सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

171 we have an approximation of the Mean planets. Treating these as the Mean planets, again obtaining the Manda and Śīghraphalas and again applying them inversely and repeating the process till constant values are obtained, we have by this method of successive approximation the Mean planets required. Comm. The method of successive approximation is clear. Verse 46. To obtain the equinoctial shadow. Convert the Ayanāṁśas into minutes of arc, and divide by the mean daily motion of the Sun; then we have the number of days before the Meṣa or Tulā Saṁ- krānti day, or before Makara and Karkaṭaka Saṁkrānts days, when the Sun will be in equinoxes or Solstices respectively. The mid-day shadow of the Sun cast by the gnomon on such an equinoctial day, will give us the equinoctial shadow required. Comm. The palabhā or equinoctial shadow as it is called is the length of the shadow cast by a gnomon taken to be of 12 units in length, (measuring the shadow also in the same units), at noon of an equinoctial day. In other words, if this shadow be of s units, clearly s / 12 = tan ϕ (Vide fig. 18). The Ayanāṁśas are the degrees of the arc of the ecliptic in between the Hindu Zero point of the Zodiac and the first point of Aries which is now behind the former due to the phenomenon known as the precession of the equinoxes. They are called Ayanāṁśas because the solstices are also behind the Makara and Karkaṭaka Saṁkrānti points of the Hindu Zodiac or points which have Hindu longitudes 270° and 90° resply, by the same arc. The Hindu astro-

172 nomers came to know that the solstices are preceding by observations made with the gnomonic shadow at mid-day around the solstitial days. The day on which the maxi- mum mid-day shadow is cast by the gnomon is the Winter solstitial day whereas the day on which the mid-day gnomonic shadow is maximum in the southern direction (assuming the place to be of northern latitude and ϕ < ω) or minimum in the northern direction (ϕ > ω) is the summer solstitial day. Calculating the Sun's longitude on that day at noon, we know how far the solstitial points have preceded behind the points of the Hindu Zodiac which have Hindu longitudes 90° and 270°. The word अयनांशाः has therefore the meaning अयनविलोमगत्यंशाः where the Samāsa may be viewed as a मध्यमपदलोपी समासः Verses 47, 48. Calculating the five fundamental H sines of a point of the Zodiac pertaining to a point of the ecliptic. The declination has to be computed from the sum of the Hindu longitude of that point and the Ayanamsas. Similarly if it be required to find the time before or after the rise of a point of the ecliptic, we have to compute them from the sum of the longitude of that point and the Ayanamsas. H sin 24° × H sin λ ─────────────────── = H sin δ I R √(R² — H sin² δ) = H cos δ = Dyujyā II δ = H sin⁻¹ (H sin δ) III s ── × H sin δ = Kujyā IV 12 Kujyā × R ───────── = Charajyā V H cos δ H sin⁻¹ (Charajyā) = Charam VI Comm. (1) Let rA☉ be the ecliptic where r = Vernal Equinox, A = first point of the Hindu Zodiac,

in between r and ☉, not shown in the figure ☉ = the Sun. Let rA = a° = Ayanamsas defined before so that r☉ = (a + λ). From the spherical triangle r☉M, by Napier's rule, sin δ = (sin λ + a) sin ω which in Hindu trigono- metry becomes H sin δ = [H sin (λ + a) × sin ω] / R In Hindu Astronomy the obliquity of the ecliptic was taken to be 24°. The value of the obliquity is now 23°–27' approximately and it has been know that this has been decreasing. At the time when the Hindu Astronomers observed this, it should have been greater than 23°–27' so that if it was taken to be 24°, their observations were not far from truth. This means that the antiquity of Hindu Astronomy might be far more than what the Moderns estimate it to be. (2) How the formula I was derived in Hindu tri- gonometry was as follows. Let ♋ be the summer solstice ♋C, ☉B, the perpendiculars dropped from ♋ and ☉ on the line of intersection of the Equatorial and Ecliptic planes namely rBC. Let perpendiculars be dropped from ♋ and ☉ on the plane of the Equator. Let them be ♋N, ☉D, Join NC and DB. Then ☉B̂D = ♋ĈN = the dihedral angle between the two planes = ω; ♋C = R since in Hindu trigonometry H sin 90° = R. Also ♋N = H sin ♋E (in Hindu trigonometry) = H sin ω; ☉D = H sin ☉M = H sin δ; ☉B = H sin r☉ = H sin (λ + a) where λ is the Hindu longitude of the Sun and a = Āyanamśas. It will be seen that ☉D is a segment of the

line of intersection of the planes PCM a plane perpendi- cular to the plane of the Equator, and ☉BD a plane per- pendicular to the Ecliptic plane. Since ⚍C 11 ☉ B and ⚍N 11 ☉ D and BD☉ = CN⚍ = 90°, so the two tri- angles are congruent ∴ ☉B / ⚍C = ☉D / ⚍N ie. H sin (λ + a) = (R × H sin δ) / (H sin ω) ∴ H sin δ = [H sin (λ + a) H sin ω] / R NC = ⚍L = √(⚍C² — ⚍N²) = √(R² — H sin² δ) = H cos δ. ⚍L is called Dyujyā as explained below in note (3). Another way of looking at the similarity of the tri- angles C⚍N and B☉D is from the fact that they are formed by the intersection of parallel planes, both of which are perpendicular to the plane of the Equator. This idea will be elaborated when we explain fig. 21, wherein the so-called latitudinal triangles will be shown to be formed by the intersection of the plane of the Equator and planes of diurnal circles with the planes of the horizon and the prime vertical. In fact, the plane of a great circle and the parallel planes of the corresponding small circles form the same dihedral angle with the planes of the celestial sphere namely the planes of the meridian, horizon and prime-vertical so that right-angled triangles formed by their intersection will be all similar. Formula II is derived from the formula H sin² δ + H cos² δ = R² derived from fig. 6 from which formula, formula III follows.

176 (3) Why H cos δ is called Dyujyā in formula II is clear from fig. 20 wherein p is the celestial pole, (C) is the Fig. 20 celestial Equator and (C) is the diurnal circle of a celestial body say the ☉ ie. the Sun. Let ☉N be the perpendicular dropped from ☉ on the plane of the Equator so that ☉N = H sin ☉A = H sin δ CN = H cos δ = C☉ = radius of the diurnal circle called Dyujyā (Dyu = day) द्युवृत्तत्रिज्या = द्युज्या (Madhyamapadalōpi Samāsa) formulae IV, V and VI will be dealt with in the next chapter Triprasnādhyāya more elaborately but one has to under- stand what Charajyā is to follow the subsequent verses of this chapter so that we shall give its location and defini- tion in the light of fig. 21. Let fig. 21 represent the celestial sphere in which SEN = Horizon, Z = Zenith, N = Nadir Ez = prime vertical EQR = celestial equator.

176 Fig. 21 PP′ = Polar axis, SBS′ = the diurnal circle of a celestial body S; EW = the East-West line, SS′ = Udayāstasūtra or the join of the rising and setting points S, S′. This SS′ is evidently a diameter of the diurnal circle which is bisected by the plane of the horizon; PEP′ is called the unmandala or the Equatorial horizon. PSA is the decli- nation circle of S cutting the Equator in A. EA is called the charam whose H sine is called Charajyā. The H sine of SB in the diurnal circle is called Kujyā so that as corresponding lines in the diurnal circle and the Equator stand in the ratio H cos δ : R, Kujyā / Charajyā = (H cos δ) / R I

177 In the beginning of the Triprasnādhyāya, Bhāskara says “साक्षे देशे खगोलवलयानां, तिरश्चीनभगोलवलयानां च संपातात्त्र्यस्राणि क्षेत्राण्युत्पद्यन्ते, तान्यक्षक्षेत्रसंज्ञानि” ie. In a place having a latitude the diurnal paths of stars and planets will be inclined to the fundamental circles of the celestial sphere namely horizon, meridian and prime vertical and so their intersection gives rise to what are called latitudinal tri- angles, in which the angles would be ϕ, 90-ϕ and 90 where ϕ is the latitude. Thus in fig. 21 the projections of the triangles ESB, EDB, DSB, EDF, EBF, ESF and EQG on the meridian plane will be all triangles in which the angles will be ϕ, 90-ϕ and 90° so that they are all latitudinal triangles. These are all similar to the funda- mental gnomonic triangle gmn of fig. 18 wherein also the angles are ϕ, 90-ϕ and 90°. Here, there is one important point to be observed. We have said “their projections are all similar”. In fact the corresponding spherical triangles enumerated above are all apparently similar, though they are not viewed in modern astronomy as regular spherical triangles, because all the three sides of the triangles are not arcs of great circles. It is not possi- ble to apply Napier’s rules to these triangles for the reason mentioned above. None the less, the property of simil- arity of the projected right angled triangles is made use of in Hindu Astronomy to obtain the magnitudes of the sides of the projected triangles. EW is called the prāk-pratīchī sūtra, SS′ the Uda- yāsta sūtra; similarly if lines through F, B, D, A parallel to EW be drawn, these sūtras will intersect the meridian planes in points which constitute the projected triangles mentioned above. Let us study these triangles which we connote by the same letters with lowered indices. Thus for example E₁ S₁ B₁ is the projected triangle of ESB where of course E₁ will be the centre of the celestial sphere. The lines drawn parallel to SS′ or EW through B, D etc. will be denoted as BB′, DD′, FF′ etc. 23

178 In the triangle E₁ S₁ B₁, E₁ S₁ is called Agrajyā, S₁ B₁ Kujyā, and E₁ B₁ Krāntijyā. Of the sides of ESB. ES and EB are arcs of great circles whereas SB is the arc of a small circle. In the projected triangle E₁ S₁ B₁, E₁ S₁ will be equal to the perpendicular from S on Eω, which will be H sin (ES) and is called Agrajyā ; E₁ B₁ will be equal to the perpendicular from B on Eω which will be H sin BE and as such called Krāntijyā. But S₁ B₁ which is equal to the perpendicular from S₁ on BB' the diameter of the diurnal circle is the H sine of SB in the diurnal circle and is called Kujyā. It will be noted that the per- pendiculars from S on Eω, and BB' and the perpendicular from B on Eω do not form a triangle by themselves but by the theorem of three perpendiculars, if SL be the perpendicular from S on the plane of the unmandala ie. the great circle EBP, and if LM be perpendicular from L on Eω, SM will be perpendicular on Eω. Here the perpendicular SL will be the Kujyā, and SM the Agrajyā, whereas LM is not actually the H sine of BE but is equal and parallel to it. Similarly take the spherical triangle SAE. This is a regular spherical triangle because the three sides are arcs of great circles. Napier's rules can be applied to this triangle and we have the formula sin SA = sin SE sin SÊA or in Hindu form H sin SA = (H sin SE × H sin Ê) / R or RH sin δ = Agrajya × H cos ϕ ie. Agrajyā = (RH sin δ) / (H cos ϕ) II Again sin EA = tan SA × tan Ê where H sin EA is called Charajyā and tan E = cot ϕ so that Charajyā = tan δ tan ϕ in modern form, and the Hindu form is <u>R tan δ tan ϕ</u>

179 From I H cos δ / R = Kujyā / Charajyā so that Kujyā = H cos δ / R × R tanδ tanϕ = H sinδ tanϕ III In Triprasnādhyāya, the elements of all the eight latitudinal triangles are found by using their similarity with the fundamental gnomonic triangle. The important elements that will enter into computation are (a) Charajyā (b) Kujyā (c) Agrajyā (d) Taddhriti ie. S₁ F₁ the proje- ction of SF on the plane of the meridian (e) Sama-Sanku = E₁ F₁ = H sin EF (f) Krāntijyā = E₁ B₁ = H sin EB = H sin δ (g) Lambajyā = H sin QS = H cos ZQ = H cos ϕ (h) Akshajyā = H sin ZQ = H sin ϕ (i) B₁ D, = Ud-Vritha-Sanku = H cos ZB (j) Dinārdha-Sanku = H cos Zq. The following points will be noted. (i) In the triangle E₁ Q G₁, the projected triangle of EQG on the meridian plane, noting that E₁ is the centre of the celestial sphere E₁ Q = R, QG₁ = H cos ϕ and E₁ G₁ = H sin ZQ = H sinϕ. (ii) S₁f₁ = H sin SB + H sin fB (both the H sines pertaining to the diurnal circle. (iii) Sama Sanku is the H cosine of the Zenith- distance when the Sun or celestial body is on the prime-vertical. (iv) BD is an arc of the great circle ZB, so that the Unmandala Sanku is the H cosine of ZB. (v) Dinārdha — Sanku = H cosine ZQ. (vi) These Sankus are the H sines of altitudes or H cosines of Zenith-distances and they are in the planes of the respective great circles.

180 (vii) Lambāṁsa-chāpa is the arc of the colatitude QS where S is the South point so that Lam- bajyā is the H sine of QS or H cosine of ZQ. This Lambajyā will be seen to be the diameter of a small circle parallel to the Equator and passing through Z. Verses 49-51. A different method of obtaining chara. This chara can be had by the so-called chara-segments of the locality using a process similar to that of finding the H sines of the smaller table of nine H sines using λ/3 where λ is the Sāyana longitude of the Sun (i.e. The Hindu longitude plus the arc of ayanāṁśas is called the Sāyana longitude or the modern longitude measured from v along the ecliptic to the Sun). Find the charas of 30°, 60° and 90° of the ecliptic measured from v. Subtract the first from the second, the second from the third. Thus we have C30°, C60°-C30°, C90°-C60°, (where Cθ° signifies the chara of θ°) which are called chara-Khandas or chara-seg- ments. The equinoctial shadow multiplied by 10, 8, 3⅓ gives the approximate values of the chara-segments in Vinādīs (a sidereal day is divided into 60 nādīs and each nādī consists of 60 Vinādīs). The chara-segments thus measured in Vinādīs are rather approximate. If further exactitude is required, better take the arc in units each of which rises in ⅙th of a Vinādī (This ⅙th part of a Vinādī is known as a prāṇa i.e. the duration of the interval between two inhales of a healthy person reckoned as 4″ of time). Comm. The charas of 30°, 60° and 90° of modern longitude are the values of the arc EA, (Ref. Fig. 21) when S has longitudes 30°, 60° and 90°. We have the folmula Charajyā = R tan δ tan ϕ

181 Taking the equinoctial shadow equal to 1 Angula means tan ϕ = 1/12. As charajyā is proportional to tan ϕ, for any equinoctial shadow of s angulas, tan ϕ being equal to s/12 the charajya got above is to be multiplied by S only to obtain the charajyā in any place where the equi- noctial shadow is s angulas. Putting the modern longitu- des equal to 30° & 60°, if the corresponding declinations be δ₁, δ₂ sin δ₁ = sin 30 sin ω, sin δ₂ = sin 60 sin ω. Tak- ing ω = 24° and applying logarithmic tables log sin δ₁ = 9.6990 + 9.6093 = 9.3083 so that δ₁=11°–44′ log sin δ₂ = 9.9375 + 9.6093 = 9.5468 so that δ₂=20°–38′ Now from the formula for charajya cited above viz. H sine (chara) = R tan δ tan ϕ or sine (chara) = tan ϕ tan δ, putting tan ϕ = 1/12 and applying tables, using the values of δ got above, charajya for 30° = (tan 11°–44′) / 12 and charajyā for 60° = (tan 20°–38′) / 12 so that log (sine chara) = 9.3175 – 1.0792 for 30° and for 60° log (sine chara) = 9.5758 – 1.0792 ∴ Chara for 30° or C (30)° = 59′ and C (60°) = 1° – 48′ Converting these arcs into their rising times at the rate of 6′ per Vinadi, we have C (30°) = 10, and C (60°) = 18 Noting δ₃ = ω, sin (chara) for 90° = (tan 24°) / 12 so that log sin (C 90°) = 9.6486 – 1.0792 = 8.5694 so that C (90°) = 2°–8′ = 21⅓ Vinādis. Thus (C 30̄) = 10, C (60) – C (30) = 18 – 10 = 8 C (90°) – C (60°) = 21⅓ – 18 = 3⅓ so that the chara Segments are respectively 10, 8, 3⅓ as given by Bhāskara.

182 For a given place of equinoctial shadow equal to s″, we have to multiply 10, 8, 3⅓, by s, which will be the chara- segments for the place. The meaning of the first half of the verse 49 is as follows.—Let the equinoctial shadow for a place be 3 Angulas. Then the chara-segments for that place are 30, 24, 10. These are three in number for a longitude λ of 90°. If the longitude be 44° (say) then proceed as we have done to find sin 24°, using the method of Bhogya Khanda Sphutīkaraṇa with respect to the table of H sines namely 21, 20, 19, 17, 15, 12, 9, 5, 2. Proceeding as directed the Sphuta Bhogya Khanda for 14° is ( (3 × 14) / 30 = 7/5 ; 30 - 7/5 = 28⅗ ; (14 × 28⅗) / 30 = 2002 / 150 = 13⅓ ; 30 + 13⅓ = 43⅓ ). Proceeding according to the modern formula we have Charajyā = tan δ tan ϕ where ϕ = 3/12, and sin δ = sin 44° sin 24° log sin δ = 9.8418 + 9.6093 = 9.4511; δ = 16°-25 log tan δ = 9.4693 ∴ log sin (chara) = 9.4693 + log tan ϕ = 9.4693 + log 3/12 = 9.4693 - .6021 = 8.8672 ∴ Chara = 4°-14' = 254 / 6 = 42⅓ Vinadis whereas we have got by the Hindu method 43⅓ which is near the truth. Verse 52. To find the durations of day and night. Fifteen ghatis increased or decreased by the Chara- nadis, according as the Sun is in the northern hemi- sphere or southern, gives half the day of the locality and the difference of 30 nādis and the above half-day gives half the duration of night. Comm. Ref. fig. 21. Let S be the Sun rising in the northern hemisphere when his declination is north. Then

183

from the figure ÂPQ is the rising hour-angle = ÂPE + ÊPQ = ÂPE + 90° where 90° correspond to 15 nādīs. So we have to add ÂPE expressed in nādīs equal to the rising time of AE. Let us find the duration of the day for the place where s=3″ and when λ of s=44° ; the latitude of the place will be 14° (from tables) when λ 44°, we have found above that 42-20 Vinadis is the chara expressed in time. Hence half the day = 15-42-20 or duration of day 31-25 ; duration of night = 28-35. Note (1) In modern astronomy we have the formula cos h = — tan ϕ tan δ where h is the rising hour angle. Putting h = H + 90° where H stands for the arc EA of fig. 21 cos (90+H) = — sin h = — tan ϕ tan δ so that sin H = tan ϕ tan δ ie. R sin (Chara) = R tan ϕ tan δ which accords with the formula found before. The word chara used for EA means etymologically रविसञ्चारवशेन दिन- प्रमाणे विकारः i e. the variation in 15 ghatis of the eqiunoctial half-day on account of the Sun's variation in declination, (2) If ϕ = 0, Chara = 0 so that duration of half- day is 15 ghatis ie. on the terrestrial equator, whatever be the Sun's declination, the length of the day will be always 12 hours. (3) Let δ = 0 so that chara = 0 ie. whatever be the latitude (provided ϕ ≯ 90-δ as we shall see shortly). the day and night will be each of 12 hrs. (4) Let δ be negative, so that the arc connoting chara ie. EA will be above the horizon, and consequently the duration of half-day will be less than 6 hrs. by the time that is taken for EA to rise. This can be seen other- wise also as cos h = — tan ϕ tan δ = + ve so that h < 90° which means half-day is less than 6 hrs. (5) We could also treat the case when ϕ is negative, but as this case is not in the purview of Hindu Astro- nomers who had only India in their mind and as such were concerned primarily with positive latitudes. If, how- ever, we consider a negative latitude, when δ is + ve, cos h will be positive and if δ is — ve, cos b will be negative. This means that when the Sun is in northern latitudes, the southern latitudes will have their day less than 12 hours and when the Sun is in southern latitudes, their day will be greater than 12 hrs.

184 Fig. 22 (6) Let ϕ + δ = 20°. (Ref. fig. 22) ie. imagine the Sun to rise at N, the north point so that RN + NP = δ + ϕ = 90°. Then the Sun's diurnal path will be entirely above the horizon, which means that what is called 'perpetual day' begins for that place on that day and lasts as long as δ ⩾ 90 - ϕ. For the same place, let δ =

• (90 - ϕ) = QS. In this case the Sun sets at S, and what is called perpetual night begins and lasts till the southern declination of the Sun is greater than 90 - ϕ. The duration of perpetual day can be found as follows which applies to the perpetual night as well. Fig. 23 (Ref. fig. 23). Let A be the point at which the declination is 90 - ϕ; let S be the summer solstice and let B be the point where again the declination is equal to 90 - ϕ. So long as the Sun traces the arc AB = 2 AS =

185 (90-rA), there will be perpetual day. But sin δ = sin λ sin ω so that sin λ = sin δ / sin ω . Putting δ = 90 - ϕ, sin λ = cos ϕ / sin ω ; Sin λ = sin γA = cos AS ∴ cos AS = cos ϕ / sin ω ∴ 2 AS = 2 cos⁻¹(cos ϕ / sin ω). Supposing AS expressed in degrees and assuming the Sun goes along the ecliptic with uniform motion, since he takes 365¼ days to trace 360°, to trace 2 AS, he takes (2 AS / 360) × 365¼ days = 365¼ / 180 × cos⁻¹(cos ϕ / sin ω) which is the length of the perpetual day. Verse 52. The correction known as Chara. The daily motion of the planet being multiplied by the chara expressed in asus and divided by the asus in a day viz. 21659 and the result being subtracted from or added to the planetary position at Sunrise according as the Sun is in the northern or southern hemisphere. The result is to be added to or subtracted from the planetary position at Sunset. Comm. The mean planets computed hold good at the Sun-rise at Lanka i.e. at zero latitude; they have to be converted to hold good at the local Sun-rise. In other words in fig. 21, the mean planet computed is B which is on the Lanka horizon, whereas we have to get S, the same planet on the local horizon. The position of S is earlier than that of B by the time interval indicated by the arc SB which is measured by EA the chara because the position B on the Lanka horizon is later than the position S on the local horizon. If the mean planet moves in a day of 21659 asus by the arc denoting its daily mean motion, by how much does it move in the time of chara expressed in asus? The result is (Chara in Asus × mean daily motion) / 21659 . 24

186 This result is to be subtracted from the position of B to get the position of S. If the position B′ which is the setting position at Lanka, and if we have to get S′ the local setting position, in as much S′ is later than B′ by the same arc, we have to add the above result to the posi- tion B′ to get S′. Here the asus in a day is given to be 21659 and not 60 × 60 × 6=21600 because the mean daily motion is during the course of the Sāvana day i.e. the true solar day and not during the course of a sidereal day of 21600 asus. The Sāvana day exceeds the sidereal by the time taken by the arc moved by the Sun during that day which is very approximately 59′. Since a minute of arc of the Equator rises in an asu i.e. in 4″, so 59′ of the Sun's motion is covered in 59 asus which is to be added to 21600 asus of the sidereal day to get the Sāvana day approxi- mately. It will be noted that this correction of chara in the planetary positions is due to latitude of the place and if the latitude is zero, it need not be done. Also if δ = 0, it need not be done, for, then, the Sun or the celestial body whose δ = 0 will be rising at E itself in which case the chara EA will be zero. Verses 54, 55. The H sines of 30°, 60° and 90°, being squared and decreased by the squares of their respective declinations, the square-root of the differences being taken, and the result being multiplied by the radius, is to be divided by the respective H cosines of the declination. The first of the three results, the difference of the second and the first and the difference of the third and the second will give us the rising times of what are called the Sāyana rasis of Mesha, Vrishabha and Mithuna; their reverses will then give the rising times of the next three; then the original ones those of the next three and again then reverses those of the last three. Comm. There are twelve Rasis in the Zodiac of equal interval. Measuring from r, and taking arcs of the

187 ecliptic successively each of 30°, we have what are called Sāyana Rasis or Rasis taking the Ayanāṁsa into consi- deration or in other words measuring from r. On the other hand successive arcs each of 30° measured from the zero point of the Hindu Zodiac, constitute what are called the Nirayana Rāśis. The words Meṣa, Vṛṣabha etc. signifying the shapes of the constellations apply strictly to the Nirayana Rāśis. But nonetheless, by the conven- tion what is called Upachāra, we name the Sāyana Rāśis also by the same names. In as much as the zero-points of the Hindu Zodiacs is ahead of the modern zero-point by an arc which is the arc of Ayanamsa or accumulated pre- cession, the words Sāyana and Nirayana came into vogue. Once upon a time approximately in Varāha’s time or rather 499 A. D, the two zero-points coincided and then the Sāyana and Nirayana Rāśis were the same. Gradually on account of the phenomenon of precession r preceded, and today the distance between r and the Hindu zero- point is about 20°–30′. Of late, there has been a big controversy as to what exactly the arc is and the Calendar reform committee has adopted a value far more than what could be justified. The present author opines, that we have no right to set aside a statement of no less an astro- nomer than Varāhamihira who stated explicity ‘आश्लेषार्धात् दक्षिणमुत्तरमयनं रखेर्धनिष्ठाद्यम् , नूनं कदाचिदासीत् येनोक्तंपूर्वशास्त्रेषु, साम्प्रतमयनं सवितुः कर्कटकाद्यं मृगादितश्चाऽन्यत् उक्ताभावो विकृतिः प्रत्यक्षपरीक्षणव्यक्तिः” i.e. True it is that once upon a time, the Sun began his southern journey when he was mid-way the constellation of Asleṣa and his northern when he was at the beginning of Dhaniṣtha, because it was stated in ancient texts. (The allusion is to Vedāṅga Jyotiṣa where it was stated as such); but now the southern journey of the Sun begins from the point marking ¾ of the con- stellation of punarvasu which is the beginning point of the Karkaṭa Rāśi and his norther from the beginning of Makara i.e. from the point marking ¼th of the constellation of Uttarāṣādha; there has been a change from what was

188 stated in the ancient texts ; let people verify this by actual observation". Accepting an Ayanāṁsa which goes against the statement of this great astronomer, who said that he observed and called upon others to observe, is really unwarranted, especially when the adopted Ayanāṁsa of the Calendar reform committee goes on the basis of surmises and consensus. We shall deal with this topic in further detail in an appendix to this work, because it is a really important issue and has been wrongly solved. The importance of knowing the exact value of the arc is clear when we observe that from the correctly computed planetary positions of modern astronomy this ayanāṁsa is being subtracted to give the positions measured from the Zero-point of the Hindu Zodiac. An error in the Ayanāṁsa therefore vitiates the positions obtained by the above method. Coming to the point, the problem on hand is to obtain the rising times of the Sāyana Rāśis at a place of zero latitude i.e. what are called Laṅkōdaya times of Sāyana Rāśis (Ref. fig. 21). Let rS = 30° so that rA the equatorial arc gives the rising time of rS. We could have the magnitude of this arc by the formula derived from Napier's rules namely cos ω = tan α tan λ I. But as in Hindu trigonometry the tangent functions of angles are not used, Bhāskara gave the following formula Sin α = √(sin² λ − sin² δ) / cos δ II. This formula could be derived easily from formula I and the formulae cos λ = cos α cos δ III and sin δ = sin λ sin ω IV ; for multi- plying the right-hand sides of I and III we have cos ω cos λ = (sin α cos δ) / tan λ i.e. sin λ cos ω = cos δ sin α ∴ sin α = (sin λ cos ω) / cos δ V. But from IV sin ω =

--- ------------------ = ------------------- sin λ cos δ cos δAnd then:H·cos δ as stated.? Wait, where does H·cos δ as stated.come from? Wait! Look at line 5:H sin α = R √(H sin² λ - H sin² δ) / (H cos δ) ...Wait, is line 5:H sin α = R √(H sin² λ - H sin² δ) / (H cos δ)Wait, why is thereH·cos δ as stated.? Could it be: ... / H·cos δ as stated.? Wait! Look at the vertical alignment: Line 5: H sin α = R[fraction numerator:√(H sin² λ - H sin² δ)] Wait, what is the denominator of line 5? H cos δ! Then on line 6: sin λ × 1/sin λ ...Wait, why isH·cos δ as stated.BELOW line 6? Wait, look at the line aboveH·cos δ as stated.: There is a horizontal line above H·cos δ: H̄·cos δ as stated.? Wait! Is that H·cos δ as stated.or is that a fraction line? Wait, look at the letter beforecos δ`: It has a bar above it:

190 From M drop a perpendicular ML on r ♎ so that SL will be perpendicular from S on r ♎ by the theorem of three perpendiculars. SL = H sin λ. Also ML will be equal to the perpendicular from S on the diameter of the diurnal circle of S parallel to r ♎ so that ML = SN = H sine of the arc in the diurnal circle corresponding to Kr ∴ SW² = SL² - LN² = H sin² λ - H sin² δ ∴ SN = √(H sin² λ - H sin² δ) ∴ The length of the perpendicular from K on r ♎ = R × √(H sin² λ - H sin² δ) / (H cos δ) since corresponding lines of the diurnal circle and the equator stand in the ratio of H cos δ : R (Vide fig. 20). But this ⊥ᵃʳ is H sin α = R √(H sin² γ - H sin² δ) / (H cos δ) . If α₁, α₂, α₃ be the Right ascensions of the points on the ecliptic whose modern longitudes are 30°, 60° and 90°, expressed in asus, then α₁, α₂ - α₁, α₃ - α₂ will give the rising times of the arcs of the ecliptic which stand for Sāyana Meṣa, Sāyana Vriṣabha and Sāyana Mithuna. The rising times of the next three Rasis will be the same in reverse order since Karkata is symmetric with Mithuna with respect to the Equator and similarly Simha and Kanya symmetric with Vriṣabha and Meṣa. The next three are again symmetric with Meṣa, Vriṣabha and Mithuna and the last three with Mithuna, Vriṣabha and Meṣa. Verse 56. The H cosines of the ends of the Rasis Karkata etc., being multiplied by the radius, and divided by the H cosines of their respective declinations, and the arcs of those H cosines being taken, subtract as before the preceding from the succeeding. Then we have the rising times of the Rasis beginning with Karkata.