A = 2033, C= 43 For Sukra Kendra, D=1" 14° 2’, K=7* 20° 9° (4-4) -CxD+K

2 2033 | 3 «2033 ~ 43x1F 14° 2'4 7% 20°9° 5 181 " —409° 47'14" = 310°12' 46" (adding 720°) ## Page 45

Sloka 14 : Mean longitude of Sani. The method is as explianed below : (i) Divide ahargana A by 30. The result will be in degrees. (ii) Divide ahargana A by 156. The result will be in minutes of arc (kalas). Convert it into degrees by dividing by 60. (iii) Add step (i) and step (ii). (iv) Subtract the product of cakra(C) and dhruvaka (D) from step (iii). (v) Add ksepaka K to (iv). . AY (A) i.e., Mean longitude of Sani = (a) (4) OES Example : A= 1521, C=8, D= 7" 15° 49’ = 225°42' and Ksepaka K= 9" 15° 21’ = 285° 21’ P AY A Yl :. Mean longitude of Sani = (é) (a) -CxD+K 7 os + ee} — 8 x 225°42'+285° 21’ “ 30 156 x60 ## Page 46

= -1469 °-3875 = 330°-6125 (adding 5 x360°)= 11% 0° 36' 45" Modern example : Given date : 11th August 1998 For the given date, A = 2033, C= 43 For Sani, D= 7% 15° 42’ andK= 9" 15° 21° E AY . «. Mean longitude of Sani = (é) (Ss) | -~CxD+K 5 (=) + ans ) —43x7715°42'+9"15°21' ~}K 30 156x 60. =- 9351°.7661324 = 8.2338676 ( adding 26x 360") = 8°14' 2” Sloka 15 : Mean daily motions of planets Mean daily motions of planets (madhyama gati) are given in Table 1.6

Bodies Sun Moon Candrocca Rahu Kuja Budha Guru Sukra Sani Kendra Kendra ee 790 6 e 311865 (minutes) hel AL li 26 24 0 0 0 (seconds) ## Page 47

Sloka 16 : The values of the Sun and the Moon’s mandocca (Candrocca) obtained from this text are equivalent to those according to the Surya Siddhanta (SS). By subtracting 9’ from the Moon (as per this text) we get as per SS. The positions Kuja, Guru, Rahu are in accordance with the Aryapaksa. Budha kendra (Budha’s Sighra anomaly) is in accordance with the Brahma paksa. By adding 5° to Sani (obtained from this text), we get that according to the Arya Siddhanta. The Sukra (sighra) kendra (of this text) is half of the sum of those obtained from the Arya and Brahma paksas (i.e., the average of the latter two).

The expressions for the mean longitudes of the Sun, the Moon, the Moon’s apogee and node, given in Table 1.6 are now derived using the revolutions of these bodies and the numbers of civil days in a Mahayuga of 432 x104 years. (i) Mean longitude of the Sun The mean longitude of the Sun is given by az(a-4-_4 ) ~atra x1 81972+349°.683 70 150%60 where A is the ahargana according to GL. Mean motion of the sun in a Cakra 1 Cakra = 4016 days Number of civil days in a Mahayuga = 1577917828 according to Surya Siddhanta. Number of revolutions of the Sun in a Mahayuga = 4320000 4320000 rev. ; . . O18 17, .. Mean daily motion of the Sun 1577917828 ## Page 48 $36 GRAHALAGHAVAM OF GANESA DAIVAJNA Therefore, in a Cakra Sun’s motion = 4016 x 59'8"-17 = 3958 °-180756 = 10-99494654 revolutions Subtracting 11", we get dhruvaka = - 1° -8192444 = ~1°49'9" (approx) The ahargana derived mean motion of the Sun is given by 4 x59'8"-17 Multiplying and dividing by 70, we get 10xAx59'8"-17 _ 4139'32"x A 70 7 70 _ 68°59'32"xA - 70 By adding and subtracting 28” , the mean motion of the Sun _ Ax (68° 59’ 32" + 28"- 28"), , { 69° -28" 7 70 = 70 _ Ax69° _ Ax28" _ Ax(69°+1°-1°) _ Ax28" "70 70 70 70 ## Page 49

A Ax70 A Ax28" A _ Foxx Ax A HAA _ 70x60%60 9, 7 70 70 70 oa ——— A A =A 70 7 Tso x60 > degrees. (ii) Mean longitude of the Moon The mean longitude of the Moon is given by A=Ax14—-14 xf -—4___Cakra x3°-76972 +3491 in deg. 17 140 x60 where A is the ahargana according to GL. Mean daily motion of the Moon = 790'34"-9 = 13° 10’ 34".9 Mean motion of the Moon in a Cakra = 4016 x13°-176361 = 52916 °-26622 = 146". 9896284 Subtracting 147 rev., we get Dhruvaka = — 3° .7337785 (taken as 3°-76972 by GL). =-3°44'01 6 The mean longitude of the Moon = 4 x13°10'34"-9 Multiplying and dividing by 17 we get Ax17x13°10'34"-9 _ 4x223°59'53" 17 - 17 ## Page 50

ee 17 - 17 “17 17 _ AxQ24° 4149-14") x7" Ax238° x14? xT” es y Ty 17 i7 2 Axi14? AxT" ——LUrr——r— . A = Axi4e _ AXI4" _ 1020 %60 7 Axia? A = Axl4® — > ~ 14571429 x60 le Note : In the last term the text has made an approximation viz. 779 gq, (iii) Mean longitude of the Moon’s apogee (mandocca) The mean longitude of the Moon’s apogee is given by u-44-4 9 + aycgp 7272-75 x Cakra +167.55 in degrees where A is the ahargana according to GL. Number of civil days in a Mahayuga = 1577917828 ‘Number of revolutions of the Moon’s apogee = 488203 according to the Sirya Siddhanta 488203 revns. Mean daily motion of the Moon’s apogee = 71577917828 = 6'40".98 ## Page 51

Mean motion of the Moon’s apogee in a cakra 1 cakra = 4016 days Mean daily motion of Moon’s apogee = 6'40”.98 . Therefore, in 1 cakra, the mean motion of the Moon’s apogee = 4016 x 6'40".98 = 447°.3154667 = 1.242542963 revns. Subtracting 2 revolutions we get Dhruvaka = — 272 °.6845333 (taken as 272°.75 by GL) =~ 272 °41'04".32 The mean longitude of the Moon’s apogee is given by A x6’40".98

Ax6'40".98x9 _ Ax1°0'8" _ A® + Ax8 _ A® + A’ 9 9 9 9x60 9 67.5 In the last term, GL takes 70 instead of 67.5 in the denominator. This re- sults in a maximum error of 2’. (iv) Mean longitude of Rahu

A, A

## Page 52

where A is the ahargana A according to GL

310.8

= 212°.848 = 212°50' = Dhruvaka which is the same as given in GL The mean motion of the Moon’s node (Rahu) is given by 4 x 3'10’.8 for the ahargana A.

Ax3'108 x19 Ax 1°0'25" A Ax2ST Ae AXDS! 19 =. 19 * «19 19 * 19 x60 a = 19 * 45.6 x 60 GL has taken 45 instead of 45.6 (v) Mean longitude of Kuja 10 ,. 104, Mean Kuja = Te a min-(C x D) deg + Kdeg where A is the ahargana according to GL, Number of revolutions of Kuja in a Mahayuga = 2296824 ## Page 53

Number of civil days in a Mahayuga = 1577917828 2296824 revns. 1577917828 -, Mean daily motion of Kuja = Mean motion of Kuja in a cakra = 4016 x 31'26".4 = 304°23'2" = — 55°36'57" = Dhruvaka (by subtracting 360°) Multiplying and dividing the ahargana derived motion by 19 we have Ax31'26"4x19 _ A x 597" 19 ~ 19 Ax (597' 21" + 2' 39" - 2' 39") 4 x600' _ Ax2'39" 19 19 19 ee Ax10 Ax600! — Ax2'39"x10 4x10" “Fo9 “19 ~ 19x10 9 Brg 9° _ A°x10 A'x10 Axl0° A'x10

(vi) Mean sighra kendra of Budha Mean Sighra kendra of Budha is given by a min—(C x D) deg. + Kdeg. 3A°+ 28 38 ## Page 54

where A is the ahargana according to GL.

Number of revolutions of Budha’s sighrocca = 17937020 Number of civil days = 1577917828 17937020 revns. 9 i i ’s $i oe = 4°52". ..Mean daily motion of Budha’s sighrocca 1577917828 Sighra anomaly = Budha’s sighrocca ~ Mean Budha Daily motion of sighra anomaly = Daily motion of Budha sighrocca — Daily motion of mean Ravi = 4°5'32".31 -59'8".17 = 3°6'24".14 Multiplying and dividing the ahargana derived motion by 28 we have Ax3°6'24".14x28 _ Ax 86 59/1592 _ A° x 87° A x 44".08 28 ~ 28 "28 28 5 A’ a(S) - Aa eoe a 3A° 28x60 28 28 io ary = UW) 2 8A A’ = 3A°* OS ~ Bali The denominator of the last term has been taken as 38 in GL. Note : In other texts Sighra kendra is obtained by calculating Budhe’s sighrocca and mean longitude separately and then subtracting the latter from the former. But the Graha Laghavam directly gives the formula to calculate the sighra kendra for Budha and Sukra (the inferior planets). ## Page 55 MADHYAMADHIKARA (CHAP. 1) 843 (vii) Mean longitude of Guru The mean longitude of Guru is given by “ae A’ £4 deg+ Kd ioaazo (C x D) deg leg where Ais the ahargana, Cis the cakra and D dhruvaka and Kthe ksepaka respectively. According to GL, here K= 212°16’ and D= 26° 18’. We have Number of civil days ina Mahayuga = 1577917828 Number of revolutions of Guruina Mahayuga = 364220 : . S64220 revi Mean daily motion of Guru = Ter7oi7qpq = 4'59"-1468 Mean motion of Guruinacakra = 4016 x 4’59" _ 4016 x 364220 ae 4 7577917828 = 333° 42' = 333° 42' — 360 = — 26°18’ sothat Dhruvaka D= 26°18’ Mean motion of Guru for ahargana A= 4'59".1468

A x 4'59".1468 x 12 A x 59’ 49".762 _ (Pag) A 12 _ 12 _ 12 _ A’ 10'238xA _ A’ 10.288KA AY “127 12 «2 12x60 ~ 12 70.32 Note : The GL has taken 70 in the denominator of last term. ## Page 56 saa GRAHALAGHAVAM OF GANESA DAIVAJNA (viii) Mean Sighra kendra of Sukra The mean longitude of Sukra’s sighrakendra is given by 3A° 3A° — . + K deg. Sot Fey TOC xD) deg. + Kdeg where A is the ahargana, C the cakra and D dhruvaka and K the ksepaka according to GL. We have in Mahayuga Number of revolutions of Sukra’s Sighrocca = 7022388 Number of civil days in a Mahayuga = 1577917828 : Mean daily motion of Sukra’s sighrocca = _7022388_ 3¢10 1577917828 1° 36° 7.74 Sighra anomaly = Sukra’s sighrocca— Mean Sukra. Here the mean Sukra is the same as the mean Sun. :. Daily motion of the sighra anomaly = Daily motion of Sukra’s sighrocca — Daily motion of mean Sun = 1° 36 7".74 - 50’ 8.17 = 36' 59°.57 The mean motion for ahargana A= A x 36' 59". 57 By multiplying and dividing the ahargana by 5 we get. Ax 36'59".57x5 = Ax184'57"2" — 3°4'57"2"x A 5 * 5 * 5 ## Page 57 MADHYAMADHIKARA (CHAP. 1) sas

+ = + 5 5 5 5 _3At AGHIST"Y 3a" 4 Ax 34" 34" =o 5x60 “5 5x60x3 °° 5 181 4|57|2 as taken in the Grahalaghavam. (viii) Mean longitude of Sani 4 A A je 44 . Mean Sani = 37 + 75-(C x D) deg. + Kdeg where A is the Ahargana, Cthe cakra and D dhruvaka and K the ksepaka according to GL. Number of civil days in a Mahayuga = 1577917828 Number of revolutions of Saniina Mahayuga = 146564 _ 146564x360° Mean daily motion of Sani = lean daily motion of Sani caqsoiTDR = 2'0".38 ‘The mean motion for ahargana A= 4 x2'0".38 Multiplying and dividing the ahargana derived motion by 30 we get Ax2'0".38 x30 60'11.6"%A — A’xI | I'.6xd 30 2-20, 30). 30x00 ## Page 58

A The second term is considered as 156 by GL 4 Age rag ie., Mean Sani = —— +—— 30 156 ## Page 59 CHAPTER 2 Ravi CanDRA SPHUTADHIKARA (True Positions of the Sun and the Moon) 2.1. Introduction In obtaining the mean positions of the Sun and the Moon, it was assumed that these bodies move in circular orbits round the earth with uniform angular velocities. However, by observations it was found that the motions are non-uniform. The procedure for calculating the major corrections to the mean positions in order to obtain the true positions, is related to the epicyclic theory which is explained in the following section. 2.2. Epicyclic theory and Mandaphala The theory is that while the mean Sun or the Moon moves along a big circular orbit (dotted in Fig.2.1), the actual (or true) Sun or Moon Fig. 2.1 ? Epicyclic Theory ## Page 60

moves along another smaller circle, called epieycle, whose centre is on the bigger circle. The bigger circle ABP with the earth E as its centre is called the kaksavrtta. Let A be the position of the mean Sun when the true Sun is farthest from the earth. The line AEP is called the apse line (or nicoccarekha) and AE is the trijya (radius) of this orbit. The epicycle, with A as centre and a prescribed radius (smaller than AE) is called the nicoccavrtta. Let the apse line PEA cut the epicycle at U and N. The two points U and N are respectively called the mandocca (apogee)and the mandanica of the Sun. Note that as the Sun moves (as seen from the earth) along the epicycle, he is farthest from the earth when he is at U and nearest when at N. The epicyclic theory assumes that as the centre of the epicycle (i.e. mean Sun) moves along the circle ABP in the direction of the signs (from west to east) with the velocity of the mean Sun, the true Sun himself moves along the epicycle with the same velocity but in the opposite direc- tion (from east to west). Further, the time taken by the Sun to complete one revolution along the epicycle is the same as that taken by the mean Sun (i.e., centre of the epicycle), to complete a revolution along the orbit. Now, in Fig 2.1, suppose the mean Sun moves from A tod’. Let 4E be joined cutting the epicycle at U’and N’ which are the current positions of the apogee (mandocca) and the mandanica. While the mean Sun is atA’, suppose the true Sun is at S on the epicycle so that U's =U'EA. Join ES cutting the orbit (i.e., circle ABP) at S’. Then 4'is the madhya (mean Sun) and S’ is spasta (or sphuta) Ravi. The differ- ence between the two positions viz, A'ES’ (or arc 4’5') is called the equa- tion of centre (mandaphala). ## Page 61 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) s49 Now, in order to obtain the true position of the Sun, it is necesary to get an expression for the equation of centre which will have to be ap- plied to the mean position. In Fig. 2.1, SC and A'D are drawn perpendicular to U'N’E and UNE respectively. The arc 44‘ (or AEA’), the angle between the mean Sun and the apogee is called the mean anomaly (mandakendra) of the Sun. We have, in the right-angled triangle 4'pE , sin AEA'=sin DEA'=A'D/A'E so that A'D=R sin AA' = R sin m (where R = A'E and m= arc AA’) is called mandakendrajya. From the similar right-angled triangles §C_4' and A'DE, we have SC /SA'=A'D/A'E so that SC =A'DxSA'/A'E Since s4' and 4’g are respectively the radii of the epicycle and the orbit, these are proportional to the circumferences of the two circles; that is, SA'/A'E =circumference of the epicycle / circumference of the orbit -, SC = (circumference of epicycle / circumference of orbit) x 4'D ## Page 62

Taking the circumference of the orbit as 360°, we have SC = (circumference of the epicycle) x (mandakendrajya) / 360°. Now, takingsc approximately the same as 4's‘, we have

= (circumference of the epicycle) x (mandakendrajya) / 360° = (r/R) (Rsin m) where R sin (m) the “Indian sine” of the anomaly m of the Sun. The maximum value of the equation of centre is r, the radius of the epicyle. By observation this can be obtained as the maximum deviation of the Sun’s position from the calculated mean position. Note that when the Sun is at his apogee or perigee, the mean and true positions coincide since sin (m) is 0 when m =0° or 180°. The maximum equation of centre for the sun was observed by Bhaskara II to be 2° 11'30"(i.e.131'.5) which is the value of r. Therefore, circumference of the epicycle of the Sun = (131.5 /3438 ) x 360° = 13°.66

Note : The same epicycle theory is applied to the Moon also. In the case of the Moon, Bhaskara II has given the maximum equation of centre as 302’. Most texts have taken the epicycles as of varying radii and not fixed. ## Page 63

RAVI CANDRA SPHUTADHIKARA (CHAP. 2) S51

Bodies Aryabhatiyam Khanda Saura siddhanta _ Sarya Siddhanta Khadyaka __(Varahamihira) Ravi 13°30" 14° 14° 13°40' to 14° Candra 31°30’ 31° 31° 31°40’ to 32° Kuja 30m 811 0a = 702 70° 72° to 75° (Budhal | 72285903152 280 28° 28° to 30° Guru ah mci = cyl 32° 32° to 33° Sukra —9.0° 0 18.0" 14° 14° 11° to 12° Sani -40°.5 to 585° 60° 60° 48° 10 49° From Table 2.1 we notice that the Khanda Khadyaka of Brahmagupta and the Saurasiddhanta (as given by Varahamihira) take the epicycles as of constant periphery (and hence radius). The Aryabhatiyam except for the Sun and the Moon and the later Sarya Siddhanta take them as varying between two limits. 2.3 Bhujantara correction The true midnight of a place differs from the mean midnight by an amount of time called “equation of time”. The equation of time is caused by (i) the eccentricity of the earth’s orbit; and (ii) the obiliquity of the ecliptic with the celestial equator. The correction to the longitude of a planet due to the part of the equation of time caused by the eccentricity of the earth’s orbit is called bhujantara. The other correction caused by the obliquity of the ecliptic is called udayantara. ## Page 64

The true midnight of a place differs from the mean midnight by an amount of time called “equation of time”. The equation of time is caused by (i) the eccentricity of the earth’s orbit; and (ii) the obiliquity of the ecliptic with the celestial equator. The correction to the longitude of a planet due to the part of the equation of time caused by the eccentricity of the earth’s orbit is called bhujantara. The other correction caused by the obliquity of the ecliptic is called udayantara. ## Page 64

While all the siddhantic texts have considered the bhujantara cor- rection the other correction-udayantara-was first introduced by Sripati (about 1025 AD) and later followed by Bhaskara-II and others. We shall discuss the bhujantara correction which is mentioned in the Surya Siddhanta. The effect of eccentricity of the earth’s orbit in the equation of the centre (mandaphala) of the sun is converted into time at the rate of 15° per hour or 6° per ghatika. This rate of conversion is due to the fact that the earth rotates about its axis at the rate of 360° in 24 hours (or 60 ghatikas.). The resulting amount in time unit is the equation of time caused by the eccentricity of the earth’s orbit. Thus, the equation of time (due to the eccentricity) = [(Equation of cente of the Sun)/ 15] hours = [(Equation of centre of the Sun)/6] ghatikas Now, to get the bhujantara correction for the Sun or the Moon or any other planet, the equation of time obtained above must be multiplied by the motion of the planet per hour or per ghatika as the case may be. That is, Bhujantara correction for a planet = [(Equation of time in hours] x (Daily motion)/24] = [(Equation of centre of the Sun)/15] x [Daily motion)/24] = [(Equation of centre of the Sun)] x (Daily motion)/360 where the factors in the numerator are in degrees and the daily motion is that of the planet. If the time unit used is ghatika, then Bhujantara correction = (Eqn. of time in ghatika) (Daily motion)/60] = (Eqn. of centre of the Sun in degrees)/6] x (Daily motion of the planet/60] = [Eqn. of centre of the Sun in degrees] x [Daily motion of the planet/360] ## Page 65 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) 853 where the daily motion of the planet is in degrees and hence the Bhujantara correction is also in degrees. However, if the daily motion of the planet is in minutes of arc, then

= (Eqn. of centre of the Sun in degrees) x (Daily motion of the planet)/21600 Further, the bhujantara correction is additive or subtractive ac- cording as the equation of centre of the Sun is so. For example, in the case of Moon, its mean daily motion is 13°.176352 or 790'.58112 - Therefore, we have (mean) bhujantara correction = (Eqn. of centre of the Sun) x 790'.58112/21600' = Eqn. of centre of the Sun/27.321674 Note : Brahamagupta takes the denominator approximately as 27 in his Khandakhadyaka. It is important to note that to obtain the actual (and not the mean) bhujantara correction of a planet, we have to use the true daily motion of the planet for the given day. Example : Find the bhujantara correction for the longitudes of the Sun and the Moon given that on a certain day True daily motion of the Sun a 59'.65 True daily motion of the Moon: 855'.23 Equation of centre of the Sun H +2°7'32" = 127'.53 Therefore, we have Gi) True bhujantara correction of the Sun = (Eqn. of centre of the Sun) x (Daily motion of the Sun)/21600 ## Page 66

= 127'.53x59'.65/21600' = 0',3521835 = 0'21" Since the equation of centre of the Sun is additive, the bhujantara correction is also additive. (ii) True bhujantara correction of the Moon = (Eqn. of centre of the Sun) x (Daily motion of the Moon)/21600 = 127'.53x855'.23/21600' = 5',0494205 = 5’3” Here also, the correction is additive since the Sun’s equation of centre is so. 2.4 Further corrections for the Moon We have applied so far an important correction namely, the equa- tion of centre (mandaphala), to the mean position of the Moon. Besides this correction, the other two corrections applied viz, desantara and bhujantara are mainly to get the true position of the moon at the true local midnight at the place of observation. However, to get the true position of the Moon at least two more important corrections will have to be applied, of course, ignoring other minor corrections due to planetary perturbations. These are : (i) Evection = (15/4) mesin (2é - 9) = 76' 26" sin (2€ - 9) where m is the ratio of mean daily motions of the Sun and the Moon, e is the eccentricity of the Moon’s orbit, § =(M - S), the elongation of the Moon from the Sun and $= M -P , the mean anomaly of the moon (P being the Moon’s perigee). (ii) Variation = 39/30" sin (2&) In the above formulae, S and M are respectively the mean longi- tudes of the Sun and the Moon. The Surya Siddhanta, being an earlier text, does not mention these corrections. However, Mafijula (932 A.D.), ## Page 67

We have applied so far an important correction namely, the equa- tion of centre (mandaphala), to the mean position of the Moon. Besides this correction, the other two corrections applied viz, desantara and bhujantara are mainly to get the true position of the moon at the true local midnight at the place of observation. However, to get the true position of the Moon at least two more important corrections will have to be applied, of course, ignoring other minor corrections due to planetary perturbations. These are : (i) Evection = (15/4) mesin (2é - 9) = 76' 26" sin (2€ - 9) where m is the ratio of mean daily motions of the Sun and the Moon, e is the eccentricity of the Moon’s orbit, § =(M - S), the elongation of the Moon from the Sun and $= M -P , the mean anomaly of the moon (P being the Moon’s perigee). (ii) Variation = 39/30" sin (2&) In the above formulae, S and M are respectively the mean longi- tudes of the Sun and the Moon. The Surya Siddhanta, being an earlier text, does not mention these corrections. However, Mafijula (932 A.D.), ## Page 67

text, does not mention these corrections. However, Majfijula (932 A.D.), Bhakara-II (1150 AD) and later Indian astronomers have recognized the evection correction in addition to the equation of centre. Besides these, the famous Orissa astronomer Samanta Candragekhara Simha discov- ered independently a fourth correction called annual equation. According to Candragekhara, (ii) Annual equation = (11'27".6) sin(Sun’s anomaly from apogee) In fact, Candraéekhara’s coefficient viz., 11'27°.6 is very close to the known modern value. ‘Tycho Brahe took the coefficient wrongly as 4°30". Sloka 1 : In this Sloka the mandakendra (anomaly) of a planet and the bhuja of mandakendra are given as follows : Mandakendra (MK) = Mandocea of the planet - Mean planet (i) If the mandakendra of the planet is less than 3 rasis (ie, 0° < MK < 90°), then MK itself is the bhuja ie., Bhuja = MK. ii) If the mandakendra is greater than 3 raéis and less than 6 rasis (ie., 90° < MK < 180°) then Bhuja = 6 rasis - MK = 180° - MK (iii) If mandakendra is greater than 6 ragis and less than 9 rasis (ie, if 180° < MK < 270°) then Bhuja = MK -6 rdasis = MK -180° (iv) If mandakendra is greater than 9 rasis and less than 12 rasis (ie., 270° < MK< 360°) then Bhuja = 12 rasis- MK = 360° - MK ## Page 68

ie., koti= 3 ragis — bhuja = 90° — bhuja 12 rasis (or 360°) have been divided into four padas (quadrants) each containing 8 rasis (90°). The I and III quadrants are visamapadas (odd quadrants) and II and IV are called samapada (even quadrants). Mandocca of the Sun = 78° = 2" 18° (taken as fixed). Sloka 2 : The method of finding the mandaphala of the Sun is explained as follows: (i) Find the mandakendra (MK) of the Sun. (ii) Find the bhuja of MK (hereafter denoted by BMK). bhuja 9 Gii) Subtract. from 20 i.e., obtain (20 - BMK/9) BMK Gv) Multiply the results of step (iii) and —)— (v) Divide the result of (iv) by 9. (vi) Subtract the result of step (v) from 57. (vii) Express the results of step (vi) and step (iv) as vikalas (seconds of are) and divide the result of step (iv) by that of step (vi)

The result is the mandaphala of the Sun.

: 9 ) 9 ie., Mandaphala of the Sun = (~- FIV Pu 7 9) 9 ## Page 69

Note : (i) If the mandakendra is within 6 ragis from Mesa (i.e., 0° < MK <180°) then the mandaphala is additive. (ii) If the mandakendra is within 6 rasis from Tula (180° < MK <360°) then the mandaphala is subtractive.

Sripati Bhatta’s expression for the jya of the mandakendra is as follows : a: atfermnfenstren: warrrsrerctrareresea: | a are@veninn fren: weg Sates saat yates doh kotibhagarahita bhihatah khandgacandrastadiya-caranonasararkadigbhih / te vyasakhandagunita vibrtah phalntu jyabhirvina’pi bhavato bhujakotijive // (180 — MK ) MK x120 ie., Mandakendra jya = 19125 _ {180 - MK) MK 4 where MK stands for the bhuja of the mandakendra oo (180- MK) MK x480 ie., Jya (ME) = 49509 —(180—MK) MK x 480 =+_9_} 9 ___ @ividing by 9 x 9 40500 2a sede 9x9 9 (oe MK 9 ## Page 70

20 Ss ME 480 oi 9 9 s00-[29 ME) BE The above derivation is based on the significant and unique formula of Bhaskara I (c.629 AD) - see Appendix-3

the parama mandaphala (i.e., maximum mandaphala) of the Sun 125° ney yrage ae v3". 125. mandakendrajya -. Mandaphala of the Sun = [x= 29-—* 125 (

MK 9 0 ## Page 71 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) 859 i.e., Mandaphala of the Sun ~ Note : The exact formula for the mandaphala of the Sun is 7 sin m where R =360°, a the periphery of the manda epicycle (degrees) and m is the Sun’s anomaly (from the apogee, mandocca). When m=90°, sin m =1 |, parama mandaphala = ni radian ## Page 72

$60 GRAHALAGHAVAM OF GANESA DAIVAJNA

a 180 _125 x—=.

5 x (taking R =360°) a 1125 2 n S57 2nx125 qos 37 degrees => a =13°.778915 According to the book, Ancient Indian Astronomy by S. Balachandra Rao, the value of a for the Sun is between 11°.80781 and 12°.31284 , based on the eccentricity of the earth’s orbit. Example : Mandocca of the Sun = 2* 18°(fixed)

Mandakendra of the Sun (MK) = Mandocca - Mean Sun 28 18° — 1% 4°13'42" 1 43°46'18"

Since 0 < MK = (43°46'18") < 90°,

## Page 73

9 9 BMK \ BMK ( aa) } 9 57 cee 9 (2 -BMK. } BMK :, Mandaphala of the Sun = 73 616558 = Gg -goongy 7 19079062 = 1°30'28 Mandaphala is additive because 0° < MK (= 43°46'18") <180°- Note : According to the formula, using the sine function, we have Mandaphala of the Sun = 7 sinm (where a=14°, R=360°, m =43°46'18") = 36 Sin (43°46'18") = 0.0269027 radian = 1°32'29" We note that the mandaphala of the Sun, without the sine-function, is very close to that obtained using the trigonometric function. The error is just 2’. Now,

1 4° 13742" 41° 30'28" = 1% 5°44'10" ## Page 74

After the manda correction, the Grahalaghavam adopts the cara correc- tion explained in Sloka 6 of this chapter. Sloka 3 : In this Sloka the method of finding the mandaphala of the Moon is given as follows : (i) Find the Mandakendra (MK) of the Moon. MK Gi) Subtract —~ from 30. MK (iii) Multiply the result of step (ii) by ca. (iv) Divide the result of step (iii) by 20 and subtract the quotient from 56. (v) Divide the result of step (iii) by that of step (iv).

i.e., Mandaphala of the Moon = Here, MK is actually the bhuja of the mandakendra. Note : In the case of the Moon, the Grahalaghavam gives three corrections in the order, cara, bhujantara and deéantara. After obtaining the Moon with these three corrections, finally the manda correction (e., the equation of centre) is applied. ## Page 75

RAVI CANDRA SPHUTADHIKARA (CHAP. 2) S63

(180 —_MK) MK x 480 We have mandakendra jya = —————~—~_—~__ 40500 — (180 - MK) MK according to Sripati Bhatta. Dividing the numberator and the denominator by 6 x6, (‘@ -MK ma 480. 6 6 dya (MK) = 40500 6x6 MK } MK 6 6

According to the Grahalaghavam the maximum mandaphalam,

5xmandakendrajya «. Mandaphala of the Moon = 720 5x 30M ts 480 6 6 —rr—™—S—C=SC—C‘=R" 120 «| 1128 -(20 1 \ Me L _ cd Me) MK 120 1125 -(20 ~4 | Me ## Page 76

i.e., Mandaphala of the Moon ~

## Page 77 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) oo ae Note : The usual formula for the mandaphala of the Moon is R sin m, When m = 90°, sin m = 1 so that max. mandaphala = radian Now, according to the Grahaléghavam. the maximum mandaphala. Parama mendaphala of the Moon = 5° ag, eg ace ers eR = 360° © => a=2nxS => a =31°.415927 According to the book Ancient Indian Astronomy by S. Balachandra Rao the value of a for the Moon lies between 36°.8085 and 42°.22932 based on the eccentricity of the Moon’s orbit. Sloka 4 : The method of finding true daily motions of the Sun and the Moon is given in this Sloka as explained below.

I. True daily motion of the Sun :

(i) Find the bhuja of the mandakendra (MK) of the Sun in degrees. (ii) Find koti of MK. (iii) Divide koti by 20. (iv) Subtract the result of step (iii) from 11. (v) Multiply the results of step (iii) and step (iv). (vi) Divide the result of step (v) by 13. This gives the ‘gatiphalam’ of the Sun in kalas (minutes of arc) ## Page 78

i.e., Gatiphalam of the Sun = ia If the mandakendra of the Sun is within 6 rasis from Karka (i.e., if 90° < MK < 270°) then add ‘gatiphalam’ to the mean daily motion of the Sun to get true daily motion. If mandakendra of the Sun is within 6 ragis from Makara (i.e., if MK is in IV or I quadrant), subtract ‘gatiphalam’ from the mean daily motion to get true daily motion of the Sun.

Mandakendra (MK) of the Sun = 43° 46'18" .. bhuja= 43° 46’ 18" Koti of MK = 90° - bhuja 90° -43° 46'18" = 46°13'42" ie koti_\ koti _ 20 ) 20 13

i _ 46° 13'42" |46°13'42" 20 20 13 = 15448413 = 1' 32" 41".43 Since MK of Ravi is in the first quadrant (i.e., within 6 rasis from Makara) ## Page 79 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) ball we have to subtract ‘gatiphalam’ from the mean daily motion of the Sun :. True daily motion of the Sun = Mean motion - gatiphalam’ = 59’ 8" -1' 32" 41".43 = 57’ 35" 18".57 Remark : The modern formula for correction to get the true motion of the Sun is given by -b AM an = con (2) where b = 14, R = 360 and the mean daily motion of the Sun, = 59'08" (based on the classical sine formula). “At

“ 360 = -1'39"54" Remark : The gatiphalam of the Sun, obtained without the use of trigo- nometric function differs from that using cos M only by 7” of arc. II. True daily motion of the Moon

Gi) Find the mandakendra MK of the Moon and its bhuja. (ii) Find koti of the mandakendra (MK); koti = 90° — bhuja. (iii) Divide koti by 20. (iv) Subtract the result of step (iii) from 11. (v) Multiply the results of steps (iv) and (iii). ## Page 80

(vi) Multiply the result of step (v) by 2. (vii) Divide the result of step (vi) by 6. (viii) Add the results of steps (vi) and (vii). A : Pott tot | (5, 2 ie, Gatiphalam of the Moon = 20 } 20 6 and True daily motion of the Moon = Mean motion of the Moon + gatiphalam If MK is within 6 rasis from Karka (ie., if the mandakendra (MK) is in II or III quadrant) add gatiphalam to the mean daily motion. If MKis within 6 ragis from Makara (i.e., if MK is in I or IV quad- rant), subtract gatiphalam from the mean daily motion.

Mandakendra of the Moon MK = 3" 25°12'17" = 115°12'17" Bhuja of MK = 180° — 115° 12' 17" = 64°47'43" = BMK :. Koti of MK = 90°-BMK = 90°-64°47'43" = 25°12/17"

0 A | .. Gatiphalam of the Moon ( 20 } 20 It 2) PG eee ra (242) 20 20 6 28'.640272 = 28'38"24” sees (1) ## Page 81 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) 869 Since MK = 115°12'17" is in the II quadrant the gatiphalam is additive.

true daily motion of the Moon = Mean motion of the Moon + gatiphalam = 790'35"+28'38724" = 819/13" Remark : From the classical sine formula for the Moon, we have cos At R

at At AA oye where A = Moon's apogee and its mean daily motion 7, = 6 4! AL b= 31°, R= 360°, <-=790'35", the Moon’s mean daily motion M = BMK = 64° 47'43", the bhuja of Mandrakendra an 3 05 (64°.47'43")[790°35"—6'41"] At 360 = 28'44"46" We observe that the difference between (1) and (2) is less than 6". Sloka 5 : In this Sloka finding palabha and carakhandas of a given place are explained as follows : I. To find palabha : The day on which the true longitude of the sayana (tropical) Sun becomes 0°0'0" (i.e. equinoctial day), determine the shadow of a 12 arigula ## Page 82

long cone (Sariku) placed on a plane surface at the noon. This shadow length is called palabha or aksabha. In other words, it is the shadow of the ganku at the equinoctial noon.

10 Multiply palabha by 10, 8 and 5 respectively. We get three khandas. They are called carakhandas which are in vikalas (seconds of arc). Example : Palabha of Almora [Long. 79° £ 40'; Lat. 29° 36';] = 6|47 angulas The three carakhandas are 10 (6/47)x10, (6|47)x8 and (6|47 x = 67.8333, 54.2666 and 22.61111 = 68", 54” and 23" Remark : The latitude 6 of a place can be obtained from the palabha of the place as follows : A Sariteu oO c ‘Shadow B (Palabha)

## Page 83 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) si

BC _ palabha AB O12 -. Palabha = 12 tan > argulas From this we have $=tan “![Palabha/12] For example, the palabha of Almora ( = 29°36’) is 12 tand = 12 tan (29°36") = 68ng agprat, Note : 1 arigula = 60 Pratyangulas. Sloka 6 : This éloka gives the method of finding cara correction. It is as follows : (i) Find sayana Ravi (tropical Sun). (ii) Find bhuja of the sayana Sun. Express it in rasis, amas, kalas and vikalas. (iii) The number which represents ragi gives the number of elapsed khandas. (iv) Find the bhogya khanda (i.e., khanda to be covered) and multiply it by the remaining amsas etc. (v) Divide the result of (iv) by 30. ## Page 84 s72 GRAHALAGHAVAM OF GANESA DAIVAJNA (vi) Add the result of (v) to the elapsed khandas. This gives the cara in seconds (") of arc .

@ if sayana Ravi (SR) is within 6 rasis from Mega (i.e., 0°<SR <180°), then cara is negative. (ii) if sayana Ravi (SR) is within 6 ragis from Tula (.e., 180°< SR < 360°) cara is positive.

(i) cara is positive if 0°<SR <180° (ii) cara is negative if 180°<SR <360°- Example : Palabha of Kasi : 5 arigulas 45 pratyanigulas = 5|45 arigulas The three carakhandas are (5/45)x10, (5/45)x8 and l45)«2 = 57", 46" and 19" Suppose sdyana Ravi = 18 23°54'10", bhuja = 1% 23° 54’ 10" The number 1 in the ragi position implies that the number of elapsed khandas = 1 :, Elapsed khanda = 57"

## Page 85 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) 873

ae 2agsail0z ee = 71099 , 0° 3 «. Cara = Elapsed khanda + 36" = 5774+36"=93" Since sayana Ravi is within 6 rasis from Mesa, the cara is negative. :, cara corrected nirayana Sun = 1° 5° 44'10"-93" = 185° 42'37" (the ayanaméa = 18° 10') Remark : The latitude of Kasi (Varandsi) is $ = 25°N19'. Therefore, palabha = 12 tan = 12 tan 25°19'=5|40 anigulas However, commentator Visvanatha has taken its value as 5/45 arig.

10 (5|40)x10, (5|40)x8 and (5|40)x—- ie, 567.766 » 45".41 and 18".92 The elapsed khanda = 56".766

## Page 86 S74 GRAHALAGHAVAM OF GANESA DAIVAJNA Remaining améas ete. = 23°54'10" Now, 2SSA10" a5" = 36".181 «, Cara = Elapsed khanda + 36".181 = 56".77 + 36".181=92".95 Therefore , cara corrected nirayana Sun = 18 5°44"10"-92".95 = 1 5° 42'37". Sloka 7 : This sloka explains the method of applying the cara, bhujantara and degantara corrections to the Moon. (1) Cara correction for the Moon : 2 x Cara os Subtract (2S) from the mean position of the Moon to get the cara corrected Moon. Here, cara must be taken in minutes (’) of arc. ie., Cara corrected Moon = Mean Moon - (2S) (2) Bhujantara correction for the Moon : (using mandaphala of the Sun). Divide the mandaphala of the Sun by 27. The result will be in degrees, minutes of arc. Add this to the cara corrected Moon. That is

mandaphala of the Sun = {Cara corrected Moon} + 7 ## Page 87 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) 875 (3) Desantara correction for the Moon : Find the yojanas of the place from the Ujjayini meridian (rekha). Divide it by 6 to get the degantara correction in kalas (minutes of arc)(approx.). (=o) Note : In modern astronomy, the equivalent of deséntara is given by “= day where ) and Ap are respectively the longitudes of a place and of the central meridian (in degrees). Let 6 be the terrestrial latitude of a place so that the R sine of its co-latitude CP is 3438 sin (90° — 4). The diameter of the earth is taken as 1600 yojanas (i.e., about 8000 miles). The maximum (equitorial) circum- ference, MC =2n(800) = 1600 x yojanas. The corrected circumference CC at a place is given by cc _ MC -CP 343 8'

From these, we get

where DMis the daily motion of a planet. ## Page 88

Grahalaghavam gives the expression for the desantara correction in kalas as one-sixth of the distance in yojanas of a place from Ujjayini. The com- mentator Viévanatha has given the distance of Kasi from the central meridian as 65 yojanas, Let r= 4000 miles ~ 800 yojanas (Fig. 2.3).

<. 7 = (800 cos 6) yojanas The circumference of the small circle through a place A is given by 2nr, = (1600 x) cos @ yojanas. Taking the circumference of the small circle as 360°, we have (1600 x cos $) yojanas = 360° 2. x yojanas =360 x/1600 x cos > (0.0716 )x cos deerees. ® (0.0716 (oonie)s in hours 15 cos} ‘ (0.0716 )x _ = Q or Rae cosh 2 ghatis.

For example, for the Moon, at Kasi,

## Page 89

(0.0716 ) 64 790'.35" Desantara correction = 6 cos (25° 36') 60 wz 11'9" we (1)

1

x64 ~10'40" =) al— The difference between (1) and (2) is just about 29". Example: Mean Moon = 6% 20° 10'24" Candrocca = 10" 14° 54' 43"

(i) Cara correction for the Moon :- Cara corrected Moon = Mean Moon — one. = 6% a0" 10° gar — 2 = (taking cara as in minutes of arc). 6 20° 10'24" — 20’ 40" = 6% 19° 49°44" ## Page 90

S78 GRAHALAGHAVAM OF GANESA DAIVAJNA.

Gi) Bhujantara correction for the Moon :

Mandaphala of the Sun = 1° 30'28" (additive)

mandaphala of the Sun = 6% 19° 49° 44rg 13028 27 27 6® 19° 49'44"+0° 3721" 6® 19° 53°05" (iii) Desantara correction for the Moon : Distance of Kaéi from Rekhapura in yojanas = 64 ~. DeSantara correction ~ (£) = 10'40" :. Degantara corrected Moon = Bhujantara corrected Moon + Desantara correction = 6 19°53'05"-10'40" = 6* 19° 42/25"

The Moon after all the three corrections = 6* 19° 42'25"

Manda correction for the Moon :

## Page 91 RAVI CANDRA SPHUTADHIKARA (CHAP. 2) S79 The thrice corrected Moon = 6* 19° 42'25" MK = Mandakendra = 3" 25° 12'18" Bhuja of MK = 180° -3" 25° 12'18" = 64° 47/42" Mandaphala of the Moon = ] [from Sloka 3] le 64 sn) 64° 47" =| 6 6 Fe ae | 20 4° 33/38" Since MK = 3" 25° 12'18"<180°, the mandaphala is additive. True Moon = Mean Moon (after three corrections) + mandaphala = 6" 19° 42'25"+4° 33'38" = 6% 24° 16 03" = 204° 1603". ## Page 92

Slokas 8 and 9 : These two glokas give the method of finding tithi, naksatra, yoga and karana. (1) To find tithi : (i) Subtract the true longitude of the Sun from the true longitude of the Moon. (ii) Divide the result of step (i) by 12. This (quotient) gives the number of elapsed tithis. (iii) The remainder gives the elapsed part of the running tithi. It will be in degrees, min., etc. (iv) Subtracting the elapsed part of the running tithi from 12, we get bhogyaméa (i.e., the portion to be covered) of the running tithi in degrees. (v) Convert the result of step (iii) into seconds of arc (vikalas). (vi) Convert the result of step (iv) into seconds of arc (vikalas). (vii) Consider the difference between true daily motions of the Moon and the Sun and convert it into seconds of arc (vikalas). (viii) Multiply the result of step (v) by 60 and divide by that of step (vii). This gives the elapsed ghatikas of the running tithi. (ix) Multiply the result of step (vi) by 60 and divide it by that of step (vii). This gives the esyaghatika (i.e., ghatikas to be covered) of the running tithi. Tithi : In the course of a lunar month, from a new moon to the next new moon, the shape and size of the Moon changes from day to day. On an améavasya day (newmoon day) the Moon is invisible as in A, (see Fig. 2.4). On the next day, a very thin “crescent” Moon B is visible, if the sky is clear, soon after the sunset in the western horizon. On the succeeding days of the Sukla paksa the brighter or visible part of the Moon keeps on growing until it is half (C) between the 7th and ## Page 93

©00908 A B H if

8th day the after the new moon. Also, each day, the Moon keeps moving up in the sky at sunset since it moves farther from the Sun at the rate of about 12° per day. When the phase of the Moon is half, it will be midway in the sky between the eastern and the western horizons. Then, each succeeding day the brighter part of the Moon grows more than half when it is said to be gibbous, D. At the end of the gukla paksa, the Moon will be fully visible, E, when the krsna paksa (dark half of the month) com- mences. In the krsna paksa, the phases of the Moon diminish (or wane) in the reverse order. From the full moon upto the 7th or 8th day, more than half the Moon is bright when it is said to be gibbous, F. Then, between saptami and astami, the Moon will be half (G). The bright portion of the Moon goes on decreasing till it is crescent (H) again, a day before the new moon day. At the end of the krsna paksa the Moon is totally invisible on the new moon day, I. The lunar month is divided into 30 parts called tithis. The bright half has 15 tithis and so too the dark half. The duration of a tithi is the time taken by the Moon to move 12° relative to the Sun. The durations of different tithis are not equal. In a paksa (fortnight), starting from the new moon or the full moon, there are 15 tithis (Table 2.2):

1. Pratipat 6. Sasti 11. Ekadasi 2. Dvitiya 7. Saptami 12. Dvadast 3. Trtiya 8. Astami 13. Trayodasi 4. Caturthi 9. Navami 14. Caturdasi 5. Paficami 10. Dasami 15. Parnima or Amavasya ## Page 94

Therefore, Tithi = [Longitude of the Moon — Longitude of the Sun] / 12° where both the longitudes are in degrees. The quotient part indicates the number of tithis completed during the lunar month, and hence the (quotient + 1) gives the currently run- ning tithi, If the running tithi is less than 15, then it is of the Sukla paksa (bright fortnight). If the tithi is greater than 15, then subtract 15 from that number and the remainder gives the running tithi in the krsna paksa. If the running tithi is 15, then it is a parnima (full moon) day and if it is 30, the day is an amavasya (new moon day). Note : If the longitude of the Moon is less than that of the Sun then add 360° to avoid the negative sign, and then divide it by 12 to get the tithi

True Sun = 1% 5° 42'37" True Moon = 6* 24° 16'3" (i) Moon — Sun = 6* 24° 1673"-1* 5° 42°37" =) 52118334264 = 168° 33'26" 168° 33'26 =1442 33" 26' ii) Ne (ii) Now, fe ## Page 95

RAVI CANDRA SPHUTADHIKARA (CHAP. 2) S83

Therefore, the number of elapsed tithis is 14 and the running tithiis 15th. That is the running tithi is Parnima. (iii) Remainder = 0° 39’ 26" implies that the elapsed part of the running tithi, viz. Parnima = 0° 33' 26". (iv) Now, 12° — 0° 33'26" = 11° 26'34", the bhogyamsa. (v) The elapsed part 0° 3326" = 2006" (vikalas), the gata. (vi) The balance, 11° 26'34" = 41194" (vikalas), the esya. (vii) True daily motion of the Moon - True daily motion of the Sun 819'0" -57'36" 761'24" = 45684" - _ a. 2006 x60 (viii) Elapsed ghatikas of the running tithi = 15684 = 28h ggveh 41194 x60 (ix) Esyagatika of the running tithi = Saree gh Se ke

Thus, we have

## Page 96 ssa GRAHALAGHAVAM OF GANESA DAIVAJNA Esya ghatikas (i.e., balance ghatikas) of parnima = 548" g¥sh ..Total duration of the tithi = 28 39” . 548% og”sh — 568h gaveh (3) To find naksatra : (i) Find the true longitude of the Moon for the given time. Convert it into kalas (minutes of arc). (ii) Divide the result ofstep (i) by 800 kalas. (The extent of a naksatra = 13° 20’ =800') (iii) The quotient in step (ii) gives the number of elapsed naksatras [see Table 2.3]. (iv) The remainder in step (ii) gives the elapsed part of the running naksatra. (v) Subtract the result of step (iv) from 800. This gives esya (i.e., kalds to be covered in the running naksatra), Express the results of (iv) and (v) in vikalas. (vi) Consider the true daily motion of the Moon and express it in vikalas (seconds of arc). (vii) Multiply the result of step (iv) by 60 and divide it by that of step (vi). This gives the (elapsed) ghatikas of the running naksatra. (viii) Multiply the result of step (v) by 60 and divide it by the result of step (vi). This gives esya ghatikas of the running naksatra (i.e., ghatikas to be covered in the running naksatra). (ix) The sum of the results of steps (vii) and (viii) gives the total duration of the running naksatra. ## Page 97

Table 2.3 Nakgatras and their range of nirayana longitudes No. Naksatra From To 1. ASsvini oro" 13°20" 2. Bharani 13°20' 26°40' 3. Krttika 26°40’ 40°00" 4, Rohini 40°00! 53°20" 5. Mrgasira 53°20" 66°40" 6. Ardré 66°40" 80°00" te Punarvasu 80°00" 93°20' 8. Pusya 93°20" 106°40° 9. Aslesa 106°40' 120°00' 10. ‘Makha (or Magha) 120°00' 133°20' nie Pubba (Parva Phalguni) —133°20' 146°40' 12, Uttara (Uttara Phalguni) 46°40’ 160°00' 13. Hasta 160°00" 173°20' 14. Citta (or Citra) 173°20" 186°40' 15. Svati 186°40' 200°00" 16. Visakha 200°00" 213°20' 17. Anuradha 213°20' 226°40' 18. dyestha 226°40' 240°00' 19. Mala 240°00' 253°20' 20. Parvasadha 253°20' 266 °40' 21. Uttarasadha 266°40' 280°00' 22, Sravana 280°00' 293°20' 23. Dhanistha 293°20' 306°40' 24. Satabigaj 306°40' 320°00' 25. Parvabhadra 320°00' 333°20' 26. Uttarabhadra 333°20' 346°40' PYG Revati 346°40' 360°00'

Gi) True nirayana Moon = 6* 24° 15'03" = 12255'03" «a 12255'3" 15 , 255'03 800° 800° ## Page 98

(iii) The quotient = 15 implies that the number of elapsed naksatras is 15 and the 16th naksatra, Visakha (see Table 2.3) is running. (iv) The remainder = 255'03" implies that the elapsed part of the running naksatra (Visakha) = 255'03" = 1530" (v) Now, 800’-255'03" = 54457" = 32697" (vikalas) (vi) True daily motion of the Moon = 8190" = 49140" 15303 x 60 ii HEE gh = ygeh 4yveh (vii) We have “55 18.6849828h = 18h 41 ie., elapsed (gata) ghatikas of Visakha ~ 1885 41veh 32697 x 60 5 gh ~ gggh 55veh 49140 39.923077! 398" 55) (viii) i.e., balance (esya) ghatikas of Visakha = 398 55veh (ix) -, Duration of Visakha = 188h 41¥8 + 39h 55veh = 58eh g6veh. (3) The find karana : (i) Multiply the number of elapsed tithi by 2 and divide it by 7. This gives the karana starting with Bava. Ahalf tithiis called a karana i.e., it is of an angular distance of 6° between the Sun and the Moon. There are totally 11 karanas. Of these 4 are inmovable (sthira) and 7 are movable (cara). In particular, 4 half- tithis viz. the second half of krsna paksa caturdasi, two halves of amAvasya, and the first half of the Sukla pratipat are the sthira, karanas named as Sakuni, Catuspada, Naga and Kimstughna in that order. ## Page 99

Then from the second half of the sukla pratipat onwards we have the movable (cara) karanas viz. Bava, Balava, Kaulava, Taitila, Gara, Vanij and Visti (or Bhadra) in that order repeating the cycle 8 times. 14 Example 1 : Elapsed tithi = 14. Now, 2 7% 4 (quotient), remainder =0 i.e., the last karana in the cycle of 7 karanas (having completed 4 cycles) namely Visti (or Bhadra). Example 2 : March 21, 1990 at 54 30™ a.m. (IST). We have true Moon, M = 262°10'0” and the true Sun, S = 336°23'13". Therefore, (M-S) = 262° 10'-336°23'13" = 285° 46'47"/(after adding 369° to avoid the negative sign). Now, (M- S)/6 = 47.629953. Since this number is greater than 7, removing the multiples of 7, we get the remainder 5.629953. The integral part of the number is 5. Counting 5 starting with Bava in the list of moving (cara) karanas, we get Gara as the running karana. Note : Since the cycle of 7 cara karanas starts with the second half (and not the first half) of éukla pratipat, the integral part of the quotient represents the run- ning (and the not the elapsed) karana. (4) To find yoga : Add the nirayana longitudes of the Sun and the Moon and convert the sum into kalas (minutes of arc). Divide this sum by g00'. In the thus obtained result the integer part gives the elapsed yoga and the fractional part represents the elapsed portion of the running (current) yoga (see Table 2.4). ## Page 100

$88 GRAHALAGHAVAM OF GANESA DAIVAJNA

Table 2.4 Yogas

1. Viskambha 15. Vajra 2. Priti 16. Siddhi 3. Ayusman 17. Vyatipata 4. Saubhagya 18. Varlyan 5. Sobhana 19. Parigha 6. Atiganda 20. Siva 7. Sukarma 21. Siddha 8. Dhrti 22, Sadhya 9. Sala 23. Subha 10. Ganda 24, Sukla 11. Vrddhi 25. Brahma 12. Dhruva 26. Indra 13. Vyaghata 27. Vaidhrta 14. Harsana (or Vaidhrti) Note : If the sum of the nirayana longitudes of the Sun and the Moon (in degrees) exceeds 360°, then subtract 360° from the sum, convert into min- utes and then divide that figure by 800. Example 1 : Sum of the longitudes of the Sun and the Moon = 7* 29° 57'40" = 14397'40". Dividing this by g00’, we get 17.9970. This means 17 yogas have elapsed and the 18th one viz. Varlyan is currently running. Example 2 : On March 21, 1990 at 5 30™ a.m. (IST), the nirayana longitude of the Sun, S = 336° 23’ and that of the Moon, M = 262° 10' (neglecting seconds of arc). ;, S + M = 598° 33'. Since the sum exceeds 360°, subtracting 360°, we get S + M= 238° 33’. Converting into kalas, we have S + M = 14,313’. Dividing this by 800’, we get 17.89125. This means that the 17 yoga (Vyatipata) is over and the 18 one viz. Varlyan is running. ## Page 101

PANCATARA SPASTADHIKARA (True Positions of Star-Planets) Slokas 1 to 5 : In the first five Slokas the Sighraphalas of the five taragrahas viz. Kuja, Budha, Guru, Sukra and Sani are explained. The Sighrankas of the five planets at intervals of 15° from 0° to 180° of the Sighrakendras are given as tabulated below :

Planets 0 1 2 3 4 5 6 7 8 9 0 nu \Kuja 0 58 117 174 228 279 325 365 393 400 368 249 0 [Budha 0 41 81 117 150 178 199 212 212 195 155 89 0 Guru 0 25 47 68 85 98 106 108 102 89 66 36 0 Sukra 0 63 126 186 246 302 354 402 440 461 443 326 0 Sani 0 15 28 39 48 54 57 57 53 45 33 18 O Sloka 6 : In the case of the superior planets viz. Kuja, Guru and Sani, the Sighrakendra = Sighrocca — mean planet = m, say, where Sighrocca is the mean Sun for all the three planets. To find the Sighranika of a planet we have the following procedure : (1) (a) Divide the Sighrakendra (if m < 180°) by 15 and find the quo- tient q (arika) and let the remainder be r. (b) If m > 180° subtract m from 360° (i.e. find 360° — m) and divide (360° — m) by 165 to get q (arika) and the remainder r. We call (360° - m) as bhuja for convenience. ## Page 102

(2) Find the Sighrankas, from Table 3.1, corresponding to the quo- tient (i.e., from the column headed by q against the planet and also from the next column [i.e., (q+ 1)" column] and take the difference between the latter and the former Sigrankas. (3) Multiply the above difference of Sighrarikas by the remainder r [obtained in (1) above] and divide by 15. (4) Add the value obtained in (3) to the Sighrarika obtained in item (2) above under the column headed by q. The result obtained by dividing by 10 is the required Sighraphala. Note that the differ- ence between two successive columns (headed by q + 1 and q) may be negative also. In that case, subtract the result of (3) from that of (2) under q. (5) If the Sighrakendra m is less than 180°, the Sighraphala is posi tive. On the otherhand , if m > 180°, then the Sighraphala is negative. Example : Suppose, on a certain day, mean Kuja = 117 3° 53'25" =MP and mean Sun = 11" 17°32'53" = MS which is the Sighrocca of Kuja. .. Sighrakendra (i.e., Sighra anomaly) of Kuja = Sighrocca — mean Kuja = MS- MP= 11" 17° 39'53" -11% 3°53'25" = 0° 13° 39" 28" = SK. Here, SK < 180°. We shall now find the Sighraphala : (1) Dividing SK by 15, quotient q = 0 (aka) and remainder r = 13° 39'28". (2) From Table 3.1, the entries against Kuja under the columns headed by qand q+1lie., 0 and 1 are respectively 0 and 58. Therefore, ## Page 103 PANCATARA SPASTADHIKARA (CHAP. 3) S91 the difference between the above Sighrankas = 58 - 0 (i.e., the latter — the former) = 58. (3) Multiplying the result of (2) by r [from (1)] and dividing by 15, we get 58 x13°39'28"/15 = 52°48'36". Adding 52° 48’ 36” to the entry under qi.e., to 0 we get 52° 48 36". Dividing the above Sighranka by 10, we get 5°16'52"=SE , Sighraphala. 5) Since m < 180°, the Sighraphala is additive. According to the Grahalaghavam, we get the true position of a taragraha, from its mean position, by successively applying two types of corrections viz.the Sighra and the Manda samskaras to be explained shortly. Slokas 7 and 8 : In these two élokas the Mandatikas of the taragrahas are given as shown in the following table :

Planets 0 1 2 3 4 5 6 Kuja 0 29 57 85 109 fot 130 Budha 0 12 21 28 33 35 36 Guru 0 4 27 39 48 55 57 Sukra 0 06 11 13 14 15 15 Sani 0 19 40 60 17 89 93 ## Page 104 soz GRAHALAGHAVAM OF GANESA DAIVAJNA According to the Grahalaghavam, the Mandoccas of the five star-planets are as follows :

Planets Mandoccas Kuja 4® = 120° Budha Te = 210° Guru 68 = 180° Sukra 3 = 909 Sani 8" = 240° Mandakendra of Planet = Mandocca — Mean Planet. Sloka 9 : The method of finding the Mandaphala in the case of the five star-planets is explained as follows : (i) Divide the bhuja of the Mandakendra by 15 and find the quotient q. Let the remainder be r. (ii) Find the Mandanka from Table 3.2 corresponding to the quotient (i.e., from the column headed by q against the planet). Also find the next Mandatika [i.e., (q+ 1) column] and take the difference between the latter and the former Mandatikas. (iii) Multiply the above difference with the remainder r (obtained in step (i) above) and divide by 15. (iv) If the latter Mandanka is less than the former Mandarika, add the result of step (iii) to the latter Mandarika. If latter Mandarika is greater than the former Mandanka, sub- tract the result of step (iii) from the latter Mandarika. ## Page 105

(v) Divide the result of step (iv) by 10 to obtain the Mandaphala. Note : If Mandakendra (MK) is less than 180°, the Mandaphala is posi- tive. On the other hand, if MK > 180°, then the Mandaphala is negative.

Now, Here,

Mean Kuja = 118 3953/25"

MK, Mandakendra = Mandocca — Mean Kuja = 4% 11% 3°53/25" = 146°6'35”" Bhuja of MK = 180°-146°6'35” = 33°53/25" 33°53'25" 3°53'25" Pee a 15 15 quotient q = 2, remainder r = 3°53'25”

The entry against Kuja under the column headed by 2,

The entry against Kuja under the column headed by (q+ 1) ie., 3,

## Page 106 soa GRAHALAGHAVAM OF GANESA DAIVAJNA Difference = 85 — 57 = 28. 28 x3°53'25" Now, ——— + 57 = 64°15'42" . Mandaphala = a = 6°25'34"

Sloka 10 : The method of applying the Mandaphala and the Sighraphala to attain true position of a planet is explained as follows : (i) Find the Sighraphala as explained earlier using mean planet (obtained from Ahargana). Add (or subtract, as the case is ) half of the Sighraphala to (or from) the mean planet. This gives the half-Sighra corrected planet. (ii) Find the Mandaphala for the half-Sighra corrected planet. Add (or subtract, as the case is) Mandaphala to (or from) the mean planet (ob- tained from the Ahargana) which gives the Manda corrected planet. (iii) Find the second Sighrakendra using the first Sighrakendra and the Mandaphala. Second Sighrakendra = First Sighrakendra + Mandaphala. (iv) Find the second Sighraphala using the second Sighrakendra. (v) Add (or subtract, as the case is) the second Sighraphala to (or from) the Manda-corrected planet. This gives the true planet. Example : We shall find the true positions of the planets for the example considered earlier viz., Monday, May 15, 1612 (G). ## Page 107 PANCATARA SPASTADHIKARA (CHAP. 3) S95 I Finding true position of Kuja : Mean Kuja = 9% 29°55'13" Sighrocea of Kuja = Mean Sun = 1* 4°13'42" () First Sighra correction : SK = Sighrakendra of Kuja = Sighrocca - Mean Kuja 17 4°13'42"-9* 29°55'13" " 94°18'29"

Ne _ 15 15 +5 so that q= 6 and r= 4°18'29". Therefore, from Table 3.1, we have Gatanka = 325 and esyanka = 365 Their difference = 365 — 325 = 40 Remainder _ x difference 4°18'29" : = x:

40 = 11[29|17 Gatanka + 11)29|17 = 325 +11[29|17 = 3362917 ## Page 108

336|29|17 10

The Sighraphala is positive since SK = 94° 18'29"<180°- : 33° 38'56" Half-Sighraphala = = 16° 49'28" ‘Therefore, the half-Sighra corrected Kuja = Mean Kuja + Half-Sighraphala = 9% 29° 55'13"+16° 49'28" = 10°16° 4441" (2) Manda correction :

Mandocca of Kuja = 4% 9°0’.

= Mandocca — Half-Sighra corrected planet = 4% 0°0'-10°16°44'41" = 5"13°15'19" Bhuja of MK = 180° — 5"13°15'19" = 16° 44’41" 16° 44'41" 444i" Now, ———— =14+ ial 15 15 Here, quotient q = 1 and remainder r= |’ 44’4]”

## Page 109

gatarika = 29, esyanka = 57 Difference = 57 — 29 = 28 28 x1/44|41 it? = 33|15|22 :. Mandaphala = 33|15|22 = 3/1932 10 Since MK < 180°, the Mandaphala M is positive. Manda corrected Kuja = Mean Kuja + Mandaphala = 9% 29°55'13"4+3°19'32" = 10% 03°14'45". (3) Second Sighra correction : First Sighrakendra = 3* 4°18'29" Mandaphala = 3° 19' 32" Second Sighrakendra = First SK — Mandaphala 3% 4°18'29"-3° 19°32" 3% 0°58'57" 90°58'57" _ 90° 58' 57 Gee 58' 57) 15 15 ## Page 110

i.e., quotient q = 6 and remainder r = 0°58" 57" From Table 3.1, gatarika = 325, esyanka = 365

M0x0' 38/57" 4.325 = 327/87 |12. p 327/37|12 Therefore, second Sighraphala = ———— = 32° 45'43" .. True longitude of Kuja = Manda corrected Kuja + Second Sighraphala =10% 03°14’ 45"432° 45'43"= 11% 06° 00'28"= 336° 00’ 28"

(1) First Sighra correction : Mean Budha = Mean Sun = 1* 4° 13'42" Sighrakendra of Budha SK =1* 17° 14'50" =47° 14'50" (SK of Budha is obtained in Madhyamadhikara i.e., Ch. 1) 47° 14'50" 2° 14'50" a ee Now, 15 15 ie., quotient = 3 and remainder = 2° 14'50"; From Table 3.1, gatarika = 117, esyanka = 150 and the differecnce = 33. ## Page 111

2° 14°50" x33 Now, is + 117 =12156[38 a 121|56[38 cay Sighraphala = "= 12° 11"40" Que

Half-Sighra corrected Budha = Mean Budha + Half Sighraphala =1% 4° 13742" +6° 05'50" = 40° 19°32" (2) Manda correction : Mandocca of Budha = 7" =210° 0'0" Mandakendra (MK) of Budha = Mandocca — Half-Sighra corrected Budha =210°-40°19'32" =169° 40'28"

Now, eaceeo = 81538 giisps Mandaphala = = 0° 49'34"

## Page 112

Manda corrected Budha = = 18 04° 1342" +0° 49'34" =1¥ 05° 0316" =35° 0316" (3) Second Sighra correction : First Sighrakendra =47°14'50" Mandaphalam = 0° 49'34" Second Sighrakendra = First SK - Mandaphala =47° 14'50"-0° 49'34" =46° 25'16" 46° 25'16" 1° 25'16" a 5 3 15 15 Now, ; quotient = 3 and remainder = |° 25'16" From Table 3.1, gatarika = 117, esyarika = 150, difference = 33 33 x1° 25'16" Now, eT = 1207/35 a 120|7|35 .. Sighraphala = Tee oe

Since SK <180°, Sighraphala is additive.

the true longitude of Budha = Manda corrected Budha + Second Sighraphala = 1% 95° 03'16"+12°0'46' = 47° 04! 02" ## Page 113 PANCATARA SPASTADHIKARA (CHAP. 3) S101

(1) First Sighra correction : Sighrocca of Guru = Mean Sun = 1% 4°13'42"

Mean Guru = 4" g°15'17"

i 18 4° 13'42"- 4% 8° 15'17"

"

ie., Sighrakendra argument = 360° - 265° 5825" = 94°01'35" Now, seoss : os, quotient = 6, remainder = 4° 01'35 From Table 3.1, gatarika = 106, esyarika = 108, difference = 2 24° 01'35" Now, areas +106 = 106 |32[12 106 [32 |12

= 10° 39'13" Since SK = 265° 58'25"> 180°, the Sighraphala is negative. ## Page 114 s102 GRAHALAGHAVAM OF GANESA DAIVAJNA

Half-Sighraphala = ae = -5°19'36"

4® 3° 15'17" + (-5° 19°36") = 4% g°15'17"—5° 19136" = 4R o°55'41" (2) Manda correction : R

Mandakendra (MK) = Mandocca — Half-Sighra corrected Guru = 6% 4® a°55'41"= 1% 27° 4'19"= 57° 4'19" Bhuja of MK = 57° 4"19" STAT 234 tPA quotient = 8, remainder = 12° 4’19" Now, = 15 15 From Table 3.2, gatanka = 39, esyarika = 48, difference = 9 9x 12° 04°19" cae Now, 15 39 = 46(14)35 46|14| 35 0

Mandaphala = = 4°37'27"

## Page 115 PANCATARA SPASTADHIKARA (CHAP. 3) $103 Manda corrected Guru = Mean Guru + Mandaphala = 4® 915117" 44° 37'27"= 4F 12° 52'44" (3) Second Sighra correction : First Sighrakendra = 265° 58'25"

Second Sighrakendra = First SK - Mandaphalam

Bhuja of SK = 360° -261° 20’58" = 98° 39/02" We have ae AGEs = 02" . quotient = 6, remainder = g° 39' 02". From Table 3.1, gatarika = 106, esyarika = 108, difference = 2 Now, 28° 39702" igg = 107|9)12 a 107|9|12 . Sighraphala = —T— = 10° 42'35"

:. True longitude of Guru = Manda corrected Guru + Sighraphala = 4F 12° 52'44"4(-10° 42'55")= 4% 2°9'49" = 122°9'49" ## Page 116

S104 GRAHALAGHAVAM OF GANESA DAIVAJNA

Sighrakendra of Sukra = 3% 5° 41'35"= 95° 41'35" (from Ch. 1) Mean Sukra = Mean Sun = 1% 4° 13'42" (1) First Sighra correction :

95° 41'35" 5°.41'35" 2 26+ 15° 15g We have ; quotient = 6, remainder = 5° 41'35" From Table 3.1, gatarika = 354, esyarika = 402, difference = 48

372 [12|58 Sighraphala = ighraphala 10 = +37° 1318" é, 37° 13'18" Half-Sighraphala = =e = 18° 36'39" Half-Sighra corrected Sukra = Mean Sukra + Half-Sighraphala eal 4134 2ec 183693 9n- ele 22150¢2 lee 192150421 (2) Manda correction :

Mandakendra of Sukra = Mandocca — Half-Sighra corrected Sukra ## Page 117 PANCATARA SPASTADHIKARA (CHAP. 3) S105 = 3% ~52°50'21" = 37°9'39" Bhuja of MK = 37° 9'39" 37°.9'39" 79139" a

15 15 ; quotient = 2, remainder = 7° 9’39". From Table 3.2, gatarika = 11, esyanka = 13, difference = 2 27° 9'39" Now, - 15 +1 = 1157/17 Manda corrected Sukra = Mean Sukra + Mandaphalam = 18 4°13'42"41°11'44"= 1% 5° 25 26" = 35° 25° 26" (3) Second gighra correction First SK = 95° 41'35" Second Sighrakendra = First SK - Mandaphala = 95° 41'35"-1°11'44" = 94° 29'51"

15 15 jquotient = 6, remainder = 4° 29'51". From Table 3.1, gatanka = 354, esyanika = 402 and their difference = 48 48 x 4° 29'51" Now, 15 +354 = 368 [23 [31 . 368331 . Sighraphala = — = +36" 50/21". ## Page 118

., True position of Sukra = Manda corrected Sukra + Sighraphala =1* 5° 25° 25" + 36° 50’ 21" = 72° 15’ 46" V Finding true longitude of Sani : We have, from Chapter 1, Mean Sani = 117 0° 36’ 45" Sighrecca of Sani = Mean Sun = 1" 4° 13' 42" Sighrakendra SK = Sighrocca — Mean Sani = 1 4° 13'49"- 117 0° 36'45" = 63° 36'57" 63° 36'57" 3° 3657" ey We have ~ 15 15° ; quotient = 4, remainder = 3° 36'57” From Table 3.1, gatanka = 48, esyanka = 54, difference = 6

Li 49|26|47

p ° 56! 41" Half-Sighraphala = oa 2° 28'21"

=11* 0° 36’ 45" + 2° 28°21” = 11° 03° 05'06" ## Page 119 PANCATARA SPASTADHIKARA (CHAP. 3) $107 2) Manda correction : Mandocca = g® Mandakendra MK = Mandocca - Half-Sighra corrected Sani = 8% _11% 03° 05'05"= 8% 26° 54’55" = 266° 54'55” Bhuja of MK = 266° 54'55"-180° = 86° 54'55" 86° 54'55” 11° 5455" Se st 15° 15°

11° 54155". From Table 3.2, gatarika = 89, esyarika = 93, difference = 4 A xT S455" g9 -92|10|39 15° .. Mandaphala = 92|10|39 pT = 9° 13 04"

The Mandaphala is negative since MK > 180°

118 0° 3645" + (-9° 1304") 108 21° 23'41" ## Page 120

(3) Second Sighra correction : Second SK = First SK - Mandaphala = 63° 36'57" -(-9° 1304") = 63° 36'57" + 9° 13' 04" = 72° 50'01" 72 50’ 01" 12° 50' 01" ee ad 13° 15° We have ; quotient = 4, remainder = 12° 50'01" From Table 3.1, gatanka = 48, esyanka = 54, difference = 6 Now, eu! a wus +48 =53/8|0 f 53g|0 « Sighraphala = 7 5° 18’ 48” Hence, True Sani = 10° 21° 23’ 41" +5° 18’ 48” =326° 42' 29". We shall now consider a modern example :

True positions of planets for August 11, 1998.

For the given date 11” August 1998, we have Ahargana A = 2033 and Cakra C = 43 For Kuja, Ksepaka K = 10* 7° 8’ and Dhruvaka D = 1" 25° 32' 10AY _(10A) Mean Kuja = {eae ) -(134) -onpex ## Page 121 PANCATARA SPASTADHIKARA (CHAP. 3) S109 : (ome : LEED) Sa pepereeieePry = 19 B = 64° 33/30" (1) First Sighra correction : Sighrocca of Kuja = Mean Sun = 115° 9'59" Deéantara corrected Sighrocca of Kuja = 115° 9'42"

115° 9°42"-64° 33/30" 50° 36'12”

Bhi 50° 36'12" Oesomlou Now, Zia. 50736712" _, , 5° 36112" 15 15 Vee ~. quotient = 3, remainder = 5° 36’12” From Table 3.1, gatarika = 174, esyarika = 228 and difference = 54.

. 15 + Gatatika ## Page 122

54 x 5° 36'12" s+ 174 15 = 194|10|19 194 |Lo|19 -. Sighraphala = ——"— = 19° 25'2" Half-Sighraphala = aumeo 42'31" Half-Sighra corrected Kuja = Mean Kuja + Half Sighraphala = 64° 33'30"+9° 42°31" = 74°16'01". (2) Manda correction :

Mandocea of Kuja = 4% =120°

20°~74°16'01" = 45° 43'59" Bhuja of Mandakendra = 45° 43'59" 45° 43'59" 0° 43'59" EE ee Now, == : ies 15 . quotient = 3, remainder = 0° 43'59” From Table 3.2, gatdrika = 85, esyarika = 109, difference = 24 ## Page 123

PANCATARA SPASTADHIKARA (CHAP. 3) SIL

:. Mandaphala = ar 85+(24x 0° 43' 59") /15" 10 8° 37'2" :, Manda corrected Kuja = Mean Kuja + Mandaphala = 64° 33'307+8° 37/02" = 73° 10'32". (3) Second Sighra correction :

First Sighrakendra = 50° 36'12"

:.Second Sighrakendra =50° 36'12"-8°37'2" = 41°59'10" 41° 59'10" 11° 59" 10" Now, —— = 24+ 15 15 so that quotient = 2 and remainder = 11° 59'10" From Table 3.2, gatarika = 117, esyanka = 174, difference = 57 (difference x remainder) . [ + gatarika| Sighraphala = 15 10 ## Page 124 s112 GRAHALAGHAVAM OF GANESA DAIVAJNA. 15 10 [fees ir] = 16° 15/17" :. True Kuja = Manda corrected Kuja + Sighraphala = 73°10'32"+16°15'17" = 89°25'49". II. To find true position of Budha : For the given date 11th Augutst 1998, we have A = 2033 and C = 43 Mean Budha Kendra = 197° 7'47" (obtained in Chapter 1). (1) First Sighra correction :

Mean Budha = Mean Sun = 115° 9' 42"

Bhuja of Sighrakendra = 197° 7'47"—180° = 17° 7/47" genta 2utuada Now, ——— = 1+——— 15° 15 ## Page 125 PANCATARA SPASTADHIKARA (CHAP. 3) su13 so that quotient = 1 and remainder = 2° 7’ 47" From Table 3.1, gatarika = 41, esyarika = 81, difference = 40 | eitterence x remainder) : Sighraphala = 15 + gata 10 [bexerer la] 15° 10 =4° 40'4" Since Sighrakendra = 197° 7'47"> 180°, Sighraphala is negative. -, Half-Sighraphala = —2° 20'2" Half-Sighra corrected Budha = Mean Budha + Half-Sighraphala = 115° 9'42"—2° 20'2"= 112° 49’ 40".

Mandocca of Budha = 7% = 210° Mean Budha = Mean Sun = 115° 9'42"

Mandakendra = Mandocca — Half-Sighra corrected Budha ## Page 126 sil4 GRAHALAGHAVAM OF GANESA DAIVAJNA = 210° -112° 49'40" = 97° 10'20" Bhuja of Mandakendra = 180° - 97° 10'20" = 82° 49740" 82° 4940" 7 49' 40" Now, ————_ = 5+ —_.—_ 15° 1S so that quotient = 5 and remainder = 7° 49’ 40" From Table 3.2 gatanka = 35, esyarika = 36, difference = 1 (difference x _remaindey) + gatanka| ». Mandaphala = [ 15 10 = 3°33'8"

115° 9142" +3° 338” 118° 42'50". (3) Second Sighra correcion :

## Page 127

Mandaphala = 3° 33'8" -.Second Sighrakendra = 197° 7' 47" 3° 33'8" = 193° 34'39" Bhuja of Sighrakendra = 193° 34'39"-180° = 13°34'39" 13° 34'39" 13° 34’39" pe eo Now, ——— = : 15 15 so that quotient = 0, remainder = 13° 34’39" From Table 3.1, gatanka = 0, esyarika = 41, difference = 41 [s 334/39" ( 15 = -3°42'40" 10 :. Sighraphala =

118° 42'50"-3° 42' 40" 115° 0'10". II. To find true position of Guru : For the given date 11th August 1998, A = 2033 and C= 43 Mean Guru = 330°17'57" After DéSantara correction Mean Guru = 330° 21' 40"

## Page 128

(1) First Sighra correction : Sighrakendra of Guru = Sighrocca - Mean Guru = Mean Sun ~ Mean Guru = 115° 9 42"—330° 21’ 40" = 144° 48’2" (by adding 360°) Bhuja = 144° 48'2" 144° 48/2" 9° 48'2" ae AO 9 TO Now, ———— = : 15 15 so that quotient = 9 and remainder = 9° 482” From Table 3.1 gatanka = 89, esyarika = 66, difference = — 23 [mez a +*9| 10 -. Sighraphala = = 7° 23'50"

.. Half-Sighra corrected Guru = 330° 21'40"+3° 4155" = 334° 3/35" (2) Manda correction :

Mandakendra = Mandocca — Half-Sighra corrected Guru = 180° — 334° 3'35" ## Page 129 PANCATARA SPASTADHIKARA (CHAP. 3) S117 = 205° 56’ 25" (by adding 360°) ‘Bhuja’ = 25° 56' 25” 25° 56' 25" “ee LH 5625/15 5 Now, so that quotient = 1 and remainder = 10° 56’ 25" From Table 3.2, gatanka = 14, esydnka = 27, difference = 13 13x 10° 56’ 25" .Mandaphalam=|\ 15°) | = ~ 2° 2053" 10 Mandaphala is negative since Mandakendra > 180°. Manda corrected Guru = Mean Guru + Mandaphala = 330° 21' 40" - 2° 20/53" = 328°0'47" (3) Second Sighra correction : Mandaphala = — 2° 20'53" First Sighrakendra = 144° 48! 2" Second Sighrakendra = 144° 482" +2° 20'58" = 147° 8/55"

147 855" _, 12° 8/55" a ae it 15 Now, ## Page 130

From Table 3.1, gatanka = 89, esyanika = 66 and difference d = - 23 —23 x12°8'55" We have [eae J--isprie - 89 -18)37|40 .. Sighrakendra = ——>—— =7° 2'13" True Guru = Manda corrected Guru + Sighrakendra = 328° 0'47"+7° 2'13" = 335° 3/0" IV. To find true position of Sukra For the given date 11th August 1998 we have A = 2033 and C = 43

Mean Sukra = Mean Sun = 115° 9' 46" (1) First Sighra correction : Sighrakendra of Sukra = 310° 9/46" Bhuja = 49° 50'14" 49° 50'14" 4° 50'14" ee = 3+ 15° 15° Ww, From Table 3.1, gatarika = 186, esyarika = 246 and difference = 60 60 x4 50'14” +4186 15°

Sighraphala = y= 202325".

## Page 131 PANCATARA SPASTADHIKARA (CHAP. 3) su9 Half-Sighraphala = —10° 16'3" Half-Sighra corrected Sukra = Mean Sukra + Half-Sighraphala = 115° 9'42"-10° 16'3" = 104° 53°39"

Manda correction :

Mandakendra MK = Mandocca — Half-Sighra corrected Sukra = 90° -104° 5339" = 345 6'21" (adding 360°) > 180° Bhuja = 14° 5339" 14° 53739". 14° 53'39" 222 29 15s 15° Now, From Table 3.2, gatanka = 0, esyarika = 06 and difference = 06 14 53'39" x 06 + 0 «. Mandaphala = | 1S = -0°35'44" 10

Manda corrected Sukra = Mean Sukra — Mandaphala = 114° 33'57" (3) Second Sighra correction : First Sighrakendra = 310° 9' 46" Mandaphala = — 0° 35'44" Second Sighrakendra = 310° 9' 46" +0° 35'44" = 310° 45'30" ## Page 132

‘Bhuja’ = 49° 14’ 30" 49° 14’ 30" 4° 14’ 30" eee ees

15 15 From Table 3.1, gatarika = 186, esyarika = 246 and difference = 60 60x 4° 14'30' +186 1S 10 = -20°17' 48 .. Sighraphala = being negative since SK > 180°.

= 114° 33'57" - 20° 17'48"= 94° 16’ 09".

For the given date 11th August 1998: A = 2033, C = 43 Mean Sani = 8° 14’ 2" Mean Sun = Sighrocca of Sani = 115° 9' 42" (1) First Sighra correction : Sighrakendra of Sani = Sighrocca - Mean Sani = 115° 9'42"-8° 14'2”= 106° 55’ 40” Bhuja argument = 106° 55’ 40" 106° 55’ 40" 1°55’ 40" Se eT ow, 15° 15° From Table 3.1, gaténka = 57, esyarika = 53, difference = - 4 ## Page 133 PANCATARA SPASTADHIKARA (CHAP. 3) S121 4xT 65' 40" 57 15

Now , Sighraphala = “= = 5° 38'55"

Half-Sighraphala corrected Sani = 8° 14’ 2" + 2° 49'27" = 11° 03' 29" (2) Manda correction :

Mandakendra = Mandocca — Half-Sighra corrected Sani = 240° -11° 03'29" = 228° 56'31"

48° 56'31" 3° 56’ 31" ——— = 3+ —— Now, : - 15 15) From Table 3.2, gatarika = 60, esyanka = 77, difference = 17 17x3° 5631 +60 1s «. Mandaphala = ~ 10 = —6° 26'48" Mandaphala = - 6° 26' 48" being negative since MK > 180°. Manda corrected Sani = Mean Sani + Mandaphalam = 8° 14’ 2"-6° 26’ 48” = 1° 47'14". ## Page 134

(3) Second Sighra correction :

Second Sighrakendra = 106° 55'40" +6° 26'48" = 113° 22'28" 113° 22'28" 8° 22'28" —— wo phe 15° low, From Table 3.1, gatarika = 57, esyanka = 53, difference = —4 x 8° 22'28' 457 152 Si hala = ., Sighraphala To True Sani : = Manda corrected Sani + Sighraphala = 1° 47'13"4+5° 2836" = 7°15'49". = 5° 28/36" Comparison of true positions of planets with the ephemeris values. Date : August 11, 1998 (Planets According to According to Grahalaghavam Indian Ephemeris|

Ravi 113°51’ 33” 114°18" 19”

Candra 331°55' 27" 330°02'

[Budha 115°" 10” 119°37' 01"

Gura 335°03' 0” 333°18' 07"

Sukra 94°16" 09" 93°15"

\Rahu 44°21' 03" ## Page 135 PANCATARA SPASTADHIKARA (CHAP. 3) $123 Note : The Ayandméa, according to the Grahalaghvam for the given year Sa.Sa. 1920 is (1920'-444)' =1476' = 24° 36’. According to the Indian Ephemeris, the Ayandméa = 23° 50’ 08”. We find fairly a good agreement. between the two sets of values for the positions of planets except in the case Sani and Budha. True daily motions of the five star-planets (Gatispastikaranam) Sloka 11 : This Sloka explains the method of obtaining Manda corrected motions of Sani, Kuja Guru, Budha and Sukra. (i) Divide the difference of Mandankas of Sani, Kuja and Guru by 75, 5 and 30 respectively. (ii) Multiply the difference of Mandankas of Budha and Sukra by 2 and divide by 5 respectively. (iii) Step (i) and step (ii) give Manda gatiphala of respective planets which will be in minutes of are (kalas). (iv) Add (or subtract) the Mandagatiphala to mean motion of planets to get Manda corrected motion. Note : If the Mandakendra of a planet is within 6 Rasis from Karka (ie., if the planet is in II or II quadrant) the gatiphala is additive. If the Mandakendra of the planet is within 6 Rasis from Makara, (ie., if the planet is in IV or I quadrant) then the gatiphala is subtractive. Sloka 12 : The method of obtaining true daily motions of planets is ex- vlained as follows. 1. True daily motion of Kuja : (i) Obtain the Manda-corrected motion as explained in the previous Sloka. (ii) Take the difference between Sighranikas which are obtained in find- ing the second Sighraphala. ## Page 136

(iii) Divide the difference obtained in step (ii) by 5. The result will be in minutes of are (kalas) (iv) Add (or subtract) the above kalas to Manda corrected motion. This gives the true daily motion of the planet. Note : If the first Sighranka (ie., elapsed Sighranka) is less than the second Sighranka (i.e., Sighranka to be covered) then add the result of step (iii) to the Manda corrected planet. On the otherhand, if the first Sighranka is greater than the second then subtract the result of step (iii) from the Manda corrected planet. The same rule holds in the case of other planets also. Example : In the case of Kuja, we have (i) The difference between the Mandankas = 28 (see the first example under Sloka 9) 28 | rope (ii) gatiphalam = 5 = 5' 36 (iii) Since Mandakendra is within 6 Ragis from Karka, the gatiphalam is additive. Manda corrected motion of Kuja = Mean motion of Kuja + gatiphalam = 81'36"+5'36" = 37'12" (iv) The difference between the Sighrarikas = 40 since the elapsed Sighratika = 325 and the Sighranka to be covered = 365. (v) Since the elapsed Sighranka is less than the Sighranika to be covered, 40 _ age add 37 8'0" to the Manda corrected planet. -, True motion of Kuja = Manda corrected motion of Kuja +80" = 37'12"+8'0" = 45'12". ## Page 137 PANCATARA SPASTADHIKARA (CHAP. 3) $125 2. True daily motion of Budha (i) Obtain the Manda corrected Budha’s motion as explained earlier. (ii) Consider the difference between the elapsed Sighranka and the Sighranka to be covered, which are obtained in finding the second Sighraphala. (iii) Divide the difference obtained in step (ii) by 5; the result will be in kalas, (minutes of arc). (iv) Add the result of step (iii) to the difference obtained in step (ii). (v) Add (or subtract as the case may be) the result of step (iv) to the Manda corrected motion of Budha.

12x2 _grage :.gatiphalam = The Manda corrected motion = Mean motion + gatiphalam = 59'8"+4'48" = 63'56"

Bea . ghatiphalam = 2 =6'36 Now, the difference between Sighrankas + 6'36" =33' +6'36" = 39°36" -. True motion of Budha = manda corrected motion +39’ 36" = 63'56" +39'36" = 103732” 8. True Motion of Guru : (i) Obtain the Manda corrected motion of Guru as explained earlier. (ii) Obtain the difference of Sighrarikas which are obtained in finding the second Sighraphala. ## Page 138

(iii) Divide the above difference by 3. The result will be in kalas. (iv) Add (or subtract) the above result to the manda corrected motion of Guru.

The difference of the Mandanikas = 9 atiphalam = 2-0 18” gatip. = 30 The Manda corrected motion = Mean motion — gatiphalam = 5’-0'18" =4'42" Now, the difference between Sighrarikas = 2 and gatiphalam = 37° 40 True motion of Guru = The Manda corrected motion + 0'40" = 4'42"+0'40" = 5/22" 4. True motion of Sukra : (i) Obtain the Manda corrected motion of Sukra as explained in the pre- vious Sloka. (ii) Obtain the difference between the Sighrankas of the second Sighraphala. (iii) Divide the above difference by 4. The result will be in kalas. (iv) Add (or subtract) the result of step (iii) to the Manda corrected mo- tion of Sukra. Example : The difference between the Mandarikas = 2 2 -.gatiphalam = 2*==0'48" ## Page 139 PANCATARA SPASTADHIKARA (CHAP. 3) $127 The Manda corrected motion of Sukra = Mean motion of Sukra - gatiphalam = 59/08" -0' 48" = 58'20".

48 ». gatiphalam = ~~ =12'0" True motion of Sukra = Manda corrected motion of Sukra + 12'0" = 58'20"+12'0" =70'20" 5. True daily motion of Sani : (i) Obtain the Manda-corrected motion of Sani as explained earlier. (ii) Obtain the difference of the Sighrarikas of the second Sighraphala (iii) Multiply the above difference by 2 and divide it by 5. The result will be in kalas. (iv) Add (or subtract) the above kalas to the Manda-corrected motion of Sani. This gives the true motion of Sani. Example : The difference between the Mandanikas = 4 4 . pati = = =0'03" .. gatiphalam = 7 Now, the Manda corrected motion = Mean motion + gatiphalam = 2'+0'03" = 2'03"

2 _arage Now, ghatiphalam = 6 x ==2'24 ## Page 140

..True daily motion of Sani = Manda corrected motion + 2'24" = 2'03°+2'24" = 4°27" Sloka 13 : In the case Kuja and Sukra, due to large differences in their Sighrankas an additional correction, in each case, to the Sighraphala is given. In the process of determining the second Sighra correction, when the Sighrakendra is divided by 15°, let the quotient be Q and the remainder R. Now, between R and 15-R, take the lesser one and call it R’. Divide R' by 5 in the case of Kuja and by 3 in the case of Sukra. The result in degrees etc. should be added to or subtracted from the Sighra corrected planet according as the Sighraphala is additive or subtractive. In the case of the other three taragrahas viz. Budha, Guru and Sani, this second difference is negligible and hence omitted. Example: Considering the example worked out earlier for Kuja, we have

Between f° 4'57") and 15°- te 4 s7*) Le., 13° 55'03", the former is less. Dividing the lesser value by 5, we get 1° 4'57"/5 =12'59". Since the Sighraphala is additive, we add 12'59" to the Sighra corrected Kuja to get: the true Kuja = 336° 00'28" +12'59" = 336°13'27". In the case of Sukra, the Sighrakendra = 94° 29' 52". Dividing by 15° the quotient is 6 and the remainder is 4° 29’ 52”. Between this value and its difference from 15° viz., 10° 30'08", the former is less. Dividing the lesser ## Page 141 PANCATARA SPASTADHIKARA (CHAP. 3) $129 value by 3, we get 4° 29’52"/3 =1° 29'57’, Since the Sighraphala is ad- ditive, by adding 1° 29’ 57" to the Sighra corrected Sukra, we get corrected true Sukra 72° 15'46"+1°29'57" = 73° 45/43". Sloka 14 : While finding the final Sighranka of Kuja, Budha and Sukra, the remainders obtained must be multiplied by 10 and then divided re- spectively by 7, 7 and 3; the result be added respectively to 35, 97 and 53. This gives the corrected gatiphala (of Kuja, Budha and Sukra). These must be accepted and not the one obtained earlier. Example : Suppose the second Sighranka of Kuja is 11 and the remain- der, 11°50’0”. Now, multiplying the remainder by 10 and dividing by 7, 10 iganyam Se A sai . we get 11|50 cae 16'54"17" as the Sighra gatiphala. This is subtractive since the Sighra gatarika and the esyanka are respectively 249 and 0 corresponding to the quotients 11 and 12 and hence decreasing. There- fore, subtracting 51'54”17” from the already obtained Manda spastagati of 25/43”, we get the final true daily motion of Kuja as 25'43"-51'54"17" =-26'11"17"" (retrograde since negative). Sloka 15 : Retrograde motion of planets is explained. The Sighrakendras of the planets for (the commencement of) retrograde motion (vakragati) are 163 °,145 °,125 °, 167 °,113°. The retrogression continues till the Sighrakendras become 360° minus the above values (i.e, 197°, 215°, 235°, 193°, 247°) respectively for Kuja, Budha, Guru, Sukra and Sani.

The star-planets move from west to east, relative to the fixed stars, as seen from the earth, due to their natural motion. However, during cer- ## Page 142

tain periods each of these planets appears to move backwards, i.e, from east to west. Their celestial longitudes keep on decreasing instead of in- creasing, day by day, for some time. This apparent backward motion is called vakragati (retrograde motion). The phenomenon of retrograde motion is caused by the difference in the angular velocities of the earth and the planet, i, the relative velocity. ‘This phenomenon is demonstrated in Fig. 3.1. In Fig. 3.1 the motion of Mars (Kuja) relative to the earth is shown in the heliocentric model. The earth’s linear speed is 18.5 miles per second while that of Mars is 3.5 miles less, i.e, 15 miles per second. As the earth over- takes Mars, the latter appears to move backwards, when seen from the earth. The direct motion of Mars eastward is shown at positions 1, 2 and 3, the retrograde at 4 and 5 westward and again direct motion eastward at 6 and 7. The rule for determining the retrograde motion of a planet is given in the Saryasiddhanta as follows : The retrograde motion (vakragati) of the different star-planets commences when the Sighrakendra (ie., Sighra anomaly), in the fourth process of determining true positions, is as follows : Kuja 164°, Budha 144°, Guru 130°, Sukra 163°, Sani 115° That is, the retrograde motion of Kuja, for example, commences when its

## Page 143 PANCATARA SPASTADHIKARA (CHAP. 3) S131 The point at which the motion of a planet changes from direct to retro- grade is called a “stationary point”. The planet remains retrograde for some days and then, again, its motion changes from retrograde to direct. This point of change is the second stationary point. At both the station- ary points the planet has no apparent motion (i.e., the relative velocity is zero). Bhaskara II (b. 1114 A.D.) in his Siddhanta Siromani (Spastadhyaya) mentions that at the maximal point the (angular) velocity of the planet vanishes : ae weet we wet aaniewarnas faery yatra grahasya paramam phalam tatraivagati phalabhavena bhavitavyam /

In Fig. 3.2 let M be the mean planet, P be the true planet on the epicycle of radius p, E be the earth and S the Sun, If n is the mean daily motion of the Sun, let t be the number of days since the Sun S was at the first point of Mesa, PM[K =0 and PEM = E, then the celestial longitude L of the planet is given by L = nt-6+E where ntis the longitude of the Sun. Therefore, aL dé dE 2,8 a” dt dt ay Let PM = p and EM = r, where the radii p and r are constants. In Fig. 3.2, we have MA = pcos@ and PA=psin@. Therefore, EA= EM + MA =r + pcos@ and hence ## Page 144

EA ee so that £ =tan Differentiating this expression with respect to t we get a |p? + xp cos 6] ir? + p?+2rp cos 0 - (2)

a, (4 _ (8) ) F (? +rpc0s0) 7 +p? +2rp cos) If nis the mean daily motion of the Sun and n’ is that of the planet and a is a suitable constant, then 0=(n —n')(t +a) de so that ai" -n') wees (4)

## Page 145 PANCATARA SPASTADHIKARA (CHAP. 3) $133

ay |np? +n'r? +rp(n +n")oos0 dt [? +p? +2rpcos6] At the stationary point where the retrograde motion begins, 7

_(p? +n'r?) pr(ntn') so that cos0= For example, in the case of Kuja, considering the mean values, we have n =0°.98560265 , n’=0°.5240193 , p=233°.5, r=360°. Here, p and rare taken as peripheries of the planet’s Sighra epicycle and of the mean orbit, which are proportional to their radii. Substituting these values, we ar- rive at. _ (0.98560265 (233.5)° +0.5240193 (360)°] cos 8 = [233 .5) (360) (1.5096219 )] ~153737.271_+67912.901]__ =12650.17 _ gc gesgy = [126898.82 } * 126898 82 Therefore, @ = 180° — cos™! (0.9586391 ) = 163°.4636 The Saryasiddhanta has taken this value as 164° while GL takes it as 163°. The other stationary point is given by 360° - 6, noting that. ## Page 146

cos 0 = cos (360° - 8) In the above example, since 9 =164° according to the Saryasiddhanta, the second stationary point is 360° - 164° =196°. This means that Kuja will be retrograde during the peirod when its Sighrakendra (or Sighra anomaly) lies between 164° and 196°. Similarly, the corresponding lim- its for other planets can be calculated. Remark : According to modern astronomy, the stationary value of the angle 0 is given by cos 0 = where a is the mean distance of the planet from the Sun and b is the mean distance of the earth from the Sun. Now, taking b as one astro- nomical unit (a.u.) we have cos 0 = > E oy «| where a is the mean distance of the planet from the Sun, in astronomical unit (a.u.). Note : 1 astronomical unit (a.u.) = Earth’s mean distance from the Sun. The stationary points 9, given in Table 3.5, are those at which the re- ## Page 147 PANCATARA SPASTADHIKARA (CHAP. 3) $135 spective planets change their motion from direct to retrograde, i.e., the beginning of the retrograde motion (vakrarambha). The other stationary points, where the retrograde motion ends (vakratyaga), are given by (60° - 0). It is noteworthy that Bhaskara II gives the correct value @ = 167° for Sukra in his Karana kutahalam. For comparison, we tabulate the critical values of the Sighrakendras for the retrograde motion of the planets according to the different texts :

Planet|Sarya- | Brahma- Bhaskara II Graha- Modern Siddhanta gupta andLalla = laghavam \Kuja | 164° 164° 163° 163° 163°. 215 Budha | ; 44° 146° 145° 145° 144°.427 Guru | 130° 125° 125° 125° 125°.565

Sani | 115° 116° 113° 113° 114°.466 Note : Brahmagupta has given the stationary value @ for Guru as 125° in his Brahmasphuta siddhanta and as 130° in his Khandakhadyaka. Sloka 16 : This sloka tells the Sighrakendra of udayasta (rising and set- ting) of Kuja, Guru and Sani. When the Sighrakendra of Kuja, Guru and Sani becomes 28°, 14° and 17° respectively, these planets rise in the east. Subtracting the above Sighrakendra values from 360° we get the Sighrakendra of setting of these planets in the west respectively. That is Kuja, Guru and Sani, set in the west when their Sighrakendras become 33%, 346 and 343° re- spectively. ## Page 148

Sloka 17 : This Sloka gives the Sighrakendra of udayasta of Budha and Sukra. When the Sighrakendra of Budha and Sukra become 50° and 24° respec- tively they rise in the west and when their Sighrakendras become 155° and 177° they set in the west. Similarly, when Sighrakendras for Budha and Sukra become respectively 205° and 283° they rise in the east and for 310° and 336° they set in the east. Sloka 18 : The Sloka explains the number of days for the commencement of or the elapsed or balance of the phenomena of retrogression, rising and setting. (1) Find the Sighrakendra (SK) of the planet for the given day. (2) Find the difference between the above Sighrakendra and the presribed Sighrakendra of retrograde motion or of rising or setting of the concerned as desired. (3) This difference is multiplied by 2 for Kaja, divided by 3 for Budha, divided by 9 and added to itself (the difference) for Guru, multiplied by 10 and divided by 6 for Sukra and the difference divided by 1 (i.e., the difference itself) for Sani gives the gata or esya (elapsed or balance) days of retrogression or rising or setting as the case many be. (ii) If the current SK is greater than the prescribed SK for the particular phenomenon, then the planet will be undergoing the said phenonemon and the number of days elapsed since the commencement of the same is as explained above. Sloka 19 : Now, for Budha and Sukra the phenomena of retrogression, rising and setting and the number of days of the same are explained. Thirty two days after the setting in the east Budha rises in the west, 32 days later Budha becomes retrograde, vakri, 3 days after that sets in the west, 16 days later Budha rises in the east and becomes direct (margi) and 32 days after this the planet sets in the east. ## Page 149 PANCATARA SPASTADHIKARA (CHAP. 3) $137 Similarly, Sukra, 2 months after setting in the east rises in the west, 8 months later becomes retrograde and 34 month (i.e., about 22% days) after this sets in the west; after this /, month (7% days) later Sukra rises in the east. Further, 3, month (22 4 days) later becomes direct and 9 months after this, Sukra sets in the east. Thus, the cycles for Budha and Sukra repeat. Sloka 20 : Now, rising, retrogression and setting days for Kuja, Guru and Sani are explained. Koja rises (in the east) four months after setting (in the west). The planet becomes retrograde ten months after rising. Kuja resumes direct motion two months later and sets ten months after that. Guru rises one month after his setting. He becomes retrograde four-and- a-quarter months later. After four months of becoming retrograde, Guru becomes direct. Finally, he sets four-and-a-quarter months later. Sani rises, becomes retrograde, resumes direct motion and sets respec- tively after 54, 74, 94 and 74 months successively (i.e. rises 5{ months after setting etc.). ## Page 151 CHAPTER 4 TRIPRASNADHIKARA (Three Problems of Direction, Place and Time) This chapter deals with issues connected with direction, place and time (dik, dega and kala). The shadow of the gnomon (Sanku) occupies an important role in determining the altitude of the Sun, the direction, the time, Sun’s declination and the latitude of a place. In this chapter, be- sides these, the ascending point of the ecliptic at the eastern horizon (Lagna, Ascendant), the rising and setting of the Sun etc. are discussed. Sloka 1 : The durations of the rising of Mesa, Vrsabha and Mithuna at Lanka are respectively 278, 299 and 323 in vighatis, These figures writ- ten in the reverse order are the durations of risings of the next three ragis. That is, the durations of risings of Karkataka, Simha and Kanya are respectively 323, 299 and 278. The durations of the rising (udayamana) of Mesa, Vrsabha and Mithuna, diminished by the cara khandas are the durations of the ris- ings of these three ragis at one’s place [see Ch. 2, SI. 5]. The durations of the risings of Karkataka, Simha and Kanya in- creased by the carakhandas are the durations of the risings of these three rasis at one’s place. The durations of risings of these six rasis, that is from Mesa to Kanya written in the reverse order give the durations of the risings of the six rasis from Tula to Mina [see Table 4.1]. ## Page 152 si40 GRAHALAGHAVAM OF GANESA DAIVAJNA

Rasis Durations of risings Mesa 278 4] Vrsabha 299 Mithuna 323 Karkataka 323 Simha 299 Kanya 278 Tula 278 Vréscika 299 Dhanu 323 Makara 323 Kumbha 299 Mina 278 The sum of the durations of the risings (udayamanas) of the six rasis at a place is 1800 palas (vig.) The sum of the durations of the risings of all 12 rasis is 3600 palas {60 palas = 1 ghati]. The durations of the risings of raéis for Kaéi The palabha of Kaéi = 5|45 arigulas.

The durations of the risings of rasis are given in Table 4.2 : ## Page 153 TRIPRASNADHIKARA (CHAP. 4) a

Mesa 278 - 57 = 221 Mina Vrsabha 299 - 46 = 253 Kumbha Mithuna 323 - 19 = 304 Makara Karkataka 323 + 19 = 342 Dhanu Simha 299 + 46 = 345 Vrscika Kanya 278 + 57 = 335 Tula Note : According to the Siryasiddhanta the durations of the risings of rasis (starting from Mesa) are respectively 1670, 1795 and 1935 asus where 6 asus = 1 pala (vighati or vinadi) Therefore, 1670 asus = 278.33 palas 1795 asus = 299.17 palas 1935 asus = 322.5 palas Now, 1670 + 1795 + 1935 = 5400 asus = 900 palas According to the Grahalaghavam, we have 278 + 299 + 323 = 900 palas. Slokas 2 and 8 : These slokas explain the method of finding lagna (ascendant) at a given time. It is as follows : (i) Determine the sayana Ravi at the given time. (ii) Find the amSas to be covered (bhogyamSa) by dividing the true posi- tion (in degrees) of the Sun by 30 and subtracting the remainder from 30°. (iii) Multiply the above remainder (degrees) by the udayamana (duration of rising) of that particular rasi and divide the product by 30. The result gives the bhogya kala (the duration to be covered) in palas (vighatis) of that lagna. ## Page 154 s142 GRAHALAGHAVAM OF GANESA DAIVAJNA (iv) Express the given time (after sunrise) in palas and subtract the bhogya kala (in palas) from it. (v) Subtract the total duration (in palas) of the Suddha rasis (i.e. already completed lagna rasis) from the result of item (iv). (vi) Bhuktaméa (ellapsed portion) of lagna

* Udayamana of the current lagna rasi (vii) Adding the above bhuktaméas to the suddha raéis (i.e., completed rasis) we get sayana spasta lagna. Subtract the ayanaméa to get the nirayana Jagna. Example : Given time (Istakala) = 10%" 30” from sunrise at Kasi Ayanaméa = 18° 0! (i) Sdyana Sun at the given time = 54° 2' 40" 54° 2" 40" 24° 2' 40" Gi) Dividing by 80, = 1+ 30° i.e., quotient = 1 and remainder = 24° 2’ 40” The amfas to be covered = 30° - 24° 2'40" = 5°57’ 20" in the 2nd ragi i.e., Vrsabha. This implies that the Sun is in the Vrsabha rasi, with remainder ## Page 155 TRIPRASNADHIKARA (CHAP. 4) 8143 5° 57’ 20". At Kasi, the duration of rising of Vrsabha rasi = 253 palas. 5° 57’ 20" x 253 .. Bhogyakala = = 50.225 palas ~50 palas Now, the given time = 108 308 = 630 palas (after sunrise). (iii) The given time — bhogyakala = 630 - 50 = 580 palas. (iv) The duration of the rising udayamana of Mithuna = 304 palas. Now, 580 — 304 = 276 palas The duration of the rising of Karkataka = 342 palas. which cannot be subtracted from 276 (and hence the aguddha rasi). 276x30° _ 8280° 342342 We have = 24° 12'37" The number of elapsed rasis from Mesa = 3 Sayana lagna = 3° 424° 12'37" = 3% 24° 12'87" and hence nirayana lagna = Sayana lagna — Ayanamsa = 3 24° 12/37" - 18° 10' = 8% 6° 237" ie, Karkataka 6° 2'37". Sloka 4 : This Sloka explains the method of finding the lagna when the time to be covered is greater than the given time (i.e., if bhogyakala > istakala) and also gives the method of finding the time from Jagna. The procedure is as follows. ## Page 156 si44 GRAHALAGHAVAM OF GANESA DAIVAJNA (i) Find the true longitude of the (sayana) Sun at the given time. (ii) Find the time to be covered (bhogyakala) (as explained in Slokas 2 & 3). (iii) If the time to be covered is greater than the given time, multiply the given time by 30 and divide by the duration of the rising of the rasi in which the sayana Sun lies. (iv) Add the quotient obtained in step (iii) to the true (sayana) longitude of the Sun. This gives the sdyana lagna at the given time. Example : The given time = 0" 40". At this instant, we have

54° 02' 40" _ | , 24 02' 40" Ne a 30 30° Here, the quotient = 1 and the remainder = 24° 02' 40" The amfas to be covered = 30° - 24° 02' 40" = 5° 57’ 20" This implies that the Sun is in the Vrsabha rasi with remainder 5° 57’ 40". The duration of rising of the Vrsabha rasi is 253 palas. 5° 57’ 40" x 253 Now, 30° = 50.225185 palas ~ 50 palas ie, Bhogyakala = 50 palas. The given time (Istakala) = 40 palas. ## Page 157 ‘TRIPRASNADHIKARA (CHAP. 4) S145 Since Bhogyakala > Istakala, multiplying the given time by 30 and divid- ing by 253 we get 40 x 30° = 4° 44! 35" 253 Sayana Ravi is in Vrsabha rasi 5° 43'15” Sayana lagna = Sayana Ravi + 4° 44'35" = 17 5° 43'15"+ 4° 44'35" = 17 10° 27'50" Nirayana lagna = 40° 27'50"- 18° 10’ = 22°17'50" (ie., in Mesa) Finding the time from lagna : (i) Find the bhogyaméa of the sdyana Ravi. (ii) Find the covered portion of the lagna (i.e., bhuktakala of the lagna). (iii) Take the sum of the results of step (i) and step (ii). (iv) Consider the duration of rising of the rai that lies between the sayana Ravi and the sdyana lagna. Add this to the sum obtained in step (iii). This gives the required time in palas. Divide it by 60 to convert the time into ghatis ete. Example : Nirayana lagna = 3° 6° 937" Sayana lagna = 3° 24° 12'37" i.e., Sayana lagna is in Karka rasi with elapsed portion = 24° 12'37". The duration of rising (udayamdna) of Karka rasi = 342 vig. at Kasi. ## Page 158

. 24° 19’ 87" x 342 Now, the elapsed time = >" palas = 276 palas. Similarly the bhogyakala of Ravi = 50 palas. Now, 276+50=326 palas. We note that the sayana lagna is in Karka rasi and the sayana Ravi is in Vrsabha rasi. We have Mithuna ragi in between these two. The durations of the rising of Mithuna = 304 palas. Now, 326 + 304 = 630 palas i.e., 108" 30% i.e., the required time at the given lagna = 10%" 30% from the sunrise. Sloka 5 : The method of finding the time from Jagna when the sayana Ravi and sayana lagna are in the same raéi, is explained as follows : 1. When the Sun and lagna are in the same rasi, consider the difference between the Sun and the lagna. 2. Multiply the above difference by the duration of rising of the raéi in which both the Sun and the lagna are present. 3. Divide the product obtained in step (2) by 30. This gives the time in palas at the given lagna. Divide it by 60 to convert it into ghatis. Note : If lagna < the Sun, then subtract from 60 the time obtained in step (3). Example 1 : Séyana lagna = 1" 98° 37'50" Sayana Ravi = 1” 23° 53'15" (i) Lagna - Ravi = 4° 44'35" ## Page 159 TRIPRASNADHIKARA (CHAP. 4) $147 We note that Ravi and lagna both are in Vrsabha rasi. The duration of rising of Vrsabha = 253 palas. 4° 44’ 35" x 253 30° (ii) Now, palas = 40 palas. :, The time = 40 palas = 0&" 40%

The given time after sunrise = 59 ghatis. True longitude of Ravi at this instant = 1° 6° 39'07".

17 6° 39'07"+6" +18° 10 24° 49' 07" Bhogyakala = 59 palas (obtained as explained earlier) The given time = 59 ghatis; the duration of half day = 338" 10" Therefore, the given time from the sunset 598! _ 338h 10% 258) 59% 1550 palas Now, we have the given time from the sunset — Bhogyakala = (1550 — 59) palas = 1491 palas

## Page 160

S148 GRAHALAGHAVAM OF GANESA DAIVAJNA

Sayana Ravi = 17 24° 49'7" Here, Lagna < Ravi Ravi - Lagna = 7° 1'56" Given time = 59 ghatis Balance = 1 ghati Duration of a day = 60 ghatis -. 60 — Balance = 59 ghatis (of the previous day) ite., the time from sunrise = 59 ghatis. Note : (i) When the Sun and Lagna (both sayana) are in the same rasi,

the given time = 30 (ii) When Lagna < Sun, the Sun will be below the horizon. Then the given istakala, since it is before the next sunrise and after the sunset of the previous day, it should be subtracted from 60 ghatis of the previous day. Sloka 6 : This Sloka tells about the Ayanagola (hemisphere), dina ratri mana (duration of day and night) and aksamfa (latitude). (i) If the sayana planet is within 6 rasis from Mesa to Kanyé, (i.e., from 0° to 180°) then it is said to be in the uttaragola (northern hemisphere). (ii) If the sayana planet is within 6 rasis from Tula to Mina (i.e., from 180° to 360°) then it is said to be in the daksinagola (southern hemi- sphere). (iii) If the sayana Ravi is from Karka to Dhanu, then that period is called ## Page 161 TRIPRASNADHIKARA (CHAP. 4) S149 Daksinayana (southern course). If the séyana Ravi is from Makara to Mithuna, then that period is called uttarayana (northern course). In fact, the daksinayana of the Sun is from about June 22 to December 22 and the uttardyana is from December 22 to June 22 (approxly.). (iv) When Ravi is in the uttaragola, adding carapalas to 15 ghatis, we get dinardham (half of the duration of the day time i.e., from the sunrise to noon). When Ravi is in the daksinagola, subtracting the carapala from 15 gh. we get the dinardham. (v) Subtracting dinardham from 308", we get ratryardha (one half of the duration of the night). Multiplying dinardham or ratryardham by 2, we get dinamana (i.e., duration of the day-time) or ratrimana (i.e., duration of the night-time) respectively. th (vi) Multiply the palabha of one’s place by 5, and subtract + of the square of the palabha from the above product to get aksaméa (stitude) of one's place. (palabha)* ie., AksamSa = 5 x palabha- 7 Remark : Let p be the palabha and 4 be the latitude (aksamsa) of a place. Then from the right-angled triangle ABC (Fig. 4.1), we have A

## Page 162

tan = a where AB = 12 argulas, the height of the Sanku (gnomon). —tan-1 {2 p=tanr' (2)

oP _(/i2pP , =f 04 (in radian) 180| p __—p® =n [12 3x18 +7. (in degrees) 180 0) Jee)? mx 3x 12 where the second term is of order p3. In GL the second term viz. ao appears to be a small error. Thus, to a first approximation, the formula 5p for the latitude, is correct. Example : Consider palabha, p = 2°"% 46?"* = 2.7666 anig. 2 By the GL formula, aksa $ = 5p aT _ 50(2.7666) — (2.7666)" 10 = 13°.0679 = 13° 04' ## Page 163 TRIPRASNADHIKARA (CHAP. 4) SI5l In fact, the palabha and aksa refer to Bangalore.

When sayana Ravi is in the uttaragola, carapala = 93?%8 _ 18h 33? (i) Dinardham = 158" 4 carapala

" (ii) Dinamanam = Dinardhamx 2 = 164 33? x2 = 338" 06? (iii) Ratryardha = 308" — Dinardham = 308 _ 168 33? = 138 277 (iv) Ratrimanam — = 2x Ratryérdham 2x 138 27? = 268" 54? W (v) Palabha = aM. 54Pr. a2

45 5x G45) OF = 25° 26/42" 10 Sloka 7 : This Sloka explains the nata and the unnata. It also gives the method of finding aksakarna. ## Page 164 S152 GRAHALAGHAVAM OF GANESA DAIVAJNA When the given time is before the noon it is pirva kapala and when it is after the noon, pascimakapala. (i) Unnatakala : The time after sunrise in ghatis before noon. After the noon, the time remaining for the sunset is unnata (see Fig. 4.2). (ii) Natakala : The balance after subtracting the unnatakdala from the dinardham.

Fig. 4.2 Natakala and Unnatakala Note : Zenith distance = 90° — Altitude

Example : Suppose the given time = 10£* 30" from sunrise. Since the given time is before noon, it is parvakapala. Therefore, Unnatakala = 108 30'8 , Dinardham = 168 33°8 (given) Natakéla = Dinardham - Unnatakala = 168 33° 108 30% — = 68! 93% (iii) To find Akgakarna : alabha)? By adding 12 to —_, we get Aksakarna in arigulas. ## Page 165 TRIPRASNADHIKARA (CHAP. 4) ae 2 12+ (alabhay” ang. ie, Aksaki = ie sakarna 5

- Aksakarna = 2 Hae = 1308 19Pr

Fig. 4.3 Aksakarna (ak)? =12? + p* =12? 1+ (p12)? where ak = aksakarna and p = palabha 2 _ifi+i Boy _ 2 12? -. ak=12h1+(p/13]# 212+ p? / 24 GLhas taken p?/25 instead of p® / 24 with a small error. ## Page 166 sis4 GRAHALAGHAVAM OF GANESA DAIVAJNA Sloka 8 : This Sloka explains the method of finding hara and sama in order to get the hara (denominator). Then dividing the bhajya (numera- tor), explained in the next gloka, we obtain the palakarna (or aksakarna). From this the istacchaya is got from the corresponding right angled tri- angle. (1) To find hara : (i) Divide carapala by 5. (ii) If sayana Ravi is in the uttaragola, then add 114 to the quotient ob- tained in step (i). The sum is called hara. (iii) If sa4yana Ravi is in the daksinagola, then subtract the quotient ob- tained in step (i) from 114. This result gives hara. (2) To find sama : (i) Add ghatikardham (i.e., 30% ) to nata ; take the square of the sum. (ii) Divide the result of (i) by 2. The resulting quotient is called sama. (30 + nata)” ie, Sama= 2 (3) To find hara : Consider the difference between hara and sama. Divide this difference by Aksakarna. The result is called hara. ; _ hara — sama ie, bara = ~aksakarna Note : If nata (in gh.) is greater than 138" 30? then take nata — 13|30. Multiply this difference by 4. Subtract the product from sama obtained earlier in (2) to get true sama. ## Page 167

‘TRIPRASNADHIKARA (CHAP. 4)

Example : Cara = 93 palas.

93 (1) Hara = 114 + aE = 14+ = 132\36 (2) Nata = 6|3 ghatis (30% +nata)? — (0% +6|3%)? _ 2 7 2 7 Sama = 33)” 2 (3) Hara is given by : hara - sama hara = ~skeakarna 132|36— 2127 = 13]19 = 8|20 = 21|27 S155 Sloka 9 : This sloka explains the method of finding the bhajya (dividend) and the istacchéya (from the numerator and the divisor), the shadow of the gnomon at the given instant. (1) To find bhajya : (i) Divide cara by 10 x palabha. Consider the square of the resulting quotient and multiply the square by 2. (ii) Add 1/5 of the result obtained in step (i) to itself. (iii) Add 114 to the sum obtained in step (ii). This result is called bhajya (dividend or numerator). ## Page 168

( cara J 2 OnRraa lara cara } 10 x palabha am ie, Bhajya =12 es an (a ‘palabha 3 2 = Ox2f—F8@_} sing 5” ” 10x palabha (2) To find istacchaya : (i) Divide bhajya obtained as above by hara. The resulting quotient is called istakarna or palakarna. (ii) Subtract 12? from the square of palakarna. (iii) Take the square root of the above difference. The result obtained is called istacchaya.

Example: Palabha = 5|45 argulas

2 6 cara i) bhajya = ~x2| 7? | +114

## Page 169 TRIPRASNADHIKARA (CHAP. 4) $137 2 6 93 = +x 2)/——_| +114. =5* far r20)16 (ii) hara = 8|20 (obtained in the previous Sloka) «. istacchaya = ( 2 120|16 = eal ~ 144 = 759|22 angulas Sloka 10 : Finding aksakarna from istacchdya and hence nata is explained in this Sloka as follows. (i) Add 12° to (chaya)®. Take the square root of this sum. The result ob- tained gives aksakarna. ie, aksakarna = ,/12 +(chay: (ii) Divide bhajya (obtained earlier) by the above aksakarna. The quo- tient gives ista hara. bhajya bhaj aksakarna ~ .[(chaya)? +12? (iii) Multiply istahara by akgakarna. Subtract the product from madhyahara (i.e., hara, obtained earlier). The result gives sama. ie, ista hara = ie., sama = madhyahara ~ (istahara x karna) (iv) Multiply the sama by 2, and take the square root. This gives nata. ## Page 170

Subtract 14 ghatf (.e., 30 palas) to get true nata. ie, nata= 2xsama True nata = (2 x sama) —30 palas. Note : If 2x sama is greater than 194 (ie., if nata is greater than 13|30) then, nata = Pxsama = 194 + (2x sama) True nata = nata — ¥% ghati = nata — 30 palas

(i) Istacchaya = 7|59|22 angulas akgakarna = \(chaya)? +12? = ¥(7|59|22)? +12? = 14|25 angulas = Karna (ii) Bhajya = 120|16 Bhajya 120|16 Karna ~ 14\46 = 8120 istahara = ## Page 171 TRIPRASNADHIKARA (CHAP. 4) 8159 (iii) Madhyahara = hara = 182|36 and akgakarna = 1319 sama = madhyahara —(istahara x aksakarna) = 132|36-(8|20x13|19) = 21/33 (iv) nata = 2x sama = 2x (21|33) = 6|33 gh. Subtracting % ghati (30 palas) we get True nata = (6|33- 30) palas = 6|03 ghatis. Example : When given nata > 1330 ghatis, finding istacchaya and con- versely finding the nata. Suppose nata = 15|10 ghatis Adding 14 ghatti.e., 30 palas, we get 15{108" +300 = 15|40 ghatis _ 5)40)? oe sama = 122|48 nata — 13/30 = 15|10-13|30=1|40 Spasta sama = sama — (4x1|40) = 122|43-6|40 = 116|3 hara = 18236 and aksakarna = 13|19 argulas (obtained earlier) ## Page 172

hara - spasta sama hara = aksakarna 182 |36 - 116|3 2 arte = aja We have, bhajya = 120|16 : bhajya — 120/16 ; istakarna = hara 7 ii = 97|29 arigulas istacchaya = (97|29)* 12” = 96|44|30 arigulas

Conversely, finding nata :

Karna = (chaya )? + 12? = y(96|44|30)* +12? = 97|29 ang. Bhajya = 120|16 (given) bhajya 120|16 hara= lama = “oTp9 = aja palakarna (aksakarna) = 13|19 arig. madhyahara (hara) = 132|36 Sama = madhyahara — {(palakarnax hara)] ## Page 173 ‘TRIPRASNADHIKARA (CHAP. 4) S161 = 13236 - [(3|19) x aj14)] = 1216)11 Now, 2 sama = 2x (116|11) = 232|22 > 194 Since 2 x sama >194 we proceed as folows : (2x sama) ~ 194 nata = [ F] +(2x sama) le % 232/22 194 V [es + 232/22 = 15/40 gh. True nata = 15/40 gh. — 30 palas = 1510 ghatis. Sloka 11 : This Sloka explains the method of finding kranti (declination) by using kranti khandas. It is as follows : The nine Krantikhandas are 40, 40, 37, 34, 30, 25, 18, 12, 4. (i) Divide the bhujaméa of séyana Ravi by10.The quotient gives the num- ber of elapsed khandas called gatakhandas. (ii) Multiply the remainder obtained in step (i) by the khanda to be covered (i.e., esyakhanda). Divide the product by 10. (iii) Add the quotient obtained from step (ii) to the sum of all elapsed khandas. (iv) Divide the above sum by 10. The result gives the kranti (declination). Sumof ie, Kranti = | €atakhandas + Remainder Esyakhanda 10 10 (see Table 4.3)

## Page 174 s162 GRAHALAGHAVAM OF GANESA DAIVAJNA Ayandméa = 18° 10' Sayana Ravi = 1 24° 9' 41" = 54° 2/41" Bhujaméa of sayana Ravi = 54° 2' 41" Bhujamsa 54° 2'41" 4° aay" Now, 7 = = 5+ 10 10° 10° remainder = 4° 2’ 41", the quotient = 5 This implies that the number of gatakhandas is 5. ite. the khandas which are over are 40, 40, 37, 34 and 30. Esyakhanda, the khanda to be covered = 25 Esyakhanda x Remainder 25° x 4° 2' 41" 10° - 10° = 10° 6' 42" Sum of the gatakhandas = 40 + 40 + 37 + 34 + 30 = 181 ‘Sum of khand Esyakhanda 10 (181° + 10° 6’ 42") Se Se eae 10 Note : For finding the declination kranti of a heavenly body, the argu- ment (2) is the sayana longitude of the body. For intervals of 10° of 2, the Grahalaghavam gives the declinations (6) as in Table 4.3. ## Page 175 TRIPRASNADHIKARA (CHAP. 4) oe

a 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 6 0 40 80 117 151 181 206 224 236 240 Difference 40 40 387 34 30 25 18 12 4 The modern formula is sinéd=sine sind. for a body on the ecliptic (latitude B= 0). Now, if §, « and A are in degrees, converting them into radians, we have for , =90°, sin ( 5max.y «| = sin (5% 180 180

Binary gq a OXE 180 180 ie, 8max * © * 24° sin 8 = 24° sin 4 for small 8. Here, is the séyana longitude, § the declination and ¢ the obliquity of the ecliptic. Ganeéa Daivajfia has multiplied the RHS by 10 (to avoid fractions) and then the final value is divided by 10. Thus his expression is § = 240 sin 1. (to be divided by 10 later). Of course, for sinA he has used an approximation formula. Sloka 12 : This Sloka gives a method for finding approximate declination (sthala kranti) for easy computation by using laghu khandas. It is as follows. ## Page 176

At intervals of 15°, the six kranti khandas known as laghu khandas are given as 6, 6, 5, 4, 2, 1. (i) Divide the bhujéméa of sayana Ravi by 15. The quotient gives the number of gatakhandas. (ii) Consider the product of the remainder obtained in step (i) and the esyakhanda. Divide this product by 15. (iii) Add the sum of all the gatakhandas to the above quotient. This sum is the sthila kranti.

ie, Kranti =| gatakhayda + 15 Example: Sayana Ravi = 17 24° 2' 41", bhujamsa = 54° 241" bhujaméa 54° 2' 41" 9° 2! 41" FF AY 2 3+25- 15 15) 15) Here, quotient = 3 and remainder = 9° 2' 41” This implies, that the number of gatakhandas is 3. They are 6, 6 and 5. esyakhanda = 4. Sum of

15 24x 4" = (646 45°) + ( ) is = 17° + 2°24'43" = 19° 24’ 43" ## Page 177 TRIPRASNADHIKARA (CHAP. 4) S165

The formula for finding declination is given by §=sin~! (sine sind) for the Sun. If ) =54° 2’ 41" and ¢=24° we get 8=sin™ (sin24’ sin54° 2' 41") = 19° 18’ 22" The declination obtained using krantikhandas is 19° 06’ 40". The approximate declination obtained using laghukhandas is 19° 24' 43". Sloka 13 : This Sloka explains the method of finding the bhujéméa of Ravi when kranti is known by reverse process as follows. (i) Subtract the sum of elapsed khandas from sthilakranti (kranti ob- tained using laghukhandas). (ii) Multiply the above difference by 15. ii) Divide the above product by esyakhanda. (iv) Take the product of the number of elapsed khandas and 15. (v) Add the result of step (iii) to that of step (iv). This sum gives the bhujaméa of Ravi. That is bhujamsa (ee = Sum of )« i: = ___gset Mhapds (no. of elapsed khandas x15) isyakhanda Example : Kranti obtained using laghukhandas, sthalakranti = 19° 24' 43" and the number of the elapsed khandas = 3 Sum of the elapsed khandas =6+6+5=17

(19° 24' 43"-17)x 15

4 + (3x15) = 54° 2' 41" ## Page 178

Sloka 14 : The method of finding krnti by knowing only the dina mana (duration of day time) without the knowledge of the position of Ravi is ex- plained as follows. (i) Subtract 15 ghatis from dinardham (in gh.). Multiply this difference by 60 to get cara (in palas). (ii) Add ¥/, of carapala to itself, and divide the sum by palabha. (iii) Add 25 kalas to the resulting quotient. This gives the krantiof the sun. ( curses carapala+ ————— ie, Kranti= +25 kalas palabha a 2x Tapa! 95 Kalas 8° palabha Note : (i) If dinardham > 158" , then the krantiis north. (ii) If dinardham < 158" , then the kranti is south. Example: Dinardham = 168" 338 carapala = (dinardham — 158") x 60 = (168 33% — 158) x60 = 98 palas Palabha = 5|45 arigulas (given) 9 carapala ——— +25’ palabha ~ Kranti= ## Page 179 TRIPRASNADHIKARA (CHAP. 4) S167

Since dinardham = 168" 33" > 158, the krantiis north. Sloka 15 : The methods of finding natamga and unnataméa using dinardham and kranti, without using kranti khandas are explained in this Sloka as follows. (1) Determine kranti and aksaméa. Then nataméa = kranti + aksaméa= 8+ $, the zenith distance. Here, for northern latitude , the negative sign and for the southern 9, positive sign are taken. In other words, natamsa = 8-6 where $ is +” or ~* according as the place is in the northern or southern hemisphere. Similarly, the declination may also be positive or negative. (2) Unnatamsa = [90° — nataméal, the altitude. (3) To find kranti without using krantikhandas : (i) Find the bhujaméa of sayana Ravi. Divide it by 10. (ii) Subtract the result of step (i) from 18°. Multiply the difference by the result of step (i). ii) Divide the result of step (ii) by 72. (iv) Subtract the result of step (iii) from 4° 30’. (v) Divide the result of step (ii) by the result of step (iv). The quotient obtained gives the kranti (in degrees). ## Page 180

10 10 i( = hele x [bhalames }} a3¢-4 te 0 72 (1s 7 inane bhujaméa ie, Kranti= We shall express the above formula in a simpler form :

Let 10 - Then Kranti = is =e degrees. yep _las’ 2a] pR Example : (1) Bhujaéméa of sayana Ravi = 54° 2' 41"

10 10 oor gq oor aq ey 1g° 54° 2.41") 54° 241 10 10 4° 30'- 72 [w-% zit) gar

(2) Kranti 5 = 19° 06" 40" north AksdmSa $ = 25° 26’ 42" in the northern hemisphere Nataméa = Kranti-aksiméa =8 - = 19° 6' 40" - 25° 26' 42" = 6° 20’ 2” south. ## Page 181 TRIPRASNADHIKARA (CHAP. 4) $169 (3) Unnataméa = 90° - nataméa = 90° 6° 20'2" = 83° 3958" To find parakhya of nataméa : (i) Divide nataméa by 6. Add the quotient to the square of itself. Divide the resulting sum by 2. (ii) Subtract the result of step (i) from 114. The resulting difference is called parakhya. (zeta) al (nasal 6 6 ie, pardkhya = 114- 5

». parakhya = 14-~——— = 119° 5455"

The declination (kranti) obtained by using laghukhandas of unnatam$a is known as para (or parakhya) of unnataméa. Example : We have unnataméa = 83° 39' 58"; its bhujamSa = 83° 39'58”" 83° 39' 58" 8° 39' 58" ——_ = 5+$ —————_ Now, ——— : 15) 15) Here, quotient = 5 and remainder = 83° 39' 58”. ## Page 182

S170 GRAHALAGHAVAM OF GANESA DAIVAJNA

the number of gatakhandas = 5. They are 6, 6, 5, 4 and 2. Esyakhanda = 1. The sum of the gatakhandas = 6+6+5+4+2=23°. Therefore, para i.e., the kranti of the unnataméa is given by Sumof —_, Eyvathanda xremainder

1° x 8° 39'58" 4h ee 15° = 23° i.e., pard of unnataméa = 23° 84’ 39° To find yantrabhaga from nata : (i) Add , ghati to the nata. Multiply the sum by 15 and divide the product by dinardham. (ii) Multiply the square of the result obtained in step (i) by parakhya of nataméa. (iii) Divide the result of step (ii) by 114 and add 228 to the resulting quo- tient. (iv) Subtract from the above result, the parakhya multiplied by 2 and con- sider the square root of the difference obtained. (v) Multiply the result of step (iv) by 6 and subtract 3 from this product. The result gives yantrabhaga.

and since '/, ghati= 30 palas, then , (q2mm18).» 2 + 228>-2p|x6-3 in degrees. 114 a :

## Page 183 TRIPRASNADHIKARA (CHAP. 4) ol Example: Nata n= 6" 03" (obtained in Sloka 10) dinardham d= 168" 33° parakhya of nataméa p= 112° 54'55" ¥, ghati = 30 palas 2 2 (a8) + 228; - 2p|x6-3 114

% 2 epee) x15] x112°54'55" 16/33 + 228) - 2x 112° 54'55"|x 6-3 114 = 33° 32’ To find nata using yantrabhaga : (i) Add 8 to the yantrabhaga, and divide the sum by 6. Take the square of the result. (ii) Add parakhya multiplied by 2 to the above square. (iii) Subtract 228 from the result of step (ii). (iv) Multiply the difference obtained in step (iii) by 114 and divide the prod- ## Page 184

uct by parakhya (of nataméa). (v) Consider the square root of the above result. Multiply it by dinardham. (vi) Divide the result of step (v) by 15 and subtract 30 palas from it. This gives nata (in ghatis). ie.,if p= parakhya, d=dinaradham and y= yantrabhaga, then Y 2 2 \ (2) +p) 8 x14 : d nata= 5 or -30 Example : Suppose Yantrabhaga = 33|32|0, parakhya p = 112° 54’55" and dinardham d= 16|33 gh.

33|32+30)° ys Ae) 42x 112" 54’ 55" 298 |x 114 : 16/33 : ef 112° 54’ 55” 15 nata = 0 = 68h ggvis Sloka 16 : This sloka gives a different method for finding istakarna (aksakarna or palakarna) from unnatam (in gh.) at a given time. It is as follows : ## Page 185 TRIPRASNADHIKARA (CHAP. 4) one) (i) Consider the product of unnatam at a given time and 90°. Divide this product by dinardham. (ii) Considering the bhujéméa of the quotient obtained in step (i), find kranti using laghukhandas. (iii) Multiply the kranti by parakhya (or para which refers to the diclination of unnataméa obtained using laghukhandas) (iv) Divide the number 6912 by the result of step (iii). This gives the aksakarna or istakarna in angulas.

(i) Multiplying unnatam by 90 and dividing by dinardham we get 0|30) x90 00/80) x90 _ 945 _ soe 5597 dinardham 16|33 (ii) bhujaméa of the above = 57° 5’ 59"; (iii) kranti of this, using laghukhandas,is given by Kranti = 20° 13" 35’ (obtained as explained in Sloka 13) (iv) parakhya = 23° 34' 39" (obtained earlier) Now, parakhya x krénti= 23° 24' 39" x 20° 13’ 35" = 476|53|12 6912 (v) palakarna = 476)53}12 = 14|29 anigulas Sloka 17 : The inverse process of finding unnatam from palakarna is ex- plained as follows. (i) Divide the number 6912 by palakarna ## Page 186 si74 GRAHALAGHAVAM OF GANESA DAIVAJNA (ii) Divide the result of step (i) by parakhya (declination of unnataméa) The result gives kranti. (iii) Find bhujaméa of the kranti obtained above as given by Sloka 13. (iv) Multiply the bhujéméa by dinardham and divide by 90. The result gives unnatam. dinardham x bhujaméa of kranti Le, unnatam = 90 6912/ ) where kranti= (*'?patakaroa ) parakhya Example: Dinardham = 16|33 gh. and Palakarna= 14|29 ang. Parakhya = 23° 34'39" (0% Kranti = 4)29) _ og aa 7" 23° 34' 39" The bhujaméa for the above kranti = 57° 9'15" ( obtained by inverse pro- cess) 16|33 x 57° 9" 15" 90° Unnatam = = 108" gore Sloka 18 : The method of finding unnatakala when the unnat&méa ob- tained from yantrabhaga is known is explained as given below. (i) Find the declination (kranti) of unnataméa obtained from yantrabhaga. Multiply this kranti by 24 and divide by parakhya (declination of unnataméa). ## Page 187 TRIPRASNADHIKARA (CHAP. 4) S175 (ii) Find the bhujaméa of the above result and multiply it by dinardham and divide the product by 90. The result gives unnatakala. For the parva kapala (eastern hemisphere) the unnatakala obtained will be the elapsed ghatis (gataghati) and for the pascima kapala (western hemi- sphere) the unnatakala is the remaining ghatis to be covered. Example: — yantrabhaga = 55/45|48 declination of the above = 19° 52/13" parakhya = 23° 34'39" declinatio nx 24 @9|52{18)x 24 : oe parakhya =~ 23)34|39 = 20° 13’ 25' bhujaméa from the above krdnti= 57° 05' 52" bhujaméa xdinardham _ 67|05|52) x(16|33) Now, m7 = 35 unnatakala = 108" 39" Since we have obtained unnatakala in parva kapéla, it gives the elapsed ghatis. Sloka 19 : This Sloka explains the inverse process of finding yantrabhaga using unnataghati. It is explained below. (i) Multiply unnataghaff (i.e., unnata in ghatis) by 90°, and divide it by dinardham. (ii) Determine the kranti of the above quotient using laghukhandas. (iii) Multiply the kranti and parakhya and divide by 24. ## Page 188

(iv) The bhujamé&a for the above result is called yantrabhaga. Example: Unnatam = 102" 308 , dinardham = 168" 33% unnatam x 90 108 30% x 90° ” dinardham 168. g3Me = 57° 5'58" (ii) bhuja of the above = 57° 5’58” kranti (of this using laghukhandas) = 20° 13’ 35" (iii) parakhya = 23° 34' 39” parakhya x kranti _ 23° 34' 39" x 20° 13' 35" 24 7 24 = 19° 52'13" (iv) bhujéméa for the result of (iii) [See Sloka 13] yantrabhaga = 55° 45' 48” Sloka 20 : The method of finding istakarna from yantrabhaga and the in- verse process of finding yantrabhaga from istakarna is explained in this

(i) Find the declination of the given yantrabhaga. Divide 288 by the decli- nation of yantrabhaga obtained above. The result gives the istakarna. 288 (ii) Divide 288 by the given istakarna. Find bhujaméa of the resulting quo- tient. This gives yantrabhaga. 288 ie, yantrabhaga = bhujaméa of (a) ## Page 189

‘TRIPRASNADHIKARA (CHAP. 4) S17

(i) given yantrabhaga = 55|45|48 Kranti (declination) of the yantrabhaga = 19° 52’ 12" 288 istakarna = Jo 59 19° = 19 54 38 arigulas

288

bhujaméa for (19|52|12) = 55|45| 48 ie, yantrabhaga = 55|45| 48 Remark : Here the maximum declination (parama kranti) is 24° . Therefore, 24x12

288 and Istakarna= Yantrabhaga Sloka 21 : Now, determination of directions is explained. On an even level ground a circle is drawn. The sanku (gnomon) is fixed at the centre. The line joining the entry and exit points of the shadow (on the circle) determines the west and the east directions respectively. The perpendicular line, through the centre, to the east-west line provides the north and south directions. ## Page 190 S178 GRAHALAGHAVAM OF GANESA DAIVAJNA Sloka 22 : A different method of finding bhuja in order to know the direc- tion (dik) is explained as follows. (i) Find the declination of the Sun at the given time and multiply it by palakarna. (ii) Multiply the above product by chayakarna. (iii) Divide the product obtained in step (ii) by 350, to get bhuja in the direc- tion of the Sun. (iv) The above bhuja is added to or subtracted from the product 2 x palabha. This gives the corrected bhuja (spasta bhuja). Note : In step (iv) take the sum if palabha and bhuja are in the same direc- tion and the difference if they are in different directions. Example: Given time from sunrise = 108” 30%" Sayana Sun at the given time = 17 5° 59! 41" Declination (kranti) of the Sun = 19° 6’ 40" Palakarna = 13/19 angulas Chayakarna = 14/25 angulas We have (Kranti x Palakarna) x Chayakarna = (19]6|40) x (18|19)x (14)25) = 3668|59|8 : . _ 3668|59|8 ie, bhuja = —————— 350 = 10/28 palabha = 5|45 arigulas (given) ## Page 191 TRIPRASNADHIKARA (CHAP. 4) $179 Since saéyana Ravi and palabhd are in different directions, we have spasta bhuja = (2 x palabha)- bhuja = (2x 5]45) - 10/28 = 102 (south) Sloka 28 : A different method of finding digaméa in order to find dik is explained as follows. (i) Multiply the difference between dinamana and 30 by 11. This result is positive or negative according as dinamana is greater than or less than 30 ghatis respectively. (ii) Find the kranti (declination) of yantrabhaga using laghukhandas. This declination is always negative (south). (iii) If the results of step (i) and step (ii) have the same sign, consider their sum ; if they are in different signs take the difference between them. (iv) Multiply the result obtained in step (iii) by 8, and divide this product by the declination of the difference between 90° and yantrabhaga. (v) The bhujaméa of the above result is called digaméa. i.e.,digamga = Bhujéméa of [(dinamana - 30)x11+ declination of yantrabhaga]x8 declination of (90° - yantrabhaga)

(® {(dinamana) ~ 30} «11 = (33%{6"** — 30% )xc11 4" 3\6x 11 +34|6 ## Page 192

S180 GRAHALAGHAVAM OF GANESA DAIVAJNA

(ii) yantrabhaga = 55° 45' 48" (given) declination of the above = - 19° 5213" which is always taken south. Gii) Now, 34|6 -19|52|13 = 14/13/47 since the results of (i) and (iii) are of opposite signs, (iv) declination of (90° - 55|45|48) 5 13|24|44 (14|13|47) x 8 and "y3)24]44 = 8/2915. (v) DigamSa bhujéméa of 8|29|15 = 21° 18" (as explained in Sloka 13). Sloka 24 : This Sloka explains the method of finding dik (directions) from digaméa. Tariya yantra is an instrument with a circular base, calibrated in degrees, and two mutually perpendicular strips (indicating the east-west and north-the south lines). For fixing the (four) directions, circular base of this instrument is placed horizontally on an even ground surface. Then, with the foot of the sariku (gnomon) coinciding with the centre of the tariya yantra, measure an angle equivalent to the digamSa (obtained from earlier Sloka) from the shadow line to the gariku. Thus, the line at that angle will be the actual east-west line. The direction perpendicular to it will be the north-south direction. Sloka 25 : For the fixing of nalika (tube) instrument finding the bhuja and kotiis explained. For the planet to be observed, its true declination (spasta kranti) has to be determined. The true declination is multiplied by the ista karna and pala karna (i.e., aksa karna). The product is divided by 700. The result is the required bhuja in the direction of the kranti which should be cor- rected with the southern palabha. This gives the spasta bhuja. Subtract ## Page 193 TRIPRASNADHIKARA (CHAP. 4) S181 the square of the spasta bhuja from the square of the ista chaya and take the of square-root of this difference. This gives the spasta koti. Example 1 : Samvat 1669, Saka 1534, Vaisakha Sukla Paurnima (15" tithi), Monday at 57 ghatis from the sunrise, the fixing of the nalika for observing Kuja is explained. The place of observation is Kasi.

true Sun is 1” 6° 37'12" and true Kuja is 11 6° 37' 29". Karna = 11|48|40, Kranti= 23° 44’ 59" (south), Sara(south) = 46}14|34 ang. Subtracting 3° (ie., 90°) from true Kuja, we get 8% 6° 3729". For this, the Kranti 8 = 23° 47' 29" (south). Aksdméa (latitude) of kasig = 25° 26’ 42" (north). Nataméa = 8 - 9 = 49° 14' 11" (south) Drkkarma = 118'44". The drkkarma corrected Kuja = 117 8° 3613". For this position of Kuja, the Kranti= 1{17|30 Sara corrected Krantii.e., spasta kranti = 3|1|33 Istaghati = 57 gh., dinamanam = 33|10 gh. Ravibhogyakala = 59 gh., Lagna= 0|15|23| 21

dinagatakala of Kuja = 4|29 gh. ## Page 194

$182 GRAHALAGHAVAM OF GANESA DAIVAJNA

Cara of the corrected Kuja = 6

Corrected cara = 14.

Dinamanam = 29|32 gh.

Unnataméa = 61° 31’ 45" (complement of NatamSa) Parakhya = 21|12|14

Multiplying this by 90 and dividing by dinardham (i.e., 14|46 gh.) we get phala = 27)19| 37 degrees. Kranti of the above phala = 10° 42' 36". Multiplying this kranti by parakhya, we get 227) 6912, we have 5|37. Dividing this by Istakarna =30|26 Aksakarna = 13|19 (Aksa karna x (Istakarna )x kranti

= (13|19) (30|26) (3|1|33) = 1/45 an, oa |45 angulas Since the kranti is south, the bhuja is also south. Palabha =5|45 angulas. ## Page 195 TRIPRASNADHIKARA (CHAP. 4) $183 Palabha corrected spasta bhuja = 1|45+ 5|45=7|30 arigulas. Squaring the spasta bhuja, we get 56/15. Ista chaya = ,/( = ,/926j11-144 = 27|58 arigulas (Chaya? = 782|11

= 26|56 arigulas Sloka 26 : Now, fixing of the nalika is explained. Knowing all the directions on an even level ground, according as the planet is in western or eastern hemisphere, the angle equal to the koti is mea- sured westward or eastward from the east-west line. From this, equal to the bhuja, towards the north or south, the length is measured. The line joining the end points of the karna and bhuja gives the chaya. (This is the hypotenuse). The line joining end of the chaya and the top of the gariku is the direction for the nalika yantra to observe the planet in the sky. The line joining the top of the Sariku, fixed at the central point, and the end of the chaya (shadow) gives the direction for fixing the nalika to observe the planet’s reflection in water. ## Page 196 CHAPTER 5 CANDRAGRAHANADHIKARAH (Lunar Eclipse) Sloka 1 : This loka explains the method of finding the true longitude of planets at a given time (istakéla). In this case we consider the bodies, Ravi, Candra and Rahu. (® Multiply the motion of the planet which is expressed in kalas by gata ghati or esyaghati (as the case is) and divide by 60. (ii) Correspondingly add (or subtract) the result of (i) to (or from) the mean longitude. This gives the mean longitude of the heavenly body. Note : While finding the mean position of a body, if the given time (istkala) is after the sunrise, then the gata (elapsed) time in ghatis must be used in (i) above and added in (ii). On the otherhand, if the given time is before the sunrise, then the esyaghati (to be covered) must be subtracted in item (ii). From the mean positions of the Sun and the Moon, applying the prescribed equations, the true positions are determined. From these the instant of true full-moon (gukla parvanta) is found out as follows : Subtract the true longitude of the Moon from that of the Sun, divide the difference by the difference between the true daily motions of the Moon and the Sun and then multiply the quotient by 60 to get the ghatis to be covered for the gukla parvanta from the istakala. That is, if S and M are the true longitudes of the Sun and the Moon at time ty gh. (istakala) and DS and DM are their true daily motions then the instant of the (Sukla) parvanta is given by ## Page 197

S+6%_M E +s * «| ghatis. At the instant of the parvanta, the true longitudes of the Sun, the Moon and the ascending node (Rahu) of the Moon are determined. Example : Vikrama Samvat 1677, Salivahana Saka 1542 Margasira Sukla Parnima Wednesday. Cakra = 9, Ahargana = 636.

At the sunrise, we have Mean Sun = 8* 0° 8'59" Mean Moon = 1° 95° 1957" Mean Rahu = 7 28° 25' 27" According to that year’s Paficariga, the end of Purnima tithiis at 38|11 gh. (after the sunrise). The true postions of the heavenly bodies at that instant are

True Sun = 8 9° 9'96" =S

True daily motion of the Sun, DS = 61'11" True daily motion of the Moon, DM = 824’ 05" Therefore, the instant of parvanta tp is given by ## Page 198 CANDRAGRAHANADHIKARAH (CHAP. 5) S187 where f, =38|11 gh. (after the sunrise) 33' 20" ty = 38/11+ x60 =40|48 gh. (after sunrise). 762' 54" ie. 28% 37%8 after the istakala, 38|11 gh. Now, at the instant of Parvanta, we have true Sun = 8* 0° 12' 06" true Moon = 2” 9° 19'01" and Rahu = 7 28° 23/18" Sloka 2 : This sloka explains the method of finding Sara (latitude) of the Moon and the possibility of an eclipse. (i) Find the true longitudes of the Sun, the Moon and Rahu at the Parvanta. (ii) Consider (Sun ~ Rahu). This result is called virahvarka. If this is less than 14° then there is a possibility of an eclipse. If this is greater than 14°, then there is no chance of an eclipse. (iii) Divide the above difference (virahvarka) by 7 and multiply it by 11. The result obtained is known as Sara (latitude) of the Moon which will be in angulas. The direction of Sara is the same as that of virahvarka. (i.e., if virahvarka is less than 180°, Sara is the +¥¢ and otherwise —°) . 5 pe : ie, Sara = > x (Sun ~ Rahu)

120sin0 = ue 8 35 ## Page 199

Let S and R be the true longitudes of the Sun and Rahu. Then we know that latitude B = 270’ sin (S- R) 270 x 72 W « oS (S-R) x Ug- = 120x38 ¢ ) 7 (S-R) Example: At the parvanta, we have True Sun = 8% 0° 12'6" and Rahu = 7° 98° 93'18" Virdhvarka = Sun — Rahu = 0F 1° 48' 48"

Bhuja of virahvarka = 1° 48' 48" <14°.

Now, we have y ll Ol oar Ea Sara = = (Sun — Rahu) = ol (1 48’ 48") = 2|50 arigulas Since virahvarka is in the uttaragola (ie., less than 180°), the gara is also in the uttaragola and hence positive. Sloka 3 : This sloka gives the methods of finding Saryabimbam (Sun’s diameter), Candrabimbam (Moon's diameter) and bhichayabimbam (diameter of the earth’s shadow). (i) Sun’s (angular) diameter, DS-55 Suryabimbam = |——z—— + 10] angulas where DS is the Sun’s true daily motion in kalas. ## Page 200 CANDRAGRAHANADHIKARAH (CHAP. 5) 5189 (ii) Moon’s (angular) diameter, DM Candrabimbam = —_- arigulas where DM is the true daily motion of the moon in kalas. (iii) The angular diameter of the earth’s shadow cone, Bhachayabimbam = (z x Moon's diameter) + (3x Moon's diameter) - | ang. According to the commentator Visvanatha, the expressions for the angular diameters of the Sun, the Moon and the earth’s shadow cone are as follows : 2DS (1) The Sun’s diameter, ds “TT anigulas DM (2) The Moon’s diameter, 2m 7a angulas (3) Diameter of the earth’s shadow cone d,= [Pugs 4+32- 3

Therefore, we have the diameters of the Sun, the Moon and the earth’s shadow given by ## Page 201

(1) ds =11|7 angulas, (2) dm =11|8 angulas and (3) d, =28|10 arigulas. Sloka 4 : This Sloka explains manaikya khanda and grasa. In the case of a lunar eclipse the earth’s shadow is called chadaka (the eclipser) and the Moon is called chadya (the eclipsed).

Manaikya khanda = ; (Chadaka bimbam + Chadya bimbam] [shadow diameter + Moon’s diameter] Nie Also, grasa (the covered portion) is given by grasa = Manaikya khanda - Sara and khagrasa is given by khagrasa = grasa — Moon's diameter. Note : ( If it is not possible to subtract Sara from manaikya khanda (i.e., when Sara > manaikyakhanda), then there will be no eclipse. (ii) If the grasa is greater than the chadya bimbam (ie Moon’s diameter) then the eclipse is total.

Example : Suppose the earth shadow’s diameter

the moon’s diameter, Chadya bimbam = 11|8 angulas. Then 28/10 +11|8 (i) Manaikya khanda = |———5 ~~ | = 19|39 angulas ## Page 202 CANDRAGRAHANADHIKARAH (CHAP. 5) S191 Sara = 2|50 angulas (ii) Grasa = Manaikya khanda - gara = 19/39 - 2|50 = 16|49 arigulas (iii) Khagrasa = grasa ~ Moon’s diameter = 16/49 - 11|7 = 5/42 angulas In this case, grasa > Moon’s diameter. Therefore the lunar eclipse is total. Sloka 5 : This Sloka explains the method of finding sthiti (the half duration of the eclipse) and marda (the half duration of totality). It is as given below. To find sthiti : (i) Add gara to the manaikya khanda and multiply the sum by 10. (ii) Multiply the above result by grasa and take the square-root of the product. al (iii) Subtract, gn of the result obtained in step (ii) from itself. (iv) Di obtained is called spasta sthiti and it will be in ghatis, ide the above value by the diameter of the Moon. The result That is, if grasa = g, manaikya khanda = m, Sara = s and the Moon's diameter = d,,, then

(e+ m)10x gf [e+m10xg]% 5 y 5 gx Ms +m)x10 x8] 2 5 = : a gh. ## Page 203

$192 GRAHALAGHAVAM OF GANESA DAIVAJNA

(i) Consider half of the difference between the diameters of the earth’s shadow and of the Moon. (ii) Add Sara to the above and multiply it by khagrasa. (iii) Multiply the result of step (ii) by 10 and take the square root. th (iv) Add ; of the result obtained above to itself. (v) Divide the above result by the Moon’s diameter. This gives marda in ghatis. That is, if the diameter of earth’s shadow = d,, khagrasa = kg, Sara = s and the Moon’s diameter = d,, then

[(* ri } + +} x10x ue)" [{ “atm uw re) + «foe dm 8 {(¢ sim) sf rox] ao ghatis Im

The next important step is to determine the instants of the beginning and the end of a lunar eclipse as also of the maximum obscuration. For this, we need to find the duration of the first half and ## Page 204 CANDRAGRAHANADHIKARAH (CHAP. 5) $193 the second half of the total duration of the eclipse. This is explained in

% : << 4A Ecliptic re Een Y \- a

Moon's Orbit

A half-duration is the time taken by the Moon, relative to the Sun, so that the point A in the figure moves through OA. We have AX P= (dy +yP-pP OA? =0 X- AX 2 = (OE + EX \) where

OE = d, = Semi-diameter of the shadow

B, = AX, = Latitude of the Moon (viksepa or Sara) When the Moon’s centre is at X,, we have Half-durati (a1 + da)? Be ‘alsduration = (Moon! sdaily motion — Sun’ s daily motion ) Since the actual moment of the beginning of the eclipse, and hence ## Page 205 S194 GRAHALAGHAVAM OF GANESA DAIVAJNA the Moon’s latitude then, are not known, the above formula is used iteratively. By a similar analysis, the half-duration of maximum obscuration (or totality as the case may be) is given by . vid BY Half-duration of max. obsn. = (Dit ~ DS) where DM and DS are the true daily motions of the Moon and the Sun, respectively. The, thus obtained half-duration of the eclipse and the maximum obscuration are : (1) subtracted from the instant of the opposition to get the first moments; and (2) added to the instant of the opposition to obtain the last moments. Finally, the magnitude (pramanam) of the eclipse is given by

Amount of obscuration(Grasa )

Obviously, if the magnitude is greater than or equal to 1, then the eclipse is total; otherwise, it is partial. It is also clear, from Fig. 5.1, that if the sum of the angular semi- diameters of the Moon and the shadow is less than the latitude of the Moon, there will be no eclipse.

Example: Sara, s = 2|50 arigulas

## Page 206 CANDRAGRAHANADHIKARAH (CHAP. 5) S195 grasa, g = 16|48 arigulas. khagrasa, kg = 5/41 anigulas. Moon’s diameter, dj, = 11(7 arigulas

1, 5 [Gs 10xg)2 v. Sthiti = = [oem 10a) ghatis = 4/36 ghatis m Vv — /2 AG tm) 4) 10a] 6 rs a ghatis = 1|54 ghatis Sloka 6 : The methods of finding sparsa sthiti, (first half duration of the eclipse), mokga sthiti (second half duration of the eclipse) and sammilana marda (first half of totality) and moksa marda or unmilana (second half of totality) are explained as follows. (i) Find the bhuja of vyagu (true longitude of the Sun — true longitude of Rahu). Multiply this bhuja of vyagu by 2. Express the product in palas (vighatis). Note : The word vyagu (vi + agu) means true Sun — Rahu. Here, agu means Rahu. If vyagu is in the even quadrant (II or IV ice., 90° <vyagu <180° or 270° <vyagu <360°) then the above product is respectively subtracted from and added to sthiti in ghatis to get sparsa sthiti and moksa sthiti. ## Page 207

ie., sparga sthiti = sthiti— 2 x bhuja of vyagu moksa sthiti = sthiti + 2 x bhuja of vyagu Carry out the same operations by considering marda in the place of sthiti. This gives the sammilana marda and the moksa marda or unmilana marda. ie, | sammilana marda = marda - 2 x bhuja of vyagu mokga marda = marda + 2 x bhuja of vyagu If vyagu is in the odd quadrant (I or Ill i.e., O<vyagu <90° or 180° <vyagu < 270°) then 2x bhuja is added to sthiti to get sparsa sthiti and it is subtracted from sthiti to get moksa sthiti. The same rule holds in the case of sammilana and moksa marda also considering marda instead of stithi. Note : Vyagu is also called virahvarka. Example: Sthiti= 4|36 ghatis

bhuja of vyagu = 1° 48' 48" 2x bhuja = 3|37|36 ~ 3 vighatis marda = 1|54 ghatis Since the vyagu is in first quadrant, sparsa sthiti = First half-duration = sthiti+ 2x bhuja ## Page 208 CANDRAGRAHANADHIKARAH (CHAP. 5) $197 = 48h ggvig , gvie = 4/39 ghatis moksa sthiti = Second half-duration = sthiti- 2x bhuja 48h ggvis _ gv = 4|88 ghatis sammilana marda = marda + 2x bhuja = 18h 54vig 4 vig = 18h pig moksa marda = marda - 2x bhuja = 19 54% — gif = 1/51 ghatis Sloka 7 : This Sloka explains about the madhyakdla (parvanta or middle of the eclipse), sparsakala (beginning of the eclipse) and moksakala (end of the eclipse) as follows. (i) The parvanta is the middle of the eclipse or madhyakala. (ii) By subtracting sparga sthiti from the madhyakala, we get sparsakala ie, sparsakala = madhyakala — sparsa sthiti (iii) By adding the moksa sthiti to the madhyakala, we get the moksakala. ie., mokgakala = madhyakala + moksa sthiti (iv) Similarly, for a total elipse, we have sammilanakala = madhyakéla - sammilana marda and unmilanakala = madhyakala + unmilana marda

Example : We have already obtained

## Page 209

Sparéa sthiti = 4|39 ghatis Moksa sthiti = 4|33 ghatis Sammilana marda = 1|57 ghatis Unmilana marda = 1|51 ghatis Now, therefore we get Spargakala = 36|09 ghatis

21 ghatis Sammilanakala = 38|51 ghatis and Unmilanakala = 42/39 ghatis Sloka 8 : This Sloka explains the method of finding grasa at a given time (istaghati). (i) Multiply grasa by istaghati and divide the product by sparsa sthiti. (ii) Add 1|15 arigula to the above result. This gives grasa at istakala. grasa xistaghati ie, asa at istakdla = a Bey grasa at tepakele sparsa sthiti +1|15 in arigulas This is the grasa at istakala after sparsakala (before madhyakala). If moksa sthithi is used in the place of sparéa sthiti then we get grasa at igtakéla before moksakéla (after madhyakéla).

Suppose the given time, istaghati = 2 gh. after spargakala grasa = 16|48 angulas ## Page 210 CANDRAGRAHANADHIKARAH (CHAP. 5) S199 sparsa sthiti = 4|39 ghatis 2 5 2x 16/48 grasa at 2 gh. after sparsakéla = +1)15 angulas 4|39 " 8|28 angulas Sloka 9 : This gloka explains the method of finding ayanavalanam. (i) In the case of a lunar eclipse, consider the true longitude of the Sun at the parvanta. Subtract 90° from it. This gives tribhona Ravi. In the case of a solar eclipse add 90° to the true position of the Sun at the darganta (conjunction). (ii) Add ayanaméa to (i) to get séyana tribhona Ravi. Find bhuja of this. (iii) Considering 7, 5, 1 as carakhandas, find gatakhanda, bhogyakhanda and the remainder for the above result (as explained in Sloka 5 of Chapter 2).

Then, dyanavalanam is given by

Bhogyakhanda x Remainder

Example: (Nirayana) Ravi at parvanta = 8” 9° 12'16" Ayandméa = 18° 18'

## Page 211

Bhuja of sdyana tribhona Ravi = 11° 29' 44" Gatakhanda = 0, bhogyakhanda = 7 and remainder = 11° 29’ 44” 7x 1P 29' 44" +0 = 2/40 arigulas 30 |40 argu -. Ayana valanam = | Since tribhona Ravi is in the uttaragola (i.e., less than 6 8) the valanam is also in the uttaragola. Notes : In Fig. 5.2 YSA is the ecliptic and 74p is the celestial equator and ¢ is the obliquity of the ecliptic. The position of a celestial body is represented by S on the ecliptic. Ayana valanam is the deflection of a point due to the obliquity of the ecliptic i.e., the angle subtended at the point by arc KP where K and P i sy ## Page 212 CANDRAGRAHANADHIKARAH (CHAP. 5) S201 are respectively the poles of the ecliptic and the celestial equator. In the figure, from spherical triangle SKP, we have sin, _ sin SKP sine sin SP But SKP =90° - 2 and SP =90° —8 where 4 and § are respectivly the celestial longitude and declination of the body. sine cosh sin dy = cos5 where v; is the ayana valanam. Sloka 10 : Now, computation of the aksa valanam is explained. (i) Divide the mean natakala by 5 to get the result in rasis ete. (ii) Taking the bhuja of (i) and using the khandas 7, 5, 1 (as explained in the previous loka) find the valanam. (iii) The valanam, obtained in (ii), is multiplied by the palabha of the place and then divided by 5. This gives the (spasta) aksa valanam. This aksavalanam is north or south according as the nata is eastern or western. (iv) If the ayana valanam and the aksa valanam are in the same direction, then take their sum (which will have the same direction). If these are in the opposite directions, then take their difference (whose direction will be that of the valanam with higher magnitude). (v) Divide the above combined valanam by 6 which gives the valananighri, in the above obtained direction, of the commencement of the eclipse. ## Page 213

Example : Madhya nata = 2\27 (east). (i) Dividing by 5, we get o* 14° 42' 0". (ii) Bhuja = 14° 49". Using the bhogya khanda 7, we have 14° 42'x 7/30 = 3|25. (iii) palabha = 5|45 ang. Multiplying 3|25 by 5|45 and dividing by 5, we get 3|56. This is (spasta) aksa valanam. This is in the north direction since the nata is in the east. (iv) The (previously obtained) ayana valanam = 2|40 north. (iv) Since both ayana and aksa valanams are in the north (the same direction), adding them, we get the combined (samskrta) valanam = 3|56 + 2|40 = 6|36, north (v) Dividing 6|36 by 6, we get 1|06 north. This is the valanarighri (or valana carana). Sloka 11 : Now, determination of grasarighri and khagrasanghri is explained. (First half of Sloka 11) : (i) Multiply grasamana by 60 and divide by manaikya khanda (sum of the diameters of the chadya and the chadaka). Take the square root of the above result. This gives the grasanghri in ang. (ii) Add 1|30 arig. to the khagrasa to get. khagrasanighri. Example : Grasa = 16/48 arigulas and Manaikyakhanda = 19|38 argulas. 16|48 x 60 (i) Therefore, Grasanighri = |~Tafgg_ ave: = 7|9[55 ang. ## Page 214 CANDRAGRAHANADHIKARAH (CHAP. 5) S203 (ii) Khagrasa = 5|41 angulas Adding 1|30 ang. to the khagrasa, we get khagrasanghri = 7|11 argulas. (Second half of Sloka 11) : Draw a circle and mark the north, east, south and west directions. Divide this circle into 82 parts (through the centre). If the (combined) valana is south, then starting from the direction of the sara proceed in the clockwise direction and measure the amount of the valanarighri. This gives the middle of the eclipse. On the otherhand, if the valana is north, then proceed in the anticlockwise direction and measure the amount of the valananghri. Similarly, in the opposite direction the khagrasa or bimba Sesa takes place. Sloka 12: Now, the determination of the directions of the beginning and the end of an eclipse (and of the totality) is explained. From the middle of the eclipse, in the eastern direction the amount of the grasanghri is measured to obtain the direction of the sparga (beginning) of the eclipse. The same in the western direction gives the direction of the end (moksa) of the eclipse in the case of a lunar eclipse. In the case of a solar eclipse the opposite of the said directions must be taken. The same is the case for a total eclipse (khagrasa) to obtain its beginning (sammilana) and the end (unmilanam) starting with khagrasa point. ## Page 216 CHAPTER 6 SURYAGRAHANADHIKARAH (Computation of Solar Eclipse) Slokas 1 and 2 : These Slokas explain the method of finding Jambana. It is as follows : (i) Find the sayana lagna at the kysna parvanta (i.e. darsanta). Subtract 3 rasis from it. The result is called sdyana vitribha lagna (or tribhona lagna). (ii) Find the kranti (declination) of the above sayana vitribha lagna. Add aksam/Sa to (or subtract from) the kranti. This result is called nataméga. ie., natdmSa = kranti + aksaméa or _nataméa = 8-6 where § = declination and $ = latitude of the place. (iii) Divide the natamésa by 22. Square the resulting quotient. (iv) If the square obtained in step (iii) is less than 2, then add 12 to it. This result is called hara. (v) On the otherhand, if the result of step (iii) is greater than 2, then subtract 2 from it. Divide the difference by 2. Add this to the square ob- tained in step (iii) and add 12 to the result. This is called hara. ## Page 217

(zeae y 2 améa 2 || 9a | natamsa } : 22 a2 ie, hdra= ( 22 2 To find lambana (effect of parallax on the longitude] : (i) Consider the difference between sayana vitribhalagna and the Sun. Divide this difference by 10. (ii) Multiply the difference between the above result and 14 by the result of step (i). (iii) Divide the result of (ii) by hara. The resulting quotient is called Jambana. ; 7 x= ( Hicihalagna - the Sun ie, if i then, lombana = i422 > bara Note : (i) In x, its bhuja must be taken; (ii) If vitribha lagna > the Sun, add Jambana to darsanta to get spasta darsanta (corrected darsanta). Remark : Lambana is the effect of parallax on the longitude and nati is that on the latitude of the Moon. Example : Vikrama samvat 1677, Salivahana Saka 1532. Margasira krsna amavasya, Wednesday, Cakra = 8, Varsagana = 90 and Ahargana = 1005.

## Page 218 SURYAGRAHANADHIKARAH (CHAP. 6) S207 At the mean sunrise, we have Mean Sun = 8 5° 39' 25”, Mean Moon:= 87 1° 10'33” Rahu = 2” 11° 4159” and Candrocca = 8” 17° 7' 21" According to that year’s paficariga the end of amavasya tithi is at 12/36 ghatis (after the sunrise). At this instant the true positions of the heav- enly bodies are as follows. True Sun = 8” 5° 25'57"=S, True Moon = 8% 5° 20'41"=M Candrocca = g® 17° 8°45" and Rahu = 2” 11° 41'19” True daily motion of the Sun, DS = 61'15" True daily motion of the Moon, DM =726' 30" Therefore, the instant of parvanta, ty is given by S-M ty=[to+ emt «00] gh. where t, =12|36 ghatis (after the sunrise) 516" t, =|12|36 + ——_-—__ x 60 P f | 726° 30" ~61' 15" © ee = 12[36 +0|28 ghatis = 13/4 ghatis i.e., 08" 28% after the istakala, 12|36 gh. Now, at this instant of parvanta (or darsanta). ## Page 219

True Sun = g* 5° 26/25", True Moon = 8* 5° 2620”

Virdhvarka = Sun - Rahu = 5” 93° 45'7" Now, darsanta = 13|4 ghatis ; lagna at darganta = 11" 9° 46'17" Vitribha lagna = lagna - 3 rasis = 11" 9° 46117" 3% = 8 2° 46'17"

aksamsa $ = 25° 26' 42" (north) for Kasi natéméa = kranti - aksiméa = 3-4 = — 23° 38'10" - 25° 26' 42" = — 49° 4'52” ite. 49° 4'52” (south) ee (4G natamsa 5 Now, (= 22 ) = [ 99 | = (2{13|51) = 4[58|37 a) nataméa\ ; Since (a) = 4|58|37>2, subtracting 2 from the square and di- 4|58|37-2 viding this by 2, we get = 1[29|18 ## Page 220 SURYAGRAHANADHIKARAH (CHAP. 6) S209 wen \2 ae 2 hara = (Patamsa re nataméa } glee 22 2 22 = (4{58|37+1/29|18 +12) gh. ~ 27412 ghatis = 18|27 ghatis To find Jambana : Vitribha lagna = 8* 29° 46'17"

2° 40'8" x= 10

Let = 0|16

(14-x)x (14-016) 0|16 a a | ye = O|11 ghatis Since vitribha lagna < the Sun, subtract Iambana from darsanta to get the spasta darsanta. ie., spasta darganta = 138 94¥& _ 8h 1128 = 128). 5gvie Sloka 3 : This loka explains the method of finding the Jambana corrected vyagu and the lambana corrected nataméa. It is as follows. (i) Multiply lambana by 13. The resulting product will be in kalas. (ii) The above result is added to or subtracted from vyagu (Sun — Rahu) depending on whether the lambana is additive or subtractive respectively. (iii) Find Sara of the above lambana corrected vyagu. ## Page 221

(iv) Multiply lambana by 6. (v) If lambana is negative, subtract the result of step (iv) from vitribha Jagna. If lambana is positive, add the result of step (iv) to vitribha lagna. (vi) Find the kranti (declination) of the lambana corrected lagna. (vii) Now, the Jambana corrected nataméa is obtained from the declina- tion 8 of the corrected lagna + aksdméa ie., 8-6 where 8 is the declination of the corrected lagna and ¢ is the latitude of the place. Example: We have vitribha lagna = 8* 9° 46'17", lambana = ~0|11 ghatis and vyagu = 5° 23° 45/07" -. (i) lambana corrected vyagu = vyagu - 13 x lambana 5F 93° 45' 07" — (13x 0|1) = 5® 23° 45’ 07" - 2" 23" = 5% 93° 49" 44” (ii) Sara for the lambana corrected vyagu il 1 Orrpyu = 7 * (bhuja of corrected vyagu) = ~~ x (6° 17'16") = 958 arigulas Since vyagu < 180", Sara is positive.

## Page 222

Since Jambana is negative, subtract 6 x lambana from vitribha lagna to get the corrected vitribha lagna. ie., corrected vitribha lagna = vitribha lagna — 6 x lambana = 8F 9° 46'17"-1° 06’ = 8% 1° 40'17" The kranti of the corrected lagna = 23° 34’ 35" (south) (obtained as explained in Chapter 4) AksamSa = 25° 26' 42" (north) for Kasi corrected natamga = kranti of corrected Jagna — aksamsa = — 23° 3435" - 25° 26’ 42" = 49° 01'17" (south) Sloka 4 : The method of determining nati as well as the sthiti is explained in this Sloka as follows : (i) Divide the lambana corrected nataméa by 10. (ii) Subtract the above result from 18. (iii) Multiply the results of step (i) and step (ii). The product will be in kalas. (iv) Subtract the result obtained in step (iii) from 6° 18’. (v) Divide the result of step (iii) by that of step (iv). The result obtained is called nati. Thus if lambana corrected nataméa is denoted by In, then ## Page 223

nati = 6° 18"- (8 - 2) Le 10/10 (vi) Using the above nati, we can get corrected Sara (or spasta Sara) by adding Sara to (or subtracting from) nati. ie., spasta Sara = nati + sara Note : Here the sara of the lambana corrected vyagu is to be considered. Using the spasta Sara find manaikyakhanda, grasa and khagrasa or khacchanna.

lambana corrected nataméa In = 49° 117" 10 }10 —T nn) = a6 6° 18’-|{1g- 2] 2 10) 10 :. Spasta Sara = -12|16 +9|53 angulas = -2|23 angulas. nati = (Note : nati is -ve since natdméa is south). To find sthiti : The method of finding sthiti is the same as that explained in Chapter 5 (computation of lunar eclipse). We have

## Page 224 SURYAGRAHANADHIKARAH (CHAP. 6) S213 Sun’s diamater, d, =11|8 arigulas Moon’s diameter, dm, =9|49 arigulas Manaikyakhandam, m =10|28.5 angulas grasa, g=8|6 arigulas (approximately) [(e-+m)x 10x g]2 5 .. sthiti = > 6 dy, ghatis = 2|44 ghatis (Sloka 5 in Chapter 5). Sloka 5 : This Sloka explains the method of finding the sparga and mokga kalas of the eclipse. It is as follows. (i) Multiply sthiti by 6. The result will be in degrees, minutes, seconds. (ii) Subtract the result of step (i) from the parvanta tribhona lagna (PTL) to get the spargakala tribhona lagna (STL) and adding the same result to the tribhona lagna (PTL) we get the moksakala tribhona lagna (MTL). (iii) Find lambana using the sparsakala tribhona lagna and the moksa kala tribhona lagna. These are called sparsakala lambana and moksakala lambana respectively. (iv) The sthiti is added to and subtracted from the sparga lambana and the moksa lambana to get the corrected sparga and moksa sthitis respec- tively. (v) Subtract the corrected sparSa sthiti from darsanta.This gives the sparsa kala in ghatis. (vi) Add moksa sthiti to darsanta. This gives the moksakala in ghatis. ## Page 225 s2i4 GRAHALAGHAVAM OF GANESA DAIVAJNA. Example : We have the sthiti = 2|44 ghatis (Sayana) vitribha (or tribhona) lagna = 8* 2° 46'17" Now, 6 x sthiti = 6 x (2/44) =16° 24’ (i) Spragakala tribhona lagna = gF 9° 46'17"-16° 24’ R 16° 29°17" (ii) Moksa kala tribhona lagna = 8" 9° 4617" + 16° 24’ = 8” 19° 10'17" To find sparga lambana : Declination of the sparsakala tribhona lagna, § = ~ 21° 24'39" AksamSa = 25° 26' 42"

As oy, WE naté méa 1| ( natamsa Hara = | ——=— _| + | ———_| -2 +12 Now, Hara ( 22 } 4( 22 } 2 2 46|51/21)" 1 | (46|51/21 -( 22 } 4 ro |e ore ## Page 226 SURYAGRAHANADHIKARAH (CHAP. 6) S215

Ravi at parvanta = 8” 5° 26' 25"

True motion of Ravi = 61'15"

Sthiti = 28 44%

:.Ravi at sparga kala = 8 5° 96’ 25" -9' 47" = 8” 5° 23°38" SparSakala vitribha Lagna = 7" 16° 2217" Ravi — Lagna = 19° 01'21" Now, th = of Bhuja of (Ravi - Lagna) = 1°54’ and 14° —1° 54" =12° 6! 1g 6 x 1° 54" _ 22149 - = 18h. y7¥8 Hara 17/48 SparSa lambana = (Numerical Value) ie., Sparga Iambana = - 18" 17% To find moksa lambana :

Declination of the above = - 23° 42’ 28"=6 (Numerical value)

## Page 227

Nataméa = 5 ~ $ = - 49° 9' 10"

Ravi at Moksakala = 8" 5° 26’ 25"+ 2/47" = 8" 5° 29/12" where 2/47" is the motion of Ravi in 28* 44" (sthiti). Now, Ravi — lagna = 0 13° 41'5" th jo of Bhuja of (Ravi ~ lagna) = 122 and 14 ~1|22 =12|38 12/38 x 1/22 Now, 229851122 _ 1288 D2 ay sou > Hara = ~—«18)28— ie., Mokga lambana = 08" 56%é To find sparga and moksa kalas : We have darganta = 138" 94% and sthiti = 28" 446 -.darganta — sthiti = 138" 94vié 98h qqvis — = 1 98h govig Sparga lambana = — 18" 178 Now, spargakéla = 108) 20% — 18 17% = g&h ogre ## Page 228 SURYAGRAHANADHIKARAH (CHAP. 6) 8217 We have darsanta + sthiti = 138" g4vié +28! 4qvi@ = 158 gre Moksa lambana = 08" 56” -. Moksakala = 158" 488 4.0% 56% = 168) gavie Summary of the eclipse : Beginning of the eclipse: 98" 93% Middle of the eclipse: 138" o4vi End of the eclipse: 168" 44vis Sloka 6 : This Sloka explains the method of finding sammilanakala and unmilanakala (in the case of a total eclipse). The colour of the eclipse is also explained. (i) To find sammilana and unmilana kalas : Consider marda in the place of sthiti and proceed as explained previously to get sammilana kala and unmilana kala. (ii) Colour of the eclipse : If the grasa of the lunar eclipse is small (alpagrasa, less than 14 ) then the colour of the eclipse is smoke colour (dhimra varna). If the grasa of the lunar eclipse is half (i.e., ardhagrasa) then the colour of the eclipse is black (krsna varna) and the colour of the eclipse will be brown (pinigalavarna) if the grasa is full (sampirna grasa). ## Page 229

In the case of a solar eclipse whether the grasa is small or half or full the solar eclipse will be only of black colour. Further, if the grasa of a total eclipse is less than one arigula, the eclipse should not be predicted since the small eclipsed portion cannot be recognised due to the powerful solar rays. Sloka 7 : This Sloka explains the method of finding grasa at a given time (istakéla) (1) Multiply the istaghati (ie., given time in ghatis) by 2 x grasa. (2) Divide the above product by the difference of sparga and moksa ghatis. (3) Add 4 arigula to the result obtained in step (2). This gives the grasa at the istakala. Example : Let istaghati = 1 ghati and grasa = 8|6 angulas (1) Now, 2 x grasa x istaghati = 2x1x8|6 = 16|12 (2) Spagakala = 98" 038 and Moksakala = 168" 44%# The difference = 78" 418 (i.e., duration of the eclipse) (3) We have

16|12 1 ‘ ; "ai * 2 = 2|6 +0|30 [- 4 angula = 0|30 angulal = 2|36 arigulas = 298 367.478 ## Page 230 SURYAGRAHANADHIKARAH (CHAP. 6) S219 (Note : 1 angula = 60 pratyangulas). To find valanam and parilekha : We have lambana corrected darganta = 128" 53¥ié The Sun at parvanta = 8” 5° 26'14" Adding 3 ragis we get 117 5° 2614" Adding ayandméa 18° 08' to the above, we get 117 23° 34’ 14" Now, as explained in Ch. 5, Slokas 9 and 10, we have (i) Ayana valanam = 1|30 (south) and dinardham = 138" 3vié nata = 0|10, aksa valanam = 0|14 (north) and palabha = 5|45 angulas |45x0|14 1]20 a - corrected aksa valanam = —— =0|16 (north)

1|30 -0|16 =1|14 (south) __ 14 Now, valanarighri = ~—— =0|12 (south) grasa = 8|6 arigulas ## Page 231

y meme fer Ve Gi) Parilekha = ee a 8|6 x 60 * | "10)28 iY : = (16|26)'2 = 6/49 (also called channanghri). ## Page 232 CHAPTER 7 MASAGANADHIKARAH (Accumulated Lunar Months) Sloka 1 : Now, computations of lunar and solar eclipses are presented, with masagana (heap of lunar months) by obtaining the true positions of the Sun, vyagu, tithi, bimba (diameters) and grasa using simple and in- genuous techniques. Sloka 2 : This sloka gives the ksepaka of the Sun, vyagu, vrtta (ie., candrakendra) and varadika of tithi as follows :

Vyagu 11” 07° 18" Candrakendra o” 14° 51' Tithi (varadika) af 4geh q5vig ‘The above values are the mean positions at the mean fullmoon following the epochal date [March 19, 1520 (J)] ie. on April 2, 1520 (J), Monday at 48* 47, Remark : In the chapter Madhyamadhikara the epochal mean positions (ksepakas) are as follows : Mean Sun : 11 19° 41’, Mean Moon : 11” 19° 06" Moon’s apogee (Mandocca) : 5¥ 17°33’ and = Rahu:: 0” 97°38" ## Page 233

;. Vipata = Mean Sun — Rahu = 10822°06', Candra kendra = Mean Moon ~ Mandocca = 6"1°33' Varadika = 24 at the mean sunrise at Ujjayini. These are the mean positions at the mean sunrise on Saka 1442 Caitra Pratipat (March 19, 1520, Monday). Now, since the mean sunrise is not the time of the darganta (amavasya), we shall find out the mean instant of the conjunction of the Sun and the Moon. The difference between the mean Sun and the mean Moon = 11F 19° 41’ 117 19° 06" = 35’ = 2100". The difference in the mean daily motions of the Moon and the Sun = 790’ 35" ~ 59’ 08" = 731’ 27" = 43887". Therefore, the time required (after sunrise) for the mean conjunction is 2100 43887 x 60 = 2 gh. 52 vig. .. Varadika of Amavasya = 2! 2e 52¥@

60 The motion of the Sun in 2 gh. 52 vig. is Thus at the mean conjunction of the Moon and the Sun (i.e., at the in- stant of the New Moon) we have the mean Sun = the mean Moon = 17" 19°44" the Moon’s apogee (mandocca) = 5®17°33"

= Mean Moon - Moon's mandocea = 6"02°11'. Vipata (Vyagu) = Mean Sun — Rahu = 19" 99°99" ## Page 234 MASAGANADHIKARAH (CHAP. 7) 8223 Pakgika Calanam : (Motion of the Sun etc. in the course of a half lunar month) : The length of a mean lunar month = 29753059. :. Half lunar month (paksa) = 14¢,765295 The mean motions of the bodies in a paksa are (i) Sun : oF 14° 33° (ii) Moon : 6¥ 14° 33° (iii) Vrtta (mandakendra) : 6® 12° 54’ (i.e., Moon’s motion — Mandocca’s motion = 6¥ 14° 33’ —'1° 39" = 6 19° 54") (iv) Vipata’s (i.e., Sun - Rahu) = 0” 15° 20’ which can be obtained as follows: Sun’s mean daily motion : 59’ 08" Rahu’s mean daily motion : - 3/11" Vipata’s daily motion : 62'19" [= 59" 08' - (- 3'11")] Hence vipata’s paksika gati _ 62'19" x 144,765295 0 degrees = 07 15° 20' ## Page 235

(v) Varadi paksika for tithi : We have paksa = 14.765295 days. Dividing the number of days by 7 and taking the remainder, we get 0¢ 45% 55". Adding the paksika motions to the relevant ksepakas (epochal positions) we get, at the succeeding full moon: Mean Sun = 0"4°21', Vipata (Vyagu) = 11°7°29', Candra kendra (= Vrttam) = 0"14°51' ,

Sloka 3 : This Sloka gives the dhruvaka of the mean Sun, vyagu, candrakendra and vara ete. of tithi as follows.

Vyagu Uae aby Candrakendra oF 1°16"

Dhruvaka is the residual motion in a cakra after removing completed revolutions. Sloka 4 : This gloka explains the method of finding positions of the Sun and the vyagu using masagana. It is as given below : (i) Find the masagana for the given date. (ii) Divide the masagana by 67 and multiply it by 2. (iii) Subtract the result obtained in step (ii) from the masagana. This gives Ravi in rasis etc. (iv) Take the masagana as raéis (for Vyagu). Divide the masagana by 3. Subtract this from the masagana to get the result in améas. (v) Divide the masagana by 4. The result will be in kalas. ## Page 236 MASAGANADHIKARAH (CHAP. 7) $225 (vi) The results of steps (iv) and (v) give the amgas and kalas of vyagu. Example: Samvat= 1669, Saka = 1534 Kartika Sukla Paurnima, Thursday i.e. Nov.8, 1612(G). For the given date, Cakra is 8 and masagana is 57. The masagana is arrived at as follows : (i) Salivahana Saka Varga — 1442 = 1534 - 1442 i.e, elapsed years from the epochal year = 92 (ii) Cakra = Quotient of 92/11=8 (iii) Sega (remainder) = 4 (iv) Sesa x 12=48 months (v) Months elapsed till the beginning of the given month (i.e., Kartika) = 7 (ie., from Caitra to ASvayuja) (vi) Adding results of (iv) and (v), we get 48 + 7 = 55 (vii) Adhikamasas = 2 (viii) Therefore the total elapsed Masagana = 48 + 7+ 2 = 57 To find Ravi : (i) Masagana x 267 = 114|67 ragis etc. = 1* 21° 2’ 41" Gi) Ravi = Masagana — item (i) = 57% _17 91° 2'41" = 55% 8° 57/19" Removing the multiples of 12 (i-e., 48), we get ## Page 237

$226 GRAHALAGHAVAM OF GANESA DAIVAJNA

Ravi = 7” g°57'19"

(i) Masagana = 57, :, Rasis = 57 i.e. 9®(removing 48%) (ii) Dividing the masagana by 3, quotient = 19 (iii) Dividing the masagana by 4, we get 14.25 kalas i.e., 14 kalas and 15 vikalas. (iv) Masagana — item (ii) = 57° - 19° = 38° (in raéis) (v) Vyagu = item (i) + item (iv) + item (iii) = 9F384'5" = 10"8°14'15" Sloka 5 : Now, the method of finding the vrtta (i.e., moon’s mandakendra, anomaly) and varadi (weekday etc.) of the tithi is explained. (i) Divide masagana by 14. (ii) Removing the multiples of 14, consider the remainder and divide it by 7. (iii) Subtract the result of (ii) from the remainder. yh (iv) Add 10 of the masagana (in degrees) to item (iii). This gives the vrttam (this is to be corrected as explained in the next Sloka).

(i) Dividing masagana by 14, the remainder is 1 considered as rasis. (ii) Remainder/7 = 1/7 rasi = 08 4° 17°08" ## Page 238 MASAGANADHIKARAH (CHAP. 7) $227 (iii) Remainder - Remainder/7 = 1% _o% 4° 17:08" = 08 25° 42' 52" (iv) Masagana/10 + item (iii) = 5° 42’ +07 25° 49'52” = 17 1° 24’ 52" Thus, Vrttam = 1" 1° 24'52" (this has to be further corrected with dhruva and ksepaka, explained in the next Sloka).

(i) Divide the masagana by 14 and consider the remainder. (ii) Add %4 of the remainder to itself. (iii) Multiply the masagana by 10. (iv) Divide the product by 327. (v) Add the result of (iv) to (ii). This gives the varddika (weekday etc.) of the tithi. Example : Masagana = 57 (i) Masagana/14 : Quotient = 4, Remainder (ii) Remainder + Remainder/2 = +4) = 17 3088 ## Page 239 $228 GRAHALAGHAVAM OF GANESA DAIVAJNA (iii) Masagana x 10 = 57x10=570 (iv) Dividing (iii) by 327, we get 14 4484. ggvig: (v) Adding (ii) and (iv), we get 19308417448 357% = 34 148 358 Thus, the varadika (weekday etc.) of the tithi = 3% 148h g5vig (This has to be further corrected with dhruva and ksepaka as explained in the next Sloka). Sloka 6 : Here, corrections with dhruvaka and ksepaka are explained. (i) In the case of Ravi, subtract (cakra x dhruvaka) from and add ksepaka to the masagana derived Ravi (obtained from Sloka 4). (ii) In the case of vipata (or vyagu), vrtta (Moon’s mandakendra) and varadika (weekday etc.) of the tithi add (cakra x dhruvaka) and ksepaka to the corresponding masagana derived results.

(Ravi: Masagana derived Ravi = 7P 8° 57/19" Subtract Dhruvaka x Cakra : - 08 13° 20° We get 6 95° 37/19"

## Page 240 Gi) (iii) (iv) MASAGANADHIKARAH (CHAP. 7) $229

Corrected Mean Ravi :

Vyagu : Masagana derived vyagu =

Add (Cakra x Dhruvaka ) :

Add ksepaka :

Corrected vyagu :

Vrtta (moon’s mandakendra) :

Masagana derived vrtta =

Add (Cakrax Dhruvaka) :

Corrected vrtta : Varadika of tithi : Masagana derived : Add (Cakra x Dhruvaka ) : Add ksepaka :

6” 29° 58'19" 10" 08°14'15” +8” 09° 36’ +117 07°18" 5 25°08'15" 1® o1 24" 52” + oF og° 48" + oF 14 51 1® 95° 03" 52" 3¢ 14eh ge +6 «16 48 my = (CH iat 2088 gg" ## Page 241

Removing the nearest multiple of 74 we get Varadika of tithi : 54 20% 0g”# (5 indicates Thursday) Sloka 7 : This loka gives the motions in a paksa of the Sun, vipata, vrtta and varadika of tithi. It is as given below. The motion of the Sun = 0° 14° 33'0" The motion of the vipata = 0% 15° 20'0" The motion of the vrtta = 6” 12° 54'10" The motion of the varadika of tithi = 07 45855" ‘The derivations of the above are shown in our explanation of Sloka 2. Sloka 8 : The motions of Ravi, vipata, vrtta and varadika of tithi in six months are given as below : The motion of Ravi = 5” 24° 38" The motion of vipata = 6%4°01' The motion of vrtta = 5 4° 53° The motion of dinadi (varadi) of tithi = 27118" oi Sloka 9 : Now, for the given (ista) tithi, finding of dinadi (weekday etc.) of the tithi, Ravi and vrtta (Moon’s mandakendra) is explained. (i) From the tithi (before or after parnimanta i.e., in the Sukla or krsna paksa) yf 64th of it is subtracted. This gives the motion of the tithi in varddika (weekday etc.) ## Page 242 MASAGANADHIKARAH (CHAP. 7) — ie., Motion of tithi = Tithi - Tithi/64 (ii) Ravi’s motion = Tithi - Tithi/34 (iii) Vrttam motion = [Tithi - Tithi/93] 13 Derivation of the above expressions Duration of a lunar month = 29% 314" sov% Therefore, we have (i) Civil days of the given tithi Tithi ——*x - o0¢ a1@* 50") 106310 3600 Tithi x 10631 10800 __ Tithi (10800 - 169) - 10800 " Tithi - (169/10800) Tithi = Tithi-Tithi/64 (ii) Ravi : Mean daily motion = 59'08" Civil days of Tithi = Tithi - Tithi/64 = ((63|64) x Tithi] ## Page 243

-. Motion of Ravi for the given Tithi (63/64) x Tithi x 59! 08" % xTithi (28) = Bx rithiys( SEF 56881 | ith 3600) ~ 64 900) ~ 57600 57600-1719 7, 1719) = =|1- Tith Jee ( an ua = ees ith | eh 33.51 34 (iii) Vrttam, Moon’s mandakendra : We have Moon’s daily motion — Mandocca’s daily motion = 790'35"-6' 41" = 783'54" ie., Vratta’s daily motion = 783' 54" 63 tegen Now, therefore the motion of vrtta for 1 (one) tithi is given by 64 x 783! 54! For the required tithi (istatithi), the motion of vrtta is 63 4; 63 [47034 ]° ithi x —— 54" — Tithix ——x|——— Tithi x x 783" 54 i ix Se | ## Page 244 MASAGANADHIKARAH (CHAP. 7) $233 eine

(iii) Vrttam, Moon’s mandakendra : We have Moon’s daily motion — Mandocca’s daily motion = 790'35"-6' 41" = 783'54" ie., Vratta’s daily motion = 783' 54" 63 tegen Now, therefore the motion of vrtta for 1 (one) tithi is given by 64 x 783! 54! For the required tithi (istatithi), the motion of vrtta is 63 4; 63 [47034 ]° ithi x —— 54" — Tithix ——x|——— Tithi x x 783" 54 i ix Se | ## Page 244 MASAGANADHIKARAH (CHAP. 7) $233

Tithi x (12.8609) = a x13

93 93 = [ren - Rah 13 93 Sloka 10 : Now, for finding the true Sun and the true tithi, the mandakendras of the Sun and the Moon will be obtained. The khandas to find mandakendraphalam are as follows : 17, 16, 14, 12, 9, 5 and 2. (i) Divide the bhujaméa of vrtta by 13. The resulting quatient gives the total number of elapsed or gata khandas. (ii) Multiply the remainder obtained in step (i) by esyakhanda (the khanda to be covered) and divide the product by 13. (iii) Add the above result to the sum of the gatakhandas. The result gives mandakendraphalam of Candra. (iv) In the case of Ravi, considering the mandakendra of Ravi,follow the procedure of (i) to (iii) above. Then divide the result of (iii) by 2. Subtract th 4 of this result from itself. This gives the Raviphalam. Example: —Vrtta = 1” 95° 3'52" (see example under Sloka 6).

## Page 245

Bhuja 55° 3/52" ° 3°52" (i) Now, ee a 13 ae 13) This implies that the number of gatakhandas is 4. They are 17, 16, 14 and 12. Their sum = 17+ 16 + 14+ 12 = 59

Esyakhanda (or Bhogyakhanda) = 9

13 13 (ii) a2 Gii) Sum of gatakhandas + item (ii) = 59° +2° 7/17" = 61° 717" Candra’s mandakendraphalam (vrttaphalam) = + 61° 7'17" (additive) Note : (i) If vrttam is within 6 rasis from Mesa (i.e., if 0° < vrttam < 180°) then mandakendraphalam is additve.

ie., if (180° < vrttam < 360°) then mandakendraphalam is negative. (ii) The mandakendraphalam defined here is different from the mandaphalam (eqn. of centre) used in the Spastadhikara. However, the former is a multiple of the latter.

Mandocca of Ravi = 2” 18° =78° (from example under Sloka 6). Ravikendra = Mandocca — Ravi ## Page 246 MASAGANADHIKARAH (CHAP. 7) 8235 " 78° -6% 29° 58'19" = 7% 18° 1' 41" (adding 12”) 228° 1' 41" Bhuja of Ravikendra = 228° 1'41"-180° = 48°1'41" Bhuja 48° 141" gai" @ te =e a 13 13 18 This implies that the number of gatakhandas = 3 They are 17, 16 and 14. Their sum = 17 + 16 + 14 = 47

Remainder = 9° 1' 41"

Esyakhanda = 12

(ii) Sum of the gatakhandas + 7 = 474 2xO UA = 55 oy oF ii) Take half of item (ii) : 55° 20'0" 2 = 27° 40'0" (iv) Subtract ¥," of (iii) from (iii) : ## Page 247

ar 40 ot — 27 400" _ 99° 93" 20° Since Ravi = 6* 29° 58'19">180° the above result is subtractive. Therefore,

Note : Here also, the “mandakendraphalam” of Ravi is different from the mandaphalam (equation of centre) explained in Spastadhikara.

= 61° 07'17" + (— 23° 03’ 20") 38° 03! 57” This is the combination of the two phalams (phala-dvaya-samskrti). Sloka 11 : This Sloka gives the method for finding hara (for determining tithi). (i) Find bhogyakhanda (or esyakhanda) of vrtta. Divide it by 6 and the result is taken in degrees (amas) etc. (ii) Divide the amgas obtained in step (i) by 3 and add 30° to it. (iii) Subtract the result of (i) from that of (ii). This gives hara. ie. Hara = (20 + sss) _ Bhogyakhanda 8 6 Example :Vrtta = 1° 25° 3' 52", Bhogyakhanda of vrtta = 9 (obtained in the previous Sloka) ## Page 248 MASAGANADHIKARAH (CHAP. 7) S237

5 9 Bern @ 6 = = Vso 7 ama obtained in(i) 4. 1°, anc Gi =? 90° 5 =z 780° = 20'+ 30° = 30° 20° (iii) .. Hara = step (ii) — step (i) = 30° 20'-1° 30’ = 28° 50’ Now, Cara must be obtained (as explained in Ravicandra spastadhikara, Slokas 5 and 6) from the true longitude of the Sun. Example: We have, Nirayana Ravi : 6” 29° 58'19" Ayandméa : 18° 10' and hence Sayana Ravi : 7* 18° 08'19" and Cara = + 84 Sloka 12 : The method of obtaining true (spasta) tithi is explained : (i) Consider phalasamskrti (i.e., Raviphalam + Candraphalam, obtained in Sloka 10). Multiply this by 10 and divide by hara. The result will be in ghatis. Gi) Find cara. If cara is positive, then consider it as negative. If cara is negative consider it as positive (i.e., take the opposite sign). (iii) Consider the desantara distance in yojanas. Subtract Y, of desantara distance from itself. The result is called desaéntaraphalam. Cara is subtracted from the desantaraphalam, given by desantara in yojanas reduced by ¥%, of it (taken positive or negative according as the place is to the east or west of the Ujjaiyini meridian). This has to be added ## Page 249

to or subtracted from (according as it is +" or -"*) the result obtained in item (i). Example: We have, from the example under Slokas 10 and 11 Phalasamskrti = 438° 3'57" and hara = 28|50

wy ara = pao = * 1h? (ii) We have Cara = + 84 (iii) Degantara yojanam = 64 yojanas. Now, 64- s =48 (iv) Taking the opposite sign for the cara (i.e., — 84), we have — Cara + 48 = - 84 + 48 = - 36 Phalatraya samskrti = 13’ 12" — 0’ 36" = 12’ 36" (in kalas) (v) Tithi = 54 208" ge (mean tithi varadika obtained under Sloka 6). The phalatraya sanskrti 12136, considered as ghatis, being additive, with the mean varadi of tithi gives true vardadi of tithi as 5432844". (vi) We have obtained earlier (in example under Sloka 6): Nirayana Mean Ravi: 6* 29° 58" 19°

## Page 250 MASAGANADHIKARAH (CHAP. 7) $239

(vii) Vyagu (obtained earlier): 5% 95° 08'15" (According Visvanatha) :

Phala-traya-samskara : +1236"

Sloka 13 : The method of obtaining true Sun and true vyagu is explained as follows. (i) Multiply Raviphalam by 4. The result will be in kalas. (ii) Divide the above result by 24. (iii) Add the results of step (i) and step (ii). (iv) The result of step (iii) is added to or subtracted from Ravi accordingly to get spasta Ravi. Similarly the same result of step (iii) is added to or subtracted from vyagu to get spasta vyagu. (v) Add (2 - 4) to hara. Divide this by 3. The result gives the diameter of the Moon in angulas. The moon’s diameter hara + (2- ¥) ie, | Candrabimbam = 7 angulas. Example : We have Raviphalam = 23' 3" 20" (i) Raviphalamx 4 = 92'13" 20" ## Page 251

.. Raviphalam x 4 Gi) “oq = 8°50" 33" (iii) 92/13" 20” + 350" 33" = 96'3"53" = 1°36'3"53” (iv) Ravi (obtained earlier from Sloka 12): 7” 0° 10'55” Since (iii) is subtractive, we get Spasta Ravi = 7% 0° 10'55" - (1° 36’ 3" 53”) ie. True Ravi = 6% 28°34'52" (neglecting 53”) (iv) (Earlier obtained) Vyagu= 5 95° 20'51" Subtracting item (iii) : -1° 36' 3" (neglecting 53”) True Vyagu : 5¥23°44'48" (v) Now, we have

hara x ‘b- ¥) oe _ 28\50x(2-%) . _ 3030 7 3 arg. = arig. = 10|10 arigulas ig IB ie., Diameter of the Moon = 10|10 anigulas ## Page 252 MASAGANADHIKARAH (CHAP. 7) S241 Sloka 14 : This Sloka explains the method of finding the diameters of the Sun and earth’s shadow.

1 (i) Consider (u -3) = 10|50 anigulas (ii) Divide bhogya (agrima) khanda by 40. (iii) According as the Ravi (manda) kendra is from 90° to 270° (II or III quadrant) or from 270° to 90° (IV or I quadrant) item (ii) must be added to or subtracted from item (i). This gives

op A

(iv) The diameter of the earth’s shadow cone

Hara-5 , Agrimakhan da

where +" sign or —” sign is taken according as Ravi (manda) kendra is between 270° and 90° (i.e., IV or I quadrant) or between 90° and 270° (.e., IT or IM quadrant). Example : Agrimakhanda (i.e., the last of the elapsed khandas) of the Sun’s manda kendra is 12 (obtained in Soka 10). Therefore, 12 (1) Ravibimbam = (20/50) - = 10|32 arigulas Note that the Ravi (manda) kendra = 228° 1' 41" (lying in III quadrant). ## Page 253

(2) Bhabhabimbam : Here, hara = 28|50.

= (28/505) + 260-8 -2 = 0950) + _ 23/50 12 5 50 15 50 = 23|50+1]85-Ol14 = 25|11 argulas. Sloka 15 : Here the possibility (or otherwise) of an eclipse is explained. As explained earlier, the tithi etc. (i.e., tithi, ara, bimba etc.) are used for determining an eclipse. After the occurrence of an eclipse, a fort- night earlier or later than six months hence (i.e., 54 or 64 months) or a fortnight hence, the possibility of an eclipse is to be found out. If the bhujamSa is less than 15° an eclipse is possible. Further, if the (south- ern) bhujaméa of the vyagu is less than 8°, then the possibility of a solar eclipse has to be considered. If the duration of the day is greater than that of the tithi (new moon), then a solar eclipse can be considered. On the otherhand, if the tithi (full-moon) ends after the sunset, a lunar eclipse can be considered. Sloka 16 : This Sloka explains the method of finding grasa (or channam) in the case of a lunar eclipse as follows. (i) Subtract 3|20 from hara and multiply the difference by 4. (ii) Divide the above result by 9. (iii) Subtract the bhujaméa of vyagu from the result of step (ii) and mul- tiply the difference by 11. (iv) Divide the result of step (iii) by 7. This gives channam or grasa in arigulas in the case of a lunar eclipse. ## Page 254 MASAGANADHIKARAH (CHAP. 7) $243 (hara -3|20) 4 ie, Grasa (or Channam) = )~___g_ - bhuja of weer x 2 Note : If the bhuja of vyagu is greater than the result of step (ii) then there is no eclipse. Example : We have, hara = 28|50 (i) (28|50-3]20)x 4 = 102|0 102|0 (ii) 9 = 11/20 (iii) Vyagu bhujaméa = 6{15[12

(iv) grasa = 5[4]48x11 _ 55|52[48 . “aa = = 1/58 axigulas 7 Sloka 17 : This sloka gives the method of finding grasa (or Channam) in the case of a solar eclipse as follows. (i) Find darsanta and natam at darsanta. (ii) Divide natam by 4. The result will be in rasis. (iii) In the case of parva (i.e., eastern) nata subtract 7 xnata from the a) Sun, and in the case of pascima (i.e., western) nata add ae nata to the ## Page 255

Sun. This gives the nata corrected Sun. (iv) Find the kranti (declination) of the nata corrected Sun. o-$) (v) Now, natamSa = kranti +aksaméa (vi) Divide natamsa by 6. (vii) Add the result of step (vi) to vyagu bhujaméa if natamsa and vyagu are in the same direction. Otherwise, subtract the result of step (vi) from vyagu bhujaméa. (viii) Subtract the result of step (vii) from 7 and divide it by 2. Then .) , (7 — step (vii) (ix) Grasa = (7 - step (vii)) + Tosa oa wb) in arigulas in the case of a solar eclipse. Note : If it is not possible to subtract the result of step (vii) from 7. (i.e., step (vii) > 7) then there is no eclipse Example : Smavat 1669, Saka 1534, Vaisakha krsna 30 (amavasya), Wednesday, Cakra 8, Masagana = 51. End of the amavasyé i.e., aménta = 2640 gh. (ie. May 30, 1612 (G) A.D., Wednesday). We have dinardham = 16|48 gh. @) nata = 9|52 in the west (obtained as explained earlier) nata _ 952 4 4 Gi) ja ## Page 256 MASAGANADHIKARAH (CHAP. 7) $245 (iii) nata corrected Ravi = 4” 5° 26’ 34" (iv) kranti of the nata corrected Ravi = 13° 52'22" (north) =& (v) aksémSa = 25° 26' 42" = -.natamsa (8-4) = 13° 52'22"- 25° 26’ 42" = ~11° 34’ 20" ie, 11° 34'20" (south) (vi) ees = noeeae =1° 55’ 43" (south) (vii) vyagu bhujamSa = 7° 59' 36" (north) Now 7° 59' 36" — 1° 55’ 43" =6° 3'53" (viii) 7° - 6° 3'53” = 0° 56'7" tix 2 ae =0° 28" 3" <. Grasa = 0|56|7 +0|28|3 = 1|24 angulas Sloka 18 : This sloka explains the determination of the parvesa (Lord of the parva). (i) Find the mean vyagu using the masagana. Divide the number in the rasi position by 12. This result is called vyagu madhya paryaya gana (ii) Multiply the above result by 2. (iii) If vyagu is greater than 180° add 1 to it. (iv) Add cakra to the above result and then divide by 7. The remainder gives the parvesa as follows : ## Page 257 s246 GRAHALAGHAVAM OF GANESA DAIVAJNA Remainder Parvesa 1 Brahma 2 Candra 3 Indra 4 Kubera 5 Varuna 6 Agni

Note : The remainder indicates gataparvesa; to get the present (vartamana) parveSa, add 1 to it. Sloka 19 : Now, the determination of the Moon’s position from that of the Sun is explained. (i) Multiply the tithi number by 12. The result will be in degrees. (ii) Add the above result to the spasta Ravi. (iii) Multiply hara by 24 and add 62 to the product. This gives candragati (ie., daily motion of the Moon) in kalas. i.e., Candragati = (hara x 24) + 62 (iv) Sarya gati (daily motion of the Sun) = 59’ 08" (v) Find the tithikala using the Sun and the Moon obtained above. (vi) Also find the gata (elapsed) and the gamya (balance) ghatis of the naksatra and yoga. Example : Thithi = 15, Ravi: 6* 28° 34’52" and hara = 28|50 (i) Spasta Candra = (15 x 12)" + 6” 28° 34’ 52" = oF 28° 34'52” (removing 12 rasis) (ii) Candragati = 28|50 24 + 62 = 754’. ## Page 258 CHAPTER 8 GRAHANADVAYA SADHANADHIKARAH (Eclipses from Paficanga - Shortcut Method) Sloka 1 : Now, the computations of the two eclipses (solar and lunar) by a shorter method based on the data from the paficdniga (almanac) viz., parvanta (end of the lunar fortnight), postions of the Sun, the Moon and Rahu, gata (elapsed) and egya (balance) ghatis of tithi and naksatra, duration of the day (from sunrise to sunset) are explained. Example : We have the following data from paficaniga : Samvat year = 1669, Saka year = 1534 Vaisakha Sukla Parnima, Monday [i.e. May 14-15, 1612 (G)] The elapsed portion of Pairnima gataghati = 2|23 gh. The balance portion of Parnima esyaghati = 54|20 gh.

Sum of the gata and the esya ghatis,

gataghatis of the Anuradha nakgatra = 20|4 gh. ## Page 259

esyaghatis of Anuradha = 38|35 gh. Sum of the gata and esya ghatis of Anuradha naksatra = 20|4 + 38)35 = 58/39 gh.

6 gh. Ravi at parvanta = 1” 6° 34'37" Rahu at parvanta = 1” 14°18'11" Virahvarka = (Sun — Rahu) = 11” 22° 1626" Sloka 2 : This loka explains the method of finding channam (grasa) as follows : (i) Subtract 7 from tithiyoga ghati (i.e., the sum of the gata and the esya ghatis of the tithi) (ii) Divide 627 by the result of step (i). The result will be in améas. (iii) Subtract bhujaméa of vyagu from item (ii). (iv) Add 75 of item (ii) + and 5 of item (ii) to item (iii) i., take the d item (ii) + + item Gi) + 4 item (ii, sum : item (ii) + 7g item (ii) + 5 item (ii).

& 627 ‘i =| - bhujaméa of vya ie, Let la oI vyagu || ## Page 260 GRAHANADVAYA SADHANADHIKARAH (CHAP. 8) S249

Example : Tithiyoga = 56|43 gh. and Vyagubhuja = 7° 43/34” Then x=|—627 __7° 43"34"| = 4|53|7 56/43 -7|0 grasa (channam) = * ++ = 7|37|59 arigulas Sloka 3 : This Sloka gives the formula for finding diameters of the Moon and earth’s shadow in the case of lunar eclipse as follows : (1) The diameter of the Moon, 695

(2) The diameter of the earth’s shadow 1322, 5 2 __ arigulas Bhabhabimbam = ———— tithiyoga —10

695 (1) Candrabimbam = —-—-— 56/43 +6 =11|4 arigulas 1322 (2) Bhubhabimbam = ——~—-— 56|43-10 =28|17 arigulas. ## Page 261

Sloka 4 : This Sloka gives the method of finding grasa using nakgatra ghati.

(i) Subtract 10 from naksatrayoga ghati (sum of gata and esya gatis of naksatra) (ii) Divide 610 by item (i). (iii) Subtract bhujamSa of vyagu from item (ii). (iv) Multiply item (iii) by 11. (v) Divide item (iv) by 7. This gives grasa in angulas. 610 —__ 610 Bajamsaofvyagu jt an Naksatrayoga ghati-10 0 7°84 7 ang

Example : Naksatrayoga ghati = 58|36 gh.

610 © 4g'gan|y it ge [aose-? 43 34 }4 = 7|34 anigulas Remark : We had obtained earlier (from loka 2) grasa = 7|37|59 arigulas using tithi yoga ghatis. Vigévanatha in his commentary gives a correction (samskara) to the bhabhabimbam (diameter of the shadow cone) as follows : Consider the following values in pratyarigulas for six rasis : 11, 16, 20, ## Page 262

16, 11, 0. If Ravi is within 6 raéis from Mesa then add the corresponding value for that rasi to the bhabhabimbam (obtained earlier from Sloka 3). If Ravi is in a rasi among the six rasis from Tula then subtract the cor- responding values. Example : The bhabhabimbam obtained earlier : 28|17 angulas. Ravi = 1% 6° 34'37" i.e., in the second rasi from Mesa. The corresponding value in pratyarigulas is 16. Therefore, corrected bhabhabimbam = 2817 +0|16 = 28|33 arigulas. Sloka 5 : This loka gives the method of finding the candrabimbam (di- ameter of the Moon) and the bhabhabimbam (diameter of earth's shadow) using naksatrayoga ghati. It is as follows : (1) Candrabimbam = —_—®49____ angulas nakgatrayoga ghati 1255 Pan pA piu (2) Bhabhabimbam = naksatrayoga ghati ~ 14 angulas

649 (1) Candrabimbam = 3836 = 114 angulas 255 (2) Bhabhabimbam = s@]3g—14 = 28/8 avigulas ## Page 263

Vigvanatha’s suggested correction (under Sloka 4) yields : Bhabhabimbam = 28|8+0|16 = 28|24 arigulas Remark : Visvanatha gives another method for obtaining bhabhabimbam (and its correction) : 1200

1200 = 58]36-16 = 28/10 angulas :. Corrected bhabhabimbam = 28|10+0|16 = 28|26 arigulas Sloka 6 : This sloka explains the method of finding grasa (channam) of solar eclipse using tithi ghati. It is as follows. (i) Divide 170 by tithighati. ii) Add 4 to item (i). (iii) Subtract vyagusphuta bhujaméa from item (ii) [see note below]. (iv) Multiply item (iii) by 11, and divide the product by 7. This gives grasa of Ravi in angulas. 170 . u po Ga (cero) one stearate a anigulas Example : On a solar eclipse day, Tithighati = 64|49

## Page 264 GRAHANADVAYA SADHANADHIKARAH (CHAP. 8) $253 170 ererasr| , IL Grasa of Ravi = [eirses)-» 56 ss x <> = 7|20|58 angulas

Another method of finding grasa of Ravi using naksatraghati is as given below : (i) Divide 233 by naksatraghati. The result will be in amgas. (ii) Add 3 to item (i). (iii) Subtract vyagu sphuta bhujamsa from item (ii). (iv) Multiply item (iii) by 11, and divide by 7. This gives grasa of Ravi in angulas. ie, Grasa of Ravi 233 i =| ere huta bhuja| x — le } vyagu sphuta us x5 Example : On a solar eclipse day, Naksatraghati = 65/56 233 oer agt |x i <. Grasa = (ae-3)~ 56'45 }a = 7|12|35 angulas We notice a small difference of only about 8 pratyangulas. Note : To find vyagusphuta bhujaméa used above : For the given date : Saka 1432, Margasira krsna amavasya Wednesday, we have the following : gata tithi ghati = 51|50, esya ghati = 12|59 ## Page 265 $254 GRAHALAGHAVAM OF GANESA DAIVAJNA tithiyoga ghati = 64/49, naksatrayoga ghati = 65|56 [gataghati = 13|54 and esyaghati = 52|2] dinamanam = 26|4 gh. Ravi at the end of the tithi = 8” 5° 26’ 20" Rahu = 2" 11° 41'18", vyagu = 5® 93° 45'2", and natam = 1°30"

p01 50y 5 al nata corrected Ravi = Ravi ome natam = Kranti of the nata corrected Ravi, § = — 23° 43’ 40", Aksamsa = 25° 26’ 42' N NatémSa = 5-4} = 49° 10’ 22" (south) Vyagu bhujaméa = 6° 14'58" Sphuta vyagu bhujamsa = a x natam& — vyagubhuja = 3x49" 10°22" 6" 14'58" = 1° 56’ 45" Sloka 7 : This Sloka gives the method of finding the saryabimbam (diam- eter of the Sun) as follows : (i) Add 12° to Ravi. Consider the bhuja of it. ## Page 266 GRAHANADVAYA SADHANADHIKARAH (CHAP. 8) $255 (ii) Divide item (i) by 3. The result will be in arigulas, pratyangulas ete. Consider the number in the angula position. (iii) Consider 11- é = 10|50 angulas (i.e. mean siryabimbam) (iv) If Ravi is within 6 ragis from Mesa (i.e., 0° < Ravi <180°) subtract item (ii) from item (iii). If Ravi is within 6 raégis from Tuld (i.e., 180° < Ravi < 360°) add item (ii) to item (iii). This gives siryabimbam in arigulas.

(1) Adding 12° to Ravi, we get 8” 5° 26’ 20" + 12°= 8” 17° 2620" and its bhujamSa = 77° 26' 20" bhujamsa _ 77° 26' 20° Ce {= 25|48|46 arigulas Now, the number in the arigula position = 25 1 (3) LG = 10|50 angulas (4) Since Ravi is within 6 ragis from Tula, we have

## Page 268 CHAPTER 9 UDAYASTADHIKARAH (Rising and Setting of Planets) In this chapter the rising and setting of the Sun, the Moon and the planets are discussed. On the first day (pratipat) of the bright fortnight whether the Moon is visible or not is determined.

(i) Find the Sun, the Moon, Rahu and virahvarka (vyagu) at the end of the pratipat. Add 12° to both, the true Sun and the vyagu. (2) Find the cara of virahvarka (i.e., of vyagu) (8) Divide the cara by 56. This result is called prathamaphalam. If the vyagu is in the uttaragola, the above result is positive and if the vyagu is in the daksinagola the result is negative. cara of vyagu i rrathamaphalam = Le., Pi tp | 56 Note : Vyagu is said to be in the uttaragola (northern hemisphere) if 0° < uyagu <180° and in the dakginagola (southern hemisphere) if 180° < vyagu < 360°. Example : The given date is as follows : Saka 1532, Magha Sukla pratipat, Saturday ie., Jan. 15, 1611 (G). ## Page 269 $258 GRAHALAGHAVAM OF GANESA DAIVAJNA For the above date, Cakra = 8, Ahargana = 1036 Sravana naksatra ghati = 28|25, Siddhayoga ghati = 40|8 Mean Sun = 97 6° 12'37", Mean Moon = 9% 19° 38' 33" Candrocca = 8® 20° 54'28", Rahu = 2 10° 3' 25" As given in the paficdriga, Tithighati = 7 gh. (ie., the end of pratipat after sunrise) At this instant we have Mean Sun = 9F 6° 19/31", Mean Moon = 9% 21° 10/47" Rahu = 2% 19° 3'3", Cara = 106’ True Sun = 9¥ 7° 9' 44", True Moon = 9¥ 18° 52"12" True daily motion of the Sun = 61'10"

True motion of the Moon = 735'01"

We shall find the instant of the ending of the pratipat. After the instant of new moon, for the end of pratipat, the moon has to cover 12° away from the Sun. Now, True Moon — True Sun = 11° 49’ 28". To complete 12° the Moon has to cover, relative to the Sun, 12° - 11° 49’ 28" = 10' 32". The difference in the daily motions of the Moon and the Sun 735’ 01" -61'10" = 673’ 51". Therefore, the balance of time taken till the end of pratipat is ## Page 270 UDAYASTADHIKARAH (CHAP. 9) ce 10’ 32” x 608" ~ 56 vig. 67351" Adding this to the paficaniga given tithi ghati, we get End of pratipat = 7gh +0|56gh = 7|56 gh

At the end of pratipat, we have

True Rahu = 2 10° 3'1", vyagu (virdhvarka) = 6® 97° 0' 40" Adding 12° to both true Ravi and vyagu, we get respectively 9” 19° 341" and 7* g° 0! 40". For this, cara = 70. Note : In the text it is printed as 6” in place of 9” in the case of the Sun. Now, we have prathamaphalam = eu a15|o Since vyagu is in the daksinagola, the prathamaphalam is negative. ie, | Prathamaphalam = -1/15|0 Slokas 2 and 3 : These two élokas explain the method of finding dvitiyaphalam, trtiyaphalam and caturthaphalam as follows. ## Page 271

(1) To find dvitiyaphalam (the second result) : (i) Consider satribha sayana Ravi. (ie., Nirayana Ravi + 3" + Ayanaméa)

(ii) Multiply the cara and the magnitude of prathamaphalam. (iii) Multiply the palabha by 2. Square the product. (iv) Divide the result of step (ii) by that of step (iii). This result is called dvitiyaphalam. |Prathamaphalam| x Cara ie, Dvitiyaphalam = (2 x palabha)* It is positive if Ravi and the Vyagu are in different golas, and negative if they are in the same gola. (2) To find trtiyaphalam (the third result) : (i) Find the udayamana of sasadbha sdyana Ravi [i.e., sayana Ravi + 6 rasis]. (ii) Take the difference of 300 and udayamana of sasadbha sayana Ravi. (iii) Divide the above difference by 25. The result is positive if the udayamana of sasadbha sayana Ravi is greater than 300 vig. and it is negative if the udayamana of sasadbha sadyana Ravi is less than 300 vig. Udayamana -300 25 ie, Trtiyaphalam ## Page 272 UDAYASTADHIKARAH (CHAP. 9) — (3) To find caturthaphalam (the fourth result) : (i) Consider the difference between the dinamanam and the end of the tithi. (ii) Divide the above difference by 5. This gives caturthaphalam. dinamanam -endof thetithi ie, Caturthaphalam = -

Find prathama, dvitlya, trtlya and caturthaphalas. Add all the four. If the resulting sum is positive, then Candra (Moon) is visible. If the sum is negative, Candra is not visible.

Satribha saéyana Ravi = 1" 7° 11' 41", Cara of the above = 68'

|prathamaphalam| x cara Dvitiyaphalam = ep (@ x palabha® (afi5|0)x 68" = “(exsa5y? = * ojagss Since vyagu and Ravi are in different golas, dvitlyaphalam is positive.

Udayamana of the rasi of the above = 345 vig. i.e., udyamana of Simha rasi for Kasi] ## Page 273

Udayamana -300

Trtiyaphalam = 25

Dinamanam = 26|28 ghatis The end of the pratipat = 7|56 ghatis dinamana -endof the tithi

6|28 —7|56 _ eed +3/42)14 5 Since dinamana > end of the tithi, the caturthaphala is positive. To find Candrodaya on sukla pratipat : Sum of the four phalas = -1[15|0 +0|38|33+1|48|0+3|42|14 = + 4|53/47 Since the sum of all the four phalas is positive, the Moon is visible. (i.e., Candrodaya will have taken place before the end of the pratipat). Sloka 4 : This Sloka explains the method of finding the rising and setting of the planet Guru (Jupiter) by using masagana. It is as follows : (i) Add cakra to masagana. (ii) Divide cakra by 18. ## Page 274 UDAYASTADHIKARAH (CHAP. 9) ey) (iii) Subtract item (ii) from item (i). (iv) Multiply item (iii) by 2. Add 10 masas and 11 days to the product. (v) Divide item (iv) by 27. Consider the remainder. (vi) Subtract the above remainder from 27. (vii) Divide the above difference by 2. The result will be in rasis, améas etc. (viii) Subtract 15° from item (vii). (ix) Find the bhuja of item (viii). (x) Divide item (ix) by 12. If the result is within 6 rasis from Tuld (i.e., > 180°) it is negative. If the result is within 6 rasis from Mesa (i.e., < 180°) the result is positive. (xi) The result of item (x) is added to or subtracted from item (vii) accordingly. (xii) Consider the above result. Subtract 15 days from it. This gives the setting of Guru. (xiii) Add 15 days to the result of (xi). This gives the rising of Guru. Note : Here masas are counted starting from Caitra. Example : For the given date, Saka 1532, cakra = 8, masagana = 25. Completed years from the epoch (ie., Sa. $4. 1442) = 90. (i) Cakra + Masagana = 8 + 25 = 33 cakra 8 = 2% = om igd o7eh gyrig 13 13 0” 187 27 41 (ii) (iii) item (i) — item (ii) = 33" - 0™ 18¢ 278 41¥i8- = 32” 117 328" 190g ## Page 275 $264 (iv) v) (vi) GRAHALAGHAVAM OF GANESA DAIVAJNA [item (iii) x 2]+10" 11% = [(82]11/32|19) x 2] +10" 114 = 75 944 g4eh ggvie item (iv) 75|04|04|38 21|04/04|38 BS ceeded bd yr, ai 27 27 27 ie., the remainder = 21|04/04|38 27 — remainder = 27 -21)04\04|38 = 5)25|55|22 (vii) Dividing item (vi) by 2, we get (viii) (ix) (x) (xi) (xii) 5|25|55|22 plzelaalaz i = 2F 97°57’ 41" item (vii) - 15° = 2” 97°57’ 417-15" = 27 12°57'41" Bhuja of 2” 19° 57'41" = 72° 57'41" Bhuja _ 7257/41 boo 27 a 7 448 Now, the sum of the results of items (vii) and (x) gives 227[57|41+6|4|48 = 9” 4¢ g&h novi item (xi) — 15 days = 3" 4¢ 98" ogvig _15¢ = 9m 194 98h oovig ## Page 276 UDAYASTADHIKARAH (CHAP. 9) a Here masas are elapsed counted from Caitra. Therefore the above result tells that Guru sets after 2” 19% 28" 29%8 from Caitra ie., on Jyestha krsna paficami. (xiii) item (xi) +15 days = 3™ 44 28h ggvig , 154 = 3™ 194 28h ogvig This implies that Guru rises after 3” 194 2& 99% from Caitra. (ie., on Asadha Krsna Paficami). Note : Guru is also called Brhaspati and Mantri. Sloka 5, 6, 7 : These three Slokas explain the method of finding, asta (setting) and udaya (rising) of Sukra, as follows : (a) To find the rising of Sukra in the west and setting in the east : (1) Multiply cakra by 17. (2) Divide cakrax17 by 45. Add the quotient which will be in masas etc. to item (1). (3) Add masagana to item (2). (4) Add 64” to the product 5 x item (3). (5) Divide item (4) by 99. Consider the remainder. (6) Subtract item (5) from 99 months. (7) Divide item (6) by 5. The result will be in masas etc. (8) Consider the result of item (7). Add 36 days to it. This gives the rising of Sukra in the west. (pagcimodaya of Sukra). (9) Subtract 36 days from result of item (7). This gives the setting of Sukra in the east (purvasta). ## Page 277

Example : Masagana from Caitra = 25, cakra = 8 (1) 17x cakra = 17x8=136 months (2) RGIS 4 (17 cara) = 8 4 136 = 139" 04 408 (3) Masagana + item (2) = 25 + 139" 04 40% = 164 04 408 (4) 64 +[5x item (3)] = 64" + (& x 164" 07 408) = gg4™ 3¢ 2088 m ad ongh m gd oggh (5) 884” 3° 20 = 8+ 22 3% 20 99 99 ie. remainder = 92” 3% 998 (6) 99” — item (5) = 99” -92 34 208 = 6” 26% 408 item (6) _ 6” 264 408 @ 5 5 5 ee (8) item (7) + 36 days = 1” 114 208 +367 = 2174 208 This implies that Sukra rises in the west after the lapse of 2” 174 208" from Caitra. i.e. , on Jyestha krsna trtiya. (9) item (7) - 36 days = 1™ 117 208 - 367 = 0” 5% 208" This implies that Sukra sets in the east after 54 20%" from the begin- ning of Caitra, i.e., on Caitra Sukla sasti. ## Page 278

To find the rising of Sukra in the east (parvodaya of Sukra) and the setting in the west (pagcimasta of Sukra) (1) Consider the masas etc. obtained previously in step (7). (2) If the above is less than 9 months and 27 days, add 9” 272 to it. If it is greater than 9” 27%, subtract 9” 974 from it. (3) Consider item (2). Add 4 days to it. This gives rising of Sukra in the east. (4) Subtract 4 days from item (2). This gives setting of Sukra in the west.

(1) Masas etc. obtained previously in item (7) : 1™ 114 2084. (2) Since it is less than 9” 27%, adding 9” 274 we get 1 117 208" +9" 274 = 11” 084 208” (8) item (2) + 4 days = 11” 08? 208% + 44 = 11" 124 208" This implies that Sukra rises in the east after the lapse of 11” 12¢ 2087 from Caitra i.e., on Phalguna Suddha trayodasi. (4) item (2) - 4 days = 11” 8? 208" - 4 days = 11” 044 208" This implies that Sukra sets in the west after the lapse of 11’ 04% 208" from Caitra ie, on Phdlguna suddha paficami. ## Page 279

Sloka 8 : This Sloka explains the method of finding parivartana kala (recurrence time) of the rising (udaya) and setting (asta) of Guru and Sukra. That is, by knowing the asta and udaya to find the next udaya and asta of Guru and Sukra. (1) In the case of Sukra, by adding 19 months and 24 days (tithis) to the rising and setting times we get again the next rising and setting of Sukra. Similarly by subtracting the same (19’247), the earlier rising and set- ting of Sukra are obtained. (2) In the case of Guru, the period to be added or subtracted is 13” 157.

(i) The interval between successive rising and setting or between two suc- cessive risings (or settings) of planets given above are only mean values. In reality these intervals vary from year to year based on true positions of the Sun and the planets. For example, the setting (asta) of Guru in 1995 was on December 6. The next setting of the same planet was January 8, 1997. The interval (parivartana) between these two successive settings of Guru is given by (1997-1-8) — (1995-12-6) = 17 1" 24 = 13” 24. The interval given in the text is 13” 15% . Similarly, the interval between successive risings is given by 1999-12-30 and 1997-2-2 is 17 1” 2 ie, 13” 24. (ii) The interval given in the text, 13” 154 is actually 13.5 lunar months which is equivalent to 13 months, 8 days of the civil calendar. Sloka 9 : This Sloka gives the method of finding sara of Candra as fol- lows. (1) Consider the bhuja of vyagu. If bhuja is in the first rasii.e., less than ## Page 280 UDAYASTADHIKARAH (CHAP. 9) azo) 30°, add 5x bhuja to bhuja of vyagu. This gives the Sara in arigulas. (2) If the bhuja of vyagu is in the second ragi (i.e., 30° < bhuja < 60°) add 47 to bhuja to get the Sara in ang. (3) If the bhuja is in the third radi (ie., 60° < bhuja < 90°) then add 77 sl to 5x bhuja to get the Sara in ang. Example : Candra = 5° 26°20", Rahu = 2” 11° 4118" Vyagu = 5% 93° 4502", _ bhuja of vyagu = 6° 14'58" Since the bhuja of vyagu is in the first rasi we have bhuja

= 6° 14'58"+ se = 9[22|27 argulas Since vyagu is in the uttaragola (i.e., < 6 rasis), the Sara is also in the uttaragola (i.e., positive or nothern). Sloka 10 : This Sloka gives another method of finding the sara (of the Moon) using khandas. It is as below : (1) The Sarakhandas are 0, 16, 15, 14, 18, 11, 9, 7, 4, 1. Using these khandas for (Moon — Rahu) find Sara by following the same ## Page 281

procedure which is adopted to find the kranti using khandas. Example : Vyaguvidhu i.e., (Moon — Rahu) = 5% 93° 45' 9" (Note : The moon is also called Vidhu.)

@ 6° 14'58' =0+8 14'58' 10 remainder = 6° 1458" This implies that the gata (elapsed) khanda = 0 and esyakhanda = 16 (from Table 9.1 of khandas) (2) Esyakhanda xremainder 16x 6° 14'58" 10 7 10 99° 59' 28" = 9975928" _ 9159 anigulas 10 [59 angu (3) gatakhanda + item (2) = 0+9|59 arigulas ive., Sara = 9|59 arigulas Explanation : The khandas for finding the Sara of the Moon given in GL are as follows :

Vyagu bhuja O° | 10° | 20°} 30°} 40°} 50°} 60° | 70°} 80°| 90° Difference 1615 14 #13 «#11 9 7 4 1 Sara(ang) O | 16 | 31 | 45 | 58 | 69 | 78 | 85 | 89 | 90 ## Page 282 UDAYASTADHIKARAH (CHAP. 9) al 30°: Sara =0+16+15+14=45 angulas For Vyagu= 160°:Sara =45+13+11+9 =78angulas 90°: Sara =78 +7 +4+1=90 ani gulas Note : The actual formula for gara is Sara = 90 sin (Vyagu) angulas (Note that 90 arig. = 270'). For example for Vyagu = 30°, 60° and 90°, we get from the above for- mula Sara = 45, 78 and 90 angulas. According to GL, the Sara of the Moon is obtained as follows. (® For 0° < bhuja < 30°, Sara = 90 sin (vyagu) ve, Sara ~ 90° x| 2x eyagul = 24 xovagu = 2 yya .e., ara = 90° x|—2 _x _— £ ; - 120x385 78"! = 35 3g Wwasu [Ref. GL, Tripragnadhikara, Sl. 22 : 120sind = 9 =. sind~—2—0 where 9 is in radians]. 35 120 x 35 (ii) For 30° < bhuja < 60°

bhuja + (78 - 45) ¢ 30 = 45 + (1.1) bhuja ## Page 283 s272 GRAHALAGHAVAM OF GANESA DAIVAJNA GL takes it as 47 + bhuja (approxly). (iii) For 60° < bhuja < 90°,

bhuja + (90-78 : 30 12 = 78+— je 30 Dhuia GL takes this as 7+ ; xbhuja. Remark : The following table provides a comparison between the approxi- mated values of Sara as per GL and the actual values using the formula, Sara = 90° sin (vyagu) arigulas.

Vyagu 0° 10° «20° 30° 40° 50°60", 70°— 80° 90° GL 0 16 St 45) 58) 69: cy = tele 233) 90 Actual 0 15.63 30.78 45 57.85 68.94 77.94 84.57 88.63 90 Sloka 11 : This Sloka gives the direction of udaya (rising) and asta (set- ting) of the planet. (1) The planet, whose longitude and daily motion are less than those of the Sun, rises in the east. (2) The planet having greater longitude and daily motion than those of the Sun, rises in the west. (3) The planet whose longitude is greater than the longitude of the Sun, and whose daily motion is less than that of the Sun will set in the west. ## Page 284 UDAYASTADHIKARAH (CHAP. 9) oa (4) The planet whose daily motion is greater than that of the Sun and whose longitude is less than that of the Sun will set in the east. Sloka 12 : This Sloka gives the kalaméas of planets. Table 9.3 : Kalaméas of planets Planets | Ca Ku Bu Gu Su Sa Va Bu Va. Su Kalamga | 12° 17° 13° 11° 09° = 15° 12° 08° Note : If Budha and Sukra are in retrograde motion (vakragati), subtract 1 from their usual (direct motion) kalaméas (as shown in Table 9.3). Sloka 13 : This Sloka gives the patas of the five planets (taragrahas) as below:

| Planets | Budha Kuja Guru Sukra_— Sani | Pataméa | 20° 40° 80° 60° 100° | | In the case of Budha and Sukra subtract sighrakendra (obtained using ahargana) from pataméa to get the corrected patamsa. Sloka 14 : The method of finding sighrakarna of planets is explained in this Sloka as follows : (1) The six khandas using which we can find sighrakarna are: 1, 2, 3, 4, 4, 2. (2) If the Sighrakendra of the planet is greater than 6 rasis, subtract it from 12 rasis. (3) The number in the rasi position gives the total number of elapsed khandas. Find the sum of all the gata (elapsed) khandas. ## Page 285

(4) Now, we have ; Esyakhandax Remainder Sighrakendraphalam = Sum of the elapsed + 30 khandas (5) Divide the result of (4) by 1, 2, 4, 1, 7 and subtract the quotients from 18, 15, 13, 19 and 12 respectively. The result gives sighrakarna of Kuja, Budha, Guru, Sukra and Sani respectively. Example : Given date is Saka 1534, Vaisakha Sukla Parnima (refer to the example under Sloka 10 in Chapter 3). For the five tara grahas (star- planets) we have their sighrakendras (anomaly of conjunction) as given in Table 9.5.

Planet Sighrakendra Kuja 3* 01° 04'57" Budha 1® 16° 25°17" Guru 8* 21° 20'58" Sukra 3 04° 59’ 52" Sani 2 92° 50’ 00" The above values are the second sighrakendras while finding the true positions of planets. (1) To find the sighrakarna of Kuja : Sighrakendra = 3" ° 4°57" (i) In the rasi position, we have 3. This implies that the first three khandas are over.

(i) Sum of the elapsed khandas = 6. (ii) Remainder = 1° 4'57", - Esyakhanda = 4 ## Page 286 UDAYASTADHIKARAH (CHAP. 9) cD Remainder x Esyakhand (iv) Sum of the elapsed khandas + a Vo 4'57"x4 LAST x4 30 =6 = 6|8|39 (v) Dividing by 1 (in the case of Kuja), subtracting the result from 18, we get the Sighrakarna of Kuja as 6|g|39

= 18-6|a|39 = 11[51]21 (2) To find the Sighrakarna of Budha : Sighrakendra of Budha = 1” 16° 25'17" Gatakhanda=1, Esyakhanda=2, and Remainder = 16° 25'17"

Esyakhanda x remainder

ey a loecoaS 2\5|41 30 Dividing the above by 2 and subtracting the resulting quotient from 15 we get the sighrakarna of Budha. 2 2|5|41 ie, Sighrakarna of Budha = 15 -"— = 13)57|10 (3) To find the sighrakarna of Guru :

## Page 287

Sighrakendra phalam = Sum of the gatakhandas + Esyakhanda xremainder _ ajgln1 30 7 7j9l11 Sighrakarna of Guru = 13-28 5 1if12|42 (4) To find the Sighrakarna of Sukra :

Sighrakendra = 3° 4° 59'52"

Sighrakendrat phalam = 6|39|58

6|39|58 yee 1 (5) To find the sighrakarna of Sani :

Sighrakendra = 2¥ 9° 50'0"

3i17\0 = 11/31/52 7

Sloka 15 : This Sloka explain the method of finding the sara (latitude) of Kuja, Budha, Guru, Sukra and Sani. It is as follows. (1) Consider the difference between the manda corrected planet and its pata. (2) Find the declination (kranti) of the above result. [Here the nirayana planet from which the pata is subtracted has to be considered]. ## Page 288 UDAYASTADHIKARAH (CHAP. 9) 8277 (3) In the case of Budha and Sukra consider [Manda corrected planet — (Pata — Sighrakendra)] (4) Multiply the declination obtained in step (2) by 23. (5) Divide the result of step (4) by Sighrakarna. This gives the Sara of the planet in anigulas. 1 (6) In the case of Kuja, subtract q of the Sara from the Sara to get the corrected Sara. (7) In the case of Guru, divide Sara (from step 5) by 2 to get the corrected Sara. (8) Multiply the Sara in arigulas by 3 to get the same in kalas. Divide Sara in kalas by 60 to get it in amSas. (9) The gara will be north or south according as the (manda corrected planet — pata) is in the northern or southern hemisphere. (10) The declinations of the above planets are to be corrected with their Saras. (1) To find Sara of Kuja : Manda corrected Kuja = 10" 3° 8’ 45", Pata of Kuja = 1% 10° Gi) Manda corrected Kuja - Pata = 10” 3° g'45"- 1" 10° = 8* 93° 8' 45" Gi) Declination of the above = 23° 4333" (south) (obtained using Krantikhandas as explained in Chapter 4). ## Page 289 $278 (iii) (iv) (v) (vi) (2) qi) ii) (iti) (iv) GRAHALAGHAVAM OF GANESA DAIVAJNA Sighrakarna of Kuja = 11|51|21 28 x (declination) = 23 x(23/43|33) = 545° 41’ 39" 545° 41' 39"

= 46)1/38 angulas

Corrected Sara :

sara

46|1|38 4 = 46|1/38 - = 34|31|14 angulas (south)

Manda corrected Budha = 1 5° 3/15", Pata = 08 20° Sighrakendra of Budha obtained from Ahargana : 1” 17° 14'50" Budha — (Pata ~ Sighrakendra) = 17 5° 315" -(0F 20° -17 17° 1450") = 2” 2° 18' 05" Declination of the above = 21° 0'51" 23x declination = 21° 0'51"x 23 = 483|19|33 ## Page 290 (v) (vi) (3) @ (ii) (iii) (iv) (v) (vi) (vii) (4) UDAYASTADHIKARAH (CHAP. 9) —

483{19|33 Sara of Budha = ——)— 13|57|10 = 34|88|24 arigulas (north)

Manda corrected Guru = 4" 12° 19' 44", Pata = 2% 20° Guru — Pata = 17 99° 12" 44" Declination of (Guru ~ Pata) = 18° 49'11" (north) 23x 18° 49°11" = 432[51|13

7 432)51|13 Sara of Guru = Tapa = 38|36|26

38|36|26 . SF 19{18|13 ang. (north)

To find gara of Sukra :

## Page 291 $280 @ Gi) iii) (iv) (v) (5) @ (ii) (iii) (iv)

GRAHALAGHAVAM OF GANESA DAIVAJNA

Sighrakendra of Sukra = 3” 5° 41'35"

Sukra — (Pata — Sighrakendra) = 17 5° 95’ 25"— (2 — 3% 5° 41'35") = 27 11°07'0" Declination of the above = 22° 32'2" (north) 28 x (22° 322") = 518|16|46