पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)

IV. 22 IV. THREE PROBLEMS 95 [मध्याह्नच्छाया] अपमोनयुताऽक्षज्यां त्रिज्यात्कृतिविशेषमूलेन । छिन्द्याद् द्वादशगुणितां लब्धा माध्याह्निकी छाया ॥ २२ ॥ Sine zenith distance 22. Subtract the Sun’s declination from the latitude (of the place), if the decli- nation is north, and add if it is south. The midday Sun’s Z.D. is got. Find its sine and multiply by twelve. Divide this by the root of the difference of the squares of the radius and sine ZD. The mid-day shadow is obtained in aṅgulas. The following are the rules: i. Degrees of zenith distance = Latitude ∓ Sun’s declination (the upper sign being used for north declination and the lower for the south.) ii. Mid-day shadow = 12 × sin ZD ÷ √120² − sin² ZD (where 120 is written for radius). It must be noted that the incompleteness mentioned in connection with the second formula if the previous work is found here too. Example 6. The latitude of a place is 25° 9′. The Sun’s declination is 11° 44′ , south, (the Sun being in the part of the ecliptic beginning from Libra). Find the mid-day shadow. The declination being south, ZD = 25° 9′ + 11° 44′ = 36° 52′. Sin ZD = 72′. ∴ Midday shadow = 12 × 72 ÷ √120² − 72² = 12 × 72 ÷ 96 = 9 aṅgulas. The rules are explained thus: From the previous rule, Latitude = zenith distance ± declination, (+ for north declination, and − for south declination), we have, zenith distance = latitude ∓ declination, (for north and south declinations, respectively). From this sine zenith distance is got. Using this, the shadow is obtained from the previous rule, sin ZD = shadow × 120 ÷ √shadow² + 144. Squaring both sides, sin² ZD = shadow² × 120² ÷ (shadow² + 144). sin² ZD × (shadow² + 144) = shadow² × 120² sin² ZD × shadow² + 144 sin² ZD = shadow² × 120² 120². shadow² − sin² ZD. shadow² = 144 sin² ZD shadow² = 144 sin² ZD ÷ (120² − sin² ZD) shadow = 12 sin ZD ÷ √120² − sin² ZD. 22. Quoted by Utpala on BS 2, p. 61. 22a. A.D. अपनोन. A.C.D. U. ॰क्षज्या c. A.C.D. गुणिता b. A. तांत्रिकृति; C.D. U. तत्तिर्ज्याकृति. A. मूला d. A. माध्याह्नकी

96 PAÑCASIDDHĀNTIKĀ IV. 23 [लम्बज्या दिनव्यासश्च] विषुवज्ज्याऽऽयामार्थवर्गविश्लेषमूलमवलम्बकः | क्रान्तित्रिज्याकृत्योरन्तरपदं द्विगुणं दिनव्यासः || २३ || Sine Co-latitude and Day-diameter 23. Square the sine of latitude and deduct from the square of the radius. Its square root is the ‘sine of co-latitude’, (its arc being the ‘co-latitude’). Square the sine of declination, deduct from the square of the radius and find its root. Twice the result is the ‘day diameter’. Now, we have (i) sine co-latitude = √(radius² − sin² latitude) (ii) Day-diameter = 2 × √(radius² − sin² declination) Example 7 (a). sin lat. = 72. Find sin co-lat, and its arc, viz. the co-lat. sin co-lat. = √(120² − 72²) = 96'. Arc 96' = 53° 8' = co-latitude. Example 7 (b). The Sun is at the end of the sign Aries. Find the day-diameter. The Sun’s longitude = rāśi. 1-0-0. Sine rāśi. 1-0-0 = 60'. ∴ sin declination = 60' (60+1)/150 = 24' 24". The day-diameter = 2 × √(120² − 24' 24"²) = 2 × 117' 30" = 235'. In the right angled triangle having the radius as the hypotenuse and the sine of latitude as the base, the sine of the co-latitude stands as the perpendicular or lamba. Therefore it is called lambajyā. By the analogy with co-sine for sine, co-tangent for tangent, and co-secant for secant, the term co- latitude for latitude, has been invented for 90'−latitude, for convenience of expression. Therefore: (since base² + perpendicular² = hypotenuse²), sin²lat + sin² co-lat = radius². From this, sin² co-lat = radius² − sin² lat. ∴ sin co-lat = √(radius² − sin² lat. As for the day-diameter, by the diurnal rotation of the earth on its axis, the Sun apparently moves round the earth every day in a circular path, at a distance from the celestial equator equal to the latitude, with the axis of the earth perpendicular to the plane of the circle. This circle is called the diurnal circle or day-circle and its diameter, the day- diameter. (See this shown in Fig.6.) The diameter can be measured thus: see Fig.9. [Fig. IV. 9: Circle with vertical diameter NP-SP, horizontal diameter CQ, center E; parallel chord SB above CQ with perpendicular SD to CQ, meeting NP-SP at A] 23. Quoted by Utpala on BS 2, p. 60. 23a. A. विवच्छायामात्यार्द्ध b. A. मूलवले लबः; C. मूलभवो लब्धः c-d. A. ॰क्रान्तिज्यात्रिज्याक्रांत्यन्तरपदं; C-D. ॰कृत्यन्तरात् पदाद् दिनव्यासः (D. पदद्द्विदिनव्यासः) d. A. द्विदिन

IV. 25 IV. THREE PROBLEMS 97 NPCSPQ is the stellar sphere, with centre E, CQ is the celestial equator, SC is the declination of the Sun S, SEC = degrees of declination, SB is the diurnal circle, with the straight line SB as its dia- meter, and SA as its radius. Suppose the sphere is cut into equal halves, with the cross section NP C SP Q E exposed to view and the axis NP E SP forming a diameter. SE is the radius, and SD (= AE) = sin declination. Then, SA = √(SE² − SD²). But SA = half day-diameter. ∴ day-diameter = 2 SA = 2 √(radius² − sin² declination). अजवृषमिथुनापक्रमजीवाः (षड्घ्नाः स्यु) 'वेद-मुनि-वसवः' | त्र्यष्ट'तिथि' षट्का(ष्टक)विकलाऽभ्यधिका [:] परिज्ञेयाः || २४ || 24. The sines of declinations of the points of the ecliptic ending Aries, Taurus and Gemini are 24' 24", 42' 15", and 48' 48". We shall show these to be correct by computing them. The sine declination of the end of Aries, i.e. rā. 1-0-0 has been derived in example 7(b) to be 24' 24". The sine of declination of the end of Gemini, i.e. rāśi 3-0-0, has been shown to be 48' 48", (the maximum) in the example above. So we shall derive here only sine declination of the end of Taurus, i.e. rāśi 2-0-0. Sine rāśi. 2-0-0 = 103' 55" (from tables). The sine of its declination by IV.16 is, 103' 55" × (60 + 1)/150 = 41' 34" + 41" 34''' = 42' 15"34'''. Here, though 34''' is greater than half a second, the author has omitted it and given 42' 15", to the nearest quarter minute. [पञ्चत्रिंशत्] त्र्यष्टकस्वरूपधृ[तिसंयु]ता क्रमाद् द्विशति | पञ्चाष्टक'तिथि'विकलाधिकौ वृषा(न्यौ) दिनव्यासः || २५ || 25. The respective day-diameters are, in the minutes parts: 200 + 35, 200 + 24, and 200 + 19, with 40" and 15" added to the second and third, (i.e. the day- diameters are, 235', 204' 40" and 219' 15"). Of these, the day-diameter of the end of Aries has been worked out in Example 7 (b). We shall derive the other two. The day-diameter for the Sun at the end of Taurus = 2 √(120² − sin² declination of the end of Taurus), = 2 √(120² − 42' 15"²) = 224' 38". 24b. A.C.D. षड्घ्नास्तु 25a. A. om पञ्चत्रिंशत् c. A. षट्काष्ट्; D. षट्काष्ट[क]विकला b. A. ॰धृता क्रमा. C.D. द्विशती d. A. ॰धिका प d. A. ॰धिको. A. वृषांत्यौ

98 PAÑCASIDDHĀNTIKĀ IV. 26 But the author gives 224′ 40″ as being more convenient to use. The day-diameter at the end of Gemini = 2√(120² − 48′ 48″²) = 219′ 15″, which is the same as given by the author. The missing part of the text, (pañcatriṁśat), has been found out by computation. (tisaṁyu) has been guessed as being necessary to supply the meaning, which is clear. [चरः] व्या(स)क्रान्तिज्याघ्नी विषुवज्ज्या लं[ब]कद्युदैर्घ्यहृता । तच्चापकलात्र्यंशश्चरखण्डविनाडिकाः स्पष्टाः ॥ २६ ॥ Cara 26. Multiply the sine of latitude by 240′ and by the sine of declination. Divide by the sine of co-latitude and by the day-diameter. Find the arc of the sine obtained in minutes—(This arc is called half-cara)-and divide by 3. The result are the accurate minutes of cara, (which might be called ‘day-difference’). From the cara we can obtain the cara-intervals, (or cara differences). This is the formula: (i) Sine half-cara = 240′ × sine latitude × sine declination ÷ (sine co-latitude × day-diameter) From this the half-cara arc is got. Then, (ii) Cara, i.e. day-difference in vināḍīs = minutes of half-cara ÷ 3. In III.12 the author gave a rule for the cara-vināḍīs to be used in North-India and its neighbour- hood and said that he would give the general rule later in the Chedyaka section. This is it. Further, in the rule of III.10, the interval of the vināḍīs were given for long intervals in degrees, like whole signs, and the value obtained can only be rough. This rule can give accurate values. The reading perhaps is ‘cara-piṇḍa’ for which the scribe has written ‘cara-khaṇḍa’ by mistake. Example 8. The sine of latitude of a place is 72′, and the sine of co-latitude 96′. The Sun is at the end of Mithuna, with the sine of its declination 48′ 48″. The day-diameter for the day is 219′ 15″. Find the cara- vināḍīs. By the formula, sine half-cara = 240′ × 72′ × 48′ 48″ ÷ (96′ × 219′ 15″) = 40′ 4″. Arc 40′ 4″ = half-cara = 19° 31′ = 19 × 60′ + 31′ = 1171′. Cara-vināḍīs = 1171/3 = 390, i.e. nāḍīs 6-30. The work is thus explained: (See fig.10) 26a. A. व्यासः क्रान्ति b. A. ज्यालक

IV. 26 IV. THREE PROBLEMS 99 Fig. IV. 10 In the stellar sphere CEC is the celestial equator, NP and SP being the north and south poles. NESP is the Unmaṇḍala or horizon of a place on the equator. Z is the zenith of the place, N and S being the north and south points and E is the east point. DsD is the day-diameter of the Sun, (s), in the northern hemisphere, making the declination sd. D₁s₁D₁ is the day-diameter of the Sun, (s₁), in the southern hemisphere, making the declination sd. D S D is the day-diameter of the Sun (s₁), in the southern hemisphere, making the declination s₁d₁. s and s₁ are the rising points of the Sun as seen from the place, NsEs₁S being its horizon. The altitude of the North Pole. N NP = angle NE NP, is the latitude, which is equal to SE SP, from which it is seen that for places in the northern hemisphere, the Unmaṇḍala is raised from the horizon by this angle in the north, and depressed by this angle in the south. As the Sun, in its diurnal circuit, takes exactly half a day to move from the eastern Unmaṇḍala to the western, the day-time is longer when the Sun’s declination is north, for it has to travel, after rising, an arc in the diurnal circle (equivalent to the great circle arc dE) to reach the Unmaṇḍala and an equal time while setting. The time is less when the declination is south, because before rising it has to travel less by an arc equiva- lent to Ed₁ to reach the horizon from the unmaṇḍala (and an equal time less while setting). dE and Ed₁ are the arcs of half-cara. Therefore when the declination is north, the time corresponding to 2 DE in the day-difference, (the day time being greater than 30 nāḍikās by this amount,) and when it is south the time equivalent of 2 Ed₁ is the day-difference, (the day-time being less than 30 nāḍikās by this amount). So we have to calculate dE, and Ed₁. In Δ dEs, right angled at d, by fundamental formula III, sin dE = Radius × sin Sd × Cos sEd ÷ (Cos sd × sin sEd). But, sd is the declination and sEd = 90° − N E NP= 90° − latitude. ∴ sin half-cara = 120′ × sin dec × cos (90°− lat.) ÷ {(Cos dec × sin (90° − lat))}

100 PAÑCASIDDHĀNTIKĀ IV. 28 = 120' sin dec × sin lat ÷ {(Cos dec × sin (90° – lat)} = 120' sin dec × sin lat ÷ (day-radius × sin co-latitude) = 240' sin dec × sin lat ÷ (day-diameter × sin co-latitude). From this the arc dE is got. Ed₁ for south declination is got in the same way, from △ s₁E d₁. From dE or Ed₁, the cara-vināḍīs are got thus: For the whole circle of 360° or 21600 minutes of arc, there are 60 × 60 = 3600 vināḍīs. ∴ For the arc of half cara in minutes there are 3600 × arc of half-cara ÷ 21600 = arc of half cara/6 vināḍīs. The whole cara-vināḍīs are twice this, and equal to 2 × minutes of half-cara/6 = minutes of half-cara/3. As we have said, these are added to 30 nāḍikās to find the day-time, when the declination is north, i.e. when the Sun is in six signs from Aries. Those vināḍīs are subtracted when the Sun is in the south, i.e. in the six signs Libra etc. The part of the formula, sin declination × sin-latitude ÷ sin co-latitude, is called ‘Earth sine’, (kṣitijyā), in Hindu astronomical works, which is required to be multiplied by the radius and divided by the day-diameter to get sin half-cara. In certain works the half-cara itself is called cara. [चराद् अक्षानयनम्] चरखण्डः'(ख)पक्षां'शज्याघ्नमद्व्यर्ह्वसमुद्धरेत् 'खजिनै': । द्विः कृत्वा तद्वर्गात् क्रान्तिज्याकृतियुतान्मूलम् ॥ २७ ॥ तेन विभजेत् स्थितज्यां व्यासार्धगुणामवाप्तमक्षज्या । नवतेरक्षोनायाः क्रमशो ज्या लम्बको भवति ॥ २८ ॥ Latitude from Cara 27. Divide the vināḍis of cara by twenty and find the sine of the resulting degrees. Multiply the day-diameter by this, and divide by 240. Put the result in two places. In one place square it and add the square of the sine of declination and find its root. 28. Multiply the result kept in the other place by the radius, and divide by this root. The result is the sine of latitude. Its arc is the latitude. 90' minus latitude is the co-latitude, and its sine, sine co-latitude. The following is the work to be done: i. The vināḍis of cara ÷ 20 = degree of half-cara. Find its sine. ii. Sine half-cara × day-diameter ÷ 240 = sine x. (This is earth-sine or kṣitijyā). 27-28. Quoted by Utpala on BS, 2, p.60. 27b. A. °महस° c. A. व्यावृद्धिं कृत्वा; C. भूजीवां कृत्वा तत् d. A. मूलम् 28a. A. थितिज्यां; C.D. क्षितिज्यां. A. पक्षज्या c. A. नवतेरक्षोसोनाया

IV. 28 IV. THREE PROBLEMS 101 iii. Earth-sine × 120 ÷ √sin² earth-sine + sin² dec = sin lat. From this the latitude is found iv. 90° — latitude = co-latitude. Its sine. co-lat. Example 9. At a certain place on a certain day, the vināḍis of cara are 390 1/3. The day-diameter is 219' 15". Find the latitude of the place, and sine co-latitude. The sine of declination required for the formulae is, by (IV.23), √ 120² — (219' 15"/2)² = 48' 48". i. Degree of half-cara = 390 1/3 ÷ 20 = 1171/(3 × 20) = 19° 31'. From this, sine half-cara = 40' 4". ii. (Earth)-sine = 40' 4" × 219' 15" ÷ 240' = 36' 36". iii. Sin lat.= 36' 36" × 120' ÷ √36' 36"² + 48' 48"² = 120 ÷ √1 +16/9 = 120' × 3/5 = 72'. From this, lat = 36° 52'. iv. Co-latitude = 90° — 36° 52' = 53° 8'. From this sine co-latitude = 96'. The rules are thus derived: a. From the rule, vināḍikās of cara = minutes of half-cara ÷ 3. By transposing, we have: Minutes of half-cara = vināḍikās of cara × 3. Degrees of half-cara = vināḍikās of cara × 3/60 = vināḍikās of cara/20, which is (i). b. From the rule, sine half-cara = sin lat.× 240 × sin dec ÷ (sin co-lat x day-diameter), we get; Sin dec = sin half-cara × sin co-lat.× day-diameter ÷ (240 × sin lat) = sin co-lat × earth-sine ÷ sin lat. Using this in (iii) above, we have: Sin lat = earth-sine × 120' ÷ √earth-sine² + sin² co-lat × earth-sine² ÷ sin²lat. 120' ÷ √sin²lat + sin²co-lat ÷ sin²lat. = earth-sine × 120' ÷ (earth-sine) √1 + sin² co-lat = 120' ÷ √sin² lat + sin² co-lat ÷ sin²lat. sin²lat = 120' ÷ √1/sin²lat = √ sin²lat = sin lat, thus proving (iii). From this the latitude is got. Then, ∵ latitude + co-latitude = 90°, Co-latitude = 90° — latitude. It should be noted that of the sin declination and the day-diameter required in the rules, one is sufficient, because the other can be got from that. As for the word khaṇḍa, meaning 'interval' or 'dif- ference', we have already said that it is piṇḍa ('the whole') we get first, and thence the khaṇḍa. As for the reading, we have corrected, carathaṇakapakṣāṁśa, into carakhaṇḍakhapakṣāṁśa, making ka into kha, because 'twenty' is required here as the divisor. This is the only correction we have made. But TS, followed by NP, have made several corrections, not realising that if Bhaṭṭotpala's reading is adopted no other correction would be required.

102 PAÑCASIDDHĀNTIKĀ IV. 30 [लङ्कोदयराशिमानम्] [राशिज्या] ऽपक्रमज्या(कृ)तिवि(श्ले)षमूल[हत] वि(स्ता)रात् । द्यु(व्या)स(ह)ता(च्चापं) 'दिग्घ्नं' राश्यु(द्ग)मविनाड्यः ॥ २९ ॥ 'वसुमुनिपक्षा' 'व्येकं शतत्रयं' 'त्रिद्विकाग्नय' [श्राङ्का] (त्) । परतस्त एव वामाः षडुत्क्रमात्ते तुलाद्यर्धे ॥ ३० ॥ Rt. ascensional difference 29. Square the sine of the longitude of a point on the ecliptic, and deduct from it the square of the sine of the declination of the point. Find its root, multiply it by the diameter and divide by the day-diameter. Find the arc of the resulting sine in degrees. Multiply the degrees by 10. The Right ascension of the point is obtained in vināḍīs. deducting the right ascension of the next rāśi from that of the previous, the right ascentional difference of the rāśis are obtained. 30. The vināḍis of right ascentional difference for the three signs from Meṣa are 278, 299 and 323. In the next quadrant they are the same in the reversed order, viz. 323, 299 and 278. In the half of ecliptic beginning from Libra, the difference are those of the first half, taken in the reverse order. The formula is: Sin Right ascension = 240' × √(sin²longitude − sin² dec) ÷ day-diameter. The degrees of right ascension multiplied by 10, are the vināḍis of right ascension. The differences as calculated, are, for Aries etc. 278, 299, 323, 323, 299, 278, 278, 299, 323, 323, 299, 278. Now, what is the meaning of saying that in the second half the differences are in the reverse order of those in the first half, when reversing the order does not make any difference? True. But the author must have meant this statement for ascensional difference in general, for, then, owing to the subtraction and addition of half day-differences (carārdha) in the first and second quadrants, the reverse order becomes different. Further, the vināḍis mentioned here are sidereal and not mean solar, because the vināḍis per degree are obtained by dividing the time of a full revolution by 360, and the time of a full revolution of the stellar sphere is a sidereal day, and not a mean solar day which is the time of the diurnal revolution of the mean Sun. 30. Quoted by Utpala on BS 2, p.61. 29a. A. भपक्रमज्या; C. मेषाद्यपक्रमज्या; D. भापक्रमज्या b. A. क्रतिविशेषमूलविस्तारात्; C. कृतिविशेषमूलगुणविस्तरात्; D. कृतिविश्लेषमूल [गणिताद्] विस्तारात् c. A. द्युद्वासहताचाप; C.D. द्युव्यासहताच्चापं 30b. A. ०काग्रयश्चाजान्; C.D. ०काग्रयश्चाजात्. U. श्राङ्काः C.A. वाभाः d. A. षड्गक्रमास्ते नुताद्यर्द्धे

IV. 30 IV. THREE PROBLEMS 103 Example 10. Find the right ascensions of the points of the ecliptic ending Aries, Taurus, and Gemini, i.e. longitudes 30°, 60° and 90°. From them find their respective differences. Sin 30° = 60′, sin 60° = 103′55″ and sin 90° = 120′. Sin dec. of the points ending Aries etc. are, respectively, 24′24″, 42′15″ and 48′48″. The respec- tive day-diameters are 235′, 2244′38″, and 219′15″. (a) For the point 30°, sin Rt. asc = √60′² – 24′ 24″² × 240 ÷ 235 = 54′ 49″ × 240 ÷ 235 = 55′ 59″. Its arc = 27° 49′. Multiplying by 10, the vināḍis of Rt. asc. are 27° 49′ × 10 = 278. (b) For the point 60°, sin Rt. asc. = √103′ 55″² – 42′ 15″² × 240 ÷ 224′ 38″ = 94′ 57″ × 240′ ÷ 224′ 38″ = 101′ 26″. Its arc = 57° 42′. The vināḍis of Rt. asc. = 57° 42′ × 10 = 577. (c) For the point 90°, sin Rt. asc. = √120′² – 48′ 48″² × 240′ ÷ 219′ 15″ = 109′ 37″.5 × 240′ ÷ 219′ 15″ = 120′. Its arc = 90°. The vināḍis of Rt. asc. arc. 90° × 10 = 900. The difference for Gemini = Rt. asc. for 90° – Rt. asc for 60° = 900 – 577 = 323 The difference for Taurus = Rt. asc. for 60° – Rt. asc. for 30° = 577 – 278 = 299. As the Rt. asc. of the first point of Aries is zero, the difference for Aries = Rt. asc. for 30° – Rt. asc. for 0° = 278 – 0 = 278. All these are the same as given by the author. This is how the formula is arrived at: The time taken by each sign of the ecliptic, beginning from Aries, to rise above the eastern horizon, for an observer on the equator, is in vināḍis 278, 299, etc., and their total is the time taken by any point to rise, after the rising of the First point of Aries. This is represented by the arc of the celestial equator (called the Rt. asc.) measured from the First point of Aries, and we have to find this arc. In Fig. 11, r is the First point of Aries. P is the point on the ecliptic of which the time of rising is required, and Pd is the declination of the point, equal to the arc of the horizon from the east point to the rising point. dr is the arc on the celestial equator, called the Right-ascension of the point P, which is required to be found. From the fundamental formula iv, Sin Rt. asc. = sin dr = sin Pd × cos Prd × Radius ÷ (Cos Pd × sin Prd) = sin Pr × cos Prd ÷ cos Pd (∵ by the fundamental formula ii, sin Prd = sin Pd × radius ÷ sin Pr.) = sin Pr × √Radius² – Radius². sin²Pd ÷ sin² Pr ÷ cos Pd = sin Pr × Radius √sin² Pr – sin² Pd ÷ (sin Pr × cos Pd) = Radius × √sin² Pr – sin² Pd ÷ cos Pd = 120′ × √sin² long. – sin² dec. ÷ 1/2 day-diameter = 240′ × √sin² long. – sin² dec. ÷ day-diameter. Fig. IV. 11 The arc of this is the Rt. asc. As there are 3600 vināḍis for a Rt. asc. of 360°, for the Rt. asc. got, the time is, Rt. asc. × 3600 ÷ 360 = Rt. asc. × 10. Then by subtracting the vināḍis pertaining to the Rt. asc. of the beginning of the sign from that of the end of the sign, the differences are got.

104 PAÑCASIDDHĀNTIKĀ IV. 31 Because the sine of the longitude and the sine of the declination (which itself varies as the sine of the longitude) decrease in the second quadrant in the reverse order of the increase in the first, and this increase and decrease are repeated in the third and fourth quadrants, the differences of vināḍīs follow the same course. [राश्युदयः] चरदलकालक्षीणास्त्रयस्त्रयः संयुताः प्रतीपैस्तैः । उदयर्क्षतुल्यकालेन यान्ति तत्सप्तमाश्चास्तम् ॥ ३१ ॥ Rising Signs 31. Take the differences of Rt. asc. of three signs at a time. From the first triplet subtract the differences of half-caras, one by one, taken in the given order. Add the half-cara differences one by one, taken in the reverse order, to the second triplets. To the third triplet add the half-cara differences taken in the given order. From the fourth triplet subtract the half-cara differences one by one, in the reverse order. The vināḍis of the rising signs, called the ascen- sional differences, as seen from any place, are obtained. The seventh from the rising signs set during the same time as the signs themselves rise. The ascensional differences for the several signs are as follows: Aries : 278 − half-cara difference for Aries Taurus : 299 − half-cara difference for Taurus Gemini : 323 − half-cara difference for Gemini Cancer : 323 + half-cara difference for Gemini Leo : 299 + half-cara difference for Taurus Virgo : 278 + half-cara difference for Aries Libra : 278 + half-cara difference for Aries Scorpio : 299 + half-cara difference for Taurus Sagittarius : 323 + half-cara difference for Gemini Capricorn : 323 − half-cara difference for Gemini Aquarius : 299 − half-cara difference for Taurus Pisces : 278 − half-cara difference for Aries It can be noted that the ascensional differences for the six signs, Libra etc., are those of the six signs Aries etc. taken in the reverse order, as mentioned by us earlier. It should also be noted that signs Aries etc. mentioned here are sāyana. For nirayana meṣa etc. (reckoned from the first point of Aśvinī) the differences, obviously, will be different, and there will not be this symmetry about the first point of Meṣa or Tulā. Also, we have already said that the vināḍis are sidereal. Note also, that for places on the equator, the ascensional differences are those given in IV.30 itself, because the 31. Quoted by Utpala BS, 2, p.61. b. A. प्रतीपैस्ते 31a. A.C.D. चरकालदशक्षीणा; (C.D. दल) d. A. नयन्ति. B. ॰माश्वास्तान्

IV.31 IV. THREE PROBLEMS 105 cara is zero there, the day-time being always 30 nāḍīs there. The Sanskrit name 'Laṅkodaya' itself suggests this, Laṅkā representing a place on the equator. Example 11. At a certain place the equinoctial shadow of a twelve-unit gnomon is 5 units. Find the ascen- sional differences of the twelve āsis. (sāyana). By III.10 the cara-vināḍīs – differences for the place, pertaining to Aries, Taurus and Gemini, are 5 × (20, 16½, 6¾) = 100, 82½, 33¾. The half-cara differences are, respectively, 50, 41, 17 vināḍīs. in the southern hemisphere it is the other way. It is called Unmaṇḍala because it is raised in one’s Aries : 278 − 50 = 228 Libra : 278 + 50 = 328 Taurus : 299 − 41 = 258 Scorpio : 299 + 41 = 340 Gemini : 323 − 17 = 306 Sagittarius: 323 + 17 = 340 Cancer : 323 + 17 = 340 Capricorn : 323 − 17 = 306 Leo : 299 + 41 = 340 Aquarius : 299 − 41 = 258 Virgo : 278 + 50 = 328 Pisces : 278 − 50 = 228 The procedure is thus explained: The horizon of a place on the equator (i.e. zero latitude) appears raised towards the north pole to a person in the northern hemisphere on account of the elevation of the pole as we go north and submerged towards the submerged south-pole. To a person in the southern hemisphere it is the other way. It is called Unmaṇḍala because it is raised in one’s own hemisphere. The Right ascensional differences having reference to the horizon of zero latitude, i.e. the unmaṇḍala. But what we want are the ascensions, i.e. risings from the horizon of the place. Therefore the risings are earlier when the declination of the rāśi is north, (for places in the northern hemisphere), by the time the Sun takes to move from the horizon to the unmaṇḍala along the diurnal circle, and later by the same time when the declination is south. It has been explained that this time is equal to the half-cara vināḍīs. So, with reference to the points of the triplet Aries, Taurus and Gemini, whose declination is north, the half-cara has to be deducted. As the declination increases, rāśi by rāśi, the differences of half-cara have to be subtracted one by one, until the maximum half-cara is reached. There the declination decreases as it has increased, still being north, and the half-cara which has to be deducted decreases in the same manner. So the differences are added in the reverse order in the second triplet, i.e. Cancer, Leo and Virgo. In the next triplet, viz. Libra, Scorpio and Sagittarius, the south declination increases, i.e. the additive half-cara increases, and to the half-cara differences are again added, in the regular order, because in the third triplet the south declination increases in the same manner as the north declination in the first triplet. Then in the fourth triplet, i.e. Capricorn, Aquarius and Pisces, the south declination decreases, i.e. the additive cara decreases, and so the differences have to be deducted. (All this can be seen clearly on a globe). From the explanation it can be seen that for places in the southern hemisphere, the risings of the rāśīs are those of their seventh in the northern hemisphere. As great circles intersect one another, the part of the ecliptic above the horizon is always half a great circle, and therefore the distance between the rising point and the setting point of the ecliptic is always six signs, as also that of the celestial equator. Therefore the change in the Rt. asc. of the setting point of the ecliptic is equal to that of the rising point, with the result that the time of the setting of a sign seventh from the rising point is that of the rising point.

106 PAÑCASIDDHĀNTIKĀ IV. 33 [उन्नतकालः] इष्टोत्तरगोलापक्रमांशकज्यां 'खभास्करा'भ्यस्ताम् । हृत्वाऽक्षजीवया तच्चापादुदयेन तत्कालः ॥ ३२ ॥ तस्मिन् दिनकृत् कुरुते सममण्डलसंश्रयं दिनाद्यर्धे । तावच्छेषे परतो न तुलादिषु विद्यते चैतत् ॥ ३३ ॥ Time to reach the Prime vertical 32. When the Sun is within 6 signs from Aries, (i.e. when the Sun's declina- tion is north), multiply the sine of the declination by 120' and divide by the sine of the latitude, (the place being presumed to be north of the equator also). The sine of the Sun's altitude at Prime vertical, (śama-śaṅku), is got. Find its arc. Treat this arc as part of the ecliptic, and find its Rt. ascension in vināḍīs. 33. This is the time taken by the Sun to reach the Prime vertical in the fore- noon after crossing the unmaṇḍala, and the time remaining to reach it after reaching the Prime vertical, in the afternoon. The Sun does not touch the Prime vertical when it is in the six signs beginning from Libra, (i.e. when the declination is south), (as seen from places in the northern hemisphere). The following is the work asked to be done: (i) Sin altitude at Prime vertical = 120' × sin dec ÷ sin lat. (ii) Sin rt. asc. = √(sin² alt − sin² dec) × 240' ÷ day − diameter. Find the arc of this. (iii) Arc in degrees × 10 = time in vināḍīs to reach the prime vertical from the unmaṇḍala (or vice versa in the afternoon) (iv) Add the total half-cara vināḍīs if the time from sunrise, (or to set, if afternoon) is wanted. Here, the author has not mentioned the work of ii-iv explicitly, intending to give it subsequently. But it is clear that he is giving the time connected with the prime vertical, and that too, not the time before noon or afternoon, but the time from sunrise or to sunset. But it is not mentioned whether the rising or setting is with reference to the horizon of the place or to the unmaṇḍala. But as the rt. ascension in the manner of computing the Laṅkodaya is clearly meant, rising or setting with refer- ence to the unmaṇḍala alone seems to be in the author's mind, for the time with reference to that alone can be got. So to get the time from actual sunrise or sunset, the half-cara has got to be added, (section iv of the work), though this is not mentioned by the author. The half-cara has already been given, and need not be computed afresh. 32-33 Quoted by Utpala on BS. 2, p.41. b. A.ज्या. A.तस्कराभ्यस्तां; D.भास्करव्यस्तां c. A.हताक्ष. A.जीवजात 33b. A.संश्रया. A1.दिनाद्यर्द्धे; A2.दिनाधर्धूं; U.दिनाद्ये वा d. C.यत्कालः d. A1.चैतन्न

IV. 33 IV. THREE PROBLEMS 107 It may be mentioned in this connection that TS understand here only the work upto finding the sine of altitude at Prime vertical. As for the time, they say it is equal to the time taken by the Sun to reach the altitude found out, when the question is how to find this very time. It should also be noted that the work upto finding the sine of rt. ascension mentioned in (i) and (ii) can be done easily, thus: Work (iv) presupposes the knowledge of sin half-cara. Using that, sin rt. ascension mentioned in (ii) = sin half-cara × sin² colat ÷ sin² lat. = sin half-cara × 144 ÷ square of equinoctial shadow. If the sin rt. ascension obtained is greater than 120', then, even when the Sun's declination is north, the Sun does not touch the prime vertical. We shall explain this later. Example 12. On a certain day, the longitude of the Sun is rāśi 1-15. The latitude of the place (north of equator) is 30°. (The equinoctial shadow is 6 aṅgulas 55.7 vyaṅgulas). When, after sunrise, does the Sun cross the prime vertical at that place, on that day. We require the sine of declination and sine half-cara for the given time and place. Sin dec = sin 1ʳ 15° × 61/150 = sin 45° × 61/150 = = 84' 51" × 61/150 = 34' 30".3. The day-diameter = 2 × √(120² − 34' 30".3²) = 229' 51".4 Sin half-cara = 240' × sin lat × sin dec ÷ (sin co.lat. × day-diameter) = 240' × 60' × 34' 30".3 ÷ (103' 55" ×229' 51".4) = 20' 48". Half-cara = arc of 20' 48" = 9° 59'. Half-cara vināḍīs arc 9° 59' × 10 = 100 = nā.1-40. All this is supposed to be known already. Now for the computation of the time: (i) sin altitude = 34' 30".3 × 120' ÷ 60' = 69' 1". (ii) sin rt. asc = √69' 1"² − 34' 30".3² × 240' ÷ 229' 51".4 = 62' 24". Its arc is 31° 21'. (iii) The corresponding time = 31° 21' × 10 = 313 vināḍis = nā. 5-13. (iv) The time of crossing the prime vertical after sunrise = nā. 5-13 + nā. 1-40 = nā. 6-53. This is for the forenoon. For the afternoon, deducting this time from the time of sunset, nā. 33- 20, the time of crossing is nā. 33-20 − nā. 6-53 = nā. 26-27. Now, according to the short-cut in the place of (i) and (ii), Sin rt. asc. = sin half-cara × sin² colat ÷ sin² lat. = 20' 48" × 103' 55"² ÷ 60'² = 20' 48" × 3. = 62' 24". (See this obtained by the regular rule). Or, sin rt. asc. = sin half-cara × 144 ÷ equinoctial shadow = 20' 48" × 144 ÷ (6 aṅg. 55.7 vyaṅg.)² = 20' 48" × 3 = 62' 24", as already obtained. The rules are explained as follows, supposing the place to be north of the equator. (For places south of the equator also the same can be used, interchanging the directions north and south, wherever they occur.) See Fig. 12.

108 PAÑCASIDDHĀNTIKĀ IV. 33 Fig. IV. 12 In this figure of the sky-sphere, Z is the zenith, and NP is the north pole. D₁D₁, DD, etc. are four diurnal circles, on which four positions of the sun, S₁, S, etc are indicated. D₂M₂D₂ is a part of the unmaṇḍala, visible. In all the diurnal circles, the Sun S₁ etc. rising at D₁ etc. moving westward, moves a little south, little by little, until it reaches the meridian point M₁ etc., where the ‘southing’ is equal to the latitude, N NP, and then proceeds to move westward, moving north little by little, setting in the west at a point having the same amplitude as the rising point, (assuming that the declination does not change). On the two equinoxes, the Sun rises due east (D₂) and sets due west (D₂) southing on the meridian by ZM₂ (= N NP = latitude), and thus is always south of the prime vertical. So, when the declination is south, the diurnal circle (D₃D₃) is always south of the prime vertical and so the Sun (S₃) never touches the prime vertical. Even when the declination S₁S₂ (= M₁M₂)is greater than the latitude (ZM₂) then the Sun is always north of the prime vertical, the diurnal circle D₁S₁;M₁D₁ being north of it. It is this that was referred to by us as the case not mentioned by the author, viz. the case of the declination being north, but still not crossing the prime vertical, the case that is possible in the southern part of India. There is only one case left, that of the Sun’s declination being north, but less than the latitude, (e.g. the Sun moving on the diurnal circle D U S MD), in which alone the Sun crosses the prime vertical as at S. The time by which the Sun rising at D describes the part of the diurnal circle, DS, is to be found. Here there are two parts, the time from D to U which is the half-cara, and the time from U to S, i.e. the time after crossing the unmaṇḍala, which alone, we have said, has been mentioned explicitly by the author, and for which alone the rules of computation have been given by him. That is why we have said that the two times should be combined to get the time after sunrise. Of these, the method for computing the half-cara has been explained already. Therefore we shall explain the second part alone. The time to move from U to S in the diurnal circle is clearly the time to move from D₂ to S₂ on the celestial equator, and given by the arc D₂ S₂ which is to be got by solving the spherical triangle

IV. 34 IV. THREE PROBLEMS 109 SS₂D₂, right angled at S₂. SS₂ is the declination. Angle S₂D₂S = ZM₂ = latitude. Therefore, from the fundamental formula IV, sin D₂S₂ = Cos S₂D₂S × sin SS₂ × radius ÷ (sin S₂D₂S.cos SS₂) = Cos lat × sin dec × radius ÷ (sin lat.× cos dec.) = sin colat × sin dec × radius ÷ (sin lat.× day-diameter/2) = sin colat × sin dec × 240′ × (sin lat.× day-diameter) (From sin D₂S₂ arc D₂S₂is found and converted into time at 10 vināḍīs per degree, as mentioned before.) We shall prove the author’s method by showing that his formula is equal to this. The author’s formula is: Sin D₂S₂ = √sin² alt. at prime-vertical − sin² dec × 240 ÷ day-diameter = √sin² dec × 120² ÷ sin² lat − sin² dec × 240 ÷ day-diameter (∵ sin D₂S = sin SS₂ ÷ sin S₂D₂S, by fundamental formula II) = √(sin² dec (120² − sin² lat) ÷ sin² lat × 240 ÷ day-diameter. = √sin² dec. sin² colat ÷ sin² lat × 240 ÷ day-diameter). = sin dec × sin colat × 240 ÷ (sin lat × day-diameter) This is identical with the formula derived by us. (The author himself will be giving this form in the next verse.) We shall now show how the formula for the condensed work is got. The formula for half-cara is: Sin half-cara = sin dec × sin lat × 240′ ÷ (sin colat × day-diameter) Multiplying the numerator and the denominator of the formula arrived at by (sin lat × sin colat), we have, Sin Rt. asc. = sin declination × sin lat × sin² colat × 240 ÷ (sin² lat × day-diameter × sin colat) = sin half-cara × sin² colat ÷ sin²lat, given by us. Again, sin² colat ÷ sin² lat. = 120′ × 12 ÷ equinoctial hypotenuse² ÷ (120′ × equinoctial shadow ÷ equinoctial hypotenuse)² = 12²/equinoctial shadow² = 144 ÷ square of equinoctial shadow. So this can be substituted for sin² colat ÷ sin² lat. It must be noted that if the declination is greater than latitude, i.e. if sin dec > sin lat, then sin colat > day-diameter. Therefore sin rt. asc. > 120′, for which there is no arc, which means that at no altitude, or at no time does the Sun cross the prime vertical. This is what was referred to earlier and here shown mathematically. ‘(ख) जीन’घ्नी क्रान्तिज्या लम्बघ्नी ध्रुवगु(ण)हृदै(र्घ्यहृता) । तच्चाप(स्य) ‘रसां’शः सक[T]लः (स) दि(वस)वृद्ध्यर्धः ॥ ३४ ॥ 34. Multiply sine declination by 240 and again by sin co-latitude and divide by the product of the sine of latitude and day-diameter. Find its arc in degrees and divide by six. (The time in nāḍīs, taken by the Sun to move from

110 PAÑCASIDDHĀNTIKĀ IV. 35 the unmaṇḍala to the prime vertical is got.) Add to it the time of half-cara. This is the time from sunrise for the Sun to reach the prime vertical. The following is the work: (i) Sin (arc corresponding to time from unmaṇḍala to prime vertical) = 240 ′ × sin dec × colat ÷ (sin lat × day-diameter) (ii) The arc in degrees of (i) is to be got. Dividing by 6, the time in nāḍīs is got. (iii) The time got by (ii) + the half-cara is the time after sunrise, for the Sun to cross the prime vertical. Note that the formula here given is what we arrived at earlier, as what the author’s formula reduces to in verses 32-33. Then, why is this repetition? In the previous two verses, the work was not given clearly and fully. Here it is clear and full. Now for the reading: From the words khajinaghnī krāntijyā lambaghnī, it is clear that the product of two sines must be the divisor. Therefore, we have corrected dhruvaguna dyudairghyahṛtā into dhruvaguna-dyudairghya-hatā, which is otherwise also a better reading. Other small corrections have been made according to the idea intended to be expressed, and according to syntax. Thus it is clearly seen that in the work sin colat appears as part of the numerator, and sin lat. as part of the denominator, from which it can be seen clearly that the formula is concerned with finding the time of the Sun’s rise from unmaṇḍala to the prime vertical, and not the half-cara. The mention of the half-cara here is just to say that it should be added to find the whole time. However, both TS and NP have been misled by the mention of the expression ‘half-cara’ into thinking that the formula itself is to find the half-cara, with the result that they take the numerator as the denominator, and the denominator as the numerator, not realising that by their interpreta- tion the rule for half-cara would be a repetition, because in IV. 26 also the same has been given, and in the same form, which NP, too, have, noticed and observe: “This in fact, is only a repetition of IV. 26. It is here out of place.” (pt.II, p.43). But it may be asked whether the work according to our interpretation is not a repetition of the work of the previous two verses. We say the work as given here is clear, succinct and full. But then what is the use of the two previous verses? The work there given is easy to explain on the basis of the rule for the Rt. ascension of the ecliptic point, gone before. Or, that method perhaps is that of the Paulīśa, the author giving the same in a better form here. The example on this has already been worked out in Example 12. [समशङ्कुः तच्छाया च] उत्तरगोलेऽर्कज्या काष्ठा(न्त)गुणा ध्रुवज्यया भक्ता | ताः शङ्कुलिप्तिकाऽऽख्यास्ताभिः सममण्डल(च्छा)या || ३५ || 34a. A1. षजिनघ्नी; A2. त्रजिनघ्नी d. C.D. सकल b. C.D. लम्बहता ध्रुवगुणा. A. हितात् A. दिनवृद्ध्यर्द्धः; C. दिवसवृद्ध्यर्द्धः; c. A. तच्चापंश D. दिन[वि]वृद्ध्यर्धः

IV. 35 IV. THREE PROBLEMS 111 Great gnomon (Sama-śaṅku) and its shadow 35. When the Sun is in the northern hemisphere, (i.e. in the six signs, Aries etc.), multiply the sine of the longitude of the Sun by the sine of the maximum declination, (i.e. by 48′ 48″), and divide by the sine of latitude. The minutes so obtained are called the minutes of the ‘Great gnomon’ or Śaṅku, (i.e. sine of altitude), (and in this case, the sine of Prime vertical altitude). From this the shadow of the Sun on the prime vertical must be calculated. (i) Sin prime vertical altitude = sin Sun’s long × 48′ 48″ ÷ sin latitude. This is the Great gnomon, and the radius is the Great hypotenuse. The square root of the square of the hypotenuse lessened by the square of the gnomon is the shadow. Therefore the Great shadow = √radius² – sin ² prime vertical alt. Therefore, by the similarity between the Great shadow and the shadow triangles, we have the proportion, Great gnomon: Great shadow :: Twelve unit gnomon: shadow. From this, the required, (ii) Shadow = 12 × √120² – sin² prime vertical alt. ÷ sin prime vertical altitude. Example 13. The longitude of the Sun is rāśi 1-0. The latitude is 30°. Find the Great gnomon of the Sun at prime vertical, and thereby the gnomonic shadow at that time. (i) The Great gnomon = sin prime vertical altitude = Sin Sun’s longitude × 48′ 48″ ÷ sin latitude = 60′ × 48′ 48″ ÷ 60′ = 48′ 48″. (ii) Shadow = 12 × √120² – 48′ 48″² ÷ 48′ 48″ = 12 × 109′ 38″ ÷ 48′ 48″ = 12 × 109 19/30 ÷ (61/150) = 1644 – 30 ÷ 61 = 26 units and 58 parts, aṅgulas and vyaṅgulas The equation (i) can be written as, Sin prime vertical alt. = sin Sun’s long. × sin max. dec. ÷ sin lat. = sin Sun’s long. × sin max. dec × radius ÷ (sin lat × radius) [Fig. IV. 13] = (sin Sun’s long. × sin max. dec ÷ radius) × (radius × sin lat.) Here, it can be shown that sin Sun’s long × sin max. dec ÷ radius = sin dec., thus: Sin Sun’s long. × sin max. dec ÷ radius = sin Sun’s long. × 48′ 48″ ÷ 120′ = sin Sun’s long. × 61/150 = sin dc. (by IV. 16). Or, from Fig. 13, thus: In the triangle right-angled at R, rS is the Sun’s long. and SR is the declination of the Sun. SrR is the maximum declination. By fundamental formula II, sin rS × sin SrR ÷ radius = sin SR. ∴ sin Sun’s long × sin max. dec ÷ radius = sin dec. 35. Quoted by Utpala on BS 2, p.42 35b. A. काष्ठान्तरगुणा d. A. मण्डलछाया; U. मण्डले छाया

112 PAÑCASIDDHĀNTIKĀ IV. 37 Now we shall show that sin prime vertical alt = sin dec × radius ÷ sin lat. In Fig. 12, SZD₂ is the prime vertical, and the part D₂S is the altitude of the Sun S, and the sine of the altitude is to be found. But, sin D₂S = Cos SZ, since D₂Z = 90°, and SZ = 90° – D₂S. Observe the triangle S Z NP, right angled at Z. Here, Z NP is the co-latitude. S NP = 90° – SS₂, (∵ NP S₂ = 90°). Now, from fundamental formula i, cos SZ = cos S NP × radius ÷ cos Z NP. ∴ sin Pv alt = cos (90° – dec) × radius ÷ cos co-lat = sin dec × radius ÷ sin (90° – colat) = sin dec × radius ÷ sin lat. As stated earlier, the Sun crossing the prime vertical can occur, if at all, only when it is in the northern hemisphere, (of course for north-latitudes) and this is mentioned in the verse by uttara- gole. It should be noted that it is this sin pv. alt that is asked to be derived in IV. 32 by the statement: iṣṭottaragolāpakramāṃśakajyāṃ khabhāskarābhystāṃ hṛtvākṣajīvayā, which can be seen by comparing the work. Only, the name Sama-śaṅku (i.e. sin pv.alt.) is not mentioned there. So, the arrangement would have been better if the author had first given the formula for sin pv.alt, and then given the time of crossing the prime vertical by either IV. 34 or IV. 32-33, and, last of all, the shadow of the Sun on prime vertical. But the great transcend all restriction! Or, there is plenty of all sorts of errors committed by scribes in this part of the text, as we have reason to think. [गणकस्य योग्यता] सममण्डलले(खा)संप्रवेशवेलाः करोति योऽर्कस्य । तत्प्रत्ययं च जनयति जानाति स भास्करं सम्यक् ॥ ३६ ॥ वर्षेण भगणमर्को यदि भुङ्क्ते किं त(तो) यथेष्टदिनैः । अज्ञोऽप्येवं गणयति किं न रविं लोष्टरेखाभिः ॥ ३७ ॥ Astronomer's qualifications 36. Only he is fit to be called an expert astronomer knowing the problems dealing with the Sun, who can compute the time of the Sun crossing the prime vertical, and prove his method mathematically and graphically. 37. Even a person with very little knowledge can, by using pieces of pot- sherds, and strokes tackle (by means of computation) problems like finding the Sun's motion in a desired number of days, given the motion is twelve rāśis per year. The idea is that anybody can tackle problems depending on mere proportion. Only an expert can understand how to solve difficult problems like computing the time of the Sun's crossing the prime vertical, and prove the soundness of his method by means of graphical representations. 36. Quoted by Utpala on BS 2, p.42 b. A. वेला; P. U. वेलां. A. करोतियोर्कस्य a. A. लेषा सं० 37b. A. तयो

IV. 39 IV. THREE PROBLEMS 113 [शङ्कुच्छाया सममण्डलं च] कृतदि(ग्ग्र)हणे वृत्ते रेखां पूर्वापरां यदा छाया । प्रविशति सम्यक्छङ्कोस्सममण्डलगस्तदा सूर्यः ॥ ३८ ॥ Gnomonic shadow and the prime vertical 38. On a circle with the east-west line drawn, and the directions marked, (according to IV.19), the time when the gnomonic shadow perfectly coin- cides with the east-west line is the time of the Sun crossing the prime vertical. The idea is that if this time is found by measuring instruments, compared with the computed time and the agreement shown, people will acquire faith in the method. It can be shown that when the Sun is on the prime vertical, the gnomonic shadow must be along the east-west line. The prime vertical is the vertical great circle of the sky-sphere, passing through the east-west points and the zenith, and therefore the east-west line forms the intersection of this vertical plane and the plane of the horizon which is horizontal. As the gnomon standing vertical and also the Sun on the prime vertical lie in the vertical plane, the shadow (intercepted by the hori- zontal plane) must also lie on the vertical plane, and therefore must fall on the east-west line, which is the intersection of the two planes. [अग्रा-दिग्ज्या] इष्टक्रान्तिज्या(घ्न)व्यासशकललम्बकांशमुष्णांशुः । समपूर्वापररेखामतीत्य यात्यस्तमुदयं वा ॥ ३९ ॥ Agrā : Sine amplitude 39. Multiply the Sun's declination by the radius and divide by the sine of co- latitude, and find the sine (of the amplitude of the rising or setting point, called Agrā). At a point distant by this amount from the east-west line (accord- ing to the declination, north or south) the Sun rises or sets. Agrā, (i.e. sine amplitude) = sin dec × radius ÷ sin colat. Find the arc of this sine. By an angle equal to this from the east to west point does the Sun rise or set on the horizon. Example 14. The latitude of a place is 60°. The longitude of the Sun is rāśi 4-0. Find the direction of rising or setting of the Sun. First, sin declination is to be found. As the Sun is in the second quadrant, Sin rāśi 4-0 = Sin rāśi 2-0 = 103' 55". Sin dec = 103' 55" × 61/150 = 42' 15", and this declination is north. Sin amplitude = 42' 15" × 120' ÷ sin (90° - 60°) = 42' 15" × 120' ÷ 60' = 84' 30".

38. Quoted by Utpala on BS 2, p.41. 38a. A. कृतिदिग्रहणे c. A. शङ्कुः 39a. A. ज्याघ्ना; D. ज्याघ्नं b. A. व्यासकल; D. व्यासशक (लं) लम्बभक्तमुष्णांशुः d. A1. मतीत्या; A2. मलिप्त corrected to मलीप्त ।

114 PAÑCASIDDHĀNTIKĀ IV. 40 The arc of 84' 30" is 44° 46'. Therefore the Sun rises at a point 44° 46' north of the east point, and sets at a point 44° 46" north of the west point, (assuming that the declination has not changed). The formula for amplitude is got thus: See Fig. 12. Take S₁ as the Sun on the diurnal circle north of equator. Then S₁S₂ is the declination of the Sun, D₁D₂ is the amplitude of sunrise. From the figure it can be seen, D₁D₂ = 90° − ND₁ Therefore sin D₁D₂ = Cos ND₁. From the right angled triangle ND NP in the figure, Cos ND can be got thus: By the fundamental formula I, Cos ND₁ = 120' × Cos D₁ NP ÷ Cos N NP. But, Cos D₁ NP = Cos S₁ NP = Cos (90° − S₁S₂) = sin S₁S₂ = sin S₁S₂ Cos N NP = sin (90° − N NP) = sin colat. ∴ Sin amplitude = 120' × sin dec ÷ sin colat. When the Sun is south of the celestial equator, (e.g. S₃ in the figure), D₃ is the rising point, and D₂D₃ is the amplitude. Its sine is got thus, from the triangle, D₂D₃E, right-angled at E. By the funda- mental formula II, Sin D₂D₃ = sin ED₃ × radius ÷ sin angle D₃D₂E, Here, ED₃ is the declination. D₃D₂E = 90° − ED₂Z = 90° − lat. ∴ sin amplitude = sin declination × radius ÷ sin (90° − lat) = sin dec × radius ÷ sin co-lat., which is the formula given. [अग्राया अक्षानयनम्] तेन हता ऽखार्कऽघ्नी क्रान्तिज्या लम्बकोऽस्य [य] (च्चा) पम् | तेन नवतिर्विहीना (यच्छेषं) तेऽक्षभागाः स्युः || ४० || Latitude from Agrā 40. Multiply sine declination by 120 and divide by the sine of amplitude. The sine of co-latitude is got. Find its arc in degrees. Deduct the degrees from 90. The remainder are the degrees of latitude. Now, sin co-lat = 120' × sin dec. ÷ sin amplitude. From this, the arc, co-lat is got. 90° − colatitude = latitude, as already stated in (IV. 28). From the formula of the previous verse, Sin amp = radius × sin dec ÷ sin co-lat, Sin colat = radius × sin dec ÷ sin amp. = 120' × sin dec ÷ sin amp. From the amplitude of the setting Sun also, the latitude can thus be found. The amplitude can be marked on a circle with the directions already marked by the observation of sunrise or sunset. 40a. A. हृता. A. खार्कघ्नी b. A1. कोस्य श्रापम्; (A2. स्प corrected to स्य) c-d. A. हीना छयेघतेक्षभागाः