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

258 Comm. This formula is a special case of the formula A = S + B since the H sine of the meridian zenith distance is the Śaṅku-Bhuja at noon. From fig. 39 (Hṛti × bhuja of a latitudinal triangle) / (Karṇa of a latitudinal triangle) = O₁a = Dinārdha Śaṅkutala. The operation of sign has been already explained. Verse 38. An alternative method. The meridian zenith distance of the Sun can be had also by the formula (Hṛti ± Taddhṛti) B.L.T. / K.L.T. where B.L.T. and K.L.T. are the bhuja and karṇa of any latitudinal triangle. Comm. (Ref. figures 43 and 21). Let E₁ be the centre of the armillary sphere so that QE₁R is the diameter of the celestial equator which is on the median plane. Let S₁ F₁ S be the diameter of the diurnal circle of the Sun, which is also on the meridian plane so that S₁ F₁ is the Taddhṛti, S₁ S is the Hṛti and S the position of the Sun on the meridian. H sin z = SM = SF₁ sin F̂₁ = SF₁ sin ϕ = (Hṛti − Taddhṛti) × s/K where s/K can be replaced by B.L.T. / K.L.T. (s=equinoctical shadow and k the Viṣuvat-Karṇa). In the Southern sphere, H sin z = S′N = S′F₁′ sin ϕ = (S′S₂ + S₂F₁′) sin ϕ = (Hṛti + Taddhṛti) × s/k. Verse 39. Still another way of obtaining the m. z. d. (meridian-zenith-distance). R − H versin (altitude) = H sin z

259 Fig. 43 This formula gives not only the meridian zenith distance but H sine of the zenith-distances of the Sun when he is on the Koṇa-Vṛtta or prime-vertical or unmaṇḍala. Comm. From fig. 43, H sin z = SM = LE₁ = E₁s − sL = R − H versine (Ss) = R − H versine (altitude) as given. Since R − H versine (altitude) = R − {R − H cos (90 − z)} = R − (R − R sin z) = H sin z, so this formula applies wherever the Sun be. This is almost begging the question as H sine of z is being sought through H versine of (90 − z).

260 First half of the verse 40. To obtain the shadow S and K the Chāyakarṇa of any shadow (H sin z × 12) / (H cos z) = S and (R × 12) / (H cos z) = K. Comm. (Ref. fig. 44). (12. H sin z) / (H cos z) = 12 tan z = S. Also 12 / K = cos z = (H cos z) / R so that (12 R) / (H cos z) = K. The Hindu method of looking at this through the similarity of ΔS OM☉ and Ogn the gnomonic triangle. is as follows. ☉M is called Mahā-Śaṅku ie. H cos z ; ☉L is Dṛkjya or H sin z = OM. 12 / (H cos z) = S / (H sin z) so that S = 12 H sin z/H cos z. Also, On / O☉ = 12 / (H cos z) ie. K / R = 12 / (H cos z) ∴ K = (12 R) / (H cos z) . It will be noted that fig. 44 pertains to any vertical plane. Second half of Verse 40. The Dinārdha-Karṇa is equal to (R × k) / Hṛti where k is the Viṣuvat-Karṇa. Comm. The formula is derived through twice apply- ing the rule of three or what is the same, through the similarities of two sets of triangles From fig. 39, O₁A / Aa = Hṛti / Dinārdha-Śaṅku = k / 12 (a) and from fig. 44 On / O☉ = 12 / Dinārdha-Śaṅku = K / R (b) where K is the required Chāyākarṇa. Dividing (a) by (b) Hṛti / 12 = k / 12 × R / K ∴ K = kR / Hṛti

261 Fig. 44 Verse 41. Alternate method of obtaining K 101530/H sin λ = para (say) where λ is the Sāyana longitude of the Sun ; then, (Para × k) / s = K where K is un-mandala-Karṇa First half of Verse 42. To obtain K when the Sun is on the prime-vertical— Para × s/k = Samavṛttakarṇa. Comm. From fig. 19, from the similarity of triangles BD ☉ and CMA, B ☉ / CA = ☉ D / AM ie. ☉ D = (B ☉ × AM) / CA ie. H sin δ = (H sin λ H sin ω) / R (a) Then consider the similarity of the first and the sixth latitudinal triangles ; then Unmandala Sanku / Krāntijyā = s / k (b) where s is the equinoctial shadow and k the Viṣuvat- Karṇa. Again taking that ☉ the Sun lies on the unman-

262 dala in figure 44, (H cos z) / 12 = R / K = (Unmandala Śaṅku) / 12 (c) Eliminating Krāntijyā and Unmandala Śaṅku from (a), (b) and (c) (Unmandala Śaṅku) / (H sin λ H nis ω/R) = s / k ∴ (12R / K) / (H sin λ H sin ω / R) = s / k ie. 12R² / (KH sin λ H sin ω) = s / k ∴ K = (12R² × k) / (s H sin λ H sin ω). Here 12R² / (H sin ω) = (12 × 3438²) / 1397 = 101531 ; but Bhāskara has taken 101530 taking a more correct value of R. Then 101530 / (H sin λ) is symbolized as para so that para × k / s = K = Unmandala Karṇa. Regard- ing the Samavṛttakarṇa, in the place of (b) above we have (Sama-Śaṅku) / Krāntijyā = k / s (b') by the similarity between the first and the fifth latitudinal triangles. Equation (c) holds good with respect to any H cos z and the corres- ponding K since 12 R = K × Śaṅku and 12 R is a cons- tant. Noting therefore R / K' = (Sama-Śaṅku) / 12 (c') eliminating Krāntijyā and Sama-Śaṅku among (a), (b'), (c'), we shall have K = (12 R² s) / (k H sin λ H sin ω) = para × s / k as stated. Second half of Verse 42. To obtain the Dinārdha- karṇa from the Unmandalakarṇa. (Un-mandalakarṇa × Charajyā) / Antyā = Dinārdhakarṇa. Comm. We have equation (c) above stating 12 R = K × Śaṅku. (c) But

263 Iṣṭa Śaṅku / Iṣṭa Hṛti = cos φ = constant = Dinārdha Śaṅku / Hṛti = Sama-Śaṅku / Taddhṛti = Unmaṇḍala Śaṅku / Kujyā (d) (23) Again by virtue of the proportionality of Iṣṭa Hṛti / Iṣṭāntyā = Hṛti / Antyā = Kujyā / Charajyā (e) (24) We have Ishta-Śaṅku / Ishtāntyā = Dinārdha Śaṅku / Antyā = Unmaṇḍala Śaṅku / Charajyā ∴ Iṣṭa Karṇa × Iṣṭāntyā = Dinārdha Karṇa × Antyā = Unmaṇḍala Karṇa × Charajyā (f) (25) ∴ Dinārdha Karṇa = (Unmaṇḍala Karṇa × Charajyā) / Antyā as stated in the verse. Verse 43. (Unmaṇḍala Karṇa × Kṣitijyā) / Hṛti = (Sama Vṛtta Karṇa × Taddhṛti) / Hṛti = Dinārdha Karṇa. Comm. From (c) and (d) above Dinārdha Karṇa × Hṛti = Sama Karṇa × Taddhṛti = Unmaṇḍala Karṇa × Kujyā (g). Khitijyā is the same as Kujyā. From this the statement follows : Verse 44. The ancient Achāryas found the gnomonic shadows when the Sun is on the meridian, prime-vertical and the Kona-Vṛtta (ie, Vertical when the northern or southern Hindu azimuths are 45°) by different methods. I consider him to be the very Sun illuminating the lotus- faces of aitronomers, if anybody could give a method to find the shadow in any required direction, which holds good in all cases universally.

264 Comm. Evident. Verse 45. Definition of Dikjyā H sin (azimuth). The angle between any vertical and the Prime-Vertical measured on the horizon is what is called Digamsa and its H́ sine is known as Dik-jyā either in the Eastern hemi- sphere or the Western. Comm. In modern astronomy azimuth is measured along the horizon from the north point towards the east point round the horizon. In Hindu Astronomy however, the azimuth is measured from the East point on either side and from the West point also on either side specifying whether it is north or south. Verse 46 and first half of 47. To obtain the gnomonic shadow in any arbitrary direction. Assume Rs / (H sin a) as the equinoctial shadow and obtain the H sine of the corresponding latitude L. Then the product of that H sin L and H sin δ divided by H sin ϕ will give H sin D where D is a hypothetical decli- nation. With the new L and this D, as the hypothetical latitude and declination, obtain the meridian zenith distance by the formula Z + D = ϕ, and through this m.z d. obtain the shadow, which will be the shadow in the required direction namely 12 tan (ϕ ± D). Comm. Let gL be the gnomonic shadow on the equinoctial day in a given direction given by a° Digamsa (the Hindu azimuth) and let gN be the shadow in the same direction on any day. (fig. 45) We know that the extremity of the gnomonic shadow on the equinoctial day traces a straight line parallel to the East-West line Eω at a distance of the equinoctial shadow s because the Equatorial plane passing through the foot of the gnomon

265 Fig. 45 and that passing through the top of the gnomon being parallel planes cut the horizontal plane in parallel straight lines. (This will be also proved analytically subsequently). Hence LM = s. Now from the figure LM / gL = AB / gA = H sin a / R ∴ gL = H sin a / Rs I This gL is spoken of as Iṣṭa-Drikmandala palabhā because it is the shadow on the equinoctial day in any vertical. LN is the increment in the shadow on account of declination and we have to compute this and correlate gL and LN. For this refer to figs. 46 and 47. In fig. 46, QRT is the equator, so that when the Sun is on the equator on the equinoctial day in the direction given by ZS, ZT is the zenith-distance. Let ZS be the zenith- distance of the Sun in the same direction on any day From the analogy of finding H sin δ from H sin λ, from this figure 34

266 H sin SR = (H sin ST × H sin T̂) / R II and H sin φ = (H sin ZT × H sin T̂) / R III so that (H sin SR) / (H sin φ) = (H sin ST) / (H sin ZT) ∴ H sin ST = (H sin SR / H sin φ) × H sin ZT. Noting that SR = δ and putting ST = D H sin D = (H sin δ / H sin ϕ) × H sin ZT. [Diagrams: Fig. 46 and Fig. 47 showing spherical triangles with vertices P, Q, N, R, S, T, D, δ] Fig. 46 Fig. 47 The same formulae are derivable from fig. 47 also; only in fig. 46 while there is a decrement in the shadow of the day as compared with the shadow on the equinoctial day, in fig. 47, there is an increment. This is seen from the decrease and increase of ST in the zenith-distance ZT of the equinoctial day in the given direction. Now corre- lating fig. 45 with figures 46 and 47, the shadow gL pertains to the zenith-distance ZT on the equinoctial day whereas the shadows gN pertains to the zenith-distance on the day concerned in the same direction. We have, S / √(12² + S²) = (H sin z) / R so that RS / √(12² + S²) = H sin z where S is the shadow at any instant when the zenith-distance is z. The process indicated by saying ‘Obtain H sin φ

267 construing Rs / (H sin a) as the equinoctial shadow', means computing H sin ZT and there from ZT from the shadow gL of fig. 45. Then the process indicated by saying “Obtain H sin D = (H sin ZT × H sin δ) / (H sin φ) and therefrom D'' means computing ST. Then clearly ZS = ZT ± ST, ie. the required zenith- distance is got by what is technically called Samskāra between ZT and ST as is stipulated between φ and δ to obtain Z from the formula Z ± δ = φ (The word Samskāra was defined as meaning addition when the directions are the same and difference when they are opposite). Then the gnomonic shadow is got from this zenith-distance using the formula S = (12 H sin z) / (H cos z) . Thus the procedure adopted by Bhāskara was con- ceived by him first having Fig. 45 before him and then using figures 46 and 47. In this particular process, H sin a is given and H sin δ also, which means that it is sought to find the shadow on a given day in a given direction. Incidentally we shall find the locus of the extremity of the gnomonic shadow during the course of a day. Let in fig. 48 g represent the gnomon's foot, and S the shadow whose extremity is p. Required to find the locus of p. Take the gnomon to be of unit length so that the length of the shadow S = 12 tan z becomes tan z here. Take Eω and sn the east-west line and the north-south as the axes. Then we have x² + y² = tan² z (1). But we have from the triangle PZS sin δ = sin φ cos z + cos φ sin z sin a ie. sin δ / (cos φ cos z) = tan φ + tan z sin a = tan φ + y (2) ie. sec z / A = y + tan φ when A = cos φ / sin δ .

268 Fig. 48 But sec z = √(1 + tan² z) = √(x² + y² + 1) ∴ √(x² + y² + 1) / A = y + tan Φ which reduces to x² + y² (1 - A²) - 2A² y tan Φ + 1 - A² tan² Φ = 0 (3) From this it is evident that the locus is an ellipse or parabola or hyperbola according as A ⋚ 1; also it will be seen that the eccentricity is A. The locus is wrongly stated to be always a hyperbola in some text books. For it to be an ellipse A < 1 ie. cos Φ < sin δ ie. δ > 90 - Φ ie. Φ + δ > 90. In such latitudes and under such decli- nations, it will be an ellipse ie. at a place just north of the place where the perpetual day just begins the locus will be an ellipse. Hence in the arctic region it will be always an ellipse; and in the place just at which the perpetual day begins it will be a parabola and in the lower latitudes it will be a hyperbola, ie. it will be a parabola where the latitude Φ is given by 90 - δ. When Φ = 90°, A = cos Φ / sin δ = 0 provided δ ≠ 0. If, however, in addition δ = 0, A becomes indeterminate, but we may note then, that the Sun will be circling round the horizon on that equinoctial day at the north pole. We

269 may further note that at the north pole, the altitude of the Sun is always δ so that the length of the shadow cast is always equal to cot δ and this will be infinite when δ = 0. If now φ = 90°, and δ ≠ 0, though A = cos φ / sin δ be comes zero, A tan φ will not be zero because A tan φ = cos φ / sin δ × tan φ = sin φ / sin δ = 1 / sin δ (∵ φ = 90°). On the other hand A² tan φ = A × A tan φ = / sin δ = 0 because A = 0. Thus the term containing y in eqn. (3) vanishes. ∴ The equation reduces to x² + y² = A² tan² φ - 1 = cosec² δ - 1 = cot² δ (ie. +ve) ie. at the north pole the locus will be a circle with radius cot δ. When A = ∞ the locus √(x² + y² + 1) / A = y + tan φ becomes y = - tan φ which means that the hyperbola degenerates into the straight line which is parallel to the east-west line and is in the north at a distance of tan φ ie. s, the equinoctial shadow since the length of the gnomon is taken to be unity. In particular when A tan φ = 1 ie. φ = δ, the constant in (3) is zero, so that the locus passes through the foot of the gnomon as is also evident from the fact that the Sun passes through the zenith. Taking a northern latitude say 17°, the loci of the extremity of the shadow are shown in fig. 48A (page 270) on important days when δ=ω, when δ=0, when δ = - ω, when δ = φ. The maximum mid-day shadow is tan (φ + ω), when δ = - ω taking the gnomon's length to be unity; this shadow is cast north of the gnomon along the south-north line through the gnomon, on Dec. 23rd of the year. The minimum length of the mid-day shadow occurs when δ = φ, the shadow being zero and being at the foot of the gnomon, the Sun being then just overhead. The maximum shadow cast south of the gnomon at mid-day is tan (ω - φ).

270 Shadow on Dec. 23 No Shadow in this direction on Dec 23 Immediately before 21st March B C A On 21st March Immediately after 21st March No Shadow in this direction on 22nd June W B C A E JUNE when δ=φ On 22nd JUNE Fig. 48A Showing the locus of the extremity of the shadow on different days at a latitude of 17° Note. OA, OB are the shadows computed by Bhāskara when the Sun is on the prime-vertical. In the method of finding the shadow under verse 46, we perceive Bhāskara's genius in (1) looking upon the shadow as being made up of two segments namely that due to φ and that due to the declination (2) in conceiving what he calls Iṣṭa-drik-mandala palabhā, and Iṣṭa-drik- mandala Krānti and (3) in deriving the equation H sin D = (H sin ZT × H sin δ) / (H sin φ) Latter half of verse 47 and verse 48. Something to be noted. In computing the shadow in a given direction, there may be two shadows at times in the northern hemi- sphere. When H sin a < Agrā and there will be none in the southern. To compute the second shadow we have to take 180 − L also as the latitude where L is the latitude computed, and proceed in the same way as we have done before.

271 Fig. 49 Comm. This too exhibits Bhāskara's genius. (Ref. fig. 49). Let MQR₂ be the equator whose pole is p. Let T₁ S₁ Z S₂ T₂ be the circle of azimuth a (Hindu azimuth). Let SS₁ S₂ be the diurnal circle of the Sun cutting the above circle of azimuth at S₁ and S₂, so that ZS₁ and ZS₂ are the two solutions giving the two zenith-distances which give two shadows in the given direction. H sin MS = Agrā ; evidently MẐS > MẐS₁ ie. H sin a < Agrā as stipulated. ZT₁ and ZT₂ give the zenith-distances in the given direction when the Sun is on the equator. S₁ T₁ and S₂ T₂ are the decrements in the zenith-distances on account of declination δ (= S₂ R₂ or S₁ R₁). If MẐS₁ were greater than MẐS ie. if H sin a > Agrā, we would have lost the position S₁ ie. we would have had only one shadow

272 in the afternoon in the given direction and no shadow in the morning. Analytically, the event of having two shadows arises on account of the following circumstance. When we are asked to find Iṣṭākṣajyā from Rs / (H sin a) the shadow in the given direction on the equinoctial day (Ref. gL fig. 45) the L for this given value of the shadow is given by H sin L = RS / √(12² + S²) where S = Rs / (H sin a). We know sin θ = a has two solutions, θ₁ and 180 − θ₁. Hence L will have two values L₁ and 180 − L₁. So, Bhāskara has asked us to compute D₁ and D₂ from L₁ and L₂ and thus have the two solutions. Fig. 50

273 Note (1) When Bhāskara said 'If H sin a < Agrā' he had in mind evidently the azimuth circle MZN which cuts the diurnal path A Q₁ R₁ at S₁ and S₂. At S₁ the azimuth EZM < EZA so that he stipulated that H sin a should be less than the Agrā. But, let the diurnal path of the Sun be Q₂ R₂ where Q₂ falls in between z and p. In such a case we know that the azimuth does not take all values but has a maximum where the vertical touches the diurnal path at T. From PTZ where T is a right angle, taking PZT = 90 - a, a being the Hindu azimuth we have by Napier’s rule sin PT = sin ZP sin (90 - a) or sin (90 - δ) = sin (90 - φ) sin 90 - a or cos δ = cos φ cos a. If a has a lesser value than is given by this equation, the diurnal path does not cut the azimuth circle ie. if cos a > cos δ / cos φ, there will be no shadow in the given direction even though the situation satisfies Bhāskara’s condition namely H sin a should be less than Agrā. Bhāskara has overlooked this case. This may be seen analytically also as follows. We have from the spherical triangle PZS, sin δ = sin φ cos z + cos φ sin z sin a = A cos z + B sin z (say). We know, the maximum value of A cos z + B sin z is √(A² + B²) which is here √(sin² φ + cos² φ sin² a) = √(sin² φ + cos² φ - cos² φ cos² a) = √(1 - cos² φ cos² a). Thus there will be no solution for z if the quantity on the left hand side namely sin δ > the above max. value ie. if sin δ > √(1 - cos² φ cos² a) ie. if sin² δ > 1 - cos² φ cos² a ie. if cos² δ > cos² φ cos² a ie. if cos δ > cos φ cos a ie. if cos a > cos δ/cos p as derived above. Hence even if H sin a > Agrā, there need not be a shadow at all in the given direction. In other words when the Hindu azimuth given is very small and when the decli- nation is too great north or south, there may not be a 35

274 shadow in the given direction. Bhāskara gives an example where he gets two shadows on a day taking the moments when the Sun is on the prime-vertical. In fact having this case of the East-West shadows alone, he conceived that two shadows could be had in a given direction under particular conditions. He chooses a place of s = 5″ ie. a place of latitude 22° – 37′ (Bhāskara often gives this latitude which night indicate that he was probably residing in that latitude which passes through approximately Itarsi). He takes a day when the Sun’s declination is given by H sin δ = 780 ie. δ = 13° – 7′. Then the Sama-Śanku is given by R sin δ ───────── (comparing the second and the fifth latitudinal sin φ triangles Sama-Śanku Krantijyā H sin δ ──────────── = ─────────── = ───────── R H sin φ H sin φ RH sin δ R sin δ ∴ Sama-Śanku = ────────── = ─────────) H sin φ sin φ 3438 × 780 = ────────── = 2028 approximately. I 1322 – 18 R sin δ Also Agrā = ───────── = 845. cos φ Knowing the Sama-Śanku, the East-West shadow may be taken to be determined. Now Bhāskara proceeds to show that at the time of having the second shadow also, in the same East-West direction, the Śanku will be the same Sama-Śanku itself. For this, proceeding according to the Rs method indicated in the verse, “Taking ─────── to be the H sin a equinoctial shadow etc.” we have Rs 3438 × 5 ─────── = ──────────── = what is called Kha-hara Rāsi. H sin a 0

275 Taking this to be the equinoctial shadow H sin L = Rs / √(12² + s²) = R itself (Dealing this way with Kha-hara Rās'is is prohibited in modern mathematics but Bhāskara adds at the end of the commentary that dealing with them cautiously does not effect computations which is of course true, for when the equinoctial shadow is infinity φ = 90° so that H sin φ = R as got). Hence L = 90° and 180° - 90° = 90° = L'. Then H sin D = (R × 780) / (1322 - 18) = same as Sama-S'anku obtained in I = H cos z so that D = 90 - z. Now from the equation z + δ = φ, z = φ - δ = L' - D = 90° - (90 - z) where z is the zenith-distance when the Sun is on the prime-vertical. ∴ The zenith-distance is again the same z. In other words, the second zenith-distance is also that when the Sun is on the prime-vertical. Bhāskara has given this example just to obtain the second shadow as well and he has chosen the event of the Sun being on the prime- vertical to show that the procedure indicated by him may be verified to hold good. Verses 49, 50. Alternate method to find the shadow. Let R² s² + H sin² a × 12² = prathama where s = equinoctial shadow and a the Hindu azimuth. Let Anya = RsA where A is the Agrā. Divide the prathama and Anya by (H sin² a - A²) and still call them prathama and Anya. Then K = √(Adya + Anya²) ± Anya where K is the Chayā- Karṇa. Comm. Let K be the required Chayā-Karṇa. Then Karnāgrā = KA / R = s + b where b is the bhuja.

276 ∴ b = KA / R - s. But (H sin a) / b = R / S where S is the shadow so that S = bR / (H sin a) = (KA / R - s) R / (H sin a) = √(K² - 12²) ∴ (KA - sR) = H sin a √(K² - 12²) ie. K² A² + s² R² - 2 AsRK = H sin² a (K² - 12²) ∴ K² (A² - H sin² a) - 2 AsRK = - s² R² - 12² H sin² a ∴ K² (H sin² a - A²) + 2 AsRK = 12² H sin² a + s² R² I This quadratic is of the form ax² + 2bx = c ie. x² + 2 (b/a) x = c/a II Here 'c'/a is called Adya and b/a Anya. The solution of the above equation is given by x = - b/a ± √(b²/a² + c/a) ie. - Anya ± √(Anya² + Adya). III When b = KA / R + s, putting - s in the place of s in I, K = Anya ± √(Anya² + Adya) IV Out of the four solutions given by III and IV we have taking the positive solutions K = √(Anya² + Adya) ± Anya as stated. Verse 51. If H sin a < A, then in the northern hemisphere ie. where δ is north, ± √(Anya² - Adya) + Anya = K. Comm. We have initially put Anya = H sin² a - A². If H sin a < A, them to avoid a negative value for the

277 Anya, we could put Anya = A² - H sin² a. As a matter of fact in verse 50, we are asked to take H sin² ~ A², as Bhāskara wanted that the second case also be included. Thus putting Anya = A² - H sin² a, equation I becomes K² (A² - H sin² a) - 2 AsRK = - (Adya) so that K = Anya ± √(Anya² - Adya) as given. Verse 52. The Bhuja is to be obtained through Karṇāgra from the equation a = b + s (already proved) and Rb / S = H sin a {ie. S = bR / H sin a as already proved}. This H sin a will be the same in the case of obtaining two values of K ie. two shadows one in the morning and the other in the afternoon, (the only difference being that they will be on alternate sides of the East-West line). Verses 53 and 54. Obtaining the shadow when time is given. In the two previous examples the magnitude of the shadow was obtained in a given direction ; now we shall obtain the same when time is given. The word unnata stands for the time that has elapsed after Sun-rise or that which is the balance of the day time. The unnata sub- tracted from half-the-day gives what is called Nata. The H sine of the unnata minus Chara or increased by the Chara according as the Sun's declination is north or south, is called Sūtra ; this multiplied by the H cos δ and divided by the radius, is called Kalā. Comm. (Ref fig. 51) The time measured by the arc MN, that is the time in between the moment when the Sun S is on the horizon and the moment when he is at L is called the unnata ie. the time measured after rising and the time measured by the arc NQ ie. the time in between