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

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.

191 Comm. This is clear from fig. 24. If S be the end of Vṛiṣabha, S' in the figure denotes the end of Karkaṭa them from the right-angled triangle SAP, ^ sin S'A = sin SA = sin APS × sin PS = sin A'K × cos SK sin S'A ∴ sin A'K = ————— . But sin S'A = cos S'♈ cos SK and cos SK = cos δ where δ is the declination at S. A'K converted into time gives the rising time of AS' ie. Karkaṭa. cos S' ∴ the rising time of Karkaṭa = ———— cos δ In tbe Hindu form, it will be Karkaṭa-anta-Koṭijyā Karkaṭa-Rāsi-Udayakāla = ————————————— × R Karkaṭa-anta-Dyujyā as stated. cos ♈S In modern terms sin A'K = cos ♈K = ———— cos δ from the formula cos λ = cos α cos δ. Thus, virtually the formula is a statement of the formula cos λ = cos α ^ cos δ. Aslo the formula sin AS' = sin P sin PS is parallel to the formula sin δ = sin λ sin ω which we proved already from Hindu methods. Verse 57. Still an alternative method. The H sines of Meṣa etc. being multiplied by H cos ω divided by their respective H cos δ's and the arcs thereof being subtracted as before the preceding from the succeeding we have the rising times of Meṣa etc. Comm. From figure 24, ^ SN SN = SL cos LSN and ————— = perpendicular from cos SK K on ♈

192 ∴ (SL cos ω) / (cos δ) = H sin rK. SL = H sin rS ∴ H sin rK = (H sin rS × H cos ω) / (H cos δ) We proved the above in a modern way. The Hindu concept is derived from the similarity of SML and ACM′ where C is the centre of the sphere and M′ is the foot of the perpendicular from A on the plane of the equator ∴ LM / CM′ = SL / CA ∴ LM = (H cos ω × H sin rS) / R Since LM = SN. LM divided by H cos δ and multiplied by R gives H sin rK ∴ H sin rK = [(H cos ω × H sin rS) / R] × [R / (H cos δ)] = = (H cos rS × H cos ω) / (H cos δ) i.e. H sin α = (H sin λ × H cos ω) / (H cos δ) Here H cos ω is called Trigṛha-dyu-maurvī because it is the H cosine of the declination of λ when λ = 90°. Verses 58, 59. The magnitudes of the rising times. Those rising times are 1670, 1793, 1937 ; these in the same and reverse orders diminished or increased by their respective Chara segments which are also in the same and reverse orders give the rising times of the Sāyana Rasis beginning from Meṣa for the locality. The Rasis from Tulā are in a reverse direction i.e. as the Meṣa is proje- cting upwards above the horizon, Tulā will be projecting below the horizon so that, the time taken by Meṣa to rise is exactly the time taken by Tulā to set. Comm. We shall compute the rising times of Sāyana Rasis for Lanka first i.e. for zero latitude using modern methods from the formula tan α = cos ω tan λ

198 log tan α = log cos ω + log tan λ; Put λ₁ = 30, and λ₂ = 60 and take ω = 24°; Let the corresponding α's be α₁, α₂ log tan α₁ = log cos 24 + log tan 30° (1) log tan α₂ = log cos 24 + log tan 60° (2) log tan α₁ = 9.9607 + 9.7614 = 9.7221 ∴ α₁ = 27° - 48' log tan α₂ = 9.9607 + 10.2386 = 10.1993 ∴ α₂ = 57° - 42' At the rate of 1 asu for 1', α₁ = 1668 asus α₂ = 3462; α₂ - α₁ = 1794 and since α₃ = the right ascension of 90° Longitude = 90°, α₃ = 5400 so that α₃ - α₂ = 1938. These are given by Bhāskara as 1670, 1793, 1937, the first exceeding by 2 asus, the second and third each less by one asu and the total according with the total. The rising times of Karkaṭa etc. will be 1937, 1793, 1670, 1670, 1793, 1937, 1937, 1793, 1670 respectively. Let us then find the rising times of these Sāyana Rasis at a locality say of latitude 13°. Refer to fig. 21. Let rS represent Meṣa so that the rising times of rE is equal to that of rS. But rE = rA - AE. We have seen rA = 1670 using Bhāskara's value. Sin EA = tan 13° tan δ₁ where δ₁ is the declination of S where rS = 30°. Sin δ₁ = sin 30° sin 24°. log sin δ₁ = 9.6990 + 9.6093 = 9.3083 ∴ δ₁ = 11° - 44' ∴ log sin EA = log tan 13° + log tan 11° - 44 = 9·3634 + 9.3175 = 8.6809 ∴ EA = 2° - 45' ∴ rE = 1670 - 165 = 1505 asus. Similarly for λ = 60°, putting α₂, δ₂ in the place of α₁, δ₁ and proceeding as before rE = rA₁ - A₁E; rA₁ = 3463; sin EA₁ = tan 13° tan δ₂ sin δ₂ = sin 60° sin 24° 25

194 ∴ log sin δ₂ = 9.6093 + 9.9375 = 9.5468 ∴ δ₂ = 20° – 30′ ∴ log sin EA₁ = 9.3634 + 9.5758 = 8.9392 ∴ EA₁ = 4° – 59′ = 299′ ∴ rE = 3463 – 299 = 3164 asus ∴ Rising time of Sāyana Vṛṣabha for the locality = 3164 – 1505 = 1659 asus. rA₂ = 5400, sin EA₂ = tan 13 tan δ₂ = tan 13 tan 24°. ∴ log sin EA₂ = 9.3634 + 9.6486 = 9.0120 ∴ EA₂ = 5° – 54′ = 354 asus ∴ rE = 5400 – 354 = 5046 ∴ Rising time of Mithuna is 5046 – 3164 = 1882 asus. Before we proceed to find the rising times of Karka- taka, Simha and Kanyā, we shall cast our previous proce- dure into the Hindu form. In fig. 21, let rS be the Sāyana Mesha. The rising time of rS is measured by rE, because when r is at E, Meṣa is just about to rise and when r is in the position indicated, the extremity of Meṣa namely S is rising. So it means that as rS of the ecliptic has risen, a portion rE of the Equator has risen. As time is measured by the arc of the equator which rises with a uniform speed, we measure the rising time of rS by the arc rE; but rE = rA – AE. rA is the Equatorial rising time of rS, because when A is at E, S will be at B i.e. A and S will then be on the equatorial horizon EB simul- taneously. Hence rA = 1670 as proved before and stated by Bhāskara. EA is the chara for 30°. The chara for one angula or inch (inch is here used technically, and does not mean what it means in ordinary parlour) as has been stated by Bhāskara and proved by us is 10 Vinadis. (Vide page 181)

195 But we have taken 13° as our latitude, so that tan ϕ = .2309. Since s/12 = tan ϕ = .2309 ∴ s = 2.7708; let us take this as 2.8″ so that the charas 10, 8, 3⅓ found for one inch are to be multiplied by 2.8 to give the local charas. They are in Vinādis, 28, 22.4, 9⅓ or in asus 168, 134·4, 56; for convenience let us take 134.4 as 134, so that the charas are 168, 134, 56. Thus the rising time of Sāyana Meṣa at this locality is 1670 − 168 = 1502 asus i.e. 250 Vinādis = 4-10 Nādis. Then let rS now represent 60°, instead of subtracting EA from rA to get the combined rising time of Meṣa and Vṛiṣabha, the Hindu practice is to subtract the chara pertaining to Vṛiṣabha from the equatorial rising of Vrishabha i.e. rE = 1793 − 134 = 1659 as got before. Here it must be noted that EA° is the chara not pertaining to Vṛiṣabha alone but to Meṣa and Vṛiṣabha put together. That is why for ease, the chara to Meṣa, the increase in chara for Vṛiṣabha, and the increase in chara for Mithuna as well as their individual equatorial rising times are given. The increments in the charas are called chara-khandas just as the increments in the H sines are called Jyā-khandas (khands means segments). Simil- arly the rising time of Sāyana Mithuna is equal to 1937 − 56 = 1881 asus = 313.5 Vinadis = 5-14 Nadis. Now with respect to Karkataka, its equatorial rising time is 1793, for, from fig. 25, the equator at the equatorial place being prime Vertical, if rM be Meṣa, its time of rising is given by rE where E is the foot of the declination circle of M. Similarly if MV represents Vṛiṣabha, when V comes to the horizon, N the foot of the declination circle comes to the horizon. Thus the rising time of any arc of the ecliptic at an equatorial place is given by the correspond- ing arc of the equator, which is intercepted between the declination circles of the ends of the arc. So from fig. 26 if r ed ≃ be the equator, rED. ≃ the ecliptio, A the Ayana or Summer solotice, P the pole rE, ED, DA etc. the Sāyana Rasis Meṣa etc. a, b, c etc. the feet of the

196 Fig. 25 Fig. 26 declination circles of A, B, C etc., since PAE is secondary both to the ecliptic and equator (i.e. perpendicular circle) spherical triangles PAD, PAB are congruent; PBC, PDE are congruent and PC ♎ is congruent with PE ♈. Hence ab = ad i.e. rising times of Karkataka and Mithuna are equal; bc = de i.e. those of Simha and Vriṣabha are

197 equal and similarly those of Meṣa and Kanyā. It will be noted that the equatorial risings alone are equal in the above cases but not at any other place, for in a place with some latitude when a point like B (fig. 26) comes to the horizon, the foot of its declination circle namely b will not be on the horizon and there arises the chara in bet- ween, which has to be taken into account, and be subtract- ed from or added to the equatorial rising time as the case may be. Now regarding the chara-khanda of Karkaṭa it is again 56; why it should be so is not proved by Bhāskara but merely stated, nor any commentator took the pains to prove. It can be proved as follows. Let in fig. 26, δ₁, δ₂, δ₃ be the declinations at the ends of Vṛṣabha, Mithuna and Karkaṭaka respectively. We know δ₁ = δ₃. The chara-segments for Mithuna and Karkaṭaka i.e. when Mithuna and Karkaṭaka are rising are to be proved to be equal, here 56 asus. Their expressions are tan ϕ tan δ₂ − tan ϕ tan δ₁ and tan ϕ tan δ₃ − tan ϕ tan δ₂ i.e. tan ϕ (tan δ₂ − tan δ₁) and tan ϕ (tan δ₃ − tan δ₂). Since δ₁ = δ₃ we perceive that they are equal but of opposite signs. So Bhāskara says rightly “अपचीयमानत्वात् धनम्” i.e. because of negative sign, the chara-segment of Karka- taka while being subtracted will be rendered positive’. Hence the rising times of Karkaṭaka, Simha and Kanya will be respectively 1937+56, 1793+134, 1670+168 asus or 1993, 1927, 1838 asus or 332, 321, 306 Vinadis or 5-32, 5-21 and 5-6 nadis. Thus, in as much as the rising times of Meṣa to Kanyā are 1670−168, 1793−134, 1937−56, 1937+56, 1793+134, 1670+168 their total is 30 nadis as should be expected because the equator bisects the ecliptic between ♈ and ♎ and the equatorial interval between ♈ and ♎ is 30 nadis. It will be noted that while the equatorial rising times of Meṣa to Kanyā are symmetrical as 1670, 1793, 1937, 1937, 1793, 1670, their rising times at any other place are not like that but

198 constitute a different kind of symmetry as 1670—168, 1793—134, 1937—55, 1937＋55, 1793＋134, 1670＋168. We have now to comment upon the statement “तुलादितोऽमी च विलोमसंस्थाः ” which means that the rising times from Tulā to Mīna are in the reverse order i.e. the rising time of Tulā equals that of Kanyā; that of Vris- chika equals that of Simha and so on the rising time of Mīna equalling that of Meṣa. Thus the rising times of Tulā to Mīna being in the reverse order are 1670＋168, 1793＋134, 1937＋55, 1937—55, 1793—134, 1670—168 for the aforesaid locality. Why it should be so can be easily seen from the fact that Kanyā and Tulā are sym- metric with respect to the line ♈ ♎ which bisects the ecliptic (Ref. fig. 27). Or again we can see this in another Fig. 27 way; the chara-segments are successively (tan ϕ tan δ₁ — tan ϕ tan 0), (tan ϕ tan δ₂—tan ϕ tan δ₁), (tan ϕ tan ω— tan ϕ tan δ₂), (tan ϕ tan δ₂—tan ϕ tan ω), tan ϕ tan δ₁— tan ϕ tan δ₂ (tan ϕ tan 0—tan ϕ tan δ₁), (tan ϕ tan δ₁— tan ϕ tan 0), (tan ϕ tan δ₂—tan ϕ tan δ₁), (tan ϕ tan ω— tan ϕ tan δ₂), (tan ϕ tan δ₂—tan ϕ tan ω) (tan ϕ tan δ₁— tan ϕ tan δ₂), (tan ϕ tan 0—tan ϕ tan δ₁) where δ₁ = declination of 30°, and δ₂ that of 60°. These are there as found before 56, 134, 168, —168, —134, —56 upto Kanyā. But the remaining, though apparently are 56, 134, 168, —168, —134, —56 must be taken with a reverse sign because the rising point of the ecliptic will be to the south of the east point and δ will be negative from 180° to 360° longi- tude. Hence the chara Segments are 56, 134, 168, —168, —134, —56, —56, —134, --168, 168, 134, 56 so that from Tulā onwards they are in the reverse order as Bhāskara