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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

IV. 43 IV. THREE PROBLEMS 115 The sine of amplitude to be used in the formula can be got by measuring the arc of amplitude, or the distance of the point, from the east-west line. Example 15. Sine declination is 42′ 15″. Sine amplitude is 84′ 30″. Find the latitude. From the formula, sin. colat = 120′ × 42′ 15″ ÷ 84′ 30″ = 60′. Colatitude arc of this, i.e. 30°, latitude = 90° − 30° = 60°. [इष्टकालच्छाया] तत्कालचरविनाडीद्विदशांशं द्विष्टमजतुलाद्येषु । [षड्घ्नी] भ्यो नाडीभ्यो जह्यात् संयोजयेच्चाऽपि ॥ ४१ ॥ तज्ज्या स्थितज्यया संयुता विसंयोजिताऽ [जतु] लाद्येषु । अविशोधने (च) जीवा षड्घ्नीनामे [व] कर्तव्या ॥ ४२ ॥ एवं कृत्वा हन्यात् द्युव्यासेनाऽवलम्बकघ्नेन । छिंद्यात् ‘खखाऽष्टवस्वश्विभिः’ फलं शङ्कुलिप्ताख्यम् ॥ ४३ ॥ Shadow at desired time 41. To find the gnomonic shadow caused by the Sun at any time: Take the cara in vināḍis and divide by 20. Degrees of half-cara are obtained. Place the degrees in two places. Convert the time from sunrise in nāḍis into degrees by multiplying by 6. From these degrees, deduct or add the half-cara degrees according as the sun is in the six signs beginning with Meṣa or in the six signs beginning with Tula, respectively. 42. Find the sine of the resulting degrees, and add or subtract this from the sine of the half-cara kept apart in the second place, according as the Sun is in the 6 signs Meṣa etc., or in the six signs Tulā etc. (The result is a sine. If the half-cara degrees cannot be deducted from the time converted into degrees, then simply find the sine of the degrees of sine, and take it for further work.) 43. Multiply this sine by the sine of colatitude and the day-diameter and divide by 28,800. The result is sine altitude of the Sun. 41-44. Quoted by Utpala on BS 2, p.61. 41b. A.सदशांशां दिष्टमज c. A.षघ्नाभ्यो 42a. A.तज्या स्थिज्यया b. A.योजिताद्येषु; C.D. योजिताजतुलाद्येषु c. A.°नेन झीवा; C-D. U.°नेन जीवा d. A.षन्नानामेषकत्र्तव्या (A2. षड्घ्नी) 43a. A.कृत्वा हन्या b. A.द्युव्योमेनाव; U. लम्बघ्नेन A. Haplographical omission of 43 c-d, 44 and 45a-b: लम्बकघ्नेन [... लम्बकघ्नेन] छिंद्यात् Hence they are added here from Utpala's quotaton thereof. 10

116 PAÑCASIDDHĀNTIKĀ IV. 44 [तत्कृतिविना (कृ) तानां 'खखवेदसमुद्रशीतरश्मीनाम्' । पदमर्कघ्नं शङ्कङ्गङ्गुलाऽऽख्यलिप्तोद्धृतं छाया ॥ ४४ ॥] 44. Square this and deduct from 14,400. Take its square root, multiply this by twelve, and divide by sine altitude. The result is the length of the shadow of the twelve-digit gnomon. The following are the steps in the work: (i) Sine altitude = {sine (degrees of the ∓ degrees of half-cara) ± sine half-cara} × sin colat × day-diameter ÷ 28,800. (Here, of ∓ or ±, the upper sign should be taken for the 6 signs Meṣa etc., and the lower for the 6 signs Tulā etc.) (ii) The shadow = 12 × √14,400 − sin ² altitude ÷ sin altitude Example 16 (a). At a certain place where the sine of the co-latitude (i.e. cos. lat.) is 103′ 55″, when the Sun is in the 6 signs from Meṣa on a particular day, the cara is 200 vināḍis, and the day-diameter is 229′ 51′. Find the length of the shadow at 8 nāḍīs from Sunrise. (i) Degrees of half-cara = 200 ÷ 20 = 10°. Degrees of time = 8 × 6 = 48°. As the Sun is in the six signs from Meṣa, deducting 10° from 48°, we get 38°. Sine 38° = 73′ 35″. The sine of the half-cara, i.e. sin 10° = 20′ 50″. Adding the two signs, (since the Sun is from Meṣa), 73′ 35″ + 20′ 50″ = 94′ 25″. Sine altitude = 94′ 25″ × 229′ 51″ × 103′ 55″ ÷ 28,800 = 78′ 19″ (ii) The shadow = 12 × √14,400 − 78′ 19″² ÷ 78′ 19″ = 13 aṅg 56 vyaṅ. Example 16 (b). At the same place, on the same day, find the shadow at one nāḍī after sunrise. (i) The degrees of half-cara (already found) = 10°. The degrees of time = 1 × 6 = 6°. The half- cara degrees have to be deducted, but cannot be deducted, being greater. Therefore, taking the sine of the 6° alone, we have 12′ 32″. Sin altitude = 12′ 32″ × 229′ 51″ × 103′ 55″ ÷ 22,800 = 10′ 24″. (ii) shadow × 12 × √14,400 − 10′ 24″² ÷ 10′ 24″ = 137 aṅgulas 57 vyaṅgulas. But it should be mentioned here, that the author's instruction for the case when the degrees of half-cara cannot be deducted from the degrees of time, will give only a rough result. This will not matter much in places where the degrees of half-cara is small, as in India, and therefore given by the author. For correctness, the following instruction is to be followed. If the degrees of half-cara cannot be deducted from the degrees of time, deduct the degrees of time from the degrees of half-cara, find its sine, and deduct this from the sine of half-cara. This sine should be multiplied by sin colat. etc. and sine altitude is to be got. Because this will not produce much differences in our country, the author has not given this detail. (Even if the cara is 5 nāḍīs the difference in sin alt. will be only 15′.) Further, the measurement of long shadows cannot be accurate, and any inaccuracy caused by the author's rough work will be submerged in the inaccuracy of measurement. 44a. C.D.विनाशकृतानां

IV. 44 IV. THREE PROBLEMS 117 We have mentioned that the author’s rough procedure is indicated only when the Sun is in the six signs from Meṣa, because only then have we to deduct the degrees of half-cara, and the question, what is to be done when the half-cara is greater, arises. As for the subtraction of sine half-cara in the six signs from Tulā, that will always be less, and the question cannot arise. TS have not understood the author here, and say something unconnected and useless. (See their commentary p.25, and English Translation, pp.34-35). The rules, (for the Sun in the northern hemisphere) can be deriyed thus: (see fig. 14.) Z = Zenith P = North pole E = East point S = Sun DD′ = Diurnal Circle AS = Altitude of the Sun ZS = zenith distance of the Sun. EP = Unmaṇḍalam Sin AS = Sin altitude of the Sun = Śaṅkuliptās (or Mahā Śaṅku or Great gnomon)/120 [Fig. IV 14] By the well-known formula of the spherical triangle, Sin altitude of the Sun = sin AS = cos ZS = cos S P. cos Z P + sin S P. sin Z P. cos PZ. Here, using the tabular sines, sin S P = dyujyā/120 = diameter of the diurnal circle/240. sin Z P = sin co-latitude = lambajyā/120. cos S P = sin (90° − SP) = sin declination of the Sun = krāntijyā/120. cos Z P = sin latitude = akṣajyā/120. cos S PZ = sin S PE = sin (D PS − D PE) = sin (degrees of the taken time − degrees of half-cara) ∴ Śaṅkuliptās (i.e. Great gnomon) = sin declination × sin latitude ÷ 120 + day-diameter × sin co-latitude × sin (degrees of taken time − degrees of half-cara) ÷ 28,800. = day-diameter × sin colatitude × {sin (degrees of taken time − degrees of half-cara) + sin decli- nation × sin latitude × 240 ÷ (day-diameter × sin colatitude)} ÷ 28,800 = day-diameter × sin colat. {sin (degrees of taken time − degrees of half-cara) + sin half-cara} ÷ 28,800 = rule (i) applied to Sun in the northern hemisphere. In the same manner, the rule can be proved for the Sun in the southern hemisphere, but here the degrees of half-cara is first to be added (instead of being subtracted) to the degrees of time, and sin half-cara is to be subtracted instead of being added, because here, S PE = D PS + D PE, and these changes have to be made accordingly.

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118 PAÑCASIDDHĀNTIKĀ Rule (ii) is derived from the Great gnomon thus: The radius itself being the Great hypotenuse, and the Great gnomon and the Great shadow (this is 120 cos altitude or 120 sin zenith distance) are the sides of the right angled triangle, we have: Great shadow = √(120² – Great gnomon²) Then the shadow of the 12 digit gnomon is found by the proportion: Great gnomon: Great shadow :: 12 digit gnomon : shadow, and we get the rule (ii), shadow = 12 × √14,400 – Great gnomon² ÷ Great gnomon. In this connection, it may be noted that later authors like Bhāskarācārya II give different terms to different sections of the work. For instance they call sin (degrees of taken time ∓ degrees of half- cara) as Sūtram. Sūtram ± half-cara is called by them Iṣṭāntyā. They call Iṣṭāntyā × day-diameter ÷ 240 as Iṣṭahṛti. Then from Iṣṭahṛti the Śaṅkuliptā is obtained by the proportion: 120: sin colat :: Iṣṭahṛti : Śaṅkuliptā, by the similarity of the akṣākṣetras. [छायातः इष्टकालनाड्यः] [छाया द्वादशकृत्योर्योगान्मूलेन लम्बकघ्नेन ।] 'खखवस्व(श्वि)मुनी(न्दून्') (वि) भज्य लब्धा प्रथमजीवा ॥ ४५ ॥ त(द्युक्रान्ति)ज्याघ्नी विषुवज्ज्या ल(म्ब)कोद्धृता स्थाप्या | प्रथमज्या विश्ले(ष्या) (मे)षाद्ये (ऽन्यत्र) संयु(क्ता) ॥ ४६ ॥ तत्स्थि(त) जीवे गुणिते 'खजिनैर्धु (व्या)सभाजिते चापे । युतवियुतेऽ(ज)तुलादिषु ष(ड्ढृ) तो नाडिका लब्धा ॥ ४७ ॥ Time after sunrise 45. Square the shadow measured in digits, add 144, and get its square root. Multiply this by the sine of co-latitude and by this product divide 1,72,800. The quotient is called the 'First sine'. 46. Now, multiply the sine of declination of the Sun by the sine of latitude and divide by the sine of co-latitude. (Let us call this by its actual name, the Earth-sine.) Place this Earth-sine in two places. In one place, subtract this from or add this to the 'first sine', according as the Sun is in the northern or south- ern hemisphere. 47. This result, and the Earth-sine, are each to be multiplied by 240 and divided by the day-diameter. These are two sines. Find the arcs of each of these. When the Sun is in the northern hemisphere add the two arcs. Other- wise subtract one from the other. Divide the result by 6. The result is the time in nāḍīs after sunrise. 45c. A. वखश्व. A. मुनीन्द्रात्; C. मुनीन्दोः; D. मुनींदु 47a. A. तस्थिति 46a. A. तद्युज्याक्तांत्रीज्याघ्नी b. A. खजिनेधुद्यासभाजिते c. A. विश्लेषा c. A. वियुते च d. A. सेषाद्येनात्र संयुत d. A. षड्ढृतो; C.D. षड्ढृता. A. लब्धा

IV.47 IV. THREE PROBLEMS 119 The following are the steps in the work to be done: (i) The ‘First sine’ = 1,72,800 ÷ (sin colatitude × √144 + shadow²) (ii) The Earth-sine (which is to be placed in two places) = sin latitude × sin declination ÷ sin co- latitude. (iii) sine I = (‘First sine’ ∓ Earth-sine) × 240 ÷ day-diameter (iv) sine II = Earth-sine × 240 ÷ day-diameter. (v) Find arc I and arc II of sin I and sin II The desired time in nāḍīs = arc I/6 ± arc II/6. In (iii) and (v) the upper sign is to be taken for the Sun in the six signs from Aries, i.e. for the Sun in the northern hemisphere; otherwise the lower sign is to be taken. It must be added here, in accordance with what was said in the same context in getting the shadow from the time, that if the ‘First sine’ is less than the Earth-sine and therefore the Earth-sine cannot be deducted in (iii), the ‘First sine’ is to be deducted from the Earth-sine, and the result, i.e. sine I, is to taken as negative. Then in (v) the nāḍīs got from this, viz. arc I/6, are also negative, and therefore deducted from arc II/6, to get the time. Here too, if the latitude of the place is not too high, the reverse of the author’s method in the context can be used without any appreciable error, though this has not been mentioned here by the author. This is the work to be done: Here the Earth-sine is greater than the ‘First sine’; omit the Earth-sine and do (iii) and (v) with the ‘First sine’ alone, i.e. multiply the ‘First sine’ by 240, divide by the day-diameter, get the arc of this, and divide by 6 and thus to get the nāḍīs after sunrise. The following points must be noted here. In the work of computing the nādīs from the shadow, as the exact time is not known, the exact Sun and therefrom the exact declination cannot be known, and we have to use the declination of the Sun at sunrise or sunset. There may be a small error on account of this. This can be avoided by repeating the work using the declination of the Sun for the computed time. It has not been specifically mentioned by the author because this can be inferred by the computer. Secondly, the author has given all this for places in the northern hemisphere in the forenoon. For places in the southern hemisphere and the afternoon, changes have to be made in the work, which have not been given by the author. It must also be noted that the ancients con- sidered the computation of the time from the shadow or the shadow from the time as very impor- tant because this was the best means available to them of knowing the times of births and muhūrtas. Example 17 (a) For a place (in the northern hemisphere) sin lat. is 60′, and therefrom sin colat is 103′ 55″. On a particular day the sin declination is 34′ 30″, (the sun being in the 6 signs from Aries) and therefore the day-diameter is 229′ 51″. Find the time from sunrise if the shadow of the 12 digit gnomon is 13 aṅg 50 vyaṅgulus. (i) ‘First sine’ = 1,72,800 ÷ (103′ 55″ × √13 14/15² + 144 = 1,72,800 ÷ (103′ 55″ × 18.389) = 90′ 26″. (ii) Earth-sine = 60′ × 34′ 30″ ÷ 103′ 55″ = 19′ 55″ (iii) Sine I = (90′ 26″ − 19′ 55″) 240 ÷ 229′ 51″ = 73′ 38″ (iv) Sine II = 19′ 55″ × 240 ÷ 229′ 51″ = 20′ 48″

120 PAÑCASIDDHĀNTIKĀ IV. 47 (v) Arc I = 38° 3′. Arc II = 9° 59′. The time from sunrise in nāḍis = 38° 3′ / 6 + 9° 59′ / 6 = 8 nāḍis. (Note that this work is the inverse of example 15 (a). There, 8 nāḍis were given, and the shadow 13 aṅg 56 vyaṅg was computed. Here, for the shadow 13 aṅg 56 vyaṅg, the nāḍis amounting to 8 have been computed). Example 17 (b) For the same place, on the same day, find the time when the shadow is 137 aṅg 57 vyaṅg. (i) ‘First sine’ = 1,72,800 ÷ (103′ 55″ × √(144 + 137 57/60² = 12′ 1″. (ii) Earth-sine = 60′ × 34′ 30″ ÷ 103′ 55″ = 19′ 55″. (iii) Sine I = (12′ 1″ − 19′ 55″) × 240 ÷ 229′ 51″ = − 8′ 15″ (iv) Sine II = 19′ 55″ × 240 ÷ 229′ 51″ = 20′ 48″. (v) Arc I = − 3° 57′, Arc II = 9° 59′. The time from sunrise = − 3° 57′/6 + 9° 59′/6 = 1 nāḍi. (Note that is the inverse of Example 15 (b). There the shadow 137 aṅg 57 vyaṅg was computed for one nāḍi from sunrise. Here for the same shadow the time one nāḍi is computed.) We shall do the same by the inverse operation of the work previously given by the author: The ‘First sine’, computed is 12′ 1″. The Earth-sine computed is 19′ 55″, and greater than the ‘First sine’. Therefore taking the ‘First sine’ alone, 12′ 1″ × 240 ÷ 229′ 51″ = 12′ 33″. The arc of this = 6° 2′. Dividing by 6, the time obtained is one nāḍi and 1/3 vinādi, and neglecting the negligible 1/3 viṇāḍi, we see the same time is got. For proof of the rules here given, we shall derive these from the rules for the shadow given the time, as the operation is practically the inverse of the operation given there. In the previous work, rule (ii) gives: 12 × √(14,400 − sin² altitude) ÷ sin altitude = shadow. ∴ 144 × (14,400 − sin² alt.) = sin² alt. = shadow². ∴ 144 × 14,400 = sin² alt. × shadow² ± 144 sin² alt. = sin² alt. (shadow² + 144). ∴ 12 × 120 = sin alt. × √(shadow² + 12²). ∴ 12 × 120 ÷ √(shadow² + 12²) = sin alt. = 12 × 120 × 120 × sin colat. ÷ (120 × sin colat. × √(shadow² + 12²) = ‘First sine’ × sin colat. ÷ 120, (because, 12 × 120 × 120 ÷ (sin colat × √(shadow² + 12²)) = 1,71,800 ÷ (sin colat × √(shadow² + 12²)) = ‘First sine’ as given). Similarly, in the previous rule (i), sin alt. = {sine (degrees of time ∓ degrees of half-cara) ± sin half-cara} × sin colat. × day-diameter ÷ 28,800, = ‘First sine’ × sin colat. ÷ 120. ∴ ‘First sine’ × 240 ÷ day-diameter = {sin (degrees of time ∓ degrees of half-cara) ± sin half- cara}. ∴ ‘First sine’ × 240 ÷ day-diameter ∓ sin half-cara = sin (degrees of time ∓ degrees of half-cara). ∴ ‘First sine’ × 240 ÷ day-diameter ∓ Earth-sine × 240 ÷ day-diameter = sin (degrees of time ∓ degrees of half-cara). ∴ (‘First sine’ ∓ earth-sine) × 240 ÷ day-diameter = sin (degrees of time ∓ degrees of half-cara) = sin (degrees of time after the Sun has touched the unmaṇḍala)

IV. 48 IV. THREE PROBLEMS 121 From the sin degrees of time, and thence by dividing by 6, the time in nāḍīs after the Sun has touched the unmaṇḍala is obtained. The addition or subtraction of the half-cara to this gives the time from sunrise, to obtain which sin half-cara is got from the Earth-sine, and then its arc, viz the degrees of half-cara. षड्घ्नेऽथ स्वद्युमिते छिन्ने सद्धादशैर्विमाध्याह्नैः । छायाङ्गुलैर्गतास्ता नाड्यः प्राक् पृष्ठतः शेषाः ॥ ४८ ॥ Time for sunset 48. Or roughly, multiply the duration of daytime in nāḍīs by 6, and divide by the shadow increased by 12 and decreased by the midday shadow of date. The time from sunrise is got in the forenoon, and the time to elapse for sunset is obtained in the afternoon. The shadows mentioned here are those of the twelve-digit gnomon and not the shadows of a person measured by his foot. The rule is the time in nāḍīs = 6 × daytime in nāḍīs ÷ (shadow + 12 − mid-day shadow). Example 18. Given the duration of daytime, nāḍīs 33-20, and mid-day shadow, 2 aṅg 50 vyaṅg. Find the time when the gnomonic shadow is 13 aṅg 56 vyaṅg. The time = 6 × 33 1/3 ÷ (13 14/15 + 12 − 2 5/6) = 200 ÷ 23 1/10 = nāḍīs 8-37. The data given in the example are for the place and day in Example 16 (a), and we must get nāḍīs 8, as the time. But we get nāḍīs 8-37. From this we can have an idea of the roughness of this method. Evidently VM wants us to use this rule if we feel that this accuracy is sufficient, for, this is easy to use, provided the daytime and the midday shadow are tabulated beforehand and kept ready. The rule may be explained in the manner we explained the similar rule with Vāsiṣṭha Siddhānta. Let us assume, time = x × day-time ÷ (shadow − mid-day shadow + y), where x and y are two con- stants to be determined. (The daytime occurs as a multiplier in the rule because, other things being equal, the time must vary with the daytime. For the deduction of the mid-day shadow from the shadow, see the explanation in the Vāsiṣṭha.) At noon the shadow is equal to the mid-day shadow of date, and the time is daytime/2. Therefore we have: x × day time ÷ (mid-day shadow − mid-day shadow + y) = daytime/2. ∴ 2x × daytime = daytime × y. ∴ 2x = y. Therefore, whatever be the multiplier for the daytime, twice that is the constant additive to the shadow, as in the author's rule here, 6 and 12, respectively. Only so far can we go in the explanation 48. Quoted by Utpala on BS 2, p.62 48a. A. षट्प्रोथवा द्युमाने; C.D. षड्घ्नेऽथवा द्युमाने c. A1. गतास्था; A2. गतास्थे b. A. ०दशे विमध्याह्ने d. A. नाद्यः. A. प्रष्टतो