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

391 Second half of verse 21 and first half of verse 22. H cos λ × H sin ω ───────────────── = H sin θ where θ is the Āyana Valana H cos δ The direction of this Valana is that of the hemisphere north or south in which the Moon lies. Comm. This is the formula we have already derived. Regarding the direction of the Āyana Valana, the conven- tion is that it is to be considered north, if the Moon be in the northern hemisphere, otherwise south. The reason is that at the time of a lunar eclipse, the Moon being in opposi- tion, if he be north, the Sun will be south of the Equator, and the line BA of fig. 80 representing the Ecliptic which is roughly the join of the Sun and the Moon will be north of the line JI which is parallel to the equator. Thus the direction of the angle IMA gives the direction of the Āyana Valana. Latter half of verse 22 and verse 23. Sphuta Valana. The Hsine of the sum or difference of the two Valanas according as they are of the same or opposite directions, multiplied by the sum of the angular radii of the Moon and Rāhu, and divided by the radius gives the Hsine of the Sphuta Valana. Those who said that the Valana is proportional to the Hversine, do not know spherical geometry properly. Comm. The direction of the Akṣa Valana, was defined in verse 20 that it is north if the hour angle is east, other- wise south. The meaning of this convention is that the diameter of the Moon's disc parallel to the Equator when the hour angle is East, is north of the diameter which is parallel to the prime-vertical. Thus combining the two conventions regarding the directions of the Valanas, it is clear that if both the Valanas are north, the line MI is north of MG, and MA is north of MI (fig. 80) so that the

392 Sphuta Valana is equal to the sum of the angles G M̂ I and I M̂ A. Suppose MA is south of MI either falling within the angle GMI or south of MG, then clearly the Sphutr Valana G M̂ A = G M̂ I - A M̂ I or A M̂ I - G M̂ I as the case may be, which is obtained as the difference of the two Valanas. Having obtained G M̂ A as the Sphuta Valana, H sin GMI multiplied by (P+r) and divided by R gives Hsine of the Sphuta Valana to be represented in a circle of radius equal to P+r. This latter convention of repres- enting the Sphuta Valana in a circle of radius P+r is only a convention. The expression H sin (Sphuta Valana) × P+r --------------------------- gives us RN of fig. 87, R where GMA is Sphuta Valana. In other words, we are to draw fig. 87 to show the point of first contact namely A in relation to MG the line parallel to the prime-vertical. Fig. 87

393 Verse 24. Conversion of liptas into what are called Angulas. H cos z of the eclipsed body at the moment of eclipse being divided by the radius and the result being added to 2½ gives the number of liptas per angula. The time elapsed after the rise of the body being divided by the rising hour angle (both being expressed in the same units of time) and the result being added to 2½ also gives the same. Comm. While a parilekha or a geometrical drawing of the eclipse is attempted at, the problem arises as to how many liptas or minutes of arc giving the measure of the disc are to be taken to be equivalent to one angula. For example, suppose the diameter of the disc is 30'. With what radius shall we draw the disc on a board or paper? In this behalf, a convention based on observations is being mentioned. The discs of the Sun and the Moon are observed to be big at rise and small when they are on the meridian. So, taking the measure of the disc to be 30' for example, if the eclipse takes place at the rise of the disc, it is laid down to draw the disc with a radius of 15/2½ ie. 6 angulas at the rate of 2½ liptas per angula. {The word angula here mentioned might not be what we take it to be today in our daily parlance as one inch. The gnomon or S'anku was taken in those days to be of a length equal to 12 angulas. Bhāskara mentions in the beginning of Lilavati that 8 Yavas are together equal to one Angula, twentyfour angulas to one hasta, four hastas to one danda and 2000 dandas to one Kros'a, and 4 Kros'as to one Yojana. Also a vamsa is equal to ten hastas. This system discloses that, a Yojana equals 5 modern miles according to Bhāskara's estimate of the diameter of the Earth as compared with its modern estimate. (The method given by Bhāskara as to how the diameter of the Earth could be measured is found to be quite scientific as men- tioned by us before). 50

394 33 modern inches or angulas are equal to 80 angulas of Bhāskara as per the above. At this rate the gnomon's modern length would be just 5″.} Reverting to our subject, we are asked to represent the disc of 30 liptas when the eclipse takes place at noon by 30/3½ angulas counting at the rate of 3½ liptas per angula. Then the question arises as to what should be the correspondence between the liptas and angulas when the eclipse takes place in between the rise of the Moon (or Sun) and its noon. The directive is that one angula = 2½′ + (H cos z) / R = 2½ + cos z I This means, supposing Z=zenith-distance of the body to be 60°, one angula is to be taken to be equal to 2½ + cos 60 = 2½ + ½ = 3′ or 3 liptas. The reason given by Bhāskara reiterating what Sri- pati said in that behalf, as to why the Moon's or Sun's disc appears to be big at the moment of rising and small on the meridian, is that the disc is immersed in its own rays at noon and rendered small in appearance, whereas, most of the rays are swallowed by the earth or its atmos- phere at the moment of rising, making the disc appear large and easily visible. Note. Bhāskara gives the proof of the above formula I as follows. Since at the time of rise, we are taking 2½ liptas of the measure of the disc to be equal to one angula and while the disc is on the meridian, 3½ liptas are to be taken as one angula, there is an increase of one lipta for an increase of H cos z from zero at the horizon to a value equal to the Radius. So, the argument adduced is 'If for an increase of H cos z equal to the radius, there is an increase

395 of one lipta in addition to 2½, what should the increase be for an arbitrary H cos z '? The result is (H cos z / R) × 1 = cos z. Note 2. As finding cos z at the time of an eclipse implies additional arithmetical calculation, and as we have already with us the data of (1) the time elapsed after rise of the celestial body (Moon or Sun) till the moment of the eclipse and (2) half the duration of the day of the body ie. the rising hour-angle converted into time at the rate of 6° per nadi, so a rough formula is given using h in the place of z. The rule of three now used, is 'If for an unnata equal to the dinārdha or rising hour-angle converted into nadis, we have an increase of 1 lipta per angula (over and above 2½ liptas) what shall we have for an arbitrary un- nata ?' The result added to 2½ liptas, gives the formula one angula = 2½ + Unnata / Dinārdha . Dinārdha corresponds to H, the rising hour-angle and Unnata corresponds to (H-h) where h is the hour-angle at the time. Bhāskara uses the word 'Angula-liptas' meaning thereby the liptas that are to be taken to be equal to one angula while drawing the parilekha of the disc at the time of its eclipse. Verse 25. Converting Valana etc. into Angulas. The measures of the Valana (defined above) or the Sara ie. the celestial latitude of the Moon or the Rāhu Bimba or the Bhuja (defined) are to be converted into Angulas at the rate given by the above formula. While drawing a figure of the solar eclipse, the celestial latitude of the Moon is to be drawn in its own direction whereas in a lunar eclipse, it has to be drawn in the opposite direction. Comm. The first part is clear. Regarding the second statement, in as much as the centre of the Rāhu Bimba lies

at the foot of the Moon's celestial latitude, the latter has to be drawn in the opposite direction, since in the pari- lekha, the centre of the Moon's disc is to be at the centre. Verses 26 to 29. How to depict the eclipse in drawing. Draw a circle with radius equal to that of the radius of the disc of the eclipsed body and also a circle of radius equal to r+p, the sum of the radii of the eclipsed and eclipsing bodies; let directions (east etc.) be marked in the figure. In the outer circle, draw the Valanajya or the Hsine of the Sphutavala with respect to the East point, Valanajyā pertaining to the moment of first contact. In the case of the Moon, the Valanajyā pertaining to the moment of first contact should be marked from the East point and that pertaining to the moment of last contact should be marked from the West point. In the case of the Sun the reverse is to be done. If the Valana is south, it should be marked in the clockwise direction, otherwise anticlockwise. Having marked the Valanajyā in the form of a Hsine, draw the line joining the centre to the top of the Valanajyā, ie. to the point of intersection of the Hsine with the outer circle. The celestial latitude of the Moon is to be drawn from this top of the Valanajyā in the form of Hsine again. If the latitude pertains to the moment of first contact, it should be drawn from the top of the Valanajyā pertaining to that moment, and if it pertains to the moment of last contact, it should be laid off from the top of the Valanajyā pertaining to the moment of last contact. The celestial latitude pertaining to the middle of the eclipse should be drawn from the centre along the line of Valanasūtra or the line joining the centre to the top of the Valanajyā. Taking the extremities of these latitudes, circles are to be drawn with the radius of the eclipsing body to depict the eclipse at the respective moments.

397 Fig. 88 Comm. Let M₁, M₂, M₃ be the positions of the centre of the Moon's disc at the moment of first contact, at the middle of the eclipse and the moment of last contact respectively. Draw a circle with radius M₁ R₁ = r+p which is called the Manaikyārdha Vritta. Let M₁ E₁ represent the Eastern direction known as the Sama- mandalaprāchī. Draw E₁ V₁ equal to Hsine of the Valana, so that M₁ V₁ is the Krānti-Vritta-prāchī ie. the point of intersection of the Ecliptic with the Eastern horizon. Draw N₁ R₁ perpendicular to M₁ V₁ in the form of Hsine, which is the latitude (Vikṣepa) of the Moon at the moment of first contact. With R₁ as centre draw the Grāhaka- Vritta with p as centre ; this circle represents the Rāhu- Bimba. Similarly let M₃ be the centre of the Moon's disc at the moment of last contact. Draw a circle with M₃ as centre and r+p as the radius which is the Manaikyārdha- Vritta. Let M₃ W₃ be the direction to the West, the Samamandalapratichī. From W₃ draw W₃ V₃ the Hsine of the Valana, so that M₃ V₃ is now the Krāntimandala- pratīchī ie. the point of intersection of ecliptic with the western horizon. Let, the Vikṣepa N₃ R₃ be drawn as a Hsine of the Mānaikyārdha Vritta. With R₃ as centre and radius p, draw the Rāhu-Bimba. Let M₂ be the centre of the Moon's disc at the middle of the eclipse. Let M₂ S₂ represent the South with respect to the prime-vertical. From S₂, draw the Hsine of the Valana S₂ V₂ so that M₂ V₂ is the Krānti-Vritta-Dakshiṇā ie. the South with respect to the Ecliptic. Now the

398 Vikṣepa or the celestial latitude of the Moon M₂ R₂ is to be drawn along this Valanasūtra M₂ V. With R₂ as centre and radius p drawn the Rāhu-Bimba. Fig. 89 Depiction of Fig. 88, keeping the Moon fixed The slight flaw in this figure is that ML₁ L₃ implied as the path of the Moon is taken to be parallel to the ecliptic R₁ R₂ R₃ the path of the eclipsing body the Grāhakamārga, in as much as latitudes are drawn perpendi- cular to ML₁ L₃. This figure depicts a total eclipse of the Moon. If M coincides with R₂ at the middle moment of the eclipse, then the eclipse is called central. The duration of a central eclipse will be on the average the time that the Moon's disc takes to cross the diameter of the Rāhu-Bimba with its relative velocity. Hence the mean duration of a central eclipse is Average diameter of Rāhu + Average diameter of the Moon ────────────────────────────────────────────────────────── Relative velocity of the Moon with respect to the shadow = [(81+64) × 24] / [790′-35″ — 59′-8″] hrs = 145/732 × 24 = 290/61 = 4 hrs–45 minutes approximately.

399 Verse 30 and first half of 31. Geometrical depiction of the eclipse at the beginning and end of totality and also of the magnitude of the eclipse. The Bhuja is to be laid from the centre of the Moon along its Valanasūtra or the line indicating the direction of the ecliptic; the latitude is to be drawn from the end of the Bhuja and perpendicular to the Bhuja. The hypo- tenuse is to be drawn from the centre of the Moon. Taking the point of intersection of the latitude (Kōti) and the hypotenuse, as centre and radius p equal to that of the eclipsing body, if circles be drawn, from these circles could be known the points where totality begins and ends as well as the magnitude of the eclipse at any given moment. Or these could be found in another way as follows. Comm. The method given above for depicting the phases of an eclipse geometrically, could be applied for any moment during the course of the eclipse and depends upon before-hand computed Bhuja and Kōti. Refer to fig. 90. Let M be the centre of the Moon's disc. Mark Eω the East-west line drawn through M. Compute the Valana for the required moment, either for the moment when totality begins or for that when totality ends or for Fig. 90

400 any arbitrary moment whatsoever. With this Valana primarily laid in the Manaikyārdha Vritta, decide the Krānti Vrittaprachī or the East-West direction of the eclipse. Thus in the figure V₁ V₂ is this direction. Then lay off the computed Bhuja along this V₁ V₂ from M, say MA for the Sammīlana moment or the moment when totality begins or MC for the Unmīlana moment or the moment when totality ends or MD for an arbitrary moment. Draw AR₁ or CR₂ or DE equal to the latitude at the particular moment, perpendicular to the Valana- sūtra. In the figure drawn the Valanasūtra is shown to be the same. This does not mean it will be the same throughout. It will be changing because the position of the Ecliptic changes from moment to moment. So Bhās- kara uses the word ie. 'the respective Valanasūtra'. Also the latitudes will be differing from moment to moment as well as the Bhujas both of which are to be computed for any moment along with the Valana. (The method of computing the Bhuja was given in verse 15). Computing the respective Karṇas or the hypotenuses from the formula K = √(Bhujā² + Kōti²), (Kōti is here the lati- tude) with centre M and radius equal to the Karṇas, if arcs be drawn to cut the latitudes, the points of intersection would be no other than R₁, R₂ or E. Join MR₁, MR₂ and ME. With centres R₁ and R₂ and radii equal to p - r, (where p is the radius of the Rāhu-Bimba, and r the radius of the Moon's disc) if circles be drawn, they just touch the Moon's disc at F and G which are the points where totality begins and ends respectively. With centre E and radius P, if a circle be drawn, that will show what amount of the disc is shadowed as well as the measure of the magnitude of the eclipse (defined in verse 11). Note. In the above commentary and figure we have depicted MD as the Iṣṭa-Bhuja or the Bhuja at a given moment, taking a moment prior to the Unmī- lanakāla, for showing the magnitude of the eclipse.

401 If a moment in between the Sammīlana and Unmīlana were taken, the then Bhuja and Koṭi could be no doubt computed, but the question of magnitude of the eclipse does not arise as the entire disc has been plunged in the shadow. Second half of verse 31 verse 32 and first half of verse 33. Alternative method of depicting the eclipse geo- metrically. Joining the upper end of the latitude of the middle moment of the eclipse to those of the first and last contacts, we have what are called the Pragrahamārga and Mokṣamārga ie. the path of the centre of the eclipsing body from the first contact to the middle moment of the eclipse and that from the middle moment to the last contact. The lengths of these paths could be computed and they could be drawn before hand. Then with the centre of the Moon as centre and radius equal to p—r, if a circle be drawn, it cuts the paths described above each in one point. With those points as centre and radii equal to p, if circles be drawn, they will touch the Moon’s disc each in one point which are respectively the points of Sammīlana and Unmīlana. Comm. In as much as the latitude of the Moon differs from moment to moment, the Pragrahamārga and the Mokṣamārga are separated to achieve a little more accuracy than could be got by joining the upper extre- mities of the initial and final latitudes. The remaining statement is evident, for, at the moments of Sammīlana and Unmīlana, the distance between the centres of the eclipsing body and the eclipsed will be p—r, so that the points of intersection of the Pragrahamārga and Mokṣa- mārga with the circle whose centre is the centre of the eclipsed body and radius p—r will give the centre of the eclipsing body. 51

402 Latter half of verse 33. To know the magnitude of the eclipse at any given moment during the course of the eclipse. Let the product of the time elapsed from the moment of first contact and the length of the path of the eclipsing body traced from the moment of the first contact to the middle of the eclipse divided by the time between the moment of first contact and the middle of the eclipse, be x. Similarly let the product of the time before the end of last contact and the path of the eclipsing body traced between the middle moment of the eclipse and the moment of last contact divided by the time between the middle moment and the moment of last contact be y. Lay off x and y units of length from the first and last points of the path of the eclipsing body along the path respectively. Then we get the points of the centre of the eclipsing body at the required moments. With these points as centre and radius p, if circles be drawn, they represent the eclipsing body. The length of the diameter of the eclipsed body shaded, gives the magnitude of the eclipse called grāsa. Comm. Here rule of three is applied namely “If during time T₁ or T₂ a path equal to l₁ or l₂ in length is traced what length will be traced in times t₁ or t₂?”, where T₁ and T₂ are the times called Sparsa-Sthiti-Khanda and Mokṣa-Sthiti-Khanda respectively, l₁ and l₂ are the times elapsed from the moment of first contact or before the moment of last contact and t₁, t₂ are the times from the beginning of the eclipse and before the end of the eclipse respectively. Then x and y give the points where the centre of the eclipsing body lies. Verse 35. Given the magnitude of the eclipse at any time to obtain the time elapsed after the first contact. The time taken by the centre of the eclipsing body to move through the segment of the path of the eclipsing

408 body which lies between the position of the eclipsing body at the moment of first contact and the point of intersection with the path of the eclipsing body of the circle drawn with the centre of the eclipsed body as centre and radius equal to the difference of p+r -g where g is the magni- tude of the eclipse (grāsa) at the moment, or similarly the time taken by the centre of the eclipsing body to move through a similar and equal segment of the path of the eclipsing body on the other side, gives the time elapsed after the moment of first contact or the time before the moment of last contact. Comm. This is the converse of the above problem. The method is clear being based on rule of three as above. Both the problems could be algebraically expressed as follows. Let T, t, l, g, and k, stand respectively for the Sthiti-Khanda ie. the time between the moment of first contact to the middle of the eclipse or the time between the middle moment to the moment of last contact ; (2). the time elapsed after the moment of first contact or the time before the moment of last contact, as the case may be ; (3) the length of the Pragrahamārga or Mokṣamārga ; (4) the grāsa which is defined as p+r-k ; (5) the Karṇa whose expression is √(B²+β²), B being the Bhuja defined and β the latitude of the Moon. Then the following working is stipulated (a) If in time T, a path of length l is traced, what will be traced in t? The result is lt/T (b) Then B = l - lt/T (c) B² + β² = K² (d) p+r-k=g. Thus combining all the steps {l (T-t) / T}² + β² = (p+r-g)² ie. l² (T-t)² + β² T² = T² (p+r-g)² I given t, this equation gives g and given g it gives t.

404 Again the following relation holds good between T and l, l² = (p+r)² – β² II and l / (m₁–s₁) = T with the nomenclature already employed which means T = √(p+r)²–β² / (m₁–s₁) III In the above working, the fundamental elements are p, r, β, m₁ and s₁ with which the other elements could be worked out. Replacing the other elements from equation I, we have {(p+r)²–β²} { √(p+r)²–β² / (m₁–s₁) – t }² + β² (p+r²–β²) / (m₁–s₁)² = (p+r)²–β² / (m₁–s₁)² (p+r–g)² ie. {(p+r)²–β²} [√(p+r)²–β² – t (m₁–s₁)]² + β² (p+r²–β²) = (p+r²–β²) (p+r–g)² ie. {√p+r²–β² – t (m₁–s₁)}² + β² = (p+r–g)² IV Putting t=0 in this equation, we have (p+r)² = (p+r–g)² ie. g=0 which means at the moment of first contact, the grāsa is zero. Again putting t = T ie. t = √(p+r)²–β² / (m₁–s₁) ie. t (m₁ – s₁) = √p+r²–β² we have β² = p+r–g² ie. g = p+r–β which gives the grāha at the middle of the eclipse which was defined as the Sthagita. In equation IV which we may take as a funda- mental equation, the two unknowns are t and g one of which being given the other could be got. Verse 36. The colour of the eclipse. When less than half the disc of the Moon is eclipsed, the colour will be what is called Dhumra ie. of the colour

405 of smoke; when the disc is half eclipsed, the colour is black; when more than half is eclipsed, the colour would be a blend of black and red and when the entire disc is eclipsed, the colour will be what is called pisanga or reddish-brown. Comm. Clear. Verse 37. When declare the occurrence of an eclipse. When even one-sixteenth of the diameter of the Moon's disc is shadowed, the eclipse will not be visible in as much as the shadowed portion is covered by the illaminating rays of the disc. In the case of the Sun, when even one-twelfth of the diameter is shadowed, the eclipse will not be visible for the same reason. Hence we shall not declare the occurence of an eclipse upto the shadowing of the discs to the extents stated above. Verses 38 and 39. Examples which disclose the invalidity of construing Valana in terms of Hversine instead of Hsine. When the Sun is in the zenith, the Ecliptic being vertical, the Valana is clearly seen to be the Agra of (☉+90) where ☉ is the longitude of the Sun. If you could show that the Valana will be the same on the basis of Hversine-formula, then I would accept that what Lallāchārya postulated in his work Śiṣya-Dhī-Vṛddhida is correct. Again, in a place of latitude 90—ω, ω being the obli- quity of the Ecliptic (ω is taken to be 24°), when the Sun being situated in Meṣa, Vṛṣabha, Mīna or Kumbha, the Moon contacts him from the south at the moment of a solar eclipse, in as much as the Ecliptic coincides with the horizon. In this circumstance, how could the Valana be equal to R, as made out by the Hversine-formula.

406 Comm. Lallāchārya gave the Valana in terms of the following verses “स्पर्शादिकालजनितोत्क्रमशिञ्जिनीभिः, क्षुण्णाक्षभा पलभवश्रवणेन भक्ता, चापानि पूर्वनतपश्चिमयोः फलानि, सौम्येतराणि समवेहि पृथक् क्रमेण; ग्राह्यात् सराशित्रितयाद् भुजज्या व्यस्ता ततः प्राग्वदप- क्रमज्या...” Verses 23, 25 Chandragrahaṇādhikāra, wherein he formulated the Valana in terms of Hversine in the place of Hsine. The reason for his slip, we have already explained. Now Bhāskara gives two glaring examples to substantiate his formula and to show up the flaw in Lallāchārya's formulation. In the first example, where the Ecliptic takes the form of a Vertical, the Sun being in the zenith, the Spaṣṭa Valana which is the angle between the Ecliptic and the prime-vertical is the same as the arc between the East point and the intersection of the Ecliptic with the horizon known as Lagna. Since the Sun is then in the zenith, the longitude of the Lagna is (90 + ☉) so that the said arc is the Agra of the point whose longitude is 90 + ☉ as stated. Hence Spaṣṭa Valanajyā = sin A = sin δ/cos ϕ where A is the agrā (using Napier's rule from triangle PNL where L is the Lagna N the north-point and P the celestial pole). In the Hindu form, this is given by H sin V = H sin A = (R H sin δ) / (H cos ϕ) where δ is the declination of a point of the Ecliptic whose longitude is (90 + ☉). But Lallāchārya's formula gives the Valanajyā as Hvers δ, δ being the declination of a point of the Ecliptic whose longitude is (90 + λ), λ being the longitude of the Eclipsed body ignoring the latitude. In other words, in the case of the lunar Eclipse when the Moon is in the zenith his Valanajyā = Hvers δ (δ having the above value) the Ākṣa Valanajyā here being zero. Since (R H sin δ) / (H cos ϕ) ≮ Hvers δ, the mistake committed by Lallāchārya is evident even supposing H cos ϕ = R when we ignore the latitude ie. take ϕ to be zero.

407 In the second example cited by Bhāskara the Ecliptic coincides with the horizon, the pole of the Ecliptic being in the zenith. Then in a Solar Eclipse the Moon eclipses the Sun from the south showing that H sin V = R. That H sin V = R is also evident from the fact that the Ecliptic makes 90° with the prime-vertical, having coincided with the horizon. But here according to Lallāchārya’s formula, H sin ξ = Hvers 90° = R and Āyana Valanajyā is Hvers δ, where δ is the declination of a point whose longitude is 90° more than ☉. If ☉ = 30°, 60° H sin θ = R/2 sin ω/R or (√3/2) (R sin ω)/R ie. (sin ω)/2 or √3/2 sin ω. Evidently the sum of the two Valanas Āyana and Ākṣa cannot be 90° as is also vouchsafed from geometry. So, here also, the flaw is evident. Note 1. Śrīpatyāchārya also followed Lallāchārya vide verses 18, 19, 20 Chandragrahaṇādhyāya, Siddhānta Śekhara. It will be noted that the commentator of Siddhānta Śekhara, while reiterating Bhāskara’s stand as the correct one, himself commits a mistake in saying “सममण्डलीय नतांशज्यास्थाने नतकाकोत्क्रमज्या गृहीता” In fact sin ξ = (sin ϕ sin h) / cos μ = (sin ϕ sin z) / cos δ as proved by us before. The commentator cited above overlooked that sin ξ could be also equal to (sin ϕ sin h) / cos μ , wherein natakāla also is implied. Note 2. It will be noted that even Pṛthūdakāchārya, while commenting on Brahmasphuṭa Siddhānta, ignored Brahmagupta and followed Lallāchārya blindly. Note 3. The formula given by Lallāchārya and followed by Pṛthūdaka as well as by Śrīpati is very rough besides containing the flaw cited, in as much as both μ and δ are taken to be zero, which are not so,

SURYAGRAHANĀDHIKĀRA Verse 1. In as much as the observer situated on the surface of the Earth and as such elevated by the radius of the Earth from the centre there of, perceives not the Sun and the Moon having the same longitude at the moment of conjunction, to be in the same line of sight, heyt being depressed unequally having different orbits, so I proceed to elucidate what are called Lambana and Nati ie. parallax in longitude and latitude, on which account they are not in the same line of sight. Fig. 91 Comm. (Refer fig. 91) Let E be the centre of the Earth, M and S the centres of the discs of the Moon and

the Sun. Let A be the position of an observer on the surface of the Earth, elevated by the radius EA from E. Let M and S be in the same line of sight as seen from E. But as seen from A, AS and AM are respectively the lines of sight to the Sun and the Moon. Evidently these lines of sight differ the Moon being depressed more than the Sun. If a line AS' be drawn which is parallel to the central line of sight namely EMS, we find that the Sun is depressed by the angle S'AS whereas the Moon is depressed by the angle S'AM'. These angles differ because the orbits of the Sun and Moon differ. Here the angle S'AS will be very very small, its magnitude being in truth just about 8" only. But the angle S'AM' will be sufficiently large since the Moon is very near the Earth compared with the Sun. Taking ES and AS to be almost parallel due to the largeness of the Sun's distance, the angle SÂM will be almost equal to AM̂S so that we could consider that the Moon is depressed from AS the line of sight to the Sun by the angle SÂM' = AM̂E. This angle AME is called the geocentric parallax of the Moon and the angle AŜE that of the Sun M'ÂS = angle of depression of the Moon over and above that of the Sun = M'AS' - M'AS = EM̂A - EŜA = geocentric parallax of the Moon minus geocentric parallax of the Sun. Verse 2. The presence or absence as well as the positiveness and negativeness of the parallax in longitude. Compute the Lagna at the moment of conjunction of the Sun and the Moon. There will be no parallax in longitude when the Sun is situated at the point called Vitribha or the point whose longitude is ≡ L - 90°, L being the longitude of the Lagna ie. the ascendant which is the point of intersection of the Ecliptic with the 52

410 horizon. If the Sun's longitude falls short of the longitude of the Vitribha or exceeds it, there will be parallax in longitude which will be positive in the former case and negative in the latter. Fig. 92 Comm. (Ref. fig. 92) Let SN be the horizon, Z the zenith and VA the Ecliptic. A is the ascendant or Lagna. Let V be the point called Vitribha which is 90° behind A. Strictly speaking V is called Vitribhalagna or lagna from which three Rāśis or 90° are subtracted (Bha = Rāśi. त्रिभिर्विरहितम् वित्रिभम्; वित्रिभम् च तत् लग्नम् च वित्रिभलग्नम् ie. a point whose longitude is got by subtracting three Rāśis from that of the Lagna). Let ZV be the vertical of V so that ZV̂A = 90°. It will be seen that AV = 90° as follows. Let A' be the point where the Ecliptic intersects the horizon on the west. One will construe that the Ecliptic is bisected by the meridian; but it is not so. Spherical triangles AVZ and A'VZ being right-angled at