ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

132 GREEK AND INDIAN METHODS from the Greek method in this case also. As the triangle KLS is difficult to show in the diagram, it is shown in its projection on the meridian plane in Burgess's translation of the "Sūrya- siddhānta,"(page 232)and also in Wilkinson and Bāpūdeva Śāstrī's translation of the 'Siddhānta Śiromaṇi,' p. 175. This has led Braunmühl to assume that the Indian method of arriving at the equation of ascensional difference and some other equations of spherical astronomy has its origin in the Analemma of Ptolemy. A careful study, however, does not justify the identification of Indian methods with the graphic method of the Analemma, which is deduced from the projections of the position of a heav- enly body on the meridian prime vertical and the horizon. It is being presently shown that what was done out of difficulty in drawing the figures properly has been taken by Braunmühl as a Greek connection. Problem III¹ :— To find the “Time-altitude” Equation If from any point S on the diurnal circle a perpendicular be drawn to the Udyāsta-Sūtra spoken of before, this perpendi- cular is called the cheda or ‘iṣṭahṛti.’ The perpendicular from S on the horizon is called ‘Śaṅku’² the sine of the altitude. The line joining the foot of the ‘Śaṅku’ and that of the perpendicular on the ‘Udayāsta-Sūtra’ goes by the name of ‘Śaṅkutala’ and this Śaṅkutala lies to the south of the ‘Udayāsta-Sūtra’ during the day. In this figure (Fig 8) if AA' be the ‘Udayāsta-Sūtra’ or the intersection of the diurnal circle and the horizon, and S a point on the diurnal circle denoting a position of the Sun, SK, SL perpendiculars on AA' and the horizon respectively; SL is called the ‘Śaṅku,’ SK the ‘cheda’ and LK, the ‘Śaṅkutala’. In this triangle KSL, the angle KSL was recognised to be the latitude of the station. Thus the triangle SKL is not taken in its projection on the meridian plane. The side SK is taken 'as formed of two parts.

1. Āryabhaṭa could not arrive at the true equation. Cf. Gola 28. The correct rules occur in Pañcasiddhāntikā, IV, 42, 44; Brahmasphuṭasiddhānta, III, 36-38, 26-40; Sūryasiddhānta, III, 34-35.
2. Bhāskara says : ग्रहस्थानाल्लम्बः शंकुः । तस्यतलमुदयास्तसूत्राद्दक्षिणतो भवति ॥ “Gola, VIII-39-41, Āryabhaṭa uses the term शङ्क्वग्रम्” Gola, 29.

TIME-ALTITUDE EQUATION 133 Let CC' be the line of inter- section of the diurnal circle and the 'six o'clock 'circle EPW. Let SK cut CC' in M. Then. SK = SM + MK Here SM, the 'sine' in the diurnal circle of the complement of the hour angle is given a distinct name 'Kāla'¹ and MK as explained before is known by the name Fig. 8 'Kujyā.' This 'Kāla' is constructed from the point S in the diurnal circle. Thus the triangles like SKL were not taken in their projections on the meridian plane as Braunmühl would suggest. From the triangle KSK, we get, 'Cheda' : 'Śaṅku' = R : R cos ϕ where ϕ is the latitude of the observer; 'Śaṅku' is here = R cos Z, Z being fhe Sun's zenith distance. ∴ 'cheda' = (R cos Z × R) / (R cos ϕ) Now 'Cheda' = radius of the diurnal circle + Kujyā - versed sine of the hour-angle in the diurnal circle O' B + O' V - BR, = R cos δ + (R sin δ × R sin ϕ) / (R cos ϕ) - (R vers H × R cos δ) / R As in the previous problem, Kujyā = SK = (R sin δ × R sin ϕ) / (R cos ϕ) or (R cos Z × R) / (R cos ϕ) = (R cos δ / R) { R + (R sin δ × R sin ϕ) / (R cos ϕ) × R / (R cos δ) - R vers H } The above equation simplified becomes cos Z = sinδ sin ϕ + cos δ cos ϕ cos H. In this connection we consider the altazimuth equation by the Indian method.

1. Bhāskara's Grahagaṇita, VIII, 55. O' is the middle point of CC' or it is the centre of the diurnal circle ABB'.

134 GREEK AND INDIAN METHODS ¹Problem IV :— The Altazimuth Equation Indiad Method Let α denote the azimuth of the Sun from the south. In the same triangle SKL in the same figure, we have, LK : SL = R sin ϕ : R cos ϕ or, ‘Śaṅkutala’ : ‘Śaṅku’ = R sin ϕ : R cos ϕ ∴ ‘Śaṅkutala’ = (R cos Z × R sin ϕ) / (R cos ϕ) Now ‘Śaṅkutala’ is made up of two parts, namely, ‘Bāhu’ and ‘Agrā’, of which the former is the distance of L from the observer’s East-West line; the ‘Agrā’ has been already found. Here ‘Bāhu’ = (R sin Z × R cos α) / R and ‘Agrā’ = (R sin δ × R) / (R cos ϕ) ∴ ‘Śaṅkutala’ = ‘Bāhu’ + ‘Agrā’ or (R cos Z × R sin ϕ) / (R cos ϕ) = (R sin Z × R cos α) / R + (R sin δ × R) / (R cos ϕ) or R sinδ = (R cos ϕ / R) ((R cos Z × R sin ϕ / R cos ϕ) - (R sin Z × R cos α / ϕ)) which is easily seen to be equivalent to sin δ = cos Z sin ϕ - sin Z cos ϕ, cos α Greek Method Ptolemy² has also a method of finding the Sun’s altitude at any hour of the day. His method is as follows :— (i) He would find by means of his tables for the times of risings of the signs of the zodiac, the orient ecliptic point. (ii) He would then find the culminating point of the ecliptic. (iii) He would finally apply Menelaus’s theorem in spherics thus :— Fig. 9 Let ASC be any position of the ecliptic, (Fig. 9) NZC the

1. The equivalent of this, in a particular case, is first found in Brāhmasphuṭasiddhānta, Ch. III, 54-56 Cf. Sūryasiddhānta, III, 28-31, also Bhāskara Grahaganita, IX, 50-52.
2. Manitius, ibid, pp. 118, 19.

PTOLEMY'S ANALEMMA 135 meridian, NAMH the horizon, Z, the zenith and S the Sun. Here the celestial longitudes of C, S and A are taken to be known; hence ZC and CH are also known. Now take ZCS for the triangle and HMA to be the trans- versal ; we then have by Menelaus's theorem. (sin ZH / sin HC) × (sin CA / sin AS) × (sin SM / sin MZ) = 1 or sin SM = (cos CZ × sin AS) / sin CA It is thus clear that Ptolemy had no direct method for connecting the Sun's altitude and the hour-angle. This method is workable for the problem “given time, find the altitude” but is not workable in the converse problem ; besides, the calcula- tion of the longitudes of A and C is very cumbrous. Again, when EA has been found out, taking ZHM for the triangle and CSA for the transversal, we get, (sin HA / sin Am) × (sin MS / sin SZ) × (sin ZC / sin CH) = 1, whence and thence HM, the azimuth can be found. The method is here also cumbrous, there being no direct connection between altitude and azimuth ; besides the time-element is not avoided. The Analemma of Ptolemy and the Indian Method. When the Sun's declination is zero and his hour-angle, is H, Zeuthen¹ following the method of the ‘Analemma’ of Ptolemy, as explained by Braunmühl² has deduced the following equations : (1) cos Z = cos H. cos ϕ (2) tan α = tan H / sin ϕ To these two, Heath following Braunmühl, adds (3) ³tanZQ = tan H / cos ϕ ─────────────────────────────────────────────────────────────

1. Heath, Greek Mathematics, Vol. II, pp. 290-91. Zeauthen, Bibliotheca Mathematica, 13, 1900, pp. 23-27.
2. Braunmuhl , ibid, pp. 12-13.
3. The Indian form of this equatiom is R Sin ZQ = (R Sin H × R) / [√(R² - R²cos²H × R² Sin²ϕ) / R] Bhāskara's, Golādhyāya, Com. on VIII, 67.

136 GREEK AND INDIAN METHODS where Z is the zenith and Q is the point of intersection of the prime vertical and its secondary passing through the Sun and the north-south points. Zeuthen¹ points out that later in the same treatise Ptolemy finds the arc 2β described above the horizon by a star of given declination δ' by a procedure equivalent to the formula. (4) cos β=tan δ' tan ϕ. With regard to the 'Analemma' of Ptolemy. it may be noted, as Heath² says, that "the procedure amounts to a method of graphically constructing the arcs required as parts of an auxiliary circle in one plane." Many things may be, in practice, done graphically far more easily than by the theoreti- cal method. Besides, no theoretical calculations occur in the 'Analemma'. Zeuthen², following the method of this work, has deduced in the general case, the two equations. (5) cos Z=(cos δ, cos H+sin δ. tan ϕ) cos ϕ. cos δ.sin H (6) tan α=———————————————————————————————————————— sin δ ————— +(cos δ.cos H+sin δ.tan ϕ) sin ϕ cos δ These equations are suggested to a modern reader from a study of the figures in the 'Analemma.' But neither in this work nor in the 'Syntaxis' are they to be found. With regard to the first four formulae, it is possible that they were recognised by Ptolemy. With regard to the last two, Zeuthen³ remarks "mais le texte nen contient rien,' and they were certainly not recog- nised by Ptolemy. Besides the tangent function is wholly absent in Greek trigonometry. They are also different in form from those arrived at by the Indian method as explained before. Thus, it is clear that the Indian methods are in no way connected with the method of the 'Analemma.' Even taking for granted that the Indians followed a method of projection much allied to the method of the Analemma' there is no adequate reason for assuming that their method is derived from any Greek source. Analogy and precedence do not neces- sarily constitute originality—there is still the chance of a remoter origin from which both the systems drew their inspiration. The method of the 'Analemma,' as has been already stated, presents a —————————————————————————————————————— 1, 2, Bjornbo, loc. cit. p. 86. 3. Zeuthen, loc. cit. p. 27.

ANGLE BETWEEN ECLIPTIC AND MERIDIAN 137 graphical method for constructing the Sun's altitude and azimuth from the hour angle when the Sun's declination is zero but such a graphical method is generally complex as compared with the elegant Indian method. An astronomer who constructs and uses an armillary sphere to arrive at his equations in spherical astro- nomy and who has not a well-developed spherical astronomy at his command must have to draw perpendiculars from the positions of the heavenly body, not only on the meridian plane, the hori- zon or on the prime vertical, as the occasion arises, but also on the line of intersection of the diurnal circle with the horizon. Hence Braunmühl's statement that the Indian methods of spheri- cal astronomy have their origin in the 'Analemma's, in spite of his admitting that Indians were first to utilise its methods, is rather far-fetched and tends to take away the honour from the great Indian astronomers, who devised the beautiful methods. The 'Analemma' as it now exists is a Latin translation from an Arabic version of the original Greek¹. We may reasonably doubt that the Arabic version was greatly influenced by the ancient Indian system. We now pass on to the consideration of other allied or similar problems in the two systems of astronomy. Problem V— To find the Angle between the Ecliptic and the Meridian Indian Method² Let ♈SA be thee cliptic, ♈CE the equator, E the east-point of the horizon (Fig 10). Cut off SH=90° and draw the great circle HEAP' cutting the meri- dian P'SCH at the points P' and H. The aim is to find AP' but it is enough to find EA since AP' is the complement of EA'. Both Āryabhaṭa and Brahmagupta were unable to find EA correctly. Let P be the celestial pole and let PAE' be

1. On the influence of the ancient Indians on Arab mathematics and astronomy; see Alberuni's India, translated by Dr. E. Sachau, Vol. II, p. 304.
2. Āryabhaṭa, Gola, 45; BrSpSi, IV. 17; Sūrya-siddhānta, IV. 25; Bhāskara's Golādhyāya, VIII, 21-74, first example in his own commentary.

138 GREEK AND INDIAN METHODS the secondary to the equator cutting it at E'. Both the above astronomers were content with the idea that AE = AE', or that AE = the declination of the point A of the ecliptic which is 90° ahead of S in the above figure. This idea continued till the time of Bhāskara II (1150 A. D.) who found out the correct equation. He recognised that CS, the declination of S = PP'; P'EH is then the horizon of the station whose north geographical latitude is CS. Also, the ‘sine’ of EA is the ‘Agrā’ or the sine of the amplitude of the point A for the latitude CS. ∴ R sin EA = (R sin AE' × R) / (R cos CS) = [R sin (90° + γS) × R sin ω] / R × R / (R cos CS) or R sin EA = [R sin (90° + l) × R sin ω] / (R cos δ) where l stands for γS and δ for CS. Greek Method : We give below the Ptolemy's method in a slightly modified form¹. Let SHA be the triangle and γCE be the transversal ; then we have, (sin SC / sin CH) × (sin HE / sin EA) × (sin Aγ / sin γS) = 1 or (sin δ / cos δ) × (sin 90° / sin EA) × [sin (90° + l) / sin l] = 1 ∴ sin EA = [sin δ × sin (90° + l)] / (cos δ × sin l) , which is readily transformed into Bhāskara's equation. The originality of Bhāskara would be readily admitted. Problem VI-- To find the Angle between the Ecliptic and the Horrizon Indian Method : (A) Āryabhaṭa's method. It consists of the following² steps :-- (1) Determination of the orient point of ecliptic. (2) Finding the sine of its amplitude.

1. Manitius, ibid, Book I, pp. 104-06.
2. Āryabhaṭa, Gola, 33 : Sūryasiddhānta, V. 5-6.

ANGLE BETWEEN ECLIPTIC AND HORIZON 139 (3) Determination of the culminating point of the ecliptic from the hour-angle of the Sun. (4) Finding the declination of the culminating point of the ecliptic. Having obtained the above elements, his rule can be follow- ed thus : In this Fig. 11 NZH is the meridian, HMEAN the horizon, CN'A the ecliptic. If N' be the nonagesimal or the highest point of the ecliptic, the altitude of N' is the inclination of the ecliptic to the horizon. Let ZN'M be the vertical through N', meeting the horizon at M. When the time is given, the longitudes of A and C can be found out, from which CZ the zenith distance of C and EA the amplitude of the orient ecliptic point can be determined. Fig. 11 Here HM=EA. According to Āryabhaṭa, R sin CN' = (R sin CZ × R sin HM) / R and R sin ZN' = √((R sin CZ)² - (R sin CN')²) This is only an approximate rule. As expressed here, R sin ZN' = (R sin CZ × R cos HM) / R approximately. = (¹R sin CZ × R cos HM × R) / (R × R cos CN') accurately. = (R sin CZ × R cos HM) / (R cos CN') . (B) The method of Brahmagupta² : Brahmagupta would also first determine the orient ecliptic

1. This correction was perhaps first noticed by Raṅganātha (1603 A. D.) in his commentary in the Sūryasiddhānta.
2. BrSpSi. V 3.

140 GREEK AND INDIAN METHODS point A. Then he subtracts 90° from the longitude of A. Thus having the longitude of N', he next finds the part of the day elapsed of N' ; from which by the time-altitude equation discussed above, he finds ZN'. This is of course more accurate than that of Āryabhaṭa. Bhāskara¹ here follows Brahmagupta. Greek Method : Let the ecliptic CN'A cut the lower half of the meridian at F. Ptolemy takes AK along the ecliptic=90° and AR along the horizon=90°; then the great circle passing through R and K passes through the nadir Z'. Now take Z'FK for the triangle and ANR for the transversal, then by Menelaus's theorem.² (sin FN / sin NZ') × (sin Z'R / sin RK) × (sin KA / sin AF) = 1 ∴ sin RK = (sin FN / sin AF) = (cos FZ' / sin AC) = (cos CZ / sin AC) = (sin CH / sin AC) or sin MN' = sin CH / sin AC. Here Ptolemy's equation is simpler than that of Āryabhaṭa; hence they must be independent of each other. [Figure 12] Problem VII:— To find the Angle made by the Vertical through any Point of the Ecliptic with the Latter This problem is considered by Ptolemy but it is not consider- ed separately in Indian Astronomy, but from the rule for parallax in longitude, the rule for its calcula- Fig. 12 tion can be deduced. Indian Method : In Fig. 12 S represents the true position of the Sun and S' the Sun's position as depressed by parallax. N'SA is the ecliptic. If from S', S'Q be drawn perpendicular to the ecli- ptic, then, if P is the horizontal parallax,

1. Grahaganita; XII, 3-4.
2. Manitius, ibid, pp. 110-111.

ANGLE MADE BY THE VERTICAL 18 SQ = SS' × (R cos S'SQ / R) = (P × R sin ZS / R) (R cos S'SQ / R) ¹= (P / R) √[(R sin ZS)² – (R sin ZN')²] ²= (P / R²) × R sin N'S × R cos ZN', where N' is the nonagesimal Thus R cos S'SQ is seen to be = (R sin N'S × R cos ZN') / (R sin ZS) The Indian method is fully described by Bhāskara in his 'Golādhyāya. VIII, 12-25. The truth of the Indian rule for R cos S'SQ is easily seen from the spherical triangle ASN, where A is the pole of the ecliptic. Greek Method : ³Ptolemy takes SK and SL 90° each, along the vertical circle ZSEK and the ecliptic N'SA. The great circle through K and L cuts the horizon at R which is the pole of the vertical circle. He takes SKL for the triangle and EAR for the trans- versal, then (sin SE / sin EK) × (sin KR / sin LR) × (sin LA / sin AS) = 1 or sin LR = (cos ZS × cos AS) / (sin ZS × sin AS) or cos S'SQ = cot ZS × cot AS = tan SE × cot AS. The Indian and the Greek rules are altogether different both in form and method. There can, therefore, be no question of any connection between them. Problem VIII :— To convert the Celestial Longitude of a Heavenly Body into its Polar Longitude If σ be the position of a (Fig.13), γK and σK are the celestial longitude and the celestial latitude, respectively : γM and σM are the polar longitude and polar latitude : γN and σV are the right ascension and declination of the star. Indian Method : All Indian astronomers attempt at finding MK which, sub-

1. Āryabhaṭa, Gola, 34; Pañcasiddhāntikā, IX, 22 BrSpSi, XI, 23.
2. BrSpSi. V, 4-5 ; Sūryasiddhānta, V, 7-8 Bhāskara, Grahagaṇita, XII, 4.
3. Manitius, ibid, p. 119.

140 GREEK AND INDIAN METHODS point A. Then he subtracts 90° from the longitude of A. Thus having the longitude of N', he next finds the part of the day elapsed of N' ; from which by the time-altitude equation discussed above, he finds ZN'. This is of course more accurate than that of Āryabhaṭa. Bhāskara¹ here follows Brahmagupta. Greek Method : Let the ecliptic CN'A cut the lower half of the meridian at F. Ptolemy takes AK along the ecliptic = 90° and AR along the horizon = 90°; then the great circle passing through R and K passes through the nadir Z'. Now take Z'FK for the triangle and ANR for the transversal, then by Menelaus's theorem.² (sin FN / sin NZ') × (sin Z'R / sin RK) × (sin KA / sin AF) = 1 ∴ sin RK = (sin FN / sin AF) = (cos FZ' / sin AC) = (cos CZ / sin AC) = (sin CH / sin AC) or

ANGLE MADE BY THE VERTICAL 141 SQ = SS' × (R cos S'SQ / R) = (P × R sin ZS / R) × (R cos S'SQ / R) ¹= (P / R) √((R sin ZS)² - (R sin ZN')²) ²= (P / R²) × R sin N'S × R cos ZN', where N' is the nonagesimal. Thus R cos S'SQ is seen to be = (R sin N'S × R cos ZN') / (R sin ZS) The Indian method is fully described by Bhāskara in his 'Golādhyāya. VIII, 12-25. The truth of the Indian rule for R cos S'SQ is easily seen from the spherical triangle ΠZS, where Π is the pole of the ecliptic. Greek Method : ³Ptolemy takes SK and SL=90° each, along the vertical circle ZSEK and the ecliptic N'SA. The great circle through K and L cuts the horizon at R which is the pole of the vertical circle. He takes SKL for the triangle and EAR for the trans- versal, then (sin SE / sin EK) × (sin KR / sin LR) × (sin LA / sin AS) = 1 or sin LR = (cos ZS × cos AS) / (sin ZS × sin AS) or cos S'SQ = cot ZS × cot AS = tan SE × cot AS. The Indian and the Greek rules are altogether different both in form and method. There can, therefore, be no question of any connection between them. Problem VIII :— To convert the Celestial Longitude of a Heavenly Body into its Polar Longitude If σ be the position of a (Fig.13), γK and σK are the celestial longitude and the celestial latitude, respectively ; γM and σM are the polar longitude and polar latitude ; γN and σV are the right ascension and declination of the star. Indian Method : All Indian astronomers attempt at finding MK which, sub-

1. Āryabhaṭa, Gola, 34; Pañcasiddhāntikā, IX, 22 BrSpSi, XI, 23.
2. BrSpSi, V, 4-5; Sūryasiddhānta, V, 7-8 Bhāskara, Grahagaṇita, XII, 4.
3. Manitius, ibid, p. 119.

142 GREEK AND INDIAN METHODS tracted from, or added to, γK the celestial longitude, gives γM the polar longitude. According to Āryabhaṭa¹, MK = (σK × R vers γK × R sin ω) / R² . Brahmagupta² makes a dis- tinct improvement on Āryabhaṭa and gives his rule for finding the projection MK on the celestial equator. If P be the celestial pole, PKH the secondary to the equator, Brahmagupta says that, Fig- 13 NH = (σK × R sin (γK+90°) × R sin ω) / R If from σ, σR is drawn perpendicular to PKH, it is evident that, R sin σ R = (R sin σ K × R sin σ KR) / R According to Āryabhaṭa and Brahmagupta, as explained before, R sin σKR = (R sin (γK+90°) × R sin ω) / R Hence Brahmagupta intends that, NH = σR = (σK × R sin σKR) / R which is rather a big assumption. He then directs the finding of the part of the ecliptic of which σR or NH is the projection on the equator thus approximately to MK. Āryabhaṭa, Brahmagupta³ and the modern Sūryasiddhānta take the declination σN = σK + KH where σK is small. They do not consider the case where σK is large. Bhāskara alone gives us fairly correct rules for this trans- formation of co-ordinates.

1. Āryabhaṭa, Gola, 36.
2. BrSpSi X, 17.
3. BrSpSi, X, 15, Sūryasiddhānta, II, 58.

CELESTIAL AND POLAR LONGITUDE 143 In order to find σN, he would multiply σK by (R cos σKP) / R ; according to him, σN = (σK × R cos σKP) / R + KH¹ This is a decided improvement on Brahmagupta's corres- ponding rule. The declination σN obtained would be very nearly accurate. Having obtained σN, Bhāskara² then directs the finding of NH, thus, NH = (σK × R sin σKP) / (R cos σN) He then directs the finding of MK on the ecliptic of which NH is the projection by means of the times of rising of the signs of the zodiac on the equator. Thus, the Indian methods show a beginning and develop- ment only. The Greek method as given by Ptolemy is mathe- matically accurate. Greek Method³ To transform the celestial longitude and celestial latitude to right ascension and declination. Let the great circle ΠσK meet the equator at Δ. Ptolemy would then from the given value of γK, find γΔ and ΔK by using his tables for the rising of signs of the zodiac on the equa- tor. He then takes ΠPσ for the triangle and γNΔ Q for the transversal. The Menelaus' Equation, then, is (sin ΠQ / sin QP) × (sin PN / sin Nσ) × (sin σΔ / sin ΔΠ) = 1² Here ΠQ = 90° + ω, QP = 90°, PN = 90°; σΔ = σK + KΔ. ΠΔ = 90° + KΔ, whence Nσ is obtained. He next takes PNQ for the triangle and ΠσΔ for the trans- versal,

144 GREEK AND INDIAN METHODS It is almost needless to say that neither in the method nor in the rules is there any agreement between the Indian and Greek spherical astronomy in the solution of this problem. Kaye’s view¹ As to Kaye, it appears that he has not been able to find a method in the translation of the Sūyrasiddhānta by Burgess. The figures of his paper referred to before do not show the “Akṣakṣet- ras” even in their projections on the meridian place. He refers to Braunmühl’s History of Trigonometry but does not appear to have been able to follow him in his “Methode der indischen Trigonometrie.” Kaye, however, is not slow in belittling Indian trigonometry when he says :—The Indian astronomers employed the sine function principally and the versed sine occasionally ; they never employed the tangent function; and generally, but not always, preferred to employ the sine of the complementary angle rather than the cosine functions.’’ It is evident that Kaye never understood the meaning of the Indian functions of ‘sine’ and ‘cosine.’ These functions are fully explained by Bhāskara² when he says :— “Of that point the dis- tance from the east-west line is the sine and the distance of the point from the north-south line is the “cosine”. [Fig. 14: Circle with points N, S, E, W; P₁, P₂, P₃, P₄; N₁, N₂, N₃, N₄; M₁, M₂, M₃, M₄, A] Fig. 14 In (Fig. 14), of the arc AP₁, P₁M₁ is the “sine” and P₁N₁ is the “cosine” of AP₂, P₂M₂ is the “sine” and P₂N₂ is the “cosine”; of AP₃, P₃M₃ is the “sine” and P₃N₃ is the “cosine”; etc. It is evident that a better definition of these fun- ctions was never given. We have thus seen that some of the solutions of Āryabhaṭa

1. J. A. S. B., N. S., XV, p. 154.
2. तस्य बिन्दोः प्राच्य-परायाश्च यदन्तरं सादोर्ज्या । बिन्दोर्याम्योत्तरायाश्च यदन्तरं सा कोटिज्या ॥ —(Bhāskara, Grahagaṇita, commentary. II. 88-21)

KAYE'S VIEW 145 are imperfect, of Brahmagupta the solutions are more accurate, while those of Bhāskara are generally mathematically correct. The date of the scientific ancient Indian Astronomy is indeed 499 A. D., while that of the Syntaxis is about 150 A. D. It is by these shortcomings and differences in the methods, new ideas (e.g., the idea of the differential calculus)* and the like, that we can safely say that Indian Astronomy in its scientific form, although of a later date than the "Syntaxis" of Ptolemy, is origi- nal and not borrowed from foreign source. There is evidence that some crude form of Greek astronomy was transmitted to India and went by the name of the "Romaka" or the "Pauliśa" Siddhānta, prior to the time of Āryabhaṭa but our great Indian astronomers, Āryabhaṭa with his pupils, Varāha-mihira and Brahmagupta, had to construct a new science altogether. (This Chapter is almost a reproduction of the paper by P. C. Sengupta, as acknowledged earlier). —: o :— Reference P.C. Sengupta : The Khaṇḍakhādyaka, 1934

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CHAPTER VII Epicyclic Theory of Ancient Indians We shall give here some details of the Indian concepts regarding the motion of planets or wandering bodies among the stars. The Vedāṅga Jyotiṣa (1400 B. C. or earlier) does not speak of this, A comparison of the astronomical constants of the Greek and the ancient Indian systems, points unmistakably to the conclusion that the Indian constants as determined by Āryabhaṭa I and his successors, are almost in all cases different from those of Greeks. Indian astronomers were highly original in their con- cepts and treatment. The originality of Āryabhaṭa I and other astronomers would be seen from what we are discribing below. Apparent Motions of the Sun and Moon We have the following passages from Āryabhaṭa : All planets move in eccentrics to their orbits at the mean rates of angular motion, in the direction of the signs of the zodiac from their apogees (or aphelia) and in the opposite directions from their Śīghroccas. The eccentric circles of planets are equal to their concentrics and the centre of the eccentric is removed from the centre of the Earth. The distance between the centre of the Earth and the centre of the eccentric is equal to the radius of the planet's epicycle; on the circumference (whether of the epicycle or of the eccentric) the planet undoubtedly moves with the mean motion. Here the central idea was that undou- btedly planets moved unifomly in circles round the Earth; if the motion appeared to be variable, it was due to the fact that the centres of such circle (i. e. the con- centric circles) did not coincide with the centre of the Earth. Fig. 15

148 EPICYCLIC THEORY OF ANCIENT INDIANS Let E represent the centre of the Earth(Fig.15).APM the Sun's circular orbit or concentric; let A and P be the apogee and the perigee respectively. From EA, cut off EC equal to the radius of the Sun's epicycle. With centre C and radius equal to EA describe the eccentric A'P'S cutting AP and AP produced at P' and A'. Here A' and P' are the real apogee and perigee of the Sun's orbit. Let PM and P'S be any two equal arcs measured from P and P'. The idea is that the mean planet M and the apparent Sun S move simultaneously from P and P' in the counterclockwise direction along the concentric and the eccentric circles. They move with the same angular motion and arrive simultaneously at M and S. Here EM and CS are parallel and equal, hence MS is also equal and parallel to EC. Let SH be drawn perpendicular to EM. The angle PEM is the mean anomaly and the angle P'ES the true anomaly; the angle SEM is the equation of the centre, is readily seen to be plus (+) from P' to A' and minus (—) from A' to P'. Thus as regards the character of the equation, the eccentric circle is quite right. We now turn to exmine how far it is true as to the amount. Let the angle SEM denoted by E and the angle ∠PEM =∠P'CS=θ; EP=CP'=a; EC=MS=p, then tan E = SH / HE = p sin θ / (a - p cos θ) ∴ E = (p / a) sin θ - (p² / 2a²) sin 2θ + (p³ / 3a³) sin 3 θ......... Now the true value of E in elliptic motion is given by E = ( 2e - e³/4 ) sin θ + (5/4) e² sin 2θ + (13e³ / 12) sin 3θ*;... It we now put p / a = 2e - e³/4, as a first approximation p / a = 2e. Hence p² / 2a² = 2e², which is greater than 5/4 e² by 3/4 e². In the case of the Sun if the value of p be correctly taken the error in the coefficient of the second term becomes +3'; similarly in the case of the Moon, the corresponding error becomes +8'. *Godfray's Astronomy, p. 149.

APPARENT MOTION OF THE SUN AND MOON 149 Again if p/a=2e, what is the centre of the eccentric circle is the empty focus of the ellipse or that the ancient astronomers practically took the planets to be moving with uniform angular motion round the empty focus. This was not a bad approxi- mation. Also ES=r=EH approximately, ∴ r=a ( 1 - p/a cos θ ) but in the elliptic motion r=a (1-e cos θ).* Hence the error is not very considerable here also. This is the way in which the ancient astronomers, both Greek and Hindu, sought to explain the inequalities in the motion of the Sun and the Moon. In the case of the Moon, these astro- nomers took the coefficient 2e - e³/4 = 300' nearly; the modern value is 377' nearly. The reason for this has been pointed out to be that the Moon was observed correctly only at times of eclipses. At the eclipses of syzygies, the evection term of the Moon's equa- tion diminishes (numerically) the principal ecliptic term by about 76'. We have thus far explained the idea of planetary motion of the ancients under the eccentric circle costruction. The same, however, is explained under the epicyclic construction. Let AMP be the circular orbit of the Sun, having E the centre of the earth for the centre. (Fig. 16) Let the diameter AEP be the apse line. A the apogee and P the perigee. Let M be the mean position of the Sun in the orbit. With M as the centre, describe the epicycle UNS. Let EM cut the epicycle at N and U. Now the construction for finding S the apparent Sun is thus given:— Fig. 16 Make ∠UMS= ∠MEA, the arc US is measured clockwise whereas the arc A to M is measured counterclockwise. From this construction MS is parallel to EA. If EC be measured equal to MS, the radius of the epicycle, along EA to-

• Godfray's Astronomy, p. 149.