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

155 plained by this author in his work ‘A critical study of Ancient Hindu Astronomy’ (published by the Karnatak University) page 98. Verse. Cited from Golādhyāya. The equation of centre is to be applied to the mean planet to obtain the centre of the Śīghraepicycle; then to obtain the true position the Śīghraphala is to applied to the Mandasphutagraha; hence the two equations are mutually related so that the true position is obtained after repeated application of the two equations. Comm. Explained above. Verse 36 latter half and 37. The true daily motion of the planet is the excess of the longitude of the true planet of the next day over that of the true planet of the previous day. The Kotiphala being multiplied by the daily motion of the Manda mean anomaly and divided by the radius, and the result being added to or subtracted from the mean motion, gives what is called Mandasphutagati. Comm. Already explained before. If M₁, M₂ and S be the mean planet, Mandasphutagraha, and the true planet respectively, and if E₁ and E₂ be respectively the two equa- tions, then M₁ + E₁ = M₂, M₂ + E₂ = S so that δM₁ + δE₁ = δM₂ and δM₂ + SE₂ = δS Here δM₁ = mean daily motion, δE₁ = daily variation in the Mandaphala, δM₂ = daily motion of the Mandasphuta graha or what is the same Mandasphutagati, δE₂ = daily variation in the second equation and δS = True daily motion.

156 Lallāchārya formulates the Mandasphutagati in a different manner which is equally correct (Ref. verse 45 Spaṣtādhikāra Śiṣyadhī Vriddhida “त्रिज्याहता ग्रहगतिः मृदुकर्णभक्ता मन्दस्फुटा भवति ” i. e. (δm × R) / K = Mandasphuta- gati where δm = Madhyagati or mean motion and K is the Manda Karṇa equal to √(R² + r² ± 2 R r cos m) ∴ Mandasphutagati = (δm × R) / √(1 ± 2 (r / R) cos m + r² / R²) = δm ( 1 ± (2r / R) cos m )⁻¹⁄₂ = δm ( 1 ∓ (r / R) cos m ) neglecting the smaller term − r² / (2R²) within brackets = δm ∓ (r H cos m δm) / R² = δm ∓ (r H cos m / R) × (δm / R) = δm ∓ (Kotiphala × Mandakendragati) / — as given by Bhāskara. Verse 38. In the case of the Moon, obtaining the true Moon for a particular moment and his daily motion for the day, the ending moment of the tithi near at hand is to be computed with that daily motion, and the method of suc- cessive approximations is to be used to rectify the ending moment. In the case of the ending moment of the tithi be- ing sufficiently far away, then it does not matter even if the above daily motion is applied to get the approximate ending moment. In as much as the Moon’s daily motion is great and varies from moment to moment the motion at the moment is to be used. Comm. Strictly speaking the ending moment of every tithi is to be computed by the method of successive approxi- mation. That is why in the computation of eclipses,

157 Moon's hourly motion is given in modern almanacs. Since it is very cumbrous to use the method of successive approxi- mation to determine the ending moment of every tithi, the Hindu almanac-makers generally compute the ending moment of a tithi using the daily motion of the moon computed for the moment of Sun-rise of the day. Only for ritual purposes, the method of successive approximation is used and also in the computation of eclipses. Verse 39. Computation of the Sīghragatiphala. δl − [H sin (90−E₂) × δm] / K = δS where δl is the daily motion of the Sīghrōcca, E₂ = Sīghraphala, δm = daily motion in the Sīghra mean anomaly, K = Sīghrakarṇa, and δS the true motion of the planet. If δS is negative the planet is retrograde. Comm. This is mathematically an important verse, and the proof given by Bhāskara really reflects his genius. Before we attend to his proof, we shall give this a modern treatment. (Ref. figs. 14, 15). SA and EA' are the Fig. 14 heliocentric and geocentric directions to Aswini the Hindu Zero-point of the Zodiac; S = Sun, E = Earth; V = In- ferior planet Venus or Mercury; J = Superior planet; E = Sīghra-phala, K = Sīghra Karṇa; M = Sīghra anomaly.

188 [चित्र: Fig. 15] Fig. 15 True motion of the planets = δ (A'EV) or δ (A'EJ) But δ (A'EV) = δ (A'EV'–n̂) and δ (A'EJ) = δ (A'ES–n̂) In the case of the Inferior planet δ (A'EV') = δ (ASV) = Sīghrōcchagati and δn is Sphutakendragati where n is called Sphutakendra, m being called Madhyakendra. In the case of the Superior planet δ (A'ES) = S'īghrōcchagati because the Sun plays the part of S'īghrōccha in the case of a Superior planet and δn = Sphutakendragati as before. Hence in both the cases, Sphutagati = S'īghragati – Sphuta- kendragati. We have now to find Sphutakendragati to obtain Sphutagati, as Bhāskara remarks rightly “महामति- द्भिः केन्द्रगतिरेव स्पष्टीकृता”. In other words we have to find δn. From the figures K cosn – R cos m = r (i) Differentia- ting this we have – K sin n δn + cos n δK + R sin m δm = 0 (ii). But K² = R² + r² + 2Rr cos m so that 2δK × K = – 2Rr sin mδ m (3) Eliminating δK between (2) and (3) – K sin nδ n – (Rr sin mδ m × cos n) / K + R sin mδm = 0 i.e. K sin nδn = R sin mδm ( 1 – r / K cos n ) = (R sin mδm / K) (K – r cos n)

159 But K — r cos n = R cos E. ∴ δn = (R sin mδm × R cos E) / (K² sin n) But R sin m = K sin n Cancelling δn = (R cos Eδm) / K = (H cos E δm) / K as given by Bhāskara. In the formula H sin E = (r H sin m) / K, Bhāskara perceived the variability of both H sin m and K on the right hand side and he exclaims “ न हि केन्द्रगतिजमेव फलयोरन्तरं स्यात्, किन्त्वन्यथाऽपि अद्यतनभुजफलश्वस्तनभुजफलान्तरे त्रिज्यागुणे अद्यतनकर्णहृते यादृशं फलं न तादृशं श्वस्तनकर्णहृते, स्वल्पा- न्तरेऽपि कर्णे भाज्यस्य बहुत्वात् बह्वन्तरं स्यादित्येतदानयनं हित्वा अन्यत् महामतिमद्भिः कल्पितम्, तद्यथा केन्द्रगतिरेव स्पष्टीकृता ” i.e. “ The variation in the Sīghraphala is not entirely constituted by the variation in m but also by that in K......So leaving the method of seeking δE through the formula H sin E = (r H sin m) / K the great intelligent astronomers used the formula. Sphutabhukti = Sīghra Bhukti — Sphuta Kendra- bhukti (Bhukti means gati) wherein it was'sought to obtain the variation in Sphutakendra i.e. n̂ in the figures. We have given a proof of Bhāskara’s formula, which circumvented finding Sīghragatiphala but which sought directly Sphutabhukti, by using the modern heliocentric figures. We shall now see how Bhāskara could deal with such a tough problem. Refer fig 16. Let P₁, P₂ be two positions of the planet on two consecutive days relative to the Sīghra A₂, so that P₁E₁P₂ is the Sphuta Kendragati spoken of. It will be noted that it is not Sphutagati be- cause P₁, P₂ are positions of the planet relative to A, which is itself moving (as rightly remarked by Bhāskara). It is Sphuta Kendragati because A₂E₁P₁ and A₂E₁P₂ are the

160 Kendras on two consecutive days whereas Madhyakendragati ^ is P₁E₂P₂. Also, we have the equation. Sīghra — Sphutagraha = Sphutakendra so that Sīghra- gati — Sphutagati = Sphuta- kendragati. Hence Sphutagati = Sīghragati — Sphutakendra- Fig. 16 gati. So we have now to seek the value of P₁Ê₂P₂. Let P₁a stand for the Sīghra- phala of the first day which is equal to e f, f being the true planet of the first day. E₂ d will be parallel to E₁ f because P₁ e being parallel to E₁ E₂ and e f being equal to P₁ d, d f 11 P₁ e (parallels to E₁ E₂ cut off equal arcs on the two circles). This may be seen also as follows. Since P₁ d is taken to be equal to e f, e being the mean planet and f the true on the first day e Ê₁ f = P₁ Ê₂ d. But E₁ e 11 E₂ P₁ ∴ e E₁ f = E₁ P̂₁ E₂ ∴ P₁ Ê₂ d = E₁ P̂₁ E₂ and alter- nate angles being equal E₁ P₁ 11 E₂d. P₁ a is the H sine of P₁ d where P₂ b is the H sine P₂ d. Looking upon P₁P₂ as an increment in P₁ d i.e. looking upon the Kendragati P₁ Ê₂ P₂ as an increment in Sīghraphala, Bhāskara uses the method of Bhogyakhanda sphuti Karana to obtain the Sphutakendragati. From the figure. P₂c = P₂b — P₁a = H sin (G + δm) — H sin E H sin E H cos δm + H cos E H sin δm = ————————————————————————————————————— — H sin E R

161 Taking H cos δ M = R and H sin δm = δm P₂c = (H cos E δm) / R This is at the end of E₁ P₂ i.e. at the end of K; so to get the corresponding chord in the deferent we do Karnānupāta so that the result is (H cos E δm) / R × R / K = (H cos E δm) / K as given by Bhāskara. We have cut short Bhāskara’s method of Bhōgyakhanda Sphutīkaraṇa to make it clear to a modern student. Since P₂c is small, passing on from the H sine to the arc is not necessary, for, the H sine of a small arc is equal to the arc itself. Bhāskara’s argument, however, is as follows:— “If for 225, we have Bhōgya Khanda, what for δm?” The result is (B × δm) / 225; Then B is rectified as follows:— “When the H cosine E is equal to the radius, i.e. initially in the H sine table, the Bhōgyakhanda is 225, then what is it for H cos E?” The result is (H cos E × 225) / R which we have to substitute for B. Then if this be at the end of K what is it at the end of R?” The result is (H cos E × 225) / R × δm / 225 × R / K = (H cos E × δm) / K as given. Before we proceed to explain ‘शेषं च वक्रा विपरीतशुद्धौ’, we shall explain what Bhāskara pointed out as a mistake in Lallācharya. Verse 40. Let Mathematicians understand that what formula was given by Lallācharya for Sīghragatiphala is not correct. When the anomaly is 90° or 270°, the gati- phala vanishes and there will be gatiphala at the points where it ought to be Zero according to his formula. 21

162 Comm. Ref. verse 45, Spaṣṭādhikāra, Śiṣya Dhī- vṛddhida तद्रहिताऽऽशुभुक्तिः, त्रिज्ज्याहता स्वचलकर्णहताऽऽशुचापभोग्य ज्यया विगुणिता विहृताऽऽद्यमौर्व्या, लब्धं त्यजेत्, स्वचलतुङ्गगतेः सदैव शेषं स्फुटा भवति च ग्रहभुक्तिरेवम्" i.e. ( शीघ्रगति-मन्दस्फुटगति ) × R Śīghraphala Bhōgyakanda --------------------------- × ----------------------- K 225 = Sīghragatiphala. Here the quantity within the brackets is δm. What Lallācharya had in his mind is as follows. ‘If for the Ādya Khanda 225 we have H cos E = R, what shall we have for the Bhōgya Khanda of the Sīghraphala ?’ The result would be [ H sin (E + 225) - H sin E ] × R / 225 = ( H sin E H cos 225 + H cos E H sin 225 / R - H sin E ) × R / 225 Taking H cos 225 = R and H sin 225 = 225 it would be H cos E × 225 / R × R / 225 = H cos E. So Lallācharya’s formula would become (δm / K) × H cos E as given by Bhās- kara. The charge levelled at Lallācharya is due to the fact that by the word ‘Āsu-chāpa’ by which Lallācharya meant ‘आशुफलचाप’ Bhāskara meant ‘आशुकेन्द्रचाप’. When the Kendra = 90 or 270, the Bhōgyakhanda being Zero, the Sīghragatiphala would be Zero. Also where it ought to be zero namely (90 + H sin⁻¹r) & (270 - H sin⁻¹r) it would not be zero. If Lallācharya had really meant what Bhāskara allributed to him, the formula would be (H cos mδm / K) and it is very unlikely that Lallācharya would have meant this wrong formula, for, even if K were taken by him to be steady, δ (r H sin m / K) = (r H cos mδm / K) and Lallāchar- ya’s formula does not contain ‘r’. The only non-rigorous part in Lallācharya’s formula is at the point where he took

163 H̄ cos 225 = R and H sin 225 = 225 which is rather crude. Bhāskara of course improved on this crudeness by taking δm to be an increment in E. In the commentary under this verse, Bhāskara, having misinterpreted Lallācharya's phrase आशुचाप as meaning आशुकेन्द्रचाप and not as आशुफलचाप which was in the mind of Lallācharya, goes on recounting examples where the wrong formula attributed to him would give wrong results. So, we need not enter into those details. Now we shall correlate Bhāskara's formula with its modern counterpart. Assuming coplanar heliocentric cir- cular orbits for planets, let us see at what points two planets appear mutually stationary, that is, have a Zero relative angular velocity before they appear mutually retrograde. Let S = Sun, E = Earth, J = Jupiter, u = Earth's linear velocity, v = Jupi- ter's linear velocity r and R the orbital radii of the Earth and Jupiter respe- ctively. Let EE' and JJ' be perpendiculars to EJ so that when the relative velo- city of Jupiter with respect to the Earth perpendicular to EJ is Zero, Jupiter will appear stationary as seen from the Earth. This Fig. 17 means that u cos θ + v cos ξ = 0 I ∴ u / v = - cos ξ / cos θ II But from triangle ESJ, R cos ϕ + K cos θ = r III and r cos ϕ + K cos ξ = R IV

164 From III & IV cos ξ / cos θ = (r cos ϕ − R) / (R cos ϕ − r) V Equating cos ξ / cos θ from II and V − u / v = (r cos ϕ − R) / (R cos ϕ − r) so that (ru + Rv) / (rv + Ru) = cos ϕ If m be the Śīghra anomaly m = 180 − ϕ so that cos m = − (ru + Rv) / (rv + Ru) VI Now we shall show that Bhāskara's formula accords with this Spaṣṭagati = Śīghragati − (H cos E δm) / K . As per Bhāskara Spaṣṭagati = 0 if Śīghragati = (H cos E δm) / K VII ie. Jupiter appears stationary as seen from Earth, if Śīghra- gati = (H cos E δm) / K The angular velocities of Earth and Jupiter are respect- ively u / r and v / R so that the Sun's apparent velocity is also u / r and δm = Kēndra gati = Sun's apparent velocity minus Jupiter's heliocentric velocity = u / r − v / R ∴ Substituting in VII Śīghragati = u / r = (H cos E / K) (u / r − v / R) ∴ u / r ( (H cos E / K) − 1 ) = (H cos E / K) × v / R ie. u / r ( (H cos E − K) / K ) = (H cos E / K) × v / R

M₂ + E₂ = S I where M₂ = Mandasphuta- graha and E₂ = the Sīghraphala where from we have δM₂ + δE₂ = δS II i.e. Mandasphutagati + Sīghragati phala = Spastagati (a) Let E₂ be maximum so that δE₂ = 0, then δM₂ = δS. This means that in the helio- centric figures 11 and 12, at the points a and b,: the Sīghraphala being maximum, the Mandasphuta gati will be itself the Spashtagati. This line ab, it will be seen corres- ponds with the so called “कक्षामध्यगततिर्यग्रेखा (प्रतिवृत्तसम्पातरेखा)” i.e. the line cutting the eccentric, drawn through the centre of the deferent. (b) The planet begins to retrograde only after the Spashtagati vanishes i.e. after δM₂ + δE₂ becomes Zero. Taking δM₂ to be almost a constant since the Mand

166 negative when the planet courses the arc of the smaller segment ab, because E₂ decreases from a maximum value to zero. As a matter of fact δE₂ negatively increases along ac and again increases from a negative minum at c to Zero at b as we course along cb (vide fig. 11 & 12). Thus the planet will assume zero velocity at two points symmetrical about c and in between ab. In other words the planet will be retrograde along the arc d₁ cd₂ not entirely along ab as some have misconstrued. Regarding E₂, it is clear that it is zero at S'. then gradually increases to a maximum as the planet traces S'a, the maximum being assumed at a, then it decreases from the maximum to zero as the planet courses the arc ac. and then increases from zero to a maxi- mum at b and finally decreases from that maximum to zero at S' again. Keeping the Earth constant it will be noted that an Inferior planet always goes anticlockwise whereas a Superior planet always goes clockwise. Also it will be seen that the Sīghraphala is positive as the planet courses the arc Sac, whereas it is negative as it courses cbS'. (c) The values of the spashtagati at S' and C will be respectively putting H cos E = R in the formula. Spashtagati = Sīghragati — (H cos Eδm) / K . Here for Sīghragati we may put U and U — V for δm, and putting K = R + r at S' and R — r at C, the values of the spash- tagati would be [U — R (U — V)] / (R + r) and [U — R (U — V)] / (R — r) ie. (RV + ru) / (R + r), (RV — rU) / (R — r) respectively. (RV + rU) / (R + r) > (RV — rU) / (R — r) if R²V — rRV + RrU — r²U > R²V — RrU + rRV — r²U

167 ie. if rR (U - V) > Rr (V - U). U - V is + ve and equal to V - U. So the positive velocity at S' of the planet will be equal to its negative or retrograde velocity at C. A quan- tity assuming values K, O, - K,O,K must have a value numerically less than K in between. Thus the velocity direct or retrograde at any point of the orbit is less than the numerical value of the velocity at S' and C. Verse 41. Retrograde motion. The planets Mars, Mercury, Jupiter, Venus and Saturn will be retrograde when the Sīghra anomaly assumes values 163, 145, 125, 165 and 113 respectively and the direct motion again ensues at (360 - 163), (360 - 145), (360 - 125), 360 - (165) and (360 - 113) respectively. Comm. Since we have had the formula VI cos m = - ( (rU + Rv) / (rV + RU) ) for stationary points, noting that cos (180 - θ) = - cos θ = cos (180 + θ) the stationary points are symmetrically situated with respect to S' of figures 11 and 12. As mentioned before substitut- ing the values of U, V, r, R for all the planets we can prove the veracity of Bhāskara's statement. Verse 42. Heliacal rising and setting of planets. Mars rises heliacally in the East by 28°, Jupiter by 14°, Saturn by 17° of Sīghra anomaly and set heliacally in the west by degrees which are the differences of the above and 360° respectively. Comm. The Sun's velocity being greater than that of the Superior planets, the Sun overtakes them so that they set in west and rise in the East. When these planets are

168 situated within particular limits from the Sun, they will be invisible in the rays of the Sun. As these superior planets will be near the Sun, near the moment of conjunction, they will not be seen at conjunction and within particular limits from the position of the Sun. The limits cited above mostly depend upon their respective brilliance and to some extent upon their distances from the Sun too. Their brilliance again depends upon the extent of their gibbosity. The formula given for the phase in modern astronomy is phase = (1 + cos EpS) / 2 and noting that Êps = E₂ = the Śīghraphala, phase = (1 + cos E₂) / 2. Hence the Supe- rior planets are always gibbous ie. the disc illuminated will be always greater than ½. Though at conjunction E₂ = 0 and the entire discs of the major planets will be illumina- ted, we cannot see them as they are immersed in the rays of the Sun. As they emerge out of conjunction gradually E₂ will be increasing so that the discs will be illuminated lesser and lesser gradually. But K decreasing, the brilli- ance will not be so much effected. Along the arc acb (figs. 11, 12) the planets gradually gain in illumination and will be brightest when they are in opposition both as cos E in- creases and K decreases. In other words the Superior planets appear more and more brilliant when they are retrograding, being most brilliant at ‘C’. The spherical radii of Jupiter Saturn and Mars being in decreasing order, their brilliance will be in decreasing order so that they will be rising at distances from the Sun which are in increasing order. The inverse square law of courses works here but combining the two factors (1) the spherical radius of the planet and (2) the inverse square law, we may take it that Bhāskara’s numbers cited above namely 28°, 17° and 14° for Mars, Saturn and Jupiter respectively accord with truth, for, these numbers are given by Bhāskara or his authority Brahmagupta only after observing the planets rising heliacally.

169 That the positions of the planets while setting or rising heliacally will be situated symmetrically with respect to the Sun, goes without saying. But one thing. In the chapter called Udayāstamayā- dhyāya, the degrees known as Kālāṁśas which are given as the arcs between the planets and the Sun for heliacal rising or setting, are different from the numbers given above, for, the latter are the values of the Śīghra anomaly. In the case of Mars when the Śīghra anomaly is 28°, the Śīghraphala will be 11°, so that the apparent distance of the planet will be 28° - 11° = 17° which are the Kālāṁśas for Mars. Similarly in the case of Jupiter, when the Śīghra anomaly is 14°, the Śīghraphala will be 3°, so that the distance between the planet and the Sun as seen from the earth will be 11°, which are given as Jupiter's Kālā- ṁśas. Also in the case of Saturn, the Śīghraphala for 17° of anomaly will be 2°, so that the Kālāṁśas would be 15°. Verse 43. Mercury and Venus rise in the West by 50° and 24° of Śīghra anomaly respectively, and set in the West by 155°, and 177° respectively. They rise in the East by 205° and 183° of Śīghra anomaly and set there by 310° and 336° respectively. Comm. Regarding the Inferior planets, they rise heliacally in the East after Inferior conjunction and then they are retrograde. They attain gradually the maximum elongation in the East and after they revert to direct motion, their elongation gradually decreases. They then set in the East and heliacally rise thereafter in the West. There again their elongation attains a maximum value; then they begin to retrograde and gradually set in the West only to rise in the East. This is all clear from the heliocentric figure 11. When the Śīghra anomalies of Mercury and Venus happen to be respectively 50° and 24°, their Śīghraphalas would be 13° and 11° respectively, so that they are themselves the Kālāṁśas, in as much as in 22

170 the case of the Inferior planets, the Sighraphala will be itself their elongation eastern or western. Then they rise in the West, being near Superior conjunction. When again their Sighra anomalies equal respectively 155° and 177°, the same Sighraphalas will arise so that they set heliacally in the West. Then as the Sighra anomalies attain the symmetrical values on the other side ie. (360- 155) and (360-177) ie. 205° and 183° the Sighraphalas being the same, they rise in the East. Again when they attain the values (360-50) and (360-24) ie. 310° and 336°, they set in the East on account of the same Sighraphalas or Kālāmsas. Verse 44. When the Sighra anomalies have parti- cular values, to decide when the planets rise or set helia- cally, we have to take the difference of those particular values and the numbers given above for the respective Sighra anomalies, convert them into minutes of arc and divide the results by the daily motion in the Sighra ano- malies in minutes of arc. Then we have the number of days in which the rising or setting takes place thereafter. Comm. Suppose it is required when Mercury rises in the West. Suppose we want to compute this on a parti- cular day when Mercury's Sighra anomaly is x°. Then because we know that Mercury rises in the West when his Sighra anomaly is 50, we have to calculate by how many days the difference | x-50 | of the Sighra anomaly would be covered. Let the daily motion of the Sighra anomaly be y' per day, Then by rule of three (| x-50 | × 60) / y will be the number of days before or after as the case may be for Mercury to rise in the West. Verse 45. To obtain the Mean planet knowing the True. Assume the True planet to be the Mean; compute the Manda and Sighraphalas and applying them inversely,

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.