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

297

298 | Cara Cāpa | Arc of the celestial Equator bet. the East point and foot of the declination circle | | | | | Carajyā | H sine of the above or the corresponding line of Kujya in the plane of the celestial Equator | 13 | 13-17 | (R · H sin φ · H sin δ) / (H cos φ · H cos δ) | R tan φ tan δ | | Natakāla | Hour angle h | | | | | | Unnata | Time elapsed after rise | | | | | | Śara | Corresponding line of phala in the Equatorial plane | 29 | 58 | H vers (h) | R (1 — cos h) | | Phala | ⊥ᵃʳ from the culminating point on a line through the celestial body parallel to Udayāstastūtra | 30 | | (H vers h · H cos δ) / R | R cos δ (1 — cos h) | | Ūrdhwa | Orthogonal projection of phala on the plane of the meridian | 31 | 59 | (H vers h · H cos φ · H cos δ) / R² | R cos φ cos δ (1-cosh) |

299 Yet (R sin δ) / (sin φ) will have a value greater than R ie. even though the Sama-S'anku is never born so to say, it has a magnitude. ‘तत्कथमिदं वन्ध्यासुतवत्’ Bhāskara exclaims with respect to the magnitudes of Sama-S'anku and Taddhṛti as well, both of which are not there, yet, both of which have magnitudes greater than R. So he says “those magnitudes of the Sama-S'anku and Taddhṛti are like the sons of a barren lady”. Then he says ‘तदपि प्रदर्श्यते’ ie. ‘We shall show how they arise.’ Here he uses his intuition of the principle of geo- metrical continuity. Even when the diurnal circle does not cut the prime-vertical, their planes intersect, out- side the sphere and the perpendicular dropped from the point of intersection on the plane of the horizon is the Sama S'anku which has a magnitude greater than R. Similarly the Taddhṛti could be seen what it is now. These magnitudes can enter into a proportion like the I in verse 22, and do help us to get the other real magnitudes like the Unmandala S'anku etc. Verses 66, 67 and first half of 68. To obtain the time from the shadow. Iṣṭāntyakā = (U.K. × Carajyā) / I.K. = (D.K. × Antyā) / I.K. = (k × R²) / (R cos δ × I.K.) ; Rvers⁻¹ (Antyā − I. A.) = h Dinārdha − h = Unnatakāla where K = Karṇa, k = Vishuvat Karṇa, I.A. = Iṣṭāntya. Comm. We had under verse 62. Iṣṭa-Karṇa × Iṣṭa-Hṛti = D. K. × Hṛti = S. K. × Taddhṛti = U. K. × Kujyā multiplying throughout by R / (R cos δ)

300 Iṣṭa-Karṇa × I. A. = D. K. × Antyā = U. K. × Carajyā so that I.A. = (D.K. × Antyā) / I.K. – (U.K. × Carajyā) / I.K. I But U.K. = 12 R / U.S. (verse 40) ∴ U.K. × Carajyā = (12R × Carajyā) / U.S. = (12R × Kujyā × R) / (U.S. × H cos δ) since Carajyā = (Kujyā × R) / (H cos δ) But Kujyā and U.S. are the Karṇa and Koṭi of the seventh latitudinal triangle so that Kujyā / U.S. = k / 12 Comparing with the first fundamental latitudinal triangle. ∴ U.K. × Carajyā = (12R² × k) / (H cos δ 12) = kR² / (H cos δ) Hence substituting in I I.A. = (U.K. × Carajyā) / I.K. = kR² / (H cos δ × I.K.) Thus we have proved the first part of the statement. Having obtained I.A., from fig. 52 we have Antyā – I.A. =(FQ–AN)=CQ. The Utkrama Cāpa of CQ = NQ = h and Dinārdha – h = Unnatakāla. Let us see what this procedure means in practice. Since on any day at any place, φ and the declination of the Sun are known we can compute all the magnitudes given in the verse or more easily H cos δ so that from the formula Iṣṭāntyā = 12R² / (H cos δ × I.K.) where K = √(S² + 12), the shadow being observed Iṣṭāntya could be computed in no time. Also the Antyā of the day R+(H tan φ tan δ) can be computed so that the segment CQ can be got. The inverse Hversine of this is h. The arc CQ above was symbolized as Sara.

301 Bhāskara’s proof of I.A. = kR² / (H cos δ × I.K.) proceeds from first principles as follows :—(i) If by k we have 12 as the Koṭi what have we for R? The result is H cos z, Mahāśaṅku. ∴ Mahā Śaṅku = 12 R / I.K. From Mahā Śaṅku we pass on to Iṣṭa Hṛti with which it forms a latitudinal triangle. If by the gnomon of 12 units we have k the Viṣuvat Karṇa, what have we by Mahā Śaṅku? The result is (12 R / I.K.) × (k / 12) = kR / I.K. Again from the Iṣṭa-Hṛti we pass on to I.A. by multiplying by R / (H cos δ) so that I.A. = kR² / (H cos δ × I.K.) as given. Second half of verse 68. The inverse Hversine if a quantity x greater than R, is 5400 + H sin ⁻¹ (θ) when x - R = θ. Comm. Since Hvers (90 + θ) = R + H sin θ = x (say) 90 + θ = Hvers ⁻¹ (R + H sin θ) = But 90° are equal to 5400 asus and θ = H sin ⁻¹ (H sin θ) = H sin ⁻¹ (x - R) = त्रिज्याधिकभागस्यक्रमचापम् ∴ 5400 + त्रिज्याधिकभागक्रमचापम् = Utkrama Cāpa of a त्रिज्यादिक quantity. Verse 69. Alternate method of obtaining the time that has elapsed after Sunrise noting the shadow S. Subtract Charajyā from or add it to Iṣṭāntyā according as δ is north or south. Obtain inverse H sine of the remainder and add the Caracāpa to this inverse H sine. Then we have the unnatakāla by converting the result into time.

302 Comm. Ref. fig. 52. I.A. = AN. I.A. - Carajyā = B.N. Hvers⁻¹ (BN) = arc EN. Arc EN + Caracāpa = EN + EM = MN. This converted into time is evidently the Unnatakāla because the arc MN of the equator is the arc intercepted between the feet of the declination circles at rising and at the time concerned M being the foot of the rising declination circle and N the foot of that at the time in question. The convention of signs is clear. Verse 70 and first half of 71. To obtain the Sun's longitude from the shadow S. The gnomonic shadow at noon, being multiplied by R and divided by K, the inverse H sine of the result gives the meridian zenith-distance. This being decreased or increased by the latitude gives the Sun's declination according as the extremity of the shadow is north or south. From the declination, we have the Sun's longitude by the formula H sin δ = (H sin λ H sin ω) / R . Comm. We have from the triangle formed by the gnomon and the shadow S, S / K = sin z or SR / K = H sin z ∴ H sin⁻¹ (SR / K) = z. Since we are directed to take the mid-day shadow, we have the meridian zenith-distance and from the formula z + δ = φ we have δ. If the extremity of the shadow be north, the Sun is south of the zenith, and then φ ~ z = δ. The word वियुक्ताः is used to signify difference which is positive. If z > φ then the declination is south and vice-versa. If the extremity of the shadow is on the south, the Sun is on the north of the zenith. in which case φ + z = δ. Second half of verse 71. To obtain φ from δ.

308 If the zenith-distance and the declination are of the same direction, their difference, otherwise their sum will be the latitude. [Fig. 58] Comm. Suppose the zenith- distance is north and decli- nation also north, then clearly from fig. 58, in this position S₁ of the Sun QS₁ — ZS₁ = ϕ = δ — Z (1) Again in the position S₃ of the Sun, zenith-distance is south and declination is also south ; so, here also difference gives ϕ ie. ZS₃ — QS₃ = Z — δ = ϕ. (3) In the position S₂, how- ever, Z is south and δ is north, so that their sum is equal to ϕ. It will be noted that the Hindu convention of signs does not contemplate negative declination and also it uses the word 'difference' to signify the positive difference, as for example, in the first two cases δ ~ Z is taken as ϕ. The modern formula Z + δ = ϕ applies universally with the convention that δ is +ve if north, ϕ is +ve if south. Verses 72, 73. To obtain the Bhuja from the shadow. Karṇa Vrittāgrā = (A × K) / R where A = Agrā and K the Chāyākarṇa. This Karnāgrā is to be taken as belonging to the opposite hemisphere to the Sun. Calling this Karnāgra as a and the equinoctial shadow as s, a ⨦ s according as δ is south or north gives the Chāyābhuja b. Thus a ⨦ s = b. If the extremity of the shadow be on the north and δ be north, then b + a = s ; if δ be south b ~ a = s. If b be north, b ~ s = A, otherwise ie. if south b + s = a. (R × a) / K = Agrā and (Agrā × Koṭi of latitudinal triangle) / (Karṇa of the latitudinal triangle) = H sin δ.

304 Comm. These verses are very important and the contents have been already elucidated under verses 13–17. We shall elucidate the convention of signs in more detail both from the modern point as well as from the Hindu traditional point. First we shall discuss the modern. We have the formula A = S + B where A = Agrā, S = Sankutala and B, Sanku-bhuja. Let us confine ourselves to north latitudes alone, for, south latitudes did not concern the Hindu astronomers. Then treating north declination as positive and also north Hindu azimuth as positive the above formula holds universally. (Ref. fig. 59) Let the figure represent the meridian plane. Let S₁, S₂, S₃ be the projections of the Sun's positions in their diurnal circles on to the meridian plane. Let A, B. be the Fig, 59

305 projections of the rising points of the Sun on the same plane. Let perpendiculars be dropped from S₁, S₂, S₃ on the plane of the horizon. Let o be the centre of the sphere. Let ns be the north-south line. In position S₁, S₁C = Śanku-Bhujā, KA = Śankutala, oA = Agrā, so that A = S + B (1) In the position S₂. S₂D = Śanku-Bhuja, LA = Śanku-tala, oA = Agrā, so that B + A = S; but here B is negative, a the azimuth being south so that writ- ing - B for B, - B + A = S ∴ A = S + B again. In position S₃, OB = A, MB = S, OM = B so that A + S = B; but here A is +ve, δ being south and B is negative a being south; hence writing - A and - B for A and B A + S = - B or A = S + B again. This shows that with the convention cited above A = S + B holds good universally. Now let us consider the situation with respect to the Karnāgrā. Each of the three quantities a, b, s have now opposite directions. If δ be north, the Sun will be on the north of Equator, but the extremity of the shadow will now be on the south of the Equinoctial shadow line (E.S.L.) ie. the line which is parallel to the East-west line at a distance of the equinoctial shadow s, and which is tbe locus of the extremity of the shadow on the equinoctial day; thus when the Agrā is considered to be positive being on the north of the East point, the Karnāgrā, though it is on fhe south will have to be considered positive. Similarly when the azimuth of the Sun a is on the north of the prime-vertical and is so considered to be positive, the extremity of the shadow being on the south of the East-west line and has a negative azimuth, the bhuja is still to be considered positive. Again the Śankutala being always south of the Udayāsta Sūtra being considered south and positive, the corresponding quantity into which it gets converted on the horizontal dial namely the equinoctial shadow s will be north of the East-west line and will be considered 39

306 positive. In other words in the equation a = b + s, a is positive when δ is north, b is +ve when the azimuth of the Sun is north of the East point, and s is always positive. Since in north latitudes, Śaṅkutala will be always south of the Udayāstasūtra and considered positive, the E.S.L. will be on the north of the East-west line so that s is considered positive. We should have had to consider s negative in southern latitudes, as per the above convention but as the Hindu astronomers did not have to concern themselves with south latitudes, the question of sign for s did not arise except taking it as always posi- tive. Hence the equation a = b + s will hold universally with the same conventions of sign which we stipulated with respect to the equation A = B + S. The foregoing analysis is on the modern lines. [चित्र: वृत्त के अंतर्गत W-E व्यास, E·S·L समानांतर रेखा, तथा बिन्दु O, C, P, R, M, L, A, N, B, K युक्त ज्यामितीय आरेख] Fig. 60

307 Now let us see how the convention of signs is stipulated in Hindu astronomy with respect to the equation a = b + s. We have said that ‘s’ is always north of the East-west line and considered positive. Regarding ‘a’, it is said by Bhāskara व्यस्तगोला which means that when δ is north and the Sun is said to be in northern hemisphere, a is said to belong to the southern hemisphere. Also when the Sun is on the north of the prime-vertical and Śankubhuja is considered north, the Chāyābhuja being south of the East-west line is considered south. With these conventions of directions (we say of directions, and not signs because the Hindu astronomers do not speak of signs but only of directions) it is stipulated in Hindu astronomy that quantities of like directions are to be added, otherwise their difference is to be taken ‘तुल्यदिशोर्योगः, भिन्नदिशोः अन्तरम्’. This convention stipul- ating addition or difference is technically called ‘संस्कार, Samskāra’. That is why it is said simply “पलच्छायया सौम्यया संस्कृता” ie. ‘Samskāra (on the aforesaid lines) is to be effected between a and s to get bhuja b’. Here it may be reiterated that the word ‘अन्तरम्’ ie. ‘difference’ is used in its restricted sense namely that the positive difference alone is to be taken. Thus the ‘antaram’ of 8 and 5 is 3 as well as that of 5 and 8 is also 3. Then it might be asked how to decide the direction of the bhuja, if we were to take s as equal to a ~ b and not a-b or b-a. The answer is that between a and s whose directions are known as per the aforesaid convention, equating their difference ie. a ~ s to b, we have to take b as having that direction which is indicated by that quantity either a or s which has a larger numerical value. Thus while on modern lines we take a = b + s to hold universally with the convention of signs which we have agreed to on modern lines namely that a is +ve if δ is north and b is positive if the Hindu azimuth is north of the East point and s is always +ve, we take on the Hindu lines a ± s = b with the conventions stipulated with respect

308 to directions (not of signs) namely that a is south if δ is north, s is always north and b is to be taken to belong to that direction to which the numerically bigger quantity of a and s belongs in the case of difference. Also it is to be taken to belong to that direction of a and s when both of them have the same direction. With this convention in mind, Bhāskara clarifies the convention by citing examples. (1) S=5, δ is north, Agrā=916'-48'' K=30 so that a = KA / R = (916⅘ × 30) / 3438 = 8 units (south, because it should be taken to be व्यस्तगोला ie. belonging to the direction opposite to that of δ). Question. "What is b and of what direction?" Answer. b = a +~ s = 8 (south) ~ 5 (north). (We are taking the difference because Samskāra is to be construed as addition of quantities having the same direction and difference of quantities of opposite direction ∴ b = 3 and is on the south because the numerically bigger quantity of 8 and 5 belongs to south. Q. 2. δ is north S=5, A=916'-48'' ; K=15, 'what is b and in what direction?' Answer. a = KA / R = 4; b = a +~ s = 4 +~ 5 = 4~5 (here a is south δ being north and s is north so that difference is stipulated as above) = 1 (north because 5 belongs to north. We add here two more examples to illustrate addition by saying that δ is south in the above examples so that a is north in both the examples. Hence in (1) b=8+5=13 (north) and in (2) b=4+5=9 (north). Refering to Fig. 60, we see there three cases depicted namely the extremi-

309 ties of the shadows being A, B and C. In the first case AL = Karnāgra = a (Karnāgra is the distance of the extremity of the shadow from the E.S.L. namely K.L.R. whereas bhuja is the distance of the same from the East- west line namely PMN) AM = bhuja and ML = s so that b=a+s addition being justified since both a and s are of the same direction namely north. In the second case BK = Karnāgra, BN = bhuja and NK = s so that b=s-a, the difference being justified because a is south and s north. Here we have taken the difference as s-a and not a-s because s is numerically greater and being oriented north, the bhuja is north. In the third case, PR=s, CP=b and CR=a so that b=a-s, the difference being justified be- cause s is north and a south. Also we have taken the difference as a-s and not s-a because a is numerically greater and as such lends its direction namely 'south' to the bhuja. Thus in the three examples cited, addition is prescribed between a and s only when the extremity of the shadow is to the north of E.S.L. In the case of A=S+B or B=A-S also, addition is prescribed only when δ is south, which means that the corresponding a ie. AL is north. In fact the prescription of addition or differ- ence accord in the cases of both the equations either A=B+S or a=b+s. Verses 74 and 75. Hereafter questions are being set and answered on diurnal problems. Seeing the shadow of the gnomon, the azimuth and longitude of the Sun or seeing two shadows with their respective directions, whoever knows the equinoctial shadow of the place, I consider him as the Garuda or Eagle who could overcome the false pride of puffed up snakes of astronomers. Given that when K=30 units, the bhuja is 3 units south, and when K=15, the bhuja is 1 unit north, com- pute the latitude, or again given H sin δ = 846 and given K and b of a shadow, compute the equinoctial shadows.

310 Comm. The questions are clear the second verse illustrating the first. Verse 76. Answer of the first question. (b₁ K₂ ⁺~ b₂ K₁) / (K₂ ~ K₁) = s according as the bhujas are of the same or opposite directions. Comm. Suppose b₁ and b₂ are of the same direction so that b₁ = a₁ - s and b₂ = a₂ - s, taking the modern convention of signs. But a₁ = (K₁A) / R and a₂ = (K₂A) / R ∴ b₁ = (K₁A) / R - s and b₂ = (K₂A) / R - s ∴ (b₁ + s) / K₁ = A / R = (b₂ + s) / K₂ ∴ K₂ b₁ - K₁ b₂ = s (K₁ - K₂) ∴ s = (K₂ b₁ - K₂ b₂) / (K₁ - K₂) If, however b₁ and b₂ are of opposite directions ie. of opposite signs, writing - b₂ for b₂, we have s = (K₂ b₁ + K₁ b₂) / (K₁ - K₂) . Here we have chosen to follow the modern convention of signs; otherwise we have to con- sider four alternatives, for, bhujas of the same direction might mean both of the type OC (fig. 60) or both of the type of OB or one of the tyye OB and one of the type of OA or both of the type OA. Verses 77 and 78. Answer to the second question. Let Laghu ≡ L = ((KH sin δ) / R)² ; 12² (L - b²) ≡ Ādya ; Para = 12² b. Let Ādya and Para be divided by L ~ 12² ; call them still Ādya and Para; then √(Para² + A) ± Para = s according as b is north or south.

311 Comm. The data are H sin δ, K and b; since K and b are given a is known. Thus from the triangle PZS (fig. 61) we have sin δ = sin ϕ cos z + cos ϕ sin z sin a, all quantities except ϕ are known. Solving this trigonometrical equation which is of the form a cos ϕ

• b sin ϕ = c, we can have ϕ. Fig. 61 We shall now see how it is solved by Bhāskara. Let s be the equinoctial shadow. Then a = b + s = KA / R = (KH sin δ) / (H cos ϕ) I But 12 / K = 12 / √(s² + 12²) = (H cos ϕ) / R so that H cos ϕ = 12 R / √(s² + 12²). Substituting in I b + s = (KH sin δ √(12² + s²)) / 12 R which reduces to 12² R² (b + s)² = K² H sin² δ (12² + s²) ie. s² (12² R² − K² H sin² δ) + 2b 12² R² s = 12² {(H sin² δ) K² − b² R²} ie. s² (12² − (K² H sin² δ) / R²) + 2. 12² b. s) = 12² ((K² H sin² δ) / R² − b²) Here (K² H sin² δ) / R² is symbolized as L 12² ((K² H sin² δ) / R² − b²) put as Ādya and 12² b is put as para.

312 So the equation reduces to s² (12² - L) + 2. Para. s = Ādya ; Divide throughout by 12² - L and put again Para / (12² - L) as Para and Ādya / (12² - L) = Ādya Then the equation reduces to s² + 2 Para s = Ādya ; completing the square (s + Para)² = Para² + Ādya ∵ s + Para = √(Para² + Ādya) ∴ s = √(Para² + Ādya) - Para as one solution. We have taken to start with a = b + s which holds good according to the Hindu convention when b is south ; if, however b is north a = b ~ s so that (b ~ s)² = s² + b² - 2bs. So in the equation we have to write -s for s, so that we have now s² - 2 Para s = Ādya ie. (s - Para)² = Para² + Ādya ∵ s = √(Para² + Ādya) + Para as the second solution. Verse 79. When the Sun's longitude is 135°, the shadow of the gnomon is 12 units and west. What is the latitude? Comm. Here is a method of obtaining the latitude of the place by observing the gnomon's shadow when the Sun is on the prime-vertical. Verse 80. Answer to the question above. (12 R) / K = H cos z ; s = (12 H sin δ) / √(Sama-Sanku² - H sin² δ) Comm. Solution in modern terms. S = 12 ∴ tan z = 1 ∴ z = 45; but when the Sun is on the prime-vertical, we have by Napier's rule Sin δ = sin ϕ cos z = sin ϕ √2. But since λ = 135°

often handles vertical fractions in plain text either as: [L..] triangle (H sin δ) / x = s / 12 where x is the Koṭi OR line-by-line exact layout. Wait! Let's examine how the lines flow: If we look at: triangle H sin δ / x = s / 12 where x is the Koṭi Let's check the rest of the page: - ∴ s = (12 H sin δ) / x. But we are given that the Sama- - Śaṅku ie. - H cos z = R cos z = R / √2 because z = 45° when the - shadow equals the length of the gnomon. - and H sin δ = (H sin λ H sin ω) / R = (H sin 135 H sin ω) / R - = (H sin 45 H sin ω) / R = R/√2 × (H sin ω) / R = (H sin ω) / √2 - ∴ x = √(Sama-Śaṅku² - H sin² δ) = √( R²/2 - (H sin² ω)/2 ) - `= (H cos ω) / √2 ∴ s = (12 H sin δ × √2) / (H cos

314 that moments I would reckon you as one who could be well compared with the goad that could be applied to the head of the wild elephants of puffed up astronomers. Verses 82, 83. Answer to the first question. Assume the H sine of the Unnatakāla to be Iṣṭa Hṛti in the first place. Multiply it by 12 s and divide by k² the square of the Viṣuvatkarṇa. Then you get an approximate value of H sin δ. Then compute with this, H sin δ, Charajyā etc. and thereby obtain a more correct value of the Iṣṭa Hṛti. Multiply this by the H sin δ got before and divide by the first Hṛti assumed. Then we have a nearer approximation of H sin δ. Repeat the process till a stationary value has been reached. That will be the correct H sin δ. Comm. We know that H sine of the Unnatakāla is nearly the Iṣṭāntyakā. It will be noted that Iṣṭāntyakā is the sum of two H sines namely (1) Carajyā (2) H sine of Unnatakāla minus Chara. The second H sine is called Sūtra or H cos h. (Vide page 278). Thus Sūtra + Carajya = Iṣṭāntyaka whereas H sine (Unnatakāla) = H sine of the Cāpas of Carajyā and Sūtra. In other words H sin (Unnatakāla) = H sin (H sin⁻¹ Carajyā + H sin⁻¹ (Sūtra)). Iṣṭa Hṛti is H cos δ Iṣṭāntyaka × ----------- . But we do not know H sin δ so R that Iṣṭa Hṛti could not be got. So we will not be far from truth in assuming the given Unnatakāla to be Iṣṭa Hṛti itself ie. Taddhṛti here, as the Sun is on the prime- vertical. The formula for Taddhṛti is R² H sin δ R². H sin δ K² H sin δ ----------------- = --------------------- = ------------ H cos ϕ H sin ϕ R. 12 s 12 s ----- × R × --- k k

315 ∴ (Taddhṛti × 12 s) / k² = sin δ. Thus assuming H sine of the given Unnatakāla to be Taddhṛti and multiplying it. by 12 s and dividing by k² we have the value of H sin δ. But this is approximate because the given Unnatakāla is not exactly Taddhṛti but only an approximate value. From this H sin δ, compute H cos δ, Charajyā, Kujyā and through the process indicated in verse 54 namely “Subtract the Characāpa from the Unnatakāla. The H sine of the result is called Sūtra. Multiply the Sūtra by H cos δ and divide by R; then we have Kalā. Add Kujyā to Kalā; we get Iṣṭa-Hṛti”, we obtain a more correct value of Taddhṛti. Then here we may cut short the process as follows namely ‘ If by the assumed Taddhṛti we had the previous H sin δ, what shall we have for this more approximate Taddhṛti ’? The result will be a more approximate value of H sin δ. Again form the Taddhṛti with this H sin δ and so repeating the process till we have a stationary value, we have the correct value of H sin δ. Note. This is a beautiful example of the method of successive approximations which is a modern technique but which was so much in vogue and favourite with the Hindu astronomers. (It will be noted how to cut short the method). Verses 84, 85. Answer to the second question. Obtain 12² R²/(R² − H sin² h) s² + 1 and divide R² by this and take the square root which gives H sin δ. Then (R × H sin δ) / (H sin ω) gives H sin λ whose Cāpa gives the longitude of the Sun. Comm. The H sine of the given Natakāla is H sin h and R² − H sin² h = H cos² h. Let H sin δ be x, which is required to be found. Then R² − x² = H cos² δ; H cos h = Sūtra and

316 (Sūtra × H cos δ) / R = Kalā = (H cos h . H cos δ) / R = ∴ (H cos h . √(R² — x²)) / R But Kalā is the Koti of the fifth latitudinal triangle of which H sin δ is Bhuja. Hence (Kalā × H sin ϕ) / (H cos ϕ) = H sin δ = x ie. (H cos h √(R² — x²)) / R × (H sin ϕ) / (H cos ϕ) = x; but (H sin ϕ) / (H cos ϕ) = s / 12 ∴ Squaring both sides [(R² — H sin² h) (R² — x²)] / R² × s² / 12² = x² ∴ 12² R² x² = s² R² (R² — H sin² h) — s² x² (R² — H sin² h) ie. x² {(12² R² + s² (R² — H sin² h)} = s² R² (R² — H sin² h) ∴ x² = [s² R² (R² — H sin² h)] / [12² R² + s² (R² — H sin² h)] = R² / [(12² R²) / (s² (R² — H sin² h)) + 1] ∴ x = R / √[(12² R²) / (s² (R² — H sin² h)) + 1] = H sin δ as given. From H sin δ, the method of obtaining λ is clear from the formula (H sin λ H sin ω) / R = H sin δ. In the given numerical example h = 5 nādīs = 360° / 12 = 30° since 60 nadis of time correspond to 360°. Thus H sin h = R / 2 ; the remaining work follows. Verse 86. Another question. When the Sun is on the prime-vertical the gnomonic shadow is noted to be 16 inches. The Unnatakāla is 8 nādis. If you could give the H sin δ and s, I shall con- sider you nothing short of one who is an adept in solving the totality of the diurnal problems. Verses 87 and 88. Answer to the question.