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

IV. 9 IV. THREE PROBLEMS 81 शेषज्याः 'स्वरतिथयो' 'गुण-शिव-धृति'भिश्च 'विंशतिः' सहिता । 'पञ्चनरकं' 'शतार्धं त्रिसमेतं' 'षष्टि'रिति लिप्ता ॥ ६ ॥ सैकाऽजे पञ्चाशत् 'पञ्चाष्टकं' 'पञ्चवर्गवेदा' श्च । 'त्रिंशच्चतुर्भिरधिका' 'षट्पञ्चाशच्छराः' शून्यम्' ॥ ७ ॥ 6-7. The other sines are the following: In the first sign, the minutes parts are successively 7, 15, 20 + 3, 20 + 11, 20 + 18, 5 × 9, 50 + 3, and 60. The seconds, respectively, are: 50 + 1, 5 × 8, 25, 4, 30 + 4, 56, 5 and 0. Thus we have for the first sign –

82 PAÑCASIDDHĀNTIKĀ IV. 11 Adding the minutes and seconds, and 60' for the end of the first sign, the sines are:-

IV. 14 IV. THREE PROBLEMS 83 there is one second more. By this mistake the 24th, i.e. the radius, has become 120' 1", and even this obvious mistake they have failed to note. The promised sine-intervals are here given: मुनयोऽब्जे व्येकान्ते 'रसत्रयं' 'पञ्च(कौ)' 'कृता' (श्र) गवि । 'शिखि-पक्षचन्द्र-शून्याः' द्वित्रिमिथुने कला ज्या [नाम्] ॥१२ ॥ मेषे विकलार्धशतं सैकं 'व्येकेन्द्रि(ये)-श्वरं त्रिंशत् । (द्वा)विंशतित्रिवर्गः ------------------------------ ॥ १३ ॥ ---------------------------------------------------- । ------ 'खगुणकृतार्णवयमनवकसमुद्रशिखिवर्गैः ॥ १४ ॥ 12. Of the intervals the minute parts are, in the first sign, 7, 7, 7, 7, 7, 7, 7, 6; in the second sign: 6, 6, 6, 5, 5, 4, 4, 4; and in the third sign, 3, 3, 2, 2, 1, 1, 0, 0. 13-14. The seconds in the first sign are, 50 + 1, 50 − 1, 50 − 5, 50 − 11, 30, 22, 9, (and here is a break resulting in the loss of the 4th foot of the 13th verse and the first two feet with 4 mātrās of the 3rd foot of the 14th verse. From the values of the sines we can compute that the seconds in the 8th interval must be 55, which must have been given in the missing part. By examining the remaining part we can construct the meaning of this verse thus). The seconds of the intervals in the second sign are, respectively 10 × (4, 2, 0, 4, 2, 5, 3, 0)

• (0, 3, 4, 4, 2, 9, 4, 9), added each to each. TS have not been able to see that in what is left of the 14th verse, the digits in the unit places of the eight numbers giving the seconds of the intervals of the second sign are given. This is because they have made mistakes in the 14th and 16th sines resulting in an error of + 1" in the 14th interval, − 1" in the 15th and + 1" again in the 16th. This has prevented them from finding the correct values by comparison, so that they make many changes in the readings here, saying that the text is very corrupt, here. It may be seen that not a single word is wrong here. 12a. A. मुनयोज्ये; B. गुनयोज्ये to 14c. D suggests for 13d [पञ्चाशच्च विषयसंयुक्तम्] b. A. त्रयं को; B. त्रयपञ्चको;. C. त्रयं (त्रिः) शशः C. indicates the gap by dots as done by A. कृताच्चे गवि; B. कृता-गेवि; C. कृताब्धी गवि; us. D suggests for 14: D. कृताग्निगवि ख[समुद्र] गुण [द्विकृताः] .......... c. A. शिखिपकृत चन्द्रः; B. शिखिपक्ष कृतार्णव [द्वि] यमनवे [न्द्रि] यसमुद्रशिखिवर्गे[शाः] || d. C. द्वौद्विमिथुने. A-B. ज्या |; C. ज्यार्द्धे; D. ज्या [सु] 13a. A. विकलार्द्धशत° d. A. समुद्रा b. B1. सैकां; B2. सैव्यं; B1.2. वर्गे; B2. वर्गे A1.B1. °न्द्रिय स्वरं C. खगुणकृतार्णवयमनवसकसमुद्रा शिखिवर्गैः (?) || c. A1.B2. द्विविंशति 13d. A.B. Unindicated gap of 13d. upto खगुण of 14c.

84 PAÑCASIDDHĀNTIKĀ IV. 15 ‘मनुविषयतिथिरसा:’ स्युस्त्रिगुणाः पञ्चाष्टकं ‘स्वरोपेतम्’ | सप्तदशनवपञ्चकं षोडश चेति क्रमान्मिथुने || १५ || 15. The seconds of the intervals in the third sign, are, 3 × 14, 3 × 5, 3 × 15, 3 × 6, 5 × 8 + 7, 17, 9 × 5 and 16. The intervals can be got by deducting the previous sines from the succeeding ones, and com- pared with the author’s concern for correctness, they are correct. It is to be noted that the intervals of TS also come correct in the third sign, because there is a uniform error of 1″ in all sines. The intervals are as tabulated:

IV. THREE PROBLEMS 85 Fig. IV. 3 AK = sine AG (neglecting the first sign) because, EF, HJ, JK, GF, are all equal, as already men- tioned. Therefore, for an arc or angle in the second quadrant, (3 to 6 signs), deduct it from six signs and get the sine of the remainder. For that in the third quadrant, (6 to 9 signs), deduct six signs and find the sine of the remainder. For that in the fourth quadrant, deduct it from twelve signs and find the sine of the remainder. Example 2 (a). Find the sine of the arc or angle equal to 4 signs. It is in the second quadrant. Therefore, sine 4 signs = sine 6 signs − 4 signs = sine 2 signs = 103′ 55″. Example 2 (b). Find the sine of signs 7-11-15. This is in the third quadrant. Therefore sine of sign 7-11-15 = sine (7-11-15 − 6-0-0) = sine 1-11- 15 = 79′ 7″. Example 2 (c). Find the sine of signs 9-15-0. This is in the 4th quadrant. Therefore sine of signs 9-15-0 = sine (12 signs − sign 9-15-0) = sine 2-15-0 = 115′ 55″. Declination From here to the end of the chapter, problems based on the solution of spherical triangles are dealt with, being problems involving position, time and direction. As a preliminary, the declina- tion of the Sun and the Moon are required, which are given first. Stellar sphere (Bhagola) The ancient astronomers speak of three spheres, the Terrestrial sphere or Earth sphere on which we live, the Stellar sphere or the Sphere of the stars, and the Sky sphere or the Sphere of the sky. We shall describe the terrestrial sphere in connection with the Sāura chap. XIII-XV. Of the other two, we shall now describe the stellar sphere, a knowledge of which is immediately required. It is the apparent sphere on which the stars appear to be fixed, and form a frame of reference for

86 PAÑCASIDDHĀNTIKĀ IV. 15 the motion of bodies like the Sun, Moon, planets etc. Actually the stars are at widely different dis- tances from us and are moving in various directions at speeds of several miles per hour. But with all this they appear to be, and can be represented, as being fixed on a sphere, because of their enormous distances from us, so much so that we are practically viewing the same sphere as people several generations ago did. But the ancients believed that the stars were luminous bodies fixed on the under-surface of a sphere of radius only sixty times that of the orbit of the Sun (actually the earth) round the earth, with the centre of the sphere at the earth's centre. This sphere seems to be rotating about once a day, by the actual rotation of the earth, about once a day. Fig. IV. 4 On this sphere (Fig. 4) there is an important great-circle called the ecliptic (Krānti-vṛtta) (EC), marked by the twenty-seven asterisms, Aśvinī etc., on which the Sun (S) moves, (actually appears to move on account of the motion of the earth), completing a revolution once a year. The Moon and the planets move in orbits inclined to the ecliptic at small angles. Their longitudes are reckoned along the ecliptic from a fixed point called the first point of Meṣa or Aśvinī. Latitudes are measured on secondaries to the great circle, meeting at a point called the pole of the ecliptic (Kadamba). Another great circle called the Celestial Equator (AOB) cuts the ecliptic at r, called the First point of Aries or Vernal Equinox point, which, instead of being fixed, has a slow westward motion on the ecliptic. At the period of the authors of the first Siddhāntas, this point coincided with the first point of Meṣa, this being the reason why that particular point was taken by them for reckoning from. The angle between the two great circles (SrR) is called the Obliquity of the ecliptic. It was about 24° at the time of Varāhamihira, and now it is about 23° 27'. The declination (SR,) (Krānti) of a body like the Sun, is measured along the secondary passing through the body, (NPSRSP,) called the Declina- tion circle, all the secondaries meeting at the celestial poles, the poles of the stellar sphere, (SPNP), on the axis joining which the sphere apparently rotates. The declination is found by solving the spherical right angled triangle SrR. Spherical triangles were in general solved by the Hindu astronomers using the properties of plane right angled triangles formed by the sections of the sphere along the great circle arcs forming the spherical triangle (The

IV. 16 IV. THREE PROBLEMS 87 Greeks solved them by an extension of Manelau's Theorem to figures on the sphere.) We shall con- tent ourselves with giving the formulae for solution and refer to them whenever necessary by way of proof. Let ABC (fig. 5) be a spherical triangle, right angled at C, and R the radius of the sphere. I. Cos AB = Cos AC × Cos BC ÷ R II. sin BC = sin AB × sin A ÷ R III. Cos A = R × Cos AB. sin AC/sin AB. Cos AC IV. Cos A = Cos BC × sin B ÷ R V. Cos AB = R. Cos A × Cos B/(sin A × sin B) These, together with the general identities, cosθ = sin (90° – θ), sin² θ + Cos² θ = 1, will suffice for explaining the formulae occurring in the text. [Fig. IV. 5] [रविचन्द्रयोः क्रान्तिः] जीवाऽध्यर्धशतां' (शघ्नै) काषष्टि (र्दि) नेशकाष्ठा (ज्या) । चन्द्रस्य सविक्षेपस्तदपक्र(मो) राशिपादे(भ्यः) ॥ १६ ॥ Declination of the Sun and the Moon 16. The sine of Sun's declination is found by multiplying the sine of its longitude by 61 and dividing by 150. Its arc is the declination. The declination of the Moon found thus is the mean declination. Its true declination is the mean declination plus latitude. The intervals of declinations for intervals of quarter signs are given (in the next two verses, 17-18). This is the formula: sin dec. = sin sāyana long. × 61/150. From this the arc forming the declina- tion is found by using the tables, and then the true declination of the Moon, using this. It must be noted that when the longitude reckoned from r (i.e. sāyana long.) is within 6 signs, the declination is North, and when more than 6 signs it is South. In the case of the Moon, if the declination and the latitude are of the same direction they should be added to get the true declination. If they are of different directions, their difference is the true declination, its direction being that of the greater. The author has not mentioned this because it is obvious. Example 3 (a). Find the maximum declination of the Sun. Obviously, the maximum sine of the Sun's longitude (i.e. when the Sun is 3 signs or 9 signs) will give the maximum declination. It is therefore given by sin dec. = 120' × 61 ÷ 150 = 48' 48". Its arc, 24° is the maximum declination. 16a. A. जीवाध्यार्द्ध०; B. जीवा व्या-र्द्ध; C. जीवाव्यर्द्ध; D. [साङ्कुलिप्ता] . A.B. काष्टान्तः; C. काष्ठान्तः; D. जीवाव्यध्यर्ध. B. सिताशा; A.C.D. शतांशाः D. काष्टान्तः b. A. सैकाःषष्टि; B. सैका षष्ठि; C. सैका षष्टि; d. A.B.C.D. तदपक्रम. A.C. ०पादेन्यः

88 PAÑCASIDDHĀNTIKĀ IV. 17 Example 3 (b). Sāyana Sun is rāśi 4-7-30. Find its declination. Sin 4ʳ 7° 30′ = Sin (6ʳ-0-0 – 4ʳ-7-30) = sin 1ʳ-22-30 = 95′ 12″. Sin Declination = 95′ 12″ × (60 + 1) ÷ 150 = 95′ 12″ × (2/5 + 2/5 × 60) = 38′ 5″ + 38″ = 38′ 43″. Its arc, 18° 50′, is the declination. Since the sāyana Sun is within 6 rāśis, the declination is north. Example 3 (c). The sāyana Moon is 9ʳ0°0′. Its latitude is 4°N. Find its declination. The sāyana longitude being 9ʳ0° 0′, the mean declination is the maximum, south, i.e. 24°S. Its lat. is 4°N, i.e. of opposite direction. ∴ the true declination is 24° – 4° = 20° S. South because South is greater. The author uses the word kāṣṭhā to signify declination, which is uncommon. Sometimes this word itself is used to mean sine declination. The rule for sin declination is explained thus: Our siddhāntas take the maximum declination to be 24°. As the maximun declination occurs when the sāyana longitude is 3 signs, the angle between the ecliptic and the celestial equator (i.e. the obliquity of the ecliptic) also is 24°. In Fig. 5, take AB and AC as parts of the ecliptic and celestial equator. Then, A = 24° and AB is the sāyana longitude. BC is the declination wanted. By formula II under the present verse, sin dec. = long. × sin 24° ÷ 120′. But sin 24° ÷ 120′ = 48′ 48′′ ÷ 120′ = 61/150. Hence, sin dec = sin long × (60 + 1)/150, which is the given rule. Now for the direction: See Fig. 4. At r the Sun moving along the ecliptic crosses the celestial equator, and passes from South to North. As great circles bisect one another, till the longitude is 6 sings it moves north of the celestial equator, for which declinations are reckoned, and then moves south of it. Therefore for longitude 0 to 6 rāśis, the declination is North, and for longitude 6 to 12 rāśīs, it is South. As the Moon and other planets move in their own orbits inclined to the ecliptic, the declinations computed from their longitudes reckoned along the ecliptic are only approximate. To get correct declinations, their distances north or south from the ecliptic points, called their ‘latitudes’, should be combined in the proper manner as instructed. But the result by thus adding or subtracting will be only approximate, because the latitudes are directed towards the pole of the ecliptic (Kadamba), while the declinations are directed towards the Celestial Pole. The maximum error that can occur thus is about 24′. Combined with the error in latitude due to other factors like proportion by degrees of the argument of latitude (advocated by the Pauliśa) instead of the sine etc., the error will be considerable. Now, for the readings. For syntactical purposes, and getting the proper meaning, śatārmśāśsaikā has been corrected as śatāṁśaghnaikā, ṣaṣtidineśa as ṣaṣṭirdineśa, kāṣṭhānta as kāṣṭhā jyā, apakramarāśi as apakramo rāśi and pādenyaḥ as pādebhyaḥ. As for TS, they have not touched this verse and the next two, saying, in so many words, that they cannot interpret them. NP’s interpretation of the three verses has also been affected by the highly corrupt text. लिप्ताशत [क] म (शीत्या) मेषे 'त्रिख' यु (क्त) 'मिन्द्रिय' 'मनू' (न) म् । गवि ('मनु') 'भव' 'मुनि' 'रूपै' - श्र [तु] गुणैः संयुतं च शतम् ॥ १७ ॥

IV. 18 IV. THREE PROBLEMS 89 नवतिस्त्रियुता षष्टिश्चत्वारिंश'च्छिवा'श्च मिथुना(न्ते) । मेषाद् गताऽऽगतमुदग्दक्षिणतोऽदस्तूलादिषु च ॥ १८ ॥ 17-18. The (promised) intervals of declinations in minutes for intervals of quarter-signs, are, in Sāyana Meṣa: 180 + 3, 180 + 0, 180 − 5, 180 − 14, in Sāyana Vṛṣabha, 100 + 4 × 14, 100 + 4 × 11, 100 + 4 × 7, 100 + 4 × 1, and in Sāyana Mithuna, 90, 63, 40 and 11. (Thus the intervals are 183, 180, 175, 166, 156, 144, 128, 104, 90, 63, 40, 11.) These are to be added successively to get the declinations from Meṣa (Aries) to Mithuna (Gemini). Then from Kar- kaṭaka (Cancer) to Kanyā (Virgo), these should be deducted in the reverse order, until at the end of Kanyā, the declination is zero. These declinations from Meṣa to the end of Kanyā are north. Then from Tulā (Libra) to the end of Dhanus (Sagittarius), the south declinations increase in the given order, and from Makara (Capricorn) to Mīna (Pisces) the south declinations decrease in the reverse order, until at the end of Mīna the declination is zero again.) Example 4. Find the declination of the ecliptic point ending (Sāyana) Capricorn. The end of Capricorn is rāśi 10-0-0. This falls between rāśis 6 and 12. ∴ The declination is south. The declination ending Sagittarius (i.e. beginning Capricorn) is 24° S. The declination at the end of Capricorn is 24° − 11′ − 40′ − 63′ − 90′ = 24° − 3° 24′ = 20° 36′S. The author has perhaps computed the declination by applying the formula of verses 16 and got the intervals by deducting the previous from the next. Or, these verses are taken in toto from the original Pauliśa and given here, for there are small differences from the computed values. Or, the differences are scribal errors. Both are given hereunder for comparison:

90 PAÑCASIDDHĀNTIKĀ IV. 19

19. (Plant a gnomon at the centre of) a circle having a diameter equal to four times the gnomon. Mark the two points where the shadow of the gnomon enters the circle and emerges from it. The line joining the points is the east- west line. The line drawn perpendicular to this by means of equal intersecting circles, is the north-south line. Though the east-west line is first asked to be drawn and the north-south next, as perpendicular bisector to the east-west, it will be better if the north-south line is first drawn by means of equal intersecting circles with the two points as centres. Then using the points of intersection of the north-south line and the original circle as centres, by the same means, the perpendicular bisector forming the east-west line can be drawn, which will pass through the centre as required. On the other hand, if the east-west line is first drawn by joining the first two points, another line parallel to it and passing through the centre is to be drawn as the desired east-west line. 19a. A. संकुश्रतुविस्तारे; C. शङ्कइगुलविस्तारे in V. 9c. as noted by the scribes. c. A. अपेरैद्री; C.D. अपेरैन्द्री Thus, B1 adds in the margin, प्रति पत्र एक; d. A. यवाश्च; C.D. यवैश्च B2. adds, प्रतिपत्र एके, and B3. adds प्रति नूं ∙ B. After कार्ये one leaf missing upto दिवा पत्र एक. And all add अग्रो नास्ति ।

IV. 21 IV. THREE PROBLEMS 91 The gnomon should be twelve units in length, not necessarily digits; no harm will result, pro- vided all measurements are given in the same units. The circle also can be of any desired diameter, not exactly four gnomons in length. We are not sure whether the word for ‘four’ occurs at all in the text, it is so corrupt in that part. We can only say it cannot be śaṅkvaṅgula as corrected by TS; the letters are so different. Finding the directions in the manner described is explained thus: The North is directed towards the north pole of the earth. Corresponding to this is the celestial north pole, (from which we can find the north, if we can only observe it correctly, and therefrom the other directions). At mid-day the Sun is on the meridian, and at equal times before and after, its altitudes and directions are equal, provided its declination does not change. As the Sun’s position is thus symmetrical, before and after noon, with the meridian as the line of symmetry, the gnomonic shadow is symmetrical with the north-south line (which corresponds to the meridian) as the line of symmetry. Therefore if two gnomonic shadows, one in the morning and one in the evening, of equal lengths, are marked on a horizontal surface, the bisector of the angle between the two shadows is the line of symmetry, and therefore the north-south line. From this the east-west line, which is its perpendicular bisector, is drawn. The circle, asked to be drawn, serves the purpose of marking the equal shadows. By the same symmetry, the ends of the shadows are at equal distances from the east-west line, and so the line drawn between them is also east-west, being parallel to the east-west line drawn. Therefore the author asks us to draw the east-west line formed by joining the two points first, and proceed. We have said that the Sun’s declination must be the same, i.e. does not change during the interval. But actually it changes. So if we do the work at a time of the year when the change in declination is very little, then the directions found will be nearly accurate. This happens near the solstices, and there- fore the work should be done when the sun is near the solstices. Methods to find the directions accu- rately even when the declinations are changing rapidly, are given by writers like Vaṭeśvara, Parameśvara etc., and also explained by Govindasvāmin in his commentary on the Mahābhāskarīya, III.1. [छायातः अक्षानयनम्] विषुवद्दिन [सममध्य] छायावर्गात् सवेदकृतुरूपात् | मूलेन शतं विंशं विषुवच्छायाहतं छिन्द्यात् || २० || लब्धं विषुवज्जीवा चाप [म] तोऽक्षोऽ [थवैव] मिष्टदिने | मेषाद्यपक्रमयुतस्तुलादिषु विवर्जितः स्वाक्षः || २१ || Latitude from Shadow 20. Measure the mid-day shadow on the day when the Sun is at the equinoxes (the equinoctial shadow), square it, add 144, and find the square root. By this divide the product of the shadow into 120. 21. The result is the sine of the latitude of the place, called Viṣuvajjīvā (or Viṣuvajyā). Its arc is the latitude. Or, do this work on any day and get the arc. If the Sun is in the six signs from sāyana-Meṣa, i.e. if the Sun’s declination is north, add the declination to the arc, the latitude is got. If the Sun is in the six

92 PAÑCASIDDHĀNTIKĀ IV. 21 signs from sāyana-Tulā, i.e. if the declination is south, subtract the declina- tion from the arc, the latitude is got. That is: i. Sine latitude = 120' × equinoctial shadow ÷ √(144 + equinoctial shadow²). The arc from this is the latitude. ii. Sine south zenith distance of the Sun, (SZD) = 120' × mid-day shadow ÷ √(144 + midday-shadow²). The arc of this is the SZD. Using SZD, Latitude = SZD±declination ('plus' should be used if the declination is north, and minus if south.) Example 5 (a). At a certain place the equinoctial shadow is 5 units. Find the latitude of the place. Sin. lat. = 5 × 120' ÷ √(5² + 144) = 600' ÷ 13 = 46' 9''. Arc of 46' 9'' = 22° 37'. The latitude is 22° 37'. Example 5 (b). At a place when the Sun is at the end of sāyana Tulā, the mid-day shadow is found to be 9 units. Find the latitude of the place. From the formula, (the mid-day-Sun's) sin SZD = 9 × 120' ÷ √(9² + 144) = 1080'/15 = 72'. SZD = arc of 72' = 36° 53'. The declination of the Sun at the end of Libra is 11° 44'S (from 16-18). Taking the minus sign, since the declination is south, the latitude = 36° 53' − 11° 44' = 25° 9'. Note: Rule (i) can be used everywhere, while rule (ii) should be used only if the midday sun is south of the zenith. If it is north, having north zenith distance, (NZD), declination = NZD = latitude. But the work being a Karaṇa, the author intends it to be used only in North India, where the midday zenith distance is always south, and hence this has not been mentioned by him. Further the author envisages only north latitudes by his formulae. Sky-sphere (Khagola) Here onwards, explanations require a knowledge of the sky-sphere (khagola) with the stellar sphere imposed on it. Therefore we shall describe the sky sphere. Hindu astronomers describe it as the ‘Casket Boundary of our universe’ (Brahmāṇḍa-kaṭāha-sampuṭa), and marked by the penetration of sunlight. Beyond that there is no sunlight. The measure of a great circle on the sky-sphere is said to be the number of yojanas the Moon, or the Sun or any planet moves in a kalpa (yuga according to the followers of Āryabhaṭa), — though, really, the sphere is only illusory and supposed to have an indefinite radius. As the stellar sphere also is enormous, we can take the surfaces of the two spheres sliding on each other, and forming spherical triangles by arcs on each intersecting those on the other. The problems will entail the solution of these triangles. (The formulae for solution have already been given). This will be understood by examining Fig. 6. 20-21. Quoted by Utpala on BS 2. pp. 59-60. 20a. A.सममधृष्टो; C.दिनमध्याह्नो                          d. A.C.D.छिंद्यम्     b. A. om-रूपात्र-om.                            12b. A. चापतोक्षो. A.C.D. ॰थवा यथेष्टदिने     c. A. शते. A.D.विंशात्                            d. A. खोक्षः

IV. 21 IV. THREE PROBLEMS 93 Fig. IV. 6 NESWZ is the sky-sphere. It is the sky as seen by an observer on the earth who fancies himself stationary, though taking part in the rotation of the earth, and thinks that the stellar sphere is rotating on the axis joining the celestial poles, NP, SP. NESW is the horizon marked by the cardinal points, North, East, South, West. (Note that the east and west points are interchanged so as to appear as we see them when looking up at the sky.) Z is the zenith, corresponding in the sky to the observer's position on the earth, and is the point where the line from the centre of the earth, through the observer, joins the sky-sphere. NNPZOS is the meridian. EZW is called the prime vertical. ENP is part of what is called unmaṇḍalam. ZSH is part of the vertical circle from the zenith to the horizon passing through the sun, moon, etc. (Note that the prime vertical is the vertical circle passing through the east and west points, and that the meridian itself can be considered as a vertical circle passing through the North and South points.) ZS is the zenith distance of S, and HS is the altitude. S₁SS₂ is the diurnal circle, the apparent path of S daily, due to earth's rotation. The stellar sphere (Fig.4) can be recognised here by the celestial equator ErROW, by the ecliptic EcrSC, by the north celestial pole NP, by the position of the Sun S, etc. and by the declination circle. NPSR, of which SR is the arc of declination. Because of the position of the observer on the earth with reference to the terrestrial North pole, the celestial North pole (NP) seems lifted up along the meridian from the north-point (N) so that its altitude is equal to the latitude of the place, and by this the celestial equator is depressed southward by the same amount from the prime vertical.

94 PAÑCASIDDHĀNTIKĀ IV. 21 Therefore the latitude of the place = NNP = ZO. (What we have said is for places in the northern hemisphere, i.e. north latitudes. In the southern hemisphere, i.e. at places of south latitudes, SP is lifted up from S, and the celestial equator is depressed northward by the same amount.) The complement of ZO, OS, is called the co-latitude (Lamba). Thus in triangles formed by great-circle- arcs of the stellar sphere and the sky-sphere, the latitude is involved directly or indirectly. The for- mulae for the solution of these triangles have been already given. Now, the two formulae for latitude can be proved by using the meridian, thus: see Fig.7 Ob:observer N: north point S: south point NS₂ZS₁OS₀S₃: The meridian Z: zenith O: point of intersection of meri- dian and celestial equator. S = S₂, S₁, S₀, S₃,: four positions of the mid-day sun. OS: Sun's declination Fig. IV. 7 As already described, OZ = latitude. On the equinoctial days at mid-day the Sun, S₀ is at O. ∴ ZS₀ (the south zenith distance of the Sun) = ZO = latitude (first formula). On other days, the Sun may be (i) south of O, (S₃), or (ii) north of O but south of Z, (S₁), (iii) north of O and north of Z, (S₂). i. Here, the latitude = OZ = S₃Z − S₃O = the south zenith distance of the Sun − the declina- tion. (second part of second formula). ii. Here the latitude = OZ = ZS₁ + S₁O = the south zenith distance of the sun + the declination (first part of the second formula). iii. Here the latitude = ZO = S₂O − S₂Z = the declination − the north zenith distance of the Sun. (This case is not given by the author). The zenith distance of the midday Sun used in the formulae is to be found thus: see Fig.8. EG = gnomon of 12 units ET = The midday shadow, TG = The shadow hypotenuse, ZGS = the zenith distance = angle TGE. Sin zenith distance (ZD) = sin ZGS = sin TGE = TE × 120′ ÷ TG = Shadow × 120′ ÷ Shadow hypotenuse = Shadow × 120′ ÷ √(shadow² + gnomon²) = Shadow × 120′ ÷ √(shadow² + 144)., (where shadow is in the units taken). From sin ZD, arc ZD is found. Fig. IV. 8

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